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Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual

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KirkMcdowells

The document discusses problems related to fluid mechanics, focusing on calculating boiling temperatures at various altitudes and pressures, pressure changes during airplane descent, and absolute pressures in different systems. It includes methodologies and solutions for determining pressures in liquids, pressures for various altitudes, and specific buoyancy calculations. Various equations and data tables are referenced to solve practical scenarios in fluid mechanics.

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Problem 3.2 [Difficulty: 2] Given: Pure water on a standard day Find: Boiling temperature at (a) 1000 m and (b) 2000 m, and compare with sea level value. Solution: We can determine the atmospheric pressure at the given altitudes from table A.3, Appendix A The data are Elevation (m) p/p o p (kPa) 0 1.000 101.3 1000 0.887 89.9 2000 0.785 79.5 We can also consult steam tables for the variation of saturation temperature with pressure: p (kPa) T sat (°C) 70 90.0 80 93.5 90 96.7 101.3 100.0 We can interpolate the data from the steam tables to correlate saturation temperature with altitude: Elevation (m) p/p o p (kPa) T sat (°C) 0 1.000 101.3 100.0 1000 0.887 89.9 96.7 2000 0.785 79.5 93.3 The data are plotted here. They show that the saturation temperature drops approximately 3.4°C/1000 m. Variation of Saturation Temperature with Pressure 88 90 92 94 96 98 100 70 75 80 85 90 95 100 105 Absolute Pressure (kPa) Saturation Temperature(°C) 2000 m 1000 m Sea Level Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual Full Download: https://alibabadownload.com/product/fox-and-mcdonalds-introduction-to-fluid-mechanics-9th-edition-pritchard-so This sample only, Download all chapters at: AlibabaDownload.com Problem 3.1 3.1
Problem 3.3 [Difficulty: 2] Given: Data on flight of airplane Find: Pressure change in mm Hg for ears to "pop"; descent distance from 8000 m to cause ears to "pop." Solution: Assume the air density is approximately constant constant from 3000 m to 2900 m. From table A.3 ρSL 1.225 kg m 3 ⋅= ρair 0.7423 ρSL⋅= ρair 0.909 kg m 3 = We also have from the manometer equation, Eq. 3.7 Δp ρair− g⋅ Δz⋅= and also Δp ρHg− g⋅ ΔhHg⋅= Combining ΔhHg ρair ρHg Δz⋅= ρair SGHg ρH2O⋅ Δz⋅= SGHg 13.55= from Table A.2 ΔhHg 0.909 13.55 999× 100× m⋅= ΔhHg 6.72 mm⋅= For the ear popping descending from 8000 m, again assume the air density is approximately constant constant, this time at 8000 m. From table A.3 ρair 0.4292 ρSL⋅= ρair 0.526 kg m 3 = We also have from the manometer equation ρair8000 g⋅ Δz8000⋅ ρair3000 g⋅ Δz3000⋅= where the numerical subscripts refer to conditions at 3000m and 8000m. Hence Δz8000 ρair3000 g⋅ ρair8000 g⋅ Δz3000⋅= ρair3000 ρair8000 Δz3000⋅= Δz8000 0.909 0.526 100× m⋅= Δz8000 173m= Problem 3.2 3.2
Problem 3.4 [Difficulty: 3] Given: Boiling points of water at different elevations Find: Change in elevation Solution: From the steam tables, we have the following data for the boiling point (saturation temperature) of water Tsat ( o F) p (psia) 195 10.39 185 8.39 The sea level pressure, from Table A.3, is pSL = 14.696 psia Hence Tsat ( o F) p/pSL 195 0.707 185 0.571 From Table A.3 p/pSL Altitude (m) Altitude (ft) 0.7372 2500 8203 0.6920 3000 9843 0.6492 3500 11484 0.6085 4000 13124 0.5700 4500 14765 Then, any one of a number of Excel functions can be used to interpolate (Here we use Excel 's Trendline analysis) p/pSL Altitude (ft) 0.707 9303 Current altitude is approximately 9303 ft 0.571 14640 The change in altitude is then 5337 ft Alternatively, we can interpolate for each altitude by using a linear regression between adjacent data points p/pSL Altitude (m) Altitude (ft) p/pSL Altitude (m) Altitude (ft) For 0.7372 2500 8203 0.6085 4000 13124 0.6920 3000 9843 0.5700 4500 14765 Then 0.7070 2834 9299 0.5730 4461 14637 The change in altitude is then 5338 ft Altitude vs Atmospheric Pressure z = -39217(p/pSL) + 37029 R2 = 0.999 2500 5000 7500 10000 12500 15000 0.55 0.60 0.65 0.70 0.75 p/pSL Altitude(ft) Data Linear Trendline Problem 3.3 3.3
Problem 3.9 [Difficulty: 2] Given: Data on tire at 3500 m and at sea level Find: Absolute pressure at 3500 m; pressure at sea level Solution: At an elevation of 3500 m, from Table A.3: pSL 101 kPa⋅= patm 0.6492 pSL⋅= patm 65.6 kPa⋅= and we have pg 0.25 MPa⋅= pg 250 kPa⋅= p pg patm+= p 316 kPa⋅= At sea level patm 101 kPa⋅= Meanwhile, the tire has warmed up, from the ambient temperature at 3500 m, to 25oC. At an elevation of 3500 m, from Table A.3 Tcold 265.4 K⋅= and Thot 25 273+( ) K⋅= Thot 298K= Hence, assuming ideal gas behavior, pV = mRT, and that the tire is approximately a rigid container, the absolute pressure of the hot tire is phot Thot Tcold p⋅= phot 354 kPa⋅= Then the gage pressure is pg phot patm−= pg 253 kPa⋅= Problem 3.4 3.4
Problem 3.5 [Difficulty: 2] Given: Data on system Find: Force on bottom of cube; tension in tether Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ Δp ρ g⋅ h⋅= where h is measured downwards The absolute pressure at the interface is pinterface patm SGoil ρ⋅ g⋅ hoil⋅+= Then the pressure on the lower surface is pL pinterface ρ g⋅ hL⋅+= patm ρ g⋅ SGoil hoil⋅ hL+( )⋅+= For the cube V 125 mL⋅= V 1.25 10 4− × m 3 ⋅= Then the size of the cube is d V 1 3 = d 0.05m= and the depth in water to the upper surface is hU 0.3 m⋅= Hence hL hU d+= hL 0.35m= where hL is the depth in water to the lower surface The force on the lower surface is FL pL A⋅= where A d 2 = A 0.0025m 2 = FL patm ρ g⋅ SGoil hoil⋅ hL+( )⋅+⎡⎣ ⎤⎦ A⋅= FL 101 10 3 × N m 2 ⋅ 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.8 0.5× m⋅ 0.35 m⋅+( )× N s 2 ⋅ kg m⋅ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0.0025× m 2 ⋅= FL 270.894N= Note: Extra decimals needed for computing T later! For the tension in the tether, an FBD gives ΣFy 0= FL FU− W− T− 0= or T FL FU− W−= where FU patm ρ g⋅ SGoil hoil⋅ hU+( )⋅+⎡⎣ ⎤⎦ A⋅= Problem 3.5 3.5
Note that we could instead compute ΔF FL FU−= ρ g⋅ SGoil⋅ hL hU−( )⋅ A⋅= and T ΔF W−= Using FU FU 101 10 3 × N m 2 ⋅ 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.8 0.5× m⋅ 0.3 m⋅+( )× N s 2 ⋅ kg m⋅ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0.0025× m 2 ⋅= FU 269.668N= Note: Extra decimals needed for computing T later! For the oak block (Table A.1) SGoak 0.77= so W SGoak ρ⋅ g⋅ V⋅= W 0.77 1000× kg m 3 ⋅ 9.81× m s 2 ⋅ 1.25× 10 4− × m 3 ⋅ N s 2 ⋅ kg m⋅ ×= W 0.944N= T FL FU− W−= T 0.282N=
Problem 3.6 [Difficulty: 2] Given: Data on system before and after applied force Find: Applied force Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ p patm ρ g⋅ y y0−( )⋅−= with p y0( ) patm= For initial state p1 patm ρ g⋅ h⋅+= and F1 p1 A⋅= ρ g⋅ h⋅ A⋅= (Gage; F1 is hydrostatic upwards force) For the initial FBD ΣFy 0= F1 W− 0= W F1= ρ g⋅ h⋅ A⋅= For final state p2 patm ρ g⋅ H⋅+= and F2 p2 A⋅= ρ g⋅ H⋅ A⋅= (Gage; F2 is hydrostatic upwards force) For the final FBD ΣFy 0= F2 W− F− 0= F F2 W−= ρ g⋅ H⋅ A⋅ ρ g⋅ h⋅ A⋅−= ρ g⋅ A⋅ H h−( )⋅= F ρH2O SG⋅ g⋅ π D 2 ⋅ 4 ⋅ H h−( )⋅= From Fig. A.1 SG 13.54= F 1000 kg m 3 ⋅ 13.54× 9.81× m s 2 ⋅ π 4 × 0.05 m⋅( ) 2 × 0.2 0.025−( )× m⋅ N s 2 ⋅ kg m⋅ ×= F 45.6N= Problem 3.6 3.6
Problem 3.7 (Difficulty: 1) 3.7 Calculate the absolute pressure and gage pressure in an open tank of crude oil 2.4 𝑚 below the liquid surface. If the tank is closed and pressurized to 130 𝑘𝑘𝑘, what are the absolute pressure and gage pressure at this location. Given: Location: ℎ = 2.4 𝑚 below the liquid surface. Liquid: Crude oil. Find: The absolute pressure 𝑝 𝑎 and gage pressure 𝑝 𝑔 for both open and closed tank . Assumption: The gage pressure for the liquid surface is zero for open tank and closed tank. The oil is incompressible. Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 The density for the crude oil is: 𝜌 = 856 𝑘𝑘 𝑚3 The atmosphere pressure is: 𝑝 𝑎𝑎𝑎𝑎𝑎 = 101000 𝑃𝑃 The pressure for the closed tank is: 𝑝 𝑡𝑡𝑡𝑡 = 130 𝑘𝑘𝑘 = 130000 𝑃𝑃 Using the hydrostatic relation, the gage pressure of open tank 2.4 m below the liquid surface is: 𝑝 𝑔 = 𝜌𝜌ℎ = 856 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 2.4 𝑚 = 20100 𝑃𝑃 The absolute pressure of open tank at this location is: 𝑝 𝑎 = 𝑝 𝑔 + 𝑝 𝑎𝑎𝑎𝑎𝑎 = 20100 𝑃𝑃 + 101000 𝑃𝑃 = 121100 𝑃𝑃 = 121.1 𝑘𝑘𝑘 The gage pressure of closed tank at the same location below the liquid surface is the same as open tank: 𝑝 𝑔 = 𝜌𝜌ℎ = 856 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 2.4 𝑚 = 20100 𝑃𝑃 The absolute pressure of closed tank at this location is: 𝑝 𝑎 = 𝑝 𝑔 + 𝑝 𝑡𝑡𝑡𝑡 = 20100 𝑃𝑃 + 130000 𝑃𝑃 = 150100 𝑃𝑃 = 150.1 𝑘𝑘𝑘
Problem 3.8 (Difficulty: 1) 3.8 An open vessel contains carbon tetrachloride to a depth of 6 𝑓𝑓 and water on the carbon tetrachloride to a depth of 5 𝑓𝑓 . What is the pressure at the bottom of the vessel? Given: Depth of carbon tetrachloride: ℎ 𝑐 = 6 𝑓𝑓. Depth of water: ℎ 𝑤 = 5 𝑓𝑓. Find: The gage pressure 𝑝 at the bottom of the vessel. Assumption: The gage pressure for the liquid surface is zero. The fluid is incompressible. Solution: Use the hydrostatic pressure relation to detmine pressures in a fluid. Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 The density for the carbon tetrachloride is: 𝜌𝑐 = 1.59 × 103 𝑘𝑘 𝑚3 = 3.09 𝑠𝑠𝑠𝑠 𝑓𝑓3 The density for the water is: 𝜌 𝑤 = 1.0 × 103 𝑘𝑘 𝑚3 = 1.940 𝑠𝑠𝑠𝑠 𝑓𝑓3 Using the hydrostatic relation, the gage pressure 𝑝 at the bottom of the vessel is: 𝑝 = 𝜌𝑐 𝑔ℎ 𝑐 + 𝜌 𝑤 𝑔ℎ 𝑤 𝑝 = 3.09 𝑠𝑠𝑠𝑠 𝑓𝑓3 × 32.2 𝑓𝑓 𝑠2 × 6 𝑓𝑓 + 1.940 𝑠𝑠𝑠𝑠 𝑓𝑓3 × 32.2 𝑓𝑓 𝑠2 × 5 𝑓𝑓 = 909 𝑙𝑙𝑙 𝑓𝑓2 = 6.25 𝑝𝑝𝑝
Problem 3.8 [Difficulty: 2] Given: Properties of a cube floating at an interface Find: The pressures difference between the upper and lower surfaces; average cube density Solution: The pressure difference is obtained from two applications of Eq. 3.7 pU p0 ρSAE10 g⋅ H 0.1 d⋅−( )⋅+= pL p0 ρSAE10 g⋅ H⋅+ ρH2O g⋅ 0.9⋅ d⋅+= where pU and pL are the upper and lower pressures, p0 is the oil free surface pressure, H is the depth of the interface, and d is the cube size Hence the pressure difference is Δp pL pU−= ρH2O g⋅ 0.9⋅ d⋅ ρSAE10 g⋅ 0.1⋅ d⋅+= Δp ρH2O g⋅ d⋅ 0.9 SGSAE10 0.1⋅+( )⋅= From Table A.2 SGSAE10 0.92= Δp 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.1× m⋅ 0.9 0.92 0.1×+( )× N s 2 ⋅ kg m⋅ ×= Δp 972Pa= For the cube density, set up a free body force balance for the cube ΣF 0= Δp A⋅ W−= Hence W Δp A⋅= Δp d 2 ⋅= ρcube m d 3 = W d 3 g⋅ = Δp d 2 ⋅ d 3 g⋅ = Δp d g⋅ = ρcube 972 N m 2 ⋅ 1 0.1 m⋅ × s 2 9.81 m⋅ × kg m⋅ N s 2 ⋅ ×= ρcube 991 kg m 3 = Problem 3.9 3.9 these equations:
Problem 3.1 [Difficulty: 2] Given: Data on nitrogen tank Find: Pressure of nitrogen; minimum required wall thickness Assumption: Ideal gas behavior Solution: Ideal gas equation of state: p V⋅ M R⋅ T⋅= where, from Table A.6, for nitrogen R 55.16 ft lbf⋅ lbm R⋅ ⋅= Then the pressure of nitrogen is p M R⋅ T⋅ V = M R⋅ T⋅ 6 π D 3 ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= p 140 lbm⋅ 55.16× ft lbf⋅ lbm R⋅ ⋅ 77 460+( )× R⋅ 6 π 2.5 ft⋅( ) 3 × ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × ft 12 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ×= p 3520 lbf in 2 ⋅= σcπDt pπD2 /4 To determine wall thickness, consider a free body diagram for one hemisphere: ΣF 0= p π D 2 ⋅ 4 ⋅ σc π⋅ D⋅ t⋅−= where σc is the circumferential stress in the container Then t p π⋅ D 2 ⋅ 4 π⋅ D⋅ σc⋅ = p D⋅ 4 σc⋅ = t 3520 lbf in 2 ⋅ 2.5 ft⋅ 4 × in 2 30 10 3 × lbf⋅ ×= t 0.0733 ft⋅= t 0.880 in⋅= Problem 3.10 3.10
Problem 3.11 (Difficulty: 2) 3.11 If at the surface of a liquid the specific weight is 𝛾0, with 𝑧 and 𝑝 both zero, show that, if 𝐸 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐, the specific weight and pressure are given 𝛾 = 𝐸 �𝑧+ 𝐸 𝛾0 � and 𝑝 = −𝐸 ln �1 + 𝛾0 𝑍 𝐸 �. Calculate specific weight and pressure at a depth of 2 𝑘𝑘 assuming 𝛾0 = 10.0 𝑘𝑘 𝑚3 and 𝐸 = 2070 𝑀𝑀𝑀. Given: Depth: ℎ = 2 𝑘𝑘. The specific weight at surface of a liquid: 𝛾0 = 10.0 𝑘𝑘 𝑚3. Find: The specific weight and pressure at a depth of 2 𝑘𝑘. Assumption:. Bulk modulus is constant Solution: Use the hydrostatic pressure relation and definition of bulk modulus to detmine pressures in a fluid. Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Definition of bulk modulus 𝐸𝑣 = 𝑑𝑑 𝑑𝑑 𝜌� = 𝑑𝑑 𝑑𝑑 𝛾� Eliminating dp from the hydrostatic pressure relation and the bulk modulus definition: 𝑑𝑑 = −𝛾 𝑑𝑑 = 𝐸𝑣 𝑑𝑑 𝛾 Or 𝑑𝑑 = −𝐸𝑣 𝑑𝑑 𝛾2 Integrating for both sides we get: 𝑧 = 𝐸𝑣 1 𝛾 + 𝑐 At 𝑧 = 0, 𝛾 = 𝛾0 so: 𝑐 = −𝐸𝑣 1 𝛾0
𝑧 = 𝐸𝑣 1 𝛾 − 𝐸𝑣 1 𝛾0 Solving for 𝛾, we have: 𝛾 = 𝐸𝑣 �𝑧 + 𝐸𝑣 𝛾0 � Solving for the pressure using the hydrostatic relation: 𝑑𝑑 = −𝛾𝛾𝛾 = − 𝐸𝑣 �𝑧 + 𝐸𝑣 𝛾0 � 𝑑𝑑 Integrating both sides we to get: 𝑝 = −𝐸𝑣 ln �𝑧 + 𝐸𝑣 𝛾0 � + 𝑐 At 𝑧 = 0, 𝑝 = 0 so: 𝑐 = 𝐸𝑣 ln � 𝐸𝑣 𝛾0 � 𝑝 = −𝐸𝑣 ln �𝑧 + 𝐸𝑣 𝛾0 � + 𝐸𝑣 ln � 𝐸𝑣 𝛾0 � = −𝐸𝑣 ln �1 + 𝛾0 𝑧 𝐸𝑣 � For the specific case ℎ = 2 𝑘𝑘 𝛾0 = 10.0 𝑘𝑘 𝑚3 𝐸𝑣 = 2070 𝑀𝑀𝑀 The specific weight: 𝛾 = 𝐸𝑣 �𝑧 + 𝐸𝑣 𝛾0 � = 2070 × 106 𝑝𝑝 �−2000 𝑃𝑃 + 2070 × 106 𝑃𝑃 10 × 103 𝑁 𝑚3 � = 10100 𝑁 𝑚3 = 10.1 𝑘𝑘 𝑚3 Pressure: 𝑝 = −𝐸𝑣 ln �1 + 𝛾0 𝑧 𝐸𝑣 � = −2070 × 106 𝑃𝑃 × ln �1 + 10000.0 𝑘𝑘 𝑚3 × � −2000 𝑚 2070 × 106 𝑃𝑃 �� = 20100 𝑘𝑘𝑘
Problem 3.12 (Difficulty: 2) 3.12 In the deep ocean the compressibility of seawater is significant in its effect on 𝜌 and 𝑝. If 𝐸 = 2.07 × 109 𝑃𝑃, find the percentage change in the density and pressure at a depth of 10000 meters as compared to the values obtained at the same depth under the incompressible assumption. Let 𝜌0 = 1020 𝑘𝑘 𝑚3 and the absolute pressure 𝑝0 = 101.3 𝑘𝑘𝑘. Given: Depth: ℎ = 10000 𝑚𝑚𝑚𝑚𝑚𝑚. The density: 𝜌0 = 1020 𝑘𝑘 𝑚3. The absolute pressure: 𝑝0 = 101.3 𝑘𝑘𝑘. Find: The percent change in density 𝜌% and pressure 𝑝%. Assumption: The bulk modulus is constant Solution: Use the relations developed in problem 3.11 for specific weight and pressure for a compressible liquid: 𝛾 = 𝐸 �𝑧 + 𝐸 𝛾0 � 𝑝 = −𝐸 ln �1 + 𝛾0 𝑧 𝐸 � The specific weight at sea level is: 𝛾0 = 𝜌0 𝑔 = 1020 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 = 10010 𝑁 𝑚3 The specific weight and density at 10000 m depth are 𝛾 = 𝐸 �𝑧 + 𝐸 𝛾0 � = 2.07 × 109 �−10000 + 2.07 × 109 10010 � 𝑁 𝑚3 = 10520 𝑁 𝑚3 𝜌 = 𝛾 𝑔 = 10520 9.81 𝑘𝑘 𝑚3 = 1072 𝑘𝑘 𝑚3 The percentage change in density is 𝜌% = 𝜌 − 𝜌0 𝜌0 = 1072 − 1020 1020 = 5.1 % The gage pressure at a depth of 10000m is: 𝑝 = −𝐸 ln �1 + 𝛾0 𝑧 𝐸 � = 101.3 𝑘𝑘𝑘 − 2.07 × 109 × ln �1 + 10010 × (−10000) 2.07 × 109 � 𝑃𝑃 = 102600 𝑘𝑘𝑘
The pressure assuming that the water is incompressible is: 𝑝𝑖𝑖 = 𝜌𝜌ℎ = 1020 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 10000 𝑚 = 100062 𝑘𝑘𝑘 The percent difference in pressure is: 𝑝% = 𝑝 − 𝑝0 𝑝0 = 102600 𝑘𝑘𝑘 − 100062 𝑘𝑘𝑘 100062 𝑘𝑘𝑘 = 2.54 %
Problem 3.12 [Difficulty: 4] Given: Model behavior of seawater by assuming constant bulk modulus Find: (a) Expression for density as a function of depth h. (b) Show that result may be written as ρ = ρo + bh (c) Evaluate the constant b (d) Use results of (b) to obtain equation for p(h) (e) Determine depth at which error in predicted pressure is 0.01% Solution: From Table A.2, App. A: SGo 1.025= Ev 2.42 GPa⋅ 3.51 10 5 × psi⋅== Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) (Definition of Bulk Modulus) Ev dp dρ ρ = Then dp ρ g⋅ dh⋅= Ev dρ ρ ⋅= or dρ ρ 2 g Ev dh= Now if we integrate: ρo ρ ρ 1 ρ 2 ⌠⎮ ⎮ ⎮ ⌡ d 0 h h g Ev ⌠ ⎮ ⎮ ⌡ d= After integrating: ρ ρo− ρ ρo⋅ g h⋅ Ev = Therefore: ρ Ev ρo⋅ Ev g h⋅ ρo⋅− = and ρ ρo 1 1 ρo g⋅ h⋅ Ev − = (Binomial expansion may be found in a host of sources, e.g. CRC Handbook of Mathematics) Now for ρo g⋅ h⋅ Ev <<1, the binomial expansion may be used to approximate the density: ρ ρo 1 ρo g⋅ h⋅ Ev += In other words, ρ ρo b h⋅+= where b ρo 2 g⋅ Ev = Since dp ρ g⋅ dh⋅= then an approximate expression for the pressure as a function of depth is: papprox patm− 0 h hρo b h⋅+( ) g⋅ ⌠ ⎮ ⌡ d= papprox patm− g h⋅ 2 ρo⋅ b h⋅+( )⋅ 2 =→ Solving for papprox we get: Problem 3.13 3.13
papprox patm g h⋅ 2 ρo⋅ b h⋅+( )⋅ 2 += patm ρo g⋅ h⋅+ b g⋅ h 2 ⋅ 2 += patm ρo h⋅ b h 2 ⋅ 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ g⋅+= Now if we subsitiute in the expression for b and simplify, we get: papprox patm ρo h⋅ ρo 2 g⋅ Ev h 2 2 ⋅+ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ g⋅+= patm ρo g⋅ h⋅ 1 ρo g⋅ h⋅ 2 Ev⋅ + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅+= papprox patm ρo g⋅ h⋅ 1 ρo g⋅ h⋅ 2Ev + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅+= The exact soution for p(h) is obtained by utilizing the exact solution for ρ(h). Thus: pexact patm− ρo ρ ρ Ev ρ ⌠ ⎮ ⎮ ⌡ d= Ev ln ρ ρo ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Subsitiuting for ρ ρo we get: pexact patm Ev ln 1 ρo g⋅ h⋅ Ev − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⋅+= If we let x ρo g⋅ h⋅ Ev = For the error to be 0.01%: Δpexact Δpapprox− Δpexact 1 ρo g⋅ h⋅ 1 x 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ Ev ln 1 x−( ) 1−⎡⎣ ⎤⎦⋅ −= 1 x 1 x 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ ln 1 x−( ) 1−⎡⎣ ⎤⎦ −= 0.0001= This equation requires an iterative solution, e.g. Excel's Goal Seek. The result is: x 0.01728= Solving x for h: h x Ev⋅ ρo g⋅ = h 0.01728 3.51× 10 5 × lbf in 2 ⋅ ft 3 1.025 1.94× slug⋅ × s 2 32.2 ft⋅ × 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × slug ft⋅ lbf s 2 ⋅ ×= h 1.364 10 4 × ft⋅= This depth is over 2.5 miles, so the incompressible fluid approximation is a reasonable one at all but the lowest depths of the ocean.
Problem 3.14 [Difficulty: 3] Air H D Air H – y y y Given: Cylindrical cup lowered slowly beneath pool surface Find: Expression for y in terms of h and H. Plot y/H vs. h/H. Solution: Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) p V⋅ M R⋅ T⋅= (Ideal Gas Equation) Assumptions: (1) Constant temperature compression of air inside cup (2) Static liquid (3) Incompressible liquid First we apply the ideal gas equation (at constant temperature) for the pressure of the air in the cup: p V⋅ constant= Therefore: p V⋅ pa π 4 ⋅ D 2 ⋅ H⋅= p π 4 ⋅ D 2 ⋅ H y−( )⋅= and upon simplification: pa H⋅ p H y−( )⋅= Now we look at the hydrostatic pressure equation for the pressure exerted by the water. Since ρ is constant, we integrate: p pa− ρ g⋅ h y−( )⋅= at the water-air interface in the cup. Since the cup is submerged to a depth of h, these pressures must be equal: pa H⋅ pa ρ g⋅ h y−( )⋅+⎡⎣ ⎤⎦ H y−( )⋅= pa H⋅ pa y⋅− ρ g⋅ h y−( )⋅ H y−( )⋅+= Explanding out the right hand side of this expression: 0 pa− y⋅ ρ g⋅ h y−( )⋅ H y−( )⋅+= ρ g⋅ h⋅ H⋅ ρ g⋅ h⋅ y⋅− ρ g⋅ H⋅ y⋅− ρ g⋅ y 2 ⋅+ pa y⋅−= ρ g⋅ y 2 ⋅ pa ρ g⋅ h H+( )⋅+⎡⎣ ⎤⎦ y⋅− ρ g⋅ h⋅ H⋅+ 0= y 2 pa ρ g⋅ h H+( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ y⋅− h H⋅+ 0= We now use the quadratic equation: y pa ρ g⋅ h H+( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ pa ρ g⋅ h H+( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 4 h⋅ H⋅−− 2 = we only use the minus sign because y can never be larger than H. Problem 3.14 3.14
Now if we divide both sides by H, we get an expression for y/H: y H pa ρ g⋅ H⋅ h H + 1+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ pa ρ g⋅ H⋅ h H + 1+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 4 h H ⋅−− 2 = The exact shape of this curve will depend upon the height of the cup. The plot below was generated assuming: pa 101.3 kPa⋅= H 1 m⋅= 0 20 40 60 80 100 0.2 0.4 0.6 0.8 Depth Ratio, h/H HeightRatio,y/H
Problem 3.16 [Difficulty: 2] patmA pbaseA Cover Given: Data on water tank and inspection cover Find: If the support bracket is strong enough; at what water depth would it fail Assumptions: Water is incompressible and static Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ Δp ρ g⋅ h⋅= where h is measured downwards The absolute pressure at the base is pbase patm ρ g⋅ h⋅+= where h 16 ft⋅= The gage pressure at the base is pbase ρ g⋅ h⋅= This is the pressure to use as we have patm on the outside of the cover. The force on the inspection cover is F pbase A⋅= where A 1 in⋅ 1× in⋅= A 1 in 2 ⋅= F ρ g⋅ h⋅ A⋅= F 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 16× ft⋅ 1× in 2 ⋅ ft 12 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × lbf s 2 ⋅ slug ft⋅ ×= F 6.94 lbf⋅= The bracket is strong enough (it can take 9 lbf). To find the maximum depth we start with F 9.00 lbf⋅= h F ρ g⋅ A⋅ = h 9 lbf⋅ 1 1.94 × ft 3 slug ⋅ 1 32.2 × s 2 ft ⋅ 1 in 2 × 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × slug ft⋅ lbf s 2 ⋅ ×= h 20.7 ft⋅= Problem 3.15 3.15
Problem 3.18 [Difficulty: 2] Given: Data on partitioned tank Find: Gage pressure of trapped air; pressure to make water and mercury levels equal Solution: The pressure difference is obtained from repeated application of Eq. 3.7, or in other words, from Eq. 3.8. Starting from the right air chamber pgage SGHg ρH2O× g× 3 m⋅ 2.9 m⋅−( )× ρH2O g× 1× m⋅−= pgage ρH2O g× SGHg 0.1× m⋅ 1.0 m⋅−( )×= pgage 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 13.55 0.1× m⋅ 1.0 m⋅−( )× N s 2 ⋅ kg m⋅ ×= pgage 3.48 kPa⋅= If the left air pressure is now increased until the water and mercury levels are now equal, Eq. 3.8 leads to pgage SGHg ρH2O× g× 1.0× m⋅ ρH2O g× 1.0× m⋅−= pgage ρH2O g× SGHg 1× m⋅ 1.0 m⋅−( )×= pgage 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 13.55 1× m⋅ 1.0 m⋅−( )× N s 2 ⋅ kg m⋅ ×= pgage 123 kPa⋅= Problem 3.16 3.16
Problem 3.20 [Difficulty: 2] Given: Two-fluid manometer as shown l 10.2 mm⋅= SGct 1.595= (From Table A.1, App. A) Find: Pressure difference Solution: We will apply the hydrostatics equation. Governing equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) d z Assumptions: (1) Static liquid (2) Incompressible liquid Starting at point 1 and progressing to point 2 we have: p1 ρwater g⋅ d l+( )⋅+ ρct g⋅ l⋅− ρwater g⋅ d⋅− p2= Simplifying and solving for p2 p1− we have: Δp p2 p1−= ρct g⋅ l⋅ ρwater g⋅ l⋅−= SGct 1−( ) ρwater⋅ g⋅ l⋅= Substituting the known data: Δp 1.591 1−( ) 1000× kg m 3 ⋅ 9.81× m s 2 ⋅ 10.2× mm⋅ m 10 3 mm⋅ ×= Δp 59.1Pa= Problem 3.17 3.17
Problem 3.22 [Difficulty: 2] Given: Two fluid manometer contains water and kerosene. With both tubes open to atmosphere, the difference in free surface elevations is known Ho 20 mm⋅= SGk 0.82= (From Table A.1, App. A) Find: The elevation difference, H, between the free surfaces of the fluids when a gage pressure of 98.0 Pa is applied to the right tube. Solution: We will apply the hydrostatics equation. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid When the gage pressure Δp is applied to the right tube, the water in the right tube is displaced downward by a distance, l. The kerosene in the left tube is displaced upward by the same distance, l. Under the applied gage pressure Δp, the elevation difference, H, is: h H A B l l H0 H1 A B Δp H Ho 2 l⋅+= Since points A and B are at the same elevation in the same fluid, their pressures are the same. Initially: pA ρk g⋅ Ho H1+( )⋅= pB ρwater g⋅ H1⋅= Setting these pressures equal: ρk g⋅ Ho H1+( )⋅ ρwater g⋅ H1⋅= Solving for H1 H1 ρk Ho⋅ ρwater ρk− = SGk Ho⋅ 1 SGk− = H1 0.82 20× mm⋅ 1 0.82− = H1 91.11 mm⋅= Now under the applied gage pressure: pA ρk g⋅ Ho H1+( )⋅ ρwater g⋅ l⋅+= pB Δp ρwater g⋅ H1 l−( )⋅+= Problem 3.18 3.18
Setting these pressures equal: SGk Ho H1+( )⋅ l+ Δp ρwater g⋅ H1 l−( )+= l 1 2 Δp ρwater g⋅ H1+ SGk Ho H1+( )⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = Substituting in known values we get: l 1 2 98.0 N m 2 ⋅ 1 999 × m 3 kg 1 9.81 × s 2 m ⋅ kg m⋅ N s 2 ⋅ × 91.11 mm⋅ 0.82 20 mm⋅ 91.11 mm⋅+( )×−[ ] m 10 3 mm⋅ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×= l 5.000 mm⋅= Now we solve for H: H 20 mm⋅ 2 5.000× mm⋅+= H 30.0 mm⋅=
Problem 3.24 [Difficulty: 2] Given: Data on manometer Find: Gage pressure at point a Assumption: Water, liquids A and B are static and incompressible Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ Δp ρ g⋅ Δh⋅= where Δh is height difference Starting at point a p1 pa ρH2O g⋅ h1⋅−= where h1 0.125 m⋅ 0.25 m⋅+= h1 0.375m= Next, in liquid A p2 p1 SGA ρH2O⋅ g⋅ h2⋅+= where h2 0.25 m⋅= Finally, in liquid B patm p2 SGB ρH2O⋅ g⋅ h3⋅−= where h3 0.9 m⋅ 0.4 m⋅−= h3 0.5m= Combining the three equations patm p1 SGA ρH2O⋅ g⋅ h2⋅+( ) SGB ρH2O⋅ g⋅ h3⋅−= pa ρH2O g⋅ h1⋅− SGA ρH2O⋅ g⋅ h2⋅+ SGB ρH2O⋅ g⋅ h3⋅−= pa patm ρH2O g⋅ h1 SGA h2⋅− SGB h3⋅+( )⋅+= or in gage pressures pa ρH2O g⋅ h1 SGA h2⋅− SGB h3⋅+( )⋅= pa 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.375 1.20 0.25×( )− 0.75 0.5×( )+[ ]× m⋅ N s 2 ⋅ kg m⋅ ×= pa 4.41 10 3 × Pa= pa 4.41 kPa⋅= (gage) Problem 3.19 3.19
Problem 3.20 (Difficulty: 1) 3.20 With the manometer reading as shown, calculate 𝑝 𝑥. Given: Oil specific gravity: 𝑆𝑆 𝑜𝑜𝑜 = 0.85 Depth: ℎ1 = 60 𝑖𝑖𝑖ℎ. ℎ2 = 30 𝑖𝑖𝑖ℎ. Find: The pressure 𝑝 𝑥. Assumption: Fluids are incompressible Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference over a difference in elevation (h): ∆𝑝 = 𝜌𝜌ℎ Repeated application of this relation yields 𝑝 𝑥 = 𝑆𝑆 𝑜𝑜𝑜 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ1 + 𝛾 𝑀ℎ2 The specific weight for mercury is: 𝛾 𝑀 = 845 𝑙𝑙𝑙 𝑓𝑓3 The pressure at the desired location is 𝑝 𝑥 = 0.85 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × � 60 12 � 𝑓𝑓 + 845 𝑙𝑙𝑙 𝑓𝑓3 × � 30 12 � 𝑓𝑓 = 2380 𝑙𝑙𝑙 𝑓𝑓2 = 16.5 𝑝𝑝𝑝
Problem 3.21 (Difficulty: 2) 3.21 Calculate 𝑝 𝑥 − 𝑝 𝑦 for this inverted U-tube manometer. Given: Oil specific gravity: 𝑆𝑆 𝑜𝑜𝑜 = 0.90 Depth: ℎ1 = 65 𝑖𝑖𝑖ℎ. ℎ2 = 20 𝑖𝑖𝑖ℎ. ℎ3 = 10 𝑖𝑖𝑖ℎ. Find: The pressure difference 𝑝 𝑥 − 𝑝 𝑦. Assume: The fluids are incompressible Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference over a difference in elevation (h): ∆𝑝 = 𝜌𝜌ℎ Starting at the location of the unknown pressure px, we have the following relations for the hydrostatic pressure: 𝑝 𝑥 − 𝑝1 = 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ1 𝑝1 − 𝑝2 = −𝑆𝑆 𝑜𝑜𝑜 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ3 𝑝2 − 𝑝 𝑦 = −𝛾 𝑤𝑤𝑤𝑤𝑤(ℎ1 − ℎ2 − ℎ3) Adding these three equations together 𝑝 𝑥 − 𝑝 𝑦 = 𝛾 𝑤𝑤𝑤𝑤𝑤(ℎ2 + ℎ3) − 𝑆𝑆 𝑜𝑜𝑜 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ3
The pressure difference is then 𝑝 𝑥 − 𝑝 𝑦 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × (10 + 20) 12 𝑓𝑓 − 0.9 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 10 12 𝑓𝑓 = 109.2 𝑙𝑙𝑙 𝑓𝑓2 = 0.758 𝑝𝑝𝑝
Problem 3.22 (Difficulty: 2) 3.22 An inclined gage having a tube of 3 mm bore, laid on a slope of 1:20, and a reservoir of 25 mm diameter contains silicon oil (SG 0.84). What distance will the oil move along the tube when a pressure of 25 mm of water is connected to the gage? Given: Silicon oil specific gravity: 𝑆𝑆 𝑜𝑜𝑜 = 0.84. Diameter: 𝐷1 = 3 𝑚𝑚. 𝐷2 = 25 𝑚𝑚. Depth: ℎ 𝑤𝑤𝑤𝑤𝑤 = 25 𝑚𝑚. Slope angle: 1: 20. Find: The distance 𝑥 of the oil move along the tube. Assumption: Fluids are incompressible Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference over a difference in elevation (h): ∆𝑝 = 𝜌𝜌ℎ We have the volume of the oil as constant, so: 𝐴 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟∆ℎ = 𝐴 𝑡𝑡𝑡𝑡 𝑥 or ∆ℎ 𝑥 = 𝐴 𝑡𝑡𝑡𝑡 𝐴 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐷1 2 𝐷2 2 = 9 625 When a pressure of 25 𝑚𝑚 of water is connected with the gage we have: 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ 𝑤𝑤𝑤𝑤𝑤 = 𝑆𝑆 𝑜𝑜𝑜 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ
ℎ = ℎ 𝑤𝑤𝑤𝑤𝑤 𝑆𝑆 𝑜𝑜𝑜 = 29.8 𝑚𝑚 Using these relations, we obtain, accounting for the slope of the manometer: ℎ = ∆ℎ + 𝑥 √202 + 12 = � 9 625 + 1 √202 + 12 � 𝑥 ℎ = ∆ℎ + 𝑥 √401 = � 9 625 + 1 √401 � 𝑥 𝑥 = ℎ � 9 625 + 1 √401 � = 463 𝑚𝑚
Problem 3.26 [Difficulty: 2] Given: Water flow in an inclined pipe as shown. The pressure difference is measured with a two-fluid manometer L 5 ft⋅= h 6 in⋅= SGHg 13.55= (From Table A.1, App. A) Find: Pressure difference between A and B Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid (3) Gravity is constant Integrating the hydrostatic pressure equation we get: Δp ρ g⋅ Δh⋅= Progressing through the manometer from A to B: pA ρwater g⋅ L⋅ sin 30 deg⋅( )⋅+ ρwater g⋅ a⋅+ ρwater g⋅ h⋅+ ρHg g⋅ h⋅− ρwater g⋅ a⋅− pB= Simplifying terms and solving for the pressure difference: Δp pA pB−= ρwater g⋅ h SGHg 1−( )⋅ L sin 30 deg⋅( )⋅−⎡⎣ ⎤⎦⋅= Substituting in values: Δp 1.94 slug ft 3 ⋅ 32.2× ft s 2 6 in⋅ ft 12 in⋅ × 13.55 1−( )× 5 ft⋅ sin 30 deg⋅( )×−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × lbf s 2 ⋅ slugft⋅ × ft 12 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ×= Δp 1.638 psi⋅= Problem 3.23 3.23
Problem 3.28 [Difficulty: 2] Given: Reservoir manometer with vertical tubes of knowm diameter. Gage liquid is Meriam red oil D 18 mm⋅= d 6 mm⋅= SGoil 0.827= (From Table A.1, App. A) Find: The manometer deflection, L when a gage pressure equal to 25 mm of water is applied to the reservoir. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: Δp ρ g⋅ Δh⋅= Beginning at the free surface of the reservoir, and accounting for the changes in pressure with elevation: patm Δp+ ρoil g⋅ x L+( )⋅+ patm= Upon simplification: x L+ Δp ρoil g⋅ = The gage pressure is defined as: Δp ρwater g⋅ Δh⋅= where Δh 25 mm⋅= Combining these two expressions: x L+ ρwater g⋅ h⋅ ρoil g⋅ = Δh SGoil = x and L are related through the manometer dimensions: π 4 D 2 ⋅ x⋅ π 4 d 2 ⋅ L⋅= x d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 L= Therefore: L Δh SGoil 1 d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = Substituting values into the expression: L 25 mm⋅ 0.827 1 6 mm⋅ 18 mm⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = (Note: s L Δh = which yields s 1.088= for this manometer.) L 27.2 mm⋅= Problem 3.24 3.24
Problem 3.29 [Difficulty: 2] Given: A U-tube manometer is connected to the open tank filled with water as shown (manometer fluid is Meriam blue) D1 2.5 m⋅= D2 0.7 m⋅= d 0.2 m⋅= SGoil 1.75= (From Table A.1, App. A) Find: The manometer deflection, l Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid D1 D2 d Integrating the hydrostatic pressure equation we get: Δp ρ g⋅ Δh⋅= When the tank is filled with water, the oil in the left leg of the manometer is displaced downward by l/2. The oil in the right leg is displaced upward by the same distance, l/2. Beginning at the free surface of the tank, and accounting for the changes in pressure with elevation: patm ρwater g⋅ D1 D2− d+ l 2 +⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅+ ρoil g⋅ l⋅− patm= Upon simplification: ρwater g⋅ D1 D2− d+ l 2 +⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ ρoil g⋅ l⋅= D1 D2− d+ l 2 + SGoil l⋅= l D1 D2− d+ SGoil 1 2 − = l 2.5 m⋅ 0.7 m⋅− 0.2 m⋅+ 1.75 1 2 − = l 1.600m= Problem 3.25 3.25
Problem 3.26 (Difficulty: 2) 3.26 The sketch shows a sectional view through a submarine. Calculate the depth of submarine, y. Assume the specific weight of the seawater is 10.0 𝑘𝑘 𝑚3. Given: Atmos. Pressure: 𝑝 𝑎𝑎𝑎𝑎𝑎 = 740 𝑚𝑚 𝐻𝐻. Seawater specific weight:𝛾 = 10.0 𝑘𝑘 𝑚3. All the dimensional relationship is shown in the figure. Find: The depth 𝑦. Assumption: Fluids are incompressible Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference over a difference in elevation (h): ∆𝑝 = 𝜌𝜌ℎ Using the barometer reading with 760 mm as atmospheric pressure, the pressure inside the submarine is: 𝑝 = 840 𝑚𝑚 760 𝑚𝑚 × 101.3 × 103 𝑃𝑃 = 111.6 × 103 𝑃𝑃
However, the actual atmosphere pressure is: 𝑝 𝑎𝑎𝑎𝑎𝑎 = 740 𝑚𝑚 760 𝑚𝑚 × 101.3 × 103 𝑃𝑃 = 98.3 × 103 𝑃𝑃 For the manometer, using the hydrostatic relation, we have for the pressure, where y is the depth of the submarine: 𝑝 = 𝑝 𝑎𝑎𝑎𝑎𝑎 + 𝛾𝛾 + 𝛾 × 200 𝑚𝑚 − 𝛾 𝐻𝐻 × 400 𝑚𝑚 𝑦 = 𝑝 + 𝛾 𝐻𝐻 × 400 𝑚𝑚 − 𝛾 × 200 𝑚𝑚 − 𝑝 𝑎𝑎𝑎𝑎𝑎 𝛾 The specific weight for mercury is: 𝛾 𝐻𝐻 = 133.1 𝑘𝑘 𝑚3 So we have for the depth y: 𝑦 = 111.6 × 103 𝑃𝑃 + 133.1 × 1000 𝑁 𝑚3 × 0.4 𝑚 − 1000 𝑁 𝑚3 × 0.2 𝑚 − 98.3 × 103 𝑃𝑃 1000 𝑁 𝑚3 𝑦 = 6.45 𝑚
Problem 3.27 (Difficulty: 1) 3.27 The manometer reading is 6 in. when the tank is empty (water surface at A). Calculate the manometer reading when the cone is filled with water. Find: The manometer reading when the tank is filled with water. Assumption: Fluids are static and incompressible Solution: Use the hydrostatic relations for pressure When the tank is empty, we have the equation as: ℎ 𝑀𝑀 ∙ 𝑆𝑆 𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∙ 𝛾 𝑤𝑤𝑤𝑤𝑤 = 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ 𝑆𝑆 𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 13.57 ℎ = ℎ 𝑀𝑀 ∙ 𝑆𝑆 𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 150 𝑚𝑚 × 13.57 = 2.04 𝑚 When the tank is filled with water, we assume the mercury interface moves by 𝑥: 𝛾 𝑤𝑤𝑤𝑤𝑤(ℎ 𝑡𝑡𝑡𝑡 + ℎ + 𝑥) = 𝛾 𝑤𝑤𝑤𝑤𝑤 ∙ 𝑆𝑆 𝑚𝑚𝑚𝑚𝑚𝑚𝑚(ℎ 𝑀𝑀 + 2𝑥) (3 𝑚 + 2.04 𝑚 + 𝑥) = 13.57(0.15𝑚 + 2𝑥) Thus 𝑥 = 0.115 𝑚 The new manometer reading is: ℎ 𝑀𝑀 ′ = ℎ 𝑀𝑀 + 2𝑥 = 0.15 𝑚 + 2 × 0.115 𝑚 = 0.38 𝑚
Problem 3.30 [Difficulty: 2] Given: Reservoir manometer with dimensions shown. The manometer fluid specific gravity is given. D 5 8 in⋅= d 3 16 in⋅= SGoil 0.827= Find: The required distance between vertical marks on the scale corresponding to Δp of 1 in water. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dz ρ− g⋅= (Hydrostatic Pressure - z is positive upwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid h x Integrating the hydrostatic pressure equation we get: Δp ρ− g⋅ Δz⋅= Beginning at the free surface of the tank, and accounting for the changes in pressure with elevation: patm Δp+ ρoil g⋅ x h+( )⋅− patm= Upon simplification: Δp ρoil g⋅ x h+( )⋅= The applied pressure is defined as: Δp ρwater g⋅ l⋅= where l 1 in⋅= Therefore: ρwater g⋅ l⋅ ρoil g⋅ x h+( )⋅= x h+ l SGoil = x and h are related through the manometer dimensions: π 4 D 2 ⋅ x⋅ π 4 d 2 ⋅ h⋅= x d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 h= Solving for h: h l SGoil 1 d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = Substituting values into the expression: h 1 in⋅ 0.827 1 0.1875 in⋅ 0.625 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = h 1.109 in⋅= Problem 3.28 3.28
Problem 3.32 [Difficulty: 3] Given: Inclined manometer as shown. D 96 mm⋅= d 8 mm⋅= Angle θ is such that the liquid deflection L is five times that of a regular U-tube manometer. Find: Angle θ and manometer sensitivity. Solution: We will apply the hydrostatics equations to this system. Governing Equation: dp dz ρ− g⋅= (Hydrostatic Pressure - z is positive upwards) Assumptions: (1) Static liquid (2) Incompressible liquid x Integrating the hydrostatic pressure equation we get: Δp ρ− g⋅ Δz⋅= Applying this equation from point 1 to point 2: p1 ρ g⋅ x L sin θ( )⋅+( )⋅− p2= Upon simplification: p1 p2− ρ g⋅ x L sin θ( )⋅+( )⋅= Since the volume of the fluid must remain constant: π 4 D 2 ⋅ x⋅ π 4 d 2 ⋅ L⋅= x d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 L⋅= Therefore: p1 p2− ρ g⋅ L⋅ d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 sin θ( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Now for a U-tube manometer: p1 p2− ρ g⋅ h⋅= Hence: p1incl p2incl− p1U p2U− ρ g⋅ L⋅ d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 sin θ( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ ρ g⋅ h⋅ = For equal applied pressures: L d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 sin θ( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ h= Since L/h = 5: sin θ( ) h L d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= 1 5 8 mm⋅ 96 mm⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= θ 11.13 deg⋅= The sensitivity of the manometer: s L Δhe = L SG h⋅ = s 5 SG = Problem 3.29 3.29
Problem 3.33 [Difficulty: 3] Given: Data on inclined manometer Find: Angle θ for given data; find sensitivity Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ Δp ρ g⋅ Δh⋅= where Δh is height difference Under applied pressure Δp SGMer ρ⋅ g⋅ L sin θ( )⋅ x+( )⋅= (1) From Table A.1 SGMer 0.827= and Δp = 1 in. of water, or Δp ρ g⋅ h⋅= where h 25 mm⋅= h 0.025m= Δp 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.025× m⋅ N s 2 ⋅ kg m⋅ ×= Δp 245Pa= The volume of liquid must remain constant, so x Ares⋅ L Atube⋅= x L Atube Ares ⋅= L d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⋅= (2) Combining Eqs 1 and 2 Δp SGMer ρ⋅ g⋅ L sin θ( )⋅ L d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Solving for θ sin θ( ) Δp SGMer ρ⋅ g⋅ L⋅ d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= sin θ( ) 245 N m 2 ⋅ 1 0.827 × 1 1000 × m 3 kg ⋅ 1 9.81 × s 2 m ⋅ 1 0.15 × 1 m ⋅ kg m⋅ s 2 N⋅ × 8 76 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= 0.186= θ 11 deg⋅= The sensitivity is the ratio of manometer deflection to a vertical water manometer s L h = 0.15 m⋅ 0.025 m⋅ = s 6= Problem 3.30 3.30
Problem 3.34 [Difficulty: 4] Given: Barometer with water on top of the mercury column, Temperature is known: h2 6.5 in⋅= h1 28.35 in⋅= SGHg 13.55= (From Table A.2, App. A) T 70 °F= pv 0.363 psi⋅= (From Table A.7, App. A) Find: (a) Barometric pressure in psia (b) Effect of increase in ambient temperature on length of mercury column for the same barometric pressure: Tf 85 °F= Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ− g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) h2 Water vapor h1 Water Mercury Assumptions: (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: Δp ρ g⋅ Δh⋅= Start at the free surface of the mercury and progress through the barometer to the vapor pressure of the water: patm ρHg g⋅ h1⋅− ρwater g⋅ h2⋅− pv= patm pv ρwater g⋅ SGHg h1⋅ h2+( )⋅+= patm 0.363 lbf in 2 ⋅ 1.93 slug ft 3 ⋅ 32.2× ft s 2 ⋅ lbf s 2 ⋅ slug ft⋅ × 13.55 28.35× in⋅ 6.5 in⋅+( )× ft 12 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ×+= patm 14.41 lbf in 2 ⋅= At the higher temperature, the vapor pressure of water increases to 0.60 psi. Therefore, if the atmospheric pressure were to remain constant, the length of the mercury column would have to decrease - the increased water vapor would push the mercury out of the tube! Problem 3.31 3.31
Problem 3.36 [Difficulty: 3] Given: Water column standin in glass tube Δh 50 mm⋅= D 2.5 mm⋅= σ 72.8 10 3− × N m = (From Table A.4, App. A) Find: (a) Column height if surface tension were zero. (b) Column height in 1 mm diameter tube Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) Δhp Δhc Δh Δhc πDδ θ Mg = ρgV ΣFz 0= (Static Equilibrium) Assumptions: (1) Static, incompressible liquid (2) Neglect volume under meniscus (3) Applied pressure remains constant (4) Column height is sum of capillary rise and pressure difference Assumption #4 can be written as: Δh Δhc Δhp+= Choose a free-body diagram of the capillary rise portion of the column for analysis: ΣFz π D⋅ σ⋅ cos θ( )⋅ π 4 D 2 ⋅ ρ⋅ g⋅ Δhc⋅−= 0= Therefore: Δhc 4 σ⋅ ρ g⋅ D⋅ cos θ( )⋅= Substituting values: Δhc 4 72.8× 10 3− × N m ⋅ 1 999 × m 3 kg ⋅ 1 9.81 × s 2 m ⋅ 1 2.5 × 1 mm ⋅ kg m⋅ N s 2 ⋅ × 10 3 mm⋅ m ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ×= Δhc 11.89 mm⋅= Therefore: Δhp Δh Δhc−= Δhp 50 mm⋅ 11.89 mm⋅−= Δhp 38.1 mm⋅= (result for σ = 0) For the 1 mm diameter tube: Δhc 4 72.8× 10 3− × N m ⋅ 1 999 × m 3 kg ⋅ 1 9.81 × s 2 m ⋅ 1 1 × 1 mm ⋅ kg m⋅ N s 2 ⋅ × 10 3 mm⋅ m ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ×= Δhc 29.71 mm⋅= Δh 29.7 mm⋅ 38.1 mm⋅+= Δh 67.8 mm⋅= Problem 3.32 3.32
Problem 3.38 [Difficulty :2] Fluid 1 Fluid 2 σπDcosθ ρ1gΔhπD2 /4 Given: Two fluids inside and outside a tube Find: (a) An expression for height Δh (b) Height difference when D =0.040 in for water/mercury Assumptions: (1) Static, incompressible fluids (2) Neglect meniscus curvature for column height and volume calculations Solution: A free-body vertical force analysis for the section of fluid 1 height Δh in the tube below the "free surface" of fluid 2 leads to F ∑ 0= Δp π D 2 ⋅ 4 ⋅ ρ1 g⋅ Δh⋅ π D 2 ⋅ 4 ⋅− π D⋅ σ⋅ cos θ( )⋅+= where Δp is the pressure difference generated by fluid 2 over height Δh, Δp ρ2 g⋅ Δh⋅= Hence Δp π D 2 ⋅ 4 ⋅ ρ1 g⋅ Δh⋅ π D 2 ⋅ 4 ⋅− ρ2 g⋅ Δh⋅ π D 2 ⋅ 4 ⋅ ρ1 g⋅ Δh⋅ π D 2 ⋅ 4 ⋅−= π− D⋅ σ⋅ cos θ( )⋅= Solving for Δh Δh 4 σ⋅ cos θ( )⋅ g D⋅ ρ2 ρ1−( )⋅ −= For fluids 1 and 2 being water and mercury (for mercury σ = 375 mN/m and θ = 140o, from Table A.4), solving for Δh when D = 0.040 in Δh 4− 0.375× N m ⋅ lbf 4.448 N⋅ × 0.0254m in × cos 140 deg⋅( )× s 2 32.2 ft⋅ × 1 0.040 in⋅ × ft 3 1.94 slug⋅ × 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 × 1 13.6 1−( ) × slugft⋅ lbf s 2 ⋅ ×= Δh 0.360 in⋅= Problem 3.33 3.33
Problem 3.40 [Difficulty: 2] Water Given: Water in a tube or between parallel plates Find: Height Δh for each system Solution: a) Tube: A free-body vertical force analysis for the section of water height Δh above the "free surface" in the tube, as shown in the figure, leads to F ∑ 0= π D⋅ σ⋅ cos θ( )⋅ ρ g⋅ Δh⋅ π D 2 ⋅ 4 ⋅−= Assumption: Neglect meniscus curvature for column height and volume calculations Solving for Δh Δh 4 σ⋅ cos θ( )⋅ ρ g⋅ D⋅ = b) Parallel Plates: A free-body vertical force analysis for the section of water height Δh above the "free surface" between plates arbitrary width w (similar to the figure above), leads to F ∑ 0= 2 w⋅ σ⋅ cos θ( )⋅ ρ g⋅ Δh⋅ w⋅ a⋅−= Solving for Δh Δh 2 σ⋅ cos θ( )⋅ ρ g⋅ a⋅ = For water σ = 72.8 mN/m and θ = 0o (Table A.4), so a) Tube Δh 4 0.0728× N m ⋅ 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.005× m⋅ kg m⋅ N s 2 ⋅ ×= Δh 5.94 10 3− × m= Δh 5.94 mm⋅= b) Parallel Plates Δh 2 0.0728× N m ⋅ 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.005× m⋅ kg m⋅ N s 2 ⋅ ×= Δh 2.97 10 3− × m= Δh 2.97 mm⋅= Problem 3.34 3.34
p SL = 101 kPa R = 286.9 J/kg.K ρ = 999 kg/m3 The temperature can be computed from the data in the figure. The pressures are then computed from the appropriate equation. From Table A.3 Agreement between calculated and tabulated data is very good (as it should be, considering the table data are also computed!) Atmospheric Pressure vs Elevation 0.00000 0.00001 0.00010 0.00100 0.01000 0.10000 1.00000 0 10 20 30 40 50 60 70 80 90 100 Elevation (km) PressureRatiop/pSL Computed Table A.3 Problem 3.35 3.35
z (km) T (o C) T (K) p /p SL z (km) p /p SL 0.0 15.0 288.0 m = 1.000 0.0 1.000 2.0 2.0 275.00 0.0065 0.784 0.5 0.942 4.0 -11.0 262.0 (K/m) 0.608 1.0 0.887 6.0 -24.0 249.0 0.465 1.5 0.835 8.0 -37.0 236.0 0.351 2.0 0.785 11.0 -56.5 216.5 0.223 2.5 0.737 12.0 -56.5 216.5 T = const 0.190 3.0 0.692 14.0 -56.5 216.5 0.139 3.5 0.649 16.0 -56.5 216.5 0.101 4.0 0.609 18.0 -56.5 216.5 0.0738 4.5 0.570 20.1 -56.5 216.5 0.0530 5.0 0.533 22.0 -54.6 218.4 m = 0.0393 6.0 0.466 24.0 -52.6 220.4 -0.000991736 0.0288 7.0 0.406 26.0 -50.6 222.4 (K/m) 0.0211 8.0 0.352 28.0 -48.7 224.3 0.0155 9.0 0.304 30.0 -46.7 226.3 0.0115 10.0 0.262 32.2 -44.5 228.5 0.00824 11.0 0.224 34.0 -39.5 233.5 m = 0.00632 12.0 0.192 36.0 -33.9 239.1 -0.002781457 0.00473 13.0 0.164 38.0 -28.4 244.6 (K/m) 0.00356 14.0 0.140 40.0 -22.8 250.2 0.00270 15.0 0.120 42.0 -17.2 255.8 0.00206 16.0 0.102 44.0 -11.7 261.3 0.00158 17.0 0.0873 46.0 -6.1 266.9 0.00122 18.0 0.0747 47.3 -2.5 270.5 0.00104 19.0 0.0638 50.0 -2.5 270.5 T = const 0.000736 20.0 0.0546 52.4 -2.5 270.5 0.000544 22.0 0.0400 54.0 -5.6 267.4 m = 0.000444 24.0 0.0293 56.0 -9.5 263.5 0.001956522 0.000343 26.0 0.0216 58.0 -13.5 259.5 (K/m) 0.000264 28.0 0.0160 60.0 -17.4 255.6 0.000202 30.0 0.0118 61.6 -20.5 252.5 0.000163 40.0 0.00283 64.0 -29.9 243.1 m = 0.000117 50.0 0.000787 66.0 -37.7 235.3 0.003913043 0.0000880 60.0 0.000222 68.0 -45.5 227.5 (K/m) 0.0000655 70.0 0.0000545 70.0 -53.4 219.6 0.0000482 80.0 0.0000102 72.0 -61.2 211.8 0.0000351 90.0 0.00000162 74.0 -69.0 204.0 0.0000253 76.0 -76.8 196.2 0.0000180 78.0 -84.7 188.3 0.0000126 80.0 -92.5 180.5 T = const 0.00000861 82.0 -92.5 180.5 0.00000590 84.0 -92.5 180.5 0.00000404 86.0 -92.5 180.5 0.00000276 88.0 -92.5 180.5 0.00000189 90.0 -92.5 180.5 0.00000130
Problem 3.44 [Difficulty: 3] Given: Atmospheric conditions at ground level (z = 0) in Denver, Colorado are p0 = 83.2 kPa, T0 = 25°C. Pike's peak is at elevation z = 2690 m. Find: p/p0 vs z for both cases. Solution: Governing Equations: dp dz ρ− g⋅= p ρ R⋅ T⋅= Assumptions: (1) Static fluid (2) Ideal gas behavior (a) For an incompressible atmosphere: dp dz ρ− g⋅= becomes p p0− 0 z zρ g⋅ ⌠ ⎮ ⌡ d−= or p p0 ρ0 g⋅ z⋅−= p0 1 g z⋅ R T0⋅ − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= (1) At z 2690 m⋅= p 83.2 kPa⋅ 1 9.81 m s 2 ⋅ 2690× m⋅ kg K⋅ 287 N⋅ m⋅ × 1 298 K⋅ × N s 2 ⋅ kg m⋅ ×− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×= p 57.5 kPa⋅= (b) For an adiabatic atmosphere: p ρ k const= ρ ρ0 p p0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 k ⋅= dp dz ρ− g⋅= becomes dp ρ0− p p0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 k ⋅ g⋅ dz⋅= or 1 p 1 k dp ρ0 g⋅ p0 1 k − dz⋅= But p0 p p 1 p 1 k ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d k k 1− p p0−( ) k 1− k ⋅= hence k k 1− p k 1− k p0 k 1− k − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠⋅ ρ0 g⋅ p0 1 k − g⋅ z⋅= Solving for the pressure ratio p p0 1 k 1− k ρ0 p0 ⋅ g⋅ z⋅− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k k 1− = or p p0 1 k 1− k g z⋅ R T0⋅ ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k k 1− = (2) At z 2690 m⋅= p 83.2 kPa⋅ 1 1.4 1− 1.4 9.81× m s 2 ⋅ 2690× m⋅ kg K⋅ 287 N⋅ m⋅ × 1 298 K⋅ × N s 2 ⋅ kg m⋅ ×− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 1.4 1.4 1− ×= p 60.2 kPa⋅= Problem 3.36 3.36
Equations 1 and 2 can be plotted: 0.4 0.6 0.8 1 0 1 10 3 × 2 10 3 × 3 10 3 × 4 10 3 × 5 10 3 × Incompressible Adiabatic Temperature Variation with Elevation Pressure Ratio (-) ElevationaboveDenver(m)
Problem 3.37 (Difficulty: 2) 3.37 If atmospheric pressure at the ground is 101.3 𝑘𝑘𝑘 and temperature is 15 ℃, calculate the pressure 7.62 𝑘𝑘 above the ground, assuming (a) no density variation, (b) isothermal variation of density with pressure, and (c) adiabatic variation of density with pressure. Assumption: Atmospheric air is stationary and behaves as an ideal gas. Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 (a) For this case with no density variation, we integrate with respect to z from the ground level pressure p0 to the pressure at any height h. The pressure is 𝑝 = 𝑝0 − 𝛾ℎ From Table A.10, the density of air at sea level is 𝜌 = 1.23 𝑘𝑘 𝑚3 Or the specific weight is 𝛾 = 𝜌𝜌 = 1.23 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 = 12.07 𝑁 𝑚3 Thus the pressure at 7.62 km is 𝑝 = 101.3 𝑘𝑘𝑘 − 12.07 𝑁 𝑚3 × 7.62 × 1000 𝑚 = 9.63 𝑘𝑘𝑘 (b) For isothermal condition we have for an ideal gas: 𝑝 𝜌 = 𝑝0 𝜌0 = 𝑅𝑅 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Therefore, since ρ = γ g and g is a constant 𝑝 𝛾 = 𝑝0 𝛾0 = 101.3 𝑘𝑘𝑘 12.07 𝑁 𝑚3 = 8420 𝑚 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 From the hydrostatic relation we have: 𝑑𝑑 = −𝛾𝛾𝛾 𝑑𝑑 𝑝 = − 𝛾 𝑝 𝑑𝑑
� 𝑑𝑑 𝑝 𝑝 𝑝0 = − 1 8420𝑚 � 𝑑𝑑 𝑧 0 ln � 𝑝 𝑝0 � = − 1 8420𝑚 𝑧 Thus the pressure at 7.62 km is 𝑝 𝑝0 = 𝑒− − 7620 𝑚 8420𝑚 = 𝑒− 0.905 = 0.4045 𝑝 = 101.3𝑘𝑘𝑘 × 0.4045 = 41.0 𝑘𝑘𝑘 (c) For a reversible and adiabatic variation of density we have: 𝑝𝑣 𝑘 = 𝑝 𝜌 𝑘 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Where k is the specific heat ratio 𝑘 = 1.4 Or, since gravity g is constant, we can write in terms of the specific weight 𝑝 𝛾 𝑘 = 𝑝0 𝛾0 𝑘 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Or the specific weight is 𝛾 = 𝛾0 � 𝑝 𝑝0 � 1 𝑘� The hydrostatic expression becomes 𝑑𝑑 = −𝛾0 � 𝑝 𝑝0 � 1 𝑘� 𝑑𝑑 Separating variables 𝑝0 1/𝑘 𝛾0 � 𝑑𝑑 (𝑝)1/𝑘 𝑝 𝑝0 = − � 𝑑𝑑 𝑧 0 Integrating between the limits p=p0 at z=0 and p = p at z = z � 𝑘 𝑘 − 1 � 𝑝0 1/𝑘 𝛾0 �𝑝 𝑘−1 𝑘 − 𝑝0 𝑘−1 𝑘 � = − 𝑧 Or � 𝑝 𝑝0 � 𝑘−1 𝑘 = 1 − � 𝑘 − 1 𝑘 � 𝛾0 𝑧 𝑝0 The pressure is then 𝑝 = 𝑝0 �1 − � 𝑘 − 1 𝑘 � 𝛾0 𝑧 𝑝0 � 𝑘 𝑘−1� = 101.3𝑘𝑘𝑘 �1 − � 1.4 − 1 1.4 � × 12.07 𝑁 𝑚3 × 7620𝑚 101.3 × 1000 𝑃𝑃 � 1.4 1.4−1� 𝑝 = 35.4 𝑘𝑘𝑘 The calculation of pressure depends heavily on the assumption we make about how density changes.
Problem 3.38 (Difficulty: 2) 3.38 If the temperature in the atmosphere is assumed to vary linearly with altitude so T = T0 - αz where T0 is the sea level temperature and α = - dT / dz is the temperature lapse rate, find p(z) when air is taken to be a perfect gas. Give the answer in terms of p0, a, g, R, and z only. Assumption: Atmospheric air is stationary and behaves as an ideal gas. Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 = −𝛾𝛾𝛾 The ideal gas relation is 𝑝 𝜌 = 𝑅𝑅 Or in terms of the specific weight, the pressure is 𝑝 = 𝜌𝜌𝜌 = 𝛾 𝑔 𝑅𝑅 Relating the temperature to the adiabatic lapse rate 𝑝 = 𝛾 𝑔 𝑅(𝑇0 − 𝛼𝛼) Inserting the expression for specific weight into the hydrostatic equation 𝑑𝑑 = − 𝑔𝑔 𝑅(𝑇0 − 𝛼𝛼) 𝑑𝑑 Separating variables 𝑑𝑑 𝑝 = − 𝑔 𝑅 𝑑𝑑 (𝑇0 − 𝛼𝛼) Integrating between the surface and any height z � 𝑑𝑑 𝑝 𝑝 𝑝0 = − 𝑔 𝑅 � 𝑑𝑑 (𝑇0 − 𝛼𝛼) 𝑧 0 Or 𝑙𝑙 � 𝑝 𝑝0 � = − 𝑔 𝑅 𝑙𝑙 � 𝑇0 − 𝛼𝛼 𝑇0 �
In terms of p 𝑝 𝑝0 = �1 − 𝛼𝛼 𝑇0 � 𝑔 𝛼𝛼�
Problem 3.46 [Difficulty: 3] Given: Door located in plane vertical wall of water tank as shown c ps a y’ y b a 1.5 m⋅= b 1 m⋅= c 1 m⋅= Atmospheric pressure acts on outer surface of door. Find: Resultant force and line of action: (a) for (b) for ps patm= psg 0.3 atm⋅= Plot F/Fo and y'/yc over range of ps/patm (Fo is force determined in (a), yc is y-ccordinate of door centroid). Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dy ρ g⋅= (Hydrostatic Pressure - y is positive downwards) FR Ap ⌠ ⎮ ⎮ ⌡ d= (Hydrostatic Force on door) y' FR⋅ Ay p⋅ ⌠ ⎮ ⎮ ⌡ d= (First moment of force) Assumptions: (1) Static fluid (2) Incompressible fluid We will obtain a general expression for the force and line of action, and then simplify for parts (a) and (b). Since dp ρ g⋅ dh⋅= it follows that p ps ρ g⋅ y⋅+= Now because patm acts on the outside of the door, psg is the surface gage pressure: p psg ρ g⋅ y⋅+= FR Ap ⌠⎮ ⎮ ⌡ d= c c a+ yp b⋅ ⌠ ⎮ ⌡ d= c c a+ ypsg ρ g⋅ y⋅+( ) b⋅ ⌠ ⎮ ⌡ d= b psg a⋅ ρ g⋅ 2 a 2 2 a⋅ c⋅+( )⋅+⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 1( ) y' FR⋅ Ay p⋅ ⌠⎮ ⎮ ⌡ d= Therefore: y' 1 FR Ay p⋅ ⌠⎮ ⎮ ⌡ d= 1 FR c c a+ yy psg ρ g⋅ y⋅+( )⋅ b⋅ ⌠ ⎮ ⌡ d⋅= Evaluating the integral: y' b FR psg 2 c a+( ) 2 c 2 −⎡⎣ ⎤⎦ ρ g⋅ 3 c a+( ) 3 c 3 −⎡⎣ ⎤⎦⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = Problem 3.39 3.39
Simplifying: y' b FR psg 2 a 2 2 a⋅ c⋅+( ) ρ g⋅ 3 a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 2( ) For part (a) we know psg 0= so substituting into (1) we get: Fo ρ g⋅ b⋅ 2 a 2 2 a⋅ c⋅+( )⋅= Fo 1 2 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1× m⋅ 1.5 m⋅( ) 2 2 1.5× m⋅ 1× m⋅+⎡⎣ ⎤⎦× N s 2 ⋅ kg m⋅ ×= Fo 25.7 kN⋅= Substituting into (2) for the line of action we get: y' ρ g⋅ b⋅ 3 Fo⋅ a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅= y' 1 3 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1× m⋅ 1 25.7 10 3 × ⋅ 1 N ⋅ 1.5 m⋅( ) 3 3 1.5× m⋅ 1× m⋅ 1.5 m⋅ 1 m⋅+( )×+⎡⎣ ⎤⎦× N s 2 ⋅ kg m⋅ ×= y' 1.9m= For part (b) we know psg 0.3 atm⋅= . Substituting into (1) we get: FR 1 m⋅ 0.3 atm⋅ 1.013 10 5 × N⋅ m 2 atm⋅ × 1.5× m⋅ 1 2 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1.5 m⋅( ) 2 2 1.5× m⋅ 1× m⋅+⎡⎣ ⎤⎦× N s 2 ⋅ kg m⋅ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×= FR 71.3 kN⋅= Substituting into (2) for the line of action we get: y' 1 m⋅ 0.3 atm⋅ 2 1.013 10 5 × N⋅ m 2 atm⋅ × 1.5( ) 2 2 1.5⋅ 1⋅+⎡⎣ ⎤⎦× m 2 ⋅ 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 3 1.5( ) 3 3 1.5⋅ 1⋅ 1.5 1+( )⋅+⎡⎣ ⎤⎦× m 3 ⋅ N s 2 ⋅ kg m⋅ ×+ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ × 71.3 10 3 × N⋅ = y' 1.789m= The value of F/Fo is obtained from Eq. (1) and our result from part (a): F Fo b psg a⋅ ρ g⋅ 2 a 2 2 a⋅ c⋅+( )⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ ρ g⋅ b⋅ 2 a 2 2 a⋅ c⋅+( )⋅ = 1 2 psg⋅ ρ g⋅ a 2 c⋅+( )⋅ += For the gate yc c a 2 += Therefore, the value of y'/yc is obtained from Eqs. (1) and (2): y' yc 2 b⋅ FR 2 c⋅ a+( )⋅ psg 2 a 2 2 a⋅ c⋅+( ) ρ g⋅ 3 a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 2 b⋅ 2 c⋅ a+( ) psg 2 a 2 2 a⋅ c⋅+( ) ρ g⋅ 3 a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ b psg a⋅ ρ g⋅ 2 a 2 2 a⋅ c⋅+( )⋅+⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅=
Simplifying this expression we get: y' yc 2 2 c⋅ a+( ) psg 2 a 2 2 a⋅ c⋅+( ) ρ g⋅ 3 a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅+ psg a⋅ ρ g⋅ 2 a 2 2 a⋅ c⋅+( )⋅+ ⋅= Based on these expressions we see that the force on the gate varies linearly with the increase in surface pressure, and that the line of action of the resultant is always below the centroid of the gate. As the pressure increases, however, the line of action moves closer to the centroid. Plots of both ratios are shown below: 0 1 2 3 4 5 0 10 20 30 40 Force Ratio vs. Surface Pressure Surface Pressure (atm) ForceRatioF/Fo 0 1 2 3 4 5 1 1.01 1.02 1.03 1.04 1.05 Line of Action Ratio vs. Surface Pressure Surface Pressure (atm) LineofActionRatioy'/yc
Problem 3.48 [Difficulty: 5] Discussion: The design requirements are specified except that a typical floor height is about 12 ft, making the total required lift about 36 ft. A spreadsheet was used to calculate the system properties for various pressures. Results are presented on the next page, followed by a sample calculation. Total cost dropped quickly as system pressure was increased. A shallow minimum was reached in the 100-110 psig range. The lowest-cost solution was obtained at a system pressure of about 100 psig. At this pressure, the reservoir of 140 gal required a 3.30 ft diameter pressure sphere with a 0.250 in wall thickness. The welding cost was $155 and the material cost $433, for a total cost of $588. Accumulator wall thickness was constrained at 0.250 in for pressures below 100 psi; it increased for higher pressures (this caused the discontinuity in slope of the curve at 100 psig). The mass of steel became constant above 110 psig. No allowance was made for the extra volume needed to pressurize the accumulator. Fail-safe design is essential for an elevator to be used by the public. The control circuitry should be redundant. Failures must be easy to spot. For this reason, hydraulic actuation is good: leaks will be readily apparent. The final design must be reviewed, approved, and stamped by a professional engineer since the design involves public safety. The terminology used in the solution is defined in the following table: Symbol Definition Units p System pressure psig Ap Area of lift piston in2 Voil Volume of oil gal Ds Diameter of spherical accumulator ft t Wall thickness of accumulator in Aw Area of weld in2 Cw Cost of weld $ Ms Mass of steel accumulator lbm Cs Cost of steel $ Ct Total Cost $ A sample calculation and the results of the system simulation in Excel are presented below. Problem 3.40 3.40
σπ tDS 4 2 SD p π
Results of system simulation:
Problem 3.50 [Difficulty: 3] FA H = 25 ft yR = 10 ft h A B z x y Given: Geometry of gate Find: Force FA for equilibrium Solution: Basic equation FR Ap ⌠ ⎮ ⎮ ⌡ d= dp dh ρ g⋅= ΣMz 0= or, use computing equations FR pc A⋅= y' yc Ixx A yc⋅ += where y would be measured from the free surface Assumptions: static fluid; ρ = constant; patm on other side; door is in equilibrium Instead of using either of these approaches, we note the following, using y as in the sketch ΣMz 0= FA R⋅ Ay p⋅ ⌠⎮ ⎮ ⌡ d= with p ρ g⋅ h⋅= (Gage pressure, since p = patm on other side) FA 1 R Ay ρ⋅ g⋅ h⋅ ⌠⎮ ⎮ ⌡ d⋅= with dA r dr⋅ dθ⋅= and y r sin θ( )⋅= h H y−= Hence FA 1 R 0 π θ 0 R rρ g⋅ r⋅ sin θ( )⋅ H r sin θ( )⋅−( )⋅ r⋅ ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d⋅= ρ g⋅ R 0 π θ H R 3 ⋅ 3 sin θ( )⋅ R 4 4 sin θ( ) 2 ⋅− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d⋅= FR ρ g⋅ R 2 H⋅ R 3 ⋅ 3 π R 4 ⋅ 8 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= ρ g⋅ 2 H⋅ R 2 ⋅ 3 π R 3 ⋅ 8 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Using given data FR 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 2 3 25× ft⋅ 10 ft⋅( ) 2 × π 8 10 ft⋅( ) 3 ×−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × lbf s 2 ⋅ slug ft⋅ ×= FR 7.96 10 4 × lbf⋅= Problem 3.41 3.41
Problem 3.42 (Difficulty: 2) 3.42 A circular gate 3 𝑚 in diameter has its center 2.5 𝑚 below a water surface and lies in a plane sloping at 60°. Calculate magnitude, direction and location of total force on the gate. Find: The direction, magnitude of the total force 𝐹. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 For the magnitude of the force we have: 𝐹 = � 𝑝𝑝𝑝 𝐴 A free body diagram of the gate is The pressure on the gate is the pressure at the centroid, which is yc = 2.5 m. So the force can be calculated as: 𝐹 = 𝜌𝜌ℎ 𝑐 𝐴 = 999 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 2.5 𝑚 × 𝜋 4 × (3 𝑚)2 = 173200 𝑁 = 173.2 𝑘𝑘 The direction is perpendicular to the gate.
For the location of the force we have: 𝑦′ = 𝑦𝑐 + 𝐼 𝑥�𝑥� 𝐴𝑦𝑐 The y axis is along the plate so the distance to the centroid is: 𝑦𝑐 = 2.5 𝑚 sin 60° = 2.89 𝑚 The area moment of inertia is 𝐼 𝑥�𝑥� = 𝜋𝐷4 64 = 𝜋 64 × (3 𝑚)4 = 3.976 𝑚4 The area is 𝐴 = 𝜋 4 𝐷2 = 𝜋 4 × (3 𝑚)2 = 7.07 𝑚2 So 𝑦′ = 2.89 𝑚 + 3.976 𝑚4 7.07 𝑚2 × 2.89 𝑚 = 2.89 𝑚 + 0.1946 𝑚 = 3.08 𝑚 The vertical location on the plate is ℎ′ = 𝑦′ sin 60° = 3.08 𝑚 × √3 2 = 2.67 𝑚 The force acts on the point which has the depth of 2.67 𝑚.
Problem 3.43 (Difficulty: 2) 3.43 For the situation shown, find the air pressure in the tank in psi. Calculate the force exerted on the gate at the support B if the gate is 10 𝑓𝑓 wide. Show a free body diagram of the gate with all the forces drawn in and their points of application located. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure and force, and the static relation for moments: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 The specfic weight for water is: 𝛾 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 The pressure of the air equals that at the surface of the water in the tank. As shown by the manometer, the pressure at the surface is less than atmospheric due to the three foot head of water. The gage pressure of the air is then: 𝑝 𝑎𝑎𝑎 = −𝛾ℎ = −62.4 𝑙𝑙𝑙 𝑓𝑓3 × 3𝑓𝑓 = −187.2 𝑙𝑙𝑙 𝑓𝑓2 A free body diagram for the gate is
For the force in the horizontal direction, we have: 𝐹1 = 𝛾ℎ 𝑐 𝐴 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 3 𝑓𝑓 × (6 𝑓𝑓 × 10 𝑓𝑓) = 11230 𝑙𝑙𝑙 𝐹2 = 𝑝 𝑎𝑎𝑎 𝐴 = −187.2 𝑙𝑙𝑙 𝑓𝑓2 × (8 𝑓𝑓 × 10 𝑓𝑓) = 14980 𝑙𝑙𝑙 With the momentume balance about hinge we have: � 𝑀 = 𝐹1ℎ 𝑐 − 𝑃ℎ − 𝐹2 ℎ 2 = 11230 𝑙𝑙𝑙 × 6𝑓𝑓 − 𝑃 × 8𝑓𝑓 − 14980 𝑙𝑙𝑙 × 4𝑓𝑓 = 0 So the force exerted on B is: 𝑃 = 933 𝑙𝑙𝑙
Problem 3.44 (Difficulty: 3) 3.44 What is the pressure at A? Draw a free body diagram of the 10 ft wide gate showing all forces and locations of their lines of action. Calculate the minimum force 𝑃 necessary to keep the gate closed. Given: All the parameters are shown in the figure. Find: The pressure 𝑝 𝐴. The minimum force 𝑃 necessary to keep the gate closed. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 The specfic weight of the water is: 𝛾 𝑤𝑤𝑤𝑤𝑤 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 The gage pressure at A is given by integrating the hydrostatic relation: 𝑝 𝐴 = 𝛾 𝑜𝑜𝑜ℎ 𝐴 = 𝑆𝑆𝛾 𝑜𝑖𝑖ℎ 𝐴 = 0.9 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 6 𝑓𝑓 = 337 𝑙𝑙𝑙 𝑓𝑓2
A free body diagram of the gate is The horizontal force F1 as shown in the figure is given by the pressure at the centroid of the submerged area (3 ft): 𝐹1 = 𝛾 𝑜𝑜𝑜ℎ 𝑐 𝐴 = 0.9 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 3 𝑓𝑓 × (6 𝑓𝑓 × 10 𝑓𝑓) = 10110 𝑙𝑙𝑙 The vertical force F2 is given by the pressure at the depth of the surface (4 ft) 𝐹2 = 𝑝 𝐴 𝐴 = 337 𝑙𝑙𝑙 𝑓𝑓2 × (4𝑓𝑓 × 10𝑓𝑓) = 13480 𝑙𝑙𝑙 The force F1 acts two-thirds of the distance down from the water surface and the force F2 acts at the centroid.. Taking the moments about the hinge: −𝐹1 × 6 𝑓𝑓−𝐹2 × 2 𝑓𝑓 + 𝑃 × 4 𝑓𝑓 = 0 So we have for the force at the support: 𝑃 = 10110 𝑙𝑙𝑙 × 6𝑓𝑓 + 13480 𝑙𝑙𝑙 × 2𝑓𝑓 4 𝑓𝑓 = 21900 𝑙𝑙𝑙
Problem 3.52 [Difficulty: 3] Given: Geometry of plane gate W h L = 3 m dF y L/2 w = 2 m Find: Minimum weight to keep it closed Solution: Basic equation FR Ap ⌠ ⎮ ⎮ ⌡ d= dp dh ρ g⋅= ΣMO 0= or, use computing equations FR pc A⋅= y' yc Ixx A yc⋅ += Assumptions: static fluid; ρ = constant; patm on other side; door is in equilibrium Instead of using either of these approaches, we note the following, using y as in the sketch ΣMO 0= W L 2 ⋅ cos θ( )⋅ Fy ⌠⎮ ⎮ ⌡ d= We also have dF p dA⋅= with p ρ g⋅ h⋅= ρ g⋅ y⋅ sin θ( )⋅= (Gage pressure, since p = patm on other side) Hence W 2 L cos θ( )⋅ Ay p⋅ ⌠⎮ ⎮ ⌡ d⋅= 2 L cos θ( )⋅ yy ρ⋅ g⋅ y⋅ sin θ( )⋅ w⋅ ⌠⎮ ⎮ ⌡ d⋅= W 2 L cos θ( )⋅ Ay p⋅ ⌠⎮ ⎮ ⌡ d⋅= 2 ρ⋅ g⋅ w⋅ tan θ( )⋅ L 0 L yy 2⌠ ⎮ ⌡ d⋅= 2 3 ρ⋅ g⋅ w⋅ L 2 ⋅ tan θ( )⋅= Using given data W 2 3 1000⋅ kg m 3 ⋅ 9.81× m s 2 ⋅ 2× m⋅ 3 m⋅( ) 2 × tan 30 deg⋅( )× N s 2 ⋅ kg m⋅ ×= W 68 kN⋅= Problem 3.45 3.45
Problem 3.54 [Difficulty: 3] Given: Gate geometry Find: Depth H at which gate tips Solution: This is a problem with atmospheric pressure on both sides of the plate, so we can first determine the location of the center of pressure with respect to the free surface, using Eq.3.11c (assuming depth H) y' yc Ixx A yc⋅ += and Ixx w L 3 ⋅ 12 = with yc H L 2 −= where L = 1 m is the plate height and w is the plate width Hence y' H L 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ w L 3 ⋅ 12 w⋅ L⋅ H L 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ += H L 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L 2 12 H L 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ += But for equilibrium, the center of force must always be at or below the level of the hinge so that the stop can hold the gate in place. Hence we must have y' H 0.45 m⋅−> Combining the two equations H L 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L 2 12 H L 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ + H 0.45 m⋅−≥ Solving for H H L 2 L 2 12 L 2 0.45 m⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ +≤ H 1 m⋅ 2 1 m⋅( ) 2 12 1 m⋅ 2 0.45 m⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × +≤ H 2.17 m⋅≤ Problem 3.46 3.46
Problem 3.56 [Difficulty: 3] Ry Rx FR Fn Given: Geometry of lock system Find: Force on gate; reactions at hinge Solution: Basic equation FR Ap ⌠ ⎮ ⎮ ⌡ d= dp dh ρ g⋅= or, use computing equation FR pc A⋅= Assumptions: static fluid; ρ = constant; patm on other side The force on each gate is the same as that on a rectangle of size h D= 10 m⋅= and w W 2 cos 15 deg⋅( )⋅ = FR Ap ⌠ ⎮ ⎮ ⌡ d= Aρ g⋅ y⋅ ⌠ ⎮ ⎮ ⌡ d= but dA w dy⋅= Hence FR 0 h yρ g⋅ y⋅ w⋅ ⌠ ⎮ ⌡ d= ρ g⋅ w⋅ h 2 ⋅ 2 = Alternatively FR pc A⋅= and FR pc A⋅= ρ g⋅ yc⋅ A⋅= ρ g⋅ h 2 ⋅ h⋅ w⋅= ρ g⋅ w⋅ h 2 ⋅ 2 = Using given data FR 1 2 1000⋅ kg m 3 ⋅ 9.81× m s 2 ⋅ 34 m⋅ 2 cos 15 deg⋅( )⋅ × 10 m⋅( ) 2 × N s 2 ⋅ kg m⋅ ×= FR 8.63 MN⋅= For the force components Rx and Ry we do the following ΣMhinge 0= FR w 2 ⋅ Fn w⋅ sin 15 deg⋅( )⋅−= Fn FR 2 sin 15 deg⋅( )⋅ = Fn 16.7 MN⋅= ΣFx 0= FR cos 15 deg⋅( )⋅ Rx−= 0= Rx FR cos 15 deg⋅( )⋅= Rx 8.34 MN⋅= ΣFy 0= Ry− FR sin 15 deg⋅( )⋅− Fn+= 0= Ry Fn FR sin 15 deg⋅( )⋅−= Ry 14.4 MN⋅= R 8.34 MN⋅ 14.4 MN⋅,( )= R 16.7 MN⋅= Problem 3.47 3.47
Problem 3.48 (Difficulty: 2) 3.48 Calculate the minimum force 𝑃 necessary to hold a uniform 12 𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠 gate weighing 500 𝑙𝑙𝑙closed on a tank of water under a pressure of 10 𝑝𝑝𝑝. Draw a free body of the gate as part of your solution. Given: All the parameters are shown in the figure. Find: The minimum force 𝑃 to hold the system. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 A free body diagram of the gate is
The gage pressure of the air in the tank is: 𝑝 𝑎𝑎𝑎 = 10 𝑝𝑝𝑝 = 1440 𝑙𝑙𝑙 𝑓𝑓2 This produces a uniform force on the gate of 𝐹1 = 𝑝 𝑎𝑎𝑎 𝐴 = 1440 𝑙𝑙𝑙 𝑓𝑓2 × (12 𝑓𝑓 × 12 𝑓𝑓) = 207360 𝑙𝑙𝑙 This pressure acts at the centroid of the area, which is the center of the gate. In addition, there is a force on the gate applied by water. This force is due to the pressure at the centroid of the area. The depth of the centroid is: 𝑦𝑐 = 12 𝑓𝑓 2 × sin 45° The force is them 𝐹2 = 𝛾ℎ 𝑐 𝐴 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 12 𝑓𝑓 2 × sin 45° × 12 𝑓𝑓 × 12 𝑓𝑓 = 38123 𝑙𝑙𝑙 The force F2 acts two-thirds of the way down from the hinge, or 𝑦′ = 8 𝑓𝑓. Take the moments about the hinge: −𝐹𝐵 𝐿 2 sin 45° + 𝐹1 𝐿 2 + 𝐹2 × 8 𝑓𝑓 − 𝑃 × 12 𝑓𝑓 = 0 Thus 𝑃 = −500 𝑙𝑙𝑙 × 6 𝑓𝑓 × sin 45° + 207360 𝑙𝑙𝑙 × 6 𝑓𝑓 + 38123 𝑙𝑙𝑙 × 8 𝑓𝑓 12 𝑓𝑓 = 128900 𝑙𝑙𝑙
Problem 3.49 (Difficulty: 2) 3.49 Calculate magnitude and location of the resultant force of water on this annular gate. Given: All the parameters are shown in the figure. Find: Resultant force of water on this annular gate. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 For the magnitude of the force we have: 𝐹 = � 𝑝𝑝𝑝 𝐴 = 𝜌𝜌ℎ 𝑐 𝐴 The pressure is determined at the location of the centroid of the area ℎ 𝑐 = 1 𝑚 + 1.5 𝑚 = 2.5 𝑚 𝐴 = 𝜋 4 (𝐷2 2 − 𝐷1 2) = 𝜋 4 ((3 𝑚)2 − (1.5 𝑚)2) = 5.3014 𝑚2 𝐹 = 999 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 2.5 𝑚 × 5.3014 𝑚2 = 129900 𝑁 = 129.9 𝑘𝑘 The y axis is in the vertical direction. For the location of the force, we have:
𝑦′ = 𝑦𝑐 + 𝐼 𝑥�𝑥� 𝐴𝑦𝑐 Where: 𝑦𝑐 = 2.5 𝑚 𝐼 𝑥�𝑥� = 𝜋(𝐷2 4 − 𝐷1 4) 64 = 𝜋 64 × ((3 𝑚)4 − (1.5 𝑚)4) = 3.7276 𝑚4 𝑦′ = 𝑦𝑐 + 𝐼 𝑥�𝑥� 𝐴𝑦𝑐 = 2.5 𝑚 + 3.7276 𝑚4 2.5 𝑚 × 5.3014 𝑚2 = 2.78 𝑚 So the force acts on the depth of 𝑦′ = 2.78 𝑚.
Problem 3.50 (Difficulty: 2) 3.50 A vertical rectangular gate 2.4 𝑚 wide and 2.7 𝑚 high is subjected to water pressure on one side, the water surface being at the top of the gate. The gate is hinged at the bottom and is held by a horizontal chain at the top. What is the tension in the chain? Given: The gate wide: 𝑤 = 2.4 𝑚. Height of the gate: ℎ = 2.7 𝑚. Find: The tension 𝐹𝑐 in the chain. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 For the magnitude of the force we have: 𝐹 = � 𝑝𝑝𝑝 𝐴 = 𝜌𝜌ℎ 𝑐 𝐴 Where hc is the depth at the centroid ℎ 𝑐 = 2.7 𝑚 2 = 1.35 𝑚 𝐴 = 𝑤ℎ = 2.4 𝑚 × 2.7 𝑚 = 6.48 𝑚2
𝐹 = 999 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 1.35 𝑚 × 6.48 𝑚2 = 85.7 𝑘𝑘 The y axis is in the vertical direction. For the location of the force, we have: ℎ 𝑝 = 2 3 × 2.7 𝑚 = 1.8 𝑚 Taking the momentum about the hinge: 𝐹�ℎ − ℎ 𝑝� − 𝐹𝑐ℎ = 0 𝐹𝑐 = 𝐹 �ℎ − ℎ 𝑝� ℎ = 85.7 𝑘𝑘 × 0.9 𝑚 2.7 𝑚 = 28.6 𝑘𝑘
Problem 3.58 [Difficulty: 4] Given: Window, in shape of isosceles triangle and hinged at the top is located in the vertical wall of a form that contains concrete. a 0.4 m⋅= b 0.3 m⋅= c 0.25 m⋅= SGc 2.5= (From Table A.1, App. A) Find: The minimum force applied at D needed to keep the window closed. Plot the results over the range of concrete depth between 0 and a. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) FR Ap ⌠ ⎮ ⎮ ⌡ d= (Hydrostatic Force on door) y' FR⋅ Ay p⋅ ⌠ ⎮ ⎮ ⌡ d= (First moment of force) ΣM 0= (Rotational equilibrium) d dA h aw b D Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface and on the outside of the window. Integrating the pressure equation yields: p ρ g⋅ h d−( )⋅= for h > d p 0= for h < d where d a c−= d 0.15 m⋅= Summing moments around the hinge: FD− a⋅ Ah p⋅ ⌠⎮ ⎮ ⌡ d+ 0= FD dF = pdA h a FD 1 a Ah p⋅ ⌠⎮ ⎮ ⌡ d⋅= 1 a d a hh ρ⋅ g⋅ h d−( )⋅ w⋅ ⌠ ⎮ ⌡ d⋅= ρ g⋅ a d a hh h d−( )⋅ w⋅ ⌠ ⎮ ⌡ d⋅= From the law of similar triangles: w b a h− a = Therefore: w b a a h−( )= Problem 3.51 3.51
Into the expression for the force at D: FD ρ g⋅ a d a h b a h⋅ h d−( )⋅ a h−( )⋅ ⌠ ⎮ ⎮ ⌡ d⋅= ρ g⋅ b⋅ a 2 d a hh 3 − a d+( ) h 2 ⋅+ a d⋅ h⋅−⎡⎣ ⎤⎦ ⌠ ⎮ ⌡ d⋅= Evaluating this integral we get: FD ρ g⋅ b⋅ a 2 a 4 d 4 −( ) 4 − a d+( ) a 3 d 3 −( )⋅ 3 + a d⋅ a 2 d 2 −( )⋅ 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= and after collecting terms: FD ρ g⋅ b⋅ a 2 ⋅ 1 4 − 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ 1 3 1 d a + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅+ 1 2 d a ⋅ 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 1( ) The density of the concrete is: ρ 2.5 1000× kg m 3 ⋅= ρ 2.5 10 3 × kg m 3 = d a 0.15 0.4 = 0.375= Substituting in values for the force at D: FD 2.5 10 3 × kg m 3 ⋅ 9.81⋅ m s 2 ⋅ 0.3⋅ m⋅ 0.4 m⋅( ) 2 ⋅ 1 4 − 1 0.375( ) 4 −⎡⎣ ⎤⎦⋅ 1 3 1 0.375+( )⋅ 1 0.375( ) 3 −⎡⎣ ⎤⎦⋅+ 0.375 2 1 0.375( ) 2 −⎡⎣ ⎤⎦⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ N s 2 ⋅ kg m⋅ ×= To plot the results for different values of c/a, we use Eq. (1) and remember that d a c−= FD 32.9N= Therefore, it follows that d a 1 c a −= In addition, we can maximize the force by the maximum force (when c = a or d = 0): Fmax ρ g⋅ b⋅ a 2 ⋅ 1 4 − 1 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= ρ g⋅ b⋅ a 2 ⋅ 12 = and so FD Fmax 12 1 4 − 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ 1 3 1 d a + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅+ 1 2 d a ⋅ 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 0.0 0.5 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Concrete Depth Ratio (c/a) ForceRatio(FD/Fmax)
Problem 3.60 [Difficulty: 2] Given: Plug is used to seal a conduit. γ 62.4 lbf ft 3 ⋅= Find: Magnitude, direction and location of the force of water on the plug. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh γ= (Hydrostatic Pressure - y is positive downwards) FR pc A⋅= (Hydrostatic Force) y' yc Ixx A yc⋅ += (Location of line of action) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on the outside of the plug. Integrating the hydrostatic pressure equation: p γ h⋅= FR pc A⋅= γ hc⋅ π 4 ⋅ D 2 ⋅= FR 62.4 lbf ft 3 ⋅ 12× ft⋅ π 4 × 6 ft⋅( ) 2 ×= FR 2.12 10 4 × lbf⋅= For a circular area: Ixx π 64 D 4 ⋅= Therefore: y' yc π 64 D 4 ⋅ π 4 D 2 ⋅ yc⋅ += yc D 2 16 yc⋅ += y' 12 ft⋅ 6 ft⋅( ) 2 16 12× ft⋅ += y' 12.19 ft⋅= The force of water is to the right and perpendicular to the plug. Problem 3.52 3.52
Problem 3.62 [Difficulty: 2] Given: Circular access port of known diameter in side of water standpipe of known diameter. Port is held in place by eight bolts evenly spaced around the circumference of the port. Center of the port is located at a know distance below the free surface of the water. d 0.6 m⋅= D 7 m⋅= L 12 m⋅= Find: (a) Total force on the port (b) Appropriate bolt diameter Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - y is positive downwards) d L D h FR pc A⋅= (Hydrostatic Force) σ F A = (Normal Stress in bolt) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Force is distributed evenly over all bolts (4) Appropriate working stress in bolts is 100 MPa (5) Atmospheric pressure acts at free surface of water and on outside of port. Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= The resultant force on the port is: FR pc A⋅= ρ g⋅ L⋅ π 4 ⋅ d 2 ⋅= FR 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 12× m⋅ π 4 × 0.6 m⋅( ) 2 × N s 2 ⋅ kg m⋅ ×= FR 33.3 kN⋅= To find the bolt diameter we consider: σ FR A = where A is the area of all of the bolts: A 8 π 4 × db 2 ⋅= 2 π⋅ db 2 ⋅= Therefore: 2 π⋅ db 2 ⋅ FR σ = Solving for the bolt diameter we get: db FR 2 π⋅ σ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 = db 1 2 π× 33.3× 10 3 × N⋅ 1 100 10 6 × × m 2 N ⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 1 2 10 3 mm⋅ m ×= db 7.28 mm⋅= Problem 3.53 3.53
Problem 3.64 [Difficulty: 3] Given: Gate AOC, hinged along O, has known width; Weight of gate may be neglected. Gate is sealed at C. b 6 ft⋅= Find: Force in bar AB Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) FR pc A⋅= (Hydrostatic Force) y' yc Ixx A yc⋅ += (Location of line of action) ΣMz 0= (Rotational equilibrium) F1 h1’ F2 L1 L2 x2’ FAB L1 Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate (4) No resisting moment in hinge at O (5) No vertical resisting force at C Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= The free body diagram of the gate is shown here: F1is the resultant of the distributed force on AO F2is the resultant of the distributed force on OC FAB is the force of the bar Cx is the sealing force at C First find the force on AO: F1 pc A1⋅= ρ g⋅ hc1⋅ b⋅ L1⋅= F1 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 6× ft⋅ 6× ft⋅ 12× ft⋅ lbf s 2 ⋅ slugft⋅ ×= F1 27.0 kip⋅= Problem 3.54 3.54
h'1 hc1 Ixx A hc1⋅ += hc1 b L1 3 ⋅ 12 b⋅ L1⋅ hc1⋅ += hc1 L1 2 12 hc1⋅ += h'1 6 ft⋅ 12 ft⋅( ) 2 12 6× ft⋅ += h'1 8 ft⋅= Next find the force on OC: F2 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 12× ft⋅ 6× ft⋅ 6× ft⋅ lbf s 2 ⋅ slug ft⋅ ×= F2 27.0 kip⋅= F1 h1’ F2 L1 L2 x2’ FAB L1 Since the pressure is uniform over OC, the force acts at the centroid of OC, i.e., x'2 3 ft⋅= Summing moments about the hinge gives: FAB L1 L3+( )⋅ F1 L1 h'1−( )⋅− F2 x'2⋅+ 0= Solving for the force in the bar: FAB F1 L1 h'1−( )⋅ F2 x'2⋅− L1 L3+ = Substituting in values: FAB 1 12 ft⋅ 3 ft⋅+ 27.0 10 3 × lbf⋅ 12 ft⋅ 8 ft⋅−( )× 27.0 10 3 × lbf⋅ 3× ft⋅−⎡⎣ ⎤⎦⋅= FAB 1800 lbf⋅= Thus bar AB is in compression
Problem 3.66 [Difficulty: 3] Given: Geometry of gate h D FR y FA y’ Find: Force at A to hold gate closed Solution: Basic equation dp dh ρ g⋅= ΣMz 0= Computing equations FR pc A⋅= y' yc Ixx A yc⋅ += Ixx w L 3 ⋅ 12 = Assumptions: Static fluid; ρ = constant; patm on other side; no friction in hinge For incompressible fluid p ρ g⋅ h⋅= where p is gage pressure and h is measured downwards The hydrostatic force on the gate is that on a rectangle of size L and width w. Hence FR pc A⋅= ρ g⋅ hc⋅ A⋅= ρ g⋅ D L 2 sin 30 deg⋅( )⋅+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ L⋅ w⋅= FR 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 1.5 3 2 sin 30 deg⋅( )+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × m⋅ 3× m⋅ 3× m⋅ N s 2 ⋅ kg m⋅ ×= FR 199 kN⋅= The location of this force is given by y' yc Ixx A yc⋅ += where y' and y c are measured along the plane of the gate to the free surface yc D sin 30 deg⋅( ) L 2 += yc 1.5 m⋅ sin 30 deg⋅( ) 3 m⋅ 2 += yc 4.5m= y' yc Ixx A yc⋅ += yc w L 3 ⋅ 12 1 w L⋅ ⋅ 1 yc ⋅+= yc L 2 12 yc⋅ += 4.5 m⋅ 3 m⋅( ) 2 12 4.5⋅ m⋅ += y' 4.67m= Taking moments about the hinge ΣMH 0= FR y' D sin 30 deg⋅( ) −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ FA L⋅−= FA FR y' D sin 30 deg⋅( ) − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L ⋅= FA 199 kN⋅ 4.67 1.5 sin 30 deg⋅( ) − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ⋅= FA 111 kN⋅= Problem 3.55 3.55
Problem 3.68 [Difficulty: 4] Given: Various dam cross-sections Find: Which requires the least concrete; plot cross-section area A as a function of α Solution: For each case, the dam width b has to be large enough so that the weight of the dam exerts enough moment to balance the moment due to fluid hydrostatic force(s). By doing a moment balance this value of b can be found a) Rectangular dam Straightforward application of the computing equations of Section 3-5 yields b D FH y mg O FH pc A⋅= ρ g⋅ D 2 ⋅ w⋅ D⋅= 1 2 ρ⋅ g⋅ D 2 ⋅ w⋅= y' yc Ixx A yc⋅ += D 2 w D 3 ⋅ 12 w⋅ D⋅ D 2 ⋅ += 2 3 D⋅= so y D y'−= D 3 = Also m ρcement g⋅ b⋅ D⋅ w⋅= SG ρ⋅ g⋅ b⋅ D⋅ w⋅= Taking moments about O M0.∑ 0= FH− y⋅ b 2 m⋅ g⋅+= so 1 2 ρ⋅ g⋅ D 2 ⋅ w⋅⎛ ⎜ ⎝ ⎞ ⎠ D 3 ⋅ b 2 SG ρ⋅ g⋅ b⋅ D⋅ w⋅( )⋅= Solving for b b D 3 SG⋅ = The minimum rectangular cross-section area is A b D⋅= D 2 3 SG⋅ = For concrete, from Table A.1, SG = 2.4, so A D 2 3 SG⋅ = D 2 3 2.4× = A 0.373 D 2 ⋅= Problem 3.56 3.56
FH b αb D FV y x m1g m2g O b) Triangular dams Instead of analysing right-triangles, a general analysis is made, at the end of which right triangles are analysed as special cases by setting α = 0 or 1. Straightforward application of the computing equations of Section 3-5 yields FH pc A⋅= ρ g⋅ D 2 ⋅ w⋅ D⋅= 1 2 ρ⋅ g⋅ D 2 ⋅ w⋅= y' yc Ixx A yc⋅ += D 2 w D 3 ⋅ 12 w⋅ D⋅ D 2 ⋅ += 2 3 D⋅= so y D y'−= D 3 = Also FV ρ V⋅ g⋅= ρ g⋅ α b⋅ D⋅ 2 ⋅ w⋅= 1 2 ρ⋅ g⋅ α⋅ b⋅ D⋅ w⋅= x b α b⋅−( ) 2 3 α⋅ b⋅+= b 1 α 3 −⎛ ⎜ ⎝ ⎞ ⎠ ⋅= For the two triangular masses m1 1 2 SG⋅ ρ⋅ g⋅ α⋅ b⋅ D⋅ w⋅= x1 b α b⋅−( ) 1 3 α⋅ b⋅+= b 1 2 α⋅ 3 −⎛ ⎜ ⎝ ⎞ ⎠ ⋅= m2 1 2 SG⋅ ρ⋅ g⋅ 1 α−( )⋅ b⋅ D⋅ w⋅= x2 2 3 b 1 α−( )⋅= Taking moments about O M0.∑ 0= FH− y⋅ FV x⋅+ m1 g⋅ x1⋅+ m2 g⋅ x2⋅+= so 1 2 ρ⋅ g⋅ D 2 ⋅ w⋅⎛ ⎜ ⎝ ⎞ ⎠ − D 3 ⋅ 1 2 ρ⋅ g⋅ α⋅ b⋅ D⋅ w⋅⎛ ⎜ ⎝ ⎞ ⎠ b⋅ 1 α 3 −⎛ ⎜ ⎝ ⎞ ⎠ ⋅+ 1 2 SG⋅ ρ⋅ g⋅ α⋅ b⋅ D⋅ w⋅⎛ ⎜ ⎝ ⎞ ⎠ b⋅ 1 2 α⋅ 3 −⎛ ⎜ ⎝ ⎞ ⎠ ⋅ 1 2 SG⋅ ρ⋅ g⋅ 1 α−( )⋅ b⋅ D⋅ w⋅⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 3 ⋅ b 1 α−( )⋅++ ... 0= Solving for b b D 3 α⋅ α 2 −( ) SG 2 α−( )⋅+ = For a right triangle with the hypotenuse in contact with the water, α = 1, and b D 3 1− SG+ = D 3 1− 2.4+ = b 0.477 D⋅= The cross-section area is A b D⋅ 2 = 0.238 D 2 ⋅= A 0.238 D 2 ⋅= For a right triangle with the vertical in contact with the water, α = 0, and
b D 2 SG⋅ = D 2 2.4⋅ = b 0.456 D⋅= The cross-section area is A b D⋅ 2 = 0.228 D 2 ⋅= A 0.228 D 2 ⋅= For a general triangle A b D⋅ 2 = D 2 2 3 α⋅ α 2 −( ) SG 2 α−( )⋅+⋅ = A D 2 2 3 α⋅ α 2 −( ) 2.4 2 α−( )⋅+⋅ = The final result is A D 2 2 4.8 0.6 α⋅+ α 2 −⋅ = The dimensionless area, A /D 2 , is plotted Alpha A /D 2 0.0 0.2282 0.1 0.2270 0.2 0.2263 0.3 0.2261 0.4 0.2263 0.5 0.2270 0.6 0.2282 0.7 0.2299 0.8 0.2321 0.9 0.2349 1.0 0.2384 Solver can be used to find the minimum area Alpha A /D 2 0.300 0.2261 Dam Cross Section vs Coefficient 0.224 0.226 0.228 0.230 0.232 0.234 0.236 0.238 0.240 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Coefficient DimensionlessAreaA/D2 From the Excel workbook, the minimum area occurs at α = 0.3 Amin D 2 2 4.8 0.6 0.3×+ 0.3 2 −⋅ = A 0.226 D 2 ⋅= The final results are that a triangular cross-section with α = 0.3 uses the least concrete; the next best is a right triangle with the vertical in contact with the water; next is the right triangle with the hypotenuse in contact with the water; and the cross-section requiring the most concrete is the rectangular cross-section.
Problem 3.70 [Difficulty: 2] Given: Geometry of dam Find: Vertical force on dam Assumptions: (1) water is static and incompressible (2) since we are asked for the force of the water, all pressures will be written as gage Solution: Basic equation: dp dh ρ g⋅= For incompressible fluid p ρ g⋅ h⋅= where p is gage pressure and h is measured downwards from the free surface The force on each horizontal section (depth d and width w) is F p A⋅= ρ g⋅ h⋅ d⋅ w⋅= (Note that d and w will change in terms of x and y for each section of the dam!) Hence the total force is (allowing for the fact that some faces experience an upwards (negative) force) FT p A⋅= Σρ g⋅ h⋅ d⋅ w⋅= ρ g⋅ d⋅ Σ⋅ h w⋅= Starting with the top and working downwards FT 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 3× ft⋅ 3 ft⋅ 12× ft⋅( ) 3 ft⋅ 6× ft⋅( )+ 9 ft⋅ 6× ft⋅( )− 12 ft⋅ 12× ft⋅( )−[ ]× lbf s 2 ⋅ slug ft⋅ ×= FT 2.70− 10 4 × lbf⋅= The negative sign indicates a net upwards force (it's actually a buoyancy effect on the three middle sections) Problem 3.57 3.57
Problem 3.72 [Difficulty: 3] Given: Parabolic gate, hinged at O has a constant width. b 2 m⋅= c 0.25 m 1− ⋅= D 2 m⋅= H 3 m⋅= Find: (a) Magnitude and line of action of the vertical force on the gate due to water (b) Horizontal force applied at A required to maintain equilibrium (c) Vertical force applied at A required to maintain equilibrium Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ΣMz 0= (Rotational equilibrium) Fv Ayp ⌠ ⎮ ⎮ ⌡ d= (Vertical Hydrostatic Force) x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= (Location of line of action) FH pc A⋅= (Horizontal Hydrostatic Force) h' hc Ixx A hc⋅ += (Location of line of action) Oy h’ B x’ x FV Ox FH y Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= (a) The magnitude and line of action of the vertical component of hydrostatic force: Fv Ayp ⌠⎮ ⎮ ⌡ d= 0 D c xρ g⋅ h⋅ b⋅ ⌠ ⎮ ⎮ ⌡ d= 0 D c xρ g⋅ D y−( )⋅ b ⌠ ⎮ ⎮ ⌡ d= 0 D c xρ g⋅ D c x 2 ⋅−( )⋅ b ⌠ ⎮ ⎮ ⌡ d= ρ g⋅ b⋅ 0 D c xD c x 2 ⋅−( ) ⌠ ⎮ ⎮ ⌡ d⋅= Evaluating the integral: Fv ρ g⋅ b⋅ D 3 2 c 1 2 1 3 D 3 2 c 1 2 ⋅− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⋅= 2 ρ⋅ g⋅ b⋅ 3 D 3 2 c 1 2 ⋅= 1( ) Problem 3.58 3.58
Substituting values: Fv 2 3 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 2× m⋅ 2 m⋅( ) 3 2 × 1 0.25 m⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 × N s 2 ⋅ kg m⋅ ×= Fv 73.9 kN⋅= To find the line of action of this force: x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= Therefore, x' 1 Fv Fvx ⌠ ⎮ ⎮ ⌡ d⋅= 1 Fv Ayx p⋅ ⌠ ⎮ ⎮ ⌡ d⋅= Using the derivation for the force: x' 1 Fv 0 D c xx ρ⋅ g⋅ D c x 2 ⋅−( )⋅ b⋅ ⌠ ⎮ ⎮ ⌡ d⋅= ρ g⋅ b⋅ Fv 0 D c xD x⋅ c x 3 ⋅−( ) ⌠ ⎮ ⎮ ⌡ d⋅= Evaluating the integral: x' ρ g⋅ b⋅ Fv D 2 D c ⋅ c 4 D c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= ρ g⋅ b⋅ Fv D 2 4 c⋅ ⋅= Now substituting values into this equation: x' 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 2× m⋅ 1 73.9 10 3 × × 1 N ⋅ 1 4 × 2 m⋅( ) 2 × 1 0.25 × m⋅ N s 2 ⋅ kg m⋅ ×= x' 1.061m= To find the required force at A for equilibrium, we need to find the horizontal force of the water on the gate and its line of action as well. Once this force is known we take moments about the hinge (point O). FH pc A⋅= ρ g⋅ hc⋅ b⋅ D⋅= ρ g⋅ D 2 ⋅ b⋅ D⋅= ρ g⋅ b⋅ D 2 2 ⋅= since hc D 2 = Therefore the horizontal force is: FH 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 2× m⋅ 2 m⋅( ) 2 2 × N s 2 ⋅ kg m⋅ ×= FH 39.2 kN⋅= To calculate the line of action of this force: h' hc Ixx A hc⋅ += D 2 b D 3 ⋅ 12 1 b D⋅ ⋅ 2 D ⋅+= D 2 D 6 += 2 3 D⋅= h' 2 3 2⋅ m⋅= h' 1.333m= Oy h’ H x’ x FV Ox FH FA y D Now we have information to solve parts (b) and (c): (b) Horizontal force applied at A for equilibrium: take moments about O: FA H⋅ Fv x'⋅− FH D h'−( )⋅− 0= Solving for FA FA Fv x'⋅ FH D h'−( )⋅+ H = FA 1 3 1 m ⋅ 73.9 kN⋅ 1.061× m⋅ 39.2 kN⋅ 2 m⋅ 1.333 m⋅−( )×+[ ]×= FA 34.9 kN⋅= Oy h’ L x’ x FV Ox FH FA y D (c) Vertical force applied at A for equilibrium: take moments about O: FA L⋅ Fv x'⋅− FH D h'−( )⋅− 0= Solving for FA FA Fv x'⋅ FH D h'−( )⋅+ L = L is the value of x at y = H. Therefore: L H c = L 3 m⋅ 1 0.25 × m⋅= L 3.464m= FA 1 3.464 1 m ⋅ 73.9 kN⋅ 1.061× m⋅ 39.2 kN⋅ 2 m⋅ 1.333 m⋅−( )×+[ ]×= FA 30.2 kN⋅=
Problem 3.74 [Difficulty: 2] Given: Open tank as shown. Width of curved surface b 10 ft⋅= Find: (a) Magnitude of the vertical force component on the curved surface (b) Line of action of the vertical component of the force Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh γ= (Hydrostatic Pressure - h is positive downwards) L x’ x FRy y Fv Ayp ⌠ ⎮ ⎮ ⌡ d−= (Vertical Hydrostatic Force) x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= (Moment of vertical force) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of wall Integrating the hydrostatic pressure equation: p γ h⋅= We can define along the surface h L R 2 x 2 −( ) 1 2 −= We also define the incremental area on the curved surface as: dAy b dx⋅= Substituting these into the force equation we get: Fv Ayp ⌠⎮ ⎮ ⌡ d−= 0 R xγ L R 2 x 2 −( ) 1 2 − ⎡⎢ ⎢ ⎣ ⎤⎥ ⎥ ⎦⋅ b⋅ ⌠ ⎮ ⎮ ⎮ ⌡ d−= γ− b⋅ 0 R xL R 2 x 2 −−( )⌠ ⎮ ⌡ d⋅= γ− b⋅ R⋅ L R π 4 ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Fv 62.4 lbf ft 3 ⋅ 10× ft⋅ 4× ft⋅ 10 ft⋅ 4 ft⋅ π 4 ×−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×⎡ ⎢ ⎣ ⎤ ⎥ ⎦ −= Fv 17.12− 10 3 × lbf⋅= (negative indicates downward) To find the line of action of the force: x' Fv⋅ Fvx ⌠⎮ ⎮ ⌡ d= where dFv γ− b⋅ L R 2 x 2 −−( )⋅ dx⋅= Therefore: x' x' Fv⋅ Fv = 1 γ b⋅ R⋅ L R π 4 ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 0 R xx γ⋅ b⋅ L R 2 x 2 −−( )⋅ ⌠ ⎮ ⌡ d⋅= 1 R L R π 4 ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 0 R xL x⋅ x R 2 x 2 −⋅−( )⌠ ⎮ ⌡ d⋅= Evaluating the integral: x' 4 R 4 L⋅ π R⋅−( )⋅ 1 2 L⋅ R 2 ⋅ 1 3 R 3 ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= 4 R 2 ⋅ R 4 L⋅ π R⋅−( )⋅ L 2 R 3 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= 4 R⋅ 4 L⋅ π R⋅− L 2 R 3 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Substituting known values: x' 4 4⋅ ft⋅ 4 10⋅ ft⋅ π 4⋅ ft⋅− 10 ft⋅ 2 4 ft⋅ 3 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= x' 2.14 ft⋅= Problem 3.59 3.59
Problem 3.76 [Difficulty: 3] Given: Dam with cross-section shown. Width of dam b 160 ft⋅= Find: (a) Magnitude and line of action of the vertical force component on the dam (b) If it is possible for the water to overturn dam Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards from free surface) Fv Ayp ⌠ ⎮ ⎮ ⌡ d= (Vertical Hydrostatic Force) FH pc A⋅= (Horizontal Hydrostatic Force) x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= (Moment of vertical force) A x’ x FH y y’ h’FV B h' hc Ixx hc A⋅ += (Line of action of vertical force) ΣMz 0= (Rotational Equilibrium) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of dam Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= Into the vertical force equation: Fv Ayp ⌠⎮ ⎮ ⌡ d= xA xB xρ g⋅ h⋅ b⋅ ⌠ ⎮ ⌡ d= ρ g⋅ b⋅ xA xB xH y−( ) ⌠ ⎮ ⌡ d⋅= From the definition of the dam contour: x y⋅ A y⋅− B= Therefore: y B x A− = and xA 10 ft 2 ⋅ 9 ft⋅ 1 ft⋅+= xA 2.11 ft⋅= Problem 3.60 3.60
Into the force equation: Fv ρ g⋅ b⋅ xA xB xH B x A− − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d⋅= ρ g⋅ b⋅ H xB xA−( )⋅ B ln xB A− xA A− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Substituting known values: Fv 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 160× ft⋅ 9 ft⋅ 7.0 ft⋅ 2.11 ft⋅−( )× 10 ft 2 ⋅ ln 7.0 1− 2.11 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × lbf s 2 ⋅ slug ft⋅ ⋅= Fv 2.71 10 5 × lbf⋅= To find the line of action of the force: x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= where dFv ρ g⋅ b⋅ H B x A− − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ dx⋅= Therefore: x' x' Fv⋅ Fv = 1 Fv xA xB xx ρ⋅ g⋅ b⋅ H B x A− − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ ⌠ ⎮ ⎮ ⌡ d⋅= 1 H xB xA−( )⋅ B ln xB A− xA A− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− xA xB xH x⋅ B x⋅ x A− − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d⋅= Evaluating the integral: x' H 2 xB 2 xA 2 −⎛ ⎝ ⎞ ⎠⋅ B xB xA−( )⋅− B A⋅ ln xB A− xA A− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− H xB xA−( )⋅ B ln xB A− xA A− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− = Substituting known values we get: x' 9 ft⋅ 2 7 2 2.11 2 −( )× ft 2 ⋅ 10 ft 2 ⋅ 7 2.11−( )× ft⋅− 10 ft 2 ⋅ 1× ft⋅ ln 7 1− 2.11 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×− 9 ft⋅ 7 2.11−( )× ft⋅ 10 ft 2 ⋅ ln 7 1− 2.11 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×− = x' 4.96 ft⋅= To determine whether or not the water can overturn the dam, we need the horizontal force and its line of action: FH pc A⋅= ρ g⋅ H 2 ⋅ H⋅ b⋅= ρ g⋅ b⋅ H 2 ⋅ 2 = Substituting values: FH 1 2 1.94× slug ft 3 ⋅ 32.2× ft s 2 ⋅ 160× ft⋅ 9 ft⋅( ) 2 × lbf s 2 ⋅ slug ft⋅ ×= FH 4.05 10 5 × lbf⋅= For the line of action: h' hc Ixx hc A⋅ += where hc H 2 = A H b⋅= Ixx b H 3 ⋅ 12 = Therefore: h' H 2 b H 3 ⋅ 12 2 H ⋅ 1 b H⋅ ⋅+= H 2 H 6 += 2 3 H⋅= h' 2 3 9⋅ ft⋅= h' 6.00 ft⋅= Taking moments of the hydrostatic forces about the origin: Mw FH H h'−( )⋅ Fv x'⋅−= Mw 4.05 10 5 × lbf⋅ 9 6−( )× ft⋅ 2.71 10 5 × lbf⋅ 4.96× ft⋅−= Mw 1.292− 10 5 × lbf ft⋅⋅= The negative sign indicates that this is a clockwise moment about the origin. Since the weight of the dam will also contribute a clockwise moment about the origin, these two moments should not cause the dam to tip to the left. Therefore, the water can not overturn the dam.
Problem 3.61 (Difficulty: 2) 3.61 The quarter cylinder 𝐴𝐴 is 10 𝑓𝑓 long. Calculate magnitude, direction, and location of the resultant force of the water on 𝐴𝐴. Given: All the parameters are shown in the figure. Assumptions: Fluid is incompressible and static Find: The resultant force. Solution: Apply the hydrostatic relations for pressure as a function of depth and for the location of forces on submerged objects. ∆𝑝 = 𝜌𝜌ℎ A freebody diagram for the cylinder is: The force balance in the horizontal direction yields thathorizontal force is due to the water pressure: 𝐹 𝐻 = 𝑃 𝐻 Where the depth is the distance to the centroid of the horizontal area (8 + 5/2 ft): 𝐹 𝐻 = 𝛾ℎ 𝑐 𝐴 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × �8 𝑓𝑓 + 5 𝑓𝑓 2 � × (5 𝑓𝑓 × 10 𝑓𝑡) = 32800 𝑙𝑙𝑙
𝑃 𝐻 = 32800 𝑙𝑙𝑙 The force in the vertical direction can be calculated as the weight of a volume of water that is 8 ft + 5 ft = 13 ft deep less the weight of water that would be in the quarter cylinder. This force is then: 𝑃𝑉 = 𝐹𝑉 − 𝑊 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 13 𝑓𝑓 × (5 𝑓𝑓 × 10 𝑓𝑓) − 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 𝜋 4 × (5 𝑓𝑓)2 × (10 𝑓𝑓) = 28308 𝑙𝑙𝑙 The total resultant force is the vector sum of the two forces: 𝑃 = � 𝑃 𝐻 2 + 𝑃𝑉 2 = �(32800 𝑙𝑙𝑙)2 + (28308 𝑙𝑙𝑙)2 = 43300 𝑙𝑙𝑙 The angle with respect to the horizontal is: 𝜃 = tan−1 � 𝑃𝑉 𝑃 𝐻 � = tan−1 � 28308 𝑙𝑙𝑙 32800 𝑙𝑙𝑙 � = 40.9° So the force acts on the quarter cylinder surface point at an angle of 𝜃 = 40.9 ° with respect to the horizontal.
Problem 3.62 (Difficulty: 2) 3.62 Calculate the magnitude, direction (horizontal and vertical components are acceptable), and line of action of the resultant force exerted by the water on the cylindrical gate 30 𝑓𝑓 long. Assumptions: Fluid is incompressible and static Find: The resultant forces. Solution: Apply the hydrostatic relations for pressure as a function of depth and for the location of forces on submerged objects. ∆𝑝 = 𝜌𝜌ℎ A free body diagram of the gate is The horizontal force is calculated as: 𝑃 𝐻 = 𝐹 𝐻 Where the depth is the distance to the centroid of the horizontal area (5 ft): 𝐹 𝐻 = 𝛾ℎ 𝑐 𝐴 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 5𝑓𝑓 × (10 𝑓𝑓 × 30 𝑓𝑓) = 93600 𝑙𝑙𝑙 𝑃 𝐻 = 93600 𝑙𝑙𝑙 The force in the vertical direction can be calculated as the weight of a volume of water that is 10 ft deep less the weight of water that would be in the quarter cylinder. This force is then:
𝑃𝑉 = 𝐹𝑉 − 𝑊 = 𝛾ℎ 𝑐 𝐴 − 𝛾∀ 𝑃𝑉 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 10 𝑓𝑓 × (10 𝑓𝑓 × 30 𝑓𝑓) − 62.4 𝑙𝑙𝑙 𝑓𝑓3 × �10 𝑓𝑓 × (10 𝑓𝑓 × 30 𝑓𝑓) − 𝜋 4 × (10 𝑓𝑡)2 × 30 𝑓𝑓� = 147000 𝑙𝑙𝑙 The total resultant force is the vector sum of the two forces: 𝑃 = � 𝑃 𝐻 2 + 𝑃𝑉 2 = �(93600 𝑙𝑙𝑙)2 + (147000 𝑙𝑙𝑙)2 = 174200 𝑙𝑙𝑙 The direction can be calculated as: 𝜃 = tan−1 � 𝑃𝑉 𝑃 𝐻 � = tan−1 � 147000 𝑙𝑙𝑙 93600 𝑙𝑙𝑙 � = 57.5°
Problem 3.63 (Difficulty: 2) 3.63 A hemispherical shell 1.2 𝑚 in diameter is connected to the vertical wall of a tank containing water. If the center of the shell is 1.8 𝑚 below the water surface, what are the vertical and horizontal force components on the shell? On the top half of the shell? Assumptions: Fluid is incompressible and static Find: The resultant forces. Solution: Apply the hydrostatic relations for pressure as a function of depth and for the location of forces on submerged objects. ∆𝑝 = 𝜌𝜌ℎ A free body diagram of the system is The force in the horizontal direction can be calculated using the distance to the centroid (1.8 m) as: 𝐹 𝐻 = 𝛾ℎ 𝑐 𝐴 = 9.81 𝑘𝑁 𝑚3 × 1.8 𝑚 × � 1 4 × 𝜋 × (1.2 𝑚)2 � = 19.97 𝑘𝑘 The force in the vertical direction is the buoyancy force due to the volume displaced by the shell: 𝐹𝑉 = 𝛾𝛾 = 9.81 𝑘𝑘 𝑚3 × 1 2 × 1 6 × 𝜋 × (1.2 𝑚)3 = 4.44 𝑘𝑘 For the top shell, the horizontal force acts at: 𝑦𝑐 = 1.8 𝑚 − 4 × 0.6 𝑚 3𝜋 = 1.545 𝑚 The horizontal force on the top half of the shell is then:
𝐹 𝐻 = 𝛾𝑦𝑐 𝐴 = 9.81 𝑘𝑘 𝑚3 × 1.545 𝑚 × 𝜋 8 × (1.2 𝑚)2 = 8.57 𝑘𝑘 The vertical force on the top half of the shell is the buoyancy force: 𝐹𝑉 = 𝑝𝑝 = 9.81 𝑘𝑘 𝑚3 × 1.8 𝑚 × 𝜋 8 × (1.2 𝑚)2 − 9.81 𝑘𝑘 𝑚3 × 1 4 × 1 6 × 𝜋 × (1.2 𝑚)3 = 7.77 𝑘𝑘
Problem 3.78 [Difficulty: 4] FV D y R A x FH F1 x y’ FB W1 W2 Weights for computing FV R/2 4R/3π WGate Given: Gate geometry Find: Force on stop B Solution: Basic equations dp dh ρ g⋅= ΣMA 0= Assumptions: static fluid; ρ = constant; patm on other side For incompressible fluid p ρ g⋅ h⋅= where p is gage pressure and h is measured downwards We need to compute force (including location) due to water on curved surface and underneath. For curved surface we could integrate pressure, but here we use the concepts that FV (see sketch) is equivalent to the weight of fluid above, and FH is equivalent to the force on a vertical flat plate. Note that the sketch only shows forces that will be used to compute the moment at A For FV FV W1 W2−= with W1 ρ g⋅ w⋅ D⋅ R⋅= 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 3× m⋅ 4.5× m⋅ 3× m⋅ N s 2 ⋅ kg m⋅ ×= W1 397 kN⋅= W2 ρ g⋅ w⋅ π R 2 ⋅ 4 ⋅= 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 3× m⋅ π 4 × 3 m⋅( ) 2 × N s 2 ⋅ kg m⋅ ×= W2 208 kN⋅= FV W1 W2−= FV 189 kN⋅= with x given by FV x⋅ W1 R 2 ⋅ W2 4 R⋅ 3 π⋅ ⋅−= or x W1 Fv R 2 ⋅ W2 Fv 4 R⋅ 3 π⋅ ⋅−= x 397 189 3 m⋅ 2 × 208 189 4 3 π⋅ × 3× m⋅−= x 1.75m= For FH Computing equations FH pc A⋅= y' yc Ixx A yc⋅ += Problem 3.64 3.64
Hence FH pc A⋅= ρ g⋅ D R 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ w⋅ R⋅= FH 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 4.5 m⋅ 3 m⋅ 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × 3× m⋅ 3× m⋅ N s 2 ⋅ kg m⋅ ×= FH 265 kN⋅= The location of this force is y' yc Ixx A yc⋅ += D R 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ w R 3 ⋅ 12 1 w R⋅ D R 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ ×+= D R 2 − R 2 12 D R 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ += y' 4.5 m⋅ 3 m⋅ 2 − 3 m⋅( ) 2 12 4.5 m⋅ 3 m⋅ 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × += y' 3.25m= The force F1 on the bottom of the gate is F1 p A⋅= ρ g⋅ D⋅ w⋅ R⋅= F1 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 4.5× m⋅ 3× m⋅ 3× m⋅ N s 2 ⋅ kg m⋅ ×= F1 397 kN⋅= For the concrete gate (SG = 2.4 from Table A.2) WGate SG ρ⋅ g⋅ w⋅ π R 2 ⋅ 4 ⋅= 2.4 1000⋅ kg m 3 ⋅ 9.81× m s 2 ⋅ 3× m⋅ π 4 × 3 m⋅( ) 2 × N s 2 ⋅ kg m⋅ ×= WGate 499 kN⋅= Hence, taking moments about A FB R⋅ F1 R 2 ⋅+ WGate 4 R⋅ 3 π⋅ ⋅− FV x⋅− FH y' D R−( )−[ ]⋅− 0= FB 4 3 π⋅ WGate⋅ x R FV⋅+ y' D R−( )−[ ] R FH⋅+ 1 2 F1⋅−= FB 4 3 π⋅ 499× kN⋅ 1.75 3 189× kN⋅+ 3.25 4.5 3−( )−[ ] 3 265× kN⋅+ 1 2 397× kN⋅−= FB 278 kN⋅=
Problem 3.80 [Difficulty: 3] Given: Cylindrical weir as shown; liquid is water Find: Magnitude and direction of the resultant force of the water on the weir Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards from free surface) dFR →⎯ p− dA →⎯ ⋅= (Hydrostatic Force) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surfaces and on the first quadrant of the cylinder D1 y x D2 h1 h2 θ Using the coordinate system shown in the diagram at the right: FRx FR →⎯ i → ⋅= A → p ⌠ ⎮ ⎮ ⌡ d− i → ⋅= Ap cos θ 90 deg⋅+( )⋅ ⌠ ⎮ ⎮ ⌡ d−= Ap sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d= FRy FR →⎯ j → ⋅= A → p ⌠ ⎮ ⎮ ⌡ d− j → ⋅= Ap cos θ( )⋅ ⌠ ⎮ ⎮ ⌡ d−= Now since dA L R⋅ dθ⋅= it follows that FRx 0 3 π⋅ 2 θp L⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d= and FRy 0 3 π⋅ 2 θp L⋅ R⋅ cos θ( )⋅ ⌠ ⎮ ⎮ ⌡ d−= Next, we integrate the hydrostatic pressure equation: p ρ g⋅ h⋅= Now over the range 0 θ≤ π≤ h1 R 1 cos θ( )−( )= Over the range π θ≤ 3 π⋅ 2 ≤ h2 R− cos θ( )⋅= Therefore we can express the pressure in terms of θ and substitute into the force equations: FRx 0 3 π⋅ 2 θp L⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d= 0 π θρ g⋅ R⋅ 1 cos θ( )−( )⋅ L⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⌡ d π 3 π⋅ 2 θρ g⋅ R⋅ cos θ( )⋅ L⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d−= FRx ρ g⋅ R 2 ⋅ L⋅ 0 π θ1 cos θ( )−( ) sin θ( )⋅ ⌠ ⎮ ⌡ d⋅ ρ g⋅ R 2 ⋅ L⋅ π 3 π⋅ 2 θcos θ( ) sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d⋅−= Problem 3.65 3.65
FRx ρ g⋅ R 2 ⋅ L⋅ 0 π θ1 cos θ( )−( ) sin θ( )⋅ ⌠ ⎮ ⌡ d π 3 π⋅ 2 θcos θ( ) sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d− ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ ⋅= ρ g⋅ R 2 ⋅ L⋅ 2 1 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= 3 2 ρ⋅ g⋅ R 2 ⋅ L⋅= Substituting known values: FRx 3 2 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1.5 m⋅( ) 2 × 6× m⋅ N s 2 ⋅ kg m⋅ ×= FRx 198.5 kN⋅= Similarly we can calculate the vertical force component: FRy 0 3 π⋅ 2 θp L⋅ R⋅ cos θ( )⋅ ⌠ ⎮ ⎮ ⌡ d−= 0 π θρ g⋅ R⋅ 1 cos θ( )−( )⋅ L⋅ R⋅ cos θ( )⋅ ⌠ ⎮ ⌡ d π 3 π⋅ 2 θρ g⋅ R⋅ cos θ( )⋅ L⋅ R⋅ cos θ( )⋅ ⌠ ⎮ ⎮ ⌡ d− ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ −= FRy ρ− g⋅ R 2 ⋅ L⋅ 0 π θ1 cos θ( )−( ) cos θ( )⋅ ⌠ ⎮ ⌡ d π 3 π⋅ 2 θcos θ( )( ) 2 ⌠ ⎮ ⎮ ⌡ d− ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ ⋅= ρ g⋅ R 2 ⋅ L⋅ π 2 3 π⋅ 4 + π 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= 3 π⋅ 4 ρ⋅ g⋅ R 2 ⋅ L⋅= Substituting known values: FRy 3 π⋅ 4 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1.5 m⋅( ) 2 × 6× m⋅ N s 2 ⋅ kg m⋅ ×= FRy 312 kN⋅= Now since the weir surface in contact with the water is a circular arc, all elements dF of the force, and hence the line of action of the resultant force, must pass through the pivot. Thus: Magnitude of the resultant force: FR 198.5 kN⋅( ) 2 312 kN⋅( ) 2 += FR 370 kN⋅= The line of action of the force: α atan 312 kN⋅ 198.5 kN⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α 57.5 deg⋅=
Problem 3.82 [Difficulty: 3] Given: Curved surface, in shape of quarter cylinder, with given radius R and width w; water stands to depth H. R 0.750 m⋅= w 3.55 m⋅= H 0.650 m⋅= Find: Magnitude and line of action of (a) vertical force and (b) horizontal force on the curved surface Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards from free surface) Fv Ayp ⌠ ⎮ ⎮ ⌡ d= (Vertical Hydrostatic Force) FH pc A⋅= (Horizontal Hydrostatic Force) x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= (Moment of vertical force) h' hc Ixx hc A⋅ += (Line of action of horizontal force) dF h HR θ Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the water and on the left side of the curved surface Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= dF h’ HR θ FV FH y’ x’From the geometry: h H R sin θ( )⋅−= y R sin θ( )⋅= x R cos θ( )⋅= dA w R⋅ dθ⋅= θ1 asin H R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ1 asin 0.650 0.750 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ1 1.048 rad⋅= Therefore the vertical component of the hydrostatic force is: Fv Ayp ⌠⎮ ⎮ ⌡ d= Aρ g⋅ h⋅ sin θ( )⋅ ⌠⎮ ⎮ ⌡ d= 0 θ1 θρ g⋅ H R sin θ( )⋅−( )⋅ sin θ( )⋅ w⋅ R⋅ ⌠ ⎮ ⌡ d= Fv ρ g⋅ w⋅ R⋅ 0 θ1 θH sin θ( )⋅ R sin θ( )( ) 2 ⋅−⎡⎣ ⎤⎦ ⌠ ⎮ ⌡ d⋅= ρ g⋅ w⋅ R⋅ H 1 cos θ1( )−( )⋅ R θ1 2 sin 2 θ1⋅( ) 4 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Problem 3.66 3.66
Fv 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 3.55× m⋅ 0.750× m⋅ 0.650 m⋅ 1 cos 1.048 rad⋅( )−( )× 0.750 m⋅ 1.048 2 sin 2 1.048× rad⋅( ) 4 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × N s 2 ⋅ kg m⋅ ×= Fv 2.47 kN⋅= To calculate the line of action of this force: x' Fv⋅ AR cos θ( )⋅ ρ⋅ g⋅ h⋅ sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d= ρ g⋅ w⋅ R 2 ⋅ 0 θ1 θH sin θ( )⋅ cos θ( )⋅ R sin θ( )( ) 2 ⋅ cos θ( )⋅−⎡⎣ ⎤⎦ ⌠ ⎮ ⌡ d⋅= Evaluating the integral: x' Fv⋅ ρ g⋅ w⋅ R 2 ⋅ H 2 sin θ1( )( ) 2 ⋅ R 3 sin θ1( )( ) 3 ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Therefore we may find the line of action: x' x' Fv⋅ Fv = ρ g⋅ w⋅ R 2 ⋅ Fv H 2 sin θ1( )( ) 2 ⋅ R 3 sin θ1( )( ) 3 ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Substituting in known values: sin θ1( ) 0.650 0.750 = x' 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 3.55× m⋅ 0.750 m⋅( ) 2 × 1 2.47 10 3 × × 1 N ⋅ 0.650 m⋅ 2 0.650 0.750 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × 0.750 m⋅ 3 0.650 0.750 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ×− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × N s 2 ⋅ kg m⋅ ×= x' 0.645m= For the horizontal force: FH pc A⋅= ρ g⋅ hc⋅ H⋅ w⋅= ρ g⋅ H 2 ⋅ H⋅ w⋅= ρ g⋅ H 2 ⋅ w⋅ 2 = FH 1 2 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 0.650 m⋅( ) 2 × 3.55× m⋅ N s 2 ⋅ kg m⋅ ×= FH 7.35 kN⋅= For the line of action of the horizontal force: h' hc Ixx hc A⋅ += where Ixx w H 3 ⋅ 12 = A w H⋅= Therefore: h' hc Ixx hc A⋅ += H 2 w H 3 ⋅ 12 2 H ⋅ 1 w H⋅ ⋅+= H 2 H 6 += 2 3 H⋅= h' 2 3 0.650× m⋅= h' 0.433m=
Problem 3.83 [Difficulty: 2] Given: Canoe floating in a pond Find: What happens when an anchor with too short of a line is thrown from canoe Solution: Governing equation: WgVF dispwB == ρ Before the anchor is thrown from the canoe the buoyant force on the canoe balances out the weight of the canoe and anchor: 11 canoewanchorcanoeB gVWWF ρ=+= The anchor weight can be expressed as aaanchor gVW ρ= so the initial volume displaced by the canoe can be written as a w a w canoe canoe V g W V ρ ρ ρ +=1 After throwing the anchor out of the canoe there will be buoyant forces acting on the canoe and the anchor. Combined, these buoyant forces balance the canoe weight and anchor weight: awcanoewanchorcanoeB gVgVWWF ρρ +=+= 22 a w a w canoe canoe V g W g W V −+= ρρ 2 Using the anchor weight, aa w a w canoe canoe VV g W V −+= ρ ρ ρ 2 Hence the volume displaced by the canoe after throwing the anchor in is less than when the anchor was in the canoe, meaning that the canoe is floating higher. Problem 3.67 3.67
Problem 3.86 [Difficulty: 4] Given: Cylinder of mass M, length L, and radius R is hinged along its length and immersed in an incompressilble liquid to depth Find: General expression for the cylinder specific gravity as a function of α=H/R needed to hold the cylinder in equilibrium for α ranging from 0 to 1. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards from free surface) dFH dFh θ dFV H = αR Fv Ayp ⌠ ⎮ ⎮ ⌡ d= (Vertical Hydrostatic Force) ΣM 0= (Rotational Equilibrium) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the liquid. The moments caused by the hydrostatic force and the weight of the cylinder about the hinge need to balance each other. Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= dFv dF cos θ( )⋅= p dA⋅ cos θ( )⋅= ρ g⋅ h⋅ w⋅ R⋅ dθ⋅ cos θ( )⋅= Now the depth to which the cylinder is submerged is H h R 1 cos θ( )−( )⋅+= Therefore h H R 1 cos θ( )−( )⋅−= and into the vertical force equation: dFv ρ g⋅ H R 1 cos θ( )−( )⋅−[ ]⋅ w⋅ R⋅ cos θ( )⋅ dθ⋅= ρ g⋅ w⋅ R 2 ⋅ H R 1 cos θ( )−( )−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ cos θ( )⋅ dθ⋅= dFv ρ g⋅ w⋅ R 2 ⋅ α 1−( ) cos θ( )⋅ cos θ( )( ) 2 +⎡⎣ ⎤⎦⋅ dθ⋅= ρ g⋅ w⋅ R 2 ⋅ α 1−( ) cos θ( )⋅ 1 cos 2 θ⋅( )+ 2 +⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ dθ⋅= Now as long as α is not greater than 1, the net horizontal hydrostatic force will be zero due to symmetry, and the vertical force is: Fv θmax− θmax Fv1 ⌠ ⎮ ⌡ d= 0 θmax Fv2 ⌠ ⎮ ⌡ d= where cos θmax( ) R H− R = 1 α−= or θmax acos 1 α−( )= Problem 3.68 3.68
Fv 2ρ g⋅ w⋅ R 2 ⋅ 0 θmax θα 1−( ) cos θ( )⋅ 1 2 + 1 2 cos 2 θ⋅( )⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⌠ ⎮ ⎮ ⌡ d⋅= Now upon integration of this expression we have: Fv ρ g⋅ w⋅ R 2 ⋅ acos 1 α−( ) 1 α−( ) α 2 α−( )⋅⋅−⎡⎣ ⎤⎦⋅= The line of action of the vertical force due to the liquid is through the centroid of the displaced liquid, i.e., through the center of the cylinde The weight of the cylinder is given by: W M g⋅= ρc V⋅ g⋅= SG ρ⋅ π⋅ R 2 ⋅ w⋅ g⋅= where ρ is the density of the fluid and SG ρc ρ = The line of action of the weight is also throught the center of the cylinder. Taking moment about the hinge we get: ΣMo W R⋅ Fv R⋅−= 0= or in other words W Fv= and therefore: SG ρ⋅ π⋅ R 2 ⋅ w⋅ g⋅ ρ g⋅ w⋅ R 2 ⋅ acos 1 α−( ) 1 α−( ) α 2 α−( )⋅⋅−⎡⎣ ⎤⎦⋅= SG 1 π acos 1 α−( ) 1 α−( ) α 2 α−( )⋅⋅−⎡⎣ ⎤⎦⋅= 0 0.5 1 0 0.2 0.4 0.6 alpha (H/R) SpecificGravity,SG
Problem *3.89 [Difficulty: 2] Given: Hydrometer as shown, submerged in nitric acid. When submerged in water, h = 0 and the immersed volume is 15 cubic cm. SG 1.5= d 6 mm⋅= Find: The distance h when immersed in nitric acid. Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) Assumptions: (1) Static fluid (2) Incompressible fluid Taking a free body diagram of the hydrometer: ΣFz 0= M− g⋅ Fbuoy+ 0= Solving for the mass of the hydrometer: M Fbuoy g = ρ Vd⋅= When immersed in water: M ρw Vw⋅= When immersed in nitric acid: M ρn Vn⋅= Since the mass of the hydrometer is the same in both cases: ρw Vw⋅ ρn Vn⋅= When the hydrometer is in the nitric acid: Vn Vw π 4 d 2 ⋅ h⋅−= ρn SG ρw⋅= Therefore: ρw Vw⋅ SG ρw⋅ Vw π 4 d 2 ⋅ h⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Solving for the height h: Vw SG Vw π 4 d 2 ⋅ h⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Vw 1 SG−( )⋅ SG− π 4 ⋅ d 2 ⋅ h⋅= h Vw SG 1− SG ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 4 π d 2 ⋅ ⋅= h 15 cm 3 ⋅ 1.5 1− 1.5 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × 4 π 6 mm⋅( ) 2 × × 10 mm⋅ cm ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ×= h 177 mm⋅= Problem 3.69 3.69
Problem 3.70 (Difficulty: 2) 3.70 A cylindrical can 76 𝑚𝑚 in diameter and 152 𝑚𝑚 high, weighing 1.11 𝑁, contains water to a depth of 76 𝑚𝑚. When this can is placed in water, how deep will it sink? Find: The depth it will sink. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 A free body diagram on the can is We have the force balance equation in the vertical direction as: 𝐹𝑏 − 𝑊𝑐𝑐𝑐 − 𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0 The buoyancy force can be calculated as: 𝐹𝑏 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑐 𝑐𝑐 = 9810 𝑁 𝑚3 × 𝜋 4 × (0.076 𝑚)2 × 𝑥 𝑚 = 44.50𝑋𝑋 𝑁 We also have: 𝑊𝑐𝑐𝑐 = 1.11 𝑁 𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 9810 𝑁 𝑚3 × 𝜋 4 × (0.076 𝑚)3 = 3.38 𝑁
Thus making a force balance for which the net force is zero at equilibrium 44.50𝑥 = 1.11 𝑁 + 3.38 𝑁 = 4.49 𝑁 𝑥 = 0.1009 𝑚 = 100.9 𝑚𝑚 So this can will sink to depth of 100.9 𝑚𝑚.
Problem 3.71 (Difficulty: 1) 3.71 If the 10 𝑓𝑓 long box is floating on the oil water system, calculate how much the box and its contents must weigh. Find: The weight of the box and its contents. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 The force balance equation in the vertical diretion: 𝐹𝐵 − 𝑊𝐵 = 0 𝐹𝐵 = 𝛾 𝑜𝑜𝑜 𝑉 + 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉 Thus 𝐹𝐵 = 0.8 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 2𝑓𝑓 × 8𝑓𝑓 × 10𝑓𝑓 + 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 1𝑓𝑓 × 8𝑓𝑓 × 10𝑓𝑓 = 12980 𝑙𝑙𝑙 So the box and its contents must weigh: 𝑊𝐵 = 12980 𝑙𝑙𝑙
Problem 3.72 (Difficulty: 2) 3.72 The timber weighs 40 𝑙𝑙𝑙 𝑓𝑓3 and is held in a horizontal position by the concrete �150 𝑙𝑙𝑙 𝑓𝑓3� anchor. Calculate the minimum total weight which the anchor may have. Find: The minimum total weight the anchor may have. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 For the buoyancy force we have: 𝐹𝑏𝑏 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑡 𝐹𝑏𝑏 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × � 6 12 𝑓𝑓� × � 6 12 𝑓𝑓� × (20 𝑓𝑓) = 312 𝑙𝑙𝑙 The weight of the timber is: 𝑊𝑡 = 𝛾𝑡 𝑉𝑡 𝑊𝑡 = 40 𝑙𝑙𝑙 𝑓𝑓3 × � 6 12 𝑓𝑓� × � 6 12 𝑓𝑓� × (20 𝑓𝑓) = 200 𝑙𝑙𝑙 At the horizontal position we take moments about the pivot: 𝐹𝑎 𝐿 + 𝑊𝑡 𝐿 2 − 𝐹𝑏𝑏 𝐿 2 = 0 𝐹𝑎 = 1 2 𝐹𝑏𝑏 − 1 2 𝑊𝑡 = 1 2 × (312 𝑙𝑙𝑙 − 200𝑙𝑙𝑙) = 56 𝑙𝑙𝑙
𝐹𝑎 = 𝐹𝑏𝑏 − 𝑊𝑎 The weight of the anchor is: 𝑊𝑎 = 𝛾𝑎 𝑉𝑎 The buoyancy force on the anchor is: 𝐹𝑏𝑏 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑎 𝛾𝑎 𝑉𝑎 − 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑎 = 56 𝑙𝑙𝑙 𝑉𝑎 = 56 𝑙𝑙𝑙 �150 𝑙𝑙𝑙 𝑓𝑓3 − 62.4 𝑙𝑙𝑙 𝑓𝑓3� = 0.64 𝑓𝑓3 So the weight is: 𝑊𝑎 = 𝛾𝑎 𝑉𝑎 = 150 𝑙𝑙𝑙 𝑓𝑓3 × 0.64 𝑓𝑓3 = 96 𝑙𝑙𝑙
Problem *3.90 [Difficulty: 3] Given: Data on sphere and weight T FB W Find: SG of sphere; equilibrium position when freely floating Solution: Basic equation FB ρ g⋅ V⋅= and ΣFz 0= ΣFz 0= T FB+ W−= where T M g⋅= M 10 kg⋅= FB ρ g⋅ V 2 ⋅= W SG ρ⋅ g⋅ V⋅= Hence M g⋅ ρ g⋅ V 2 ⋅+ SG ρ⋅ g⋅ V⋅− 0= SG M ρ V⋅ 1 2 += SG 10 kg⋅ m 3 1000 kg⋅ × 1 0.025 m 3 ⋅ × 1 2 += SG 0.9= The specific weight is γ Weight Volume = SG ρ⋅ g⋅ V⋅ V = SG ρ⋅ g⋅= γ 0.9 1000× kg m 3 ⋅ 9.81× m s 2 ⋅ N s 2 ⋅ kg m⋅ ×= γ 8829 N m 3 ⋅= For the equilibriul position when floating, we repeat the force balance with T = 0 FB W− 0= W FB= with FB ρ g⋅ Vsubmerged⋅= From references (trying Googling "partial sphere volume") Vsubmerged π h 2 ⋅ 3 3 R⋅ h−( )⋅= where h is submerged depth and R is the sphere radius R 3 V⋅ 4 π⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 3 = R 3 4 π⋅ 0.025⋅ m 3 ⋅⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 3 = R 0.181m= Hence W SG ρ⋅ g⋅ V⋅= FB= ρ g⋅ π h 2 ⋅ 3 ⋅ 3 R⋅ h−( )⋅= h 2 3 R⋅ h−( )⋅ 3 SG⋅ V⋅ π = h 2 3 0.181⋅ m⋅ h−( )⋅ 3 0.9⋅ .025⋅ m 3 ⋅ π = h 2 0.544 h−( )⋅ 0.0215= This is a cubic equation for h. We can keep guessing h values, manually iterate, or use Excel's Goal Seek to find h 0.292 m⋅= Problem 3.73 3.73
Problem *3.91 [Difficulty: 2] Given: Specific gravity of a person is to be determined from measurements of weight in air and the met weight when totally immersed in water. Find: Expression for the specific gravity of a person from the measurements. Solution: We will apply the hydrostatics equations to this system. Governing Equation: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) Fnet Fbuoy Mg Assumptions: (1) Static fluid (2) Incompressible fluid Taking a free body diagram of the body: ΣFy 0= Fnet M g⋅− Fbuoy+ 0= Fnet is the weight measurement for the immersed body. Fnet M g⋅ Fbuoy−= M g⋅ ρw g⋅ Vd⋅−= However in air: Fair M g⋅= Therefore the weight measured in water is: Fnet Fair ρw g⋅ Vd⋅−= and Vd Fair Fnet− ρw g⋅ = Now in order to find the specific gravity of the person, we need his/her density: Fair M g⋅= ρ g⋅ Vd⋅= ρ g⋅ Fair Fnet−( ) ρw g⋅ ⋅= Simplifying this expression we get: Fair ρ ρw Fair Fnet−( )= Now if we call the density of water at 4 deg C ρw4C then: Fair ρ ρw4C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ρw ρw4C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Fair Fnet−( )= SG SGw Fair Fnet−( )⋅= Solving this expression for the specific gravity of the person SG, we get: SG SGw Fair Fair Fnet− ⋅= Problem 3.74 3.74
Problem *3.93 [Difficulty: 2] Given: Geometry of steel cylinder Find: Volume of water displaced; number of 1 kg wts to make it sink Solution: The data is For water ρ 999 kg m 3 ⋅= For steel (Table A.1) SG 7.83= For the cylinder D 100 mm⋅= H 1 m⋅= δ 1 mm⋅= The volume of the cylinder is Vsteel δ π D 2 ⋅ 4 π D⋅ H⋅+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Vsteel 3.22 10 4− × m 3 ⋅= The weight of the cylinder is W SG ρ⋅ g⋅ Vsteel⋅= W 7.83 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 3.22× 10 4− × m 3 ⋅ N s 2 ⋅ kg m⋅ ×= W 24.7N= At equilibium, the weight of fluid displaced is equal to the weight of the cylinder Wdisplaced ρ g⋅ Vdisplaced⋅= W= Vdisplaced W ρ g⋅ = 24.7 N⋅ m 3 999 kg⋅ × s 2 9.81 m⋅ × kg m⋅ N s 2 ⋅ ×= Vdisplaced 2.52L= To determine how many 1 kg wts will make it sink, we first need to find the extra volume that will need to be dsiplaced Distance cylinder sank x1 Vdisplaced π D 2 ⋅ 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = x1 0.321m= Hence, the cylinder must be made to sink an additional distance x2 H x1−= x2 0.679m= We deed to add n weights so that 1 kg⋅ n⋅ g⋅ ρ g⋅ π D 2 ⋅ 4 ⋅ x2⋅= n ρ π⋅ D 2 ⋅ x2⋅ 4 1 kg⋅× = 999 kg m 3 ⋅ π 4 × 0.1 m⋅( ) 2 × 0.679× m⋅ 1 1 kg⋅ × N s 2 ⋅ kg m⋅ ×= n 5.33= Hence we need n 6= weights to sink the cylinder Problem 3.75 3.75
Problem 3.76 (Difficulty: 2) 3.76 If the timber weights 670 𝑁, calculate its angle of inclination when the water surface is 2.1 𝑚 above the pivot. Above what depth will the timber stand vertically? Find: Above what depth will the timber stand vertically. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 The buoyancy force is: 𝐹𝑏 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉 = 0.152 𝑚 × 0.152 𝑚 × 𝑥 𝑚 × 9810 𝑁 𝑚3 = 226.7𝑥 (𝑁) Take the moment about pivot we have: 𝑀 = 𝑊 × 0.5 × 3.6 𝑚 cos 𝜃 − 𝑥 2 𝑚 × 𝐹𝑏 cos 𝜃 = 0 670 𝑁 × 0.5 × 3.6 𝑚 × cos 𝜃 − 𝑥 2 𝑚 × 226.7𝑥 × cos 𝜃 = 0 Soving this equation we have: 𝑥 = 3.26 𝑚 The angle when water surface 𝑦 = 2.1 𝑚 is: 𝜃 = sin−1 � 2.1 𝑚 3.26 𝑚 � = 40.1 °
We have the following relation: 𝑥 = 𝑦 sin 𝜃 Substitute in to the momentum we have: 670 𝑁 × 0.5 × 3.6 𝑚 − 𝑦 2sin 𝜃 𝑚 × 226.7 𝑦 sin 𝜃 = 0 If the timber is vertically, we have: 𝜃 = 90° sin 90° = 1 So we have: 670 𝑁 × 0.5 × 3.6 𝑚 − 𝑦 2 𝑚 × 226.7𝑦 = 0 Solving this equation we have: 𝑦 = 3.26 𝑚 When the water surface is 𝑦 = 3.26 𝑚, the timber will stand vertically.
Problem 3.77 (Difficulty: 2) 3.77 The barge shown weights 40 𝑡𝑡𝑡𝑡 and carries a cargo of 40 𝑡𝑡𝑡𝑡. Calculate its draft in freshwater. Find: The draft, where the draft is the depth to which the barge sinks. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 For the barge floating in water we have the buoyancy force as: 𝐹𝐵 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉 = 𝑊 The weight of the barge is: 𝑊 = (40 + 40)𝑡𝑡𝑡𝑡 = 80 𝑡𝑡𝑡𝑡 × 2000 𝑙𝑙𝑙 𝑡𝑡𝑡 = 160000 𝑙𝑙𝑙 The volume of water displaced is then: 𝑉 = 𝑊 𝛾 𝑤𝑤𝑤𝑤𝑤 = 160000 𝑙𝑙𝑙 62.4 𝑙𝑙𝑙 𝑓𝑓3 = 2564 𝑓𝑓3 The volume in terms of the draft d is: ∀= 𝐴 𝑐 𝐿 = �40𝑓𝑓 + 40𝑓𝑓 + 2 × 5 8 𝑑� × 𝑑 2 × 20𝑓𝑓 = 800𝑑 + 12.5𝑑2 Thus we have the relation: 800𝑑 + 12.5𝑑2 = 2564
Solving this equation we have for the draft: 𝑑 = 3.06 𝑓𝑓
Problem *3.94 [Difficulty: 2] Given: Experiment performed by Archimedes to identify the material conent of King Hiero's crown. The crown was weighed in air and in water. Find: Expression for the specific gravity of the crown as a function of the weights in water and air. Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fb ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) Fb Mg Ww Assumptions: (1) Static fluid (2) Incompressible fluid Taking a free body diagram of the body: ΣFz 0= Ww M g⋅− Fb+ 0= Ww is the weight of the crown in water. Ww M g⋅ Fbuoy−= M g⋅ ρw g⋅ Vd⋅−= However in air: Wa M g⋅= Therefore the weight measured in water is: Ww Wa ρw g⋅ Vd⋅−= so the volume is: Vd Wa Ww− ρw g⋅ = Now the density of the crown is: ρc M Vd = M ρw⋅ g⋅ Wa Ww− = Wa Wa Ww− ρw⋅= Therefore, the specific gravity of the crown is: SG ρc ρw = Wa Wa Ww− = SG Wa Wa Ww− = Note: by definition specific gravity is the density of an object divided by the density of water at 4 degrees Celsius, so the measured temperature of the water in the experiment and the data from tables A.7 or A.8 may be used to correct for the variation in density of the water with temperature. Problem 3.78 3.78
Problem *3.96 [Difficulty: 2] Given: Balloons with hot air, helium and hydrogen. Claim lift per cubic foot of 0.018, 0.066, and 0.071 pounds force per cubic f for respective gases, with the air heated to 150 deg. F over ambient. Find: (a) evaluate the claims of lift per unit volume (b) determine change in lift when air is heated to 250 deg. F over ambient. Solution: We will apply the hydrostatics equations to this system. Governing Equations: L ρa g⋅ V⋅ ρg g⋅ V⋅−= (Net lift force is equal to difference in weights of air and gas) p ρ R⋅ T⋅= (Ideal gas equation of state) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Ideal gas behavior The lift per unit volume may be written as: LV L V g ρa ρg−( )⋅== ρa g⋅ 1 ρg ρa − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= now if we take the ideal gas equation and we take into account that the pressure inside and outside the balloon are equal: L V ρa g⋅ 1 Ra Ta⋅ Rg Tg⋅ − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= γa 1 Ra Ta⋅ Rg Tg⋅ − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= At standard conditions the specific weight of air is: γa 0.0765 lbf ft 3 ⋅= the gas constant is: Ra 53.33 ft lbf⋅ lbm R⋅ ⋅= and Ta 519 R⋅= For helium: Rg 386.1 ft lbf⋅ lbm R⋅ ⋅= Tg Ta= and therefore: LVHe 0.0765 lbf ft 3 ⋅ 1 53.33 386.1 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×= LVHe 0.0659 lbf ft 3 ⋅= For hydrogen: Rg 766.5 ft lbf⋅ lbm R⋅ ⋅= Tg Ta= and therefore: LVH2 0.0765 lbf ft 3 ⋅ 1 53.33 766.5 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×= LVH2 0.0712 lbf ft 3 ⋅= For hot air at 150 degrees above ambient: Rg Ra= Tg Ta 150 R⋅+= and therefore: LVair150 0.0765 lbf ft 3 ⋅ 1 519 519 150+ −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×= LVair150 0.0172 lbf ft 3 ⋅= The agreement with the claims stated above is good. For hot air at 250 degrees above ambient: Rg Ra= Tg Ta 250 R⋅+= and therefore: LVair250 0.0765 lbf ft 3 ⋅ 1 519 519 250+ −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×= LVair250 0.0249 lbf ft 3 ⋅= LVair250 LVair150 1.450= Air at ΔT of 250 deg. F gives 45% more lift than air at ΔT of 150 deg.F! Problem 3.79 3.79
Problem *3.98 [Difficulty: 3] Fbuoyancy Wload y Whot air Given: Data on hot air balloon Find: Maximum mass of balloon for neutral buoyancy; mass for initial acceleration of 2.5 ft/s2. Assumptions: Air is treated as static and incompressible, and an ideal gas Solution: Basic equation FB ρatm g⋅ V⋅= and ΣFy M ay⋅= Hence ΣFy 0= FB Whotair− Wload−= ρatm g⋅ V⋅ ρhotair g⋅ V⋅− M g⋅−= for neutral buoyancy M V ρatm ρhotair−( )⋅= V patm⋅ R 1 Tatm 1 Thotair − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= M 320000 ft 3 ⋅ 14.7× lbf in 2 ⋅ 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × lbm R⋅ 53.33 ft⋅ lbf⋅ × 1 48 460+( ) R⋅ 1 160 460+( ) R⋅ −⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ×= M 4517 lbm⋅= Initial acceleration ΣFy FB Whotair− Wload−= ρatm ρhotair−( ) g⋅ V⋅ Mnew g⋅−= Maccel a⋅= Mnew 2 ρhotair⋅ V⋅+( ) a⋅= Solving for Mnew ρatm ρhotair−( ) g⋅ V⋅ Mnew g⋅− Mnew 2 ρhotair⋅ V⋅+( ) a⋅= Mnew V ρatm ρhotair−( ) g⋅ 2 ρhotair⋅ a⋅− a g+ ⋅= V patm⋅ a g+ g 1 Tatm 1 Thotair −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 2 a⋅ Thotair −⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Mnew 320000 ft 3 ⋅ 14.7⋅ lbf in 2 ⋅ 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⋅ lbm R⋅ 53.33 ft⋅ lbf⋅ ⋅ s 2 2.5 32.2+( ) ft⋅ ⋅ 32.2 1 48 460+( ) 1 160 460+( ) −⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ 2 2.5⋅ 1 160 460+( ) ⋅−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ ft s 2 R⋅ ⋅= Mnew 1239 lbm⋅= To make the balloon move up or down during flight, the air needs to be heated to a higher temperature, or let cool (or let in ambient air). Problem 3.80 3.80
Problem 3.81 (Difficulty: 2) 3.81 The opening in the bottom of the tank is square and slightly less than 2 𝑓𝑓 on each side. The opening is to be plugged with a wooden cube 2 𝑓𝑓 on a side. (a) What weight 𝑊 should be attached to the wooden cube to insure successful plugging of the hole? The wood weighs 40 𝑙𝑙𝑙 𝑓𝑓3, (b) What upward force must be exerted on the block to lift it and allow water to drain from the tank? Find: The weight of the block and the force needed to lift it Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 (a) Because the wood bottom surface is in the atmosphere so the pressure on the bottom surface is zero in this case and there is no buoyancy force. The force acting on the wood cube in the vertical direction is: 𝐹𝑉 = 𝐹𝑝 + 𝐺 𝐹𝑉 = 𝛾ℎ1 𝐴 + 𝐺 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 5 𝑓𝑓 × 2𝑓𝑓 × 2𝑓𝑓 + 40 𝑙𝑙𝑙 𝑓𝑓3 × (2 𝑓𝑓)3 = 1568 𝑙𝑙𝑙
The direction of 𝐹𝑉 is downward. So we do not need any weight 𝑊 attached to the wood cube. (b) To lift the block, we need a force larger than 𝐹𝑉, so we have: 𝐹𝑢𝑢 ≥ 𝐹𝑉 = 1568 𝑙𝑙𝑙
Problem 3.82 (Difficulty: 2) 3.82 A balloon has a weight (including crew but not gas) of 2.2 𝑘𝑘 and a gas-bag capacity of 566 𝑚3 . At the ground it is (partially) inflated with 445 𝑁 of helium. How high can this balloon rise in the U.S standard atmosphere if the helium always assumes the pressure and temperature of the atmosphere? Find: How high this balloon will rise. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 At the sea level, for helium we have: 𝑝 = 101.3 𝑘𝑘𝑘 𝑇 = 288 𝐾 𝑅 = 2076.8 𝐽 𝑘𝑘 ∙ 𝐾 According to the ideal gas law: 𝜌ℎ = 𝑝 𝑅𝑅 = 101.3 𝑘𝑘𝑘 2076.8 𝐽 𝑘𝑔 ∙ 𝐾 × 288 𝐾 = 0.1694 𝑘𝑘 𝑚3 𝛾ℎ = 𝜌𝜌 = 0.1694 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 = 1.662 𝑁 𝑚3 The volume of the helium is: 𝑉ℎ = 𝑊ℎ 𝛾ℎ = 445 𝑁 1.662 𝑁 𝑚3 = 268 𝑚3 The buoyancy force is calculated by: 𝐹𝐵 = 𝛾 𝑎𝑎𝑎 𝑉𝑏 The weight of the whole balloon is:
𝑊 = 2.2 𝑘𝑘 + 𝑊ℎ We have the following table as (the helium always has the same temperature and pressure as the atmosphere): Altitude (km) Pressure (kPa) Temperature (K) ∀ (𝑚3) 𝛾 𝑎𝑎𝑎 � 𝑁 𝑚3 � 𝑊ℎ (𝑘𝑘) 𝐹𝐵 (𝑘𝑘) 𝑊 (𝑘𝑘) 6 47.22 249.2 497 6.46 0.445 3.21 2.65 8 35.70 236.3 566 5.14 0.402 2.91 2.60 10 26.50 223.4 566 4.04 0.317 2.29 2.52 When the maximum volume of the helium is reached, the volume will become a constant for helium. Equilibrium is reached as: 𝐹𝐵 = 𝑊 At 8 𝑘𝑘 we have: 𝐹𝐵 − 𝑊 = 0.31 𝑘𝑘 At 10 𝑘𝑘 we have: 𝐹𝐵 − 𝑊 = −0.23 𝑘𝑘 With the interpolation we have the height for equilibrium as: ℎ = 8𝑘𝑘 + 2𝑘𝑘 × 0.31 0.31 + 0.23 = 9.15 𝑘𝑘
Problem *3.100 [Difficulty: 3] Given: A pressurized balloon is to be designed to lift a payload of mass M to an altitude of 40 km, where p = 3.0 mbar and T = -25 deg C. The balloon skin has a specific gravity of 1.28 and thickness 0.015 mm. The gage pressure of the helium is 0.45 mbar. The allowable tensile stress in the balloon is 62 MN/m2 t M D Find: (a) The maximum balloon diameter (b) The maximum payload mass Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to mass of displaced fluid) p ρ R⋅ T⋅= (Ideal gas equation of state) πD2 ∆p/4 πDtσ Assumptions: (1) Static, incompressible fluid (2) Static equilibrium at 40 km altitude (3) Ideal gas behavior The diameter of the balloon is limited by the allowable tensile stress in the skin: ΣF π 4 D 2 ⋅ ∆p⋅ π D⋅ t⋅ σ⋅−= 0= Solving this expression for the diameter: Dmax 4 t⋅ σ⋅ ∆p = Fbuoyant Mbg Mg z Dmax 4 0.015× 10 3− × m⋅ 62× 10 6 × N m 2 ⋅ 1 0.45 10 3− ⋅ bar⋅ × bar m 2 ⋅ 10 5 N⋅ ×= Dmax 82.7m= To find the maximum allowable payload we perform a force balance on the system: ΣFz Fbuoy MHe g⋅− Mb g⋅− M g⋅−= 0= ρa g⋅ Vb⋅ ρHe g⋅ Vb⋅− ρs g⋅ Vs⋅− M g⋅− 0= Solving for M: M ρa ρHe−( ) Vb⋅ ρs Vs⋅−= The volume of the balloon is: Vb π 6 D 3 ⋅= The volume of the skin is: Vs π D 2 ⋅ t⋅= Therefore, the mass is: M π 6 ρa ρHe−( )⋅ D 3 ⋅ π ρs⋅ D 2 ⋅ t⋅−= The air density: ρa pa Ra T⋅ = ρa 3.0 10 3− × bar⋅ kg K⋅ 287 N⋅ m⋅ × 1 273 25−( ) K⋅ × 10 5 N⋅ bar m 2 ⋅ ×= ρa 4.215 10 3− × kg m 3 = Repeating for helium: ρHe p R T⋅ = ρHe 6.688 10 4− × kg m 3 = The payload mass is: M π 6 4.215 0.6688−( )× 10 3− × kg m 3 ⋅ 82.7 m⋅( ) 3 × π 1.28× 10 3 × kg m 3 ⋅ 82.7 m⋅( ) 2 × 0.015× 10 3− × m⋅−= M 638 kg= Problem 3.83 3.83
Problem *3.102 [Difficulty: 3] Given: Glass hydrometer used to measure SG of liquids. Stem has diameter D=5 mm, distance between marks on stem is d=2 mm per 0.1 SG. Hydrometer floats in kerosene (Assume zero contact angle between glass and kerosene). Find: Magnitude of error introduced by surface tension. Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) d = 2 mm/0.1 SG ∆FB y D = 5 mm Kerosene Fσ Assumptions: (1) Static fluid (2) Incompressible fluid (3) Zero contact angle between ethyl alcohol and glass The surface tension will cause the hydrometer to sink ∆h lower into the liquid. Thus for this change: ΣFz ∆Fbuoy Fσ−= 0= The change in buoyant force is: ∆Fbuoy ρ g⋅ ∆V⋅= ρ g⋅ π 4 ⋅ D 2 ⋅ ∆h⋅= The force due to surface tension is: Fσ π D⋅ σ⋅ cos θ( )⋅= π D⋅ σ⋅= Thus, ρ g⋅ π 4 ⋅ D 2 ⋅ ∆h⋅ π D⋅ σ⋅= Upon simplification: ρ g⋅ D⋅ ∆h⋅ 4 σ= Solving for ∆h: ∆h 4 σ⋅ ρ g⋅ D⋅ = From Table A.2, SG = 1.43 and from Table A.4, σ = 26.8 mN/m Therefore, ∆h 4 26.8× 10 3− × N m ⋅ m 3 1430 kg⋅ × s 2 9.81 m⋅ × 1 5 10 3− × m⋅ × kg m⋅ s 2 N⋅ ×= ∆h 1.53 10 3− × m= So the change in specific gravity will be: ∆SG 1.53 10 3− × m⋅ 0.1 2 10 3− × m⋅ ×= ∆SG 0.0765= From the diagram, surface tension acts to cause the hydrometer to float lower in the liquid. Therefore, surface tension results in an indicated specific gravity smaller than the actual specific gravity. Problem 3.84 3.84
Problem *3.103 [Difficulty:4] Given: Sphere partially immersed in a liquid of specific gravity SG. Find: (a) Formula for buoyancy force as a function of the submersion depth d (b) Plot of results over range of liquid depth Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts everywhere d Rsinθ R dθ θmax hWe need an expression for the displaced volume of fluid at an arbitrary depth d. From the diagram we see that: d R 1 cos θmax( )−( )= at an arbitrary depth h: h d R 1 cos θ( )−( )⋅−= r R sin θ( )⋅= So if we want to find the volume of the submerged portion of the sphere we calculate: Vd 0 θmax hπ r 2⌠ ⎮ ⌡ d= π 0 θmax θR 2 sin θ( )( ) 2 ⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⌡ d⋅= π R 3 ⋅ 0 θmax θsin θ( )( ) 3⌠ ⎮ ⌡ d⋅= Evaluating the integral we get: Vd π R 3 ⋅ cos θmax( )( )3 3 cos θmax( )− 2 3 + ⎡⎢ ⎢ ⎣ ⎤⎥ ⎥ ⎦ ⋅= Now since: cos θmax( ) 1 d R −= we get: Vd π R 3 ⋅ 1 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ − 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Thus the buoyant force is: Fbuoy ρw SG⋅ g⋅ π⋅ R 3 ⋅ 1 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ 3 ⋅ 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ − 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= If we non-dimensionalize by the force on a fully submerged sphere: Fd Fbuoy ρw SG⋅ g⋅ 4 3 ⋅ π⋅ R 3 ⋅ = 3 4 1 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ 3 ⋅ 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ − 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = Fd 3 4 1 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ 3 ⋅ 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ − 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = 0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 Submergence Ratio d/R ForceRatioFd Problem 3.85 3.85
Problem *3.106 [Difficulty: 4] FB W FL FU x y Given: Data on sphere and tank bottom Find: Expression for SG of sphere at which it will float to surface; minimum SG to remain in position Assumptions: (1) Water is static and incompressible (2) Sphere is much larger than the hole at the bottom of the tank Solution: Basic equations FB ρ g⋅ V⋅= and ΣFy FL FU− FB+ W−= where FL patm π⋅ a 2 ⋅= FU patm ρ g⋅ H 2 R⋅−( )⋅+⎡⎣ ⎤⎦ π⋅ a 2 ⋅= FB ρ g⋅ Vnet⋅= Vnet 4 3 π⋅ R 3 ⋅ π a 2 ⋅ 2⋅ R⋅−= W SG ρ⋅ g⋅ V⋅= with V 4 3 π⋅ R 3 ⋅= Now if the sum of the vertical forces is positive, the sphere will float away, while if the sum is zero or negative the sphere will stay at the bottom of the tank (its weight and the hydrostatic force are greater than the buoyant force). Hence ΣFy patm π⋅ a 2 ⋅ patm ρ g⋅ H 2 R⋅−( )⋅+⎡⎣ ⎤⎦ π⋅ a 2 ⋅− ρ g⋅ 4 3 π⋅ R 3 ⋅ 2 π⋅ R⋅ a 2 ⋅−⎛ ⎜ ⎝ ⎞ ⎠ ⋅+ SG ρ⋅ g⋅ 4 3 ⋅ π⋅ R 3 ⋅−= This expression simplifies to ΣFy π ρ⋅ g⋅ 1 SG−( ) 4 3 ⋅ R 3 ⋅ H a 2 ⋅−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= ΣFy π 1.94× slug ft 3 ⋅ 32.2× ft s 2 ⋅ 4 3 1 0.95−( )× 1 in⋅ ft 12 in⋅ ×⎛ ⎜ ⎝ ⎞ ⎠ 3 × 2.5 ft⋅ 0.075 in⋅ ft 12 in⋅ ×⎛ ⎜ ⎝ ⎞ ⎠ 2 ×− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × lbf s 2 ⋅ slug ft⋅ ×= ΣFy 0.012− lbf⋅= Therefore, the sphere stays at the bottom of the tank. Problem 3.86 3.86
Problem *3.108 [Difficulty: 3] H = 8 ft h = 7 ft θ = 60o Floating Sinking Given: Data on boat Find: Effective density of water/air bubble mix if boat sinks Solution: Basic equations FB ρ g⋅ V⋅= and ΣFy 0= We can apply the sum of forces for the "floating" free body ΣFy 0= FB W−= where FB SGsea ρ⋅ g⋅ Vsubfloat⋅= Vsubfloat 1 2 h⋅ 2 h⋅ tan θ⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ L⋅= L h 2 ⋅ tan θ( ) = SGsea 1.024= (Table A.2) Hence W SGsea ρ⋅ g⋅ L⋅ h 2 ⋅ tan θ( ) = (1) We can apply the sum of forces for the "sinking" free body ΣFy 0= FB W−= where FB SGmix ρ⋅ g⋅ Vsub⋅= Vsubsink 1 2 H⋅ 2 H⋅ tan θ⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ L⋅= L H 2 ⋅ tan θ( ) = Hence W SGmix ρ⋅ g⋅ L⋅ H 2 ⋅ tan θ( ) = (2) Comparing Eqs. 1 and 2 W SGsea ρ⋅ g⋅ L⋅ h 2 ⋅ tan θ( ) = SGmix ρ⋅ g⋅ L⋅ H 2 ⋅ tan θ( ) = SGmix SGsea h H ⎛ ⎜ ⎝ ⎞ ⎠ 2 ⋅= SGmix 1.024 7 8 ⎛ ⎜ ⎝ ⎞ ⎠ 2 ×= SGmix 0.784= The density is ρmix SGmix ρ⋅= ρmix 0.784 1.94× slug ft 3 ⋅= ρmix 1.52 slug ft 3 ⋅= Problem 3.87 3.87
Problem *3.112 [Difficulty: 2] Given: Steel balls resting in floating plastic shell in a bucket of water Find: What happens to water level when balls are dropped in water Solution: Basic equation FB ρ Vdisp⋅ g⋅= W= for a floating body weight W When the balls are in the plastic shell, the shell and balls displace a volume of water equal to their own weight - a large volume because the balls are dense. When the balls are removed from the shell and dropped in the water, the shell now displaces only a small volume of water, and the balls sink, displacing only their own volume. Hence the difference in displaced water before and after moving the balls is the difference between the volume of water that is equal to the weight of the balls, and the volume of the balls themselves. The amount of water displaced is significantly reduced, so the water level in the bucket drops. Volume displaced before moving balls: V1 Wplastic Wballs+ ρ g⋅ = Volume displaced after moving balls: V2 Wplastic ρ g⋅ Vballs+= Change in volume displaced ∆V V2 V1−= Vballs Wballs ρ g⋅ −= Vballs SGballs ρ⋅ g⋅ Vballs⋅ ρ g⋅ −= ∆V Vballs 1 SGballs−( )⋅= Hence initially a large volume is displaced; finally a small volume is displaced (∆V < 0 because SGballs > 1) Problem 3.88 3.88
Problem *3.113 [Difficulty: 4] Open-Ended Problem Statement: A proposed ocean salvage scheme involves pumping air into “bags” placed within and around a wrecked vessel on the sea bottom. Comment on the practicality of this plan, supporting your conclusions with analyses. Discussion: This plan has several problems that render it impractical. First, pressures at the sea bottom are very high. For example, Titanic was found in about 12,000 ft of seawater. The corresponding pressure is nearly 6,000 psi. Compressing air to this pressure is possible, but would require a multi-stage compressor and very high power. Second, it would be necessary to manage the buoyancy force after the bag and object are broken loose from the sea bed and begin to rise toward the surface. Ambient pressure would decrease as the bag and artifact rise toward the surface. The air would tend to expand as the pressure decreases, thereby tending to increase the volume of the bag. The buoyancy force acting on the bag is directly proportional to the bag volume, so it would increase as the assembly rises. The bag and artifact thus would tend to accelerate as they approach the sea surface. The assembly could broach the water surface with the possibility of damaging the artifact or the assembly. If the bag were of constant volume, the pressure inside the bag would remain essentially constant at the pressure of the sea floor, e.g., 6,000 psi for Titanic. As the ambient pressure decreases, the pressure differential from inside the bag to the surroundings would increase. Eventually the difference would equal sea floor pressure. This probably would cause the bag to rupture. If the bag permitted some expansion, a control scheme would be needed to vent air from the bag during the trip to the surface to maintain a constant buoyancy force just slightly larger than the weight of the artifact in water. Then the trip to the surface could be completed at low speed without danger of broaching the surface or damaging the artifact. Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual Full Download: https://alibabadownload.com/product/fox-and-mcdonalds-introduction-to-fluid-mechanics-9th-edition-pritchard-so This sample only, Download all chapters at: AlibabaDownload.com Problem 3.89 3.89

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Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual

  • 1. Problem 3.2 [Difficulty: 2] Given: Pure water on a standard day Find: Boiling temperature at (a) 1000 m and (b) 2000 m, and compare with sea level value. Solution: We can determine the atmospheric pressure at the given altitudes from table A.3, Appendix A The data are Elevation (m) p/p o p (kPa) 0 1.000 101.3 1000 0.887 89.9 2000 0.785 79.5 We can also consult steam tables for the variation of saturation temperature with pressure: p (kPa) T sat (°C) 70 90.0 80 93.5 90 96.7 101.3 100.0 We can interpolate the data from the steam tables to correlate saturation temperature with altitude: Elevation (m) p/p o p (kPa) T sat (°C) 0 1.000 101.3 100.0 1000 0.887 89.9 96.7 2000 0.785 79.5 93.3 The data are plotted here. They show that the saturation temperature drops approximately 3.4°C/1000 m. Variation of Saturation Temperature with Pressure 88 90 92 94 96 98 100 70 75 80 85 90 95 100 105 Absolute Pressure (kPa) Saturation Temperature(°C) 2000 m 1000 m Sea Level Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual Full Download: https://alibabadownload.com/product/fox-and-mcdonalds-introduction-to-fluid-mechanics-9th-edition-pritchard-so This sample only, Download all chapters at: AlibabaDownload.com Problem 3.1 3.1
  • 2. Problem 3.3 [Difficulty: 2] Given: Data on flight of airplane Find: Pressure change in mm Hg for ears to "pop"; descent distance from 8000 m to cause ears to "pop." Solution: Assume the air density is approximately constant constant from 3000 m to 2900 m. From table A.3 ρSL 1.225 kg m 3 ⋅= ρair 0.7423 ρSL⋅= ρair 0.909 kg m 3 = We also have from the manometer equation, Eq. 3.7 Δp ρair− g⋅ Δz⋅= and also Δp ρHg− g⋅ ΔhHg⋅= Combining ΔhHg ρair ρHg Δz⋅= ρair SGHg ρH2O⋅ Δz⋅= SGHg 13.55= from Table A.2 ΔhHg 0.909 13.55 999× 100× m⋅= ΔhHg 6.72 mm⋅= For the ear popping descending from 8000 m, again assume the air density is approximately constant constant, this time at 8000 m. From table A.3 ρair 0.4292 ρSL⋅= ρair 0.526 kg m 3 = We also have from the manometer equation ρair8000 g⋅ Δz8000⋅ ρair3000 g⋅ Δz3000⋅= where the numerical subscripts refer to conditions at 3000m and 8000m. Hence Δz8000 ρair3000 g⋅ ρair8000 g⋅ Δz3000⋅= ρair3000 ρair8000 Δz3000⋅= Δz8000 0.909 0.526 100× m⋅= Δz8000 173m= Problem 3.2 3.2
  • 3. Problem 3.4 [Difficulty: 3] Given: Boiling points of water at different elevations Find: Change in elevation Solution: From the steam tables, we have the following data for the boiling point (saturation temperature) of water Tsat ( o F) p (psia) 195 10.39 185 8.39 The sea level pressure, from Table A.3, is pSL = 14.696 psia Hence Tsat ( o F) p/pSL 195 0.707 185 0.571 From Table A.3 p/pSL Altitude (m) Altitude (ft) 0.7372 2500 8203 0.6920 3000 9843 0.6492 3500 11484 0.6085 4000 13124 0.5700 4500 14765 Then, any one of a number of Excel functions can be used to interpolate (Here we use Excel 's Trendline analysis) p/pSL Altitude (ft) 0.707 9303 Current altitude is approximately 9303 ft 0.571 14640 The change in altitude is then 5337 ft Alternatively, we can interpolate for each altitude by using a linear regression between adjacent data points p/pSL Altitude (m) Altitude (ft) p/pSL Altitude (m) Altitude (ft) For 0.7372 2500 8203 0.6085 4000 13124 0.6920 3000 9843 0.5700 4500 14765 Then 0.7070 2834 9299 0.5730 4461 14637 The change in altitude is then 5338 ft Altitude vs Atmospheric Pressure z = -39217(p/pSL) + 37029 R2 = 0.999 2500 5000 7500 10000 12500 15000 0.55 0.60 0.65 0.70 0.75 p/pSL Altitude(ft) Data Linear Trendline Problem 3.3 3.3
  • 4. Problem 3.9 [Difficulty: 2] Given: Data on tire at 3500 m and at sea level Find: Absolute pressure at 3500 m; pressure at sea level Solution: At an elevation of 3500 m, from Table A.3: pSL 101 kPa⋅= patm 0.6492 pSL⋅= patm 65.6 kPa⋅= and we have pg 0.25 MPa⋅= pg 250 kPa⋅= p pg patm+= p 316 kPa⋅= At sea level patm 101 kPa⋅= Meanwhile, the tire has warmed up, from the ambient temperature at 3500 m, to 25oC. At an elevation of 3500 m, from Table A.3 Tcold 265.4 K⋅= and Thot 25 273+( ) K⋅= Thot 298K= Hence, assuming ideal gas behavior, pV = mRT, and that the tire is approximately a rigid container, the absolute pressure of the hot tire is phot Thot Tcold p⋅= phot 354 kPa⋅= Then the gage pressure is pg phot patm−= pg 253 kPa⋅= Problem 3.4 3.4
  • 5. Problem 3.5 [Difficulty: 2] Given: Data on system Find: Force on bottom of cube; tension in tether Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ Δp ρ g⋅ h⋅= where h is measured downwards The absolute pressure at the interface is pinterface patm SGoil ρ⋅ g⋅ hoil⋅+= Then the pressure on the lower surface is pL pinterface ρ g⋅ hL⋅+= patm ρ g⋅ SGoil hoil⋅ hL+( )⋅+= For the cube V 125 mL⋅= V 1.25 10 4− × m 3 ⋅= Then the size of the cube is d V 1 3 = d 0.05m= and the depth in water to the upper surface is hU 0.3 m⋅= Hence hL hU d+= hL 0.35m= where hL is the depth in water to the lower surface The force on the lower surface is FL pL A⋅= where A d 2 = A 0.0025m 2 = FL patm ρ g⋅ SGoil hoil⋅ hL+( )⋅+⎡⎣ ⎤⎦ A⋅= FL 101 10 3 × N m 2 ⋅ 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.8 0.5× m⋅ 0.35 m⋅+( )× N s 2 ⋅ kg m⋅ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0.0025× m 2 ⋅= FL 270.894N= Note: Extra decimals needed for computing T later! For the tension in the tether, an FBD gives ΣFy 0= FL FU− W− T− 0= or T FL FU− W−= where FU patm ρ g⋅ SGoil hoil⋅ hU+( )⋅+⎡⎣ ⎤⎦ A⋅= Problem 3.5 3.5
  • 6. Note that we could instead compute ΔF FL FU−= ρ g⋅ SGoil⋅ hL hU−( )⋅ A⋅= and T ΔF W−= Using FU FU 101 10 3 × N m 2 ⋅ 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.8 0.5× m⋅ 0.3 m⋅+( )× N s 2 ⋅ kg m⋅ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ 0.0025× m 2 ⋅= FU 269.668N= Note: Extra decimals needed for computing T later! For the oak block (Table A.1) SGoak 0.77= so W SGoak ρ⋅ g⋅ V⋅= W 0.77 1000× kg m 3 ⋅ 9.81× m s 2 ⋅ 1.25× 10 4− × m 3 ⋅ N s 2 ⋅ kg m⋅ ×= W 0.944N= T FL FU− W−= T 0.282N=
  • 7. Problem 3.6 [Difficulty: 2] Given: Data on system before and after applied force Find: Applied force Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ p patm ρ g⋅ y y0−( )⋅−= with p y0( ) patm= For initial state p1 patm ρ g⋅ h⋅+= and F1 p1 A⋅= ρ g⋅ h⋅ A⋅= (Gage; F1 is hydrostatic upwards force) For the initial FBD ΣFy 0= F1 W− 0= W F1= ρ g⋅ h⋅ A⋅= For final state p2 patm ρ g⋅ H⋅+= and F2 p2 A⋅= ρ g⋅ H⋅ A⋅= (Gage; F2 is hydrostatic upwards force) For the final FBD ΣFy 0= F2 W− F− 0= F F2 W−= ρ g⋅ H⋅ A⋅ ρ g⋅ h⋅ A⋅−= ρ g⋅ A⋅ H h−( )⋅= F ρH2O SG⋅ g⋅ π D 2 ⋅ 4 ⋅ H h−( )⋅= From Fig. A.1 SG 13.54= F 1000 kg m 3 ⋅ 13.54× 9.81× m s 2 ⋅ π 4 × 0.05 m⋅( ) 2 × 0.2 0.025−( )× m⋅ N s 2 ⋅ kg m⋅ ×= F 45.6N= Problem 3.6 3.6
  • 8. Problem 3.7 (Difficulty: 1) 3.7 Calculate the absolute pressure and gage pressure in an open tank of crude oil 2.4 𝑚 below the liquid surface. If the tank is closed and pressurized to 130 𝑘𝑘𝑘, what are the absolute pressure and gage pressure at this location. Given: Location: ℎ = 2.4 𝑚 below the liquid surface. Liquid: Crude oil. Find: The absolute pressure 𝑝 𝑎 and gage pressure 𝑝 𝑔 for both open and closed tank . Assumption: The gage pressure for the liquid surface is zero for open tank and closed tank. The oil is incompressible. Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 The density for the crude oil is: 𝜌 = 856 𝑘𝑘 𝑚3 The atmosphere pressure is: 𝑝 𝑎𝑎𝑎𝑎𝑎 = 101000 𝑃𝑃 The pressure for the closed tank is: 𝑝 𝑡𝑡𝑡𝑡 = 130 𝑘𝑘𝑘 = 130000 𝑃𝑃 Using the hydrostatic relation, the gage pressure of open tank 2.4 m below the liquid surface is: 𝑝 𝑔 = 𝜌𝜌ℎ = 856 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 2.4 𝑚 = 20100 𝑃𝑃 The absolute pressure of open tank at this location is: 𝑝 𝑎 = 𝑝 𝑔 + 𝑝 𝑎𝑎𝑎𝑎𝑎 = 20100 𝑃𝑃 + 101000 𝑃𝑃 = 121100 𝑃𝑃 = 121.1 𝑘𝑘𝑘 The gage pressure of closed tank at the same location below the liquid surface is the same as open tank: 𝑝 𝑔 = 𝜌𝜌ℎ = 856 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 2.4 𝑚 = 20100 𝑃𝑃 The absolute pressure of closed tank at this location is: 𝑝 𝑎 = 𝑝 𝑔 + 𝑝 𝑡𝑡𝑡𝑡 = 20100 𝑃𝑃 + 130000 𝑃𝑃 = 150100 𝑃𝑃 = 150.1 𝑘𝑘𝑘
  • 9. Problem 3.8 (Difficulty: 1) 3.8 An open vessel contains carbon tetrachloride to a depth of 6 𝑓𝑓 and water on the carbon tetrachloride to a depth of 5 𝑓𝑓 . What is the pressure at the bottom of the vessel? Given: Depth of carbon tetrachloride: ℎ 𝑐 = 6 𝑓𝑓. Depth of water: ℎ 𝑤 = 5 𝑓𝑓. Find: The gage pressure 𝑝 at the bottom of the vessel. Assumption: The gage pressure for the liquid surface is zero. The fluid is incompressible. Solution: Use the hydrostatic pressure relation to detmine pressures in a fluid. Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 The density for the carbon tetrachloride is: 𝜌𝑐 = 1.59 × 103 𝑘𝑘 𝑚3 = 3.09 𝑠𝑠𝑠𝑠 𝑓𝑓3 The density for the water is: 𝜌 𝑤 = 1.0 × 103 𝑘𝑘 𝑚3 = 1.940 𝑠𝑠𝑠𝑠 𝑓𝑓3 Using the hydrostatic relation, the gage pressure 𝑝 at the bottom of the vessel is: 𝑝 = 𝜌𝑐 𝑔ℎ 𝑐 + 𝜌 𝑤 𝑔ℎ 𝑤 𝑝 = 3.09 𝑠𝑠𝑠𝑠 𝑓𝑓3 × 32.2 𝑓𝑓 𝑠2 × 6 𝑓𝑓 + 1.940 𝑠𝑠𝑠𝑠 𝑓𝑓3 × 32.2 𝑓𝑓 𝑠2 × 5 𝑓𝑓 = 909 𝑙𝑙𝑙 𝑓𝑓2 = 6.25 𝑝𝑝𝑝
  • 10. Problem 3.8 [Difficulty: 2] Given: Properties of a cube floating at an interface Find: The pressures difference between the upper and lower surfaces; average cube density Solution: The pressure difference is obtained from two applications of Eq. 3.7 pU p0 ρSAE10 g⋅ H 0.1 d⋅−( )⋅+= pL p0 ρSAE10 g⋅ H⋅+ ρH2O g⋅ 0.9⋅ d⋅+= where pU and pL are the upper and lower pressures, p0 is the oil free surface pressure, H is the depth of the interface, and d is the cube size Hence the pressure difference is Δp pL pU−= ρH2O g⋅ 0.9⋅ d⋅ ρSAE10 g⋅ 0.1⋅ d⋅+= Δp ρH2O g⋅ d⋅ 0.9 SGSAE10 0.1⋅+( )⋅= From Table A.2 SGSAE10 0.92= Δp 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.1× m⋅ 0.9 0.92 0.1×+( )× N s 2 ⋅ kg m⋅ ×= Δp 972Pa= For the cube density, set up a free body force balance for the cube ΣF 0= Δp A⋅ W−= Hence W Δp A⋅= Δp d 2 ⋅= ρcube m d 3 = W d 3 g⋅ = Δp d 2 ⋅ d 3 g⋅ = Δp d g⋅ = ρcube 972 N m 2 ⋅ 1 0.1 m⋅ × s 2 9.81 m⋅ × kg m⋅ N s 2 ⋅ ×= ρcube 991 kg m 3 = Problem 3.9 3.9 these equations:
  • 11. Problem 3.1 [Difficulty: 2] Given: Data on nitrogen tank Find: Pressure of nitrogen; minimum required wall thickness Assumption: Ideal gas behavior Solution: Ideal gas equation of state: p V⋅ M R⋅ T⋅= where, from Table A.6, for nitrogen R 55.16 ft lbf⋅ lbm R⋅ ⋅= Then the pressure of nitrogen is p M R⋅ T⋅ V = M R⋅ T⋅ 6 π D 3 ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= p 140 lbm⋅ 55.16× ft lbf⋅ lbm R⋅ ⋅ 77 460+( )× R⋅ 6 π 2.5 ft⋅( ) 3 × ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × ft 12 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ×= p 3520 lbf in 2 ⋅= σcπDt pπD2 /4 To determine wall thickness, consider a free body diagram for one hemisphere: ΣF 0= p π D 2 ⋅ 4 ⋅ σc π⋅ D⋅ t⋅−= where σc is the circumferential stress in the container Then t p π⋅ D 2 ⋅ 4 π⋅ D⋅ σc⋅ = p D⋅ 4 σc⋅ = t 3520 lbf in 2 ⋅ 2.5 ft⋅ 4 × in 2 30 10 3 × lbf⋅ ×= t 0.0733 ft⋅= t 0.880 in⋅= Problem 3.10 3.10
  • 12. Problem 3.11 (Difficulty: 2) 3.11 If at the surface of a liquid the specific weight is 𝛾0, with 𝑧 and 𝑝 both zero, show that, if 𝐸 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐, the specific weight and pressure are given 𝛾 = 𝐸 �𝑧+ 𝐸 𝛾0 � and 𝑝 = −𝐸 ln �1 + 𝛾0 𝑍 𝐸 �. Calculate specific weight and pressure at a depth of 2 𝑘𝑘 assuming 𝛾0 = 10.0 𝑘𝑘 𝑚3 and 𝐸 = 2070 𝑀𝑀𝑀. Given: Depth: ℎ = 2 𝑘𝑘. The specific weight at surface of a liquid: 𝛾0 = 10.0 𝑘𝑘 𝑚3. Find: The specific weight and pressure at a depth of 2 𝑘𝑘. Assumption:. Bulk modulus is constant Solution: Use the hydrostatic pressure relation and definition of bulk modulus to detmine pressures in a fluid. Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Definition of bulk modulus 𝐸𝑣 = 𝑑𝑑 𝑑𝑑 𝜌� = 𝑑𝑑 𝑑𝑑 𝛾� Eliminating dp from the hydrostatic pressure relation and the bulk modulus definition: 𝑑𝑑 = −𝛾 𝑑𝑑 = 𝐸𝑣 𝑑𝑑 𝛾 Or 𝑑𝑑 = −𝐸𝑣 𝑑𝑑 𝛾2 Integrating for both sides we get: 𝑧 = 𝐸𝑣 1 𝛾 + 𝑐 At 𝑧 = 0, 𝛾 = 𝛾0 so: 𝑐 = −𝐸𝑣 1 𝛾0
  • 13. 𝑧 = 𝐸𝑣 1 𝛾 − 𝐸𝑣 1 𝛾0 Solving for 𝛾, we have: 𝛾 = 𝐸𝑣 �𝑧 + 𝐸𝑣 𝛾0 � Solving for the pressure using the hydrostatic relation: 𝑑𝑑 = −𝛾𝛾𝛾 = − 𝐸𝑣 �𝑧 + 𝐸𝑣 𝛾0 � 𝑑𝑑 Integrating both sides we to get: 𝑝 = −𝐸𝑣 ln �𝑧 + 𝐸𝑣 𝛾0 � + 𝑐 At 𝑧 = 0, 𝑝 = 0 so: 𝑐 = 𝐸𝑣 ln � 𝐸𝑣 𝛾0 � 𝑝 = −𝐸𝑣 ln �𝑧 + 𝐸𝑣 𝛾0 � + 𝐸𝑣 ln � 𝐸𝑣 𝛾0 � = −𝐸𝑣 ln �1 + 𝛾0 𝑧 𝐸𝑣 � For the specific case ℎ = 2 𝑘𝑘 𝛾0 = 10.0 𝑘𝑘 𝑚3 𝐸𝑣 = 2070 𝑀𝑀𝑀 The specific weight: 𝛾 = 𝐸𝑣 �𝑧 + 𝐸𝑣 𝛾0 � = 2070 × 106 𝑝𝑝 �−2000 𝑃𝑃 + 2070 × 106 𝑃𝑃 10 × 103 𝑁 𝑚3 � = 10100 𝑁 𝑚3 = 10.1 𝑘𝑘 𝑚3 Pressure: 𝑝 = −𝐸𝑣 ln �1 + 𝛾0 𝑧 𝐸𝑣 � = −2070 × 106 𝑃𝑃 × ln �1 + 10000.0 𝑘𝑘 𝑚3 × � −2000 𝑚 2070 × 106 𝑃𝑃 �� = 20100 𝑘𝑘𝑘
  • 14. Problem 3.12 (Difficulty: 2) 3.12 In the deep ocean the compressibility of seawater is significant in its effect on 𝜌 and 𝑝. If 𝐸 = 2.07 × 109 𝑃𝑃, find the percentage change in the density and pressure at a depth of 10000 meters as compared to the values obtained at the same depth under the incompressible assumption. Let 𝜌0 = 1020 𝑘𝑘 𝑚3 and the absolute pressure 𝑝0 = 101.3 𝑘𝑘𝑘. Given: Depth: ℎ = 10000 𝑚𝑚𝑚𝑚𝑚𝑚. The density: 𝜌0 = 1020 𝑘𝑘 𝑚3. The absolute pressure: 𝑝0 = 101.3 𝑘𝑘𝑘. Find: The percent change in density 𝜌% and pressure 𝑝%. Assumption: The bulk modulus is constant Solution: Use the relations developed in problem 3.11 for specific weight and pressure for a compressible liquid: 𝛾 = 𝐸 �𝑧 + 𝐸 𝛾0 � 𝑝 = −𝐸 ln �1 + 𝛾0 𝑧 𝐸 � The specific weight at sea level is: 𝛾0 = 𝜌0 𝑔 = 1020 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 = 10010 𝑁 𝑚3 The specific weight and density at 10000 m depth are 𝛾 = 𝐸 �𝑧 + 𝐸 𝛾0 � = 2.07 × 109 �−10000 + 2.07 × 109 10010 � 𝑁 𝑚3 = 10520 𝑁 𝑚3 𝜌 = 𝛾 𝑔 = 10520 9.81 𝑘𝑘 𝑚3 = 1072 𝑘𝑘 𝑚3 The percentage change in density is 𝜌% = 𝜌 − 𝜌0 𝜌0 = 1072 − 1020 1020 = 5.1 % The gage pressure at a depth of 10000m is: 𝑝 = −𝐸 ln �1 + 𝛾0 𝑧 𝐸 � = 101.3 𝑘𝑘𝑘 − 2.07 × 109 × ln �1 + 10010 × (−10000) 2.07 × 109 � 𝑃𝑃 = 102600 𝑘𝑘𝑘
  • 15. The pressure assuming that the water is incompressible is: 𝑝𝑖𝑖 = 𝜌𝜌ℎ = 1020 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 10000 𝑚 = 100062 𝑘𝑘𝑘 The percent difference in pressure is: 𝑝% = 𝑝 − 𝑝0 𝑝0 = 102600 𝑘𝑘𝑘 − 100062 𝑘𝑘𝑘 100062 𝑘𝑘𝑘 = 2.54 %
  • 16. Problem 3.12 [Difficulty: 4] Given: Model behavior of seawater by assuming constant bulk modulus Find: (a) Expression for density as a function of depth h. (b) Show that result may be written as ρ = ρo + bh (c) Evaluate the constant b (d) Use results of (b) to obtain equation for p(h) (e) Determine depth at which error in predicted pressure is 0.01% Solution: From Table A.2, App. A: SGo 1.025= Ev 2.42 GPa⋅ 3.51 10 5 × psi⋅== Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) (Definition of Bulk Modulus) Ev dp dρ ρ = Then dp ρ g⋅ dh⋅= Ev dρ ρ ⋅= or dρ ρ 2 g Ev dh= Now if we integrate: ρo ρ ρ 1 ρ 2 ⌠⎮ ⎮ ⎮ ⌡ d 0 h h g Ev ⌠ ⎮ ⎮ ⌡ d= After integrating: ρ ρo− ρ ρo⋅ g h⋅ Ev = Therefore: ρ Ev ρo⋅ Ev g h⋅ ρo⋅− = and ρ ρo 1 1 ρo g⋅ h⋅ Ev − = (Binomial expansion may be found in a host of sources, e.g. CRC Handbook of Mathematics) Now for ρo g⋅ h⋅ Ev <<1, the binomial expansion may be used to approximate the density: ρ ρo 1 ρo g⋅ h⋅ Ev += In other words, ρ ρo b h⋅+= where b ρo 2 g⋅ Ev = Since dp ρ g⋅ dh⋅= then an approximate expression for the pressure as a function of depth is: papprox patm− 0 h hρo b h⋅+( ) g⋅ ⌠ ⎮ ⌡ d= papprox patm− g h⋅ 2 ρo⋅ b h⋅+( )⋅ 2 =→ Solving for papprox we get: Problem 3.13 3.13
  • 17. papprox patm g h⋅ 2 ρo⋅ b h⋅+( )⋅ 2 += patm ρo g⋅ h⋅+ b g⋅ h 2 ⋅ 2 += patm ρo h⋅ b h 2 ⋅ 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ g⋅+= Now if we subsitiute in the expression for b and simplify, we get: papprox patm ρo h⋅ ρo 2 g⋅ Ev h 2 2 ⋅+ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ g⋅+= patm ρo g⋅ h⋅ 1 ρo g⋅ h⋅ 2 Ev⋅ + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅+= papprox patm ρo g⋅ h⋅ 1 ρo g⋅ h⋅ 2Ev + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅+= The exact soution for p(h) is obtained by utilizing the exact solution for ρ(h). Thus: pexact patm− ρo ρ ρ Ev ρ ⌠ ⎮ ⎮ ⌡ d= Ev ln ρ ρo ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Subsitiuting for ρ ρo we get: pexact patm Ev ln 1 ρo g⋅ h⋅ Ev − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1− ⋅+= If we let x ρo g⋅ h⋅ Ev = For the error to be 0.01%: Δpexact Δpapprox− Δpexact 1 ρo g⋅ h⋅ 1 x 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ Ev ln 1 x−( ) 1−⎡⎣ ⎤⎦⋅ −= 1 x 1 x 2 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ ln 1 x−( ) 1−⎡⎣ ⎤⎦ −= 0.0001= This equation requires an iterative solution, e.g. Excel's Goal Seek. The result is: x 0.01728= Solving x for h: h x Ev⋅ ρo g⋅ = h 0.01728 3.51× 10 5 × lbf in 2 ⋅ ft 3 1.025 1.94× slug⋅ × s 2 32.2 ft⋅ × 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × slug ft⋅ lbf s 2 ⋅ ×= h 1.364 10 4 × ft⋅= This depth is over 2.5 miles, so the incompressible fluid approximation is a reasonable one at all but the lowest depths of the ocean.
  • 18. Problem 3.14 [Difficulty: 3] Air H D Air H – y y y Given: Cylindrical cup lowered slowly beneath pool surface Find: Expression for y in terms of h and H. Plot y/H vs. h/H. Solution: Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) p V⋅ M R⋅ T⋅= (Ideal Gas Equation) Assumptions: (1) Constant temperature compression of air inside cup (2) Static liquid (3) Incompressible liquid First we apply the ideal gas equation (at constant temperature) for the pressure of the air in the cup: p V⋅ constant= Therefore: p V⋅ pa π 4 ⋅ D 2 ⋅ H⋅= p π 4 ⋅ D 2 ⋅ H y−( )⋅= and upon simplification: pa H⋅ p H y−( )⋅= Now we look at the hydrostatic pressure equation for the pressure exerted by the water. Since ρ is constant, we integrate: p pa− ρ g⋅ h y−( )⋅= at the water-air interface in the cup. Since the cup is submerged to a depth of h, these pressures must be equal: pa H⋅ pa ρ g⋅ h y−( )⋅+⎡⎣ ⎤⎦ H y−( )⋅= pa H⋅ pa y⋅− ρ g⋅ h y−( )⋅ H y−( )⋅+= Explanding out the right hand side of this expression: 0 pa− y⋅ ρ g⋅ h y−( )⋅ H y−( )⋅+= ρ g⋅ h⋅ H⋅ ρ g⋅ h⋅ y⋅− ρ g⋅ H⋅ y⋅− ρ g⋅ y 2 ⋅+ pa y⋅−= ρ g⋅ y 2 ⋅ pa ρ g⋅ h H+( )⋅+⎡⎣ ⎤⎦ y⋅− ρ g⋅ h⋅ H⋅+ 0= y 2 pa ρ g⋅ h H+( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ y⋅− h H⋅+ 0= We now use the quadratic equation: y pa ρ g⋅ h H+( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ pa ρ g⋅ h H+( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 4 h⋅ H⋅−− 2 = we only use the minus sign because y can never be larger than H. Problem 3.14 3.14
  • 19. Now if we divide both sides by H, we get an expression for y/H: y H pa ρ g⋅ H⋅ h H + 1+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ pa ρ g⋅ H⋅ h H + 1+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 4 h H ⋅−− 2 = The exact shape of this curve will depend upon the height of the cup. The plot below was generated assuming: pa 101.3 kPa⋅= H 1 m⋅= 0 20 40 60 80 100 0.2 0.4 0.6 0.8 Depth Ratio, h/H HeightRatio,y/H
  • 20. Problem 3.16 [Difficulty: 2] patmA pbaseA Cover Given: Data on water tank and inspection cover Find: If the support bracket is strong enough; at what water depth would it fail Assumptions: Water is incompressible and static Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ Δp ρ g⋅ h⋅= where h is measured downwards The absolute pressure at the base is pbase patm ρ g⋅ h⋅+= where h 16 ft⋅= The gage pressure at the base is pbase ρ g⋅ h⋅= This is the pressure to use as we have patm on the outside of the cover. The force on the inspection cover is F pbase A⋅= where A 1 in⋅ 1× in⋅= A 1 in 2 ⋅= F ρ g⋅ h⋅ A⋅= F 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 16× ft⋅ 1× in 2 ⋅ ft 12 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × lbf s 2 ⋅ slug ft⋅ ×= F 6.94 lbf⋅= The bracket is strong enough (it can take 9 lbf). To find the maximum depth we start with F 9.00 lbf⋅= h F ρ g⋅ A⋅ = h 9 lbf⋅ 1 1.94 × ft 3 slug ⋅ 1 32.2 × s 2 ft ⋅ 1 in 2 × 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × slug ft⋅ lbf s 2 ⋅ ×= h 20.7 ft⋅= Problem 3.15 3.15
  • 21. Problem 3.18 [Difficulty: 2] Given: Data on partitioned tank Find: Gage pressure of trapped air; pressure to make water and mercury levels equal Solution: The pressure difference is obtained from repeated application of Eq. 3.7, or in other words, from Eq. 3.8. Starting from the right air chamber pgage SGHg ρH2O× g× 3 m⋅ 2.9 m⋅−( )× ρH2O g× 1× m⋅−= pgage ρH2O g× SGHg 0.1× m⋅ 1.0 m⋅−( )×= pgage 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 13.55 0.1× m⋅ 1.0 m⋅−( )× N s 2 ⋅ kg m⋅ ×= pgage 3.48 kPa⋅= If the left air pressure is now increased until the water and mercury levels are now equal, Eq. 3.8 leads to pgage SGHg ρH2O× g× 1.0× m⋅ ρH2O g× 1.0× m⋅−= pgage ρH2O g× SGHg 1× m⋅ 1.0 m⋅−( )×= pgage 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 13.55 1× m⋅ 1.0 m⋅−( )× N s 2 ⋅ kg m⋅ ×= pgage 123 kPa⋅= Problem 3.16 3.16
  • 22. Problem 3.20 [Difficulty: 2] Given: Two-fluid manometer as shown l 10.2 mm⋅= SGct 1.595= (From Table A.1, App. A) Find: Pressure difference Solution: We will apply the hydrostatics equation. Governing equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) d z Assumptions: (1) Static liquid (2) Incompressible liquid Starting at point 1 and progressing to point 2 we have: p1 ρwater g⋅ d l+( )⋅+ ρct g⋅ l⋅− ρwater g⋅ d⋅− p2= Simplifying and solving for p2 p1− we have: Δp p2 p1−= ρct g⋅ l⋅ ρwater g⋅ l⋅−= SGct 1−( ) ρwater⋅ g⋅ l⋅= Substituting the known data: Δp 1.591 1−( ) 1000× kg m 3 ⋅ 9.81× m s 2 ⋅ 10.2× mm⋅ m 10 3 mm⋅ ×= Δp 59.1Pa= Problem 3.17 3.17
  • 23. Problem 3.22 [Difficulty: 2] Given: Two fluid manometer contains water and kerosene. With both tubes open to atmosphere, the difference in free surface elevations is known Ho 20 mm⋅= SGk 0.82= (From Table A.1, App. A) Find: The elevation difference, H, between the free surfaces of the fluids when a gage pressure of 98.0 Pa is applied to the right tube. Solution: We will apply the hydrostatics equation. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid When the gage pressure Δp is applied to the right tube, the water in the right tube is displaced downward by a distance, l. The kerosene in the left tube is displaced upward by the same distance, l. Under the applied gage pressure Δp, the elevation difference, H, is: h H A B l l H0 H1 A B Δp H Ho 2 l⋅+= Since points A and B are at the same elevation in the same fluid, their pressures are the same. Initially: pA ρk g⋅ Ho H1+( )⋅= pB ρwater g⋅ H1⋅= Setting these pressures equal: ρk g⋅ Ho H1+( )⋅ ρwater g⋅ H1⋅= Solving for H1 H1 ρk Ho⋅ ρwater ρk− = SGk Ho⋅ 1 SGk− = H1 0.82 20× mm⋅ 1 0.82− = H1 91.11 mm⋅= Now under the applied gage pressure: pA ρk g⋅ Ho H1+( )⋅ ρwater g⋅ l⋅+= pB Δp ρwater g⋅ H1 l−( )⋅+= Problem 3.18 3.18
  • 24. Setting these pressures equal: SGk Ho H1+( )⋅ l+ Δp ρwater g⋅ H1 l−( )+= l 1 2 Δp ρwater g⋅ H1+ SGk Ho H1+( )⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = Substituting in known values we get: l 1 2 98.0 N m 2 ⋅ 1 999 × m 3 kg 1 9.81 × s 2 m ⋅ kg m⋅ N s 2 ⋅ × 91.11 mm⋅ 0.82 20 mm⋅ 91.11 mm⋅+( )×−[ ] m 10 3 mm⋅ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×= l 5.000 mm⋅= Now we solve for H: H 20 mm⋅ 2 5.000× mm⋅+= H 30.0 mm⋅=
  • 25. Problem 3.24 [Difficulty: 2] Given: Data on manometer Find: Gage pressure at point a Assumption: Water, liquids A and B are static and incompressible Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ Δp ρ g⋅ Δh⋅= where Δh is height difference Starting at point a p1 pa ρH2O g⋅ h1⋅−= where h1 0.125 m⋅ 0.25 m⋅+= h1 0.375m= Next, in liquid A p2 p1 SGA ρH2O⋅ g⋅ h2⋅+= where h2 0.25 m⋅= Finally, in liquid B patm p2 SGB ρH2O⋅ g⋅ h3⋅−= where h3 0.9 m⋅ 0.4 m⋅−= h3 0.5m= Combining the three equations patm p1 SGA ρH2O⋅ g⋅ h2⋅+( ) SGB ρH2O⋅ g⋅ h3⋅−= pa ρH2O g⋅ h1⋅− SGA ρH2O⋅ g⋅ h2⋅+ SGB ρH2O⋅ g⋅ h3⋅−= pa patm ρH2O g⋅ h1 SGA h2⋅− SGB h3⋅+( )⋅+= or in gage pressures pa ρH2O g⋅ h1 SGA h2⋅− SGB h3⋅+( )⋅= pa 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.375 1.20 0.25×( )− 0.75 0.5×( )+[ ]× m⋅ N s 2 ⋅ kg m⋅ ×= pa 4.41 10 3 × Pa= pa 4.41 kPa⋅= (gage) Problem 3.19 3.19
  • 26. Problem 3.20 (Difficulty: 1) 3.20 With the manometer reading as shown, calculate 𝑝 𝑥. Given: Oil specific gravity: 𝑆𝑆 𝑜𝑜𝑜 = 0.85 Depth: ℎ1 = 60 𝑖𝑖𝑖ℎ. ℎ2 = 30 𝑖𝑖𝑖ℎ. Find: The pressure 𝑝 𝑥. Assumption: Fluids are incompressible Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference over a difference in elevation (h): ∆𝑝 = 𝜌𝜌ℎ Repeated application of this relation yields 𝑝 𝑥 = 𝑆𝑆 𝑜𝑜𝑜 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ1 + 𝛾 𝑀ℎ2 The specific weight for mercury is: 𝛾 𝑀 = 845 𝑙𝑙𝑙 𝑓𝑓3 The pressure at the desired location is 𝑝 𝑥 = 0.85 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × � 60 12 � 𝑓𝑓 + 845 𝑙𝑙𝑙 𝑓𝑓3 × � 30 12 � 𝑓𝑓 = 2380 𝑙𝑙𝑙 𝑓𝑓2 = 16.5 𝑝𝑝𝑝
  • 27. Problem 3.21 (Difficulty: 2) 3.21 Calculate 𝑝 𝑥 − 𝑝 𝑦 for this inverted U-tube manometer. Given: Oil specific gravity: 𝑆𝑆 𝑜𝑜𝑜 = 0.90 Depth: ℎ1 = 65 𝑖𝑖𝑖ℎ. ℎ2 = 20 𝑖𝑖𝑖ℎ. ℎ3 = 10 𝑖𝑖𝑖ℎ. Find: The pressure difference 𝑝 𝑥 − 𝑝 𝑦. Assume: The fluids are incompressible Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference over a difference in elevation (h): ∆𝑝 = 𝜌𝜌ℎ Starting at the location of the unknown pressure px, we have the following relations for the hydrostatic pressure: 𝑝 𝑥 − 𝑝1 = 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ1 𝑝1 − 𝑝2 = −𝑆𝑆 𝑜𝑜𝑜 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ3 𝑝2 − 𝑝 𝑦 = −𝛾 𝑤𝑤𝑤𝑤𝑤(ℎ1 − ℎ2 − ℎ3) Adding these three equations together 𝑝 𝑥 − 𝑝 𝑦 = 𝛾 𝑤𝑤𝑤𝑤𝑤(ℎ2 + ℎ3) − 𝑆𝑆 𝑜𝑜𝑜 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ3
  • 28. The pressure difference is then 𝑝 𝑥 − 𝑝 𝑦 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × (10 + 20) 12 𝑓𝑓 − 0.9 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 10 12 𝑓𝑓 = 109.2 𝑙𝑙𝑙 𝑓𝑓2 = 0.758 𝑝𝑝𝑝
  • 29. Problem 3.22 (Difficulty: 2) 3.22 An inclined gage having a tube of 3 mm bore, laid on a slope of 1:20, and a reservoir of 25 mm diameter contains silicon oil (SG 0.84). What distance will the oil move along the tube when a pressure of 25 mm of water is connected to the gage? Given: Silicon oil specific gravity: 𝑆𝑆 𝑜𝑜𝑜 = 0.84. Diameter: 𝐷1 = 3 𝑚𝑚. 𝐷2 = 25 𝑚𝑚. Depth: ℎ 𝑤𝑤𝑤𝑤𝑤 = 25 𝑚𝑚. Slope angle: 1: 20. Find: The distance 𝑥 of the oil move along the tube. Assumption: Fluids are incompressible Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference over a difference in elevation (h): ∆𝑝 = 𝜌𝜌ℎ We have the volume of the oil as constant, so: 𝐴 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟∆ℎ = 𝐴 𝑡𝑡𝑡𝑡 𝑥 or ∆ℎ 𝑥 = 𝐴 𝑡𝑡𝑡𝑡 𝐴 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐷1 2 𝐷2 2 = 9 625 When a pressure of 25 𝑚𝑚 of water is connected with the gage we have: 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ 𝑤𝑤𝑤𝑤𝑤 = 𝑆𝑆 𝑜𝑜𝑜 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ
  • 30. ℎ = ℎ 𝑤𝑤𝑤𝑤𝑤 𝑆𝑆 𝑜𝑜𝑜 = 29.8 𝑚𝑚 Using these relations, we obtain, accounting for the slope of the manometer: ℎ = ∆ℎ + 𝑥 √202 + 12 = � 9 625 + 1 √202 + 12 � 𝑥 ℎ = ∆ℎ + 𝑥 √401 = � 9 625 + 1 √401 � 𝑥 𝑥 = ℎ � 9 625 + 1 √401 � = 463 𝑚𝑚
  • 31. Problem 3.26 [Difficulty: 2] Given: Water flow in an inclined pipe as shown. The pressure difference is measured with a two-fluid manometer L 5 ft⋅= h 6 in⋅= SGHg 13.55= (From Table A.1, App. A) Find: Pressure difference between A and B Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid (3) Gravity is constant Integrating the hydrostatic pressure equation we get: Δp ρ g⋅ Δh⋅= Progressing through the manometer from A to B: pA ρwater g⋅ L⋅ sin 30 deg⋅( )⋅+ ρwater g⋅ a⋅+ ρwater g⋅ h⋅+ ρHg g⋅ h⋅− ρwater g⋅ a⋅− pB= Simplifying terms and solving for the pressure difference: Δp pA pB−= ρwater g⋅ h SGHg 1−( )⋅ L sin 30 deg⋅( )⋅−⎡⎣ ⎤⎦⋅= Substituting in values: Δp 1.94 slug ft 3 ⋅ 32.2× ft s 2 6 in⋅ ft 12 in⋅ × 13.55 1−( )× 5 ft⋅ sin 30 deg⋅( )×−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × lbf s 2 ⋅ slugft⋅ × ft 12 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ×= Δp 1.638 psi⋅= Problem 3.23 3.23
  • 32. Problem 3.28 [Difficulty: 2] Given: Reservoir manometer with vertical tubes of knowm diameter. Gage liquid is Meriam red oil D 18 mm⋅= d 6 mm⋅= SGoil 0.827= (From Table A.1, App. A) Find: The manometer deflection, L when a gage pressure equal to 25 mm of water is applied to the reservoir. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: Δp ρ g⋅ Δh⋅= Beginning at the free surface of the reservoir, and accounting for the changes in pressure with elevation: patm Δp+ ρoil g⋅ x L+( )⋅+ patm= Upon simplification: x L+ Δp ρoil g⋅ = The gage pressure is defined as: Δp ρwater g⋅ Δh⋅= where Δh 25 mm⋅= Combining these two expressions: x L+ ρwater g⋅ h⋅ ρoil g⋅ = Δh SGoil = x and L are related through the manometer dimensions: π 4 D 2 ⋅ x⋅ π 4 d 2 ⋅ L⋅= x d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 L= Therefore: L Δh SGoil 1 d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = Substituting values into the expression: L 25 mm⋅ 0.827 1 6 mm⋅ 18 mm⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = (Note: s L Δh = which yields s 1.088= for this manometer.) L 27.2 mm⋅= Problem 3.24 3.24
  • 33. Problem 3.29 [Difficulty: 2] Given: A U-tube manometer is connected to the open tank filled with water as shown (manometer fluid is Meriam blue) D1 2.5 m⋅= D2 0.7 m⋅= d 0.2 m⋅= SGoil 1.75= (From Table A.1, App. A) Find: The manometer deflection, l Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid D1 D2 d Integrating the hydrostatic pressure equation we get: Δp ρ g⋅ Δh⋅= When the tank is filled with water, the oil in the left leg of the manometer is displaced downward by l/2. The oil in the right leg is displaced upward by the same distance, l/2. Beginning at the free surface of the tank, and accounting for the changes in pressure with elevation: patm ρwater g⋅ D1 D2− d+ l 2 +⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅+ ρoil g⋅ l⋅− patm= Upon simplification: ρwater g⋅ D1 D2− d+ l 2 +⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ ρoil g⋅ l⋅= D1 D2− d+ l 2 + SGoil l⋅= l D1 D2− d+ SGoil 1 2 − = l 2.5 m⋅ 0.7 m⋅− 0.2 m⋅+ 1.75 1 2 − = l 1.600m= Problem 3.25 3.25
  • 34. Problem 3.26 (Difficulty: 2) 3.26 The sketch shows a sectional view through a submarine. Calculate the depth of submarine, y. Assume the specific weight of the seawater is 10.0 𝑘𝑘 𝑚3. Given: Atmos. Pressure: 𝑝 𝑎𝑎𝑎𝑎𝑎 = 740 𝑚𝑚 𝐻𝐻. Seawater specific weight:𝛾 = 10.0 𝑘𝑘 𝑚3. All the dimensional relationship is shown in the figure. Find: The depth 𝑦. Assumption: Fluids are incompressible Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 Integrating with respect to z for an incompressible fluid, we have the relation for the pressure difference over a difference in elevation (h): ∆𝑝 = 𝜌𝜌ℎ Using the barometer reading with 760 mm as atmospheric pressure, the pressure inside the submarine is: 𝑝 = 840 𝑚𝑚 760 𝑚𝑚 × 101.3 × 103 𝑃𝑃 = 111.6 × 103 𝑃𝑃
  • 35. However, the actual atmosphere pressure is: 𝑝 𝑎𝑎𝑎𝑎𝑎 = 740 𝑚𝑚 760 𝑚𝑚 × 101.3 × 103 𝑃𝑃 = 98.3 × 103 𝑃𝑃 For the manometer, using the hydrostatic relation, we have for the pressure, where y is the depth of the submarine: 𝑝 = 𝑝 𝑎𝑎𝑎𝑎𝑎 + 𝛾𝛾 + 𝛾 × 200 𝑚𝑚 − 𝛾 𝐻𝐻 × 400 𝑚𝑚 𝑦 = 𝑝 + 𝛾 𝐻𝐻 × 400 𝑚𝑚 − 𝛾 × 200 𝑚𝑚 − 𝑝 𝑎𝑎𝑎𝑎𝑎 𝛾 The specific weight for mercury is: 𝛾 𝐻𝐻 = 133.1 𝑘𝑘 𝑚3 So we have for the depth y: 𝑦 = 111.6 × 103 𝑃𝑃 + 133.1 × 1000 𝑁 𝑚3 × 0.4 𝑚 − 1000 𝑁 𝑚3 × 0.2 𝑚 − 98.3 × 103 𝑃𝑃 1000 𝑁 𝑚3 𝑦 = 6.45 𝑚
  • 36. Problem 3.27 (Difficulty: 1) 3.27 The manometer reading is 6 in. when the tank is empty (water surface at A). Calculate the manometer reading when the cone is filled with water. Find: The manometer reading when the tank is filled with water. Assumption: Fluids are static and incompressible Solution: Use the hydrostatic relations for pressure When the tank is empty, we have the equation as: ℎ 𝑀𝑀 ∙ 𝑆𝑆 𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∙ 𝛾 𝑤𝑤𝑤𝑤𝑤 = 𝛾 𝑤𝑤𝑤𝑤𝑤ℎ 𝑆𝑆 𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 13.57 ℎ = ℎ 𝑀𝑀 ∙ 𝑆𝑆 𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 150 𝑚𝑚 × 13.57 = 2.04 𝑚 When the tank is filled with water, we assume the mercury interface moves by 𝑥: 𝛾 𝑤𝑤𝑤𝑤𝑤(ℎ 𝑡𝑡𝑡𝑡 + ℎ + 𝑥) = 𝛾 𝑤𝑤𝑤𝑤𝑤 ∙ 𝑆𝑆 𝑚𝑚𝑚𝑚𝑚𝑚𝑚(ℎ 𝑀𝑀 + 2𝑥) (3 𝑚 + 2.04 𝑚 + 𝑥) = 13.57(0.15𝑚 + 2𝑥) Thus 𝑥 = 0.115 𝑚 The new manometer reading is: ℎ 𝑀𝑀 ′ = ℎ 𝑀𝑀 + 2𝑥 = 0.15 𝑚 + 2 × 0.115 𝑚 = 0.38 𝑚
  • 37. Problem 3.30 [Difficulty: 2] Given: Reservoir manometer with dimensions shown. The manometer fluid specific gravity is given. D 5 8 in⋅= d 3 16 in⋅= SGoil 0.827= Find: The required distance between vertical marks on the scale corresponding to Δp of 1 in water. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dz ρ− g⋅= (Hydrostatic Pressure - z is positive upwards) ρ SG ρwater⋅= (Definition of Specific Gravity) Assumptions: (1) Static liquid (2) Incompressible liquid h x Integrating the hydrostatic pressure equation we get: Δp ρ− g⋅ Δz⋅= Beginning at the free surface of the tank, and accounting for the changes in pressure with elevation: patm Δp+ ρoil g⋅ x h+( )⋅− patm= Upon simplification: Δp ρoil g⋅ x h+( )⋅= The applied pressure is defined as: Δp ρwater g⋅ l⋅= where l 1 in⋅= Therefore: ρwater g⋅ l⋅ ρoil g⋅ x h+( )⋅= x h+ l SGoil = x and h are related through the manometer dimensions: π 4 D 2 ⋅ x⋅ π 4 d 2 ⋅ h⋅= x d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 h= Solving for h: h l SGoil 1 d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = Substituting values into the expression: h 1 in⋅ 0.827 1 0.1875 in⋅ 0.625 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ = h 1.109 in⋅= Problem 3.28 3.28
  • 38. Problem 3.32 [Difficulty: 3] Given: Inclined manometer as shown. D 96 mm⋅= d 8 mm⋅= Angle θ is such that the liquid deflection L is five times that of a regular U-tube manometer. Find: Angle θ and manometer sensitivity. Solution: We will apply the hydrostatics equations to this system. Governing Equation: dp dz ρ− g⋅= (Hydrostatic Pressure - z is positive upwards) Assumptions: (1) Static liquid (2) Incompressible liquid x Integrating the hydrostatic pressure equation we get: Δp ρ− g⋅ Δz⋅= Applying this equation from point 1 to point 2: p1 ρ g⋅ x L sin θ( )⋅+( )⋅− p2= Upon simplification: p1 p2− ρ g⋅ x L sin θ( )⋅+( )⋅= Since the volume of the fluid must remain constant: π 4 D 2 ⋅ x⋅ π 4 d 2 ⋅ L⋅= x d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 L⋅= Therefore: p1 p2− ρ g⋅ L⋅ d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 sin θ( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Now for a U-tube manometer: p1 p2− ρ g⋅ h⋅= Hence: p1incl p2incl− p1U p2U− ρ g⋅ L⋅ d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 sin θ( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ ρ g⋅ h⋅ = For equal applied pressures: L d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 sin θ( )+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ h= Since L/h = 5: sin θ( ) h L d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= 1 5 8 mm⋅ 96 mm⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= θ 11.13 deg⋅= The sensitivity of the manometer: s L Δhe = L SG h⋅ = s 5 SG = Problem 3.29 3.29
  • 39. Problem 3.33 [Difficulty: 3] Given: Data on inclined manometer Find: Angle θ for given data; find sensitivity Solution: Basic equation dp dy ρ− g⋅= or, for constant ρ Δp ρ g⋅ Δh⋅= where Δh is height difference Under applied pressure Δp SGMer ρ⋅ g⋅ L sin θ( )⋅ x+( )⋅= (1) From Table A.1 SGMer 0.827= and Δp = 1 in. of water, or Δp ρ g⋅ h⋅= where h 25 mm⋅= h 0.025m= Δp 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.025× m⋅ N s 2 ⋅ kg m⋅ ×= Δp 245Pa= The volume of liquid must remain constant, so x Ares⋅ L Atube⋅= x L Atube Ares ⋅= L d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⋅= (2) Combining Eqs 1 and 2 Δp SGMer ρ⋅ g⋅ L sin θ( )⋅ L d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Solving for θ sin θ( ) Δp SGMer ρ⋅ g⋅ L⋅ d D ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= sin θ( ) 245 N m 2 ⋅ 1 0.827 × 1 1000 × m 3 kg ⋅ 1 9.81 × s 2 m ⋅ 1 0.15 × 1 m ⋅ kg m⋅ s 2 N⋅ × 8 76 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 −= 0.186= θ 11 deg⋅= The sensitivity is the ratio of manometer deflection to a vertical water manometer s L h = 0.15 m⋅ 0.025 m⋅ = s 6= Problem 3.30 3.30
  • 40. Problem 3.34 [Difficulty: 4] Given: Barometer with water on top of the mercury column, Temperature is known: h2 6.5 in⋅= h1 28.35 in⋅= SGHg 13.55= (From Table A.2, App. A) T 70 °F= pv 0.363 psi⋅= (From Table A.7, App. A) Find: (a) Barometric pressure in psia (b) Effect of increase in ambient temperature on length of mercury column for the same barometric pressure: Tf 85 °F= Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ− g⋅= (Hydrostatic Pressure - h is positive downwards) ρ SG ρwater⋅= (Definition of Specific Gravity) h2 Water vapor h1 Water Mercury Assumptions: (1) Static liquid (2) Incompressible liquid Integrating the hydrostatic pressure equation we get: Δp ρ g⋅ Δh⋅= Start at the free surface of the mercury and progress through the barometer to the vapor pressure of the water: patm ρHg g⋅ h1⋅− ρwater g⋅ h2⋅− pv= patm pv ρwater g⋅ SGHg h1⋅ h2+( )⋅+= patm 0.363 lbf in 2 ⋅ 1.93 slug ft 3 ⋅ 32.2× ft s 2 ⋅ lbf s 2 ⋅ slug ft⋅ × 13.55 28.35× in⋅ 6.5 in⋅+( )× ft 12 in⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ×+= patm 14.41 lbf in 2 ⋅= At the higher temperature, the vapor pressure of water increases to 0.60 psi. Therefore, if the atmospheric pressure were to remain constant, the length of the mercury column would have to decrease - the increased water vapor would push the mercury out of the tube! Problem 3.31 3.31
  • 41. Problem 3.36 [Difficulty: 3] Given: Water column standin in glass tube Δh 50 mm⋅= D 2.5 mm⋅= σ 72.8 10 3− × N m = (From Table A.4, App. A) Find: (a) Column height if surface tension were zero. (b) Column height in 1 mm diameter tube Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) Δhp Δhc Δh Δhc πDδ θ Mg = ρgV ΣFz 0= (Static Equilibrium) Assumptions: (1) Static, incompressible liquid (2) Neglect volume under meniscus (3) Applied pressure remains constant (4) Column height is sum of capillary rise and pressure difference Assumption #4 can be written as: Δh Δhc Δhp+= Choose a free-body diagram of the capillary rise portion of the column for analysis: ΣFz π D⋅ σ⋅ cos θ( )⋅ π 4 D 2 ⋅ ρ⋅ g⋅ Δhc⋅−= 0= Therefore: Δhc 4 σ⋅ ρ g⋅ D⋅ cos θ( )⋅= Substituting values: Δhc 4 72.8× 10 3− × N m ⋅ 1 999 × m 3 kg ⋅ 1 9.81 × s 2 m ⋅ 1 2.5 × 1 mm ⋅ kg m⋅ N s 2 ⋅ × 10 3 mm⋅ m ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ×= Δhc 11.89 mm⋅= Therefore: Δhp Δh Δhc−= Δhp 50 mm⋅ 11.89 mm⋅−= Δhp 38.1 mm⋅= (result for σ = 0) For the 1 mm diameter tube: Δhc 4 72.8× 10 3− × N m ⋅ 1 999 × m 3 kg ⋅ 1 9.81 × s 2 m ⋅ 1 1 × 1 mm ⋅ kg m⋅ N s 2 ⋅ × 10 3 mm⋅ m ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ×= Δhc 29.71 mm⋅= Δh 29.7 mm⋅ 38.1 mm⋅+= Δh 67.8 mm⋅= Problem 3.32 3.32
  • 42. Problem 3.38 [Difficulty :2] Fluid 1 Fluid 2 σπDcosθ ρ1gΔhπD2 /4 Given: Two fluids inside and outside a tube Find: (a) An expression for height Δh (b) Height difference when D =0.040 in for water/mercury Assumptions: (1) Static, incompressible fluids (2) Neglect meniscus curvature for column height and volume calculations Solution: A free-body vertical force analysis for the section of fluid 1 height Δh in the tube below the "free surface" of fluid 2 leads to F ∑ 0= Δp π D 2 ⋅ 4 ⋅ ρ1 g⋅ Δh⋅ π D 2 ⋅ 4 ⋅− π D⋅ σ⋅ cos θ( )⋅+= where Δp is the pressure difference generated by fluid 2 over height Δh, Δp ρ2 g⋅ Δh⋅= Hence Δp π D 2 ⋅ 4 ⋅ ρ1 g⋅ Δh⋅ π D 2 ⋅ 4 ⋅− ρ2 g⋅ Δh⋅ π D 2 ⋅ 4 ⋅ ρ1 g⋅ Δh⋅ π D 2 ⋅ 4 ⋅−= π− D⋅ σ⋅ cos θ( )⋅= Solving for Δh Δh 4 σ⋅ cos θ( )⋅ g D⋅ ρ2 ρ1−( )⋅ −= For fluids 1 and 2 being water and mercury (for mercury σ = 375 mN/m and θ = 140o, from Table A.4), solving for Δh when D = 0.040 in Δh 4− 0.375× N m ⋅ lbf 4.448 N⋅ × 0.0254m in × cos 140 deg⋅( )× s 2 32.2 ft⋅ × 1 0.040 in⋅ × ft 3 1.94 slug⋅ × 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 × 1 13.6 1−( ) × slugft⋅ lbf s 2 ⋅ ×= Δh 0.360 in⋅= Problem 3.33 3.33
  • 43. Problem 3.40 [Difficulty: 2] Water Given: Water in a tube or between parallel plates Find: Height Δh for each system Solution: a) Tube: A free-body vertical force analysis for the section of water height Δh above the "free surface" in the tube, as shown in the figure, leads to F ∑ 0= π D⋅ σ⋅ cos θ( )⋅ ρ g⋅ Δh⋅ π D 2 ⋅ 4 ⋅−= Assumption: Neglect meniscus curvature for column height and volume calculations Solving for Δh Δh 4 σ⋅ cos θ( )⋅ ρ g⋅ D⋅ = b) Parallel Plates: A free-body vertical force analysis for the section of water height Δh above the "free surface" between plates arbitrary width w (similar to the figure above), leads to F ∑ 0= 2 w⋅ σ⋅ cos θ( )⋅ ρ g⋅ Δh⋅ w⋅ a⋅−= Solving for Δh Δh 2 σ⋅ cos θ( )⋅ ρ g⋅ a⋅ = For water σ = 72.8 mN/m and θ = 0o (Table A.4), so a) Tube Δh 4 0.0728× N m ⋅ 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.005× m⋅ kg m⋅ N s 2 ⋅ ×= Δh 5.94 10 3− × m= Δh 5.94 mm⋅= b) Parallel Plates Δh 2 0.0728× N m ⋅ 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 0.005× m⋅ kg m⋅ N s 2 ⋅ ×= Δh 2.97 10 3− × m= Δh 2.97 mm⋅= Problem 3.34 3.34
  • 44. p SL = 101 kPa R = 286.9 J/kg.K ρ = 999 kg/m3 The temperature can be computed from the data in the figure. The pressures are then computed from the appropriate equation. From Table A.3 Agreement between calculated and tabulated data is very good (as it should be, considering the table data are also computed!) Atmospheric Pressure vs Elevation 0.00000 0.00001 0.00010 0.00100 0.01000 0.10000 1.00000 0 10 20 30 40 50 60 70 80 90 100 Elevation (km) PressureRatiop/pSL Computed Table A.3 Problem 3.35 3.35
  • 45. z (km) T (o C) T (K) p /p SL z (km) p /p SL 0.0 15.0 288.0 m = 1.000 0.0 1.000 2.0 2.0 275.00 0.0065 0.784 0.5 0.942 4.0 -11.0 262.0 (K/m) 0.608 1.0 0.887 6.0 -24.0 249.0 0.465 1.5 0.835 8.0 -37.0 236.0 0.351 2.0 0.785 11.0 -56.5 216.5 0.223 2.5 0.737 12.0 -56.5 216.5 T = const 0.190 3.0 0.692 14.0 -56.5 216.5 0.139 3.5 0.649 16.0 -56.5 216.5 0.101 4.0 0.609 18.0 -56.5 216.5 0.0738 4.5 0.570 20.1 -56.5 216.5 0.0530 5.0 0.533 22.0 -54.6 218.4 m = 0.0393 6.0 0.466 24.0 -52.6 220.4 -0.000991736 0.0288 7.0 0.406 26.0 -50.6 222.4 (K/m) 0.0211 8.0 0.352 28.0 -48.7 224.3 0.0155 9.0 0.304 30.0 -46.7 226.3 0.0115 10.0 0.262 32.2 -44.5 228.5 0.00824 11.0 0.224 34.0 -39.5 233.5 m = 0.00632 12.0 0.192 36.0 -33.9 239.1 -0.002781457 0.00473 13.0 0.164 38.0 -28.4 244.6 (K/m) 0.00356 14.0 0.140 40.0 -22.8 250.2 0.00270 15.0 0.120 42.0 -17.2 255.8 0.00206 16.0 0.102 44.0 -11.7 261.3 0.00158 17.0 0.0873 46.0 -6.1 266.9 0.00122 18.0 0.0747 47.3 -2.5 270.5 0.00104 19.0 0.0638 50.0 -2.5 270.5 T = const 0.000736 20.0 0.0546 52.4 -2.5 270.5 0.000544 22.0 0.0400 54.0 -5.6 267.4 m = 0.000444 24.0 0.0293 56.0 -9.5 263.5 0.001956522 0.000343 26.0 0.0216 58.0 -13.5 259.5 (K/m) 0.000264 28.0 0.0160 60.0 -17.4 255.6 0.000202 30.0 0.0118 61.6 -20.5 252.5 0.000163 40.0 0.00283 64.0 -29.9 243.1 m = 0.000117 50.0 0.000787 66.0 -37.7 235.3 0.003913043 0.0000880 60.0 0.000222 68.0 -45.5 227.5 (K/m) 0.0000655 70.0 0.0000545 70.0 -53.4 219.6 0.0000482 80.0 0.0000102 72.0 -61.2 211.8 0.0000351 90.0 0.00000162 74.0 -69.0 204.0 0.0000253 76.0 -76.8 196.2 0.0000180 78.0 -84.7 188.3 0.0000126 80.0 -92.5 180.5 T = const 0.00000861 82.0 -92.5 180.5 0.00000590 84.0 -92.5 180.5 0.00000404 86.0 -92.5 180.5 0.00000276 88.0 -92.5 180.5 0.00000189 90.0 -92.5 180.5 0.00000130
  • 46. Problem 3.44 [Difficulty: 3] Given: Atmospheric conditions at ground level (z = 0) in Denver, Colorado are p0 = 83.2 kPa, T0 = 25°C. Pike's peak is at elevation z = 2690 m. Find: p/p0 vs z for both cases. Solution: Governing Equations: dp dz ρ− g⋅= p ρ R⋅ T⋅= Assumptions: (1) Static fluid (2) Ideal gas behavior (a) For an incompressible atmosphere: dp dz ρ− g⋅= becomes p p0− 0 z zρ g⋅ ⌠ ⎮ ⌡ d−= or p p0 ρ0 g⋅ z⋅−= p0 1 g z⋅ R T0⋅ − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= (1) At z 2690 m⋅= p 83.2 kPa⋅ 1 9.81 m s 2 ⋅ 2690× m⋅ kg K⋅ 287 N⋅ m⋅ × 1 298 K⋅ × N s 2 ⋅ kg m⋅ ×− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ ×= p 57.5 kPa⋅= (b) For an adiabatic atmosphere: p ρ k const= ρ ρ0 p p0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 k ⋅= dp dz ρ− g⋅= becomes dp ρ0− p p0 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 k ⋅ g⋅ dz⋅= or 1 p 1 k dp ρ0 g⋅ p0 1 k − dz⋅= But p0 p p 1 p 1 k ⌠ ⎮ ⎮ ⎮ ⎮ ⌡ d k k 1− p p0−( ) k 1− k ⋅= hence k k 1− p k 1− k p0 k 1− k − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠⋅ ρ0 g⋅ p0 1 k − g⋅ z⋅= Solving for the pressure ratio p p0 1 k 1− k ρ0 p0 ⋅ g⋅ z⋅− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k k 1− = or p p0 1 k 1− k g z⋅ R T0⋅ ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ k k 1− = (2) At z 2690 m⋅= p 83.2 kPa⋅ 1 1.4 1− 1.4 9.81× m s 2 ⋅ 2690× m⋅ kg K⋅ 287 N⋅ m⋅ × 1 298 K⋅ × N s 2 ⋅ kg m⋅ ×− ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 1.4 1.4 1− ×= p 60.2 kPa⋅= Problem 3.36 3.36
  • 47. Equations 1 and 2 can be plotted: 0.4 0.6 0.8 1 0 1 10 3 × 2 10 3 × 3 10 3 × 4 10 3 × 5 10 3 × Incompressible Adiabatic Temperature Variation with Elevation Pressure Ratio (-) ElevationaboveDenver(m)
  • 48. Problem 3.37 (Difficulty: 2) 3.37 If atmospheric pressure at the ground is 101.3 𝑘𝑘𝑘 and temperature is 15 ℃, calculate the pressure 7.62 𝑘𝑘 above the ground, assuming (a) no density variation, (b) isothermal variation of density with pressure, and (c) adiabatic variation of density with pressure. Assumption: Atmospheric air is stationary and behaves as an ideal gas. Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 𝑑𝑑 = −𝜌 𝑔 = −𝛾 (a) For this case with no density variation, we integrate with respect to z from the ground level pressure p0 to the pressure at any height h. The pressure is 𝑝 = 𝑝0 − 𝛾ℎ From Table A.10, the density of air at sea level is 𝜌 = 1.23 𝑘𝑘 𝑚3 Or the specific weight is 𝛾 = 𝜌𝜌 = 1.23 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 = 12.07 𝑁 𝑚3 Thus the pressure at 7.62 km is 𝑝 = 101.3 𝑘𝑘𝑘 − 12.07 𝑁 𝑚3 × 7.62 × 1000 𝑚 = 9.63 𝑘𝑘𝑘 (b) For isothermal condition we have for an ideal gas: 𝑝 𝜌 = 𝑝0 𝜌0 = 𝑅𝑅 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Therefore, since ρ = γ g and g is a constant 𝑝 𝛾 = 𝑝0 𝛾0 = 101.3 𝑘𝑘𝑘 12.07 𝑁 𝑚3 = 8420 𝑚 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 From the hydrostatic relation we have: 𝑑𝑑 = −𝛾𝛾𝛾 𝑑𝑑 𝑝 = − 𝛾 𝑝 𝑑𝑑
  • 49. � 𝑑𝑑 𝑝 𝑝 𝑝0 = − 1 8420𝑚 � 𝑑𝑑 𝑧 0 ln � 𝑝 𝑝0 � = − 1 8420𝑚 𝑧 Thus the pressure at 7.62 km is 𝑝 𝑝0 = 𝑒− − 7620 𝑚 8420𝑚 = 𝑒− 0.905 = 0.4045 𝑝 = 101.3𝑘𝑘𝑘 × 0.4045 = 41.0 𝑘𝑘𝑘 (c) For a reversible and adiabatic variation of density we have: 𝑝𝑣 𝑘 = 𝑝 𝜌 𝑘 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Where k is the specific heat ratio 𝑘 = 1.4 Or, since gravity g is constant, we can write in terms of the specific weight 𝑝 𝛾 𝑘 = 𝑝0 𝛾0 𝑘 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Or the specific weight is 𝛾 = 𝛾0 � 𝑝 𝑝0 � 1 𝑘� The hydrostatic expression becomes 𝑑𝑑 = −𝛾0 � 𝑝 𝑝0 � 1 𝑘� 𝑑𝑑 Separating variables 𝑝0 1/𝑘 𝛾0 � 𝑑𝑑 (𝑝)1/𝑘 𝑝 𝑝0 = − � 𝑑𝑑 𝑧 0 Integrating between the limits p=p0 at z=0 and p = p at z = z � 𝑘 𝑘 − 1 � 𝑝0 1/𝑘 𝛾0 �𝑝 𝑘−1 𝑘 − 𝑝0 𝑘−1 𝑘 � = − 𝑧 Or � 𝑝 𝑝0 � 𝑘−1 𝑘 = 1 − � 𝑘 − 1 𝑘 � 𝛾0 𝑧 𝑝0 The pressure is then 𝑝 = 𝑝0 �1 − � 𝑘 − 1 𝑘 � 𝛾0 𝑧 𝑝0 � 𝑘 𝑘−1� = 101.3𝑘𝑘𝑘 �1 − � 1.4 − 1 1.4 � × 12.07 𝑁 𝑚3 × 7620𝑚 101.3 × 1000 𝑃𝑃 � 1.4 1.4−1� 𝑝 = 35.4 𝑘𝑘𝑘 The calculation of pressure depends heavily on the assumption we make about how density changes.
  • 50. Problem 3.38 (Difficulty: 2) 3.38 If the temperature in the atmosphere is assumed to vary linearly with altitude so T = T0 - αz where T0 is the sea level temperature and α = - dT / dz is the temperature lapse rate, find p(z) when air is taken to be a perfect gas. Give the answer in terms of p0, a, g, R, and z only. Assumption: Atmospheric air is stationary and behaves as an ideal gas. Solution: Use the hydrostatic relation to find the pressures in the fluid Governing equation: Hydrostatic pressure in a liquid, with z measured upward: 𝑑𝑑 = −𝛾𝛾𝛾 The ideal gas relation is 𝑝 𝜌 = 𝑅𝑅 Or in terms of the specific weight, the pressure is 𝑝 = 𝜌𝜌𝜌 = 𝛾 𝑔 𝑅𝑅 Relating the temperature to the adiabatic lapse rate 𝑝 = 𝛾 𝑔 𝑅(𝑇0 − 𝛼𝛼) Inserting the expression for specific weight into the hydrostatic equation 𝑑𝑑 = − 𝑔𝑔 𝑅(𝑇0 − 𝛼𝛼) 𝑑𝑑 Separating variables 𝑑𝑑 𝑝 = − 𝑔 𝑅 𝑑𝑑 (𝑇0 − 𝛼𝛼) Integrating between the surface and any height z � 𝑑𝑑 𝑝 𝑝 𝑝0 = − 𝑔 𝑅 � 𝑑𝑑 (𝑇0 − 𝛼𝛼) 𝑧 0 Or 𝑙𝑙 � 𝑝 𝑝0 � = − 𝑔 𝑅 𝑙𝑙 � 𝑇0 − 𝛼𝛼 𝑇0 �
  • 51. In terms of p 𝑝 𝑝0 = �1 − 𝛼𝛼 𝑇0 � 𝑔 𝛼𝛼�
  • 52. Problem 3.46 [Difficulty: 3] Given: Door located in plane vertical wall of water tank as shown c ps a y’ y b a 1.5 m⋅= b 1 m⋅= c 1 m⋅= Atmospheric pressure acts on outer surface of door. Find: Resultant force and line of action: (a) for (b) for ps patm= psg 0.3 atm⋅= Plot F/Fo and y'/yc over range of ps/patm (Fo is force determined in (a), yc is y-ccordinate of door centroid). Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dy ρ g⋅= (Hydrostatic Pressure - y is positive downwards) FR Ap ⌠ ⎮ ⎮ ⌡ d= (Hydrostatic Force on door) y' FR⋅ Ay p⋅ ⌠ ⎮ ⎮ ⌡ d= (First moment of force) Assumptions: (1) Static fluid (2) Incompressible fluid We will obtain a general expression for the force and line of action, and then simplify for parts (a) and (b). Since dp ρ g⋅ dh⋅= it follows that p ps ρ g⋅ y⋅+= Now because patm acts on the outside of the door, psg is the surface gage pressure: p psg ρ g⋅ y⋅+= FR Ap ⌠⎮ ⎮ ⌡ d= c c a+ yp b⋅ ⌠ ⎮ ⌡ d= c c a+ ypsg ρ g⋅ y⋅+( ) b⋅ ⌠ ⎮ ⌡ d= b psg a⋅ ρ g⋅ 2 a 2 2 a⋅ c⋅+( )⋅+⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 1( ) y' FR⋅ Ay p⋅ ⌠⎮ ⎮ ⌡ d= Therefore: y' 1 FR Ay p⋅ ⌠⎮ ⎮ ⌡ d= 1 FR c c a+ yy psg ρ g⋅ y⋅+( )⋅ b⋅ ⌠ ⎮ ⌡ d⋅= Evaluating the integral: y' b FR psg 2 c a+( ) 2 c 2 −⎡⎣ ⎤⎦ ρ g⋅ 3 c a+( ) 3 c 3 −⎡⎣ ⎤⎦⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = Problem 3.39 3.39
  • 53. Simplifying: y' b FR psg 2 a 2 2 a⋅ c⋅+( ) ρ g⋅ 3 a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 2( ) For part (a) we know psg 0= so substituting into (1) we get: Fo ρ g⋅ b⋅ 2 a 2 2 a⋅ c⋅+( )⋅= Fo 1 2 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1× m⋅ 1.5 m⋅( ) 2 2 1.5× m⋅ 1× m⋅+⎡⎣ ⎤⎦× N s 2 ⋅ kg m⋅ ×= Fo 25.7 kN⋅= Substituting into (2) for the line of action we get: y' ρ g⋅ b⋅ 3 Fo⋅ a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅= y' 1 3 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1× m⋅ 1 25.7 10 3 × ⋅ 1 N ⋅ 1.5 m⋅( ) 3 3 1.5× m⋅ 1× m⋅ 1.5 m⋅ 1 m⋅+( )×+⎡⎣ ⎤⎦× N s 2 ⋅ kg m⋅ ×= y' 1.9m= For part (b) we know psg 0.3 atm⋅= . Substituting into (1) we get: FR 1 m⋅ 0.3 atm⋅ 1.013 10 5 × N⋅ m 2 atm⋅ × 1.5× m⋅ 1 2 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1.5 m⋅( ) 2 2 1.5× m⋅ 1× m⋅+⎡⎣ ⎤⎦× N s 2 ⋅ kg m⋅ ×+ ⎡ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ ×= FR 71.3 kN⋅= Substituting into (2) for the line of action we get: y' 1 m⋅ 0.3 atm⋅ 2 1.013 10 5 × N⋅ m 2 atm⋅ × 1.5( ) 2 2 1.5⋅ 1⋅+⎡⎣ ⎤⎦× m 2 ⋅ 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 3 1.5( ) 3 3 1.5⋅ 1⋅ 1.5 1+( )⋅+⎡⎣ ⎤⎦× m 3 ⋅ N s 2 ⋅ kg m⋅ ×+ ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ × 71.3 10 3 × N⋅ = y' 1.789m= The value of F/Fo is obtained from Eq. (1) and our result from part (a): F Fo b psg a⋅ ρ g⋅ 2 a 2 2 a⋅ c⋅+( )⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ ρ g⋅ b⋅ 2 a 2 2 a⋅ c⋅+( )⋅ = 1 2 psg⋅ ρ g⋅ a 2 c⋅+( )⋅ += For the gate yc c a 2 += Therefore, the value of y'/yc is obtained from Eqs. (1) and (2): y' yc 2 b⋅ FR 2 c⋅ a+( )⋅ psg 2 a 2 2 a⋅ c⋅+( ) ρ g⋅ 3 a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 2 b⋅ 2 c⋅ a+( ) psg 2 a 2 2 a⋅ c⋅+( ) ρ g⋅ 3 a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ b psg a⋅ ρ g⋅ 2 a 2 2 a⋅ c⋅+( )⋅+⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅=
  • 54. Simplifying this expression we get: y' yc 2 2 c⋅ a+( ) psg 2 a 2 2 a⋅ c⋅+( ) ρ g⋅ 3 a 3 3 a⋅ c⋅ a c+( )⋅+⎡⎣ ⎤⎦⋅+ psg a⋅ ρ g⋅ 2 a 2 2 a⋅ c⋅+( )⋅+ ⋅= Based on these expressions we see that the force on the gate varies linearly with the increase in surface pressure, and that the line of action of the resultant is always below the centroid of the gate. As the pressure increases, however, the line of action moves closer to the centroid. Plots of both ratios are shown below: 0 1 2 3 4 5 0 10 20 30 40 Force Ratio vs. Surface Pressure Surface Pressure (atm) ForceRatioF/Fo 0 1 2 3 4 5 1 1.01 1.02 1.03 1.04 1.05 Line of Action Ratio vs. Surface Pressure Surface Pressure (atm) LineofActionRatioy'/yc
  • 55. Problem 3.48 [Difficulty: 5] Discussion: The design requirements are specified except that a typical floor height is about 12 ft, making the total required lift about 36 ft. A spreadsheet was used to calculate the system properties for various pressures. Results are presented on the next page, followed by a sample calculation. Total cost dropped quickly as system pressure was increased. A shallow minimum was reached in the 100-110 psig range. The lowest-cost solution was obtained at a system pressure of about 100 psig. At this pressure, the reservoir of 140 gal required a 3.30 ft diameter pressure sphere with a 0.250 in wall thickness. The welding cost was $155 and the material cost $433, for a total cost of $588. Accumulator wall thickness was constrained at 0.250 in for pressures below 100 psi; it increased for higher pressures (this caused the discontinuity in slope of the curve at 100 psig). The mass of steel became constant above 110 psig. No allowance was made for the extra volume needed to pressurize the accumulator. Fail-safe design is essential for an elevator to be used by the public. The control circuitry should be redundant. Failures must be easy to spot. For this reason, hydraulic actuation is good: leaks will be readily apparent. The final design must be reviewed, approved, and stamped by a professional engineer since the design involves public safety. The terminology used in the solution is defined in the following table: Symbol Definition Units p System pressure psig Ap Area of lift piston in2 Voil Volume of oil gal Ds Diameter of spherical accumulator ft t Wall thickness of accumulator in Aw Area of weld in2 Cw Cost of weld $ Ms Mass of steel accumulator lbm Cs Cost of steel $ Ct Total Cost $ A sample calculation and the results of the system simulation in Excel are presented below. Problem 3.40 3.40
  • 57. Results of system simulation:
  • 58. Problem 3.50 [Difficulty: 3] FA H = 25 ft yR = 10 ft h A B z x y Given: Geometry of gate Find: Force FA for equilibrium Solution: Basic equation FR Ap ⌠ ⎮ ⎮ ⌡ d= dp dh ρ g⋅= ΣMz 0= or, use computing equations FR pc A⋅= y' yc Ixx A yc⋅ += where y would be measured from the free surface Assumptions: static fluid; ρ = constant; patm on other side; door is in equilibrium Instead of using either of these approaches, we note the following, using y as in the sketch ΣMz 0= FA R⋅ Ay p⋅ ⌠⎮ ⎮ ⌡ d= with p ρ g⋅ h⋅= (Gage pressure, since p = patm on other side) FA 1 R Ay ρ⋅ g⋅ h⋅ ⌠⎮ ⎮ ⌡ d⋅= with dA r dr⋅ dθ⋅= and y r sin θ( )⋅= h H y−= Hence FA 1 R 0 π θ 0 R rρ g⋅ r⋅ sin θ( )⋅ H r sin θ( )⋅−( )⋅ r⋅ ⌠ ⎮ ⌡ d ⌠ ⎮ ⌡ d⋅= ρ g⋅ R 0 π θ H R 3 ⋅ 3 sin θ( )⋅ R 4 4 sin θ( ) 2 ⋅− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d⋅= FR ρ g⋅ R 2 H⋅ R 3 ⋅ 3 π R 4 ⋅ 8 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= ρ g⋅ 2 H⋅ R 2 ⋅ 3 π R 3 ⋅ 8 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Using given data FR 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 2 3 25× ft⋅ 10 ft⋅( ) 2 × π 8 10 ft⋅( ) 3 ×−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × lbf s 2 ⋅ slug ft⋅ ×= FR 7.96 10 4 × lbf⋅= Problem 3.41 3.41
  • 59. Problem 3.42 (Difficulty: 2) 3.42 A circular gate 3 𝑚 in diameter has its center 2.5 𝑚 below a water surface and lies in a plane sloping at 60°. Calculate magnitude, direction and location of total force on the gate. Find: The direction, magnitude of the total force 𝐹. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 For the magnitude of the force we have: 𝐹 = � 𝑝𝑝𝑝 𝐴 A free body diagram of the gate is The pressure on the gate is the pressure at the centroid, which is yc = 2.5 m. So the force can be calculated as: 𝐹 = 𝜌𝜌ℎ 𝑐 𝐴 = 999 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 2.5 𝑚 × 𝜋 4 × (3 𝑚)2 = 173200 𝑁 = 173.2 𝑘𝑘 The direction is perpendicular to the gate.
  • 60. For the location of the force we have: 𝑦′ = 𝑦𝑐 + 𝐼 𝑥�𝑥� 𝐴𝑦𝑐 The y axis is along the plate so the distance to the centroid is: 𝑦𝑐 = 2.5 𝑚 sin 60° = 2.89 𝑚 The area moment of inertia is 𝐼 𝑥�𝑥� = 𝜋𝐷4 64 = 𝜋 64 × (3 𝑚)4 = 3.976 𝑚4 The area is 𝐴 = 𝜋 4 𝐷2 = 𝜋 4 × (3 𝑚)2 = 7.07 𝑚2 So 𝑦′ = 2.89 𝑚 + 3.976 𝑚4 7.07 𝑚2 × 2.89 𝑚 = 2.89 𝑚 + 0.1946 𝑚 = 3.08 𝑚 The vertical location on the plate is ℎ′ = 𝑦′ sin 60° = 3.08 𝑚 × √3 2 = 2.67 𝑚 The force acts on the point which has the depth of 2.67 𝑚.
  • 61. Problem 3.43 (Difficulty: 2) 3.43 For the situation shown, find the air pressure in the tank in psi. Calculate the force exerted on the gate at the support B if the gate is 10 𝑓𝑓 wide. Show a free body diagram of the gate with all the forces drawn in and their points of application located. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure and force, and the static relation for moments: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 The specfic weight for water is: 𝛾 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 The pressure of the air equals that at the surface of the water in the tank. As shown by the manometer, the pressure at the surface is less than atmospheric due to the three foot head of water. The gage pressure of the air is then: 𝑝 𝑎𝑎𝑎 = −𝛾ℎ = −62.4 𝑙𝑙𝑙 𝑓𝑓3 × 3𝑓𝑓 = −187.2 𝑙𝑙𝑙 𝑓𝑓2 A free body diagram for the gate is
  • 62. For the force in the horizontal direction, we have: 𝐹1 = 𝛾ℎ 𝑐 𝐴 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 3 𝑓𝑓 × (6 𝑓𝑓 × 10 𝑓𝑓) = 11230 𝑙𝑙𝑙 𝐹2 = 𝑝 𝑎𝑎𝑎 𝐴 = −187.2 𝑙𝑙𝑙 𝑓𝑓2 × (8 𝑓𝑓 × 10 𝑓𝑓) = 14980 𝑙𝑙𝑙 With the momentume balance about hinge we have: � 𝑀 = 𝐹1ℎ 𝑐 − 𝑃ℎ − 𝐹2 ℎ 2 = 11230 𝑙𝑙𝑙 × 6𝑓𝑓 − 𝑃 × 8𝑓𝑓 − 14980 𝑙𝑙𝑙 × 4𝑓𝑓 = 0 So the force exerted on B is: 𝑃 = 933 𝑙𝑙𝑙
  • 63. Problem 3.44 (Difficulty: 3) 3.44 What is the pressure at A? Draw a free body diagram of the 10 ft wide gate showing all forces and locations of their lines of action. Calculate the minimum force 𝑃 necessary to keep the gate closed. Given: All the parameters are shown in the figure. Find: The pressure 𝑝 𝐴. The minimum force 𝑃 necessary to keep the gate closed. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 The specfic weight of the water is: 𝛾 𝑤𝑤𝑤𝑤𝑤 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 The gage pressure at A is given by integrating the hydrostatic relation: 𝑝 𝐴 = 𝛾 𝑜𝑜𝑜ℎ 𝐴 = 𝑆𝑆𝛾 𝑜𝑖𝑖ℎ 𝐴 = 0.9 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 6 𝑓𝑓 = 337 𝑙𝑙𝑙 𝑓𝑓2
  • 64. A free body diagram of the gate is The horizontal force F1 as shown in the figure is given by the pressure at the centroid of the submerged area (3 ft): 𝐹1 = 𝛾 𝑜𝑜𝑜ℎ 𝑐 𝐴 = 0.9 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 3 𝑓𝑓 × (6 𝑓𝑓 × 10 𝑓𝑓) = 10110 𝑙𝑙𝑙 The vertical force F2 is given by the pressure at the depth of the surface (4 ft) 𝐹2 = 𝑝 𝐴 𝐴 = 337 𝑙𝑙𝑙 𝑓𝑓2 × (4𝑓𝑓 × 10𝑓𝑓) = 13480 𝑙𝑙𝑙 The force F1 acts two-thirds of the distance down from the water surface and the force F2 acts at the centroid.. Taking the moments about the hinge: −𝐹1 × 6 𝑓𝑓−𝐹2 × 2 𝑓𝑓 + 𝑃 × 4 𝑓𝑓 = 0 So we have for the force at the support: 𝑃 = 10110 𝑙𝑙𝑙 × 6𝑓𝑓 + 13480 𝑙𝑙𝑙 × 2𝑓𝑓 4 𝑓𝑓 = 21900 𝑙𝑙𝑙
  • 65. Problem 3.52 [Difficulty: 3] Given: Geometry of plane gate W h L = 3 m dF y L/2 w = 2 m Find: Minimum weight to keep it closed Solution: Basic equation FR Ap ⌠ ⎮ ⎮ ⌡ d= dp dh ρ g⋅= ΣMO 0= or, use computing equations FR pc A⋅= y' yc Ixx A yc⋅ += Assumptions: static fluid; ρ = constant; patm on other side; door is in equilibrium Instead of using either of these approaches, we note the following, using y as in the sketch ΣMO 0= W L 2 ⋅ cos θ( )⋅ Fy ⌠⎮ ⎮ ⌡ d= We also have dF p dA⋅= with p ρ g⋅ h⋅= ρ g⋅ y⋅ sin θ( )⋅= (Gage pressure, since p = patm on other side) Hence W 2 L cos θ( )⋅ Ay p⋅ ⌠⎮ ⎮ ⌡ d⋅= 2 L cos θ( )⋅ yy ρ⋅ g⋅ y⋅ sin θ( )⋅ w⋅ ⌠⎮ ⎮ ⌡ d⋅= W 2 L cos θ( )⋅ Ay p⋅ ⌠⎮ ⎮ ⌡ d⋅= 2 ρ⋅ g⋅ w⋅ tan θ( )⋅ L 0 L yy 2⌠ ⎮ ⌡ d⋅= 2 3 ρ⋅ g⋅ w⋅ L 2 ⋅ tan θ( )⋅= Using given data W 2 3 1000⋅ kg m 3 ⋅ 9.81× m s 2 ⋅ 2× m⋅ 3 m⋅( ) 2 × tan 30 deg⋅( )× N s 2 ⋅ kg m⋅ ×= W 68 kN⋅= Problem 3.45 3.45
  • 66. Problem 3.54 [Difficulty: 3] Given: Gate geometry Find: Depth H at which gate tips Solution: This is a problem with atmospheric pressure on both sides of the plate, so we can first determine the location of the center of pressure with respect to the free surface, using Eq.3.11c (assuming depth H) y' yc Ixx A yc⋅ += and Ixx w L 3 ⋅ 12 = with yc H L 2 −= where L = 1 m is the plate height and w is the plate width Hence y' H L 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ w L 3 ⋅ 12 w⋅ L⋅ H L 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ += H L 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L 2 12 H L 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ += But for equilibrium, the center of force must always be at or below the level of the hinge so that the stop can hold the gate in place. Hence we must have y' H 0.45 m⋅−> Combining the two equations H L 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L 2 12 H L 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ + H 0.45 m⋅−≥ Solving for H H L 2 L 2 12 L 2 0.45 m⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ +≤ H 1 m⋅ 2 1 m⋅( ) 2 12 1 m⋅ 2 0.45 m⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × +≤ H 2.17 m⋅≤ Problem 3.46 3.46
  • 67. Problem 3.56 [Difficulty: 3] Ry Rx FR Fn Given: Geometry of lock system Find: Force on gate; reactions at hinge Solution: Basic equation FR Ap ⌠ ⎮ ⎮ ⌡ d= dp dh ρ g⋅= or, use computing equation FR pc A⋅= Assumptions: static fluid; ρ = constant; patm on other side The force on each gate is the same as that on a rectangle of size h D= 10 m⋅= and w W 2 cos 15 deg⋅( )⋅ = FR Ap ⌠ ⎮ ⎮ ⌡ d= Aρ g⋅ y⋅ ⌠ ⎮ ⎮ ⌡ d= but dA w dy⋅= Hence FR 0 h yρ g⋅ y⋅ w⋅ ⌠ ⎮ ⌡ d= ρ g⋅ w⋅ h 2 ⋅ 2 = Alternatively FR pc A⋅= and FR pc A⋅= ρ g⋅ yc⋅ A⋅= ρ g⋅ h 2 ⋅ h⋅ w⋅= ρ g⋅ w⋅ h 2 ⋅ 2 = Using given data FR 1 2 1000⋅ kg m 3 ⋅ 9.81× m s 2 ⋅ 34 m⋅ 2 cos 15 deg⋅( )⋅ × 10 m⋅( ) 2 × N s 2 ⋅ kg m⋅ ×= FR 8.63 MN⋅= For the force components Rx and Ry we do the following ΣMhinge 0= FR w 2 ⋅ Fn w⋅ sin 15 deg⋅( )⋅−= Fn FR 2 sin 15 deg⋅( )⋅ = Fn 16.7 MN⋅= ΣFx 0= FR cos 15 deg⋅( )⋅ Rx−= 0= Rx FR cos 15 deg⋅( )⋅= Rx 8.34 MN⋅= ΣFy 0= Ry− FR sin 15 deg⋅( )⋅− Fn+= 0= Ry Fn FR sin 15 deg⋅( )⋅−= Ry 14.4 MN⋅= R 8.34 MN⋅ 14.4 MN⋅,( )= R 16.7 MN⋅= Problem 3.47 3.47
  • 68. Problem 3.48 (Difficulty: 2) 3.48 Calculate the minimum force 𝑃 necessary to hold a uniform 12 𝑓𝑓 𝑠𝑠𝑠𝑠𝑠𝑠 gate weighing 500 𝑙𝑙𝑙closed on a tank of water under a pressure of 10 𝑝𝑝𝑝. Draw a free body of the gate as part of your solution. Given: All the parameters are shown in the figure. Find: The minimum force 𝑃 to hold the system. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 A free body diagram of the gate is
  • 69. The gage pressure of the air in the tank is: 𝑝 𝑎𝑎𝑎 = 10 𝑝𝑝𝑝 = 1440 𝑙𝑙𝑙 𝑓𝑓2 This produces a uniform force on the gate of 𝐹1 = 𝑝 𝑎𝑎𝑎 𝐴 = 1440 𝑙𝑙𝑙 𝑓𝑓2 × (12 𝑓𝑓 × 12 𝑓𝑓) = 207360 𝑙𝑙𝑙 This pressure acts at the centroid of the area, which is the center of the gate. In addition, there is a force on the gate applied by water. This force is due to the pressure at the centroid of the area. The depth of the centroid is: 𝑦𝑐 = 12 𝑓𝑓 2 × sin 45° The force is them 𝐹2 = 𝛾ℎ 𝑐 𝐴 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 12 𝑓𝑓 2 × sin 45° × 12 𝑓𝑓 × 12 𝑓𝑓 = 38123 𝑙𝑙𝑙 The force F2 acts two-thirds of the way down from the hinge, or 𝑦′ = 8 𝑓𝑓. Take the moments about the hinge: −𝐹𝐵 𝐿 2 sin 45° + 𝐹1 𝐿 2 + 𝐹2 × 8 𝑓𝑓 − 𝑃 × 12 𝑓𝑓 = 0 Thus 𝑃 = −500 𝑙𝑙𝑙 × 6 𝑓𝑓 × sin 45° + 207360 𝑙𝑙𝑙 × 6 𝑓𝑓 + 38123 𝑙𝑙𝑙 × 8 𝑓𝑓 12 𝑓𝑓 = 128900 𝑙𝑙𝑙
  • 70. Problem 3.49 (Difficulty: 2) 3.49 Calculate magnitude and location of the resultant force of water on this annular gate. Given: All the parameters are shown in the figure. Find: Resultant force of water on this annular gate. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 For the magnitude of the force we have: 𝐹 = � 𝑝𝑝𝑝 𝐴 = 𝜌𝜌ℎ 𝑐 𝐴 The pressure is determined at the location of the centroid of the area ℎ 𝑐 = 1 𝑚 + 1.5 𝑚 = 2.5 𝑚 𝐴 = 𝜋 4 (𝐷2 2 − 𝐷1 2) = 𝜋 4 ((3 𝑚)2 − (1.5 𝑚)2) = 5.3014 𝑚2 𝐹 = 999 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 2.5 𝑚 × 5.3014 𝑚2 = 129900 𝑁 = 129.9 𝑘𝑘 The y axis is in the vertical direction. For the location of the force, we have:
  • 71. 𝑦′ = 𝑦𝑐 + 𝐼 𝑥�𝑥� 𝐴𝑦𝑐 Where: 𝑦𝑐 = 2.5 𝑚 𝐼 𝑥�𝑥� = 𝜋(𝐷2 4 − 𝐷1 4) 64 = 𝜋 64 × ((3 𝑚)4 − (1.5 𝑚)4) = 3.7276 𝑚4 𝑦′ = 𝑦𝑐 + 𝐼 𝑥�𝑥� 𝐴𝑦𝑐 = 2.5 𝑚 + 3.7276 𝑚4 2.5 𝑚 × 5.3014 𝑚2 = 2.78 𝑚 So the force acts on the depth of 𝑦′ = 2.78 𝑚.
  • 72. Problem 3.50 (Difficulty: 2) 3.50 A vertical rectangular gate 2.4 𝑚 wide and 2.7 𝑚 high is subjected to water pressure on one side, the water surface being at the top of the gate. The gate is hinged at the bottom and is held by a horizontal chain at the top. What is the tension in the chain? Given: The gate wide: 𝑤 = 2.4 𝑚. Height of the gate: ℎ = 2.7 𝑚. Find: The tension 𝐹𝑐 in the chain. Assumptions: Fluid is static and incompressible Solution: Apply the hydrostatic relations for pressure, force, and moments, with y measured from the surface of the liquid: 𝑑𝑑 𝑑𝑑 = 𝜌 𝑔 = 𝛾 𝐹𝑅 = � 𝑝 𝑑𝑑 𝑦′ 𝐹𝑅 = � 𝑦 𝑝 𝑑𝑑 For the magnitude of the force we have: 𝐹 = � 𝑝𝑝𝑝 𝐴 = 𝜌𝜌ℎ 𝑐 𝐴 Where hc is the depth at the centroid ℎ 𝑐 = 2.7 𝑚 2 = 1.35 𝑚 𝐴 = 𝑤ℎ = 2.4 𝑚 × 2.7 𝑚 = 6.48 𝑚2
  • 73. 𝐹 = 999 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 × 1.35 𝑚 × 6.48 𝑚2 = 85.7 𝑘𝑘 The y axis is in the vertical direction. For the location of the force, we have: ℎ 𝑝 = 2 3 × 2.7 𝑚 = 1.8 𝑚 Taking the momentum about the hinge: 𝐹�ℎ − ℎ 𝑝� − 𝐹𝑐ℎ = 0 𝐹𝑐 = 𝐹 �ℎ − ℎ 𝑝� ℎ = 85.7 𝑘𝑘 × 0.9 𝑚 2.7 𝑚 = 28.6 𝑘𝑘
  • 74. Problem 3.58 [Difficulty: 4] Given: Window, in shape of isosceles triangle and hinged at the top is located in the vertical wall of a form that contains concrete. a 0.4 m⋅= b 0.3 m⋅= c 0.25 m⋅= SGc 2.5= (From Table A.1, App. A) Find: The minimum force applied at D needed to keep the window closed. Plot the results over the range of concrete depth between 0 and a. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) FR Ap ⌠ ⎮ ⎮ ⌡ d= (Hydrostatic Force on door) y' FR⋅ Ay p⋅ ⌠ ⎮ ⎮ ⌡ d= (First moment of force) ΣM 0= (Rotational equilibrium) d dA h aw b D Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface and on the outside of the window. Integrating the pressure equation yields: p ρ g⋅ h d−( )⋅= for h > d p 0= for h < d where d a c−= d 0.15 m⋅= Summing moments around the hinge: FD− a⋅ Ah p⋅ ⌠⎮ ⎮ ⌡ d+ 0= FD dF = pdA h a FD 1 a Ah p⋅ ⌠⎮ ⎮ ⌡ d⋅= 1 a d a hh ρ⋅ g⋅ h d−( )⋅ w⋅ ⌠ ⎮ ⌡ d⋅= ρ g⋅ a d a hh h d−( )⋅ w⋅ ⌠ ⎮ ⌡ d⋅= From the law of similar triangles: w b a h− a = Therefore: w b a a h−( )= Problem 3.51 3.51
  • 75. Into the expression for the force at D: FD ρ g⋅ a d a h b a h⋅ h d−( )⋅ a h−( )⋅ ⌠ ⎮ ⎮ ⌡ d⋅= ρ g⋅ b⋅ a 2 d a hh 3 − a d+( ) h 2 ⋅+ a d⋅ h⋅−⎡⎣ ⎤⎦ ⌠ ⎮ ⌡ d⋅= Evaluating this integral we get: FD ρ g⋅ b⋅ a 2 a 4 d 4 −( ) 4 − a d+( ) a 3 d 3 −( )⋅ 3 + a d⋅ a 2 d 2 −( )⋅ 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= and after collecting terms: FD ρ g⋅ b⋅ a 2 ⋅ 1 4 − 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ 1 3 1 d a + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅+ 1 2 d a ⋅ 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 1( ) The density of the concrete is: ρ 2.5 1000× kg m 3 ⋅= ρ 2.5 10 3 × kg m 3 = d a 0.15 0.4 = 0.375= Substituting in values for the force at D: FD 2.5 10 3 × kg m 3 ⋅ 9.81⋅ m s 2 ⋅ 0.3⋅ m⋅ 0.4 m⋅( ) 2 ⋅ 1 4 − 1 0.375( ) 4 −⎡⎣ ⎤⎦⋅ 1 3 1 0.375+( )⋅ 1 0.375( ) 3 −⎡⎣ ⎤⎦⋅+ 0.375 2 1 0.375( ) 2 −⎡⎣ ⎤⎦⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ N s 2 ⋅ kg m⋅ ×= To plot the results for different values of c/a, we use Eq. (1) and remember that d a c−= FD 32.9N= Therefore, it follows that d a 1 c a −= In addition, we can maximize the force by the maximum force (when c = a or d = 0): Fmax ρ g⋅ b⋅ a 2 ⋅ 1 4 − 1 3 + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= ρ g⋅ b⋅ a 2 ⋅ 12 = and so FD Fmax 12 1 4 − 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 4 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ 1 3 1 d a + ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅+ 1 2 d a ⋅ 1 d a ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 − ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= 0.0 0.5 1.0 0.0 0.2 0.4 0.6 0.8 1.0 Concrete Depth Ratio (c/a) ForceRatio(FD/Fmax)
  • 76. Problem 3.60 [Difficulty: 2] Given: Plug is used to seal a conduit. γ 62.4 lbf ft 3 ⋅= Find: Magnitude, direction and location of the force of water on the plug. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh γ= (Hydrostatic Pressure - y is positive downwards) FR pc A⋅= (Hydrostatic Force) y' yc Ixx A yc⋅ += (Location of line of action) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on the outside of the plug. Integrating the hydrostatic pressure equation: p γ h⋅= FR pc A⋅= γ hc⋅ π 4 ⋅ D 2 ⋅= FR 62.4 lbf ft 3 ⋅ 12× ft⋅ π 4 × 6 ft⋅( ) 2 ×= FR 2.12 10 4 × lbf⋅= For a circular area: Ixx π 64 D 4 ⋅= Therefore: y' yc π 64 D 4 ⋅ π 4 D 2 ⋅ yc⋅ += yc D 2 16 yc⋅ += y' 12 ft⋅ 6 ft⋅( ) 2 16 12× ft⋅ += y' 12.19 ft⋅= The force of water is to the right and perpendicular to the plug. Problem 3.52 3.52
  • 77. Problem 3.62 [Difficulty: 2] Given: Circular access port of known diameter in side of water standpipe of known diameter. Port is held in place by eight bolts evenly spaced around the circumference of the port. Center of the port is located at a know distance below the free surface of the water. d 0.6 m⋅= D 7 m⋅= L 12 m⋅= Find: (a) Total force on the port (b) Appropriate bolt diameter Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - y is positive downwards) d L D h FR pc A⋅= (Hydrostatic Force) σ F A = (Normal Stress in bolt) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Force is distributed evenly over all bolts (4) Appropriate working stress in bolts is 100 MPa (5) Atmospheric pressure acts at free surface of water and on outside of port. Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= The resultant force on the port is: FR pc A⋅= ρ g⋅ L⋅ π 4 ⋅ d 2 ⋅= FR 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 12× m⋅ π 4 × 0.6 m⋅( ) 2 × N s 2 ⋅ kg m⋅ ×= FR 33.3 kN⋅= To find the bolt diameter we consider: σ FR A = where A is the area of all of the bolts: A 8 π 4 × db 2 ⋅= 2 π⋅ db 2 ⋅= Therefore: 2 π⋅ db 2 ⋅ FR σ = Solving for the bolt diameter we get: db FR 2 π⋅ σ⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 = db 1 2 π× 33.3× 10 3 × N⋅ 1 100 10 6 × × m 2 N ⋅ ⎛ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎠ 1 2 10 3 mm⋅ m ×= db 7.28 mm⋅= Problem 3.53 3.53
  • 78. Problem 3.64 [Difficulty: 3] Given: Gate AOC, hinged along O, has known width; Weight of gate may be neglected. Gate is sealed at C. b 6 ft⋅= Find: Force in bar AB Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) FR pc A⋅= (Hydrostatic Force) y' yc Ixx A yc⋅ += (Location of line of action) ΣMz 0= (Rotational equilibrium) F1 h1’ F2 L1 L2 x2’ FAB L1 Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate (4) No resisting moment in hinge at O (5) No vertical resisting force at C Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= The free body diagram of the gate is shown here: F1is the resultant of the distributed force on AO F2is the resultant of the distributed force on OC FAB is the force of the bar Cx is the sealing force at C First find the force on AO: F1 pc A1⋅= ρ g⋅ hc1⋅ b⋅ L1⋅= F1 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 6× ft⋅ 6× ft⋅ 12× ft⋅ lbf s 2 ⋅ slugft⋅ ×= F1 27.0 kip⋅= Problem 3.54 3.54
  • 79. h'1 hc1 Ixx A hc1⋅ += hc1 b L1 3 ⋅ 12 b⋅ L1⋅ hc1⋅ += hc1 L1 2 12 hc1⋅ += h'1 6 ft⋅ 12 ft⋅( ) 2 12 6× ft⋅ += h'1 8 ft⋅= Next find the force on OC: F2 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 12× ft⋅ 6× ft⋅ 6× ft⋅ lbf s 2 ⋅ slug ft⋅ ×= F2 27.0 kip⋅= F1 h1’ F2 L1 L2 x2’ FAB L1 Since the pressure is uniform over OC, the force acts at the centroid of OC, i.e., x'2 3 ft⋅= Summing moments about the hinge gives: FAB L1 L3+( )⋅ F1 L1 h'1−( )⋅− F2 x'2⋅+ 0= Solving for the force in the bar: FAB F1 L1 h'1−( )⋅ F2 x'2⋅− L1 L3+ = Substituting in values: FAB 1 12 ft⋅ 3 ft⋅+ 27.0 10 3 × lbf⋅ 12 ft⋅ 8 ft⋅−( )× 27.0 10 3 × lbf⋅ 3× ft⋅−⎡⎣ ⎤⎦⋅= FAB 1800 lbf⋅= Thus bar AB is in compression
  • 80. Problem 3.66 [Difficulty: 3] Given: Geometry of gate h D FR y FA y’ Find: Force at A to hold gate closed Solution: Basic equation dp dh ρ g⋅= ΣMz 0= Computing equations FR pc A⋅= y' yc Ixx A yc⋅ += Ixx w L 3 ⋅ 12 = Assumptions: Static fluid; ρ = constant; patm on other side; no friction in hinge For incompressible fluid p ρ g⋅ h⋅= where p is gage pressure and h is measured downwards The hydrostatic force on the gate is that on a rectangle of size L and width w. Hence FR pc A⋅= ρ g⋅ hc⋅ A⋅= ρ g⋅ D L 2 sin 30 deg⋅( )⋅+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ L⋅ w⋅= FR 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 1.5 3 2 sin 30 deg⋅( )+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × m⋅ 3× m⋅ 3× m⋅ N s 2 ⋅ kg m⋅ ×= FR 199 kN⋅= The location of this force is given by y' yc Ixx A yc⋅ += where y' and y c are measured along the plane of the gate to the free surface yc D sin 30 deg⋅( ) L 2 += yc 1.5 m⋅ sin 30 deg⋅( ) 3 m⋅ 2 += yc 4.5m= y' yc Ixx A yc⋅ += yc w L 3 ⋅ 12 1 w L⋅ ⋅ 1 yc ⋅+= yc L 2 12 yc⋅ += 4.5 m⋅ 3 m⋅( ) 2 12 4.5⋅ m⋅ += y' 4.67m= Taking moments about the hinge ΣMH 0= FR y' D sin 30 deg⋅( ) −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ FA L⋅−= FA FR y' D sin 30 deg⋅( ) − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ L ⋅= FA 199 kN⋅ 4.67 1.5 sin 30 deg⋅( ) − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ⋅= FA 111 kN⋅= Problem 3.55 3.55
  • 81. Problem 3.68 [Difficulty: 4] Given: Various dam cross-sections Find: Which requires the least concrete; plot cross-section area A as a function of α Solution: For each case, the dam width b has to be large enough so that the weight of the dam exerts enough moment to balance the moment due to fluid hydrostatic force(s). By doing a moment balance this value of b can be found a) Rectangular dam Straightforward application of the computing equations of Section 3-5 yields b D FH y mg O FH pc A⋅= ρ g⋅ D 2 ⋅ w⋅ D⋅= 1 2 ρ⋅ g⋅ D 2 ⋅ w⋅= y' yc Ixx A yc⋅ += D 2 w D 3 ⋅ 12 w⋅ D⋅ D 2 ⋅ += 2 3 D⋅= so y D y'−= D 3 = Also m ρcement g⋅ b⋅ D⋅ w⋅= SG ρ⋅ g⋅ b⋅ D⋅ w⋅= Taking moments about O M0.∑ 0= FH− y⋅ b 2 m⋅ g⋅+= so 1 2 ρ⋅ g⋅ D 2 ⋅ w⋅⎛ ⎜ ⎝ ⎞ ⎠ D 3 ⋅ b 2 SG ρ⋅ g⋅ b⋅ D⋅ w⋅( )⋅= Solving for b b D 3 SG⋅ = The minimum rectangular cross-section area is A b D⋅= D 2 3 SG⋅ = For concrete, from Table A.1, SG = 2.4, so A D 2 3 SG⋅ = D 2 3 2.4× = A 0.373 D 2 ⋅= Problem 3.56 3.56
  • 82. FH b αb D FV y x m1g m2g O b) Triangular dams Instead of analysing right-triangles, a general analysis is made, at the end of which right triangles are analysed as special cases by setting α = 0 or 1. Straightforward application of the computing equations of Section 3-5 yields FH pc A⋅= ρ g⋅ D 2 ⋅ w⋅ D⋅= 1 2 ρ⋅ g⋅ D 2 ⋅ w⋅= y' yc Ixx A yc⋅ += D 2 w D 3 ⋅ 12 w⋅ D⋅ D 2 ⋅ += 2 3 D⋅= so y D y'−= D 3 = Also FV ρ V⋅ g⋅= ρ g⋅ α b⋅ D⋅ 2 ⋅ w⋅= 1 2 ρ⋅ g⋅ α⋅ b⋅ D⋅ w⋅= x b α b⋅−( ) 2 3 α⋅ b⋅+= b 1 α 3 −⎛ ⎜ ⎝ ⎞ ⎠ ⋅= For the two triangular masses m1 1 2 SG⋅ ρ⋅ g⋅ α⋅ b⋅ D⋅ w⋅= x1 b α b⋅−( ) 1 3 α⋅ b⋅+= b 1 2 α⋅ 3 −⎛ ⎜ ⎝ ⎞ ⎠ ⋅= m2 1 2 SG⋅ ρ⋅ g⋅ 1 α−( )⋅ b⋅ D⋅ w⋅= x2 2 3 b 1 α−( )⋅= Taking moments about O M0.∑ 0= FH− y⋅ FV x⋅+ m1 g⋅ x1⋅+ m2 g⋅ x2⋅+= so 1 2 ρ⋅ g⋅ D 2 ⋅ w⋅⎛ ⎜ ⎝ ⎞ ⎠ − D 3 ⋅ 1 2 ρ⋅ g⋅ α⋅ b⋅ D⋅ w⋅⎛ ⎜ ⎝ ⎞ ⎠ b⋅ 1 α 3 −⎛ ⎜ ⎝ ⎞ ⎠ ⋅+ 1 2 SG⋅ ρ⋅ g⋅ α⋅ b⋅ D⋅ w⋅⎛ ⎜ ⎝ ⎞ ⎠ b⋅ 1 2 α⋅ 3 −⎛ ⎜ ⎝ ⎞ ⎠ ⋅ 1 2 SG⋅ ρ⋅ g⋅ 1 α−( )⋅ b⋅ D⋅ w⋅⎡ ⎢ ⎣ ⎤ ⎥ ⎦ 2 3 ⋅ b 1 α−( )⋅++ ... 0= Solving for b b D 3 α⋅ α 2 −( ) SG 2 α−( )⋅+ = For a right triangle with the hypotenuse in contact with the water, α = 1, and b D 3 1− SG+ = D 3 1− 2.4+ = b 0.477 D⋅= The cross-section area is A b D⋅ 2 = 0.238 D 2 ⋅= A 0.238 D 2 ⋅= For a right triangle with the vertical in contact with the water, α = 0, and
  • 83. b D 2 SG⋅ = D 2 2.4⋅ = b 0.456 D⋅= The cross-section area is A b D⋅ 2 = 0.228 D 2 ⋅= A 0.228 D 2 ⋅= For a general triangle A b D⋅ 2 = D 2 2 3 α⋅ α 2 −( ) SG 2 α−( )⋅+⋅ = A D 2 2 3 α⋅ α 2 −( ) 2.4 2 α−( )⋅+⋅ = The final result is A D 2 2 4.8 0.6 α⋅+ α 2 −⋅ = The dimensionless area, A /D 2 , is plotted Alpha A /D 2 0.0 0.2282 0.1 0.2270 0.2 0.2263 0.3 0.2261 0.4 0.2263 0.5 0.2270 0.6 0.2282 0.7 0.2299 0.8 0.2321 0.9 0.2349 1.0 0.2384 Solver can be used to find the minimum area Alpha A /D 2 0.300 0.2261 Dam Cross Section vs Coefficient 0.224 0.226 0.228 0.230 0.232 0.234 0.236 0.238 0.240 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Coefficient DimensionlessAreaA/D2 From the Excel workbook, the minimum area occurs at α = 0.3 Amin D 2 2 4.8 0.6 0.3×+ 0.3 2 −⋅ = A 0.226 D 2 ⋅= The final results are that a triangular cross-section with α = 0.3 uses the least concrete; the next best is a right triangle with the vertical in contact with the water; next is the right triangle with the hypotenuse in contact with the water; and the cross-section requiring the most concrete is the rectangular cross-section.
  • 84. Problem 3.70 [Difficulty: 2] Given: Geometry of dam Find: Vertical force on dam Assumptions: (1) water is static and incompressible (2) since we are asked for the force of the water, all pressures will be written as gage Solution: Basic equation: dp dh ρ g⋅= For incompressible fluid p ρ g⋅ h⋅= where p is gage pressure and h is measured downwards from the free surface The force on each horizontal section (depth d and width w) is F p A⋅= ρ g⋅ h⋅ d⋅ w⋅= (Note that d and w will change in terms of x and y for each section of the dam!) Hence the total force is (allowing for the fact that some faces experience an upwards (negative) force) FT p A⋅= Σρ g⋅ h⋅ d⋅ w⋅= ρ g⋅ d⋅ Σ⋅ h w⋅= Starting with the top and working downwards FT 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 3× ft⋅ 3 ft⋅ 12× ft⋅( ) 3 ft⋅ 6× ft⋅( )+ 9 ft⋅ 6× ft⋅( )− 12 ft⋅ 12× ft⋅( )−[ ]× lbf s 2 ⋅ slug ft⋅ ×= FT 2.70− 10 4 × lbf⋅= The negative sign indicates a net upwards force (it's actually a buoyancy effect on the three middle sections) Problem 3.57 3.57
  • 85. Problem 3.72 [Difficulty: 3] Given: Parabolic gate, hinged at O has a constant width. b 2 m⋅= c 0.25 m 1− ⋅= D 2 m⋅= H 3 m⋅= Find: (a) Magnitude and line of action of the vertical force on the gate due to water (b) Horizontal force applied at A required to maintain equilibrium (c) Vertical force applied at A required to maintain equilibrium Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards) ΣMz 0= (Rotational equilibrium) Fv Ayp ⌠ ⎮ ⎮ ⌡ d= (Vertical Hydrostatic Force) x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= (Location of line of action) FH pc A⋅= (Horizontal Hydrostatic Force) h' hc Ixx A hc⋅ += (Location of line of action) Oy h’ B x’ x FV Ox FH y Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of gate Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= (a) The magnitude and line of action of the vertical component of hydrostatic force: Fv Ayp ⌠⎮ ⎮ ⌡ d= 0 D c xρ g⋅ h⋅ b⋅ ⌠ ⎮ ⎮ ⌡ d= 0 D c xρ g⋅ D y−( )⋅ b ⌠ ⎮ ⎮ ⌡ d= 0 D c xρ g⋅ D c x 2 ⋅−( )⋅ b ⌠ ⎮ ⎮ ⌡ d= ρ g⋅ b⋅ 0 D c xD c x 2 ⋅−( ) ⌠ ⎮ ⎮ ⌡ d⋅= Evaluating the integral: Fv ρ g⋅ b⋅ D 3 2 c 1 2 1 3 D 3 2 c 1 2 ⋅− ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ ⋅= 2 ρ⋅ g⋅ b⋅ 3 D 3 2 c 1 2 ⋅= 1( ) Problem 3.58 3.58
  • 86. Substituting values: Fv 2 3 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 2× m⋅ 2 m⋅( ) 3 2 × 1 0.25 m⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 2 × N s 2 ⋅ kg m⋅ ×= Fv 73.9 kN⋅= To find the line of action of this force: x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= Therefore, x' 1 Fv Fvx ⌠ ⎮ ⎮ ⌡ d⋅= 1 Fv Ayx p⋅ ⌠ ⎮ ⎮ ⌡ d⋅= Using the derivation for the force: x' 1 Fv 0 D c xx ρ⋅ g⋅ D c x 2 ⋅−( )⋅ b⋅ ⌠ ⎮ ⎮ ⌡ d⋅= ρ g⋅ b⋅ Fv 0 D c xD x⋅ c x 3 ⋅−( ) ⌠ ⎮ ⎮ ⌡ d⋅= Evaluating the integral: x' ρ g⋅ b⋅ Fv D 2 D c ⋅ c 4 D c ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= ρ g⋅ b⋅ Fv D 2 4 c⋅ ⋅= Now substituting values into this equation: x' 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 2× m⋅ 1 73.9 10 3 × × 1 N ⋅ 1 4 × 2 m⋅( ) 2 × 1 0.25 × m⋅ N s 2 ⋅ kg m⋅ ×= x' 1.061m= To find the required force at A for equilibrium, we need to find the horizontal force of the water on the gate and its line of action as well. Once this force is known we take moments about the hinge (point O). FH pc A⋅= ρ g⋅ hc⋅ b⋅ D⋅= ρ g⋅ D 2 ⋅ b⋅ D⋅= ρ g⋅ b⋅ D 2 2 ⋅= since hc D 2 = Therefore the horizontal force is: FH 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 2× m⋅ 2 m⋅( ) 2 2 × N s 2 ⋅ kg m⋅ ×= FH 39.2 kN⋅= To calculate the line of action of this force: h' hc Ixx A hc⋅ += D 2 b D 3 ⋅ 12 1 b D⋅ ⋅ 2 D ⋅+= D 2 D 6 += 2 3 D⋅= h' 2 3 2⋅ m⋅= h' 1.333m= Oy h’ H x’ x FV Ox FH FA y D Now we have information to solve parts (b) and (c): (b) Horizontal force applied at A for equilibrium: take moments about O: FA H⋅ Fv x'⋅− FH D h'−( )⋅− 0= Solving for FA FA Fv x'⋅ FH D h'−( )⋅+ H = FA 1 3 1 m ⋅ 73.9 kN⋅ 1.061× m⋅ 39.2 kN⋅ 2 m⋅ 1.333 m⋅−( )×+[ ]×= FA 34.9 kN⋅= Oy h’ L x’ x FV Ox FH FA y D (c) Vertical force applied at A for equilibrium: take moments about O: FA L⋅ Fv x'⋅− FH D h'−( )⋅− 0= Solving for FA FA Fv x'⋅ FH D h'−( )⋅+ L = L is the value of x at y = H. Therefore: L H c = L 3 m⋅ 1 0.25 × m⋅= L 3.464m= FA 1 3.464 1 m ⋅ 73.9 kN⋅ 1.061× m⋅ 39.2 kN⋅ 2 m⋅ 1.333 m⋅−( )×+[ ]×= FA 30.2 kN⋅=
  • 87. Problem 3.74 [Difficulty: 2] Given: Open tank as shown. Width of curved surface b 10 ft⋅= Find: (a) Magnitude of the vertical force component on the curved surface (b) Line of action of the vertical component of the force Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh γ= (Hydrostatic Pressure - h is positive downwards) L x’ x FRy y Fv Ayp ⌠ ⎮ ⎮ ⌡ d−= (Vertical Hydrostatic Force) x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= (Moment of vertical force) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of wall Integrating the hydrostatic pressure equation: p γ h⋅= We can define along the surface h L R 2 x 2 −( ) 1 2 −= We also define the incremental area on the curved surface as: dAy b dx⋅= Substituting these into the force equation we get: Fv Ayp ⌠⎮ ⎮ ⌡ d−= 0 R xγ L R 2 x 2 −( ) 1 2 − ⎡⎢ ⎢ ⎣ ⎤⎥ ⎥ ⎦⋅ b⋅ ⌠ ⎮ ⎮ ⎮ ⌡ d−= γ− b⋅ 0 R xL R 2 x 2 −−( )⌠ ⎮ ⌡ d⋅= γ− b⋅ R⋅ L R π 4 ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Fv 62.4 lbf ft 3 ⋅ 10× ft⋅ 4× ft⋅ 10 ft⋅ 4 ft⋅ π 4 ×−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×⎡ ⎢ ⎣ ⎤ ⎥ ⎦ −= Fv 17.12− 10 3 × lbf⋅= (negative indicates downward) To find the line of action of the force: x' Fv⋅ Fvx ⌠⎮ ⎮ ⌡ d= where dFv γ− b⋅ L R 2 x 2 −−( )⋅ dx⋅= Therefore: x' x' Fv⋅ Fv = 1 γ b⋅ R⋅ L R π 4 ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 0 R xx γ⋅ b⋅ L R 2 x 2 −−( )⋅ ⌠ ⎮ ⌡ d⋅= 1 R L R π 4 ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 0 R xL x⋅ x R 2 x 2 −⋅−( )⌠ ⎮ ⌡ d⋅= Evaluating the integral: x' 4 R 4 L⋅ π R⋅−( )⋅ 1 2 L⋅ R 2 ⋅ 1 3 R 3 ⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= 4 R 2 ⋅ R 4 L⋅ π R⋅−( )⋅ L 2 R 3 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= 4 R⋅ 4 L⋅ π R⋅− L 2 R 3 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Substituting known values: x' 4 4⋅ ft⋅ 4 10⋅ ft⋅ π 4⋅ ft⋅− 10 ft⋅ 2 4 ft⋅ 3 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= x' 2.14 ft⋅= Problem 3.59 3.59
  • 88. Problem 3.76 [Difficulty: 3] Given: Dam with cross-section shown. Width of dam b 160 ft⋅= Find: (a) Magnitude and line of action of the vertical force component on the dam (b) If it is possible for the water to overturn dam Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards from free surface) Fv Ayp ⌠ ⎮ ⎮ ⌡ d= (Vertical Hydrostatic Force) FH pc A⋅= (Horizontal Hydrostatic Force) x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= (Moment of vertical force) A x’ x FH y y’ h’FV B h' hc Ixx hc A⋅ += (Line of action of vertical force) ΣMz 0= (Rotational Equilibrium) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts at free surface of water and on outside of dam Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= Into the vertical force equation: Fv Ayp ⌠⎮ ⎮ ⌡ d= xA xB xρ g⋅ h⋅ b⋅ ⌠ ⎮ ⌡ d= ρ g⋅ b⋅ xA xB xH y−( ) ⌠ ⎮ ⌡ d⋅= From the definition of the dam contour: x y⋅ A y⋅− B= Therefore: y B x A− = and xA 10 ft 2 ⋅ 9 ft⋅ 1 ft⋅+= xA 2.11 ft⋅= Problem 3.60 3.60
  • 89. Into the force equation: Fv ρ g⋅ b⋅ xA xB xH B x A− − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d⋅= ρ g⋅ b⋅ H xB xA−( )⋅ B ln xB A− xA A− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Substituting known values: Fv 1.94 slug ft 3 ⋅ 32.2× ft s 2 ⋅ 160× ft⋅ 9 ft⋅ 7.0 ft⋅ 2.11 ft⋅−( )× 10 ft 2 ⋅ ln 7.0 1− 2.11 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × lbf s 2 ⋅ slug ft⋅ ⋅= Fv 2.71 10 5 × lbf⋅= To find the line of action of the force: x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= where dFv ρ g⋅ b⋅ H B x A− − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ dx⋅= Therefore: x' x' Fv⋅ Fv = 1 Fv xA xB xx ρ⋅ g⋅ b⋅ H B x A− − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ ⌠ ⎮ ⎮ ⌡ d⋅= 1 H xB xA−( )⋅ B ln xB A− xA A− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− xA xB xH x⋅ B x⋅ x A− − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⌠ ⎮ ⎮ ⌡ d⋅= Evaluating the integral: x' H 2 xB 2 xA 2 −⎛ ⎝ ⎞ ⎠⋅ B xB xA−( )⋅− B A⋅ ln xB A− xA A− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− H xB xA−( )⋅ B ln xB A− xA A− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− = Substituting known values we get: x' 9 ft⋅ 2 7 2 2.11 2 −( )× ft 2 ⋅ 10 ft 2 ⋅ 7 2.11−( )× ft⋅− 10 ft 2 ⋅ 1× ft⋅ ln 7 1− 2.11 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×− 9 ft⋅ 7 2.11−( )× ft⋅ 10 ft 2 ⋅ ln 7 1− 2.11 1− ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×− = x' 4.96 ft⋅= To determine whether or not the water can overturn the dam, we need the horizontal force and its line of action: FH pc A⋅= ρ g⋅ H 2 ⋅ H⋅ b⋅= ρ g⋅ b⋅ H 2 ⋅ 2 = Substituting values: FH 1 2 1.94× slug ft 3 ⋅ 32.2× ft s 2 ⋅ 160× ft⋅ 9 ft⋅( ) 2 × lbf s 2 ⋅ slug ft⋅ ×= FH 4.05 10 5 × lbf⋅= For the line of action: h' hc Ixx hc A⋅ += where hc H 2 = A H b⋅= Ixx b H 3 ⋅ 12 = Therefore: h' H 2 b H 3 ⋅ 12 2 H ⋅ 1 b H⋅ ⋅+= H 2 H 6 += 2 3 H⋅= h' 2 3 9⋅ ft⋅= h' 6.00 ft⋅= Taking moments of the hydrostatic forces about the origin: Mw FH H h'−( )⋅ Fv x'⋅−= Mw 4.05 10 5 × lbf⋅ 9 6−( )× ft⋅ 2.71 10 5 × lbf⋅ 4.96× ft⋅−= Mw 1.292− 10 5 × lbf ft⋅⋅= The negative sign indicates that this is a clockwise moment about the origin. Since the weight of the dam will also contribute a clockwise moment about the origin, these two moments should not cause the dam to tip to the left. Therefore, the water can not overturn the dam.
  • 90. Problem 3.61 (Difficulty: 2) 3.61 The quarter cylinder 𝐴𝐴 is 10 𝑓𝑓 long. Calculate magnitude, direction, and location of the resultant force of the water on 𝐴𝐴. Given: All the parameters are shown in the figure. Assumptions: Fluid is incompressible and static Find: The resultant force. Solution: Apply the hydrostatic relations for pressure as a function of depth and for the location of forces on submerged objects. ∆𝑝 = 𝜌𝜌ℎ A freebody diagram for the cylinder is: The force balance in the horizontal direction yields thathorizontal force is due to the water pressure: 𝐹 𝐻 = 𝑃 𝐻 Where the depth is the distance to the centroid of the horizontal area (8 + 5/2 ft): 𝐹 𝐻 = 𝛾ℎ 𝑐 𝐴 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × �8 𝑓𝑓 + 5 𝑓𝑓 2 � × (5 𝑓𝑓 × 10 𝑓𝑡) = 32800 𝑙𝑙𝑙
  • 91. 𝑃 𝐻 = 32800 𝑙𝑙𝑙 The force in the vertical direction can be calculated as the weight of a volume of water that is 8 ft + 5 ft = 13 ft deep less the weight of water that would be in the quarter cylinder. This force is then: 𝑃𝑉 = 𝐹𝑉 − 𝑊 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 13 𝑓𝑓 × (5 𝑓𝑓 × 10 𝑓𝑓) − 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 𝜋 4 × (5 𝑓𝑓)2 × (10 𝑓𝑓) = 28308 𝑙𝑙𝑙 The total resultant force is the vector sum of the two forces: 𝑃 = � 𝑃 𝐻 2 + 𝑃𝑉 2 = �(32800 𝑙𝑙𝑙)2 + (28308 𝑙𝑙𝑙)2 = 43300 𝑙𝑙𝑙 The angle with respect to the horizontal is: 𝜃 = tan−1 � 𝑃𝑉 𝑃 𝐻 � = tan−1 � 28308 𝑙𝑙𝑙 32800 𝑙𝑙𝑙 � = 40.9° So the force acts on the quarter cylinder surface point at an angle of 𝜃 = 40.9 ° with respect to the horizontal.
  • 92. Problem 3.62 (Difficulty: 2) 3.62 Calculate the magnitude, direction (horizontal and vertical components are acceptable), and line of action of the resultant force exerted by the water on the cylindrical gate 30 𝑓𝑓 long. Assumptions: Fluid is incompressible and static Find: The resultant forces. Solution: Apply the hydrostatic relations for pressure as a function of depth and for the location of forces on submerged objects. ∆𝑝 = 𝜌𝜌ℎ A free body diagram of the gate is The horizontal force is calculated as: 𝑃 𝐻 = 𝐹 𝐻 Where the depth is the distance to the centroid of the horizontal area (5 ft): 𝐹 𝐻 = 𝛾ℎ 𝑐 𝐴 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 5𝑓𝑓 × (10 𝑓𝑓 × 30 𝑓𝑓) = 93600 𝑙𝑙𝑙 𝑃 𝐻 = 93600 𝑙𝑙𝑙 The force in the vertical direction can be calculated as the weight of a volume of water that is 10 ft deep less the weight of water that would be in the quarter cylinder. This force is then:
  • 93. 𝑃𝑉 = 𝐹𝑉 − 𝑊 = 𝛾ℎ 𝑐 𝐴 − 𝛾∀ 𝑃𝑉 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 10 𝑓𝑓 × (10 𝑓𝑓 × 30 𝑓𝑓) − 62.4 𝑙𝑙𝑙 𝑓𝑓3 × �10 𝑓𝑓 × (10 𝑓𝑓 × 30 𝑓𝑓) − 𝜋 4 × (10 𝑓𝑡)2 × 30 𝑓𝑓� = 147000 𝑙𝑙𝑙 The total resultant force is the vector sum of the two forces: 𝑃 = � 𝑃 𝐻 2 + 𝑃𝑉 2 = �(93600 𝑙𝑙𝑙)2 + (147000 𝑙𝑙𝑙)2 = 174200 𝑙𝑙𝑙 The direction can be calculated as: 𝜃 = tan−1 � 𝑃𝑉 𝑃 𝐻 � = tan−1 � 147000 𝑙𝑙𝑙 93600 𝑙𝑙𝑙 � = 57.5°
  • 94. Problem 3.63 (Difficulty: 2) 3.63 A hemispherical shell 1.2 𝑚 in diameter is connected to the vertical wall of a tank containing water. If the center of the shell is 1.8 𝑚 below the water surface, what are the vertical and horizontal force components on the shell? On the top half of the shell? Assumptions: Fluid is incompressible and static Find: The resultant forces. Solution: Apply the hydrostatic relations for pressure as a function of depth and for the location of forces on submerged objects. ∆𝑝 = 𝜌𝜌ℎ A free body diagram of the system is The force in the horizontal direction can be calculated using the distance to the centroid (1.8 m) as: 𝐹 𝐻 = 𝛾ℎ 𝑐 𝐴 = 9.81 𝑘𝑁 𝑚3 × 1.8 𝑚 × � 1 4 × 𝜋 × (1.2 𝑚)2 � = 19.97 𝑘𝑘 The force in the vertical direction is the buoyancy force due to the volume displaced by the shell: 𝐹𝑉 = 𝛾𝛾 = 9.81 𝑘𝑘 𝑚3 × 1 2 × 1 6 × 𝜋 × (1.2 𝑚)3 = 4.44 𝑘𝑘 For the top shell, the horizontal force acts at: 𝑦𝑐 = 1.8 𝑚 − 4 × 0.6 𝑚 3𝜋 = 1.545 𝑚 The horizontal force on the top half of the shell is then:
  • 95. 𝐹 𝐻 = 𝛾𝑦𝑐 𝐴 = 9.81 𝑘𝑘 𝑚3 × 1.545 𝑚 × 𝜋 8 × (1.2 𝑚)2 = 8.57 𝑘𝑘 The vertical force on the top half of the shell is the buoyancy force: 𝐹𝑉 = 𝑝𝑝 = 9.81 𝑘𝑘 𝑚3 × 1.8 𝑚 × 𝜋 8 × (1.2 𝑚)2 − 9.81 𝑘𝑘 𝑚3 × 1 4 × 1 6 × 𝜋 × (1.2 𝑚)3 = 7.77 𝑘𝑘
  • 96. Problem 3.78 [Difficulty: 4] FV D y R A x FH F1 x y’ FB W1 W2 Weights for computing FV R/2 4R/3π WGate Given: Gate geometry Find: Force on stop B Solution: Basic equations dp dh ρ g⋅= ΣMA 0= Assumptions: static fluid; ρ = constant; patm on other side For incompressible fluid p ρ g⋅ h⋅= where p is gage pressure and h is measured downwards We need to compute force (including location) due to water on curved surface and underneath. For curved surface we could integrate pressure, but here we use the concepts that FV (see sketch) is equivalent to the weight of fluid above, and FH is equivalent to the force on a vertical flat plate. Note that the sketch only shows forces that will be used to compute the moment at A For FV FV W1 W2−= with W1 ρ g⋅ w⋅ D⋅ R⋅= 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 3× m⋅ 4.5× m⋅ 3× m⋅ N s 2 ⋅ kg m⋅ ×= W1 397 kN⋅= W2 ρ g⋅ w⋅ π R 2 ⋅ 4 ⋅= 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 3× m⋅ π 4 × 3 m⋅( ) 2 × N s 2 ⋅ kg m⋅ ×= W2 208 kN⋅= FV W1 W2−= FV 189 kN⋅= with x given by FV x⋅ W1 R 2 ⋅ W2 4 R⋅ 3 π⋅ ⋅−= or x W1 Fv R 2 ⋅ W2 Fv 4 R⋅ 3 π⋅ ⋅−= x 397 189 3 m⋅ 2 × 208 189 4 3 π⋅ × 3× m⋅−= x 1.75m= For FH Computing equations FH pc A⋅= y' yc Ixx A yc⋅ += Problem 3.64 3.64
  • 97. Hence FH pc A⋅= ρ g⋅ D R 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ w⋅ R⋅= FH 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 4.5 m⋅ 3 m⋅ 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × 3× m⋅ 3× m⋅ N s 2 ⋅ kg m⋅ ×= FH 265 kN⋅= The location of this force is y' yc Ixx A yc⋅ += D R 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ w R 3 ⋅ 12 1 w R⋅ D R 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ ×+= D R 2 − R 2 12 D R 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ += y' 4.5 m⋅ 3 m⋅ 2 − 3 m⋅( ) 2 12 4.5 m⋅ 3 m⋅ 2 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × += y' 3.25m= The force F1 on the bottom of the gate is F1 p A⋅= ρ g⋅ D⋅ w⋅ R⋅= F1 1000 kg m 3 ⋅ 9.81× m s 2 ⋅ 4.5× m⋅ 3× m⋅ 3× m⋅ N s 2 ⋅ kg m⋅ ×= F1 397 kN⋅= For the concrete gate (SG = 2.4 from Table A.2) WGate SG ρ⋅ g⋅ w⋅ π R 2 ⋅ 4 ⋅= 2.4 1000⋅ kg m 3 ⋅ 9.81× m s 2 ⋅ 3× m⋅ π 4 × 3 m⋅( ) 2 × N s 2 ⋅ kg m⋅ ×= WGate 499 kN⋅= Hence, taking moments about A FB R⋅ F1 R 2 ⋅+ WGate 4 R⋅ 3 π⋅ ⋅− FV x⋅− FH y' D R−( )−[ ]⋅− 0= FB 4 3 π⋅ WGate⋅ x R FV⋅+ y' D R−( )−[ ] R FH⋅+ 1 2 F1⋅−= FB 4 3 π⋅ 499× kN⋅ 1.75 3 189× kN⋅+ 3.25 4.5 3−( )−[ ] 3 265× kN⋅+ 1 2 397× kN⋅−= FB 278 kN⋅=
  • 98. Problem 3.80 [Difficulty: 3] Given: Cylindrical weir as shown; liquid is water Find: Magnitude and direction of the resultant force of the water on the weir Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards from free surface) dFR →⎯ p− dA →⎯ ⋅= (Hydrostatic Force) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surfaces and on the first quadrant of the cylinder D1 y x D2 h1 h2 θ Using the coordinate system shown in the diagram at the right: FRx FR →⎯ i → ⋅= A → p ⌠ ⎮ ⎮ ⌡ d− i → ⋅= Ap cos θ 90 deg⋅+( )⋅ ⌠ ⎮ ⎮ ⌡ d−= Ap sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d= FRy FR →⎯ j → ⋅= A → p ⌠ ⎮ ⎮ ⌡ d− j → ⋅= Ap cos θ( )⋅ ⌠ ⎮ ⎮ ⌡ d−= Now since dA L R⋅ dθ⋅= it follows that FRx 0 3 π⋅ 2 θp L⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d= and FRy 0 3 π⋅ 2 θp L⋅ R⋅ cos θ( )⋅ ⌠ ⎮ ⎮ ⌡ d−= Next, we integrate the hydrostatic pressure equation: p ρ g⋅ h⋅= Now over the range 0 θ≤ π≤ h1 R 1 cos θ( )−( )= Over the range π θ≤ 3 π⋅ 2 ≤ h2 R− cos θ( )⋅= Therefore we can express the pressure in terms of θ and substitute into the force equations: FRx 0 3 π⋅ 2 θp L⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d= 0 π θρ g⋅ R⋅ 1 cos θ( )−( )⋅ L⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⌡ d π 3 π⋅ 2 θρ g⋅ R⋅ cos θ( )⋅ L⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d−= FRx ρ g⋅ R 2 ⋅ L⋅ 0 π θ1 cos θ( )−( ) sin θ( )⋅ ⌠ ⎮ ⌡ d⋅ ρ g⋅ R 2 ⋅ L⋅ π 3 π⋅ 2 θcos θ( ) sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d⋅−= Problem 3.65 3.65
  • 99. FRx ρ g⋅ R 2 ⋅ L⋅ 0 π θ1 cos θ( )−( ) sin θ( )⋅ ⌠ ⎮ ⌡ d π 3 π⋅ 2 θcos θ( ) sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d− ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ ⋅= ρ g⋅ R 2 ⋅ L⋅ 2 1 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= 3 2 ρ⋅ g⋅ R 2 ⋅ L⋅= Substituting known values: FRx 3 2 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1.5 m⋅( ) 2 × 6× m⋅ N s 2 ⋅ kg m⋅ ×= FRx 198.5 kN⋅= Similarly we can calculate the vertical force component: FRy 0 3 π⋅ 2 θp L⋅ R⋅ cos θ( )⋅ ⌠ ⎮ ⎮ ⌡ d−= 0 π θρ g⋅ R⋅ 1 cos θ( )−( )⋅ L⋅ R⋅ cos θ( )⋅ ⌠ ⎮ ⌡ d π 3 π⋅ 2 θρ g⋅ R⋅ cos θ( )⋅ L⋅ R⋅ cos θ( )⋅ ⌠ ⎮ ⎮ ⌡ d− ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ −= FRy ρ− g⋅ R 2 ⋅ L⋅ 0 π θ1 cos θ( )−( ) cos θ( )⋅ ⌠ ⎮ ⌡ d π 3 π⋅ 2 θcos θ( )( ) 2 ⌠ ⎮ ⎮ ⌡ d− ⎡ ⎢ ⎢ ⎢ ⎣ ⎤ ⎥ ⎥ ⎥ ⎦ ⋅= ρ g⋅ R 2 ⋅ L⋅ π 2 3 π⋅ 4 + π 2 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= 3 π⋅ 4 ρ⋅ g⋅ R 2 ⋅ L⋅= Substituting known values: FRy 3 π⋅ 4 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 1.5 m⋅( ) 2 × 6× m⋅ N s 2 ⋅ kg m⋅ ×= FRy 312 kN⋅= Now since the weir surface in contact with the water is a circular arc, all elements dF of the force, and hence the line of action of the resultant force, must pass through the pivot. Thus: Magnitude of the resultant force: FR 198.5 kN⋅( ) 2 312 kN⋅( ) 2 += FR 370 kN⋅= The line of action of the force: α atan 312 kN⋅ 198.5 kN⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = α 57.5 deg⋅=
  • 100. Problem 3.82 [Difficulty: 3] Given: Curved surface, in shape of quarter cylinder, with given radius R and width w; water stands to depth H. R 0.750 m⋅= w 3.55 m⋅= H 0.650 m⋅= Find: Magnitude and line of action of (a) vertical force and (b) horizontal force on the curved surface Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards from free surface) Fv Ayp ⌠ ⎮ ⎮ ⌡ d= (Vertical Hydrostatic Force) FH pc A⋅= (Horizontal Hydrostatic Force) x' Fv⋅ Fvx ⌠ ⎮ ⎮ ⌡ d= (Moment of vertical force) h' hc Ixx hc A⋅ += (Line of action of horizontal force) dF h HR θ Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the water and on the left side of the curved surface Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= dF h’ HR θ FV FH y’ x’From the geometry: h H R sin θ( )⋅−= y R sin θ( )⋅= x R cos θ( )⋅= dA w R⋅ dθ⋅= θ1 asin H R ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ1 asin 0.650 0.750 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = θ1 1.048 rad⋅= Therefore the vertical component of the hydrostatic force is: Fv Ayp ⌠⎮ ⎮ ⌡ d= Aρ g⋅ h⋅ sin θ( )⋅ ⌠⎮ ⎮ ⌡ d= 0 θ1 θρ g⋅ H R sin θ( )⋅−( )⋅ sin θ( )⋅ w⋅ R⋅ ⌠ ⎮ ⌡ d= Fv ρ g⋅ w⋅ R⋅ 0 θ1 θH sin θ( )⋅ R sin θ( )( ) 2 ⋅−⎡⎣ ⎤⎦ ⌠ ⎮ ⌡ d⋅= ρ g⋅ w⋅ R⋅ H 1 cos θ1( )−( )⋅ R θ1 2 sin 2 θ1⋅( ) 4 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Problem 3.66 3.66
  • 101. Fv 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 3.55× m⋅ 0.750× m⋅ 0.650 m⋅ 1 cos 1.048 rad⋅( )−( )× 0.750 m⋅ 1.048 2 sin 2 1.048× rad⋅( ) 4 − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × N s 2 ⋅ kg m⋅ ×= Fv 2.47 kN⋅= To calculate the line of action of this force: x' Fv⋅ AR cos θ( )⋅ ρ⋅ g⋅ h⋅ sin θ( )⋅ ⌠ ⎮ ⎮ ⌡ d= ρ g⋅ w⋅ R 2 ⋅ 0 θ1 θH sin θ( )⋅ cos θ( )⋅ R sin θ( )( ) 2 ⋅ cos θ( )⋅−⎡⎣ ⎤⎦ ⌠ ⎮ ⌡ d⋅= Evaluating the integral: x' Fv⋅ ρ g⋅ w⋅ R 2 ⋅ H 2 sin θ1( )( ) 2 ⋅ R 3 sin θ1( )( ) 3 ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Therefore we may find the line of action: x' x' Fv⋅ Fv = ρ g⋅ w⋅ R 2 ⋅ Fv H 2 sin θ1( )( ) 2 ⋅ R 3 sin θ1( )( ) 3 ⋅− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Substituting in known values: sin θ1( ) 0.650 0.750 = x' 999 kg m 3 ⋅ 9.81× m s 2 ⋅ 3.55× m⋅ 0.750 m⋅( ) 2 × 1 2.47 10 3 × × 1 N ⋅ 0.650 m⋅ 2 0.650 0.750 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × 0.750 m⋅ 3 0.650 0.750 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ×− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × N s 2 ⋅ kg m⋅ ×= x' 0.645m= For the horizontal force: FH pc A⋅= ρ g⋅ hc⋅ H⋅ w⋅= ρ g⋅ H 2 ⋅ H⋅ w⋅= ρ g⋅ H 2 ⋅ w⋅ 2 = FH 1 2 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 0.650 m⋅( ) 2 × 3.55× m⋅ N s 2 ⋅ kg m⋅ ×= FH 7.35 kN⋅= For the line of action of the horizontal force: h' hc Ixx hc A⋅ += where Ixx w H 3 ⋅ 12 = A w H⋅= Therefore: h' hc Ixx hc A⋅ += H 2 w H 3 ⋅ 12 2 H ⋅ 1 w H⋅ ⋅+= H 2 H 6 += 2 3 H⋅= h' 2 3 0.650× m⋅= h' 0.433m=
  • 102. Problem 3.83 [Difficulty: 2] Given: Canoe floating in a pond Find: What happens when an anchor with too short of a line is thrown from canoe Solution: Governing equation: WgVF dispwB == ρ Before the anchor is thrown from the canoe the buoyant force on the canoe balances out the weight of the canoe and anchor: 11 canoewanchorcanoeB gVWWF ρ=+= The anchor weight can be expressed as aaanchor gVW ρ= so the initial volume displaced by the canoe can be written as a w a w canoe canoe V g W V ρ ρ ρ +=1 After throwing the anchor out of the canoe there will be buoyant forces acting on the canoe and the anchor. Combined, these buoyant forces balance the canoe weight and anchor weight: awcanoewanchorcanoeB gVgVWWF ρρ +=+= 22 a w a w canoe canoe V g W g W V −+= ρρ 2 Using the anchor weight, aa w a w canoe canoe VV g W V −+= ρ ρ ρ 2 Hence the volume displaced by the canoe after throwing the anchor in is less than when the anchor was in the canoe, meaning that the canoe is floating higher. Problem 3.67 3.67
  • 103. Problem 3.86 [Difficulty: 4] Given: Cylinder of mass M, length L, and radius R is hinged along its length and immersed in an incompressilble liquid to depth Find: General expression for the cylinder specific gravity as a function of α=H/R needed to hold the cylinder in equilibrium for α ranging from 0 to 1. Solution: We will apply the hydrostatics equations to this system. Governing Equations: dp dh ρ g⋅= (Hydrostatic Pressure - h is positive downwards from free surface) dFH dFh θ dFV H = αR Fv Ayp ⌠ ⎮ ⎮ ⌡ d= (Vertical Hydrostatic Force) ΣM 0= (Rotational Equilibrium) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts on free surface of the liquid. The moments caused by the hydrostatic force and the weight of the cylinder about the hinge need to balance each other. Integrating the hydrostatic pressure equation: p ρ g⋅ h⋅= dFv dF cos θ( )⋅= p dA⋅ cos θ( )⋅= ρ g⋅ h⋅ w⋅ R⋅ dθ⋅ cos θ( )⋅= Now the depth to which the cylinder is submerged is H h R 1 cos θ( )−( )⋅+= Therefore h H R 1 cos θ( )−( )⋅−= and into the vertical force equation: dFv ρ g⋅ H R 1 cos θ( )−( )⋅−[ ]⋅ w⋅ R⋅ cos θ( )⋅ dθ⋅= ρ g⋅ w⋅ R 2 ⋅ H R 1 cos θ( )−( )−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ cos θ( )⋅ dθ⋅= dFv ρ g⋅ w⋅ R 2 ⋅ α 1−( ) cos θ( )⋅ cos θ( )( ) 2 +⎡⎣ ⎤⎦⋅ dθ⋅= ρ g⋅ w⋅ R 2 ⋅ α 1−( ) cos θ( )⋅ 1 cos 2 θ⋅( )+ 2 +⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ dθ⋅= Now as long as α is not greater than 1, the net horizontal hydrostatic force will be zero due to symmetry, and the vertical force is: Fv θmax− θmax Fv1 ⌠ ⎮ ⌡ d= 0 θmax Fv2 ⌠ ⎮ ⌡ d= where cos θmax( ) R H− R = 1 α−= or θmax acos 1 α−( )= Problem 3.68 3.68
  • 104. Fv 2ρ g⋅ w⋅ R 2 ⋅ 0 θmax θα 1−( ) cos θ( )⋅ 1 2 + 1 2 cos 2 θ⋅( )⋅+ ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⌠ ⎮ ⎮ ⌡ d⋅= Now upon integration of this expression we have: Fv ρ g⋅ w⋅ R 2 ⋅ acos 1 α−( ) 1 α−( ) α 2 α−( )⋅⋅−⎡⎣ ⎤⎦⋅= The line of action of the vertical force due to the liquid is through the centroid of the displaced liquid, i.e., through the center of the cylinde The weight of the cylinder is given by: W M g⋅= ρc V⋅ g⋅= SG ρ⋅ π⋅ R 2 ⋅ w⋅ g⋅= where ρ is the density of the fluid and SG ρc ρ = The line of action of the weight is also throught the center of the cylinder. Taking moment about the hinge we get: ΣMo W R⋅ Fv R⋅−= 0= or in other words W Fv= and therefore: SG ρ⋅ π⋅ R 2 ⋅ w⋅ g⋅ ρ g⋅ w⋅ R 2 ⋅ acos 1 α−( ) 1 α−( ) α 2 α−( )⋅⋅−⎡⎣ ⎤⎦⋅= SG 1 π acos 1 α−( ) 1 α−( ) α 2 α−( )⋅⋅−⎡⎣ ⎤⎦⋅= 0 0.5 1 0 0.2 0.4 0.6 alpha (H/R) SpecificGravity,SG
  • 105. Problem *3.89 [Difficulty: 2] Given: Hydrometer as shown, submerged in nitric acid. When submerged in water, h = 0 and the immersed volume is 15 cubic cm. SG 1.5= d 6 mm⋅= Find: The distance h when immersed in nitric acid. Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) Assumptions: (1) Static fluid (2) Incompressible fluid Taking a free body diagram of the hydrometer: ΣFz 0= M− g⋅ Fbuoy+ 0= Solving for the mass of the hydrometer: M Fbuoy g = ρ Vd⋅= When immersed in water: M ρw Vw⋅= When immersed in nitric acid: M ρn Vn⋅= Since the mass of the hydrometer is the same in both cases: ρw Vw⋅ ρn Vn⋅= When the hydrometer is in the nitric acid: Vn Vw π 4 d 2 ⋅ h⋅−= ρn SG ρw⋅= Therefore: ρw Vw⋅ SG ρw⋅ Vw π 4 d 2 ⋅ h⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Solving for the height h: Vw SG Vw π 4 d 2 ⋅ h⋅−⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Vw 1 SG−( )⋅ SG− π 4 ⋅ d 2 ⋅ h⋅= h Vw SG 1− SG ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 4 π d 2 ⋅ ⋅= h 15 cm 3 ⋅ 1.5 1− 1.5 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ × 4 π 6 mm⋅( ) 2 × × 10 mm⋅ cm ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 3 ×= h 177 mm⋅= Problem 3.69 3.69
  • 106. Problem 3.70 (Difficulty: 2) 3.70 A cylindrical can 76 𝑚𝑚 in diameter and 152 𝑚𝑚 high, weighing 1.11 𝑁, contains water to a depth of 76 𝑚𝑚. When this can is placed in water, how deep will it sink? Find: The depth it will sink. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 A free body diagram on the can is We have the force balance equation in the vertical direction as: 𝐹𝑏 − 𝑊𝑐𝑐𝑐 − 𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 0 The buoyancy force can be calculated as: 𝐹𝑏 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑐 𝑐𝑐 = 9810 𝑁 𝑚3 × 𝜋 4 × (0.076 𝑚)2 × 𝑥 𝑚 = 44.50𝑋𝑋 𝑁 We also have: 𝑊𝑐𝑐𝑐 = 1.11 𝑁 𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 9810 𝑁 𝑚3 × 𝜋 4 × (0.076 𝑚)3 = 3.38 𝑁
  • 107. Thus making a force balance for which the net force is zero at equilibrium 44.50𝑥 = 1.11 𝑁 + 3.38 𝑁 = 4.49 𝑁 𝑥 = 0.1009 𝑚 = 100.9 𝑚𝑚 So this can will sink to depth of 100.9 𝑚𝑚.
  • 108. Problem 3.71 (Difficulty: 1) 3.71 If the 10 𝑓𝑓 long box is floating on the oil water system, calculate how much the box and its contents must weigh. Find: The weight of the box and its contents. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 The force balance equation in the vertical diretion: 𝐹𝐵 − 𝑊𝐵 = 0 𝐹𝐵 = 𝛾 𝑜𝑜𝑜 𝑉 + 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉 Thus 𝐹𝐵 = 0.8 × 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 2𝑓𝑓 × 8𝑓𝑓 × 10𝑓𝑓 + 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 1𝑓𝑓 × 8𝑓𝑓 × 10𝑓𝑓 = 12980 𝑙𝑙𝑙 So the box and its contents must weigh: 𝑊𝐵 = 12980 𝑙𝑙𝑙
  • 109. Problem 3.72 (Difficulty: 2) 3.72 The timber weighs 40 𝑙𝑙𝑙 𝑓𝑓3 and is held in a horizontal position by the concrete �150 𝑙𝑙𝑙 𝑓𝑓3� anchor. Calculate the minimum total weight which the anchor may have. Find: The minimum total weight the anchor may have. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 For the buoyancy force we have: 𝐹𝑏𝑏 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑡 𝐹𝑏𝑏 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × � 6 12 𝑓𝑓� × � 6 12 𝑓𝑓� × (20 𝑓𝑓) = 312 𝑙𝑙𝑙 The weight of the timber is: 𝑊𝑡 = 𝛾𝑡 𝑉𝑡 𝑊𝑡 = 40 𝑙𝑙𝑙 𝑓𝑓3 × � 6 12 𝑓𝑓� × � 6 12 𝑓𝑓� × (20 𝑓𝑓) = 200 𝑙𝑙𝑙 At the horizontal position we take moments about the pivot: 𝐹𝑎 𝐿 + 𝑊𝑡 𝐿 2 − 𝐹𝑏𝑏 𝐿 2 = 0 𝐹𝑎 = 1 2 𝐹𝑏𝑏 − 1 2 𝑊𝑡 = 1 2 × (312 𝑙𝑙𝑙 − 200𝑙𝑙𝑙) = 56 𝑙𝑙𝑙
  • 110. 𝐹𝑎 = 𝐹𝑏𝑏 − 𝑊𝑎 The weight of the anchor is: 𝑊𝑎 = 𝛾𝑎 𝑉𝑎 The buoyancy force on the anchor is: 𝐹𝑏𝑏 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑎 𝛾𝑎 𝑉𝑎 − 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉𝑎 = 56 𝑙𝑙𝑙 𝑉𝑎 = 56 𝑙𝑙𝑙 �150 𝑙𝑙𝑙 𝑓𝑓3 − 62.4 𝑙𝑙𝑙 𝑓𝑓3� = 0.64 𝑓𝑓3 So the weight is: 𝑊𝑎 = 𝛾𝑎 𝑉𝑎 = 150 𝑙𝑙𝑙 𝑓𝑓3 × 0.64 𝑓𝑓3 = 96 𝑙𝑙𝑙
  • 111. Problem *3.90 [Difficulty: 3] Given: Data on sphere and weight T FB W Find: SG of sphere; equilibrium position when freely floating Solution: Basic equation FB ρ g⋅ V⋅= and ΣFz 0= ΣFz 0= T FB+ W−= where T M g⋅= M 10 kg⋅= FB ρ g⋅ V 2 ⋅= W SG ρ⋅ g⋅ V⋅= Hence M g⋅ ρ g⋅ V 2 ⋅+ SG ρ⋅ g⋅ V⋅− 0= SG M ρ V⋅ 1 2 += SG 10 kg⋅ m 3 1000 kg⋅ × 1 0.025 m 3 ⋅ × 1 2 += SG 0.9= The specific weight is γ Weight Volume = SG ρ⋅ g⋅ V⋅ V = SG ρ⋅ g⋅= γ 0.9 1000× kg m 3 ⋅ 9.81× m s 2 ⋅ N s 2 ⋅ kg m⋅ ×= γ 8829 N m 3 ⋅= For the equilibriul position when floating, we repeat the force balance with T = 0 FB W− 0= W FB= with FB ρ g⋅ Vsubmerged⋅= From references (trying Googling "partial sphere volume") Vsubmerged π h 2 ⋅ 3 3 R⋅ h−( )⋅= where h is submerged depth and R is the sphere radius R 3 V⋅ 4 π⋅ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 3 = R 3 4 π⋅ 0.025⋅ m 3 ⋅⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 1 3 = R 0.181m= Hence W SG ρ⋅ g⋅ V⋅= FB= ρ g⋅ π h 2 ⋅ 3 ⋅ 3 R⋅ h−( )⋅= h 2 3 R⋅ h−( )⋅ 3 SG⋅ V⋅ π = h 2 3 0.181⋅ m⋅ h−( )⋅ 3 0.9⋅ .025⋅ m 3 ⋅ π = h 2 0.544 h−( )⋅ 0.0215= This is a cubic equation for h. We can keep guessing h values, manually iterate, or use Excel's Goal Seek to find h 0.292 m⋅= Problem 3.73 3.73
  • 112. Problem *3.91 [Difficulty: 2] Given: Specific gravity of a person is to be determined from measurements of weight in air and the met weight when totally immersed in water. Find: Expression for the specific gravity of a person from the measurements. Solution: We will apply the hydrostatics equations to this system. Governing Equation: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) Fnet Fbuoy Mg Assumptions: (1) Static fluid (2) Incompressible fluid Taking a free body diagram of the body: ΣFy 0= Fnet M g⋅− Fbuoy+ 0= Fnet is the weight measurement for the immersed body. Fnet M g⋅ Fbuoy−= M g⋅ ρw g⋅ Vd⋅−= However in air: Fair M g⋅= Therefore the weight measured in water is: Fnet Fair ρw g⋅ Vd⋅−= and Vd Fair Fnet− ρw g⋅ = Now in order to find the specific gravity of the person, we need his/her density: Fair M g⋅= ρ g⋅ Vd⋅= ρ g⋅ Fair Fnet−( ) ρw g⋅ ⋅= Simplifying this expression we get: Fair ρ ρw Fair Fnet−( )= Now if we call the density of water at 4 deg C ρw4C then: Fair ρ ρw4C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ρw ρw4C ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ Fair Fnet−( )= SG SGw Fair Fnet−( )⋅= Solving this expression for the specific gravity of the person SG, we get: SG SGw Fair Fair Fnet− ⋅= Problem 3.74 3.74
  • 113. Problem *3.93 [Difficulty: 2] Given: Geometry of steel cylinder Find: Volume of water displaced; number of 1 kg wts to make it sink Solution: The data is For water ρ 999 kg m 3 ⋅= For steel (Table A.1) SG 7.83= For the cylinder D 100 mm⋅= H 1 m⋅= δ 1 mm⋅= The volume of the cylinder is Vsteel δ π D 2 ⋅ 4 π D⋅ H⋅+ ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= Vsteel 3.22 10 4− × m 3 ⋅= The weight of the cylinder is W SG ρ⋅ g⋅ Vsteel⋅= W 7.83 999× kg m 3 ⋅ 9.81× m s 2 ⋅ 3.22× 10 4− × m 3 ⋅ N s 2 ⋅ kg m⋅ ×= W 24.7N= At equilibium, the weight of fluid displaced is equal to the weight of the cylinder Wdisplaced ρ g⋅ Vdisplaced⋅= W= Vdisplaced W ρ g⋅ = 24.7 N⋅ m 3 999 kg⋅ × s 2 9.81 m⋅ × kg m⋅ N s 2 ⋅ ×= Vdisplaced 2.52L= To determine how many 1 kg wts will make it sink, we first need to find the extra volume that will need to be dsiplaced Distance cylinder sank x1 Vdisplaced π D 2 ⋅ 4 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ = x1 0.321m= Hence, the cylinder must be made to sink an additional distance x2 H x1−= x2 0.679m= We deed to add n weights so that 1 kg⋅ n⋅ g⋅ ρ g⋅ π D 2 ⋅ 4 ⋅ x2⋅= n ρ π⋅ D 2 ⋅ x2⋅ 4 1 kg⋅× = 999 kg m 3 ⋅ π 4 × 0.1 m⋅( ) 2 × 0.679× m⋅ 1 1 kg⋅ × N s 2 ⋅ kg m⋅ ×= n 5.33= Hence we need n 6= weights to sink the cylinder Problem 3.75 3.75
  • 114. Problem 3.76 (Difficulty: 2) 3.76 If the timber weights 670 𝑁, calculate its angle of inclination when the water surface is 2.1 𝑚 above the pivot. Above what depth will the timber stand vertically? Find: Above what depth will the timber stand vertically. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 The buoyancy force is: 𝐹𝑏 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉 = 0.152 𝑚 × 0.152 𝑚 × 𝑥 𝑚 × 9810 𝑁 𝑚3 = 226.7𝑥 (𝑁) Take the moment about pivot we have: 𝑀 = 𝑊 × 0.5 × 3.6 𝑚 cos 𝜃 − 𝑥 2 𝑚 × 𝐹𝑏 cos 𝜃 = 0 670 𝑁 × 0.5 × 3.6 𝑚 × cos 𝜃 − 𝑥 2 𝑚 × 226.7𝑥 × cos 𝜃 = 0 Soving this equation we have: 𝑥 = 3.26 𝑚 The angle when water surface 𝑦 = 2.1 𝑚 is: 𝜃 = sin−1 � 2.1 𝑚 3.26 𝑚 � = 40.1 °
  • 115. We have the following relation: 𝑥 = 𝑦 sin 𝜃 Substitute in to the momentum we have: 670 𝑁 × 0.5 × 3.6 𝑚 − 𝑦 2sin 𝜃 𝑚 × 226.7 𝑦 sin 𝜃 = 0 If the timber is vertically, we have: 𝜃 = 90° sin 90° = 1 So we have: 670 𝑁 × 0.5 × 3.6 𝑚 − 𝑦 2 𝑚 × 226.7𝑦 = 0 Solving this equation we have: 𝑦 = 3.26 𝑚 When the water surface is 𝑦 = 3.26 𝑚, the timber will stand vertically.
  • 116. Problem 3.77 (Difficulty: 2) 3.77 The barge shown weights 40 𝑡𝑡𝑡𝑡 and carries a cargo of 40 𝑡𝑡𝑡𝑡. Calculate its draft in freshwater. Find: The draft, where the draft is the depth to which the barge sinks. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 For the barge floating in water we have the buoyancy force as: 𝐹𝐵 = 𝛾 𝑤𝑤𝑤𝑤𝑤 𝑉 = 𝑊 The weight of the barge is: 𝑊 = (40 + 40)𝑡𝑡𝑡𝑡 = 80 𝑡𝑡𝑡𝑡 × 2000 𝑙𝑙𝑙 𝑡𝑡𝑡 = 160000 𝑙𝑙𝑙 The volume of water displaced is then: 𝑉 = 𝑊 𝛾 𝑤𝑤𝑤𝑤𝑤 = 160000 𝑙𝑙𝑙 62.4 𝑙𝑙𝑙 𝑓𝑓3 = 2564 𝑓𝑓3 The volume in terms of the draft d is: ∀= 𝐴 𝑐 𝐿 = �40𝑓𝑓 + 40𝑓𝑓 + 2 × 5 8 𝑑� × 𝑑 2 × 20𝑓𝑓 = 800𝑑 + 12.5𝑑2 Thus we have the relation: 800𝑑 + 12.5𝑑2 = 2564
  • 117. Solving this equation we have for the draft: 𝑑 = 3.06 𝑓𝑓
  • 118. Problem *3.94 [Difficulty: 2] Given: Experiment performed by Archimedes to identify the material conent of King Hiero's crown. The crown was weighed in air and in water. Find: Expression for the specific gravity of the crown as a function of the weights in water and air. Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fb ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) Fb Mg Ww Assumptions: (1) Static fluid (2) Incompressible fluid Taking a free body diagram of the body: ΣFz 0= Ww M g⋅− Fb+ 0= Ww is the weight of the crown in water. Ww M g⋅ Fbuoy−= M g⋅ ρw g⋅ Vd⋅−= However in air: Wa M g⋅= Therefore the weight measured in water is: Ww Wa ρw g⋅ Vd⋅−= so the volume is: Vd Wa Ww− ρw g⋅ = Now the density of the crown is: ρc M Vd = M ρw⋅ g⋅ Wa Ww− = Wa Wa Ww− ρw⋅= Therefore, the specific gravity of the crown is: SG ρc ρw = Wa Wa Ww− = SG Wa Wa Ww− = Note: by definition specific gravity is the density of an object divided by the density of water at 4 degrees Celsius, so the measured temperature of the water in the experiment and the data from tables A.7 or A.8 may be used to correct for the variation in density of the water with temperature. Problem 3.78 3.78
  • 119. Problem *3.96 [Difficulty: 2] Given: Balloons with hot air, helium and hydrogen. Claim lift per cubic foot of 0.018, 0.066, and 0.071 pounds force per cubic f for respective gases, with the air heated to 150 deg. F over ambient. Find: (a) evaluate the claims of lift per unit volume (b) determine change in lift when air is heated to 250 deg. F over ambient. Solution: We will apply the hydrostatics equations to this system. Governing Equations: L ρa g⋅ V⋅ ρg g⋅ V⋅−= (Net lift force is equal to difference in weights of air and gas) p ρ R⋅ T⋅= (Ideal gas equation of state) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Ideal gas behavior The lift per unit volume may be written as: LV L V g ρa ρg−( )⋅== ρa g⋅ 1 ρg ρa − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= now if we take the ideal gas equation and we take into account that the pressure inside and outside the balloon are equal: L V ρa g⋅ 1 Ra Ta⋅ Rg Tg⋅ − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= γa 1 Ra Ta⋅ Rg Tg⋅ − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= At standard conditions the specific weight of air is: γa 0.0765 lbf ft 3 ⋅= the gas constant is: Ra 53.33 ft lbf⋅ lbm R⋅ ⋅= and Ta 519 R⋅= For helium: Rg 386.1 ft lbf⋅ lbm R⋅ ⋅= Tg Ta= and therefore: LVHe 0.0765 lbf ft 3 ⋅ 1 53.33 386.1 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×= LVHe 0.0659 lbf ft 3 ⋅= For hydrogen: Rg 766.5 ft lbf⋅ lbm R⋅ ⋅= Tg Ta= and therefore: LVH2 0.0765 lbf ft 3 ⋅ 1 53.33 766.5 −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×= LVH2 0.0712 lbf ft 3 ⋅= For hot air at 150 degrees above ambient: Rg Ra= Tg Ta 150 R⋅+= and therefore: LVair150 0.0765 lbf ft 3 ⋅ 1 519 519 150+ −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×= LVair150 0.0172 lbf ft 3 ⋅= The agreement with the claims stated above is good. For hot air at 250 degrees above ambient: Rg Ra= Tg Ta 250 R⋅+= and therefore: LVair250 0.0765 lbf ft 3 ⋅ 1 519 519 250+ −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ×= LVair250 0.0249 lbf ft 3 ⋅= LVair250 LVair150 1.450= Air at ΔT of 250 deg. F gives 45% more lift than air at ΔT of 150 deg.F! Problem 3.79 3.79
  • 120. Problem *3.98 [Difficulty: 3] Fbuoyancy Wload y Whot air Given: Data on hot air balloon Find: Maximum mass of balloon for neutral buoyancy; mass for initial acceleration of 2.5 ft/s2. Assumptions: Air is treated as static and incompressible, and an ideal gas Solution: Basic equation FB ρatm g⋅ V⋅= and ΣFy M ay⋅= Hence ΣFy 0= FB Whotair− Wload−= ρatm g⋅ V⋅ ρhotair g⋅ V⋅− M g⋅−= for neutral buoyancy M V ρatm ρhotair−( )⋅= V patm⋅ R 1 Tatm 1 Thotair − ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅= M 320000 ft 3 ⋅ 14.7× lbf in 2 ⋅ 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 × lbm R⋅ 53.33 ft⋅ lbf⋅ × 1 48 460+( ) R⋅ 1 160 460+( ) R⋅ −⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ×= M 4517 lbm⋅= Initial acceleration ΣFy FB Whotair− Wload−= ρatm ρhotair−( ) g⋅ V⋅ Mnew g⋅−= Maccel a⋅= Mnew 2 ρhotair⋅ V⋅+( ) a⋅= Solving for Mnew ρatm ρhotair−( ) g⋅ V⋅ Mnew g⋅− Mnew 2 ρhotair⋅ V⋅+( ) a⋅= Mnew V ρatm ρhotair−( ) g⋅ 2 ρhotair⋅ a⋅− a g+ ⋅= V patm⋅ a g+ g 1 Tatm 1 Thotair −⎛ ⎜ ⎝ ⎞ ⎟ ⎠ ⋅ 2 a⋅ Thotair −⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Mnew 320000 ft 3 ⋅ 14.7⋅ lbf in 2 ⋅ 12 in⋅ ft ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 ⋅ lbm R⋅ 53.33 ft⋅ lbf⋅ ⋅ s 2 2.5 32.2+( ) ft⋅ ⋅ 32.2 1 48 460+( ) 1 160 460+( ) −⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ 2 2.5⋅ 1 160 460+( ) ⋅−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅ ft s 2 R⋅ ⋅= Mnew 1239 lbm⋅= To make the balloon move up or down during flight, the air needs to be heated to a higher temperature, or let cool (or let in ambient air). Problem 3.80 3.80
  • 121. Problem 3.81 (Difficulty: 2) 3.81 The opening in the bottom of the tank is square and slightly less than 2 𝑓𝑓 on each side. The opening is to be plugged with a wooden cube 2 𝑓𝑓 on a side. (a) What weight 𝑊 should be attached to the wooden cube to insure successful plugging of the hole? The wood weighs 40 𝑙𝑙𝑙 𝑓𝑓3, (b) What upward force must be exerted on the block to lift it and allow water to drain from the tank? Find: The weight of the block and the force needed to lift it Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 (a) Because the wood bottom surface is in the atmosphere so the pressure on the bottom surface is zero in this case and there is no buoyancy force. The force acting on the wood cube in the vertical direction is: 𝐹𝑉 = 𝐹𝑝 + 𝐺 𝐹𝑉 = 𝛾ℎ1 𝐴 + 𝐺 = 62.4 𝑙𝑙𝑙 𝑓𝑓3 × 5 𝑓𝑓 × 2𝑓𝑓 × 2𝑓𝑓 + 40 𝑙𝑙𝑙 𝑓𝑓3 × (2 𝑓𝑓)3 = 1568 𝑙𝑙𝑙
  • 122. The direction of 𝐹𝑉 is downward. So we do not need any weight 𝑊 attached to the wood cube. (b) To lift the block, we need a force larger than 𝐹𝑉, so we have: 𝐹𝑢𝑢 ≥ 𝐹𝑉 = 1568 𝑙𝑙𝑙
  • 123. Problem 3.82 (Difficulty: 2) 3.82 A balloon has a weight (including crew but not gas) of 2.2 𝑘𝑘 and a gas-bag capacity of 566 𝑚3 . At the ground it is (partially) inflated with 445 𝑁 of helium. How high can this balloon rise in the U.S standard atmosphere if the helium always assumes the pressure and temperature of the atmosphere? Find: How high this balloon will rise. Assumptions: Fluid is incompressible and static Solution: Apply the hydrostatic relations: Pressure as a function of depth ∆𝑝 = 𝜌𝜌ℎ Buoyancy force: 𝐹𝑏 = 𝜌 𝑔 𝑉 At the sea level, for helium we have: 𝑝 = 101.3 𝑘𝑘𝑘 𝑇 = 288 𝐾 𝑅 = 2076.8 𝐽 𝑘𝑘 ∙ 𝐾 According to the ideal gas law: 𝜌ℎ = 𝑝 𝑅𝑅 = 101.3 𝑘𝑘𝑘 2076.8 𝐽 𝑘𝑔 ∙ 𝐾 × 288 𝐾 = 0.1694 𝑘𝑘 𝑚3 𝛾ℎ = 𝜌𝜌 = 0.1694 𝑘𝑘 𝑚3 × 9.81 𝑚 𝑠2 = 1.662 𝑁 𝑚3 The volume of the helium is: 𝑉ℎ = 𝑊ℎ 𝛾ℎ = 445 𝑁 1.662 𝑁 𝑚3 = 268 𝑚3 The buoyancy force is calculated by: 𝐹𝐵 = 𝛾 𝑎𝑎𝑎 𝑉𝑏 The weight of the whole balloon is:
  • 124. 𝑊 = 2.2 𝑘𝑘 + 𝑊ℎ We have the following table as (the helium always has the same temperature and pressure as the atmosphere): Altitude (km) Pressure (kPa) Temperature (K) ∀ (𝑚3) 𝛾 𝑎𝑎𝑎 � 𝑁 𝑚3 � 𝑊ℎ (𝑘𝑘) 𝐹𝐵 (𝑘𝑘) 𝑊 (𝑘𝑘) 6 47.22 249.2 497 6.46 0.445 3.21 2.65 8 35.70 236.3 566 5.14 0.402 2.91 2.60 10 26.50 223.4 566 4.04 0.317 2.29 2.52 When the maximum volume of the helium is reached, the volume will become a constant for helium. Equilibrium is reached as: 𝐹𝐵 = 𝑊 At 8 𝑘𝑘 we have: 𝐹𝐵 − 𝑊 = 0.31 𝑘𝑘 At 10 𝑘𝑘 we have: 𝐹𝐵 − 𝑊 = −0.23 𝑘𝑘 With the interpolation we have the height for equilibrium as: ℎ = 8𝑘𝑘 + 2𝑘𝑘 × 0.31 0.31 + 0.23 = 9.15 𝑘𝑘
  • 125. Problem *3.100 [Difficulty: 3] Given: A pressurized balloon is to be designed to lift a payload of mass M to an altitude of 40 km, where p = 3.0 mbar and T = -25 deg C. The balloon skin has a specific gravity of 1.28 and thickness 0.015 mm. The gage pressure of the helium is 0.45 mbar. The allowable tensile stress in the balloon is 62 MN/m2 t M D Find: (a) The maximum balloon diameter (b) The maximum payload mass Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to mass of displaced fluid) p ρ R⋅ T⋅= (Ideal gas equation of state) πD2 ∆p/4 πDtσ Assumptions: (1) Static, incompressible fluid (2) Static equilibrium at 40 km altitude (3) Ideal gas behavior The diameter of the balloon is limited by the allowable tensile stress in the skin: ΣF π 4 D 2 ⋅ ∆p⋅ π D⋅ t⋅ σ⋅−= 0= Solving this expression for the diameter: Dmax 4 t⋅ σ⋅ ∆p = Fbuoyant Mbg Mg z Dmax 4 0.015× 10 3− × m⋅ 62× 10 6 × N m 2 ⋅ 1 0.45 10 3− ⋅ bar⋅ × bar m 2 ⋅ 10 5 N⋅ ×= Dmax 82.7m= To find the maximum allowable payload we perform a force balance on the system: ΣFz Fbuoy MHe g⋅− Mb g⋅− M g⋅−= 0= ρa g⋅ Vb⋅ ρHe g⋅ Vb⋅− ρs g⋅ Vs⋅− M g⋅− 0= Solving for M: M ρa ρHe−( ) Vb⋅ ρs Vs⋅−= The volume of the balloon is: Vb π 6 D 3 ⋅= The volume of the skin is: Vs π D 2 ⋅ t⋅= Therefore, the mass is: M π 6 ρa ρHe−( )⋅ D 3 ⋅ π ρs⋅ D 2 ⋅ t⋅−= The air density: ρa pa Ra T⋅ = ρa 3.0 10 3− × bar⋅ kg K⋅ 287 N⋅ m⋅ × 1 273 25−( ) K⋅ × 10 5 N⋅ bar m 2 ⋅ ×= ρa 4.215 10 3− × kg m 3 = Repeating for helium: ρHe p R T⋅ = ρHe 6.688 10 4− × kg m 3 = The payload mass is: M π 6 4.215 0.6688−( )× 10 3− × kg m 3 ⋅ 82.7 m⋅( ) 3 × π 1.28× 10 3 × kg m 3 ⋅ 82.7 m⋅( ) 2 × 0.015× 10 3− × m⋅−= M 638 kg= Problem 3.83 3.83
  • 126. Problem *3.102 [Difficulty: 3] Given: Glass hydrometer used to measure SG of liquids. Stem has diameter D=5 mm, distance between marks on stem is d=2 mm per 0.1 SG. Hydrometer floats in kerosene (Assume zero contact angle between glass and kerosene). Find: Magnitude of error introduced by surface tension. Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) d = 2 mm/0.1 SG ∆FB y D = 5 mm Kerosene Fσ Assumptions: (1) Static fluid (2) Incompressible fluid (3) Zero contact angle between ethyl alcohol and glass The surface tension will cause the hydrometer to sink ∆h lower into the liquid. Thus for this change: ΣFz ∆Fbuoy Fσ−= 0= The change in buoyant force is: ∆Fbuoy ρ g⋅ ∆V⋅= ρ g⋅ π 4 ⋅ D 2 ⋅ ∆h⋅= The force due to surface tension is: Fσ π D⋅ σ⋅ cos θ( )⋅= π D⋅ σ⋅= Thus, ρ g⋅ π 4 ⋅ D 2 ⋅ ∆h⋅ π D⋅ σ⋅= Upon simplification: ρ g⋅ D⋅ ∆h⋅ 4 σ= Solving for ∆h: ∆h 4 σ⋅ ρ g⋅ D⋅ = From Table A.2, SG = 1.43 and from Table A.4, σ = 26.8 mN/m Therefore, ∆h 4 26.8× 10 3− × N m ⋅ m 3 1430 kg⋅ × s 2 9.81 m⋅ × 1 5 10 3− × m⋅ × kg m⋅ s 2 N⋅ ×= ∆h 1.53 10 3− × m= So the change in specific gravity will be: ∆SG 1.53 10 3− × m⋅ 0.1 2 10 3− × m⋅ ×= ∆SG 0.0765= From the diagram, surface tension acts to cause the hydrometer to float lower in the liquid. Therefore, surface tension results in an indicated specific gravity smaller than the actual specific gravity. Problem 3.84 3.84
  • 127. Problem *3.103 [Difficulty:4] Given: Sphere partially immersed in a liquid of specific gravity SG. Find: (a) Formula for buoyancy force as a function of the submersion depth d (b) Plot of results over range of liquid depth Solution: We will apply the hydrostatics equations to this system. Governing Equations: Fbuoy ρ g⋅ Vd⋅= (Buoyant force is equal to weight of displaced fluid) Assumptions: (1) Static fluid (2) Incompressible fluid (3) Atmospheric pressure acts everywhere d Rsinθ R dθ θmax hWe need an expression for the displaced volume of fluid at an arbitrary depth d. From the diagram we see that: d R 1 cos θmax( )−( )= at an arbitrary depth h: h d R 1 cos θ( )−( )⋅−= r R sin θ( )⋅= So if we want to find the volume of the submerged portion of the sphere we calculate: Vd 0 θmax hπ r 2⌠ ⎮ ⌡ d= π 0 θmax θR 2 sin θ( )( ) 2 ⋅ R⋅ sin θ( )⋅ ⌠ ⎮ ⌡ d⋅= π R 3 ⋅ 0 θmax θsin θ( )( ) 3⌠ ⎮ ⌡ d⋅= Evaluating the integral we get: Vd π R 3 ⋅ cos θmax( )( )3 3 cos θmax( )− 2 3 + ⎡⎢ ⎢ ⎣ ⎤⎥ ⎥ ⎦ ⋅= Now since: cos θmax( ) 1 d R −= we get: Vd π R 3 ⋅ 1 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ − 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= Thus the buoyant force is: Fbuoy ρw SG⋅ g⋅ π⋅ R 3 ⋅ 1 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ 3 ⋅ 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ − 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= If we non-dimensionalize by the force on a fully submerged sphere: Fd Fbuoy ρw SG⋅ g⋅ 4 3 ⋅ π⋅ R 3 ⋅ = 3 4 1 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ 3 ⋅ 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ − 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = Fd 3 4 1 3 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ 3 ⋅ 1 d R −⎛ ⎜ ⎝ ⎞ ⎠ − 2 3 + ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ = 0.0 0.5 1.0 1.5 2.0 0.0 0.5 1.0 Submergence Ratio d/R ForceRatioFd Problem 3.85 3.85
  • 128. Problem *3.106 [Difficulty: 4] FB W FL FU x y Given: Data on sphere and tank bottom Find: Expression for SG of sphere at which it will float to surface; minimum SG to remain in position Assumptions: (1) Water is static and incompressible (2) Sphere is much larger than the hole at the bottom of the tank Solution: Basic equations FB ρ g⋅ V⋅= and ΣFy FL FU− FB+ W−= where FL patm π⋅ a 2 ⋅= FU patm ρ g⋅ H 2 R⋅−( )⋅+⎡⎣ ⎤⎦ π⋅ a 2 ⋅= FB ρ g⋅ Vnet⋅= Vnet 4 3 π⋅ R 3 ⋅ π a 2 ⋅ 2⋅ R⋅−= W SG ρ⋅ g⋅ V⋅= with V 4 3 π⋅ R 3 ⋅= Now if the sum of the vertical forces is positive, the sphere will float away, while if the sum is zero or negative the sphere will stay at the bottom of the tank (its weight and the hydrostatic force are greater than the buoyant force). Hence ΣFy patm π⋅ a 2 ⋅ patm ρ g⋅ H 2 R⋅−( )⋅+⎡⎣ ⎤⎦ π⋅ a 2 ⋅− ρ g⋅ 4 3 π⋅ R 3 ⋅ 2 π⋅ R⋅ a 2 ⋅−⎛ ⎜ ⎝ ⎞ ⎠ ⋅+ SG ρ⋅ g⋅ 4 3 ⋅ π⋅ R 3 ⋅−= This expression simplifies to ΣFy π ρ⋅ g⋅ 1 SG−( ) 4 3 ⋅ R 3 ⋅ H a 2 ⋅−⎡ ⎢ ⎣ ⎤ ⎥ ⎦ ⋅= ΣFy π 1.94× slug ft 3 ⋅ 32.2× ft s 2 ⋅ 4 3 1 0.95−( )× 1 in⋅ ft 12 in⋅ ×⎛ ⎜ ⎝ ⎞ ⎠ 3 × 2.5 ft⋅ 0.075 in⋅ ft 12 in⋅ ×⎛ ⎜ ⎝ ⎞ ⎠ 2 ×− ⎡ ⎢ ⎣ ⎤ ⎥ ⎦ × lbf s 2 ⋅ slug ft⋅ ×= ΣFy 0.012− lbf⋅= Therefore, the sphere stays at the bottom of the tank. Problem 3.86 3.86
  • 129. Problem *3.108 [Difficulty: 3] H = 8 ft h = 7 ft θ = 60o Floating Sinking Given: Data on boat Find: Effective density of water/air bubble mix if boat sinks Solution: Basic equations FB ρ g⋅ V⋅= and ΣFy 0= We can apply the sum of forces for the "floating" free body ΣFy 0= FB W−= where FB SGsea ρ⋅ g⋅ Vsubfloat⋅= Vsubfloat 1 2 h⋅ 2 h⋅ tan θ⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ L⋅= L h 2 ⋅ tan θ( ) = SGsea 1.024= (Table A.2) Hence W SGsea ρ⋅ g⋅ L⋅ h 2 ⋅ tan θ( ) = (1) We can apply the sum of forces for the "sinking" free body ΣFy 0= FB W−= where FB SGmix ρ⋅ g⋅ Vsub⋅= Vsubsink 1 2 H⋅ 2 H⋅ tan θ⋅ ⎛ ⎜ ⎝ ⎞ ⎠ ⋅ L⋅= L H 2 ⋅ tan θ( ) = Hence W SGmix ρ⋅ g⋅ L⋅ H 2 ⋅ tan θ( ) = (2) Comparing Eqs. 1 and 2 W SGsea ρ⋅ g⋅ L⋅ h 2 ⋅ tan θ( ) = SGmix ρ⋅ g⋅ L⋅ H 2 ⋅ tan θ( ) = SGmix SGsea h H ⎛ ⎜ ⎝ ⎞ ⎠ 2 ⋅= SGmix 1.024 7 8 ⎛ ⎜ ⎝ ⎞ ⎠ 2 ×= SGmix 0.784= The density is ρmix SGmix ρ⋅= ρmix 0.784 1.94× slug ft 3 ⋅= ρmix 1.52 slug ft 3 ⋅= Problem 3.87 3.87
  • 130. Problem *3.112 [Difficulty: 2] Given: Steel balls resting in floating plastic shell in a bucket of water Find: What happens to water level when balls are dropped in water Solution: Basic equation FB ρ Vdisp⋅ g⋅= W= for a floating body weight W When the balls are in the plastic shell, the shell and balls displace a volume of water equal to their own weight - a large volume because the balls are dense. When the balls are removed from the shell and dropped in the water, the shell now displaces only a small volume of water, and the balls sink, displacing only their own volume. Hence the difference in displaced water before and after moving the balls is the difference between the volume of water that is equal to the weight of the balls, and the volume of the balls themselves. The amount of water displaced is significantly reduced, so the water level in the bucket drops. Volume displaced before moving balls: V1 Wplastic Wballs+ ρ g⋅ = Volume displaced after moving balls: V2 Wplastic ρ g⋅ Vballs+= Change in volume displaced ∆V V2 V1−= Vballs Wballs ρ g⋅ −= Vballs SGballs ρ⋅ g⋅ Vballs⋅ ρ g⋅ −= ∆V Vballs 1 SGballs−( )⋅= Hence initially a large volume is displaced; finally a small volume is displaced (∆V < 0 because SGballs > 1) Problem 3.88 3.88
  • 131. Problem *3.113 [Difficulty: 4] Open-Ended Problem Statement: A proposed ocean salvage scheme involves pumping air into “bags” placed within and around a wrecked vessel on the sea bottom. Comment on the practicality of this plan, supporting your conclusions with analyses. Discussion: This plan has several problems that render it impractical. First, pressures at the sea bottom are very high. For example, Titanic was found in about 12,000 ft of seawater. The corresponding pressure is nearly 6,000 psi. Compressing air to this pressure is possible, but would require a multi-stage compressor and very high power. Second, it would be necessary to manage the buoyancy force after the bag and object are broken loose from the sea bed and begin to rise toward the surface. Ambient pressure would decrease as the bag and artifact rise toward the surface. The air would tend to expand as the pressure decreases, thereby tending to increase the volume of the bag. The buoyancy force acting on the bag is directly proportional to the bag volume, so it would increase as the assembly rises. The bag and artifact thus would tend to accelerate as they approach the sea surface. The assembly could broach the water surface with the possibility of damaging the artifact or the assembly. If the bag were of constant volume, the pressure inside the bag would remain essentially constant at the pressure of the sea floor, e.g., 6,000 psi for Titanic. As the ambient pressure decreases, the pressure differential from inside the bag to the surroundings would increase. Eventually the difference would equal sea floor pressure. This probably would cause the bag to rupture. If the bag permitted some expansion, a control scheme would be needed to vent air from the bag during the trip to the surface to maintain a constant buoyancy force just slightly larger than the weight of the artifact in water. Then the trip to the surface could be completed at low speed without danger of broaching the surface or damaging the artifact. Fox and McDonalds Introduction to Fluid Mechanics 9th Edition Pritchard Solutions Manual Full Download: https://alibabadownload.com/product/fox-and-mcdonalds-introduction-to-fluid-mechanics-9th-edition-pritchard-so This sample only, Download all chapters at: AlibabaDownload.com Problem 3.89 3.89