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

Transcription

1 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

2 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

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

4 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

5 (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

6 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 P a p gas 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 P a or 1.49 bar. 6

7 (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 = 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

8 Case (ii) (1.3 1) 10 5 P a L = ( kg/m 3 )(9.81 m/s 2 ) L = 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

9 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 = m 3 /kg 1 ] 1/ kpa kpa m 3 /kg 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 = K

10 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 = /3600 m/s V 2 = m/s 1 10

11 Change in Kinetic Energy KE = 1 2 m(v 2 2 V 2 1 ) 2 KE = 1 2 (14, 000 kg)( )m 2 /s 2 KE = J KE = 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 = J P E = 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

12 Solution: V 1 = 100 km/hr V 1 = /3600 m/s V 1 = m/s 1 Change in Kinetic Energy KE = 1 2 m(v 2 2 V 2 1 ) 2 KE = 1 2 (900 kg)( )m 2 /s 2 KE = J KE = 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 = J P E = 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

13 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 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 = m/s 2 13

14 HW-5 (25 points) (a) Given: 1 for writing given, find, EFD, etc., V 1 = m 3 and V 2 = m 3 A = 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 = m 3 /0.018 m 2 x 1 = m 1 x 2 = V 2 /A x 2 = m 3 /0.018 m 2 x 2 = m 1 14

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

16 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 (x ))Adx x2 x 1 dx A x2 x 1 (x ))dx W = ( P a)(0.018 m 2 )(x 2 x 1 ) [( ) x (0.018 m 2 2 ) x 2 ( )] x x 1 W = ( P a)(0.018 m 2 )( ) [( ) ( (0.018 m ) W = 125 J 1 )] p V diagram: bar 1 1 bar m m 3 V 16

17 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 = J W = 50.4 kj 1 17

18 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

19 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

20 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 = E 1 = 70 E 1 = 20 kj 1 Process b: E = E 2 E 1 E = E = 30 kj 1 E = Q W 30 = 50 W W = 20 kj 1 20

21 Process c: E = Q W 20 = Q ( 60) Q = 40 kj 1 E 2 E 1 = 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 b c d e

22 (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 m m 3 V 22

23 Work, W, W = 2 1 pdv W = Area under p V curve W = ( P a)( ) m 3 W = 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 + ( ) Q = 40 kj/kg 1 Process B: p V diagram: 1 10 bar 2 1 bar m 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

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

25 HW-8 (25 points) (a) (i) From the saturation pressure table, at p = 5 bar, T sat = 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 = C SLVM SHV C CL p = 5 bar SLVM SHV (ii) From the saturation pressure table, at p = 5 bar, T sat = 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 = bar = 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

26 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 = 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 = 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

27 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

28 T ( o C) v g (kj/kg) v v = v 2 = ( ) v 2 = 0.0 kj/kg 1 ( ) 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) v f (kj/kg) v f v f = Interpolation for v g1 : v f1 = ( ) v f1 = 0.0 kj/kg 1 ( ) T ( o C) v g (kj/kg) v g v g = v g1 = ( ) v g1 = 0.0 kj/kg 1 ( )

29 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 = m g1 = xm tot m g1 = ( kg) m g1 = 29.1 kg 1 29

30 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

31 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

32 HW-10 (25 points) Given: 1 for writing given, find, EFD, etc., Propane, m = 0.1 kg, p = 0.4 MP a V 1 = m 3, V 2 = 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 = m 3 /0.1kg v 1 = 0.13 m 3 /kg 1 v 2 = V 2 /m v 2 = m 3 /0.1kg v 2 = 0.14 m 3 /kg 1 Temperature, T 1 : At p = 0.4 MP a, v f = m 3 /kg and v g = m 3 /kg. 32

33 v 1 > v g. Thus, it is a Superheated Vapor (SHV). 1 Interpolation for T 1 : v (m 3 /kg) T ( o C) 20 T 1 30 T = ( ) T 1 = 20 + (30 20) T 1 = 22.9 o C 1 Temperature, T 2 : At p = 0.4 MP a, v f = m 3 /kg and v g = m 3 /kg. v 2 > v g. Thus, it is a Superheated Vapor (SHV). 1 Interpolation for T 2 : v (m 3 /kg) T ( o C) 40 T 2 50 Work, W, T = ( ) T 2 = 40 + (50 40) T 2 = 41.3 o C 1 W = pdv W = p dv (p = constant) W = p V 2 W = kp a( ) 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

34 Interpolation for u 1 : T ( o C) u (kj/kg) u Interpolation for u 2 : u = u 1 = ( ) u 1 = kj/kg 1 ( ) T ( o C) u (kj/kg) u u = u 2 = ( ) u 2 = kj/kg 1 ( ) Q = W + m u Q = 0.4 kj kg( ) kj/kg Q = kj 1 34

35 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 = 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

36 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 = K T 2 = o C 2 36

37 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

38 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 = 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 = kg 1 m 1 = m c1 + m t1 m 1 = m 1 = 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 = m 3 1 V 1 = V c1 + V t V 1 = V 1 = m 3 38

39 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 = m c2 = kg 1 (1.982 kg)(0.189 kj/kgk)(290 K) V c2 = 200 kp a V c2 = m 3 1 V 2 = V c2 + V t V 2 = V 2 = m 3 Work, W, Heat transfer, Q, W = p V W = p(v c2 V c1 ) W = 200 kp a( ) 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

40 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 m m 3 V 40

41 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 = 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 = 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 = C C = (400 kp a)(1.134 m 3 ) 1.28 C = p 3 V = C V 3 = C p 3 1/1.28 V 3 = /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 = W 23 = C pdv C dv 2 V dv 2 V 1.28 V 0.28 V 3 =3.35 m 3 W 23 = C (for the integration) V 2 =1.134 m 3 W 23 = ( ) 0.28 W 23 = kj 1 41

42 Work for process 3-1, W 31, W 31 = W 31 = p pdv dv W 31 = p(v 1 V 3 ) W 31 = 100 kp a( ) m 3 W 31 = 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 = 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 = kj/kg 1 Internal energy at state 2, u 2 (from ideal gas tables) u 2 = kj/kmol u 2 = kj/kmol 44 kg/kmol u 2 = kj/kg 1 Internal energy at state 3, u 3 (from ideal gas tables) Interpolation for u 3 : 42

43 T ( K) u (kj/kmol) u u = u 3 = ( ) u 3 = kj/kmol 1 ( ) Energy equation, u 3 = kj/kmol u 3 = kj/kmol 44 kg/kmol u 3 = 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)( ) kj/kg + 0 Q 12 = kj 1 Q 23 = m(u 3 u 2 ) + W 12 Q 23 = (2 kg)( ) kj/kg kj Q 23 = kj 1 Q 31 = m(u 1 u 3 ) + W 31 Q 31 = (2 kg)( ) kj/kg kj Q 31 = kj 1 Sum of work and heat transfer for the entire cycle, 43

44 Process W (kj) Q (kj) Sum

45 HW-14 (25 points) Given: 1 for writing given, find, EFD, etc., Area of inlet, A i = m = m 2 Area of exit, A e = m = 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 = kj/kgk 1 45

46 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 kp a ρ = (0.287 kj/kgk)(293 K) ρ = 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 = 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 = kg/s (1.189 kg/m 3 )(20.91 m 2 ) V e1 = 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 = m/s 1 46

47 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 Enthalpy at inlet, h i, Interpolation for h i : T (K) h (kj/kg) h i h i Enthalpy at exit for case 1, h o1, = h i = ( ) h i = kj/kg 1 ( ) h o = h i + V i 2 2 V o 2 2 h o = kj/kg h o = kj/kg 1 kj/kg kj/kg Temperature at exit for case 1, T o1, Interpolation for T o1 : h (kj/kg) T (K) 290 T o1 295 T o = T o1 = ( ) T o1 = K 1 ( )

48 Temperature of air at exit for case 1, T o1 = 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 = kp a p o1 = bar 1 Enthalpy at exit for case 2, h o2, h o = h i + V i 2 2 V o 2 2 h o = kj/kg h o = kj/kg 1 kj/kg kj/kg Temperature at exit for case 2, T o2, Interpolation for T o2 : h (kj/kg) T (K) 295 T o2 300 T o = T o2 = ( ) T o2 = K 1 Temperature of air at exit for case 2, T o2 = K = 26.9 o C. Pressure at exit for case 2, p o2, ( ) p o2 = ρrt o2 p o2 = (1.189 kg/m 3 )(0.287 kj/kgk)(299.9 K) p o2 = kp a p o2 = bar 1 48

49 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 = 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

50 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) h (kj/kg) h o h o = h o = ( ) h o = kj/kg 1 Enthalpy of air at inlet for case 1, h i1, Q = ṁ(h o h i ) ( ) h i = Q ṁ + h o h i1 = Q ṁ 1 + h o h i1 = kw kg/s h i1 = kj/kg kj/kg Temperature of air at inlet, T i1, Interpolation for T i1 : 50

51 h (kj/kg) T (K) 260 T i1 270 T i = T i1 = ( ) T i1 = K 1 Temperature of air at inlet, T i1 = K = 7.36 o C. Enthalpy of air at inlet for case 2, h i2, ( ) h i2 = Q ṁ 2 + h o h i2 = kw 1600 kg/s h i2 = kj/kg kj/kg Temperature of air at inlet, T i2, Interpolation for T i2 : h (kj/kg) T (K) 240 T i2 250 T i = T i2 = ( ) T i2 = K 1 Temperature of air at inlet, T i2 = K = 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 = ( ) Q ṁc + T i kw (550 kg/s)(2.7 kj/kg K) K T o = 233 K 1 Exit temperature of secondary liquid, T o = 233 K = 40 o C. 51

52 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

53 Enthalpy of air at inlet, h i, Interpolation for h i : T (K) h (kj/kg) h i Enthalpy of air at outlet, h o, h i = h i = ( ) h i = kj/kg 1 ( ) h o = Ẇ ṁ + h i h o = kw 1600 kg/s h o = kj/kg kj/kg Temperature of air at outlet, T o, Interpolation for T o : h (kj/kg) T (K) 295 T o 300 Gas constant for air, R, T o = T o = ( ) T o = K 1 R = R/MW ( ) R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = kj/kgk 53

54 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 = kg/m 3 1 p o = (1.189 kg/m 3 )(0.287 kj/kgk)( K) p o = kp a 1 p = p o p i p = ( ) kp a p = kp a 1 54

55 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

56 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 = kj/kg (from saturation tables). 1 Thus h 2 = kj/kg. At state 2, p 2 = 2 bar, h f = kj/kg and h g = 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 = 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 = o C. T 3 > T sat. Thus it is a Super Heated Vapor (SHV). From SHV tables, h 3 = kj/kg. 1 Enthalpy at state 4: at p 4 = 1bar, T sat = 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 kj/kg + ( m 3 /kg)( ) kp a h kj/kg 1 Enthalpy at state 5: at p 5 = 1bar, T sat = 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 kj/kg + ( m 3 /kg)( ) kp a h kj/kg 1 56

57 Thus, ṁ 2 = ṁ4(h 4 h 5 ) h 3 h 2 (2)( ) ṁ 2 = ( ) ṁ 2 = kg/s 1 57

58 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

59 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 = kj/kg 1 Enthalpy at state 2, h 2 = kj/kg 1 Enthalpy at state 3, h 3 = 1880 kj/kg 1 Enthalpy at state 4, h 4 = kj/kg 1 Thus, the net shaft output power is ẇ = (h 1 h 2 + h 3 h 4 ) ẇ = ( ) ẇ = kj/kg 2 59

60 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

61 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 = 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 kj/kg + ( m 3 /kg)( ) kp a h kj/kg 1 Enthalpy at state 2, h 2 = h g (p 2 ) = kj/kg. Enthalpy at state 3, h 3 = h f (p 3 ) = kj/kg. Thus, ṁ 2 = ṁ1(h 1 h 3 ) (h 2 h 3 ) (5 kg/s)( ) ṁ 2 = ( ) ṁ 2 = 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

62 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)( ) Ẇ = kw 1 62

63 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

64 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

65 At p 1 = 4 MP a, T sat = o C. T 1 > T sat, thus it is a Super Heated Vapor (SHV). Enthalpy at state 1, h 1 = kj/kg Enthalpy at state 2, h 2 = kj/kg Thermal efficiency, w = (h 1 h 2 ) w = w = 1066 kj/kg 1 η = w net q in 1 η = w t w p q in η = 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 = 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 kj/kg + ( m 3 /kg)( ) kp a h kj/kg 1 At p 6 = 1 bar, T sat = 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 kj/kg + ( m 3 /kg)( ) kp a h kj/kg 1 65

66 ṁ cw ṁ s ṁ cw ṁ s ṁ cw ṁ s = q w t w p h 6 h ( 4) = kg of cooling water = kg of steam 1 66

67 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 = kw. Q people = 6000 kj/hr, Q ext = 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

68 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 = = kj/hr. 1 Heat transfer to A/C unit is thus, Q in,2 = 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 = Ẇ in,a/c = 2.08 kw 1 Now, Ẇ out = Ẇin,A/C + Ẇother 2 Ẇ out = Ẇ out = 18 kw 1 Heat input to the power cycle, Q in,1 = 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 = η = 55.38% 1 Q out,1 = Q in,1 Ẇout 2 Q out,1 = Q out,1 = 14.5 kw 1 Q out,2 = Q in,2 + Ẇin,A/C 2 Q out,2 = Q out,2 = 8.75 kw 1 68

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

70 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 = Ẇ in = 700 kw hr 1 Cost of electricity = = $ Cost of electricity if resistance heating is used = = $

71 Savings in cost = $273 $105 = $ Energy balance for the A/C yields, Q out = Ẇin + Q in 1 Q in = Q out Ẇin Q in = Q in = 1120 kw hr 1 Percentage of energy coming from the environment, x = Q in Q out 1 x = x = % 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

72 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 = COP = 6 1 Energy balance of the heat engine yields COP = Q out Ẇ in Ẇ in = Q out COP Ẇ in = Ẇ in = 20 kw 1 Q in = Ẇout + Q out 1 Q in = Q in = 25 kw 1 Efficiency of the heat engine, η = Ẇout Q in η = η = 80 % 1 72

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

74 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

75 Gas constant for air, R, R = R/MW R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 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 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 = m 3 1 V 1 75

76 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( ) 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 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( ) W 41 = 676 kj 1 76

77 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 = W net = 36 kj 1 η = W net Q in η = η = 60 % 1 η rev = T h T c T c η rev = 750 η rev = 60 % 1 77

78 HW-24 (25 points) (a) At inlet to flash chamber p = 4 bar, h = kj/kg. At p = 4 bar, h f = kj/kg and h g = 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 = s = (1 x)s f + xs g s = (1 0.07)(1.7765) (6.8955) s = 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)( ) + 0.9(8.2273) s 4 = kj/(kgk) 1 (b) (i) At p 1 = 0.20 bar, s f = kj/(kgk) and s g = 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 = Enthalpy, h = (1 x)h f + xh g h = (1 0.5)(251.42) + 0.5(2608.9) h = kj/kg 1 T s diagram: 3 78

79 T CL p = constant SLVM SHV s (ii) At p = 10 bar, u g = kj/kg. Since u > u g, it is a Super Heated Vapor (SHV). 1 At u = kj/kg, s = 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 = kj/(kgk), s g = kj/(kgk). T s diagram: 3 s = (1 x)s f4 + xs g4 s = (1 0.8)( ) + 0.8( ) s = kj/(kgk) 1 79

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

81 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 = kj/(kgk). 1 Using compressed liquid tables, at state 2, s 2 = kj/(kgk). 1 Thus, (ii) Using saturated liquid approximation, Thus, s = s 2 s 1 s = s = kj/(kgk) 1 s s f (T ) 1 s kj/(kgk) 1 s kj/(kgk) 1 s = s 2 s 1 s = s = kj/(kgk) 1 (iii) Using constant specific heat, s = Cln s = 4.18ln ) 1 T ( 1 ) ( T2 s = 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) = kj/(kgk), C w = 4180 J/(kgK) = 4.18 kj/(kgk). Find: S c and S w EFD: 2 Copper Water 81

82 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 = 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 = kj/k 1 ( 310 K 298 K ) 82

83 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

84 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 ) Gas constant for air, R, R = R/MW R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = kj/kgk 1 84

85 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( ) w 12 = w 12 = kj/kg 2 T (K) s (kj/(kgk)) s o Since 2-3 is isentropic, s o Internal energy at state 3, u 3, Interpolation for u 3 : s o 3 s o 2 Rln = s o 2 = ( ) s o 2 = 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 ( ) ) ) s o 3 = (0.287)ln s o 3 = kj/(kgk) 1 ( )

86 s (kj/(kgk)) u (kj/kg) u u = u 3 = ( ) u 3 = 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 : ( ) T (K) u (kj/kg) u u = u 2 = ( ) u 2 = 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 = ( ) kj/kg w 23 = 99.2 kj/kg 1 w net = w 12 + w 23 1 w net = w net = kj/kg 1 ( )

87 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: 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

88 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 = 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 σ = kj/k 2 88 ( p2f p 2i ))

89 (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

90 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 = 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

91 From superheated tables, at inlet, h i = kj/kg and s i = kj/kgk. 2 From saturation tables, at outlet, h o = kj/kg and s o = kj/kgk. 2 Thus Q is, Q = Ẇ + ṁ(h o h i ) Q = 2000 kw + ( kg/s)( ) Q = 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 kw σ = + ( kg/s)( ) 300 K σ = kw/k 1 91

92 HW-29 (25 points) Given: State 1, p 1 = 96 kp a, T 1 = 27 o C = 300 K, V = m 3 /min = 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

93 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 = kj/kgk pv = mrt p V = ṁ a RT ṁ a = p V RT ṁ a = (96 kp a)( m3 /s) (0.287 kj/kgk)(300 K) ṁ a = 0.5 kg/s 1 From ideal gas tables, at state 1, h 1 = kj/kg and s o 1 = kj/kgk. 1 From ideal gas tables, at state 2, h 2 = kj/kg and s o 2 = kj/kgk. 1 From ideal gas tables, at state 3, h 3 = kj/kg and s o 3 = kj/kgk. 1 Thus Ẇ is, Ẇ = ṁ a (h 1 h 2 ) Ẇ = (0.5 kg/s)( ) 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

94 Enthalpy at state A: at p A = 1bar, T sat = 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 kj/kg + ( m 3 /kg)( ) kp a h A kj/kg 1 s A s f (T A ) s A kj/kgk 1 Enthalpy at state B: at p B = 1bar, T sat = 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 kj/kg + ( m 3 /kg)( ) kp a h B kj/kg 1 s B s f (T B ) s B kj/kgk 1 Entropy balance for the compressor yields, ṁ w = h 2 h 3 ṁ a h B h A ṁ w = (0.5 kg/s) ṁ w = 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) ln 96 σ = kw/k 1 94

95 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)( ) + (0.4036)( ) σ = kw/k 1 95

96 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 = 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 = W = m(u 1 u 2 ) w = u 1 u

97 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 = p r2 = p r T (K) 330 T Interpolation for u 2 : T = T 2 = ( ) T 2 = K 1 ( ) T (K) u (kj/kg) u u = u 2 = ( ) u 2 = kj/kg 1 From ideal gas tables, at state 1, u 1 = kj/kg. 1 w = u 1 u 2 w = w = kj/kg 1 ( )

98 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 ) ( T 2 = K 1 k 1 k ) w = u 1 u 2 w = C v (T 1 T 2 ) 1 w = (0.718 kj/kgk)( ) w = 204 kj/kg 1 98

99 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

100 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 = kp a and s 1 = kj/kgk 1 At state 2, p 2 = 100 kp a and s 2s = s 1 = 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 x 2s = x 2s = h 2s = (1 x 2s )h f + x 2s h g 1 h 2s = ( )(417.50) + (0.983)(2674.9) h 2s = kj/kg 1 100

101 From ideal gas tables, h 3 = kj/kg and p r3 = 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 = p r4s = p r h (kj/kg) h 4s h 4s = h 4s = ( ) h 4s = 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 = ( ( ) ) 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 ( ) ṁ a (1)(1) ṁ s = ṁ a 101

102 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

103 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 = W Ẇ rev = 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 MW

104 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

105 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 = kj/kg, s 1 = kj/kgk. 1 At state 3, from saturation tables, h 3 = kj/kg, s 3 = kj/kgk. 1 At state 2s, s 2s = s 1 = kj/kgk and p 2s = 6 kp a. At p 2s, s f = kj/kgk and s g = 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 x 2s = x 2s = h 2s = (1 x 2s )h f + x 2s h g 1 h 2s = ( )(151.48) + (0.7697)(2566.6) h 2s = 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 = (0.8)( ) h 2 = 2273 kj/kg 1 At state 4s, s 4s = s 3 = kj/kgk and p 4s = 10 MP a. At p 4s = 10 MP a, s f = 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 = ( m 3 /kg)( ) h 4s = kj/kg Alternatively, h 4s can also be found from the compressed liquid tables, Interpolation for h 4s : 105

106 s h (kj/kg) h 4s h 4s = h 4s = ( ) h 4s = 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 h 4 = h 4 = kj/kg 1 ( ) 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 = 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 = = kj/kg 1 106

107 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 = ṁ Q in = 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 = = kj/kg 1 Net heat transfer, Ẇ net = Ẇt + Ẇp Ẇ net ṁ = Ẇt ṁ + Ẇp ṁ Ẇ net = ṁ Ẇ net = kj/kg ṁ Q net = Q in + Q out Q net ṁ Q net = Q in ṁ + Q out ṁ = ṁ Q net = kj/kg ṁ 107

108 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 = η th = 32.79% 1 T p = const 1 CL 4s 4 3 p = const SLVM 2 2s SHV s 108

109 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

110 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 = kj/kg, s 1 = kj/kgk. At state 2, p 2 = 2 MP a, s 2 = kj/kgk. From superheated tables, Interpolation for h 2 : s h (kj/kg) h h = h 2 = ( ) ( ) h 2 = kj/kg 1 At state 3, p 3 = 0.3 MP a, s 3 = 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 x 3 = x 3 = h 3 = (1 x 3 )h f + x 3 h g h 3 = ( )(561.43) + (0.8924)(2724.9) h 3 = kj/kg 1 At state 4, p 4 = 6 kp a, s 4 = kj/kgk. s f < s 4 < s g, thus it is a saturated liquid vapor 110

111 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 x 4 = x 4 = h 4 = (1 x 4 )h f + x 4 h g h 4 = ( )(151.48) + (0.7554)(2566.6) h 4 = kj/kg 1 At state 5, from saturation tables, h 5 = kj/kg, v 5 = m 3 /kg. At state 6, enthalpy is, h 6 h 5 + v 5 (p 6 p 5 ) h kj/kg + ( m 3 /kg)(300 6) kp a h kj/kg 1 At state 7, from saturation tables, h 7 = kj/kg, v 7 = m 3 /kg. At state 8, enthalpy is, h 8 h 7 + v 7 (p 8 p 7 ) h kj/kg + ( m 3 /kg)( ) kp a h kj/kg 1 At state 9, from compressed liquid tables: at p = 10 MP a: Interpolation for h a : T h (kj/kg) h a h a = h a = ( ) h a = kj/kg at p = 15 MP a: Interpolation for h b : ( )

112 T h (kj/kg) h b h b = h b = ( ) h b = kj/kg at p = 12 MP a: Interpolation for h 9 : p ( ) h (kj/kg) h h = h 9 = ( ) h 9 = kj/kg 1 At state 10, from saturation tables, h 10 = kj/kg. At state 11, h 11 = h 10 = kj/kg. Energy balance for the steam generator yields, ( ) 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 = ( kg/s)( ) Q in = 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 y = y =

113 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 (0.1704)( ) = y = 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 = ( kg/s)( )( ) Q out = MW 1 η th = Ẇnet Q in 1 η th = Q in Q out 1 Q in η th = η th = 45.9% 1 113

114 T s diagram: 5 T 1 p = const p = const CL p = const 3 SHV 5 p = const SLVM 4 s 114

115 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 = 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

116 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 = kj/kg, s 1 = kj/kgk. 1 At state 3, from saturation tables, h 3 = kj/kg, s 3 = kj/kgk. 1 At state 2s, s 2s = s 1 = 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 h (kj/kg) h 2s h 2s = h 2s = ( ) h 2s = 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 = kj/kg Interpolation for s 2 : η c h 2 = h 2 = kj/kg 1 ( ) h s (kj/kgk) s s = s 2 = ( ) s 2 = kj/kgk 1 ( )

117 At state 4, h 4 = h 3 = 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 = ( kg/s)( ) 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 = ( kg/s)( )) Q in = kw Q in = tons 1 COP = Q in Ẇc 1 COP = COP = 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 = ( kg/s)( )) Q out = kw 1 Boundary temperature at which heat transfer occurs at the condenser T b = T sat (8 bar) 5 o C = = o C = 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 = ( kg/s)( ) σ = kw/k 1 117

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

119 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

120 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 = = 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 h (kj/kg) h Interpolation for s 1 : h = h 1 = ( ) h 1 = kj/kg 1 T ( ) s (kj/kgk) s s = s 1 = ( ) s 1 = kj/kgk 1 At state 3, from saturation tables, Interpolation for h 3 : ( )

121 T h (kj/kg) h Interpolation for s 3 : h = h 3 = ( ) h 3 = kj/kg 1 T ( ) s (kj/kgk) s Interpolation for p 3 : s = s 3 = ( ) s 3 = kj/kgk 1 T p (bar) p ( ) p = p 3 = ( ) p 3 = bar 1 Closest value in the superheated tables is 8 bar. Take p 3 = 8 bar At state 2s, s 2s = s 1 = kj/kgk and p 2s = p 3 = 8 bar. From superheated tables,1 Interpolation for h 2s : ( )

122 s h (kj/kg) h 2s h 2s = h 2s = ( ) h 2s = 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 = kj/kg Interpolation for s 2 : η c h 2 = h 2 = kj/kg 1 ( ) h s (kj/kgk) s s = s 2 = ( ) s 2 = kj/kgk 1 At state 4, h 4 = h 3 = kj/kg. 1 h f4 = h f ( 10 o C) h f4 = h f( 12 o C) + h f ( 8 o C) h f4 = 2 h f4 = kj/kg ( )

123 h g4 = h g ( 10 o C) h g4 = h g( 12 o C) + h g ( 8 o C) h g4 = 2 h g4 = 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 x 4 = x 4 = s f4 = s f ( 10 o C) s f4 = s f( 12 o C) + s f ( 8 o C) s f4 = 2 s f4 = kj/kgk s g4 = s g ( 10 o C) s g4 = s g( 12 o C) + s g ( 8 o C) s g4 = 2 s g4 = kj/kgk Entropy at state 4, s 4 = (1 x 4 )s f4 + x 4 s g4 s 4 = (1 0.27)( ) + (0.27)( ) s 4 = kj/kgk 1 123

124 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 = ṁ = 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 = ( kg/s)( ) Ẇ c = 3.81 kw 1 COP = Q out Ẇc COP = COP = 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 = 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 = Q in = kw 124

125 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 = ( kg/s)( ) σ evap = 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 = ( kg/s)( ) σ comp = 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 = ( kg/s)( ) σ cond = 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 = ( kg/s)( ) σ valve = kw/k

126 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

127 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 = kj/kg. 1 At state 2, from ideal gas tables, u 2 = 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 u (kj/kg) u Gas constant for air, R, u = u 4 = ( ) u 4 = kj/kg 1 R = R/MW ( ) R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 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)( ) w 12 = w 12 = kj/kg 1 127

128 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 = ( ) 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 = q 23 = 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)( ) w 34 = w 34 = kj/kg 1 From energy balance, q = u + w q 34 = (u 4 u 3 ) + w 34 q 34 = ( ) 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 = q 41 = kj/kg 1 128

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

130 HW-38 (25 points) Given: State 1: p 1 = 100 kp a = 1 bar, T 1 = 300 K, V 1 = 14 L = m 3. Q 23 = 22.7/2 = kj/kg and Q 34 = 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 = kj/kg, v r1 = 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 = v r2 = ( )

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

132 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 = kj KJ/kg kg u 3 = kj/kg 1 Interpolation for T 3 : u T (K) 1500 T Interpolation for h 3 : T = T 3 = ( ) T 3 = K 1 ( ) u h (kj/kg) 1636 h For process 3-4, h = h 3 = ( ) h 3 = kj/kg 1 ( ) W 34 = W 34 = p W 34 = p V pdv dv ( p = const) 132

133 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 kj h 4 = kj/kg kg h 4 = kj/kg 1 h T (K) 2050 T Interpolation for v r4 : T = T 4 = ( ) T 4 = K 1 ( ) h v r ( no unit) v r v r = v r4 = ( ) 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 ( )

134 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 ( v r5 = v r5 = ) v r T (K) 1160 T Interpolation for u 5 : T = T 5 = ( ) T 5 = K 1 ( ) v r u (kj/kg) u u = u 5 = ( ) u 5 = 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. ( )

135 Energy balance yields, Net work, Q 51 = U 51 Q 51 = m(u 1 u 5 ) Q 51 = ( kg)( ) Q 51 = 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 = W net = kj 1 MEP = η = W net η = Q in kj 22.7 kj/kg η = % 1 W net V 1 V 2 1 W net MEP = V 1 (1 V 2 /V 1 ) kj MEP = ( m 3 )(1 1/9) MEP = kp a 1 135

136 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

137 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 = kj/kg, p r1 = 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 = p r2s = ( ) 20 1 p r2s h (kj/kg) h 2s 660 h 2s = h 2s = ( ) h 2s = kj/kg 1 At state 3, from ideal gas tables, h 3 = 2377 kj/kg, p r3 = 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 = ( )

138 p r4s h (kj/kg) 1069 h 4s 1092 Gas constant for air, R, Mass flow rate is given as, h 4s = h 4s = ( ) h 4s = kj/kg 1 R = R/MW ( ) R = (8.314 kj/kmol K)/(28.97 kg/kmol) R = 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 = 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 Ẇ c = (59.73) 0.92 Ẇ c = MW 1 138

139 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)( )0.95 Ẇ t = MW 1 Enthalpy at state 2 is found as, Ẇ net = Ẇt + Ẇc Ẇ net = 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 h 2 = h 2 = 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)( ) Q in = MW 1 η = Ẇnet Q in 1 η = η = % 1 139

140 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