Heat Transfer Final Project, Peden

Heat Transfer Final Project, Peden

• 1.
  Drew Peden ME 321-402Final Project 11.76 Known: In a Rankine power system steam is passed over 100 thin walled tubes 0.01 m in diameter in the condenser which is a shell-and-tube heat exchanger. The steam enters as a saturated vapor at 0.51 bar and leaves as a saturated liquid. 𝑚̇ 𝑠𝑡 = 1.5 𝑘𝑔 𝑠 ; 𝑇𝑐,𝑖 = 280 𝐾 ; 𝑚̇ 𝑐𝑤 = 15 𝑘𝑔 𝑠 ; ℎ̅ 𝑜 = 5000 𝑊 𝑚2 ∙ 𝐾 Properties of Cooling Water 𝑐 𝑝 ( 𝐽 𝑘𝑔 ∙ 𝐾 ) 𝑘 ( 𝑊 𝑚 ∙ 𝐾 ) 𝜇 ( 𝑘𝑔 𝑠 ∙ 𝑚 ) Pr 4178 0.628 700 X 10−6 4.6 Assume: - Heat exchanger is well insulated (no heat loss to surroundings) - Constant Properties - Negligible tube wall thermal resistance - Fully developed flow throughout - Modeled as a reversible process Diagram: Figure 1: Rankine Cycle Figure 2: Temperature/Enthalpy Rankine Diagram
• 2.
  Find: (a) What isthe water outlet Temperature? In an ideal Rankine cycle, the heat dissipated in the condenser can be modelled as: (1) 𝑞 = 𝑚̇ (ℎ2 − ℎ3) Where 𝑚̇ is the mass flow rate of the hot working fluid; and ℎ2 and ℎ3 are the enthalpy values associated with the hot working fluid at the given pressure at points 2 and 3, in this case 0.51 bar and conveniently corresponding to the saturated liquid and saturated vapor states. These states are indicated by the numbered locations in figures 1 and 2 and can be obtained from steam tables. Plugging in values to equation 1: 𝑞ℎ𝑜𝑡 = 1.5 𝑘𝑔 𝑠 (2645.9 − 340.49) 𝑘𝐽 𝑘𝑔 = 3458.115 𝑘𝑊 𝑞ℎ𝑜𝑡 = 𝑞 𝑐𝑜𝑙𝑑 so to find the water outlet temperature we can use (2) 𝑞 𝑐𝑜𝑙𝑑 = 𝑚̇ 𝑤 𝑐 𝑝,𝑤(𝑇𝑜 − 𝑇𝑖) We know all values but 𝑇𝑜 so just need to rearrange, plug in values and solve: 𝑇𝑜 = 𝑞 𝑚̇ 𝑤 𝑐 𝑝,𝑤 + 𝑇𝑖 = 3458115 𝑊 15 𝑘𝑔 𝑠 (4178 𝐽 𝑘𝑔∙𝐾 ) + 280 𝐾 (b) What is the required tube length? The general procedure will be to find the universal heat transfer coefficient (U) defined as: (3) 𝑈 = 1 1 ℎ 𝑜 + 1 ℎ 𝑖 Where ℎ 𝑜 and ℎ 𝑖 are the convection coefficients on the outside and inside of the thin walled tubes. The outside transfer coefficient is known. In order to find the inside coefficient we must find the Nusselt number, Nu, defined as: (4) 𝑁𝑢 = ℎ𝐷 𝑘 𝑓 The Nusselt number is a function of the Reynolds and Pruitt number. The Pruitt number of the cooling water is known so the Reynolds number needs to be found. For a circular pipe with fully developed flow the Reynolds number is (4) 𝑅𝑒 𝐷 = 4𝑚̇ 𝜋𝐷𝜇 𝑇𝑜 = 335.12 𝐾 = 62° C
• 3.
  All the valuesare known so 𝑅𝑒 𝐷 = 4𝑚̇ 𝜋𝐷𝜇 = 4 ( 15 𝑘𝑔 100 𝑡𝑢𝑏𝑒𝑠 𝑠 ) 𝜋(0.01𝑚)(700 X 10−6 𝑘𝑔 𝑚 ∙ 𝑠 ) = 27,283.7 For internal flow a Reynolds number 𝑅𝑒 ≥ 2300 is associated with turbulent flow. Looking at Table 8-41 it can be seen that the appropriate equation for the Nusselt number is (5) 𝑁𝑢 𝐷 = 0.023𝑅𝑒4/5 𝑃𝑟 𝑛 Where n = 0.4 for cold to hot convection. So the Nusselt number is 𝑁𝑢 𝐷 = 0.023𝑅𝑒 4 5 𝑃𝑟 𝑛 = 0.023 (27283.7 4 5)4.60.4 = 150 The inside convection coefficient can be calculated from Equation 4 as ℎ 𝑖 = 𝑁𝑢 ∙ 𝑘 𝑓 𝐷 = 150 (0.628 𝑊 𝑚 ∙ 𝐾 ) 0.01 𝑚 = 9,420 𝑊 𝑚2 ∙ 𝐾 Now that the convection coefficients are known the universal heat transfer coefficient can be calculated from Equation 3: 𝑈 = 1 1 ℎ 𝑜 + 1 ℎ 𝑖 = 1 1 5000 + 1 9420 = 3,266 𝑊 𝑚2 ∙ 𝐾 To find the length of each tube we must use the effectiveness method from the equation (6) 𝑁𝑇𝑈( 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑢𝑛𝑖𝑡𝑠) = 𝑈𝐴 𝐶 𝑚𝑖𝑛 In order to do so we first need to find the heat capacitance C of each fluid which will give us 𝐶 𝑚𝑖𝑛 and a heat capacitance ratio that will allow to find NTU. (7) 𝐶 = 𝑚̇ 𝑐 𝑝 Because the hot fluid undergoes a phase change the temperature does not change. Specific heat is defined as 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑒𝑛𝑒𝑟𝑔𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 so 𝑐 𝑝,ℎ = ∞ 𝐶ℎ = 𝑚ℎ̇ 𝑐 𝑝,ℎ = 1.5 𝑘𝑔 𝑠 ∞ = ∞ 𝐶 𝑐 = 𝑚 𝑐̇ 𝑐 𝑝,𝑐 = 15 𝑘𝑔 𝑠 (4178 𝐽 𝑘𝑔 ∙ 𝐾 ) = 62,670 𝑊 𝐾
• 4.
  These results showthat 𝐶 𝑐 = 𝐶 𝑚𝑖𝑛 because it is smaller. The heat capacity ratio will give us our NTU: (8) 𝐶𝑟 = 𝐶 𝑚𝑖𝑛 𝐶 𝑚𝑎𝑥 = 62,670 ∞ = 0 We can now consult table 8-41 and observe that when 𝐶 𝑟 = 0, NTU is given by: (9) 𝑁𝑇𝑈 = −𝑙𝑛(1 − 𝜀) The efficiency, ε, is needed still and is given by: (10) 𝜀 = 𝑞 𝑞 𝑚𝑎𝑥 Where q is the previously found heat transfer rate and 𝑞 𝑚𝑎𝑥 is: (11) 𝑞 𝑚𝑎𝑥 = 𝐶 𝑚𝑖𝑛( 𝑇ℎ,𝑖 − 𝑇𝑐,1) = 62,670 𝑊 𝐾 (354.48 𝐾 − 280 𝐾) = 4,667.66 𝑘𝑊 So substituting in our efficiency into equation 9, our NTU becomes 𝑁𝑇𝑈 = −𝑙𝑛 (1 − 3458.115 𝑘𝑊 4.667.66 𝑘𝑊 ) = 1.35 We can finally solve for our tube length by substituting values into equation 6 and solving for L: 𝑁𝑇𝑈 = 𝑈𝐴 𝐶 𝑚𝑖 𝑛 = 𝑈𝜋𝐷𝐿 𝐶 𝑚𝑖𝑛 𝐿 = 𝑁𝑇𝑈 ∙ 𝐶 𝑚𝑖𝑛 𝑈𝜋𝐷 = (1.35) (62,670 𝑊 𝐾 ) 3,266 𝑊 𝑚2 ∙ 𝐾 ∙ 𝜋(0.01𝑚) = 824.57 𝑚 (c) With an accumulated fouling factor of 𝑅 𝑓 = 0.0003 𝑚2∙K 𝑊 , and using calculated length and inlet conditions, what mass fraction of vapor is condensed? Because the previous un-fouled system condensed all of the steam into water we can consider the former condensed amount 100%. All we need to do is find the fouled efficiency and compare the ratios. The overall heat transfer coefficient with fouling is 𝑈𝑓 = ( 1 1 𝑈 + 𝑅 𝑓 ) = ( 1 1 3266 + 0.0003 ) = 1650 𝑊 𝑚2 ∙ 𝐾 824.57 100 𝑚 𝑡𝑢𝑏𝑒 = 8.25 𝑚 𝑡𝑢𝑏𝑒
• 5.
  Our new NTUcan be found using equation 6: 𝑁𝑇𝑈 = 𝑈𝐴 𝐶 𝑚𝑖𝑛 = 1650 ∙ 𝜋 ∙ (0.01)(824.57) 62,670 = 0.682 We can find our effectiveness by rearranging equation 9: 𝜀 𝑓 = −1 𝑒 𝑁𝑇𝑈 + 1 = −1 𝑒0.680 + 1 = 0.4944 The former efficiency was 𝜀 𝑢−𝑓 = 3458.115 4.667.66 = 0.7409 So our mass fraction condensed is (d) With tube length from (b) and fouling from (c), explore the extent to which the water flow rate and inlet temperature can be varied to improve the condenser performance. Show results graphically. *Note: I am very confused with this problem, if my 𝑚̇ changes, it effects many other variables. My argument will be that as 𝑚̇ 𝑐 changes, so does Reynolds, Nusselt, convection coefficient, overall heat transfer coefficient, NTU, and finally effectiveness. Data in Appendix 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 5 10 15 20 25 30 35 Effectiveness Mass Flow Rate of Cooling Water, kg/s Mass Flow Rate of Cooling Water vs. Effectiveness 𝜀 𝑢−𝑓 𝜀 𝑓 = 0.7409 0.4944 = 0.667
• 6.
  13.55 Known: Two concentric spheresare separated by an air space. 𝐷1 = 0.8 𝑚; 𝐷2 = 1.2 𝑚; 𝑇1 = 400 𝐾; 𝑇2 = 300 𝐾 Diagram: Assume: - Steady State Find: (a) If the surfaces are black,what is the net rate of radiation exchange between the surfaces? This can be calculated with: (11) 𝑞 = 𝐴𝜎(𝑇𝑠 4 − 𝑇𝑠𝑢𝑟 4 ) Where 𝜎 ≡ 𝑡ℎ𝑒 𝑆𝑡𝑒𝑓𝑎𝑛 − 𝐵𝑜𝑙𝑡𝑧𝑚𝑎𝑛𝑛 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 5.670 𝑋 10−8 𝑊 𝑚2∙𝐾4 . Solving equation 11: 𝑞 = 𝜋0.8𝑚2 (5.670 𝑋10−8 𝑊 𝑚2 ∙ 𝐾4 .) (400 𝐾4 − 300 𝐾4) (b) What is the net rate of radiation exchange if they are diffuse and gray and 𝜀1 = 0.5 and 𝜀2 = 0.05? From Table 13.31 the heat flux for concentric spheres is (12) 𝑞12 = 𝜎𝐴( 𝑇1 4 − 𝑇2 4) 1 𝜀1 + 1 − 𝜀2 𝜀2 ( 𝑟1 𝑟2 ) 2 So 𝑞 = 1995 𝑊
• 7.
  𝑞12 = (5.670 ∙10−8) 𝜋0.82(100 𝐾) 1 0.5 + 1 − 0.05 0.05 ( 0.8 1.2 ) 2 = 191 𝑊 (c) What is the net rate of radiation exchange if 𝐷2 is increased to 20 m, with 𝜀1 = 0.5 and 𝜀2 = 0.05, and 𝐷2 = 0.8 𝑚? Can just plug in additional value to equation 12 : 𝑞12 = (5.670 ∙ 10−8) 𝜋0.82(100 𝐾) 1 0.5 + 1 − 0.05 0.05 ( 0.8 20 ) 2 = 982.6 𝑊 What error introduced if 𝜀2 = 1? Using equation 12 again we get: 𝑞12 = (5.670 ∙ 10−8) 𝜋0.82(1.75 ∙ 1010 ) 1 0.5 + 1 − 1 1 ( 0.8 20 ) 2 = 997.5 𝑊 (d) For 𝐷2 = 1.2 𝑚 and emissivities of 𝜀1 = 0.1,0.5, 𝑎𝑛𝑑 1.0, compute and plot the net rate of radiation exchange as a function of 𝜀2 𝑓𝑜𝑟 0.05 ≤ 𝜀2 ≤ 1.0. 𝑞12 = 191 𝑊 𝑞12 = 982.6 𝑊 100 − 982.6 997.5 ∙ 100 = 1.5%
• 8.
  Data in Appendix 0 50 100 150 200 250 00.2 0.4 0.6 0.8 1 q, W ε2 Rate of Radiation, ε1 = 0.1 0 200 400 600 800 1000 1200 0 0.2 0.4 0.6 0.8 1 q, W ε2 Radiation Rate, ε1 = 0.5 0 500 1000 1500 2000 2500 0 0.2 0.4 0.6 0.8 1 q, W ε2 Radiation Rate, ε1 = 1.0
• 9.
  13.112 Known: There is adouble paned window separated by atmospheric air for which the critical Raleigh number for the onset of convection is 𝑅𝑎 = 2000. 𝑇1 = 22℃ ; 𝑇2 = −20℃ Diagram: Assume: - Isothermal Surfaces - Steady State - Quiescent Air - Air is dry and at 1 atm Find: (a) What is the conduction heat flux across the air gap for the optimal spacing 𝐿 𝑜𝑝 between the panes? For this first part we are just dealing with conduction heat flux through the air so need to first find the properties of the air at the mean temperature: 𝑇 𝑚 = 𝑇1 + 𝑇2 2 = 22 + (−20) 2 = 1℃ From Table A-41 𝜈 ∙ 106 ( 𝑚2 𝑠 ) 𝛼 ∙ 106 ( 𝑚2 𝑠 ) 𝛽 ( 𝐾−1)( 1 𝑇 𝑚 ) 𝑘 ∙ 103 ( 𝑊 𝑚 ∙ 𝐾 ) 13.67 19.2 0.003648 24.3 The critical Rayleigh Number, where free convection will turn turbulent, is given as 2000. Using the definition of the Rayleigh number I think I just need to solve for x: (13) 𝑅𝑎 = 𝑔𝛽(𝑇2 − 𝑇1)𝑥3 𝛼 ∙ 𝜈 𝑥 = √ 2000(19.2 ∙ 10−6)(13.64 ∙ 10−6) 9.8 𝑚 𝑠2 (0.003648 𝐾−1)(295.15 − 253.15) 3 = 0.00704 𝑚
• 10.
  So the conductionheat flux across it can be calculated with Fourier’s Law (14) 𝑞̇ = 𝑘 ∆𝑇 ∆𝑥 Plugging in: (b) If the glass has an emissivity of 𝜀 𝑔 = 0.90, what is the total heat flux across the gap? We can look at Table 13.31 again to get the equation we need parallel planes: (15) 𝑞12 = 𝜎𝐴( 𝑇1 4 − 𝑇2 4) 1 𝜀1 + 1 𝜀2 − 1 With 𝜀1 = 𝜀2 = 0.9 and solving for heat flux instead of heat rate: 𝑞̇12 = 5.670 ∙ 10−8 𝑊 𝑚2 ∙ 𝐾4 (295.15 𝐾4 − 253.15 𝐾4) 1 0.9 + 1 0.9 − 1 = 161.3 𝑊 𝑚2 So total heat flux is the conduction from part (a) plus the radiation: (c) What is the total heat flux if a special, low emissivity coating (𝜀1 = 0.1) is applied to one of the panes at its air glass interface? What is total heat flux if both panes are coated? One Pane Assuming the other pane still has an emissivity of 0.9 we can just plug into equation 15 with our new value: 𝑞̇12 = 5.670 ∙ 10−8 𝑊 𝑚2 ∙ 𝐾4 (295.15 𝐾4 − 253.15 𝐾4) 1 0.9 + 1 0.1 − 1 = 19.5 𝑊 𝑚2 𝑞̇ = (0.0243 𝑊 𝑚 ∙ 𝐾 ) 42℃ 0.00704 𝑚 = 145 𝑊 𝑚2 𝑞̇ 𝑛𝑒𝑡 = 161.3 + 145 = 306.3 𝑊 𝑚2
• 11.
  Then add tothe conduction: Two Panes Same procedure as with one pane but both emissivities equal to 0.1 𝑞̇12 = 5.670 ∙ 10−8 𝑊 𝑚2 ∙ 𝐾4 (295.15 𝐾4 − 253.15 𝐾4) 1 0.1 + 1 0.1 − 1 = 10.4 𝑊 𝑚2 And add to conduction 𝑞̇ 𝑛𝑒𝑡 = 19.5 + 145 = 164.5 𝑊 𝑚2 𝑞̇ 𝑛𝑒𝑡 = 10.4 + 145 = 155.4 𝑊 𝑚2
• 12.
  References: 1. Bergman, Lavine,Incropera, Dewitt, “Fundamentals of Heat and mass Transfer,” Seventh Edition, 2007
• 13.
  Appendix Data for Problem11.76 (d), mass flow rate effect m new Re Nu h cold Uf NTU effect 2 3636.364 29.88046 1876.493 968.1394 2.954471 0.947894 4 7272.727 52.0249 3267.164 1240.578 1.892936 0.849371 6 10909.09 71.95886 4519.016 1386.41 1.410303 0.755931 8 14545.45 90.58061 5688.462 1479.74 1.12893 0.676621 10 18181.82 108.2837 6800.219 1545.466 0.943259 0.610643 12 21818.18 125.2876 7868.064 1594.652 0.811066 0.555616 14 25454.55 141.7313 8900.724 1633.052 0.71194 0.509309 16 29090.91 157.71 9904.188 1663.984 0.634747 0.469931 18 32727.27 173.2931 10882.81 1689.509 0.572875 0.436098 20 36363.64 188.5329 11839.87 1710.98 0.52214 0.40675 22 40000 203.4705 12777.95 1729.326 0.479762 0.381069 24 43636.36 218.1385 13699.1 1745.208 0.443821 0.35842 26 47272.73 232.5637 14605 1759.109 0.412944 0.338301 28 50909.09 246.7685 15497.06 1771.39 0.386125 0.320314 30 54545.45 260.7716 16376.46 1782.33 0.362609 0.304142 32 58181.82 274.5891 17244.19 1792.145 0.341818 0.289523 34 61818.18 288.2348 18101.14 1801.006 0.323302 0.276245
• 14.
  Data for Problem13.55 (d) ε1 = 0.1 ε1 = 0.5 ε1 = 1.0 q ε2 q, W ε2 q, W ε2 108.167603 0.05 191.012183 0.05 211.237183 0.05 142.504072 0.1 332.502217 0.1 399.003192 0.1 159.367112 0.15 441.518854 0.15 567.004059 0.15 169.389349 0.2 528.09072 0.2 718.204597 0.2 176.031483 0.25 598.502394 0.25 855.004886 0.25 180.756732 0.3 656.892487 0.3 979.368622 0.3 184.290261 0.35 706.097568 0.35 1092.91799 0.35 187.032419 0.4 748.12687 0.4 1197.00479 0.4 189.222286 0.45 784.443423 0.45 1292.76455 0.45 191.011451 0.5 816.137848 0.5 1381.1581 0.5 192.500675 0.55 844.03976 0.55 1463.0039 0.55 193.75955 0.6 868.791444 0.6 1539.00352 0.6 194.837684 0.65 890.897899 0.65 1609.76173 0.65 195.771394 0.7 910.761662 0.7 1675.80268 0.7 196.587878 0.75 928.707537 0.75 1737.58289 0.75 197.307909 0.8 945.000497 0.8 1795.5018 0.8 197.947625 0.85 959.858853 0.85 1849.91044 0.85 198.519754 0.9 973.46409 0.9 1901.11854 0.9 199.034469 0.95 985.968322 0.95 1949.40045 0.95 199.5 1 997.5 1 1995 1