MAE 110A Homework 3: Solutions 10/20/2017
3.10: For H 2O, determine the specified property at the indicated state. Locate the state on a sketch of the T-v diagram. Given a) T 140 C, v 0.5 m 3 kg b) p 30MPa, T 100 C c) p 10MPa, T 485 C d) T 80 C, x 0.75 a) p, in bar, and locate on T-v diagram b) v, in m 3 kg, and locate on T v diagram c) v, in m 3 kg, and locate on T v diagram d) p, in bar, and v, in m 3 kg, and locate on T v diagram v 8 v 9 + x v ; v 9 a) From Table A-3 at T 140 C: v 9 1.0435 10 >? m 3 kg, v ; 1.673 m 3 kg. v 9 > v > v ;, therefore the state is in the two-phase liquid-vapor region as shown in the diagram below. The pressure is the saturation pressure at 140 C From Table A-3: p 3.613 bar p p sat, 140 C 3.613 bar T 140 C b) Pressure is higher than the critical pressure as shown in the diagram below (critical properties pulled from table A-3). Therefore, the state is in the compressed liquid region. From Table A-5: v 1.0290 m 3 kg p 30 MPa 300 bar p K 220.9 bar T K 374.14 C T 100 C 1
c) The given temperature is greater than the saturation temperature at 100 bar (saturation properties from table A-3). Therefore, the state is in the superheated vapor region. Interpolating from Table A-4: y y S + x x S v [SZ v XYSZ v v XYSZ + T T XYSZ T [SZ T XYSZ T U >T V 8 U >8 V 0.03160 m? kg + 485K 480K 0.03394 m? kg 0.03160 m? kg 520K 480K 0.03189 m? kg p 10 MPa 100 bar T 485 C T K 374.14 C T sat, 1SS bar 311.1 C d) The given data tell us the state is saturated water with a certain quality, as shown in the diagram below. From Table A-2: p 0.4739 bar Using equation 3.2 and data from Table A-2 v v 9 + x v ; v 9 1.0291 10 >? m? kg + 0.75 3.407 m? kg 1.0291 10 >? m? kg 2.556 m? kg p p sat, 80 C 0.4739 bar x 0.75 T 80 C 2
3.50: Refrigerant 22 undergoes a constant pressure process within a piston-cylinder assembly from saturated vapor at 4 bar to a final temperature of 30 C. Kinetic and potential energy effects are negligible. For the refrigerant, show the process on a p-v diagram. Evaluate the work and the heat transfer, each in kj per kg of refrigerant. Given Conversions Model p, sat p, sat 4 bar 1 bar 100 kpa Negligible potential and kinetic energy effects T 30 C 1 kj 1 kpa m 3 Refrigerant is the closed system Constant pressure process Volume change is the only work done w, specific work, in kj kg q, specific heat transfer, in kj kg p-v and T-v diagrams for the process q pq W pd w rst q Q m, w W m W pd w pdv p v v From Table A-8: v v ;, 4 bar 0.0581 m? kg From Table A-9: v 0.06872 m? kg 100 kpa w 4 bar 0.06872 m? kg 0.0581 m? kg 1 bar 4.248 kj kg ΔU mδu Q pq W rst Δu q pq w rst u u q pq w rst q pq u u + w rst From Table A-8: u u ;, 4 bar 224.24 kj kg From Table A-9: u 245.73 kj kg q pq 235.73 kj kg 224.24 kj kg + 4.248 kj kg 15.738 kj kg Discussion The problem describes the working fluid, refrigerant 22, initially as a saturated vapor undergoing a constant pressure process to a final temperature. The p-v diagram indicates that the second state is within the superheated vapor region. With known states the property values can be easily determined from the tables. After the property values are evaluated the problem becomes identical to those previously solved in homework 2. 3
3.63: A closed, rigid tank filled with water, initially at 20 bar, a quality of 80%, and a volume of 0.5 m 3, is cooled until the pressure is 4 bar. Show the process of the water on a sketch of the T-v diagram and evaluate the heat transfer, in kj. Given Conversions Model p 20 bar 1 bar 100 kpa Negligible potential and kinetic energy effects x 0.8 1 kj 1 kpa m 3 Water is the closed system V 0.5 m 3 Constant volume process, v v p 4 bar Only energy transfer is via heat Q, heat transfer, in kj T-v diagram for the process p-v diagram for the process water T f 1 x 0.8 g p 20 bar p 4 bar Q rst f 2 g v v 9 + x v ; v 9 v v m Assume heat transfer is net into the system. ΔU Q rst Q rst m u u From Table A-3 at 20 bar: v v 9 + x v ; v 9 1.1764 10 >? m? kg + 0.8 0.09963 m? kg 1.1764 10 >? m? kg 0.07994 m? kg u u 9 + x u ; u 9 906.44 kj/kg + 0.8 2600.3 kj/kg 906.44 kj/kg 2261.53 kj/kg From Table A-3 at 4 bar: v v 9 + x v ; v 9 x v v 9 0.07994 m? kg 1.0836 10 >? m? kg v ; v 9 0.4625 m? kg 1.0836 10 >? m? kg u u 9 + x u ; u 9 937.44 kj/kg 0.1709 v m m v 0.5 m 3 0.07994 m? kg 6.255 kg Q rst 6.255 kg 2261.53 kj/kg 937.44 kj/kg 8282.18 kj 604.31 kj/kg + 0.1709 2553.6 kj/kg 604.31 kj/kg 4
Discussion As in the previous problem, this problem relies heavily property evaluations via the tables. In this case, the quality is needed to determine the properties of the water at each set point. The quality for the initial state is given, but the quality for the final state must be determined by using the fact that the process is constant volume. Once the necessary properties are evaluated, this problem boils down to just the first law. 5
H3.2: A frictionless piston-cylinder assembly contains 3.5 kg of H 2O initially at T 130 C and 1.1 m 3. Initially, the piston rests on stops. The mass of the piston and effect of the atmospheric pressure above are such that a pressure of the water of 400 kpa (abs) is required to move the piston, and if it moves it does so with no acceleration. The water is heated until the temperature reaches T 240 C. Assume a quasiequillibrium process with negligible kinetic and potential energy effects. Given Conversions Model T 130 C 1 bar 100 kpa Negligible potential and kinetic energy effects m 3.5 kg 1 kj 1 kpa m 3 Water is the closed system 1.1 m 3 Constant pressure process, p p T 240 C p 400 kpa W, work, in kj Q, heat transfer, in kj W pd v v 9 + x v ; v 9 v m v m v m v 1.1 m3 3.5 kg 0.314 m3 kg From Table A-2 at T 130 C: v 9 < v < v ; 1.0697 10 >? m 3 kg < 0.314 m 3 kg < 0.6685 m 3 kg saturated liquid-vapor v v 9 + x v ; v 9 v v 9 + x v ; v 9 x v v 9 v ; v 9 0.314 m3 kg 1.0697 10 >? m 3 kg 0.6685 m 3 kg 1.0697 10 >? m 3 kg 0.469 u u 9 + x u ; u 9 546.02 kj/kg + 0.469 2539.9 kj/kg 546.02 kj/kg 1481.15 kj/kg The initial state of this problem is defined by T and v to be within the saturation zone. The p-v and T-v diagrams, shown above, can now be drawn. Note the initial pressure is less than 400 kpa. 6
The volume does not begin to change until the pressure reaches 400 kpa, at which point the pressure remains constant to the final state. This allows for the work out of the system to be calculated: W pd p From Interpolation of Table A-4: v v S v v S + p p S 0.781 m 3 kg + 4.0 bar 3.0 bar 0.4646 m3 kg 0.781 m 3 kg p p S 5.0 bar 3.0 bar 0.6228 m 3 kg v m v m v m 0.6228 m 3 kg 3.5 kg 2.18 m 3 W 400 kpa 2.18 m 3 1.1 m 3 432 kj (out) From Interpolation of Table A-4: u u S 2707.6 kj kg 2713.1 kj kg u u S + p p S 2713.1 kj/kg + 4.0 bar 3.0 bar p p S 5.0 bar 3.0 bar 2710.3 kj kg ΔU Q pq W rst Q pq m u u + W rst Q pq 3.5 kg 2710.3 kj kg 1481.15 kj/kg + 432 kj 4734.0 kj 7
3.104: the specific volume, in m 3 kg, of Refrigerant 134a at 16 bar, 100 C using various methods. Given Conversions Model p 16 bar 1 bar 100 kpa Refrigerant is the closed system T 100 C v, specific volume, in m 3 kg a) Using Table A-12 b) Using Figure A-1 c) Using the ideal gas law R134a p 16 bar T 100 C pv RT Z pv RT p p p K T T T K a) From Table A-12 using the given p and T: v 0.01601 m 3 kg b) From Table A-1: p K 40.7 bar p p 16 bar p K 40.7 bar 0.393 T T 373 K T K 374 K 1.0 From Figure A-1: Z 0.86 8.314 J/mol K Z pv ZRT 0.86 1000 g/kg 373K v RT p 102.03 g/mol 16 bar 100 10? Pa bar 0.01634 m 3 kg c) 8.314 J/mol K pv RT v RT 1000 g/kg 373K p 102.03 g/mol 16 bar 100 10? Pa bar 0.0190 m 3 kg Discussion The first two values are only 2% different, with the table value being slightly lower. While the value calculated with the ideal gas law is 19% greater than the table value. This illustrates fallibility of the ideal gas law, only under certain conditions is the law considered accurate. 8
3.116: Two kg of oxygen fills the cylinder of a piston-cylinder assembly. The initial volume and pressure are 2 m 3 and 1 bar, respectively. Heat transfer to the oxygen occurs at constant pressure until the volume is doubled. Given Conversions Model p p 1 bar 1 bar 100 kpa Oxygen is the closed system k 1.35 1 kj 1 kpa m 3 Negligible kinetic and potential energy effects 2 m 3 Pressure is constant 2 4 m 3 Specific heat ratio is constant, c, c Š constant m 2 kg Oxygen is an ideal gas Q, heat transfer, in kj Q pq Oxygen W pd pv RT c Š R eq 3.47b k 1 du c Š dt W rst W for constant c Š : pd W p 1 bar 100 kpa bar 4 m 3 2 m 3 200 kj du c Š dt u u c Š T T ΔU Q pq W rst Q pq m u u + p mc Š T T + p 2 m pv RT T pv 1 bar 100 10? Pa bar 3 R 2 kg 8.314 J/mol K 1000 g/kg 32.0 g/mol 384.9K p R p R T v T v T T v v 384.9K 4 m 3 2 kg 2 m 3 2 kg 769.8K 8.314 J/mol K c Š R k 1 32.0 g/mol 0.742 kj/kg K 1.35 1 Q pq 2 kg 0.742 kj/kg K 769.8K 384.9K + 1 bar 100 kpa bar 4 m 3 2 m 3 770 kj 9
Discussion With an ideal gas, property evaluation is simplified. The equation of state relates p, v, T. With further assumption of constant specific heat, du c Š dt can be integrated to obtain u u c Š T T and property tables are not needed. 10
H3.3: A piston-cylinder device contains air initially at p 90 kpa, T 20 C, and 0.2L. The air is then compressed in a quasiequillibrium polytropic process with n 1.25 to a volume one-seventh of the original volume. Assume ideal gas behavior and negligible kinetic and potential energy effects. Given Conversions Model p 90 kpa 1 kj 1 kpa m 3 Air is the closed system 0.2 L 1 L 0.001 m 3 Negligible kinetic and potential energy effects T 20 C Air is an ideal gas n 1.25 7 0.2 7 L Q rst 0.2 L p 90 kpa T 20 C W pq Polytropic process a) p, final pressure, in kpa b) T, final temperature, in C c) W, work, in kj [magnitude and direction] d) Q, heat transfer, in kj [magnitude and direction] v v 7 v W pd p q constant p mrt a) p q constant p q p q p p q b) p q constant p q p q p p p mrt p T p T T T p p q 90 10? Pa 0.2 L 0.001 m 3 L.[ 0.2 7 L 0.001 m3 L 1.025 MPa q q q> T T q>.[ 11
T T c) W q> pd T T p p 1 n q> 20 C + 273.15 476.6K 1025 kpa 0.2 L 0.001 m 3 L 0.2 7 L 0.001 m3 L.[> 0.2 7 L 0.001 m3 L 90 kpa 0.2 L 0.001 m 3 L 1 1.25 0.0451 kj (in) Since compression W pq W 0.0451 kj d) Work is into the system, so I will assume heat transfer is out of the system. ΔU W pq Q rst Q rst W pq m u u Interpolating from Table A-22 using T 20 C 293K: u [ u S u u S + T T S T [ T S 210.49 kj/kg 206.91 kj/kg 206.91 kj/kg + 293K 290K 295K 290K 209.06 kj/kg Interpolating from Table A-22 using T 476.6 K: u?s u?s[ u u?s[ + T T?S[ T?S T?S[ 344.70 kj/kg 337.32 kj/kg 337.32 kj/kg + 476.6K 470K 480K 470K 342.19 kj/kg p mrt m p 90 10? Pa 0.2 L 0.001 m 3 L 2.14 10 RT 8.314 J/mol K >X kg 1000 g/kg 293K 29.0 g/mol Q rst 0.0451 kj 2.14 10 >X kg 342.19 kj/kg 209.06 kj/kg 0.0166 kj (out) Assumption of heat transfer out of the system is confirmed by the positive value. Discussion Since specific heats are not constant the ideal gas tables are used to find u(t). 12
3.88: In a heat-treating process, a 1-kg metal part, initially at 1075 K, is quenched in a closed tank containing 100 kg of water, initially at 295 K. There is a negligible heat transfer between the contents of the tank and their surroundings. Modeling the metal part and the water as incompressible with constant specific heats 0.5 kj/kg K and 4.4 kj/kg K, respectively. Given Conversions Model m št œ 1 kg 1 kj 1 kpa m 3 Water + metal is the closed system m tš 100 kg 1 L 0.001 m 3 Negligible kinetic and potential energy effects T, št œ 1075K Adiabatic and no work T, tš 295K Constant specific heats c št œ 0.5 kj/kg K Water and metal behave as incompressible substances c tš 4.4 kj/kg K T eq, the equilibrium temperature, in K ΔU št œ + ΔU tš 0 m št œ c št œ T šž T, št œ + m tš c tš T šž T, tš 0 T šž m št œc št œ T, št œ + m tš c tš T, tš m št œ c št œ + m tš c tš 1 kg 0.5 kj/kg K 1075K + 100 kg 4.4 kj/kg K 295K 1 kg 0.5 kj/kg K + 100 kg 4.4 kj/kg K 295.9K Discussion Physically, there is heat transfer from the metal to the water. However, since that occurs inside the defined system, it is not considered in the analysis. 13