Dishwasher. Heater. Homework Solutions ME Thermodynamics I Spring HW-1 (25 points)

HW-1 (25 points) (a) Given: 1 for writing given, find, EFD, etc., Schematic of a household piping system Find: Identify system and location on the system boundary where the system interacts with the environment and describe the direction of flow for heater, dishwasher and the shower head. EFD: Heater: 5 (1 for system boundary and 1 for each flow direction) Heater Dishwasher: 5 (1 for system boundary and 1 for each flow direction) Dishwasher 1

Shower: 5 (1 for system boundary and 1 for each flow direction) Shower Assumptions: None. Basic Equations: None Solution: See EFD. (b) Given: 1 for writing given, find, EFD, etc., Schematic of a piston-cylinder device Find: EFDs/systems for the gas and the hot plate. EFD: Gas: 4 (2 for system boundary and 1 for each flow direction) 2

Hot plate: 4 (2 for system boundary and 1 for each flow direction) Assumptions: Cylinder is insulated. Basic Equations: None Solution: See EFD. 3

HW-2 (25 points) (a) Given: 1 for writing given, find, EFD, etc., Schematic of a piston-cylinder device Find: EFDs/systems for the air and the electric resistance heater. EFD: Air: 3 (1 for system boundary and 1 for each flow direction) Electric resistance heater: 3 (1 for system boundary and 1 for each flow direction) Assumptions: None. Basic Equations: None Solution: See EFD. 4

(b) (i) True. 1 By definition, a closed system has no mass transfer. 2 (ii) True. 1 Only the mass stays constant in a closed system. 2 (iii) True. 1 If the states are specified, the temperature is fixed too since temperature depends only on the state. 2 (iv) True. 1 Mass enters and leaves the system. 2 (v) False. 1 A chemical reaction can occur within the system, changing the composition. 2 (vi) False. 1 Control volume may have mass and/or energy transfers. 2 5

HW-3 (25 points) (a) Given: 1 for writing given, find, EFD, etc., Piston mass, m = 50 kg p atm = 1 bar A = 0.01 m 2 Find: Pressure at which piston starts moving EFD: Assumptions: None. Basic Equations: None Solution: The piston begins to move when the force exerted by the gas exceeds the resisting force of the piston weight and the atmospheric pressure. F gas W eight + F atm 2 p gas A mg + p atm A p gas mg/a + p atm p gas (50 kg)(9.81 m/s2 ) 0.01 m 2 + 1 10 5 P a p gas 1.49 10 5 P a 2 (1 for the answer and 1 for units) The piston begins to move when the pressure is greater than or equal to 1.49 10 5 P a or 1.49 bar. 6

(b) Given: 1 for writing given, find, EFD, etc., p atm = 1 bar Case (i) p = 1.5 bar and ρ = 997 kg/m 3 Case (ii) p = 1.3 bar and ρ = 13.59 g/cm 3 Find: L EFD: Assumptions: None. Basic Equations: None Solution: Pressure at points a and b should be equal since they are at the same height. Case (i) L = p a = p b p a = p atm + ρgl 2 L = p a p atm ρg (1.5 1) 10 5 P a (997 kg/m 3 )(9.81 m/s 2 ) L = 5.11 m 2 (1 for the answer and 1 for units) 7

Case (ii) (1.3 1) 10 5 P a L = (13.59 10 3 kg/m 3 )(9.81 m/s 2 ) L = 0.225 m L = 22.5 cm 2 (1 for the answer and 1 for units) (c) Given: 1 for writing given, find, EFD, etc., p 1 = 250 kp a V 1 = 1.5 m 3 p 2 = 100 kp a Find: v 2 EFD: W Q Assumptions: None. Basic Equations: None Solution: Specific volume at state 1, v 1 = V 1 /m v 1 = 1.5 m 3 /3 kg v 1 = 0.5 m 3 /kg 1 8

p v diagram: 3 pv 0.5 = constant p 1 v1 0.5 = p 2 v 0.5 [ p1 v1 0.5 v 2 = 2 1 ] 1/0.5 p 2 [ (250 kp a)(0.5 m 3 /kg) 0.5 v 2 = (100 kp a) v 2 = 3.125 m 3 /kg 1 ] 1/0.5 250 kpa 1 100 kpa 2 1.5 m 3 /kg 3.125 m 3 /kg (d) (i) True. Specific volume is independent of the mass. 1.5 (ii) False. It would be at a gage pressure of 0.2 atm. 1.5 (iii) True. From zeroth law of thermodynamics. 1.5 (iv) True. Since 1 R = 0.556 K. 1.5 9

HW-4 (25 points) (a) Given: 1 for writing given, find, EFD, etc., m = 14, 000 kg V 1 = 0 km/h (Starts at rest) and V 2 = 620 km/h Z 1 = 0 m (Takes off from airport) and Z 2 = 10, 000 m g = 9.78 m/s 2 Find: KE and P E EFD: Speed = 620 km/h H = 10,000m Z Assumptions: None. Basic Equations: None Solution: V 2 = 620 km/hr V 2 = 620 1000/3600 m/s V 2 = 172.22 m/s 1 10

Change in Kinetic Energy KE = 1 2 m(v 2 2 V 2 1 ) 2 KE = 1 2 (14, 000 kg)(172.222 0 2 )m 2 /s 2 KE = 2.08 10 8 J KE = 2.08 10 5 kj 1 Change in Potential Energy P E = mg(z 2 Z 1 ) 2 P E = (14, 000 kg)(9.78 m/s 2 )(10, 000 0)m P E = 1.37 10 9 J P E = 1.37 10 6 kj 1 (b) Given: 1 for writing given, find, EFD, etc., m = 900 kg V 1 = 100 km/h (At bottom of hill) and V 2 = 0 km/h Z 1 = 0 m (At bottom of hill) and Z 2 = 50 m Find: KE and P E EFD: Speed = 0 km/h H = 50 m Speed = 100 km/h Z Assumptions: None. Basic Equations: None 11

Solution: V 1 = 100 km/hr V 1 = 100 1000/3600 m/s V 1 = 27.78 m/s 1 Change in Kinetic Energy KE = 1 2 m(v 2 2 V 2 1 ) 2 KE = 1 2 (900 kg)(02 27.78 2 )m 2 /s 2 KE = 3.47 10 5 J KE = 3.47 10 2 kj 1 Change in Potential Energy P E = mg(z 2 Z 1 ) 2 P E = (900 kg)(9.8 m/s 2 )(50 0)m P E = 4.41 10 5 J P E = 4.41 10 2 kj 1 (c) Given: 1 for writing given, find, EFD, etc., m = 900 kg V 1 = 100 km/h and V 2 = 0 km/h Z 1 = 0 m and Z 2 = 50 m Find: V 2 EFD: 12

L = 10 m 40 o Z Assumptions: Neglect air resistance and friction. 1 Basic Equations: E = Q W Solution: E = Q W U + P E + KE = 0 P E + KE = 0 KE = P E 2 1 2 m(v 2 2 V1 2 ) = mg(z 2 Z 1 ) 2 (for KE and P E expressions) V 2 = 2g(Z 1 Z 2 ) 1 V 2 = 2(9.81 m/s 2 )(10sin40 o 0)m V 2 = 11.23 m/s 2 13

HW-5 (25 points) (a) Given: 1 for writing given, find, EFD, etc., V 1 = 0.003 m 3 and V 2 = 0.002 m 3 A = 0.018 m 2 F 1 = 900 N and F 2 = 0 N P atm = 100 kp a Find: p 1, p 2, work done and the p V diagram. EFD: 2 W Assumptions: 2 Spring is linear. Friction between piston and cylinder wall can be neglected. Basic Equations: 1 W b = pdv Solution: V = Ax, where x is the distance from the piston to the bottom of the cylinder. x 1 = V 1 /A 2 x 1 = 0.003 m 3 /0.018 m 2 x 1 = 0.1667 m 1 x 2 = V 2 /A x 2 = 0.002 m 3 /0.018 m 2 x 2 = 0.1111 m 1 14

Spring force, F, varies linearly with x. When x = 0.1667 m, F = 900 N and when x = 0.1111 m, F = 0 N. Thus variation of force with x can be expressed as (x 0.1111) F = 900 (0.1667 0.1111) F = 16200(x 0.1111) N 2 Force balance on the piston gives, pa = p atm A + F 2 pa = p atm A + 16200(x 0.1111) p = p atm + 16200(x 0.1111)/A p = p atm + 16200(x 0.1111)/0.018 m 2 p = p atm + 9 10 5 (x 0.1111) 1 At state 1, p 1 = p atm + 9 10 5 (x 1 0.1111) p 1 = 1 10 5 P a + 9 10 5 (0.1667 0.1111) p 1 = 1.5 10 5 P a p 1 = 1.5 bar 2 At state 2, p 2 = p atm + 9 10 5 (x 2 0.1111) p 2 = 1 10 5 P a + 9 10 5 (0.1111 0.1111) p 2 = 1 10 5 P a p 2 = 1 bar 2 15

Work, 2 (for the integration; area under the curve is also acceptable) W = W = 2 1 x2 x 1 pdv W = p atm A (p atm + 9 10 5 (x 0.1111))Adx x2 x 1 dx + 9 10 5 A x2 x 1 (x 0.1111))dx W = (1 10 5 P a)(0.018 m 2 )(x 2 x 1 ) [( ) x + 9 10 5 (0.018 m 2 2 ) 2 2 0.1111x 2 ( )] x 2 1 2 0.1111x 1 W = (1 10 5 P a)(0.018 m 2 )(0.1111 0.1667) [( ) ( 0.1111 + 9 10 5 (0.018 m 2 2 0.1667 2 ) 0.1111 0.1111 2 2 W = 125 J 1 )] 0.1111 0.1667 p V diagram: 3 1.5 bar 1 1 bar 2 0.002 m 3 0.003 m 3 V 16

HW-6 (25 points) (a) Given: 1 for writing given, find, EFD, etc., m = 4 kg V = 1 m 3 Ẇ = 14 W t = 1 h u = 10 kj/kg Find: v, W, Q and direction of Q. EFD: 2 W Q Assumptions: 1 The tank is rigid. No change in KE and PE. No mass enters or leaves the system. Basic Equations: 1 E = Q W Solution: Specific volume is same at both the initial and final states, v 1 = v 2 = v, Work, W, v = V/m 1 v = 1 m 3 /4 kg v = 0.25 m 3 /kg 1 W = Ẇ t(since W goes into the system) 1 W = (14 W )(1 h) W = (14 W )(3600 s) W = (14 W )(3600 s) W = 50400 J W = 50.4 kj 1 17

Heat transfer, Q, E = Q W U + KE + P E = Q W U = Q W m u = Q W Q = W + m u 2(for simplifying energy equation) Q = 50.4 kj + (4 kg)(10 kj/kg) Q = 10.4 kj/kg 1 Since Q is negative, heat leaves the system. 1 (b) Given: 1 for writing given, find, EFD, etc., V = 10 V I = 0.5 A t = 30 min Find: R and W EFD: 1 W Assumptions: None Basic Equations: None Solution: Resistance, R, R = V/I 1 R = 10 V/0.5 A R = 20 ohm 1 18

Work, W, W = V It 1 W = (10 V )(0.5 A)(30 min) W = (10 V )(0.5 A)(1800 s) W = 9000 J W = 9 kj 1 (c) (i) True. 1 If there is no temperature difference, the system and the surrounding are in thermal equilibrium and hence there is no heat transfer. 1 (ii) False. 1 A closed system cannot have mass transfer across its boundary. 1 (iii) True. 1 Since E = U + KE + P E. 1 19

HW-7 (25 points) (a) Given: Data for different processes of a closed system Find: Missing entries in the table EFD: 1 W Q Assumptions: None Basic Equations: 1 E = Q W Solution: Process a: E = Q W 70 = Q ( 20) Q = 50 kj 1 E 2 E 1 = 70 50 E 1 = 70 E 1 = 20 kj 1 Process b: E = E 2 E 1 E = 50 20 E = 30 kj 1 E = Q W 30 = 50 W W = 20 kj 1 20

Process c: E = Q W 20 = Q ( 60) Q = 40 kj 1 E 2 E 1 = 20 60 E 1 = 20 E 1 = 40 kj 1 Process d: E = Q W 0 = Q ( 90) Q = 90 kj 1 E 2 E 1 = 0 50 E 1 = 0 E 1 = 50 kj 1 Process e: E = Q W E = 50 (150) E = 100 kj 1 E 2 E 1 = E E 2 20 = 100 E 2 = 80 kj 1 The table: Process Q W E 1 E 2 E 2 E 1 a +50-20 -20 50 +70 b +50 +20 20 50 +30 c -40-60 40 60 +20 d -90-90 50 50 0 e +50 +150 20-80 -100 21

(b) Given: p 1 = 1 bar, V 1 = 1.0 m 3, U 1 = 400 kj p 2 = 10 bar, V 2 = 0.1 m 3, U 2 = 450 kj Process A: Constant-pressure path from state 1 to a volume of 0.1 m 3, followed by a constantvolume path to state 2. Process B: pv = constant Find: Sketch p V diagram and find W and Q. EFD: 1 Assumptions: 1 Ignore KE and PE changes. Basic Equations: E = Q W 1 W = pdv 1 Solution: Process A: p V diagram: 1 10 bar 2 1 bar 1 0.1 m 3 1.0 m 3 V 22

Work, W, W = 2 1 pdv W = Area under p V curve W = (1 10 5 P a)(1.0 0.1) m 3 W = 9 10 4 J W = 90 kj Work goes into the system, thus W = 90 kj. 1 Heat transfer, Q, E = Q W U + KE + P E = Q W U = Q W Q = W + U 1 (for simplifying energy equation) Q = 90 + (450 400) Q = 40 kj/kg 1 Process B: p V diagram: 1 10 bar 2 1 bar 1 0.1 m 3 1.0 m 3 V pv = constant p 1 V 1 = (100 kp a)(1.0 m 3 ) p 1 V 1 = 100 kj p = 100/V 1 23

Work, W, Heat transfer, Q, W = W = W = 2 1 V2 V 1 0.1 1 pdv 100/V dv 100/V dv W = [100ln(V )] 0.1 1 1 (integration) W = 100(ln(0.1) ln(1)) W = 230 kj 1 Q = W + U Q = 230 + (450 400) Q = 180 kj 1 24

HW-8 (25 points) (a) (i) From the saturation pressure table, at p = 5 bar, T sat = 151.83 o C. Since T = T sat, it is either a Saturated Liquid or a Saturated Vapor or a Saturated Liquid Vapor Mixture (SLVM). 1 p v and T v diagrams: 2 (1 point for each) p T 5 bar CL T = 151.83 C SLVM SHV 153.81 C CL p = 5 bar SLVM SHV (ii) From the saturation pressure table, at p = 5 bar, T sat = 151.83 o C. Since T > T sat, it is a Superheated Vapor (SHV). 1 p v and T v diagrams: 2 (1 point for each) p T 5 bar CL T = 200 C SLVM SHV 200 C CL p = 5 bar SLVM SHV (iii) From the saturation temperature table, at T = 200 o C, p sat = 15.549 bar = 1.5549 MP a. Since p > p sat, it is a Compressed Liquid (CL). 1 p v and T v diagrams: 2 (1 point for each) 25

p T 5 bar CL T = 200 C SLVM SHV 200 C CL p = 25 bar SLVM SHV (iv) From the saturation temperature table, at T = 160 o C, p sat = 6.1823 bar. Since p < p sat, it is a Superheated Vapor (SHV). 1 p v and T v diagrams: 2 (1 point for each) p T 4.8 bar CL T = 160 C SLVM SHV 160 C CL p = 4.8 bar SLVM SHV (v) From the saturation pressure table, at p = 1 bar, T sat = 99.61 o C. Since T < T sat, it lies to the left of the dome. In this case, it is a sub-cooled solid 1 p v and T v diagrams: 2 (1 point for each) p T 1 bar CL T = -12 C SLVM SHV -12 C CL p = 1 bar SLVM SHV (b) Given: 1 for writing given, find, EFD, etc., Refrigerant 134a 26

V 1 = 1.5 m 3, T 1 = 10 o C T 2 = 50 o C and x 2 = 1.0 Find: Locate initial and final states on p v diagram and indicate process line. Find mass of vapor in the initial and final states. EFD: Q Assumptions: 1 No mass enters or leaves the system. The tank is rigid (i.e. volume doesn t change). Basic Equations: None Solution: p v diagram: 2 p T = 50 C CL T = 10 C SLVM SHV Specific volume at state 2, v 2, Interpolation for v 2 : 27

T ( o C) 48 50 52 v g (kj/kg) 0.015951 v 2 0.014276 v 2 0.015951 50 48 = 0.014276 0.015951 52 48 v 2 = 0.015951 + (0.014276 0.015951) v 2 = 0.0 kj/kg 1 ( ) 50 48 52 48 m g2 = V/v g2 m g2 = 1.5/0.015 m g2 = 99.2 kg 1 Since volume, V and mass is constant, specific volume is also constant, i.e. v 1 = v 2. Interpolation for v f1 : T ( o C) 8 10 12 v f (kj/kg) 0.00078873 v f1 0.00079745 v f1 0.00078873 0.00079745 0.00078873 = 10 8 12 8 Interpolation for v g1 : v f1 = 0.00078873 + (0.00079745 0.00078873) v f1 = 0.0 kj/kg 1 ( ) 10 8 12 8 T ( o C) 8 10 12 v g (kj/kg) 0.052804 v g1 0.046332 v g1 0.052804 0.046332 0.052804 = 10 8 12 8 v g1 = 0.052804 + (0.046332 0.052804) v g1 = 0.0 kj/kg 1 ( ) 10 8 12 8 28

Quality at state 1, Mass of vapor in initial state, m g1 : v 1 = (1 x 1 )v f1 + x 1 v g1 x 1 = v 1 v f1 v g1 v f1 x 1 = 0.29 1 m g1 = xm tot m g1 = 0.2936(99.2490 kg) m g1 = 29.1 kg 1 29

HW-9 (25 points) Given: 1 for solution format Water-vapor contained in a piston-cylinder device. Process 1-2: Constant-temperature to p 2 = 2p 1 Process 1-3: Constant-volume to p 3 = 2p 1 Process 1-4: Constant-pressure to v 4 = 2v 1 Process 1-5: Constant-temperature to v 5 = 2v 1 Find: sketch each process on a p V diagram; identify the work by an area on the diagram; indicate whether the work is done by, or on, the water vapor. EFD: N/A. Assumptions: None Basic Equations: None Solution: p v diagrams 16 (4 for each) p p T = constant T = constant SHV 2 SHV 3 CL 1 CL 1 SLVM SLVM p p T = constant SHV T = constant SHV CL SLVM 1 4 CL SLVM 1 5 30

Alternatively, if water vapor is assumed to be an ideal gas, we can have Process 1-2: Water vapor is compressed. So work is done on the vapor. 2 Process 1-3: There is no work in this constant volume process. 2 Process 1-4: Water vapor expands to a larger volume. So work is done by the vapor. 2 Process 1-5: Water vapor expands to a larger volume. So work is done by the vapor. 2 31

HW-10 (25 points) Given: 1 for writing given, find, EFD, etc., Propane, m = 0.1 kg, p = 0.4 MP a V 1 = 0.013 m 3, V 2 = 0.014 m 3 Find: Initial and final temperature ( o C), work (kj) and heat transfer (kj). EFD: 3 Assumptions: No mass enters or leaves the system. 1 No mass enters or leaves the system. 1 Friction between piston and cylinder is negligible. 1 Basic Equations: E = Q W 2 W b = pdv 1 Solution: Specific volume, v 1, Specific volume, v 2, v 1 = V 1 /m 1 v 1 = 0.013 m 3 /0.1kg v 1 = 0.13 m 3 /kg 1 v 2 = V 2 /m v 2 = 0.014 m 3 /0.1kg v 2 = 0.14 m 3 /kg 1 Temperature, T 1 : At p = 0.4 MP a, v f = 0.0018658 m 3 /kg and v g = 0.11393 m 3 /kg. 32

v 1 > v g. Thus, it is a Superheated Vapor (SHV). 1 Interpolation for T 1 : v (m 3 /kg) 0.1284 0.13 0.1339 T ( o C) 20 T 1 30 T 1 20 0.13 0.1284 = 30 20 0.1339 0.1284 ( ) 0.13 0.1284 T 1 = 20 + (30 20) 0.1339 0.1284 T 1 = 22.9 o C 1 Temperature, T 2 : At p = 0.4 MP a, v f = 0.0018658 m 3 /kg and v g = 0.11393 m 3 /kg. v 2 > v g. Thus, it is a Superheated Vapor (SHV). 1 Interpolation for T 2 : v (m 3 /kg) 0.1393 0.14 0.1446 T ( o C) 40 T 2 50 Work, W, T 2 40 0.14 0.1393 = 50 40 0.1446 0.1393 ( ) 0.14 0.1393 T 2 = 40 + (50 40) 0.1446 0.1393 T 2 = 41.3 o C 1 W = pdv W = p dv (p = constant) W = p V 2 W = 0.4 10 3 kp a(0.014 0.013) m 3 W = 0.4 kj 1 Heat transfer, Q, E = Q W U + KE + P E = Q W U = Q W Q = W + U Q = W + m u 2 33

Interpolation for u 1 : T ( o C) 20 22.91 30 u (kj/kg) 456.6 u 1 472.0 Interpolation for u 2 : u 1 456.6 472.0 456.6 = 22.91 20 30 20 u 1 = 456.6 + (472.0 456.6) u 1 = 461.1 kj/kg 1 ( ) 22.91 20 30 20 T ( o C) 40 41.32 50 u (kj/kg) 487.8 u 2 504.0 u 2 487.8 504.0 487.8 = 41.32 40 50 40 u 2 = 487.8 + (504.0 487.8) u 2 = 489.9 kj/kg 1 ( ) 41.32 40 50 40 Q = W + m u Q = 0.4 kj + 0.1 kg(489.94 461.08) kj/kg Q = 3.286 kj 1 34

HW-11 (25 points) Given: 3 for writing given, find, EFD, etc., m c = 13 kg, T c1 = 27 o C = 300 K, C c = 0.385 kj/kg K m w = 4 kg, T w1 = 50 o C = 323 K, C w = 4.18 kj/kg K W = 100 kj Find: Final equilibrium temperature ( o C). EFD: 5 W Assumptions: 5 No mass enters or leaves the system. KE and PE changes are negligible. Resistor mass is negligible. System is insulated. Incompressible substance. Basic Equations: E = Q W 3 Solution: 35

At equilibrium, temperature of copper and water are the same, T w2 = T c2 = T 2 1 E = Q W U + KE + P E = Q W U = W 3 (U 2 U 1 ) = W 3 m w C w (T w2 T w1 ) + m c C c (T c2 T c1 ) = W m w C w (T 2 T w1 ) + m c C c (T 2 T c1 ) = W T 2 (m w C w + m c C c ) = W + m w C w T w1 + m c C c T c1 T 2 = W + m wc w T w1 + m c C c T c1 (m w C w + m c C c ) ( 100) + (4)(4.18)(323) + (13)(0.385)(300) T 2 = (4)(4.18)) + (13)(0.385) T 2 = 322.30 K T 2 = 49.30 o C 2 36

HW-12 (25 points) Given: 2 for writing given, find, EFD, etc., Tank conditions: m t1 = 3 kg, p t1 = 500 kp a, T t1 = T t2 = 290 K Cylinder conditions: V c1 = 0.05 m 3, p c1 = p c2 = 200 kp a, T c1 = T c2 = 290 K Find: W and Q. EFD: 3 W Q Assumptions: 3 No mass enters or leaves the system. KE and PE changes are negligible. CO 2 behaves as an ideal gas. Basic Equations: pv = mrt 1 W = pdv 1 E = Q W 1 37

Solution: The only work contribution would be from the piston-cylinder assembly, since the tank is rigid. Work, W, W = pdv W = p dv The gas constant for CO 2 is Initial mass in the cylinder, m c1, Initial total mass, m 1, R = R/MW W = p V 2 R = (8.314 kj/kmol K)/(44 kg/kmol) R = 0.189 kj/kgk 1 p c1 V c1 = m c1 RT m c1 = p c1v c1 RT m c1 = (200 kp a)(0.05 m3 ) (0.189 kj/kgk)(290 K) m c1 = 0.182 kg 1 m 1 = m c1 + m t1 m 1 = 0.182 + 3.0 m 1 = 3.182 kg 1 From mass conservation, total mass remains the same, i.e. m 1 = m 2. Volume of the tank, Initial volume of the system, V 1, p t1 V t = m t1 RT V t = m t1rt p t1 (3 kg)(0.189 kj/kgk)(290 K) V t = 500 kp a V t = 0.329 m 3 1 V 1 = V c1 + V t V 1 = 0.05 + 0.329 V 1 = 0.379 m 3 38

Final mass in the tank, m t2, Final mass in the cylinder, m c2, p t2 V t = m t2 RT m t2 = p t2v t RT m t2 = (200 kp a)(0.329 m3 ) (0.189 kj/kgk)t (290 K) m t2 = 1.2 kg 1 Final volume of cylinder, V c2, Final volume of the system, V 2, p c2 V c2 = m c2 RT V c2 = m c2rt p c2 m 2 = m c2 + m t2 m c2 = m 2 m t2 m c2 = 3.182 1.2 m c2 = 1.982 kg 1 (1.982 kg)(0.189 kj/kgk)(290 K) V c2 = 200 kp a V c2 = 0.543 m 3 1 V 2 = V c2 + V t V 2 = 0.543 + 0.329 V 2 = 0.872 m 3 Work, W, Heat transfer, Q, W = p V W = p(v c2 V c1 ) W = 200 kp a(0.543 0.05) m 3 W = 98.6 kj 2 E = Q W U + KE + P E = Q W U = Q W 0 = Q W (since T is constant) Q = W 2 Q = 98.6 kj 1 39

HW-13 (25 points) Given: m = 2 kg, p 1 = 1 bar, T 1 = 300 K Process 1-2: p 2 /p 1 = 4, v 2 = v 1 Process 2-3: pv 1.28 = constant Process 3-1: p 1 = p 3 Find: Sketch p V diagram and find W and Q for each process. EFD: 1 Assumptions: 1 No mass enters or leaves the system. KE and PE changes are negligible. CO 2 behaves as an ideal gas. Basic Equations: pv = mrt 1 W = pdv 1 E = Q W 1 Solution: p V diagram 3 4 bar 2 1 bar 1 1.134 m 3 3 3.35 m 3 V 40

The gas constant for CO 2 is Volume at state 1, V 1, R = R/MW R = (8.314 kj/kmol K)/(44 kg/kmol) R = 0.189 kj/kgk 1 p 1 V 1 = mrt 1 V 1 = mrt 1 V 1 = p 1 (2 kg)(0.189 kj/kgk)(300 K) (100 kp a) V 1 = 1.134 m 3 1 V 1 = V 2 and p 2 = 4p 1 = 4 bar. For process 2-3, pv 1.28 = C Pressure at state 3, p 3 = p 1 = 100 kp a. Volume at state 3, V 3, p 2 V 1.28 2 = C C = (400 kp a)(1.134 m 3 ) 1.28 C = 469.86 p 3 V 1.28 3 = C V 3 = C p 3 1/1.28 V 3 = 469.86 1/1.28 (100 kp a) V 3 = 3.35 m 3 1 Work for process 1-2, W 12 = 0, since volume stays constant. Work for process 2-3, W 23, W 23 = W 23 = 3 2 3 W 23 = C pdv C dv 2 V 1.28 3 1 dv 2 V 1.28 V 0.28 V 3 =3.35 m 3 W 23 = C 0.28 1 (for the integration) V 2 =1.134 m 3 W 23 = 469.86 ( 3.35 0.28 1.134 0.28) 0.28 W 23 = 423.8 kj 1 41

Work for process 3-1, W 31, W 31 = W 31 = p 1 3 1 3 pdv dv W 31 = p(v 1 V 3 ) W 31 = 100 kp a(1.134 3.35) m 3 W 31 = 221.6 kj 1 Temperature at state 2, T 2, Temperature at state 3, T 3, p 2 V 2 = mrt 2 T 2 = p 2V 2 mr T 2 = (400 kp a)(1.134 m3 ) (2 kg)(0.189 kj/kgk) T 2 = 1200 K 1 p 3 V 3 = mrt 3 T 3 = p 3V 3 mr T 3 = (100 kp a)(3.35 m3 ) (2 kg)(0.189 kj/kgk) T 3 = 886.24 K 1 Internal energy at state 1, u 1 (from ideal gas tables) u 1 = 6935 kj/kmol u 1 = 6935 kj/kmol 44 kg/kmol u 1 = 157.61 kj/kg 1 Internal energy at state 2, u 2 (from ideal gas tables) u 2 = 43864 kj/kmol u 2 = 43864 kj/kmol 44 kg/kmol u 2 = 996.9 kj/kg 1 Internal energy at state 3, u 3 (from ideal gas tables) Interpolation for u 3 : 42

T ( K) 880 886.24 890 u (kj/kmol) 29017 u 3 29461 u 3 29017 886.24 880 = 29461 29017 890 880 u 3 = 29017 + (29461 29017) u 3 = 29294.1 kj/kmol 1 ( ) 886.24 880 890 880 Energy equation, u 3 = 29294 kj/kmol u 3 = 29294 kj/kmol 44 kg/kmol u 3 = 665.77 kj/kg 1 E = Q W U + KE + P E = Q W U = Q W Q = m u + W 2 Heat transfer for process 1-2, Q 12, Heat transfer for process 2-3, Q 23, Q 12 = m(u 2 u 1 ) + W 12 Q 12 = (2 kg)(996.9 157.61) kj/kg + 0 Q 12 = 1678.58 kj 1 Q 23 = m(u 3 u 2 ) + W 12 Q 23 = (2 kg)(665.77 996.9) kj/kg + 423.8 kj Q 23 = 238.46 kj 1 Q 31 = m(u 1 u 3 ) + W 31 Q 31 = (2 kg)(157.61 665.77) kj/kg 221.6 kj Q 31 = 1237.92 kj 1 Sum of work and heat transfer for the entire cycle, 43

Process W (kj) Q (kj) 1-2 0 +1679 2-3 +424-239 3-1 -222-1238 Sum +202 +202 44

HW-14 (25 points) Given: 1 for writing given, find, EFD, etc., Area of inlet, A i = 2.75 3.65 m = 10.04 m 2 Area of exit, A e = 4.14 5.05 m = 20.91 m 2 Velocity, V i1 = 22.4 m/s Velocity, V i2 = 134 m/s Pressure, p = 1 bar, temperature, T = 20 o C = 293 K Find: Mass flow rates, exit velocities, pressures and temperatures. EFD: 2 Assumptions: 2 No friction in the duct. The diffuser is adiabatic. Steady state, steady flow. 1-dimensional uniform flow. Density is constant through the diffuser. Air is an ideal gas. Basic Equations: pv = mrt 1 ṁ = ρav 1 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 Solution: Gas constant for air, R, R = R/MW R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 0.287 kj/kgk 1 45

Air density, ρ, Mass flow rate for case 1, ṁ 1, Mass flow rate for case 2, ṁ 2, Exit velocity for case 1, V e1, Exit velocity for case 2, V e2, pv = mrt p = m V RT p = ρrt ρ = p RT 1 100 kp a ρ = (0.287 kj/kgk)(293 K) ρ = 1.189 kg/m 3 1 ṁ i1 = ρa i V i1 ṁ i1 = (1.189 kg/m 3 )(10.04 m 2 )(22.4 m/s) ṁ i1 = 267.4 kg/s 1 ṁ i2 = ρa i V i2 ṁ i2 = (1.189 kg/m 3 )(10.04 m 2 )(134 m/s) ṁ i2 = 1600 kg/s 1 ṁ e1 = ρa e V e1 V e1 = ṁe1 ρa e V e1 = ṁi1 ρa e 1 V e1 = 267.4 kg/s (1.189 kg/m 3 )(20.91 m 2 ) V e1 = 10.76 m/s 1 ṁ e2 = ρa e V e2 V e2 = ṁe2 ρa e V e2 = ṁi2 ρa e 1600 kg/s V e2 = (1.189 kg/m 3 )(20.91 m 2 ) V e2 = 64.36 m/s 1 46

Exit temperature, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o 0 = ṁ i (h + ke) i ṁ o (h + ke) o h o = h i + ke i ke o h o = h i + V i 2 2 V o 2 2 2 Enthalpy at inlet, h i, Interpolation for h i : T (K) 290 293 295 h (kj/kg) 290.1 h i 295.1 h i 290.1 295.1 290.1 Enthalpy at exit for case 1, h o1, = 293 290 295 290 h i = 290.1 + (295.1 290.1) h i = 293.1 kj/kg 1 ( ) 293 290 295 290 h o = h i + V i 2 2 V o 2 2 h o = 293.1 kj/kg + 22.42 2 1000 h o = 293.3 kj/kg 1 kj/kg 10.762 2 1000 kj/kg Temperature at exit for case 1, T o1, Interpolation for T o1 : h (kj/kg) 290.1 293.3 295.1 T (K) 290 T o1 295 T o1 290 295 290 = 293.3 290.1 295.1 290.1 T o1 = 290 + (295 290) T o1 = 293.2 K 1 ( ) 293.3 290.1 295.1 290.1 47

Temperature of air at exit for case 1, T o1 = 293.2 K = 20.2 o C. Pressure at exit for case 1, p o1, p o1 = ρrt o1 p o1 = (1.189 kg/m 3 )(0.287 kj/kgk)(293.2 K) p o1 = 100.05 kp a p o1 = 1.0005 bar 1 Enthalpy at exit for case 2, h o2, h o = h i + V i 2 2 V o 2 2 h o = 293.1 kj/kg + 1342 2 1000 h o = 300.0 kj/kg 1 kj/kg 64.362 2 1000 kj/kg Temperature at exit for case 2, T o2, Interpolation for T o2 : h (kj/kg) 295.1 300.0 300.1 T (K) 295 T o2 300 T o2 295 300 295 = 300.0 295.1 300.1 295.1 T o2 = 295 + (300 295) T o2 = 299.9 K 1 Temperature of air at exit for case 2, T o2 = 299.9 K = 26.9 o C. Pressure at exit for case 2, p o2, ( ) 300.0 295.1 300.1 295.1 p o2 = ρrt o2 p o2 = (1.189 kg/m 3 )(0.287 kj/kgk)(299.9 K) p o2 = 102.34 kp a p o2 = 1.0234 bar 1 48

HW-15 (25 points) Given: 1 for writing given, find, EFD, etc., Outlet temperature, T o = 35 o C = 238 K Q = 7.39 MW ṁ 1 = 267.4 kg/s (50 mph case) ṁ 2 = 1600 kg/s (300 mph case) Secondary liquid properties: ṁ = 550 kg/s, T i = 45 o C = 228 K and C = 2.7 kj/kg K Find: Inlet temperatures for case 1 and 2, T i1 and T i2 and exit temperature of secondary fluid. EFD: Air as the system: 3 Water-ethylene glycol mixture as the system: 3 Assumptions: 4 Steady state, steady flow (SSSF). 1-dimensional uniform flow. Kinetic and potential energy effects can be ignored. No work transfer. 1-inlet and 1-outlet. Air is an ideal gas. Water-ethylene glycol mixture can be treated as an incompressible substance with constant specific heat. 49

Basic Equations: de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 2 Solution: Simplify energy equation (identical for both the systems), de dt = Q Ẇ + ṁ i (h + ke + pe) i i ṁ o (h + ke + pe) o o Q = ṁ o h o ṁ i h i ṁ o = ṁ i = ṁ (from mass conservation) 2 Q = ṁ(h o h i ) 2 Enthalpy of air at outlet, h o, Interpolation for h o : T (K) 230 238 240 h (kj/kg) 230.0 h o 240.0 h o 230.0 240.0 230.0 = 238 230 240 230 h o = 230.0 + (240.0 230.0) h o = 238.0 kj/kg 1 Enthalpy of air at inlet for case 1, h i1, Q = ṁ(h o h i ) ( ) 238 230 240 230 h i = Q ṁ + h o h i1 = Q ṁ 1 + h o h i1 = 7.39 103 kw 267.4 kg/s h i1 = 265.63 kj/kg 1 + 238 kj/kg Temperature of air at inlet, T i1, Interpolation for T i1 : 50

h (kj/kg) 260.0 265.36 270.0 T (K) 260 T i1 270 T i1 260 270 260 = 265.36 260.0 270.0 260.0 T i1 = 260 + (270 260) T i1 = 265.36 K 1 Temperature of air at inlet, T i1 = 265.36 K = 7.36 o C. Enthalpy of air at inlet for case 2, h i2, ( ) 265.36 260.0 270.0 260.0 h i2 = Q ṁ 2 + h o h i2 = 7.39 103 kw 1600 kg/s h i2 = 242.62 kj/kg 1 + 238 kj/kg Temperature of air at inlet, T i2, Interpolation for T i2 : h (kj/kg) 240.0 242.62 250.0 T (K) 240 T i2 250 T i2 240 250 240 = 242.62 240.0 250.0 240.0 T i2 = 240 + (250 240) T i2 = 242.62 K 1 Temperature of air at inlet, T i2 = 242.62 K = 30.38 o C. Exit temperature of water-ethylene glycol mixture, T o, Q = ṁ(h o h i ) Q = ṁc(t o T i ) 2 T o = T o = ( ) 242.62 240.0 250.0 240.0 Q ṁc + T i 7.39 10 3 kw (550 kg/s)(2.7 kj/kg K) + 228 K T o = 233 K 1 Exit temperature of secondary liquid, T o = 233 K = 40 o C. 51

HW-16 (25 points) Given: 2 for writing given, find, EFD, etc., Inlet conditions: T i = 20 o C = 293 K; p i = 1bar Ẇ = 3.73 MW ṁ = 1600 kg/s (300 mph case) Find: Air pressure rise, p = p o p i EFD: 3 Assumptions: 5 Steady state, steady flow (SSSF). 1-dimensional uniform flow. Kinetic and potential energy effects can be ignored. No heat transfer. 1-inlet and 1-outlet. Air is an ideal gas. Basic Equations: de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 2 Solution: Simplify energy equation, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o Ẇ = ṁ i h i ṁ o h o ṁ o = ṁ i = ṁ (from mass conservation) 2 Ẇ = ṁ(h i h o ) h o = Ẇ ṁ + h i 2 52

Enthalpy of air at inlet, h i, Interpolation for h i : T (K) 290 293 295 h (kj/kg) 290.1 h i 295.1 Enthalpy of air at outlet, h o, h i 290.1 295.1 290.1 = 293 290 295 290 h i = 290.1 + (295.1 290.1) h i = 293.1 kj/kg 1 ( ) 293 290 295 290 h o = Ẇ ṁ + h i h o = 3.73 103 kw 1600 kg/s h o = 295.43 kj/kg 2 + 293.1 kj/kg Temperature of air at outlet, T o, Interpolation for T o : h (kj/kg) 295.1 295.43 300.1 T (K) 295 T o 300 Gas constant for air, R, T o 295 300 295 = 295.43 295.1 300.1 295.1 T o = 295 + (300 295) T o = 295.33 K 1 R = R/MW ( ) 295.43 295.1 300.1 295.1 R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 0.287 kj/kgk 53

Air density (same at inlet and outlet), ρ, Pressure at outlet, p o, Pressure rise, p, p = ρrt pv = mrt p = m V RT p = ρrt ρ = p RT 2 ρ = p i RT i 100 kp a ρ = (0.287 kj/kgk)(293 K) ρ = 1.189 kg/m 3 1 p o = (1.189 kg/m 3 )(0.287 kj/kgk)(295.33 K) p o = 100.779 kp a 1 p = p o p i p = (100.779 100.0) kp a p = 0.779 kp a 1 54

HW-17 (25 points) Given: 1 for writing given, find, EFD, etc., State 1: p 1 = 9 bar, saturated liquid R134a State 2: p 2 = 2 bar State 3: p 3 = 2 bar, T 3 = 10 o C = 283 K State 4: T 4 = 25 o C = 298 K, p 4 = 1 bar, ṁ 4 = 2 kg/s, water State 5: T 5 = 15 o C = 288 K, p 5 = 1 bar Find: T 2 and refrigerant mass flow rate. EFD: For the throttling valve: 2 For the heat exchanger: 2 Assumptions: 3 Steady state, steady flow. 1-dimensional uniform flow. Kinetic and potential energy changes are negligible. No work transfer. No heat transfer. Basic Equations: de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 55

Solution: Energy balance for the throttle, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o 0 = ṁ 1 h 1 ṁ 2 h 2 ṁ 1 = ṁ 2 (from mass conservation) 2 h 1 = h 2 2 Enthalpy at state 1, h 1 = 101.64 kj/kg (from saturation tables). 1 Thus h 2 = 101.64 kj/kg. At state 2, p 2 = 2 bar, h f = 38.452 kj/kg and h g = 224.50 kj/kg. h f < h 2 < h g. Thus it is a Saturated Liquid Vapor Mixture (SLVM). 1 Hence T 2 = T sat at p 2 = 2 bar. T 2 = 10.076 o C. 1 Since refrigerant and water do not mix in the heat exchanger, mass conservation yields, ṁ 2 = ṁ 3 and ṁ 4 = ṁ 5. 2 Energy balance for the heat exchanger, de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁ o (h + ke + pe) o 0 = i ṁ i h i o ṁ o h o 0 = ṁ 2 h 2 + ṁ 4 h 4 ṁ 3 h 3 ṁ 5 h 5 0 = ṁ 2 (h 2 h 3 ) + ṁ 4 (h 4 h 5 ) ṁ 2 = ṁ4(h 4 h 5 ) h 3 h 2 2 Enthalpy at state 3: at p 3 = 2 bar, T sat = 10.076 o C. T 3 > T sat. Thus it is a Super Heated Vapor (SHV). From SHV tables, h 3 = 261.6 kj/kg. 1 Enthalpy at state 4: at p 4 = 1bar, T sat = 99.61 o C. T 4 < T sat. Thus it is a Compressed Liquid (CL). Using the approximation, h 4 h f (T 4 ) + v f (T 4 )(p 4 p sat (T 4 )) 1 h 4 104.83 kj/kg + (0.001003 m 3 /kg)(100 3.1699) kp a h 4 104.93 kj/kg 1 Enthalpy at state 5: at p 5 = 1bar, T sat = 99.61 o C. T 5 < T sat. Thus it is a Compressed Liquid (CL). Using the approximation, h 5 h f (T 5 ) + v f (T 5 )(p 5 p sat (T 4 )) h 5 62.981 kj/kg + (0.0010009 m 3 /kg)(100 1.7058) kp a h 5 63.08 kj/kg 1 56

Thus, ṁ 2 = ṁ4(h 4 h 5 ) h 3 h 2 (2)(104.93 63.08) ṁ 2 = (261.6 101.64) ṁ 2 = 0.523 kg/s 1 57

HW-18 (25 points) Given: 2 for writing given, find, EFD, etc., State 1: p 1 = 1 bar, T 1 = 300 K State 2: p 2 = 25.0 bar, T 1 = 750 K State 3: p 3 = 25.0 bar, T 1 = 1700 K State 4: p 4 = 1.0 bar, T 1 = 800 K Find: Net shaft output power (kj/kg) EFD: 5 Assumptions: 5 Steady state, steady flow. 1-dimensional uniform flow. Compressor and turbine are adiabatic. Neglect potential energy change. Neglect kinetic energy change. Air is modeled as an ideal gas. Basic Equations: de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 2 Solution: From conservation of mass, ṁ 1 = ṁ 2 = ṁ 3 = ṁ 4 = ṁ 2 58

Energy balance for the compressor and turbine gives, de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁ o (h + ke + pe) o 0 = Ẇ + i ṁ i h i o ṁ o h o Ẇ = ṁ 1 h 1 + ṁ 3 h 3 ṁ 2 h 2 ṁ 4 h 4 Ẇ = ṁ(h 1 h 2 + h 3 h 4 ) Ẇ ṁ = (h 1 h 2 + h 3 h 4 ) ẇ = (h 1 h 2 + h 3 h 4 ) 3 For ideal gas, h is a function of temperature only. Enthalpy at state 1, h 1 = 300.1 kj/kg 1 Enthalpy at state 2, h 2 = 767.3 kj/kg 1 Enthalpy at state 3, h 3 = 1880 kj/kg 1 Enthalpy at state 4, h 4 = 821.9 kj/kg 1 Thus, the net shaft output power is ẇ = (h 1 h 2 + h 3 h 4 ) ẇ = (300.1 767.3 + 1880 821.9) ẇ = 590.9 kj/kg 2 59

HW-19 (25 points) Given: 1 for writing given, find, EFD, etc., State 1: p 1 = 15 bar, T 1 = 180 o C, ṁ 1 = 5 kg/s State 2: p 2 = 4 bar, Saturated Vapor State 3: p 3 = 4 bar, Saturated Liquid State 4: p 4 = 0.08 bar, x 4 = 90% Find: Power developed by the turbine EFD: System 1: Valve and the flash chamber 3 System 2: Turbine 3 Assumptions: 3 Steady state, steady flow. 1-dimensional uniform flow. All components are adiabatic. Neglect potential energy change. 60

Neglect kinetic energy change. Basic Equations: dm dt = i ṁ i o ṁo 1 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 2 Solution: Conservation of mass for system 1, dm dt = ṁ i i o ṁ o 0 = ṁ 1 ṁ 2 ṁ 3 ṁ 3 = ṁ 1 ṁ 2 1 Energy balance for system 1, de dt = Q Ẇ + ṁ i (h + ke + pe) i i o 0 = ṁ 1 h 1 ṁ 2 h 2 ṁ 3 h 3 ṁ o (h + ke + pe) o 0 = ṁ 1 h 1 ṁ 2 h 2 (ṁ 1 ṁ 2 )h 3 0 = ṁ 1 (h 1 h 3 ) ṁ 2 (h 2 h 3 ) ṁ 2 = ṁ1(h 1 h 3 ) (h 2 h 3 ) 2 At p 1 = 15 bar, T sat = 198.29 o C, T 1 < T sat, and so it is a Compressed Liquid (CL). h 1 h 1 (T 1 ) + v f (T 1 )(p 1 p sat (T 1 )) 1 h 1 763.05 kj/kg + (0.0011274 m 3 /kg)(1500 1002.8) kp a h 1 763.6 kj/kg 1 Enthalpy at state 2, h 2 = h g (p 2 ) = 2738.1 kj/kg. Enthalpy at state 3, h 3 = h f (p 3 ) = 604.65 kj/kg. Thus, ṁ 2 = ṁ1(h 1 h 3 ) (h 2 h 3 ) (5 kg/s)(763.6 604.65) ṁ 2 = (2738.1 604.65) ṁ 2 = 0.372 kg/s 1 Energy balance for system 2, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o Ẇ = ṁ 2 h 2 ṁ 4 h 4 ṁ 4 = ṁ 2 (from mass conservation) 2 Ẇ = ṁ 2 (h 2 h 4 ) 2 61

Enthalpy at state 4, Power output from turbine, h 4 = (1 x)h f4 + xh g4 h 4 = (1 0.9)(173.84) + 0.9(2576.2) h 4 = 2336 kj/kg 1 Ẇ = ṁ 2 (h 2 h 4 ) Ẇ = (0.372 kg/s)(2738.1 2336) Ẇ = 149.6 kw 1 62

HW-20 (25 points) Given: 1 for writing given, find, EFD, etc., State 1: p 1 = 4 MP a, T 1 = 600 o C State 2: p 2 = 0.2 bar, Saturated Vapor State 5: T 5 = 15 o C State 6: T 6 = 35 o C w 34 = 4 kj/kg q 41 = 3400 kj/kg Find: η and ṁ cw /ṁ s EFD: System 1: Turbine 3 System 2: All components 3 63

Assumptions: 3 Steady state, steady flow. 1-dimensional uniform flow. Neglect potential energy change. Neglect kinetic energy change. Basic Equations: dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 2 Solution: Energy balance for system 1, de dt = Q Ẇ + ṁ i (h + ke + pe) i i ṁ o (h + ke + pe) o o Ẇ = ṁ 1 h 1 ṁ 2 h 2 ṁ 1 = ṁ 2 = ṁ 3 = ṁ 4 = ṁ s (from mass conservation) 1 Ẇ = ṁ 1 (h 1 h 2 ) w = (h 1 h 2 ) 2 64

At p 1 = 4 MP a, T sat = 250.35 o C. T 1 > T sat, thus it is a Super Heated Vapor (SHV). Enthalpy at state 1, h 1 = 3674.9 kj/kg Enthalpy at state 2, h 2 = 2608.9 kj/kg Thermal efficiency, w = (h 1 h 2 ) w = 3674.9 2608.9 w = 1066 kj/kg 1 η = w net q in 1 η = w t w p q in η = 1066 4 3400 η = 31.24% 1 Energy balance for system 2, de dt = Q Ẇ + ṁ i (h + ke + pe) i i 0 = Q in Ẇt Ẇp + ṁ 5 h 5 ṁ 6 h 6 ṁ 5 = ṁ 6 = ṁ cw (from mass conservation) 1 ṁ cw ṁ s 0 = Q in Ẇt Ẇp + ṁ cw (h 5 h 6 ) 0 = Q in ṁ s Ẇt ṁ s Ẇp ṁ s + ṁcw ṁ s (h 5 h 6 ) 0 = q w t w p + ṁcw ṁ s (h 5 h 6 ) = q w t w p h 6 h 5 2 ṁ o (h + ke + pe) o o At p 5 = 1 bar, T sat = 99.61 o C. T 5 < T sat. Thus it is a Compressed Liquid (CL). Enthalpy at state 5, h 5 h 5 (T 5 ) + v f (T 5 )(p 5 p sat (T 5 )) 1 h 5 62.981 kj/kg + (0.0010009 m 3 /kg)(100 1.7058) kp a h 5 63.1 kj/kg 1 At p 6 = 1 bar, T sat = 99.61 o C. T 6 < T sat. Thus it is a Compressed Liquid (CL). Enthalpy at state 6, h 6 h 6 (T 6 ) + v f (T 6 )(p 6 p sat (T 6 )) h 6 146.63 kj/kg + (0.001006 m 3 /kg)(100 5.629) kp a h 6 146.7 kj/kg 1 65

ṁ cw ṁ s ṁ cw ṁ s ṁ cw ṁ s = q w t w p h 6 h 5 3400 1066 ( 4) = 146.7 63.1 kg of cooling water = 27.97 kg of steam 1 66

HW-21 (25 points) (a) Yes, there are irreversibilities due to heat transfer between the steam and the air through a finite temperature difference. There could also be irreversibilities associated with friction between the fluid and the pipe walls. 2 (b) Assumptions for the gas as the system to have no irreversibilities during the process (internally reversible process): 2 No frictional effects within the system. Expansion happens slowly so that the system is in equilibrium (all properties are uniform throughout the system) during the expansion process. Assumptions for both the gas and surroundings to have no irreversibilities for the process (totally reversible process): 2 Temperature difference between the system and the surroundings is negligible. No friction within the system or between the piston and the cylinder. Expansion happens slowly so that the system and the surroundings are in equilibrium (all properties are uniform throughout) during the expansion process. (c) Given: 1 for writing given, find, EFD, etc., Q = 650 J/m 2 s, A = 50 m 2. T surr = 37 o C, T in = 22 o C, Ẇ other = 15.92 kw. Q people = 6000 kj/hr, Q ext = 18000 kj/hr, COP = 3.2 Find: Ẇ in,a/c, η and Q out,tot EFD: 1 Assumptions: 1 All systems are in steady state. 67

No mass transfer occurs across the systems. Basic Equations: dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 Solution: Total heat transfer to the house, Q in,h = 6000 + 18000 = 24000 kj/hr. 1 Heat transfer to A/C unit is thus, Q in,2 = 24000 kj/hr = 6.67 kw. COP = Q in,2 Ẇ in,a/c 1 Ẇ in,a/c = Q in,2 COP Ẇ in,a/c = 6.67 3.2 Ẇ in,a/c = 2.08 kw 1 Now, Ẇ out = Ẇin,A/C + Ẇother 2 Ẇ out = 2.08 + 15.92 Ẇ out = 18 kw 1 Heat input to the power cycle, Q in,1 = 50 650 J/s = 32.5 kw. Efficiency, Heat rejected from the power cycle, Heat rejected from the A/C, η = Ẇout Q in,1 1 η = 18 32.5 η = 55.38% 1 Q out,1 = Q in,1 Ẇout 2 Q out,1 = 32.5 18 Q out,1 = 14.5 kw 1 Q out,2 = Q in,2 + Ẇin,A/C 2 Q out,2 = 6.67 + 2.08 Q out,2 = 8.75 kw 1 68

Total rate of heat transfer to the surroundings, Q out,tot = Q out,1 + Q out,2 Q out,tot = 14.5 + 8.75 Q out,tot = 23.25 kw 1 69

HW-22 (25 points) (a) Given: Q out = 1820 kw hr, COP = 2.62, Cost of electricity = 15 cents/kw hr. Find: Savings in electricity cost, percentage of energy coming from the environment. EFD: 1 Heat Pump Assumptions: Steady state. Basic Equations: COP = Q out Ẇ in 1 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 Solution: COP = Q out Ẇ in Ẇ in = Q out COP Ẇ in = 1820 2.6 Ẇ in = 700 kw hr 1 Cost of electricity = 700 0.15 = $105. 1 Cost of electricity if resistance heating is used = 1820 0.15 = $273. 1 70

Savings in cost = $273 $105 = $168. 1 Energy balance for the A/C yields, Q out = Ẇin + Q in 1 Q in = Q out Ẇin Q in = 1820 700 Q in = 1120 kw hr 1 Percentage of energy coming from the environment, x = Q in Q out 1 x = 1120 1820 x = 61.54 % 1 (b) Given: T h1 = 727 o C = 1000 K, T c1 = 27 o C = 300 K, Q out,1 = 5 kw T h2 = 27 o C = 300 K, T c2 = 23 o C = 250 K, Q out,2 = 120 kw Heat pump is reversible. Ẇ out = Ẇin Find: Determine if the heat engine is reversible, irreversible or impossible. EFD: 2 T h1 T c2 Heat Engine Heat Pump T c1 T h2 Assumptions: Steady state. 71

Basic Equations: COP = Q out 1 Ẇ in COP rev = T h 1 T h T c η = Ẇout Q in 1 η rev = T h T c T h 1 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 Solution: Since the heat pump is reversible, its COP is, Work input of the heat pump is thus, T h COP = T h T c 300 COP = 300 250 COP = 6 1 Energy balance of the heat engine yields COP = Q out Ẇ in Ẇ in = Q out COP Ẇ in = 120 6 Ẇ in = 20 kw 1 Q in = Ẇout + Q out 1 Q in = 20 + 5 Q in = 25 kw 1 Efficiency of the heat engine, η = Ẇout Q in η = 20 25 η = 80 % 1 72

Efficiency under reversible conditions, η rev = T h T c T h 1000 300 η rev = 1000 η rev = 70 % 1 Since η > η rev, the heat engine operation is impossible. 1 73

HW-23 (25 points) Given: m = 2 kg, T 1 = T 2 = 750 K, T 3 = T 4 = 300 K, Q 12 = 60 kj, V 2 = 0.4 m 3 Find: p 1, V 1, Q and W for all processes. Find η and compare with η rev. EFD: 1 Assumptions: 1 Air behaves as an ideal gas. Quasi-equilibrium process. Potential and kinetic energy changes can be neglected. No mass enters or leaves the system. Basic Equations: W = pdv 1 pv = mrt 1 η = Ẇout Q in 1 η rev = T h T c T h 1 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 Solution: p V diagram: 3 74

Gas constant for air, R, R = R/MW R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 0.287 kj/kgk Closed system Energy balance yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o o For process 1-2, E = Q W U + KE + P E = Q W Work can also be expressed as, U = Q W 1 m u = Q W m u = Q 12 W 12 W 12 = i 0 = Q 12 W 12 (isothermal process) 1 W 12 = Q 12 W 12 = 60 kj 2 1 2 pdv mrt 1 60 = dv 1 V ( ) V2 60 = mrt 1 ln V 1 ( ) 0.4 m 3 60 kj = (2 kg)(0.287 kj/kgk)(750 K)ln V 1 = 0.348 m 3 1 V 1 75

From ideal gas law, at state 1, p 1 V 1 = mrt 1 p 1 = mrt 1 p 1 = V 1 (2 kg)(0.287 kj/kgk)(750 K) (0.348 m 3 ) p 1 = 1237 kp a 1 From ideal gas tables, u 1 = u 2 = 552 kj/kg. 1 From ideal gas tables, u 3 = u 4 = 214 kj/kg. 1 Process 2-3 is adiabatic and hence Q 23 = 0. Thus from energy balance, we have, m u = W 23 W 23 = m(u 2 u 3 ) W 23 = 2(552 214) W 23 = 676 kj 1 For process 3-4, since it is reversible, Q C = T C 2 Q H T H Q 34 = 300 K Q 12 700 K Q 34 = (60 kj) 300 K 700 K Q 34 = 24 kj From energy balance of process 3-4, we have, Q 34 = 24 kj (since Q is out of the system) 1 m u = Q 34 W 34 0 = Q 34 W 34 (isothermal process) W 34 = Q 34 W 34 = 24 kj 1 Process 4-1 is adiabatic and hence Q 41 = 0. Thus from energy balance, we have, m u = W 41 W 41 = m(u 4 u 1 ) W 41 = 2(214 552) W 41 = 676 kj 1 76

Net work, Efficiency, Carnot expression for efficiency, Thus we have η = η rev. W net = W 12 + W 23 + W 34 + W 41 W net = 60 + 676 24 676 W net = 36 kj 1 η = W net Q in η = 36 60 η = 60 % 1 η rev = T h T c T c 750 300 η rev = 750 η rev = 60 % 1 77

HW-24 (25 points) (a) At inlet to flash chamber p = 4 bar, h = 763.6 kj/kg. At p = 4 bar, h f = 604.65 kj/kg and h g = 2738.1 kj/kg. h f < h < h g. Thus it is a Saturated Liquid Vapor Mixture (SLVM). Quality at state 1, h 1 = (1 x 1 )h f1 + x 1 h g1 x 1 = h 1 h f1 h g1 h f1 x 1 = 0.07 1 s = (1 x)s f + xs g s = (1 0.07)(1.7765) + 0.07(6.8955) s = 2.13483 kj/(kgk) 1 State 4 (exit of turbine) is a Saturated Liquid Vapor Mixture (SLVM). Entropy at state 4, s 4 = (1 x)s f4 + xs g4 1 s 4 = (1 0.9)(0.59249) + 0.9(8.2273) s 4 = 7.46386 kj/(kgk) 1 (b) (i) At p 1 = 0.20 bar, s f = 0.83202 kj/(kgk) and s g = 7.9072 kj/(kgk). Since s f < s 1 < s g, it is a Saturated Liquid Vapor Mixture (SLVM). 1 Quality at state 1, s 1 = (1 x 1 )s f1 + x 1 s g1 x 1 = s 1 s f1 s g1 s f1 x 1 = 0.50 1 Enthalpy, h = (1 x)h f + xh g h = (1 0.5)(251.42) + 0.5(2608.9) h = 1430.16 kj/kg 1 T s diagram: 3 78

T CL p = constant SLVM SHV s (ii) At p = 10 bar, u g = 2582.7 kj/kg. Since u > u g, it is a Super Heated Vapor (SHV). 1 At u = 3125.0 kj/kg, s = 7.764 kj/(kgk). 1 T s diagram: 3 T CL p = constant SLVM SHV s (iii) Since quality is given, it is a Saturated Liquid Vapor Mixture (SLVM). 1 At T = 20 o C, s f = 0.30063 kj/(kgk), s g = 0.92243 kj/(kgk). T s diagram: 3 s = (1 x)s f4 + xs g4 s = (1 0.8)(0.30063) + 0.8(0.92243) s = 0.79808 kj/(kgk) 1 79

T CL p = constant SLVM SHV s (iv) At T = 20 o C, s f = 1.0421 kj/(kgk), s g = 5.0891 kj/(kgk). Since s = s g, it is a Saturated Vapor (SV). 1 Thus u = u g = 1333.1 kj/kg 1. T s diagram: 3 T CL p = constant SLVM SHV s 80

HW-25 (25 points) (a) Given: p 1 = 25 bar, T 1 = 20 o C, p 2 = 50 bar, T 2 = 40 o C. Find: s. Solution: (i) Using compressed liquid tables, at state 1, s 1 = 0.2960 kj/(kgk). 1 Using compressed liquid tables, at state 2, s 2 = 0.5705 kj/(kgk). 1 Thus, (ii) Using saturated liquid approximation, Thus, s = s 2 s 1 s = 0.5705 0.2960 s = 0.2745 kj/(kgk) 1 s s f (T ) 1 s 1 0.296480 kj/(kgk) 1 s 2 0.572400 kj/(kgk) 1 s = s 2 s 1 s = 0.572400 0.296480 s = 0.27592 kj/(kgk) 1 (iii) Using constant specific heat, s = Cln s = 4.18ln ) 1 T ( 1 ) 40 + 273 20 + 273 ( T2 s = 0.27601 kj/(kgk) 1 (b) Given: 1 for writing given, find, etc., m c = 50 kg, T ic = 350 o C = 623 K, V w = 120 L, thus m w = 120 kg and T iw = 25 o C = 298 K, C c = 385 J/(kgK) = 0.385 kj/(kgk), C w = 4180 J/(kgK) = 4.18 kj/(kgk). Find: S c and S w EFD: 2 Copper Water 81

Assumptions: 3 Closed system. Neglect KE and PE changes. Both copper and water are incompressible substances and have constant specific heats. No heat transfer or work transfer. Basic Equations: E = Q W 1 U = mc T ( 1 ) Tf S = mcln T i 1 Solution: At equilibrium, final temperature of copper and water are same. Thus T fw = T fc = T f. 1 E = Q W U = 0 U c + U w = 0 2 m c C c (T fc T ic ) + m w C w (T fw T iw ) = 0 m c C c (T f T ic ) + m w C w (T f T iw ) = 0 Entropy change of copper block, Entropy change of liquid water, T f (m c C c + m w C w ) = m c C c T ic + m w C w T iw S c = m c C c ln T f = m cc c T ic + m w C w T iw 1 m c C c + m w C w (50)(0.385)(623) + (120)(4.18)(298) T f = (50)(0.385) + (120)(4.18) T f = 310 K 1 ( Tfc T ic S c = (50)(0.385)ln ) S c = 13.436 kj/k 1 S w = m w C w ln ( Tfw T iw S w = (120)(4.18)ln ( ) 310 K 623 K ) S w = 19.8026 kj/k 1 ( 310 K 298 K ) 82

HW-26 (25 points) Given: 1 for writing given, find, etc., T 1 = 290 K, p 1 = 1 bar, p 2 = 5 bar, p 3 = 1 bar Process 1-2: pv 1.19 = const Process 2-3: Isentropic process. Find: T s diagram, T 2 and W net. EFD: 2 Assumptions: 3 Closed system. Neglect KE and PE changes. Air is an ideal gas. Basic Equations: E = Q W 1 Solution: T s diagram 3 2 T 1 3 s 83

For an ideal gas, pv = RT pv T = R = const p 1 v 1 = p 2v 2 T 1 T ( 2 ) ( ) T 2 p2 v2 = T 1 p 1 v 1 For a polytropic process, ( v1 p 1 v1 n = p 2 v2 n ) n ( ) p2 = v 2 ( v2 ) n = p 1 ( p1 ) Thus, we have T 2 T 1 = T 2 T 1 = T 2 T 1 = v 1 ( v2 v 1 ( p2 p 1 ( p2 p 1 ( p2 p 1 ) = ) ( p1 p 2 ( p1 p 2 ) ( p2 ) T 2 = T 1 ( p2 p 1 T 2 = (290 K) T 2 = 375 K 1 p 1 p 2 ) ) 1 n ) n 1 n ) 1 n 1 n n 1 n ( 5 1 ) 1 1.19 1 1.19 Gas constant for air, R, R = R/MW R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 0.287 kj/kgk 1 84

For a polytropic process, At state 2, entropy, s o 2, Interpolation for s o 2: w 12 = p 2v 2 p 1 v 1 1 n w 12 = R(T 2 T 1 ) (ideal gas) 1 n 0.287(375 290) w 12 = 1 1.19 w 12 = 128.39 kj/kg 2 T (K) 370 375 380 s (kj/(kgk)) 1.914 s o 2 1.941 Since 2-3 is isentropic, s o 2 1.914 1.941 1.914 Internal energy at state 3, u 3, Interpolation for u 3 : s o 3 s o 2 Rln = 375 370 380 370 s o 2 = 1.914 + (1.941 1.914) s o 2 = 1.9275 kj/(kgk) 1 ( p3 s = 0 1 ) = 0 p 2 s o 3 = s o 2 + Rln s o 3 = s o 2 + Rln ( p3 p ( 2 p3 p 2 ( ) 375 370 380 370 ) ) s o 3 = 1.9275 + (0.287)ln s o 3 = 1.4651 kj/(kgk) 1 ( ) 1 5 85

s (kj/(kgk)) 1.437 1.4651 1.479 u (kj/kg) 164.0 u 3 171.1 u 3 164.0 171.1 164.0 = 1.4651 1.437 1.479 1.437 u 3 = 164.0 + (171.1 164.0) u 3 = 168.8 kj/kg 1 For process 2-3, Q 23 = 0 since it is an isentropic process. 1 Internal energy at state 2, Interpolation for u 2 : ( ) 1.4651 1.437 1.479 1.437 T (K) 370 375 380 u (kj/kg) 264.4 u 2 271.6 u 2 264.4 271.6 264.4 = 375 370 380 370 u 2 = 264.4 + (271.6 264.4) u 2 = 268.0 kj/kg 1 Thus, the energy equation becomes, Net work, w net, W 23 = U 1 w 23 = u w 23 = (u 3 u 2 ) w 23 = (168.8 268.0) kj/kg w 23 = 99.2 kj/kg 1 w net = w 12 + w 23 1 w net = 128.39 + 99.2 w net = 29.19 kj/kg 1 ( ) 375 370 380 370 86

HW-27 (25 points) (a) Given: 1 for writing given, find, etc., p 1i = 4 bar, T 1i = 400 K, V 1i = 1.5 m 3 p 2i = 2 bar, T 2i = 400 K, V 2i = 1.5 m 3 Find: T f, p f and σ. EFD: 2 1 2 Assumptions: 3 Closed system. Neglect KE and PE changes. Air is an ideal gas. Piston is massless and thermally conducting. No heat or work transfer. Basic Equations: E = Q W 1 S = Q T b + σ 1 Solution: At equilibrium, T 1f = T 2f = T f and p 1f = p 2f = p f. 1 From energy balance, E = Q W U = 0 U 1 + U 2 = 0 2 m 1 (u 1f u 1i ) + m 2 (u 2f u 2i ) = 0 m 1 (u f u 1i ) + m 2 (u f u 2i ) = 0 (since T 1f = T 2f ) u f (m 1 + m 2 ) = m 1 u i1 + m 2 u i2 u f (m 1 + m 2 ) = (m 1 + m 2 )u 1 (since T 1i = T 2i ) u f = u i T f = T i T f = 400 K 1 87

Initial mass at state 1, Similarly, initial mass at state 2, Final pressure, p f = mrt f V f m 1 = p 1iV 1i RT 1i m 1 = p 1iV f /2 RT f 1 m 2 = p 2iV 2i RT 2i m 2 = p 2iV f /2 RT f 1 p f = (m 1 + m 2 )RT f p f = V f ( p1i V f /2 + p ) 2iV f /2 RT f RT f RT f p f = p 1i + p 2i 2 p f = 4 + 2 2 p f = 3 bar 2 V f (b) (i) True. 1 S = Q T b + σ σ = S 1 + S 2 ( )) ( σ = m 1 (s o 1f s o p1f 1i Rln + m 2 s o 2f s o 2i Rln p ( )) 1i ( ( )) p1f p2f σ = m 1 ( Rln + m 2 Rln p 1i p 2i σ = p ( ( )) 1iV f /2 p1f Rln + p ( ( )) 2iV f /2 p2f Rln RT f p 1i RT f p 2i σ = V ( ( )) ( ( )) f/2 p1f p2f p 1i ln + p 2i ln T f p 1i p 2i σ = 1.5 ( ( )) ( ( )) 3 3 (4)ln + (2)ln 400 4 2 σ = 0.1274 kj/k 2 88 ( p2f p 2i ))

(ii) False (the given statement is not always true). 1 (iii) True. 1 (iv) True. 1 (v) False. 1 (vi) True. 1 (vii) False. 1 89

HW-28 (25 points) Given: 1 for writing given, find, etc., p i = 6 MP a, T i = 600 o C, ṁ = 125 kg/min = 2.0833 kg/s, Ẇ = 2 MW p o = s0 kp a, saturated vapor T b = 27 o C = 300 K Find: Q and σ. EFD: 4 Assumptions: 4 Neglect KE and PE changes. Quasi-equilibrium. Steady state, steady flow. 1 dimensional uniform flow. Basic Equations: dm dt = i ṁ i o ṁo 1 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 ds dt = j Q T b,j + i ṁ i s i o ṁos o + σ 1 Solution: Energy balance for the turbine yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o Q = Ẇ + ṁ oh o ṁ i h i ṁ o = ṁ i = ṁ (from mass conservation) 2 Q = Ẇ + ṁ(h o h i ) 3 90

From superheated tables, at inlet, h i = 3658.7 kj/kg and s i = 7.169 kj/kgk. 2 From saturation tables, at outlet, h o = 2608.9 kj/kg and s o = 7.9072 kj/kgk. 2 Thus Q is, Q = Ẇ + ṁ(h o h i ) Q = 2000 kw + (2.0833 kg/s)(2608.9 3658.7) Q = 187.04 kw 1 Entropy balance for the turbine yields, ds dt = Q + T j b,j ṁ i s i ṁ o s o + σ i o σ = Q + ṁ(s o s i ) 2 T b 187.04 kw σ = + (2.0833 kg/s)(7.9072 7.169) 300 K σ = 2.162 kw/k 1 91

HW-29 (25 points) Given: State 1, p 1 = 96 kp a, T 1 = 27 o C = 300 K, V = 26.91 m 3 /min = 0.4485 m 3 /s State 2, p 2 = 230 kp a, T 2 = 127 o C = 400 K State 3, p 3 = p 2 = 230 kp a, T 3 = 77 o C = 350 K State A, T A = 25 o C, p A = 1 bar, water State B, T B = 40 o C, p B = 1 bar, water Find: Ẇ, ṁ w, σ for compressor and heat exchanger. EFD: 2 Assumptions: 3 Neglect KE and PE changes. Quasi-equilibrium. Steady state, steady flow. 1 dimensional uniform flow. No heat transfer. No work transfer for the heat exchanger. Air behaves as an ideal gas. Basic Equations: pv = mrt dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 92

ds dt = j Q T b,j + i ṁ i s i o ṁos o + σ 1 Solution: Energy balance for the compressor yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i i ṁ o (h + ke + pe) o o Ẇ = ṁ 1 h 1 ṁ 2 h 2 ṁ 1 = ṁ 2 = ṁ 3 = ṁ a (from mass conservation) 1 Ẇ = ṁ a (h 1 h 2 ) 1 Gas constant for air, R, Mass flow rate of air, R = R/MW R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 0.287 kj/kgk pv = mrt p V = ṁ a RT ṁ a = p V RT ṁ a = (96 kp a)(0.4485 m3 /s) (0.287 kj/kgk)(300 K) ṁ a = 0.5 kg/s 1 From ideal gas tables, at state 1, h 1 = 300.1 kj/kg and s o 1 = 1.703 kj/kgk. 1 From ideal gas tables, at state 2, h 2 = 401.1 kj/kg and s o 2 = 1.993 kj/kgk. 1 From ideal gas tables, at state 3, h 3 = 350.5 kj/kg and s o 3 = 1.858 kj/kgk. 1 Thus Ẇ is, Ẇ = ṁ a (h 1 h 2 ) Ẇ = (0.5 kg/s)(300.1 401.1) Ẇ = 50.5 kw 1 Entropy balance for the heat exchanger yields, de dt = Q Ẇ + i ṁ i (h + ke + pe) i o 0 = ṁ a h 2 + ṁ A h A ṁ a h 3 dotm B h B ṁ A = ṁ B = ṁ w (from mass conservation) 1 0 = ṁ a (h 2 h 3 ) + ṁ w (h A h B ) ṁ w = h 2 h 3 h B h A ṁ a 1 93 ṁ o (h + ke + pe) o

Enthalpy at state A: at p A = 1bar, T sat = 99.61 o C. T A < T sat. Thus it is a Compressed Liquid (CL). Using the approximation, h A h f (T A ) + v f (T A )(p A p sat (T A )) h A 104.83 kj/kg + (0.001003 m 3 /kg)(100 3.1699) kp a h A 104.93 kj/kg 1 s A s f (T A ) s A 0.36722 kj/kgk 1 Enthalpy at state B: at p B = 1bar, T sat = 99.61 o C. T B < T sat. Thus it is a Compressed Liquid (CL). Using the approximation, Thus, ṁ w is, h B h f (T B ) + v f (T B )(p B p sat (T B )) h B 167.53 kj/kg + (0.0010079 m 3 /kg)(100 7.38489) kp a h B 167.62 kj/kg 1 s B s f (T B ) s B 0.5724 kj/kgk 1 Entropy balance for the compressor yields, ṁ w = h 2 h 3 ṁ a h B h A 401.1 350.5 ṁ w = (0.5 kg/s) 167.62 104.93 ṁ w = 0.4036 kg/s 1 ds dt = Q + T j b,j ṁ i s i ṁ o s o + σ i o σ = ṁ a (s 2 s 1 ) 1 ( )) σ = ṁ a (s o 2 s o p2 1 Rln p ( 1 ( )) 230 σ = (0.5 kg/s) 1.993 1.703 0.287ln 96 σ = 0.0196 kw/k 1 94

Entropy balance for the heat exchanger yields, ds dt = Q + T j b,j i ṁ i s i o ṁ o s o + σ σ = ṁ a (s 3 s 2 ) + ṁ w (s B s A ) 1 ( )) σ = ṁ a (s o 3 s o p3 2 Rln + ṁ w (s B s A ) p 2 σ = (0.5 kg/s)(1.858 1.993) + (0.4036)(0.5724 0.36722) σ = 0.0153 kw/k 1 95

HW-30 (25 points) Given: 1 for writing given, find, etc., State 1, p 1 = 12 bar, T 1 = 620 K State 2, p 2 = 1.4 bar C v = 0.718 kj/kgk Find: T 2 and w using tables and with constant specific heat ratio. EFD: 3 W Assumptions: 5 Closed system. Kinetic and potential energy changes are negligible. Quasi-equilibrium. No heat transfer. Air behaves as an ideal gas. Basic Equations: E = Q W 1 Solution: Energy balance for the air yields, E = Q W U = W From ideal gas tables, at state 1, p r1 = 18.36. 1 W = m(u 1 u 2 ) w = u 1 u 2 2 96

For an isentropic process, Interpolation for T 2 : p r1 p r2 = p 1 p 2 2 p r2 = p 2 p 1 p r1 p r2 = 1.4 12 18.36 p r2 = 2.142 1 p r 1.9352 2.142 2.149 T (K) 330 T 2 340 Interpolation for u 2 : T 2 330 340 330 = 2.142 1.9352 2.149 1.9352 T 2 = 330 + (340 330) T 2 = 339.7 K 1 ( ) 2.142 1.9352 2.149 1.9352 T (K) 330 339.7 340 u (kj/kg) 235.6 u 2 242.8 u 2 235.6 242.8 235.6 = 339.7 330 340 330 u 2 = 235.6 + (242.8 235.6) u 2 = 242.6 kj/kg 1 From ideal gas tables, at state 1, u 1 = 450.1 kj/kg. 1 w = u 1 u 2 w = 450.1 242.6 w = 207.5 kj/kg 1 ( ) 339.7 330 340 330 97

Using constant specific heat assumption, Work, T 2 T 1 = ( p2 p 1 ) T 2 = T 1 ( p2 p 1 T 2 = 620 k 1 k ) ( 1.4 12 T 2 = 335.6 K 1 k 1 k ) 0.4 1.4 2 w = u 1 u 2 w = C v (T 1 T 2 ) 1 w = (0.718 kj/kgk)(620 335.6) w = 204 kj/kg 1 98

HW-31 (25 points) Given: State 1, p 1 = 2 MP a, T 1 = 440 o C, steam State 2, p 2 = 100 kp a, η t = 90%, steam State 3, p 3 = 110 kp a, T 3 = 27 o C = 300 K, air State 4, p 4 = 496 kp a, η c = 85%, air Find: ṁ s /ṁ a EFD: 3 Assumptions: 4 Neglect KE and PE changes. Quasi-equilibrium. Steady state, steady flow. 1 dimensional uniform flow. No heat transfer. No work transfer. Air behaves as an ideal gas. Basic Equations: dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 99

Solution: Energy balance for the system yields, de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁ o (h + ke + pe) o 0 = ṁ 1 h 1 + ṁ 3 h 3 ṁ 2 h 2 ṁ 4 h 4 ṁ 1 = ṁ 2 = ṁ s (from mass conservation) 1 ṁ 3 = ṁ 4 = ṁ a (from mass conservation) 1 ṁ s (h 1 h 2 ) = ṁ a (h 4 h 3 ) ṁ s = h 4 h 3 2 ṁ a h 1 h 2 Isentropic efficiency for the turbine, Isentropic efficiency for the compressor, η t = h 1 h 2 h 1 h 2s 1 h 1 h 2 = η t (h 1 h 2s ) Thus, η c = h 4s h 3 h 4 h 3 1 h 4 h 3 = h 4s h 3 η c ṁ s = 1 ṁ a η c η t ( ) h4s h 3 h 1 h 2s From superheated tables, at state 1, h 1 = 3336.3 kp a and s 1 = 7.256 kj/kgk 1 At state 2, p 2 = 100 kp a and s 2s = s 1 = 7.256 kj/kgk. s f < s 2s < s g. Thus it is a saturated liquid vapor mixture (SLVM) 1 Quality at state 2s, s 2s = (1 x 2s )s f2s + x 2s s g2s Enthalpy, h 2s, x 2s = s 2s s f s g s f 7.256 1.3028 x 2s = 7.3588 1.3028 x 2s = 0.9830 1 h 2s = (1 x 2s )h f + x 2s h g 1 h 2s = (1 0.983)(417.50) + (0.983)(2674.9) h 2s = 2636.52 kj/kg 1 100

From ideal gas tables, h 3 = 300.1 kj/kg and p r3 = 1.386. 1 Since 3-4s is isentropic, Interpolation for h 4s : p r3 p r4s = p 3 p 4 s 1 p r4s = p 4 p 3 p r3 p r4s = 496 110 1.386 p r4s = 6.2496 1 p r 6.245 6.2496 6.742 h (kj/kg) 462.2 h 4s 472.4 h 4s 462.2 472.4 462.2 = 6.2496 6.245 6.742 6.245 h 4s = 462.2 + (472.4 462.2) h 4s = 462.3 kj/kg 1 ṁ s = 1 ( ) h4s h 3 ṁ a η c η t h 1 h 2s ṁ s 1 = ṁ a (0.85)(0.9) ṁ s ṁ a = 0.303 1 ( 462.3 300.1 3336.3 2636.52 ( ) 6.2496 6.245 6.742 6.245 ) If η c = η t = 1, then ṁ s = 1 ( ) h4s h 3 ṁ a η c η t h 1 h 2s ṁ s = 1 ( ) 462.3 300.1 ṁ a (1)(1) 3336.3 2636.52 ṁ s = 0.232 1 ṁ a 101

HW-32 (25 points) Given: 1 for writing given, find, etc., ṁ = 950 kg/s, p 1 = 1.5 bar, p 2 = 1 bar and v = 10 3 m 3 /kg Find: Ẇ max EFD: 5 Assumptions: 5 Neglect KE changes. Quasi-equilibrium. Steady state, steady flow. 1 dimensional uniform flow. No heat transfer. Liquid water is treated as an incompressible substance. Basic Equations: dm dt = i ṁ i o ṁo 1 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 Solution: In the case of an internally reversible and adiabatic process, the entropy is constant. For an incompressible substance, this implies temperature is constant. The power output of an internally 102

reversible turbine can then be obtained from energy balance as, de dt = Q Ẇrev + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o 2 i o ṁ 1 = ṁ 2 = ṁ (from mass conservation) 2 Ẇ rev = ṁ(h 1 h 2 + g(z 1 z 2 )) Ẇ rev = ṁ(u 1 + p 1 v u 2 p 2 v + g(z 1 z 2 )) 2 Ẇ rev = ṁ(v(p 1 p 2 ) + g(z 1 z 2 )) (u 1 = u 2 since T 1 = T 2 ) 2 Ẇ rev = (950 kg/s)((10 3 m 3 /kg)(1.5 1) 10 5 P a + (9.81 m/s 2 )(160 ( 10)) m) Ẇ rev = 1.632 10 6 W Ẇ rev = 1.632 MW 2 Maximum power output of the hydraulic turbine is equal to the work output of the internally reversible turbine. Hence the maximum power output is 1.632 MW. 2 103

HW-33 (25 points) Given: State 1: p 1 = 10 MP a, T 1 = 480 o C. State 3: p 3 = 6 kp a, saturated liquid. η t = 0.8 and η p = 0.7 Find: Q in /ṁ, Q out /ṁ and η th. EFD: 3 Assumptions: 2 Neglect KE and PE changes. Quasi-equilibrium. Steady state, steady flow. 1 dimensional uniform flow. No work transfer for the steam generator and condenser. No heat transfer for the turbine and the pump. Pressure across condenser and boiler is constant. Basic Equations: dm dt = i ṁ i o ṁo 104

de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o Solution: At state 1, from superheated tables, h 1 = 3323.0 kj/kg, s 1 = 6.531 kj/kgk. 1 At state 3, from saturation tables, h 3 = 151.48 kj/kg, s 3 = 0.52082 kj/kgk. 1 At state 2s, s 2s = s 1 = 6.531 kj/kgk and p 2s = 6 kp a. At p 2s, s f = 0.52082 kj/kgk and s g = 8.3290 kj/kgk. s f < s 2s < s g. Thus it is a saturated liquid vapor mixture (SLVM). Quality at state 2s, Enthalpy, h 2s, s 2s = (1 x 2s )s f2s + x 2s s g2s x 2s = s 2s s f s g s f 6.531 0.52082 x 2s = 8.3290 0.52082 x 2s = 0.7697 1 h 2s = (1 x 2s )h f + x 2s h g 1 h 2s = (1 0.7697)(151.48) + (0.7697)(2566.6) h 2s = 2010.4 kj/kg 1 Isentropic efficiency of a turbine is defined as, η t = h 1 h 2 h 1 h 2s 1 h 2 = h 1 (η t )(h 1 h 2s ) h 2 = 3323.0 (0.8)(3323.0 2010.4) h 2 = 2273 kj/kg 1 At state 4s, s 4s = s 3 = 0.52082 kj/kgk and p 4s = 10 MP a. At p 4s = 10 MP a, s f = 3.3606 kj/kgk. s < s f, thus it is a compressed liquid. For a liquid pump, h 4s h 3 = v 3 (p 4 p 3 ) h 4s = h 3 + v 3 (p 4 p 3 ) h 4s = 151.48 + (0.0010065 m 3 /kg)(10000 6) h 4s = 161.53 kj/kg Alternatively, h 4s can also be found from the compressed liquid tables, Interpolation for h 4s : 105

s 0.2944 0.52082 0.5685 h (kj/kg) 93.28 h 4s 176.36 h 4s 93.28 176.36 93.28 = 0.52082 0.2944 0.5685 0.2944 h 4s = 93.28 + (176.36 93.28) h 4s = 161.9 kj/kg 1 Isentropic efficiency of a pump is defined as, Energy balance for the turbine yields, η c = h 4s h 3 h 4 h 3 1 h 4 = h 3 + h 4s h 3 η c 161.9 151.48 h 4 = 151.48 + 0.7 h 4 = 166.36 kj/kg 1 ( ) 0.52082 0.2944 0.5685 0.2944 de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o Ẇ t = ṁ(h 1 h 2 ) Ẇ t ṁ = h 1 h 2 1 Ẇ t = 3323.0 2273.0 ṁ Ẇ t = 1050 kj/kg 1 ṁ Energy balance for the pump yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i i ṁ o (h + ke + pe) o o Ẇ p = ṁ(h 3 h 4 ) Ẇ p ṁ = h 3 h 4 Ẇ p ṁ Ẇ p ṁ = 151.48 166.36 = 14.88 kj/kg 1 106

Energy balance for the steam generator yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o ṁ 1 = ṁ 2 = ṁ 3 = ṁ 4 = ṁ (from mass conservation) Q in = ṁ(h 1 h 4 ) Q in ṁ = h 1 h 4 1 Q in = 3323.0 166.36 ṁ Q in = 3156.6 kj/kg 1 ṁ Energy balance for the condenser yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o Q out = ṁ(h 3 h 2 ) Q out ṁ = h 3 h 2 Net work, Q out ṁ Q out ṁ = 151.48 2273 = 2121.5 kj/kg 1 Net heat transfer, Ẇ net = Ẇt + Ẇp Ẇ net ṁ = Ẇt ṁ + Ẇp ṁ Ẇ net = 1050 14.88 ṁ Ẇ net = 1035.12 kj/kg ṁ Q net = Q in + Q out Q net ṁ Q net = Q in ṁ + Q out ṁ = 3156.6 2121.5 ṁ Q net = 1035.1 kj/kg ṁ 107

Ẇ net Thus ṁ Q net since it is a cycle. 1 ṁ Thermal efficiency, T s diagram: 2 η th = Ẇnet/ṁ Q in /ṁ η th = 1035.12 3156.6 η th = 32.79% 1 T p = const 1 CL 4s 4 3 p = const SLVM 2 2s SHV s 108

HW-34 (25 points) Given: State 1: p 1 = 120 MP a, T 1 = 480 o C. State 2: p 2 = 2 MP a, s 2 = s 1. State 3: p 3 = 0.3 MP a, s 3 = s 2. State 4: p 4 = 6 kp a, s 4 = s 3. State 5: p 5 = 6 kp a, saturated liquid. State 6: p 6 = 0.3 MP a, s 6 = s 5 State 7: p 7 = 0.3 MP a, saturated liquid. State 8: p 8 = 2 MP a, s 8 = s 7. State 9: p 9 = 12 MP a, T 9 = 210 o C. State 10: p 10 = 2 MP a, saturated liquid. State 11: p 11 = 0.3 MP a, h 11 = h 10. Find: Q in /ṁ, Q out /ṁ and η th. EFD: 2 1 y y' 1-y-y' 1 1-y-y' y y Assumptions: 1 Neglect KE and PE changes. Quasi-equilibrium. Steady state, steady flow. 1 dimensional uniform flow. No work transfer for the steam generator and condenser. 109

No heat transfer for the turbine and the pumps. No work and heat transfer for the feedwater heaters. Pressure across condenser and boiler is constant. Basic Equations: dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o Solution: Find enthalpy at all the states: At state 1, from superheated tables, h 1 = 3295.3 kj/kg, s 1 = 6.419 kj/kgk. At state 2, p 2 = 2 MP a, s 2 = 6.419 kj/kgk. From superheated tables, Interpolation for h 2 : s 6.339 6.419 6.497 h (kj/kg) 2798.4 h 2 2877.2 h 2 2798.4 2877.2 2798.4 = 6.419 6.339 6.497 6.339 h 2 = 2798.4 + (2877.2 2798.4) ( ) 6.419 6.339 6.497 6.339 h 2 = 2838.3 kj/kg 1 At state 3, p 3 = 0.3 MP a, s 3 = 6.419 kj/kgk. s f < s 3 < s g, thus it is a saturated liquid vapor mixture (SLVM). Quality at state 3, Enthalpy, h 3, s 3 = (1 x 3 )s f3 + x 3 s g3 x 3 = s 3 s f s g s f 6.419 1.6717 x 3 = 6.9916 1.6717 x 3 = 0.8924 h 3 = (1 x 3 )h f + x 3 h g h 3 = (1 0.8924)(561.43) + (0.8924)(2724.9) h 3 = 2492.1 kj/kg 1 At state 4, p 4 = 6 kp a, s 4 = 6.419 kj/kgk. s f < s 4 < s g, thus it is a saturated liquid vapor 110

mixture (SLVM). Quality at state 4, Enthalpy, h 3, s 4 = (1 x 4 )s f4 + x 4 s g4 x 4 = s 4 s f s g s f 6.419 0.52082 x 4 = 8.3290 0.52082 x 4 = 0.7554 h 4 = (1 x 4 )h f + x 4 h g h 4 = (1 0.7554)(151.48) + (0.7554)(2566.6) h 4 = 1975.9 kj/kg 1 At state 5, from saturation tables, h 5 = 151.48 kj/kg, v 5 = 0.0010065 m 3 /kg. At state 6, enthalpy is, h 6 h 5 + v 5 (p 6 p 5 ) h 6 151.48 kj/kg + (0.0010065 m 3 /kg)(300 6) kp a h 6 151.77 kj/kg 1 At state 7, from saturation tables, h 7 = 561.43 kj/kg, v 7 = 0.0010732 m 3 /kg. At state 8, enthalpy is, h 8 h 7 + v 7 (p 8 p 7 ) h 8 561.43 kj/kg + (0.0010732 m 3 /kg)(12000 300) kp a h 8 573.98 kj/kg 1 At state 9, from compressed liquid tables: at p = 10 MP a: Interpolation for h a : T 180 210 220 h (kj/kg) 767.68 h a 945.81 h a 767.68 945.81 767.68 = 210 180 220 180 h a = 767.68 + (945.81 767.68) h a = 901.3 kj/kg at p = 15 MP a: Interpolation for h b : ( ) 210 180 220 180 111

T 180 210 220 h (kj/kg) 770.32 h b 947.43 h b 770.32 947.43 770.32 = 210 180 220 180 h b = 770.32 + (947.43 770.32) h b = 903.2 kj/kg at p = 12 MP a: Interpolation for h 9 : p 10 12 15 ( ) 210 180 220 180 h (kj/kg) 901.3 h 9 903.2 h 9 901.3 903.2 901.3 = 12 10 15 10 h 9 = 901.3 + (903.2 901.3) h 9 = 902.1 kj/kg 1 At state 10, from saturation tables, h 10 = 908.5 kj/kg. At state 11, h 11 = h 10 = 908.5 kj/kg. Energy balance for the steam generator yields, ( ) 12 10 15 10 de dt = Q Ẇ + ṁ i (h + ke + pe) i i ṁ o (h + ke + pe) o o ṁ 9 = ṁ 1 = ṁ (from mass conservation) Q in = ṁ(h 1 h 9 ) 1 Q in = (1.5 10 4 kg/s)(3295.3 902.1) Q in = 35898 MW 1 Energy balance for the closed heater yields, de dt = Q Q + i ṁ i (h + ke + pe) i o ṁ o (h + ke + pe) o 0 = ṁh 8 + ṁyh 2 ṁh 9 ṁyh 10 y = h 9 h 8 1 h 2 h 10 902.1 573.981 y = 2838.3 908.5 y = 0.17 1 112

Energy balance for the open heater yields, de dt = Q Q + i Energy balance for the condenser yields, Thermal efficiency, ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o o 0 = ṁy h 3 + ṁyh 11 + (1 y y )h 6 ṁh 7 y = h 7 h 6 + y(h 6 h 11 ) 1 h 3 h 6 y 561.43 151.77 + (0.1704)(151.77 908.5) = 2492.1 151.77 y = 0.12 1 de dt = Q Ẇ + ṁ i (h + ke + pe) i i ṁ o (h + ke + pe) o o Q out = ṁ(1 y y )(h 5 h 4 ) 1 Q out = (1 y y )(h 5 h 4 ) Q out = (1.5 10 4 kg/s)(1 0.1704 0.12)(151.48 1975.9) Q out = 19419 MW 1 η th = Ẇnet Q in 1 η th = Q in Q out 1 Q in 35898 19419 η th = 35898 η th = 45.9% 1 113

T s diagram: 5 T 1 p = const 2 9 8 10 p = const CL 6 7 11 p = const 3 SHV 5 p = const SLVM 4 s 114

HW-35 (25 points) Given: State 1: p 1 = 2 bar, saturated vapor (R134a). State 3: p 3 = 8 bar, saturated liquid. η c = 0.8 and ṁ = 7 kg/min = 0.1167 kg/s Find: Ẇ c, Q in, COP and σ for the condenser. EFD: 2 Assumptions: 2 Neglect KE and PE changes. Quasi-equilibrium. Steady state, steady flow. 1 dimensional uniform flow. No work transfer for the evaporator, condenser and expansion valve. No heat transfer for the compressor and expansion valve. Pressure across condenser and evaporator is constant. 115

Basic Equations: dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o Solution: At state 1, from saturation tables, h 1 = 244.47 kj/kg, s 1 = 0.93779 kj/kgk. 1 At state 3, from saturation tables, h 3 = 95.501 kj/kg, s 3 = 0.35414 kj/kgk. 1 At state 2s, s 2s = s 1 = 0.93779 kj/kgk and p 2s = 8 bar. At p 2s, s > s g Thus it is a superheated vapor (SHV). 1 From superheated tables, Interpolation for h 2s : s 0.918 0.93779 0.948 h (kj/kg) 267.3 h 2s 276.5 h 2s 267.3 276.5 267.3 = 0.93779 0.918 0.948 0.918 h 2s = 267.3 + (276.5 267.3) h 2s = 273.4 kj/kg 1 Isentropic efficiency of a compressor is defined as, η c = h 2s h 1 h 2 h 1 1 h 2 = h 1 + h 2s h 1 At state 2, p 2 = 8 bar and h 2 = 280.63 kj/kg Interpolation for s 2 : η c 273.4 244.47 h 2 = 244.47 + 0.8 h 2 = 280.63 kj/kg 1 ( ) 0.93779 0.918 0.948 0.918 h 276.5 280.63 286.7 s (kj/kgk) 0.948 s 2 0.980 s 2 0.948 0.980 0.948 = 280.63 276.5 286.7 276.5 s 2 = 0.948 + (0.980 0.948) s 2 = 0.961 kj/kgk 1 ( ) 280.63 276.5 286.7 276.5 116

At state 4, h 4 = h 3 = 95.501 kj/kg. 1 Energy balance for the compressor yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i i ṁ o (h + ke + pe) o o ṁ 1 = ṁ 2 = ṁ 3 = ṁ 4 = ṁ (from mass conservation) 1 Ẇ c = ṁ(h 1 h 2 ) 1 Ẇ c = (0.1167 kg/s)(244.47 280.63) Ẇ c = 4.22 kw 1 Energy balance for the evaporator yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o Q in = ṁ(h 1 h 4 ) 1 Coefficient of performance, Q in = (0.1167 kg/s)(244.47 95.501)) Q in = 17.3846 kw Q in = 4.943 tons 1 COP = Q in Ẇc 1 COP = 17.3846 4.22 COP = 4.11 1 Energy balance for the condenser yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o Q out = ṁ(h 3 h 2 ) Q out = (0.1167 kg/s)(95.501 280.63)) Q out = 21.604 kw 1 Boundary temperature at which heat transfer occurs at the condenser T b = T sat (8 bar) 5 o C = 31.33 5 = 26.33 o C = 299.33 K 1 Entropy balance for the condenser yields, ds dt = Q + ṁ i s i ṁ o s o + σ T j b,j i o σ = Q out T b + ṁ(s 3 s 2 ) 1 σ = 21.604 + (0.1167 kg/s)(0.35414 0.961) 299.33 σ = 1.3539 10 3 kw/k 1 117

T s diagram: 2 2 T 2s 3 p = const CL 4 p = const SLVM 1 SHV s 118

HW-36 (25 points) Given: Q out = 20 kw Inside temperature = 21 o C, outside temperature = 0 o C. η c = 0.82 Find: ṁ, Ẇ c, COP and COP rev. EFD: 1 Assumptions: 1 Neglect KE and PE changes. Quasi-equilibrium. Steady state, steady flow. 1 dimensional uniform flow. No work transfer for the evaporator, condenser and expansion valve. No heat transfer for the compressor and expansion valve. Pressure across condenser and evaporator is constant. 119

Basic Equations: dm dt = i ṁ i o ṁo de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o ds dt = Q j + i T ṁ i s i o ṁos o + σ b,j Solution: State 1: T 1 = 0 10 = 10 o C, saturated vapor (R134a) (the evaporator has a lower temperature than the surroundings). 1 State 3: T 3 = 21 + 10 = 31 o C, saturated liquid (the condenser has a higher temperature than the office). 1 At state 1, from saturation tables, Interpolation for h 1 : T -12-10 -8 h (kj/kg) 243.31 h 1 245.72 Interpolation for s 1 : h 1 243.31 245.72 243.31 = 10 12 8 12 h 1 = 243.31 + (245.72 243.31) h 1 = 244.515 kj/kg 1 T -12-10 -8 ( ) 10 12 8 12 s (kj/kgk) 0.93918 s 1 0.93636 s 1 0.93918 10 12 = 0.93636 0.93918 8 12 s 1 = 0.93918 + (0.93636 0.93918) s 1 = 0.93777 kj/kgk 1 At state 3, from saturation tables, Interpolation for h 3 : ( ) 10 12 8 12 120

T 30 31 32 h (kj/kg) 93.578 h 3 96.479 Interpolation for s 3 : h 3 93.578 96.479 93.578 = 31 30 32 30 h 3 = 93.578 + (96.479 93.578) h 3 = 95.029 kj/kg 1 T 30 31 32 ( ) 31 30 32 30 s (kj/kgk) 0.34789 s 3 0.35730 Interpolation for p 3 : s 3 0.34789 31 30 = 0.35730 0.34789 32 30 s 3 = 0.34789 + (0.35730 0.34789) s 3 = 0.35260 kj/kgk 1 T 30 31 32 p (bar) 7.7020 p 3 8.1543 ( ) 31 30 32 30 p 3 7.7020 31 30 = 8.1543 7.7020 32 30 p 3 = 7.7020 + (8.1543 7.7020) p 3 = 7.92815 bar 1 Closest value in the superheated tables is 8 bar. Take p 3 = 8 bar At state 2s, s 2s = s 1 = 0.93777 kj/kgk and p 2s = p 3 = 8 bar. From superheated tables,1 Interpolation for h 2s : ( ) 31 30 32 30 121

s 0.918 0.93777 0.948 h (kj/kg) 267.3 h 2s 276.5 h 2s 267.3 276.5 267.3 = 0.93777 0.918 0.948 0.918 h 2s = 267.3 + (276.5 267.3) h 2s = 273.4 kj/kg 1 Isentropic efficiency of a compressor is defined as, η c = h 2s h 1 h 2 h 1 h 2 = h 1 + h 2s h 1 At state 2, p 2 = 8 bar and h 2 = 279.74 kj/kg Interpolation for s 2 : η c 273.4 244.515 h 2 = 244.515 + 0.82 h 2 = 279.74 kj/kg 1 ( ) 0.93777 0.918 0.948 0.918 h 276.5 279.74 286.7 s (kj/kgk) 0.948 s 2 0.980 s 2 0.948 0.980 0.948 = 279.74 276.5 286.7 276.5 s 2 = 0.948 + (0.980 0.948) s 2 = 0.958 kj/kgk 1 At state 4, h 4 = h 3 = 95.029 kj/kg. 1 h f4 = h f ( 10 o C) h f4 = h f( 12 o C) + h f ( 8 o C) 2 35.922 + 41.193 h f4 = 2 h f4 = 38.558 kj/kg ( ) 279.74 276.5 286.7 276.5 122

h g4 = h g ( 10 o C) h g4 = h g( 12 o C) + h g ( 8 o C) 2 243.31 + 245.72 h g4 = 2 h g4 = 244.52 kj/kg Quality at state 4, h 4 = (1 x 4 )h f4 + x 4 h g4 x 4 = h 4 h f h g h f 95.029 38.558 x 4 = 244.52 38.558 x 4 = 0.2742 s f4 = s f ( 10 o C) s f4 = s f( 12 o C) + s f ( 8 o C) 2 0.14505 + 0.16499 s f4 = 2 s f4 = 0.15502 kj/kgk s g4 = s g ( 10 o C) s g4 = s g( 12 o C) + s g ( 8 o C) 2 0.93918 + 0.93636 s g4 = 2 s g4 = 0.93777 kj/kgk Entropy at state 4, s 4 = (1 x 4 )s f4 + x 4 s g4 s 4 = (1 0.27)(0.15502) + (0.27)(0.93777) s 4 = 0.36636 kj/kgk 1 123

Energy balance for the condenser yields, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o ṁ 1 = ṁ 2 = ṁ 3 = ṁ 4 = ṁ (from mass conservation) Q out = ṁ(h 3 h 2 ) Q out ṁ = h 3 h 2 20 ṁ = 95.029 279.74 ṁ = 0.1083 kg/s 1 Energy balance for the compressor yields, Coefficient of performance, de dt = Q Ẇ + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o Ẇ c = ṁ(h 2 h 1 ) Ẇ c = (0.1083 kg/s)(279.74 244.515) Ẇ c = 3.81 kw 1 COP = Q out Ẇc COP = 20 3.81 COP = 5.24 1 For a Carnot heat pump operating between T c = 0 o C = 273 K and T H = 21 o C = 294 K, T H COP Carnot = T H T C COP Carnot = 294 21 COP Carnot = 14 1 COP Carnot > COP because of irreversibilities associated with the system. 1 Heat transfer for the evaporator, Q in = Q out Ẇc Q in = 20 3.81 Q in = 16.19 kw 124

Entropy balance for the evaporator yields, ds dt = Q + T j b ṁ i s i ṁ o s o + σ i o Entropy balance for the compressor yields, Entropy balance for the condenser yields, Entropy balance for the valve yields, Thus σ comp > σ valve > σ evap > σ cond 1 σ = ṁ(s 1 s 4 ) Q T b σ = (0.1083 kg/s)(0.93777 0.36636) 16.19 10 + 273 σ evap = 0.0003248 kw/k 1 ds dt = Q + T j b ṁ i s i i ṁ o s o + σ o σ = ṁ(s 2 s 1 ) σ = (0.1083 kg/s)(0.958 0.93777) σ comp = 0.002191 kw/k 1 ds dt = Q + T j b ṁ i s i i ṁ o s o + σ o σ = ṁ(s 3 s 2 ) Q T b σ = (0.1083 kg/s)(0.35260 0.958) σ cond = 0.0002246 kw/k 1 ds dt = Q + T j b ṁ i s i ṁ o s o + σ i o σ = ṁ(s 4 s 3 ) σ = (0.1083 kg/s)(0.36636 0.35260) σ valve = 0.001490 kw/k 1 20 31 + 273 125

HW-37 (25 points) Given: State 1: p 1 = 1 bar, T 1 = 300 K. State 3: T 3 = 2000 K. V 1 /V 2 = 9. Otto cycle (thus, V 1 = V 4 and V 2 = V 3 ) Find: q and w for each process, η and MEP. EFD: 1 W Q Assumptions: 2 Neglect KE and PE changes. Quasi-equilibrium. Closed system. Air behaves as an ideal gas. Basic Equations: E = Q W 1 W = pdv pv = RT Solution: Process 1-2 is polytropic compression. Thus, ( ) (n 1) V1 T 2 = T 1 1 V 2 T 2 = (300 K) (9) (1.3 1) T 2 = 580 K 1 126

Process 3-4 is polytropic compression. Thus, T 4 = T 3 ( V3 T 4 = (2000 K) ) (n 1) V 4 ) (1.3 1) ( 1 9 T 4 = 1035 K 1 At state 1, from ideal gas tables, u 1 = 214.1 kj/kg. 1 At state 2, from ideal gas tables, u 2 = 419.6 kj/kg. 1 At state 3, from ideal gas tables, u 3 = 1678 kj/kg. 1 At state 4, from ideal gas tables, Interpolation for u 4 : T 1020 1035 1040 u (kj/kg) 777.6 u 4 794.5 Gas constant for air, R, u 4 777.6 794.5 777.6 = 1035 1020 1040 1020 u 4 = 777.6 + (794.5 777.6) u 4 = 790.3 kj/kg 1 R = R/MW ( ) 1035 1020 1040 1020 R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 0.287 kj/kgk For process 1-2, work is given as, W 12 = w 12 = pdv pdv w 12 = R(T 2 T 1 ) (polytropic process) 1 1 n (0.287 kj/kgk)(580 300) w 12 = 1 1.3 w 12 = 267.9 kj/kg 1 127

From energy balance, E = Q W U = Q W u = q w q = u + w 2 q 12 = (u 2 u 1 ) + w 12 q 12 = (419.6 214.1) 267.9 q 12 = 62.4 kj/kg 1 For process 2-3, w 23 = 0 since volume stays constant. From energy balance, q = u + w q 23 = (u 3 u 2 ) q 23 = 1678 419.6 q 23 = 1258.4 kj/kg 1 For process 3-4, work is given as, W 34 = w 34 = pdv pdv w 34 = R(T 4 T 3 ) (polytropic process) 1 n (0.287 kj/kgk)(1035 2000) w 34 = 1 1.3 w 34 = 923.1 kj/kg 1 From energy balance, q = u + w q 34 = (u 4 u 3 ) + w 34 q 34 = (790.3 1678) + 923.1 q 34 = 35.4 kj/kg 1 For process 4-1, w 41 = 0 since volume stays constant. From energy balance, q = u + w q 41 = (u 1 u 4 ) q 41 = 214.1 790.3 q 41 = 576.2 kj/kg 1 128

Net work, Heat input, w net = w 12 + w 23 + w 34 + w 34 w net = 267.9 + 0 + 923.1 + 0 w net = 655.2 kj/kg 1 Thermal efficiency, Specific volume at state 1, Mean effective pressure is, q in = q 23 + q 34 q in = 1258.4 + 35.4 q in = 1293.8 kj/kg 1 pv = RT v 1 = RT 1 η = w net q in η = 655.2 1293.8 η = 50.64 % 1 p 1 (0.287 kj/kgk)(300 K) v 1 = 100 kp a v 1 = 0.861 m 3 /kg MEP = w net v 1 v 2 1 w net MEP = v 1 (1 v 2 /v 1 ) 655.2 MEP = (0.861 m 3 /kg)(1 1/9) MEP = 8.561 bar 1 129

HW-38 (25 points) Given: State 1: p 1 = 100 kp a = 1 bar, T 1 = 300 K, V 1 = 14 L = 1.4 10 2 m 3. Q 23 = 22.7/2 = 11.35 kj/kg and Q 34 = 11.35 kj/kg. V 1 /V 2 = 9. Dual cycle (thus, V 3 = V 2 and V 5 = V 1 ) Find: T 3, T 4, W net, η and MEP EFD: 1 W Q Assumptions: 2 Neglect KE and PE changes. Quasi-equilibrium. Closed system. Air behaves as an ideal gas. Basic Equations: E = Q W 1 W = pdv pv = RT Solution: At state 1, from ideal gas tables, u 1 = 214.1 kj/kg, v r1 = 621.2 1 Process 1-2 is isentropic compression. Thus, v r2 v r1 = V 2 V 1 1 v r2 = v r1 V 2 V 1 v r2 = 621.2 v r2 = 69.022 1 ( ) 1 9 130

Interpolation for T 2 : v r2 69.76 69.022 67.07 T (K) 700 T 2 710 Interpolation for u 2 : T 2 700 710 700 = 69.022 69.76 67.07 69.76 T 2 = 700 + (710 700) T 2 = 702.7 K 1 ( ) 69.022 69.76 67.07 69.76 v r2 69.76 69.022 67.07 u (kj/kg) 512.3 u 2 520.2 Gas constant for air, R, u 2 512.3 520.2 512.3 = 69.022 69.76 67.07 69.76 u 2 = 512.3 + (520.2 512.3) u 2 = 514.5 kj/kg 1 R = R/MW Mass inside the system is given as, ( ) 69.022 69.76 67.07 69.76 R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 0.287 kj/kgk pv = mrt m = pv RT m = p 1V 1 RT 1 m = (100 kp a)(1.4 10 2 m 3 ) (0.287 kj/kgk)(300 K) m = 0.01626 kg 1 For process 2-3, W 23 = 0 since volume stays constant. 131

From energy balance, E = Q W U = Q W m u = Q W 1 m(u 3 u 2 ) = Q 23 u 3 = Q 23 m + u 2 u 3 = 11.35 kj + 514.5 KJ/kg 0.01626 kg u 3 = 1212.53 kj/kg 1 Interpolation for T 3 : u 1205 1212.53 1224 T (K) 1500 T 3 1520 Interpolation for h 3 : T 3 1500 1212.53 1205 = 1520 1500 1224 1205 T 3 = 1500 + (1520 1500) T 3 = 1507.9 K 1 ( ) 1212.53 1205 1224 1205 u 1205 1212.53 1224 h (kj/kg) 1636 h 3 1660 For process 3-4, h 3 1636 1212.53 1205 = 1660 1636 1224 1205 h 3 = 1636 + (1660 1636) h 3 = 1645.5 kj/kg 1 ( ) 1212.53 1205 1224 1205 W 34 = W 34 = p 4 3 4 3 W 34 = p V pdv dv ( p = const) 132

From energy balance, Interpolation for T 4 : E = Q W U = Q W Q 34 = U 34 + W 34 Q 34 = U 34 + p V 34 Q 34 = H 34 Q 34 = m(h 4 h 3 ) h 4 = Q 34 m + h 3 11.35 kj h 4 = + 1645.5 kj/kg 0.01626 kg h 4 = 2343.53 kj/kg 1 h 2315 2343.53 2377 T (K) 2050 T 4 2100 Interpolation for v r4 : T 4 2050 2343.53 2315 = 2100 2050 2377 2315 T 4 = 2050 + (2100 2050) T 4 = 2073.0 K 1 ( ) 2343.53 2315 2377 2315 h 2315 2343.53 2377 v r ( no unit) 2.555 v r4 2.356 v r4 2.555 2.356 2.555 = 2343.53 2315 2377 2315 v r4 = 2.555 + (2.356 2.555) v r4 = 2.5 no unit 1 For process 3-4, since p 3 = p 4, V 3 /V 4 = T 3 /T 4. 1 ( ) 2343.53 2315 2377 2315 133

Since process 4-5 is isentropic, Interpolation for T 5 : v r5 v r4 = V 5 v r5 V 4 v r4 = V 1 v r5 v r4 = v r5 V ( 4 V1 ) V 2 V 2 V ( 4 ) V1 V 3 V 2 V ( 4 ) V1 T 3 V 2 T 4 = v r4 v r5 = v r4 ( ) V1 T 3 v r5 = v r4 V 2 T ( 4 9 1507.9 v r5 = 2.5 1 2073 v r5 = 16.37 1 ) v r5 16.064 16.37 16.946 T (K) 1160 T 5 1140 Interpolation for u 5 : T 5 1160 16.37 16.064 = 1140 1160 16.946 16.064 T 5 = 1160 + (1140 1160) T 5 = 1153.1 K 1 ( ) 16.37 16.064 16.946 16.064 v r5 16.064 16.37 16.946 u (kj/kg) 898.2 u 5 880.7 u 5 898.2 880.7 898.2 = 16.37 16.064 16.946 16.064 u 5 = 898.2 + (880.7 898.2) u 5 = 892.1 kj/kg 1 Now, Q 12 = Q 45 = 0 since these processes are adiabatic. For process 5-1, W 51 = 0 since volume stays constant. ( ) 16.37 16.064 16.946 16.064 134

Energy balance yields, Net work, Q 51 = U 51 Q 51 = m(u 1 u 5 ) Q 51 = (0.01626 kg)(214.1 892.1) Q 51 = 11.02 kj Thermal efficiency, Mean effective pressure is, W net = Q net W net = Q 12 + Q 23 + Q 34 + Q 45 + Q 51 W net = 0 + 11.35 + 11.35 + 0 11.02 W net = 11.67 kj 1 MEP = η = W net η = Q in 11.67 kj 22.7 kj/kg η = 51.43 % 1 W net V 1 V 2 1 W net MEP = V 1 (1 V 2 /V 1 ) 11.67 kj MEP = (1.4 10 2 m 3 )(1 1/9) MEP = 937.76 kp a 1 135

HW-39 (25 points) Given: State 1: p 1 = 0.8 bar, T 1 = 280 K, V1 = 60 m 3 /s. p 2 /p 1 = 20, thus p 2 = 16 bar State 3: T 3 = 2100 K, p 3 = p 2 = 16 bar. p 4 = p 1 = 0.8 bar, η c = 92% and η t = 95%. Find: Ẇ net, Q in and η th EFD: 3 Assumptions: 2 Neglect KE and PE changes. Quasi-equilibrium. Steady state, steady flow. One dimensional uniform flow. Compressor and turbine are adiabatic. Pressure across heat exchangers are constant. No work transfer for the heat exchangers. Air behaves as an ideal gas. 136

Basic Equations: dm dt = i ṁ i o ṁo 1 de dt = Q Ẇ + i ṁ i (h + ke + pe) i o ṁo(h + ke + pe) o 1 η c = w s w act = h out,s h in h out h in η t = w act w s = h in h out h in h out,s Solution: At state 1, from ideal gas tables, h 1 = 280.1 kj/kg, p r1 = 1.0889 1 Process 1-2s is isentropic compression. Thus, Interpolation for h 2s : p r2s p r1 = p 2s p 1 1 p r2s = p r1 p 2 p 1 p r2s = 1.0889 p r2s = 21.778 1 ( ) 20 1 p r2s 20.64 21.778 21.86 h (kj/kg) 649.4 h 2s 660 h 2s 649.4 660 649.4 = 21.778 20.64 21.86 20.64 h 2s = 649.4 + (660 649.4) h 2s = 659.3 kj/kg 1 At state 3, from ideal gas tables, h 3 = 2377 kj/kg, p r3 = 2559 1 Process 3-4s is isentropic expansion. Thus, Interpolation for h 4s : p r4s p r3 = p 4s p 3 p r4s = p r3 p 4 p 3 p r4s = 2559 ( ) 1 20 p r4s = 127.95 1 ( ) 21.778 20.64 21.86 20.64 137

p r4s 123.4 127.95 133.3 h (kj/kg) 1069 h 4s 1092 Gas constant for air, R, Mass flow rate is given as, h 4s 1069 127.95 123.4 = 1092 1069 133.3 123.4 h 4s = 1069 + (1092 1069) h 4s = 1079.6 kj/kg 1 R = R/MW ( ) 127.95 123.4 133.3 123.4 R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 0.287 kj/kgk p V = ṁrt ṁ = p V RT Energy balance for the compressor yields, ṁ = p V 1 1 RT 1 ṁ = (80 kp a)(60 m3 /s) (0.287 kj/kgk)(280 K) ṁ = 59.73 kg/s 1 de dt = Q Ẇc + ṁ i (h + ke + pe) i i ṁ o (h + ke + pe) o o ṁ 1 = ṁ 2 = ṁ 3 = ṁ 4 = ṁ (from mass conservation) 1 Ẇ c = ṁ(h 1 h 2 ) 1 Ẇ c = ṁ h 1 h 2s η c 280.1 659.3 Ẇ c = (59.73) 0.92 Ẇ c = 24.62 MW 1 138

Energy balance for the turbine yields, Net work, de dt = Q Ẇc + ṁ i (h + ke + pe) i ṁ o (h + ke + pe) o i o Ẇ t = ṁ(h 3 h 4 ) 1 Ẇ t = ṁ(h 3 h 4s )η t Ẇ t = (59.73)(2377 1079.57)0.95 Ẇ t = 73.62 MW 1 Enthalpy at state 2 is found as, Ẇ net = Ẇt + Ẇc Ẇ net = 73.62 24.62 Ẇ net = 49 MW 1 η c = h 1 h 2s h 1 h 2 h 2 = h 1 h 1 h 2s η c 280.1 659.3 h 2 = 280.1 0.92 h 2 = 692.3 kj/kg Energy balance for the heat exchanger (combustor) yields, Thermal efficiency, de dt = Q in Ẇ + ṁ i (h + ke + pe) i i ṁ o (h + ke + pe) o o Q in = ṁ(h 3 h 2 ) 1 Q in = (59.73)(2377 692.3) Q in = 100.63 MW 1 η = Ẇnet Q in 1 η = 49 100.63 η = 48.70 % 1 139

HW-40 (25 points) Given: State a: p a = 26 kp a, T a = 230 K, V a = 220 m/s. ṁ = 25 kg/s p 2 /p 1 = 11 T 3 = 1400 K. p 5 = 26 kp a. η c = 85% and η t = 90%. Find: p and T at each state, Q in and V 5 EFD: 1 Assumptions: 1 Neglect PE changes. Neglect KE except at diffuser inlet and nozzle exit. Quasi-equilibrium. Steady state, steady flow. One dimensional uniform flow. Compressor and turbine are adiabatic. Nozzle and diffuser have no work or heat transfer. Pressure across combustor is constant. No work transfer for the combustor. Air behaves as an ideal gas. 140