005 first law

005 first law

• 1.
  LECTURE UNIT 005 BASICLAW OF THERMODYNAMICS 1. Zeroth Law of Thermodynamics States that when two bodies are in thermal equilibrium with a third body, the two bodies are in thermal equilibrium with each other. ILLUSTRATION: TS1 TS2 TS3 T1 = T2 = T3 2. First Law of Thermodynamics (Law of Conservation of Energy - states that energy itself cannot be created nor destroyed but only transformed from one form of energy to another. The sum of energies going out of the system must equal the sum of the energies going into the system. FIRST LAW APPLYING TO AN OPEN OR STEADY-FLOW SYSTEM ILLUSTRATION: PE1 KE1 WSF U1 n Wf1 TS PE1 KE1 U1 Wf1 Q [ Ein = Eout] PE1 + KE1 + U1 + Wf1 + Q = PE2 + KE2 + U2 + Wf2 + WSF Q = (PE1 - PE2) + (KE2 - KE1) + (U2 - U1) + (Wf2 - Wf1) + WSF Q= PE + KE + U+ Wf + WSF Steady Flow Energy Equation (SFEE) Recall: H= U+ Wf Q= PE + KE + H + WSF When: PE ˜ 0 (z1 ˜ z2) KE ˜ 0 (v1 ˜ v2) ˜ ˜ Q= H + WSF In differential form we can write: dQ = dH + dWSF NOTE: GENERAL FORMULA Q = T ds WSF = - V dP dQ = T ds dWSF = -V dP “Coming together is a beginning; keeping together is progress; working together is success.”
• 2.
  FIRST LAW APPLYINGTO CLOSED OR NON-FLOW SYSTEM ILLUSTRATION: m State 1: V Q m State 2: WNF V Q [ Ein = Eout] Initial energy stored within Initial energy entering the Final energy stored within Final energy leaving the + = + the system system the system system U1 + Q = U2 + WNF Q = (U2 - U1) + WNF Q= U + WNF Non-Flow Energy Equation (NFEE) In differential form we can write: dQ = dU + dWNF NOTE: GENERAL FORMULA Q = T ds WNF = P dV dQ = T ds dWNF = P dV PROBLEM SET: 1. A fluid at 100 psi has a specific volume of 4 ft3/lb and enters an apparatus with a velocity of 500 ft/s. Heat radiation losses in the apparatus are equal to 10 Btu/lb of fluid supplied. The fluid leaves the apparatus at 20 psia with a specific volume of 15 ft3/lb and a velocity of 1000 ft/s. In the apparatus, the shaft work done by the fluid is equal to 195,000 ft- lbf/lbm. Does the the internal energy of the fluid increases or decreases, and how much is the change? [ ] 2. Shaft work of -15 Btu/lb and heat transfer of -10 Btu/lb. What is the change in enthalpy of this system. [ ] 3. Air is contained in a vertical piston-cylinder assembly by a piston of mass 50 kg and having a face area of 0.01 m2. The mass of the air is 4 g, and initially the air occupies a volume of 5 L. The atmosphere exerts a pressure of 100 kPa on the top of the piston. Heat transfer of magnitude 1.41 kJ occurs slowly from the air to the surroundings, and the volume of the air decreases to 0.0025 m3. neglecting friction between the piston and the cylinder wall, determine the change in specific internal energy of the air, in kJ/kg. [ ] 4. A rigid steel tank contains a mixture of vapor and liquid water at a temperature of 65oC. The tank has a volume of 0.5 m3, the liquid phase occupying 30% of the volume. Determine the amount of heat added to the system to raise the pressure to 3.5 MPa. [ ] 5. In a steady flow apparatus, 135 kJ of work is done by each kg of fluid. The specific volume of the fluid, pressure and speed at the inlet are 0.37 m3/kg, 600 kPa, and 16 m/s. The inlet is 32 m above the floor, and the discharge pipe is at floor level. The discharge conditions are 0.62 m3/kg, 100 kPa, and 270 m/s. The total heat loss between the inlet and discharge is 9 kJ/ kg of fluid. In flowing through this apparatus, does the specific internal energy increases or decrease, and by how much? [ ] “People who live for self never succeed in satisfying self or anybody else.”
• 3.
  6. A fluid enters with a steady flow of 3.7 kg/s and an initial pressure of 690 kPa, an initial density of 3.2 kg/m3, an initial velocity of 60 m/s, and an initial internal energy of 2000 kJ/kg. It leaves at 172 kPa, ρ = 0.64 kg/m3, v = 160 m/s, and u = 1950 kJ/kg. The heat loss is to be found to be 18.6 kJ/kg. Find the power in kW. [ ] 7. An insulated 2-kg box falls from a balloon 3.5 km above the earth. What is the change in internal energy of the box after it has hit the earth surface? [ ] 8. A fluid at 700 kPa, with a specific volume of 0.25 m3/kg and a velocity of 175 m/s, enters a device. Heat loss from the device by radiation is 23 kJ/kg. The work done by the fluid is 465 kJ/kg. The fluid exists at 136 kPa, 0.94 m3/kg, and 335 m/s. Determine the change in internal energy. [ ] 9. A 0.3 m3 rigid tank receives paddle work at a rate of 4.5 W for 30 minutes. Initially, the gas has a density of 1.4 kg/m3. Find the specific volume at the final state and the change of specific internal energy. [ ] 10. A 25-cm cube at 0oC melts in a region where the barometric pressure is 759 mmHg. The density of liquid water at 0oC is 0.998 gm/cm3 and that of ice is 0.914 gm/cm3. Is there any work done on the ice by the surroundings? [ ] “You can get everything you want if you help others get what they want.” Zig Ziglar
• 4.
  ft lbf Q = -10 Btu (1 lbm)( 778 ) lbm Btu Q = -7780 ft - lbf 1lbm KE = 1 m (v22 - v12) = 1 [ (1000 ft/s)2 - (500 ft/s)2] 2 gc 2 lbm - ft ( 32.2 lbf - sec2 ) KE = 11645.9627 ft - lbf lbf 144 in2 ft3 lb 2 ft3 Wf = m(P2v2 - P1v1 ) = (1 lbm) 20 in 2 ( 1 ft2 )(15 lbm ) - 100 in2f ( 144 2in )(4 lbm 1 ft ) Wf = -14400 ft - lbf ft - lbf WSF = 195000 (1 lbm) lbm WSF = 195000 ft - lbf Substituting: - 7780 ft - lbf = 11645.9627 ft - lbf + U - 14400 ft - lbf + 195000 ft - lbf U = - 200025.9627 ft - lbf ( 1 Btu 778 ft - lbf ) U = - 257.1028 Btu There is a decrease in Internal Energy 6.) Given: m = 3.7 kg/s m = 3.7 kg/s P1 = 690 kPa P2 = 172 kPa ρ1 = 3.2 kg/m3 ρ2 = 0.64 kg/m3 v1 = 60 m/s v2 = 160 m/s u1 = 2000 kJ/kg u1 = 1950 kJ/kg Q = 18.6 kJ/kg Required: Power, kW Solution: Using SFEE: (Steady Flow Energy Equation) Q= PE + KE + U + Wf + WSF kg Q = - 18.6 kJ (3.7 ) kg s Q = - 68.82 kJ s kg 3.7 s KE = 1 m (v22 - v12) = 1 [ (160 m/s)2 - (60 m/s)2] 2 gc 2 kg - m 1000 N (1 )( ) N - s2 1 kN KE = 40.7 kJ s
• 5.
  WSF 121.0825 kW 9.) Given: V = 0.3 m3 WNF = - 4.5 W = - 4.5 J (30 min) ( 60 s ) = - 8100 J s 1 min ρ1 = 1.4 kg/m3 Required: a.) Specific volume at final state, v2 Assuming t1 = t2 v2 = 1 = ρ 1 1.4 kg/m3 v2 = 0.7143 m3/kg b.) Change in specific internal energy, u Using NFEE (Non-flow Energy Equation) Q= U + WNF 0 = m u + WNF Solving for m; ρ= m V m 1.4 kg/m3 = 0.3 m3 m = 0.42 kg 0 = (0.42 kg) u - 8100 J u = 19285.7143 J/kg u = 19.2857 kJ/kg