ch08_1_40

8-1 In an ideal vapor-compression refrigeration cycle using refrigerant R-134a, saturated vapor enters the compressor at a temperature of –20 ºC with a volumetric flow rate of 1.5 m3/min. The refrigerant leaves the condenser at 35 ºC, 10 bar. Determine:

a. the compressor power (in kW) b. the refrigerating capacity (in tons) c. the coefficient of performance (COP).

Approach:

An energy balance across the compressor can be used to determine its power. Likewise, an energy balance on the evaporator can be used to calculate the input heat transfer rate.

Assumptions:

1. The system is steady. 2. Potential and kinetic energy effects are negligible. 3. The compressor is isentropic. 4. No work is done in the evaporator. 5. The subcooled liquid approximation is valid.

Solution:

a) For compressor power, apply the energy and mass equations to the compressor. Assume reversible, adiabatic, negligible potential and kinetic energy effects, and steady.

ddEt

Q=2

2W m h− + +

V2g+

2

2inm h

⎛ ⎞⎜ ⎟ − +⎜ ⎟⎝ ⎠

∑ V2g+

out

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

∑ and ddmt in out

m m= −∑ ∑

Therefore, and where 1 2m m m= = 1 2( ) inW m h h W= − = − 1 1m V v= For the properties, use the R-134a tables, Appendices B-14—B-16: , ( ) ( ) 3

1 1 1 1,saturated vapor 0.1464 m / kggv T v T= = ( ) ( )1 1 1 1,saturated vapor 235.31kJ / kggh T h T= =

Because of isentropic compression, 2 1s s= , and we want to evaluate ( )2 2 2,h P s

( )1 1 1 0.9332kJ / kg Kgs s T= =

State 2 is superheated vapor, so by interpolation: 2 277.14kJ kgh = . Therefore,

( ) ( )3 31.5m / min / 0.1464m / kg 10.2kg / minm = =

( )( ) kJ 1min 1kW10.2kg 235.31 277.14 7.11kWkg 60s 1 kJ/s

W⎛ ⎞⎛ ⎞= − = −⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

7.11kWinW W= − = Answer b) To obtain the refrigerating capacity, apply the energy and mass equations to the refrigerant flowing through the evaporator. Assume steady, negligible potential and kinetic energy effects, and no work. Therefore, and 4 1m m m= =

3

( )1 4 inQ m h h Q= − =

Across the throttling value, and 4h h= ( )3 3 3,h T P → This is a subcooled liquid, so we use the subcooled liquid approximation ( )3 3 3 98.78kJ / kgfh h T =∼

Therefore, ( )kg kJ 1min 1kW10.2 235.31 98.78min kg 60s 1 kJ / s

Q⎛ ⎞⎛ ⎞ ⎛= − ⎜ ⎟⎜ ⎟ ⎜

⎝ ⎠ ⎝⎝ ⎠

⎞⎟⎠

23.2 kWinQ Q= = Answer For the refrigerating capacity in tons, 1ton 211kJ / min=

( ) 1kJ/s 60s 1ton23.2kW 6.6 tons1 kW 1min 211kJ/mininQ

⎛ ⎞⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

c) Finally, W 23.2kW 7.11kW 3.26Ref in inCOP Q= = = Answer

8- 1

8-2 Refrigerant R-134a is used in a vapor-compression refrigeration cycle. Saturated vapor at 20 psia enters the compressor, which has an isentropic efficiency of 80%, and leaves at 120 psia. Saturated liquid exits the condenser, and saturated vapor exits the evaporator. The mass flow rate is 15 lbm/min. Determine:

a. the compressor power (in hp) b. the refrigerating capacity (in tons) c. the coefficient of performance (COP).

Approach:

An energy balance across the compressor can be used to determine its power. Likewise, an energy balance on the evaporator can be used to calculate the input heat transfer rate.

Assumptions:

1. The system is steady. 2. Potential and kinetic energy effects are negligible. 3. The compressor is adiabatic. 4. No work is done in the evaporator. 5. The subcooled liquid approximation is valid.

Solution:

a) Apply the energy and mass equations to the compressor. Assume steady, adiabatic, with negligible potential and kinetic energy effects. Therefore, and 1 2m m m= = 1 ( )2 1inW m h h= −

From the R-134a tables, Appendices B-14—B-16: ( ) ( )1 1 1 1,saturated vapor 101.39Btu lbmgh P h P= =

and ( )1 1 1 0.2227 Btu / lbm Rgs s T= =

To determine we need to use the definition of isentropic efficiency: 2h

( )1 2C 2 1

1 2

/ss

h hh h h h

h h 1 2 Cη η−

= → = − −−

By interpolation in the superheat vapor region at 120 psia, knowing that 2 1s s= : ( )2 2 2, 117.3Btu / lbsh P s = m

( )2 101.39 101.39 117.3 / 0.80 121.3Btu / lbmh = − − =

( )lbm Btu Btu15 121.3 101.39 298min lbm mininW ⎛ ⎞= − =⎜ ⎟

⎝ ⎠

Btu 60min 1hp298 7.04 hpmin 1hr 2544.5Btu / hr

⎛ ⎞⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

Answer

b) Apply the energy and mass equations to the refrigerant flowing through the evaporator, and assume steady, no work, and negligible potential and kinetic energy effects: and 4 1m m m= = ( )1 4inQ m h h= −

From the throttling valve ( )4 3 3 3 40.91Btu lbmfh h h P= = = . Therefore,

( )lbm Btu Btu15 101.39 40.91 907.2min lbm mininQ ⎛ ⎞= − =⎜ ⎟

⎝ ⎠

Btu 1ton907.2 4.54 tonsmin 200Btu / min

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Answer

c) 907.2Btu/min 3.04298.7 Btu/min

inRef

in

QCOP

W= = = Answer

8- 2

8-3 A vapor-compression refrigeration cycle uses R-134a. Liquid at 1200 kPa exits the condenser at 40 ºC. The evaporator operates at a pressure of 240 kPa. The compressor isentropic efficiency is 75%. Determine the cycle coefficient of performance (COP) if the refrigerant leaves the evaporator as superheated vapor at 0, 5, 10, 15, and 20 ºC above the saturation temperature.

Approach: An energy balance across the compressor can be used to determine its power. Likewise, an energy balance on the evaporator can be used to calculate the input heat transfer rate. Their ratio is used to determine the cycle COP.

Assumptions:


Solution:

The refrigeration cycle CO is P

inRef

in

QCOP

W=

Applying the mass and energy equations to the evaporator, and assuming steady, no work, and negligible potential and kinetic energy effects: and 4 1m m m= = ( )in 1 4 1 4in inQ m h h Q m q h h= − → = = − Applying the mass and energy equations to the compressor, and assuming steady, adiabatic, and negligible potential and kinetic energy effects: and 2 1m m m= = ( )2 1 2in in inW m h h W m w h h= − → = = − 1

From the throttling valve, . ( )4 3 3 3,h h P T=

Using the subcooled liquid approximation and Appendix A-14, ( )3 3 3 106.19kJ kgfh h T≈ =

To find , and are known, so can be evaluated ( )1 1 1,h P T 1P 1T 1hTo obtain , we need to use the definition of isentropic efficiency 2h

( )1 2C 2 1 1

1 2

/ss

h hh h h h

h h 2 Cη η−

= → = − −−

where, because 2 1s s= , is obtained by interpolation from the R-134a superheated vapor table,

Appendix A-16. Below is a table of all the values with (2 2 2,sh P s )

1

,1 5.37 CsatT = −

1 ,satT T−

( )C 1T

( )C ( )1 1 1,h P T

( )kJ / kg ( )1 1 1,s P T

( )kJ / kg K 2sh

( )kJ / kg 2h

( )kJ / kg inw inq COP

0 -5.37 244.09 0.9222 277.4 288.5 44.4 137.9 3.11 5 -0.37 248.71 0.9392 283.0 294.4 45.7 142.5 3.12

10 4.63 253.03 0.9548 288.1 299.8 46.8 146.8 3.14 15 9.63 257.51 0.9709 293.6 305.6 48.1 151.3 3.15 20 14.63 262.01 0.9866 298.9 311.2 49.2 155.8 3.17

Comments:

As can be seen, with increasing superheat from the evaporator, the compressor power and heat input increase and the COP decreases. This is a consequence of the density at the compressor inlet decreasing with increasing temperature.

8- 3

8-4 An ice-making plant is designed to produce 10,000 lbm of ice each day. Liquid water enters the plant at 50 ºF and solid ice leaves it at 20 ºF; the enthalpy change for the water when it goes from the liquid to the solid is 167.4 Btu/lbm. Refrigerant R-134a enters the compressor as saturated vapor at 20 lbf/in.2 and leaves the condenser as a saturated liquid at 100 lbf/in.2. The compressor isentropic efficiency is 85%. Detemine:

a. the refrigerant flow rate (in lbm/s) b. the compressor input power (in hp) c. the cycle coefficient of performance.

Approach:

An energy balance on the evaporator can be used to calculate the refrigerant flow rate, since we know the water flow rate and can evaluate the enthalpies. Likewise, an energy balance across the compressor power can be used to determine its power.

Assumptions:

1. The system is steady. 2. Potential and kinetic energy effects are negligible. 3. The compressor and evaporator are adiabatic. 4. No work is done in the evaporator. 5. The subcooled liquid approximation is valid.

Solution:

a) Apply the mass and energy equations to the whole evaporator, and assume adiabatic, no work, and negligible potential and kinetic energy effects for both the R-134a and the water 1 4 Rm m m= = and and A Bm m m= = W ( ) ( )1 4R W A Bm h h m h h− = −

Solving for the refrigerant flow rate: 1 4

A BR W

h hm mh h−

=−

From the R-134a table Appendix B-15, ( ) ( )4 3 3 3 3,saturated liquid 36.99Btu lbmfh h P h P= = =

( ) ( )1 1 1 1,saturated vapor 101.39Btu / lbmgh P h P= =

lbm 167.4 1Day hr lbm10000 0.301Day 101.39 36.99 24 hr 3600 s sRm

⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠ Answer

b) Apply the mass and energy equations to the compressor and assume steady, adiabatic, and negligible potential and kinetic energy effects: and 2 1m m m= = ( )2 1inW m h h= − For , we need to use the definition of isentropic efficiency 2h

1 2 1 22 1

1 2

s sC

C

h h h hh h

h hη

η− −

= → = −−

With 2 1s s= , and ( )1 1 1 0.2227 Btu lbm Rgs s P= = , by interpolation in Appendix B-16

( )2 2 2, 115.64Btu / lbmsh P s =

2101.39 115.64101.39 118.16Btu / lbm

0.85h −

= − =

( )lbm Btu Btu0.301 118.16 101.39 5.05 7.14hps lbm sinW ⎛ ⎞= − = =⎜ ⎟

⎝ ⎠ Answer

c) The coefficient of performance is defined as Ref in inCOP Q W= From the energy balance on the refrigerant flowing through the evaporator ( ) ( )( )1 4 0.301lbm s 101.39 36.99 Btu lbm 19.38Btu sin RQ m h h= − = − =

19.38Btu s 3.845.05Btu sRefCOP = = Answer

8- 4

8-5 A large frozen food storage building is to be maintained at –10 ºC. The cooling load is 243 kW. An ideal R-134a vapor-compression refrigeration cycle is to be used for the cooling. Saturated vapor enters the compressor at 100 kPa, and saturated liquid leaves the condenser at 800 kPa. Water used to cool the condenser experiences a 10 ºC temperature rise. Determine:

a. the mass flow rate of the refrigerant (in kg/s) b. the power input to the compressor (in kW) c. the cycle coefficient of performance d. the water mass flow rate (in kg/s).

Approach:

Each process is evaluated with conservation of energy using the appropriate assumptions. The water mass flow rate is evaluated by applying the energy equation to the complete condenser.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the condenser or evaporator. 4. The compressor is isentropic. 5. Water is an ideal liquid with a constant specific heat. 6. The condenser is adiabatic.

Solution:

a) Applying the mass and the energy equations to the refrigerant flowing through the evaporator, and assuming steady, no work, and negligible potential and kinetic energy effects: and 4 1m m m= =

)( )1 4inQ m h h= −

Solving for . From the refrigerant R-134a table, Appendix A-15: ( 1 4/inm Q h h= −

( ) ( )11 1 1,saturated vapor 231.35kJ / kggh P h P= = ( )1 1 1 0.9395kJ / kg Kgs s P= = i

For the throttling value ( )4 3 3 3 93.42 kJ / kgfh h h P= = =

Therefore, ( )

1kJ/s243kW1 kW 1.76 kg / s

231.35 93.42 kJ kgm

⎛ ⎞⎜ ⎟⎝ ⎠= =

− Answer

b) Applying the mass and energy equation to the compressor and assuming steady, reversible, adiabatic, and negligible potential and kinetic energy effects

1 2m m m= =

1

and ( )2 1inW m h h= −

The compressor is isentropic, 2s s= , so interpolation in the superheated vapor region, Appendix A-16, at 800 kPa ( )2 2 2, 274.3kJ / kgh P s =

( )kg kJ 1kW1.76 274.3 231.35 75.6 kWs kg 1 kJ/sinW ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ Answer

c) 243kW 3.2175.6kW

inRef

in

QCOP =

W= = Answer

d) Apply the mass and energy equations to the condenser, and assuming steady, no heat transfer, no work, and negligible potential and kinetic energy effects for both the water and R-134a 2 3 Rm m m= = and A Bm m m= = W ( ) ( )2 3R W B Am h h m h h− = − d) Assuming the water acts as an ideal liquid with a constant specific heat, so that ph c T∆ = ∆ (for water

) 4.18kJ / kg Kpc = i

( ) ( )( )

( )( )2 3 1.76kg s 274.3 93.42 kJ kg kg7.62

s4.18kJ kg K 10 CR

Wp W

m h hm

c T− −

= = =∆ i

Answer

8- 5

8-6 Data from an experiment on a new R-134a vapor-compression refrigeration cycle were obtained. The motor driving the compressor consumed 2.14 hp. The refrigerant entered the compressor at 20 ºF and 20 lbf/in.2 and exited at 170 ºF and 160 lbf/in.2. Refrigerant exited the condenser at 155 lbf/in.2 as a saturated liquid, and the pressure just downstream of the expansion valve was 22 lbf/in2. Determine:

a. the compressor isentropic efficiency b. the cooling capacity (in tons) c. the cycle coefficient of performance (COP).

Approach: An energy balance across the compressor can be used to determine its power. Likewise, an energy balance on the evaporator can be used to calculate the input heat transfer rate.

Assumptions:


Solution:

a) Compressor isentropic efficiency is defined as

1 2C

1 2

h hh h

η−

=−

From the R-134a tables for a superheated vapor (Appendix B-16, by interpolation) ( )1 1 1, 105.88Btu / lbmh P T = 1 0.2323Btu / lbm Rs =

( )2 2 2, 132.42Btu / lbmh P T = 2 0.2433Btu / lbm Rs =

with 2 1s s= , ( )2 2 2, 125.63Btu / lbmsh P s =Therefore,

105.88 125.63 0.744105.88 132.42Cη

−= =

− Answer

b) For the cooling capacity, apply the mass and energy equations to the refrigerant flowing through the evaporator, assuming steady, no work, and negligible potential and kinetic energy effects: and 4 1m m m= = ( )1 4inQ m h h= −

From the throttling valve, Appendix B-15, ( )4 3 3 3 46.85Btu / lbmfh h h P= = = To obtain the mass flow rate, we apply the mass and energy equations to the compressor, and assume steady, adiabatic, and negligible potential and kinetic energy effects: and 2 1m m m= = ( )2 1inW m h h= −

( )2 1

2545Btu/hp2.14hp1 hp lbm205.2

Btu hr132.42 105.88lbm

inWm

h h

⎛ ⎞⎜ ⎟⎝ ⎠= = =

− −

( )lbm Btu Btu205.2 105.88 46.85 12,110 1.01tonhr lbm hrinQ ⎛ ⎞= − = =⎜ ⎟

⎝ ⎠ Answer

c) ( ) ( )

12113Btu/hr 2.222.14hp 2545 Btu/hr 1hp

inRef

in

QCOP =

W= =

⎡ ⎤⎣ ⎦ Answer

8- 6

8-7 A home is heated with a groundwater heat pump, which uses subterranean water at 50 ºF as the low temperature reservoir. The heat pump is designed to blow air in the residential space at 30 ºF above the thermostat set point. Heat loss from the building to the outside air is 358 W per degree Fahrenheit of temperature difference between the inside of the house and the outside air. On a winter day when the outside temperature is 20 ºF, the thermostat is set at 65 ºF. Determine the minimum possible electric power that must be supplied to the heat pump under these conditions.

Approach:

Because the minimum possible power is sought, a heat pump operating on the reverse Carnot cycle must be analyzed. Sufficient information is given to evaluate the required heat transfer rate.

Assumptions:

1. The heat pump operates on the reverse Carnot cycle.

Solution:

Using the definition of COP for a heat pump and solving for inW

Hin

HP

QW

COP=

For the minimum power, we use the definition of the COP for a heat pump operating on the reverse Carnot cycle.

H H HHP

H Lin H L

Q Q TCOP

T TW Q Q= = =

−−

The given information is: (358 )H in outQ T -T= where 358 has units of W/°F and T is in °F.

65 Fin ThermostatT T= =

20 FoutT =

o o30 F 95 F 555 RH inT T= + = =

Groundwater 50 F 510RLT = =Substituting in the absolute temperatures:

555R 12.33555R-510R

HHP

H L

TCOP

T T= = =

−

From the given information about heat loss:

( ) oo

W358 65-20 F 16,110 WFHQ ⎛ ⎞= =⎜ ⎟

⎝ ⎠

Therefore:

16,110W 1310 W12.33inW = = Answer

8- 7

8-8 In winter a building requires 94,000 kJ/hr of heat, and an ideal vapor-compression heat pump is used. R-134a enters the isentropic compressor of the heat pump at 0.4 MPa, 10 ºC and exits at 1 MPa. Saturated liquid leaves the condenser. Determine:

a. the mass flow rate of the refrigerant (in kg/s) b. the power input to the compressor (in kW) c. the cycle coefficient of performance (COP).

Approach: An energy balance across the compressor can be used to determine its power since we can evaluate the enthalpies. Likewise, an energy balance on the condenser can be used to calculate the heat output. The definition is used to determine COP.

Assumptions:

1. The system is steady. 2. Potential and kinetic energy effects are negligible. 3. The compressor is isentropic. 4. No work is done in the condenser. 5. The subcooled liquid approximation is valid.

Solution:

a) Apply the mass and energy equations to the refrigerant flowing through the condenser, and assume steady, no work, and negligible potential and kinetic energy effects: and 3 2m m m= = ( ) ( )2 3 2 3out outQ m h h m Q h h= − → = − From the R-134a tables, Appendices A-14—A-16 With , 3 2P P= ( )3 3 3 105.29kJ / kgfh h P= =

Because the compressor is isentropic, 2 1s s= , ( )1 1 1, 0.9182kJ / kg Ks P T = and ( )1 1 1, 253.35kJ / kgh P T =

We need so, by interpolation, (2 2 2,h P s ) ( )2 2 2, 272.37 kJ / kgh P s =

( )( )( )

94000kJ / hr 1hr 3600 s0.156kg / s

272.37 105.29 kJ / kgm = =

− Answer

b) Apply the mass and energy equations to the compressor, and assume steady, adiabatic, and negligible potential and kinetic energy effects: and 2 1m m m= = ( )2 1inW m h h= −

( )kg kJ 1kW0.156 272.37 253.35 2.97 kWs kg 1kJ / sinW

⎛ ⎞⎛ ⎞= − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

Answer

c) The heat pump coefficient of performance is

( )( )

( )

94000kJ / hr 1hr / 3600 s8.78

1kJ / s2.97 kW1kW

outHP

in

QCOP =

W= =

⎛ ⎞⎜ ⎟⎝ ⎠

Answer

8- 8

8-9 A vapor-compression refrigeration system with a cooling capacity of 6 tons is to be used as a heat pump to warm liquid water. The working fluid is R-134a. The water enters the condenser at 55 ºF and leaves at 80 ºF. Saturated vapor enters the compressor at 40 lbf/in.2, and superheated vapor leaves at 120 lbf/in.2, 110 ºF. Heat transfer between the compressor and the surroundings occurs at a rate of 1.0 Btu/lbm of refrigerant flowing through the compressor. Liquid refrigerant leaves the condenser at 85 ºF, 120 lbf/in2. Determine:

a. the compressor power input (in Btu/min) b. the water flow rate through the condenser (in lbm/min) c. the coefficient of performance.

Approach:

Each process is evaluated with conservation of energy using the appropriate assumptions. The water mass flow rate is evaluated by applying the energy equation to the complete condenser.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the condenser or evaporator. 4. The subcooled liquid approximation is valid. 5. Water is ideal with a constant specific heat.

Solution:

a) Apply the mass and energy equations to the compressor, and assume steady with negligible potential and kinetic energy effects: 2 1 Rm m m= = and 1 1 2 2 0C CQ W m h m h− + − =

( ) ( ) ( )1 2 2 1 2 1C C R in R R c R CW Q m h h W m h h m q m h h q= + − → = − − = − − From the R-134a tables (Appendices A-14—A-16, by interpolation, as needed): ( ) ( )1 1 1 1,saturated vapor 105.88Btu lbmgh P h P= = and ( )2 2 2, 118.92Btu / lbmh P T = To determine , apply the mass and energy equations to the evaporator, and assume steady, no work, and negligible potential and kinetic energy effects:

m

1 4 Rm m m= = and ( ) ( )1 4 1 4in R R inQ m h h m Q h h= − → = − Across the throttling value using the subcooled liquid approximation( )4 3 3 3,h h P T= → ( )3 3 3 38.99Btu lbmfh h T= =

[ ]( )

( )6 tons 200Btu / min 1ton lbm17.9

105.88 38.99 Btu lbm minRm = =−

Therefore, ( )[ ]17.9 lbm min 118.92 105.88 1.0 Btu lbm 216Btu mininW = − − = Answer b) Apply the mass and energy equations to the water flowing through the condenser, and assume steady, no work, and negligible potential and kinetic energy effects: ( ) ( )W B A out W out B AQ m h h Q m Q h h= − = → = −

For a cycle out C in inQ W Q +Q - Q = W= →∑ ∑ Btu 200 Btu/min Btu lbm Btu216 6 tons 1 17.9 1398

min 1ton lbm min minout in in CQ = W +Q - Q⎛ ⎞ ⎛ ⎞⎛ ⎞= + − =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

Assuming the water is an ideal liquid with a constant specific heat, ph c T∆ = ∆ , and using 1.0Btu / lbm Rpc = i

( ) ( )

1398Btu / min lbm55.91.0 Btu lbm R 80 55 R minWm = =

−i Answer

c) The coefficients of performance for the heat pump and for the refrigeration system are 1398Btu / min 6.47216 Btu / min

outHP

in

QCOP =

W= = Answer 1200Btu / min 5.56

216Btu / minin

Refin

QCOP =

W= = Answer

8- 9

8-10 An ideal vapor-compression heat pump cycle using R-134a is used to heat a house. The inside temperature is 22 ºC; the outside temperature is 0 ºC. Saturated vapor at 2.2 bar enters the compressor, and saturated liquid leaves the condenser at 8 bar. The mass flow rate is 0.2 kg/s. Determine:

a the power input to the compressor (in kW) b. the coefficient of performance c. the coefficient of performance if the system were used as a refrigeration cycle d. the maximum theoretical COP working between thermal reservoirs at 22 ºC and 0 ºC.

Approach: An energy balance across the compressor can be used to determine its power since we can evaluate the enthalpies. Likewise, an energy balance on the condenser can be used to calculate the heat output. The definition is used to determine COP.

Assumptions:

1. The system is steady. 2. Potential and kinetic energy effects are negligible. 3. The compressor is isentropic. 4. No work is done in the condenser. 5. The subcooled liquid approximation is valid.

Solution:

a) Apply the mass and energy equation to the compressor, and assume steady, isentropic, and negligible potential and kinetic energy effects: and 2 1m m m= = ( )2 1inW m h h= −

From the R-134a tables (Appendices A-14—A-16), ( ) ( )1 1 1 1,saturated vapor 242.7 kJ kggh P h P= =

( )1 1 1 0.9238kJ / kg Kgs s P= =

With 2 1s s= , by interpolation ( )2 2 2, 269.45kJ / kgh P s =

( )kg kJ 1kW0.2 269.45 242.7 5.35kWs kg 1 kJ/sinW ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟


b) The coefficient of performance is defined as

outHP

in

QCOP =

W

The heating rate, , is determined from mass and energy balance around the refrigerant flowing through the condenser, and assuming steady, no work, and negligible potential and kinetic energy effects:

outQ

and 3 2m m m= = ( )2 3outQ m h h= −

( ) ( )3 3 3 3,saturated liquid 93.42kJ / kgfh P h P= =

( )kg kJ 1kW0.2 269.45 93.42 35.2kWs kg 1kJ/soutQ

⎛ ⎞⎛ ⎞= − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

35.2 kW 6.585.35kWHPCOP = = Answer

c) 1 6.58 1 5.58HP Ref RefCOP COP COP= + ⇒ = − = Answer

d)( )

( ) ( )273 22 K

13.4273 22 K 273 0 K

HCARNOT HP H L

TCOP

T T+

= = =− + − +

Answer

8- 10

8-11 R-134a is used in a vapor-compression heat pump cycle in which the refrigerant enters the adiabatic compressor at 2.4 bar, 0 ºC, with a volumetric flow rate of 0.8 m3/min, and leaves at 10 bar, 55 ºC. Liquid leaves the condenser at 34 ºC. Determine:

a. the power input to the compressor (in kW) b. heating capacity of the system (in kW) c. the coefficient of performance d. the isentropic compressor efficiency.

Approach:

An energy balance across the compressor can be used to determine its power since we can evaluate the enthalpies. Likewise, an energy balance on the condenser can be used to calculate the heat output. Definitions are used to determine COP and isentropic efficiency.

Assumptions:

1. The system is steady. 2. Potential and kinetic energy effects are negligible. 3. The compressor and evaporator are adiabatic. 4. No work is done in the evaporator. 5. The subcooled liquid approximation is valid.

Solution:

a) Apply the mass and energy equations to the compressor, and assume steady, adiabatic, and negligible potential and kinetic energy effects: and where 2 1m m m= = )( 2 1inW m h h= − 1 1/m V v=From the R-134a tables (Appendices A-14—A-16,by interpolation) ( )1 1 1, 248.89kJ / kgh P T = ( ) 3

1 1 1, 0.08574m / kgv P T = ( )1 1 1, 0.9399kJ / ks P T = g

( )2 2 2, 285.78kJ / kgh P T =Therefore, ( ) ( )3 20.8m min 0.08574m / kg 9.33kg min 0.156kg / sm = = =

( )kg kJ 1kW0.156 285.78 248.89 5.75kWs kg 1 kJ/sinW ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟


b) The heating capacity is , so apply the mass and energy equations to the refrigerant flowing through the condenser. Assuming steady, no work, and negligible potential and kinetic energy effects:

outQ

and 3 2m m m= = ( )2 3outQ m h h= −

using the subcooled liquid approximation ( )3 3 3,h P T ⇒ ( )3 3 3 97.31kJ / kgfh h T =∼

( ) kJ 1kW0.156kg 285.78-97.31 29.4kWkg 1 kJ/soutQ ⎛ ⎞= =⎜ ⎟

⎝ ⎠ Answer

c) The coefficient of performance for a heat pump is defined as

29.4kW 5.115.75kW

outHP

out

QCOP =

W= = Answer

d) Isentropic efficiency is 1 2

1 2

sC

h hh h

η−

=−

Since 2 1s s= , by interpolation ( )2 2 2, 279.27 kJ / kgsh P s =

248.89 279.27 0.823248.89 285.78Cη

−= =

− Answer

8- 11

8-12 In an ideal Rankine cycle, saturated water vapor enters the turbine at 20 MPa and exits at 10 kPa. Saturated liquid exits the condenser. Determine:

a. the net work per unit mass of steam flow (in kJ/kg) b. heat input per unit mass of steam flow (in kJ/kg) c. the cycle thermal efficiency d. the heat rejection per unit mass of steam flow (in kJ/kg).

Approach:

Basic definitions combined with conservation of mass and energy are used to calculate the needed quantities.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the boiler. 4. The turbine and pump are isentropic. 5. The water is incompressible.

Solution:

a) The net work is defined as: . To determine these quantities, we apply the mass and energy equations to the turbine and pump, and assume steady, isentropic, and negligible potential and kinetic energy effects:

net T PW W W= −

) and 4 3m m m= = 2 1m m m= =

and ( 3 4TW m h h= − ( ) ( ) ( )2 1 3 4 2 1mP netW m h h W h h h h= − ⇒ = − − − Evaluating the four enthalpies using Appendices A-10—A-12: ( ) ( )3 3 3 3, 2409.7 kJ / kggh P T h P= = ( )3 3 4.9269kJ/kg Kgs s P= =

For because the turbine is isentropic, (4 4 4,h P s ) 34s s= , and state 4 is in the two-phase region

4 ,44

,4

4.9269 0.6493 0.5707.5009

f

fg

s sx

s− −

= = =

( )( )4 ,4 4 ,4 191.83kJ kg 0.570 2392.8kJ kg 1556.4kJ kgf fgh h x h= + = + =

For the isentropic pump and assuming water is incompressible, PW m h mv P= ∆ = ∆

( ) ( )[ ] ( )3 22 1 ,1 2 1 0.001010m kg 20000 10 kN m kJ kN-m 20.2kJ kgfh h h v P P− = ∆ = − = − =

Therefore, ( )m 2409.7 1556.4 20.2 833.1kJ / kgnetW = − − = Answer b) To find the heat addition, apply the mass and energy equations to the boiler, and assume steady, no work, and negligible potential and kinetic energy effects: and 3 2m m m= = ( )3 3min i inQ m h h Q h h= − ⇒ = − 2

( )2 1 2 1 pumph h h h= + − with ( )1 1 191.83kJ / kgfh h P= =

2 191.83 20.2 212.0kJ/kgh = + =

m 2409.7 212 2197.7 kJ kginQ = − = Answer

c) Cycle thermal efficiency is/ 833.1kJ / kg 0.379

2197.7 kJ / kg/net

cycle net inin

W mW Q

Q mη = = = = Answer

d) For a cycle, net net in outW Q Q Q= = −

kJ2197.7 833.1 1364.6kg

out in netout in net

Q Q WQ Q W

m m m= − ⇒ = − = − = Answer

Comments:

The exit quality from the turbine is very low and could damage the turbine. This is not a good design.

8- 12

8-13 Data from a simple Rankine cycle power plant were measured to determine actual performance. The measured steam flow rate was 6.8 kg/s. The measured conditions of the water are shown in the following table.

Device Inlet conditions Outlet conditions

Pump P1 =10 kPa; T1 = 45 ºC P2 =5.2 MPa; T2 = 46 ºC

Boiler P3 =5.1 MPa; T3 = 45 ºC P4 =5.0 MPa; T4 = 500 ºC

Turbine P5 =4.5 MPa; T5 = 500 ºC P6 =15 kPa; x6 = 0.97

Condenser P6 =15 kPa; x6 = 0.97 P7 =12 kPa; T7 = 45 ºCDetermine:

a) the heat addition (in kW) b) the net power produced (in kW) c) the heat rejection (in kW) d) the turbine isentropic efficiency e) the pump isentropic efficiency f) the cycle thermal efficiency.

Approach:

We are given information about the fluid state upstream and downstream of each device. Hence, application of conservation of mass and energy to each device allows us to calculate the heat transfer rate or power of each.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the boiler. 4. The subcooled liquid approximation is valid. 5. The pump and turbine are adiabatic.

Solution:

a) Apply the mass and energy equations to the water flowing through the boiler, and assume steady, no work, and negligible potential and kinetic energy effects: and 4 3m m m= =

)( )4 3inQ m h h= −

Evaluating the properties, is in the subcooled liquid region, so we use the subcooled liquid approximation and Appendix A-10:

(3 3 3,h P T

( ) ( ) ( )

( )

3 3 3 3 3 3 ,3

3

2

,

kJ m kN 1kJ188.45 0.00101 5100 9.593 193.6kJ / kgkg kg 1kN-mm

f f sath P T h T v P P≈ + −

⎛ ⎞ ⎛ ⎞= + − =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

State 4 is in the superheated vapor region (Appendix A-12), so ( )4 4 4, 3433.8kJ / kgh P T =

( )kg kJ 1kW6.8 3433.8 193.6 22,033kWs kg 1 kJ/sinQ ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟


b) The net work is net T PW W W= −Applying the mass and energy equations to the turbine and pump, and assuming steady, adiabatic, and negligible potential and kinetic energy effects: ( )5 6TW m h h= −

where state 5 is in the superheated vapor region, ( )5 5 5, 3439.6kJ / kgh P T = and state 6 is in the two-phase region:

8- 13

( ) ( )6 6 6 ,6 225.94 0.97 2373.1 2527.8kJ / kgf fgh h x h= + = + = Therefore,

( )kg kJ 1kW6.8 3439.6 2527.8 6200kWs kg 1 kJ/sTW ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Likewise, ( )2 1PW m h h= −

State 1 is slightly subcooled liquid, so that ( ) ( ) ( )1 1 1 1 1 1 ,1, f f sh P T h T v P P≈ + − at

( )3

1 2

kJ m kN 1kJ188.45 0.00101 10 9.593 188.45kJ / kgkg kg 1kN-mm

h⎛ ⎞ ⎛ ⎞

= + − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Likewise, for state 2 with interpolations ( ) ( ) ( ) ( )( )2 2 2 2 2 2 2 ,2, 192.6 0.001010 5200 10.14 197.9kJ / kgf f ath P T h T v P P= + − = + − =

( )kg kJ 1kW6.8 197.9 188.45 64.3kWs kg 1 kJ/sPW ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Therefore, 6200 64.3 6136kWnetW = − = Answer c) To find the output heat transfer, consider the overall cycle: in out netQ W Q Q W= ⇒ − =∑ ∑ 22033 6136 15,897 kWoutQ = − = Answer d) Turbine isentropic efficiency definition is

5 6

5 6

actT

ss

W h hh hW

η−

= =−

For isentropic behavior, 6 5s s= , and . Therefore, for ( )5 5 5, 7.0301kJ / kg Ks P T = ( )6 6 6,sh P s

By interpolation 6 66

6

7.0301 0.7549 0.8657.2536

fs

fg

s sx

s− −

= = =

And 6 6 66 6

6

225.942278.9kJ / kg

2373.1s f s

s sfg

h h hx h

h− −

= = → =

Therefore,

3439.6 2527.8 0.7863439.6 2278.9Tη

−= =

− Answer

e) Pump isentropic efficiency definition is sP

act

WW

η = . Recognizing that the liquid is incompressible,

( )

( )

( )

3

2

2 1

m kN0.001010 5200-10kg m

0.56kJ197.9-188.45kg

sP

act

W mv Pm h hW

η

⎛ ⎞⎜ ⎟

∆ ⎝ ⎠= = = =−

Answer

f) The cycle thermal efficiency is:

6136kW 0.27822033kW

netcycle

in

W=

Qη = = Answer

8- 14

8-14 A Rankine cycle power plant has a flow control valve (for use during partial load conditions) located between the boiler and the turbine. At one partial load condtion, steam leaves the boiler at 6.0 MPa, 300 ºC, flows through the valve, which drops the pressure to 4.5 MPa, then enters the turbine in which it expands to the condenser pressure of 10 kPa. Net power output is 500 MW. The turbine isentropic efficiency is 82% and that of the pump is 68%. The condensate from the condenser leaves at 40 ºC. Determine:

a. the steam flow rate (in kg/s) b. the cycle thermal efficiency c. the steam flow rate and cycle thermal efficiency if no valve is present between the boiler and

turbine.

Approach: With the net power known and states up and down stream of the turbine and pump, the energy equation is combined with the known information to calculate the mass flow. Cycle thermal efficiency is determined from its definition and the energy equation applied to the boiler.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the boiler. 4. The turbine is adiabatic. 5. The pump is isentropic.

Solution:

a) To determine the steam flow rate, we know 500MWnet T P in outW W W Q Q= − = − =We evaluate the pump and turbine work by applying the energy and mass equations, and assuming, steady, adiabatic, and negligible potential and kinetic energy effects: 5 4m m m= = 2 1m m m= =

( )4 5TW m h h= − ( )2 1PW m h h= −

Combining these expressions: ( ) ( )4 5 2 1netm W h h h h= − − −⎡ ⎤⎣ ⎦ We need the enthalpies at various locations. Using Appendices A-10—A-12, across the throttling valve,

. For , we need to use the definition of isentropic efficiency ( )4 3 3 3, 2884.2 kJ / kgh h P T= = 5h

( )4 55 4 4 5

4 5Wact

T Tss

W h hh h h h

h hη η

−= = → = − −

− s

) 4

For , we know (5 5 5,sh P s 5s s= , so we need to find ( )4 4 4,s P h . By interpolation in the superheated vapor table, . State 5 is in the two-phase region, so 4 6.1776kJ / kg Ks =

( )55 5 ,5 5 ,5

6.1776 0.6493 0.737 191.83 0.737 2392.8 1955.4kJ kg7.5009

fs s f fg

fg

s sx h h x h

s− −

= = = → = + = + =

( )( )5 2884.2 0.82 2884.2 1955.4 2122.6kJ kgh = − − = For the pump, because it is an ideal liquid (assumed) ( ) ( )[ ] ( )3 2

2 1 ,1 2 1 0.001010m kg 6000 10 kN m 1kJ 1kN-m 6.05kJ / kgs fh h v P P− = − = − = Using the definition of pump isentropic efficiency:

2 1 2 12 1

2 1

6.05 kJ kJ8.90.68 kg kg

s s sP

Pact

W h h h hh h

h hWη

η− −

= = → − = = =−

Therefore, ( ) ( )

1kJ / s500000kW1 kW 664.3kg / s

kJ2884.2 2122.6 8.9kg

m

⎛ ⎞⎜ ⎟⎝ ⎠= =

− −⎡ ⎤⎣ ⎦

Answer

8- 15

b) Cycle thermal efficiency is defined as cycle net inW Qη = . We need to evaluate the input heat transfer rate. To do so, we apply the mass and energy equations to the boiler, and assume steady, no work, and negligible potential and kinetic energy effects: and 3 2m m m= = ( )3 2inQ m h h= −

Using subcooled liquid approximation to evaluate ( ) ( )1 1 1 1, 167.57 kJ / kgfh P T h T =

( )2 1 2 1 167.57 8.9 176.5kJ / kgpumph h h h= + − = + =

( )kg kJ 1 kW664.3 2884.2 176.5 1,799,000kWs kg 1 kJ/sinQ ⎛ ⎞ ⎛ ⎞= − =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

cycle500,000kW 0.278

1,799,000kWη = = Answer

c) If no valve is present, the analysis is the same but the properties upstream and downstream of the turbine change, which changes the mass flow rate ( )3 3 3, 6.0674 kJ / kg Ks P T =

To evaluate the new , we know (5 5 5,sh P s ) 35s s= , so that the new 56.0674 0.6493 0.722

7.5009sx −= =

NEW ( )5 191.83 0.722 2392.8 1920.2kJ / kgsh = + =

NEW ( )5 2884.2 0.82 2884.2 1920.2 2093.7 kJ / kgh = − − =

NEW ( ) ( )

1kJ/s500000 kW1 kW 639.7 kg / s

kJ2884.2-2093.7 8.9 kgm

⎛ ⎞⎜ ⎟⎝ ⎠= =−⎡ ⎤⎣ ⎦

NEW ( )kg kJ 1kW639.7 2884.2 176.5 1,732,000kWs kg 1kJ/sinQ

⎛ ⎞⎛ ⎞= − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

NEW 500,000kW 0.2891,732,000kWcycleη = = Answer

Comments:

The valve creates a pressure drop that represents an unrecoverable energy loss. Hence, without the valve, the cycle efficiency increases.

8- 16

8-15 In a 2 MW Rankine cycle, saturated vapor leaves the boiler at 2 MPa, and expands in the turbine to an outlet condition of 15 kPa, 94% quality. Saturated liquid leaves the condenser. The pump is ideal. The temperature rise of the cooling water in the condenser is 10 ºC. Determine:

a. the mass flow rate of steam (in kg/s) b. the input heat transfer rate (in MW) c. the cycle thermal efficiency d. the cooling water flowrate (in kg/s).

Approach:

For the steam mass flow rate, we are given information the net power output (2 MW) and about the fluid state upstream and downstream of the pump and turbine. These two pieces of information can be combined to calculate the flow rate. Likewise, sufficient information is given for the fluid states around the boiler and condenser to calculate the input heat transfer rate, and the cooling water flow rate, respectively. Hence, we will apply conservation of mass and energy to each device.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the boiler or condenser. 4. The subcooled liquid approximation is valid. 5. The pump and turbine are adiabatic, and the pump

is isentropic. 6. The condenser is adiabatic.

Solution:

a) The net power is . Applying the mass and energy equations to the turbine and pump, and assuming steady, adiabatic, and negligible potential and kinetic energy effects:

net T PW W W= −

and 4 3m m m= = ( )3 4TW m h h= −

Likewise, for the pump ( )2 1PW m h h= −

Therefore, ( ) ( ) ( ) ( )3 4 2 1 3 4 2 1hnet netW m h h m h h m W h h h= − − − ⇒ = − − −⎡ ⎤⎣ ⎦ Evaluating the properties using Appendices A-10—A-12: ( ) ( )3 3 3,saturated vapor 2799.5kJ / kggh P h P= =

( )4 4 4 4 225.94 0.94 2373.1 2456.7 kJ / kgf fgh h x h= + = + =

Assuming the water is an ideal liquid, the across the pump ( )2 1 1 2 1P fh h h v P v P P∆ = − = ∆ = −

( )3

2

m kN 1kJ0.001014 2000 15 2.0 kJ / kgkg 1 kN-mm

h⎛ ⎞ ⎛ ⎞∆ = − =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

Therefore,

( )

1kJ/s2000kW1 kW 5.87 kg / s

kJ2799.5-2456.7 2.0kg

m

⎛ ⎞⎜ ⎟⎝ ⎠= =

−⎡ ⎤⎣ ⎦

Answer

b) Appling the mass and energy equations to the boiler, and assuming steady, no work, and negligible potential and kinetic energy effects: and 3 2m m m= = ( )3 2inQ m h h= − Evaluating the properties: ( ) ( )2 1 2 1 1 225.94 2 227.94kJ / kgf Ph h h h h P h= + − = + ∆ = + =

8- 17

( )kg kJ 1kW 1MW5.87 2799.5 225.94 15.1MWs kg 1kJ/s 1000 kWinQ

⎛ ⎞⎛ ⎞ ⎛ ⎞= − =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Answer

c) The cycle thermal efficiency is

2MW 0.13215.1MW

netcycle

in

W=

Qη = = Answer

d) Appling the mass and energy equations to the condenser, and assuming steady, adiabatic, no work, and negligible andPE∆ KE∆ for both fluids: 1 4 stm m m= = and B A Wm m m= =

4 4 1 1 0A A B Bm h m h m h m h+ − − =

( ) ( )4 1 0st W B Am h h m h h− − − =

4 1W st

B A

h hm m

h h⎛ ⎞−

= ⎜ ⎟−⎝ ⎠

Assuming the water is an ideal liquid with a constant specific heat ( )4.18kJ / kg Kpc = , B A Ph h h c T∆ = − = ∆

( )

( )

kJ2456.7 225.94kg kgkg5.87 313.3s s

=kJ4.18 10 K

kg K

Wm−

⎛ ⎞= ⎜ ⎟ ⎛ ⎞⎝ ⎠⎜ ⎟⎝ ⎠

Answer

8- 18

8-16 A 1000 MW coal-fired Rankine cycle power plant uses coal that has a heating value of 13,390 Btu/lbm. The inlet conditions to the steam turbine are 1000 psia, 750 ºF. Saturated liquid exits the condenser at a pressure of 1 psia. The boiler has a combustion efficiency of 87%, and the electric generator has an efficiency of 94%. The turbine isentropic efficiency is 91%, and the pump isentropic efficiency is 82%. Determine:

a. the overall plant efficiency b. the coal flow rate (in tons/day).

Approach:

Begin with basic definitions and use conservation of mass and energy applied to the various devices to evaluate the terms in the definitions. Sufficient information is given to evaluate all properties up and downstream of each device.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the boiler or condenser. 4. The liquid water is incompressible, and the

subcooled liquid approximation is valid. 5. The pump and turbine are adiabatic.

Solution: a) The overall plant efficiency is overall boiler cycleη η η= where

cycle net inW Qη = , , and in coal boilerQ Q η= ( ) 1000MWnet gen T PW W Wη= − = To find the mass flow rate, apply the mass and energy equations to the turbine and pump, and assume steady, adiabatic, and negligible potential and kinetic energy effects: and 4 3m m m= = 2 1m m m= = ( )3 4TW m h h= − ( )2 1PW m h h= −

Combining these expressions, we obtain: ( ) ( )3 4 2 1

net

gen

Wm

h h h hη=

− − −⎡ ⎤⎣ ⎦

Evaluating the properties (Appendices A-10-A-12, by interpolation as needed: ( ) ( )1 1 ,1 1, saturated liquid 69.74Btu / lbmfh P h P= =

For , use the subcooled liquid approximation and the definition of isentropic efficiency 2h

( ) ( )3 2

2 1 ,1 2 1 2 2

Btu ft lbf 144in. 1Btu Btu69.74 0.01614 1000 1 69.74 3.0 72.7lbm lbm 778.2ft-lbf lbmin. fts fh h v P P

⎛ ⎞ ⎛ ⎞⎛ ⎞= + − = + − = + =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

1 2 1 22 1

1 2

3.0 Btu69.74 73.40.82 lbm

s sP

P

h h h hh h

h hη

η− −

= ⇒ = + = + =−

( )3 3 3, 1357.3Btu / lbmh P T = 3 1.5412Btu / lbm Rs =

To determine , we first use (4 3 4,sh P s ) 34s s= and recognize that state 4 is in the two-phase region:

4 ,4 4 ,4 44 4

69.741.5412 0.13266 0.763 860.6Btu / lbm1.9779 0.13266 1105.8 69.74

f f ss s

g f g f

s s h h hx h

s s h h− − −−

= = = = = ⇒ =− − − −

( )3 44

3 4

1357.3 0.91 1357.3 860.6 905.3Btu / lbmTs

h hh

h hη

−= → = − − =

−

8- 19

( ) ( )

1Btu/s1,000,000kW1.055kW

2248.9 lbm / sBtu0.94 1357.3-905.3 73.4 69.74lbm

m

⎛ ⎞⎜ ⎟⎝ ⎠= =− −⎡ ⎤⎣ ⎦

To find the input heat transfer, apply the mass and energy equations to the water flowing through the boiler, and assume steady, no work, and negligible potential and kinetic energy effects:

( ) ( )3 2lbm Btu 1.055kW2248.9 1357.3 73.4 3,046, 460kW

s lbm 1Btu/sinQ m h h⎛ ⎞⎛ ⎞= − = − =⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

For the cycle thermal efficiency

1,000,000kW 0.3283,046, 460kWcycleη = =

For the overall plant thermal efficiency: ( )0.87 0.328 0.286overallη = = Answer b) To obtain the coal flow rate we use the required heat input and the energy content of the coal: in coal boiler coal coal boilerQ Q m HVη η= =Solving for the coal flow rate:

( )

1Btu/s3,046, 460kW1.055kW lbm247.9

Btu s13390 0.87lbm

coalcoal boiler

QmHV η

⎛ ⎞⎜ ⎟⎝ ⎠= = =

⎛ ⎞⎜ ⎟⎝ ⎠

lbm 3600s 24hr 1ton tons247.9 10,707s 1hr day 2000lbm day

⎛ ⎞ ⎛ ⎞⎛ ⎞⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

Answer

8- 20

8-17 For Problem P 8-12 assume that the turbine has an isentropic efficiency of 91% and the pump has an isentropic efficiency of 78%. Determine:

a. the net work per unit mass of steam flow (in kJ/kg) b. the heat input per unit mass of steam flow (in kJ/kg) c. the cycle thermal efficiency d. the heat rejection per unit mass of steam flow (in kJ/kg).

Approach:

Basic definitions combined with conservation of mass and energy are used to determine the unknowns.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the boiler or condenser. 4. The liquid water is incompressible. 5. The pump and turbine are isentropic.

Solution:

a) From the analysis in P 8-12, we know that where both work terms for this problem represent the ideal power. net T PW W W= −From the definitions of isentropic efficiency

Turbine: actT a

s

WW W

Wη η= → =ct T s

Pump: s sP act

Pact

W WW

Wη

η= → =

So that ( ) kJ 20.2 kJ kJ0.91 2409.7 1556.4 750.6kg 0.78 kg kg

netWm

= − − = Answer

b) For the heat input, ( )3 2inQ m h h= −

Incorporating the isentropic efficiency does not change , but does because of the pump isentropic efficiency, so

3h 2h

2 ,120.2191.83 217.7 kJ / kg0.78

Pf

P

hh h

η∆

= + = − =

2409.7 217.7 2192kJ / kginQm

= − = Answer

c) The cycle thermal efficiency changes because both the net work and heat input changed:

750.6kJ / kg 0.3422192kJ / kgcycleη = = Answer

d) For the heat transfer output, we start with basic information about the cycle performance: net net in outW Q Q Q= = −

out in netQ Q W= −

2192 750.6 1441.4kJ/kgout in netQ Q Wm m m

= − = − = Answer

Comment: As expected, net work and cycle thermal efficiency decrease when the isentropic efficiencies of the turbine and pump are included. Note, however, that most of the change is due to the inefficiencies of the turbine. The pump inefficiencies have little effect.

8- 21

8-18 Steam flowing at 15.9 lbm/s enters the turbine of a simple Rankine cycle power plant at 1000 psia, 800 ºF and exits at 2 psia. Saturated liquid exits the condenser. The turbine and pump are isentropic. Determine:

a. the power output of the turbine (in hp and kW) b. the power input to the pump (in hp and kW) c. the heat input (in hp and kW) d. the cycle thermal efficiency.

Approach:

Basic definitions combined with conservation of mass and energy are used to determine the needed quantities.

Assumptions:


Solution:

a) To find the power produced by the turbine, apply the mass and energy equations, and assume steady, adiabatic, and negligible potential and kinetic energy effects: and 4 3m m m= = ( )3 4TW m h h= −From the superheated vapor table for water, Appendix A-16: and ( )3 3 3, 1388.5Btu / lbmh P T = 3 1.5664 Btu lbm Rs =

Because the turbine is isentropic, 4 3s s= , and we need to evaluate ( )4 4 4,h P s . State 4 is in the two-phase region (Appendix A-11):

4 ,44

,4

1.5664 0.1745 0.7981.7448

f

fg

s sx

s− −

= = =

( )4 ,4 4 ,4 94.02 0.798 1022.1 909.4Btu / lbmf fgh h x h= + = + =

( )lbm Btu 1.055kW15.9 1388.5 909.4 8037.6kWs lbm 1Btu/sTW

⎛ ⎞⎛ ⎞= − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

Answer

( ) 1hp8037.6kW 10,779hp0.7457 kW

⎛ ⎞= =⎜ ⎟

⎝ ⎠ Answer

b) Apply the same equations and assumptions to the pump to obtain: and 2 1m m m= = ( )2 1PW m h h= −

For a reversible process for an incompressible fluid, 2 1dw v P v P h h h= − = ∆ = ∆ = −∫

( ) ( )( ) ( )( )3 2 2 22 1 ,1 2 1 0.01623ft lbm 1000 2 lbf in. 144in. 1ft 1Btu 778.2 ft-lbf 3.0Btu lbmfh h v P P− = − = − =

As with the turbine ( )( )( )15.9lbm s 3Btu lbm 1.055kW Btu s =50.3kW 67.5hpPW = = Answer c) To obtain the heat input, apply the mass and energy equations to the boiler, and assume steady, no work, and negligible potential and kinetic energy effects and 3 2m m m= = ( )3 2inQ m h h= −

( )2 1 2 1 ,1 94.02 3 97.0Btu / lbmf pumppumph h h h h h= + − = + ∆ = + =

( )( ) ( )15.9 lbm s 1388.5 97.0 Btu lbm 1.055kW Btu s 21,666 kW 29,055hpinQ = − = = Answer d) The cycle thermal efficiency is:

T 10779hp 67.5hp 0.36929055hp

net Pcycle

in in

W W WQ Q

η− −

= = = = Answer

8- 22

8-19 For Problem P 8-18, assume the turbine has an isentropic efficiency of 87% and the pump has an isentropic efficiency of 75%. All other conditions remain the same. Determine:

a. the power output of the turbine (in hp and kW) b. the power input to the pump (in hp and kW) c. the heat input (in hp and kW) d. the cycle thermal efficiency.

Approach:

Basic definitions combined with conservation of mass and energy are used to determine needed quantities.

Assumptions:


Solution:

a) From the analysis in P 8-18, we known that ( )3 4T sW m h h W= − =Using the definition of isentropic efficiency

( )8037.6kW 0.87 6992.7kW=9377.4hpactT act s T

s

WW W

Wη η= → = = = Answer

b) For the pump, ( )2 1P sW m h h W= − = Using the definition of isentropic efficiency

50.3kW 67.1kW=89.9hp0.75

s sP act

Pact

W WW

Wη

η= → = = = Answer

c) For the heat input ( )3 2inQ m h h= −

The enthalpy does not change; but does change, so 3h 2h

2 ,1394.02 98.0Btu / lbm

0.75pump

fP

hh h

η∆

= + = + =

( )( ) ( )15.9 lbm s 1388.5 98.0 Btu lbm 1.055kW Btu s 21,650 kW 29,033hpinQ = − = = Answer d) The cycle thermal efficiency is

9377.4 89.9 0.32029033cycleη −

= = Answer

Comment:

As expected, when the pump and turbine are not isentropic, net work and cycle thermal efficiency decrease. Note, though, that almost all of the change is due to the turbine inefficiency. In Rankine cycles, because the back work ratio is small, inefficiencies in the pump do not have a significant impact on the overall cycle performance.

8- 23

8-20 An ideal Rankine cycle uses one stage of reheat. Steam enters the high-pressure turbine at 10 MPa, 550 ºC, expands to 1 MPa where it is extracted and routed to a reheater where the steam temperature is raised to 500 ºC. The steam is then expanded in the low-pressure turbine to the condenser pressure of 20 kPa. Saturated liquid exits the condenser. The net power produced by the plant is 100 MW. Both turbine stages and the pump are isentropic. Determine:

a. the mass flow rate of the steam (in kg/s), b. the cycle thermal efficiency.

Approach:

We are given the net power produced and have enough information to analyze the power of the turbine and pump. Combining these two facts allows us to calculate the steam flow rate.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the boiler or reheater. 4. The subcooled liquid approximation is valid. 5. The pump and turbine are isentropic.

Solution:

a) To find the flow rate, apply the mass and energy equations to the turbine and pump, and assume steady, adiabatic, reversible, and negligible potential and kinetic energy effects: and 6 5 4 3m m m m m= = = = 2 1m m m= =

)

and ( ) (3 4 5 6TW m h h m h h= − + − ( )2 1PW m h h= −

Because , then 100MW= Wnet T PW W= −)( ) ( ) (3 4 5 6 2 1h

netWm

h h h h h=

− + − − −

Evaluating the properties using Appendices A-10—A-12:

( )11 1 251.4kg / kJfh h P= =

( ) [ ]1

3

2 1 2 1 2

m kN 1 kJ251.4 0.001017 10000 20 261.5K 1 kN-m kgmfh h v P P

⎛ ⎞ ⎛ ⎞+ − = + − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

∼ kJ

( )3 3 3, 3500.9kJ / kgh P T = 3 6.7561kJ / kg Ks = i With 4 3 6and 5s s s= = s superheated vapor, so by interpolation ( )4 4 4,h P s → 4 2858.8kJ kgh =

( )5 5 5, 3478.5kJ / kgh P T = 5 7.7622kJ / kg Ks = i

( )6 6 6 67.7622 0.832, 0.979

7.0766h P s x −

→ = = and ( )6 251.4 0.979 2358.3 2560.2kJ / kgh = + =

( ) ( ) ( )

1kJ/s100,000kWkg1 kW 64.5s3500.9-2858.8 3478.5 2560.2 261.5 251.4 kJ kg

m

⎛ ⎞⎜ ⎟⎝ ⎠= =

+ − − −⎡ ⎤⎣ ⎦ Answer

b) The definition of cycle thermal efficiency is cycle net inW Qη = . Apply the mass and energy equations to the boiler and reheater, and assume steady, no work, and negligible potential and kinetic energy effects:

( ) ( )

( ) ( )

3 2 5 4

kg kJ 1kW64.5 3500.9 261.5 3478.5 2858.8 248,910kWs kg 1k

in boiler reheaterQ = Q +Q m h h m h h= − + −

⎛ ⎞⎛ ⎞= − + − =⎡ ⎤ ⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠J/s

cycle100,000kW 0.402248,910kW

η = = Answer

8- 24

8-21 For Problem P 8-20, the both stages of the turbine have isentropic efficiencies of 85%, and the pump isentropic efficiency is 78%. All other conditions remain the same. Determine:

a. the mass flow rate of the steam (in kg/s) b. the cycle thermal efficiency, and compare the efficiency to that calculated in P 8-20.

Approach:

The same approach is used as in P 8-20. The only difference is the definition of isentropic efficiency is now incorporated.

Assumptions:


Solution:

a) Using the information from problem P 8-20, and incorporating the definitions of isentropic efficiency:

Turbine 1: ( )3 44

3 4

kJ3500.9 0.85 3500.9 2858.8 2955.1kgT

s

h hh

h hη

−= → = − − =

−

Turbine 2: ( )5 66

5 6

kJ3478.5 0.85 3478.5 2560.2 2697.9kgT

s

h hh

h hη

−= → = − − =

−

Pump: 1 22

1 2

251.4 261.5 kJ251.3 264.30.78 kg

sP

h hh

h hη

− −= → = − =

−

( ) ( ) ( )

1kJ/s100,000kWkW 76.1 kg / s

3500.9-2955.1 3478.5 2697.9 264.3 251.4 kJ kgm

⎛ ⎞⎜ ⎟⎝ ⎠= =

+ − − −⎡ ⎤⎣ ⎦ Answer

b) For the new cycle thermal efficiency:

( ) ( )kg kJ 1kW76.1 3500.9 264.3 3478.5 2955.1 286,100kWs kg 1 kinQ

⎛ ⎞⎛ ⎞= − + − =⎡ ⎤ ⎜ ⎟⎜ ⎟ ⎣ ⎦⎝ ⎠ ⎝ ⎠J/s

100,000kW 0.349286,100kW

netcycle

in

WQ

η = = = Answer

Comments:

For the same net power output, a larger mass flow rate is required when non-isentropic pumps and turbines are used since less energy is extracted in the turbine. In addition, cycle thermal efficiency drops significantly when the isentropic efficiencies are taken into account.

8- 25

8-22 A Rankine cycle has three turbine stages with two reheats between the stages. Superheated vapor leaves the boiler at 30 MPa, 550 ºC; the vapor leaves the first reheater at 5 MPa, 500 ºC and the second reheater at 0.5 MPa, 400 ºC. The condenser pressure is 0.05 bar, and saturated liquid exits the condenser. Total mass flow is 2.5×106 kg/hr. Determine:

a. the net power (in kW) b. the cycle thermal efficiency.

Approach:

Beginning with the definition of net power, conservation of mass and energy can be applied to the turbines and pumps to evaluate each power term. Cycle thermal efficiency is determined from its definition.

Assumptions:


Solution:

a) For net power, we know , , ,net T A T B T C PW W W W W= + + −

)4

Applying the mass and energy equations to each turbine and the pump, and assuming steady, isentropic, and negligible potential and kinetic energy effects: (, 3T AW m h h= − ( ), 5T BW m h h= − 6

)8

(, 7T CW m h h= − ( )2 1PW m h h= −

Evaluating the properties using Appendices A-10—A-12: ( )1 ,1 1 137.82kJ / kgfh h P= =

State 2 is a subcooled liquid ( )2 ,1 ,1 2 1f fh h v P P→ = + −

( )3

2 2

kJ m kN 1kJ kJ137.82 0.001005 30000 5 137.82 30.1 167.9kg kg 1kN-m kgm

h⎛ ⎞ ⎡ ⎤

= + − = + =⎜ ⎟ ⎢ ⎥⎣ ⎦⎝ ⎠

( )3 3 3, 3275.4 kJ kgh P T = 3 6.0342kJ / kg Ks = For isentropic turbines, 4 3s s= , 6 5s s= , and 8 7s s= State 4 is superheated vapor, so by interpolation ( )4 4 4, 2827.3kJ kgh P s =

( )5 5 5, 3433.8kJ / kgh P T = 5 6.9759kJ / kg Ks =

State 6 is superheated vapor, so by interpolation ( )6 6 6, 2818.0kJ / kgh P s =

( )7 7 7, 3271.9kJ / kgh P T = 7 7.7938kJ / kg Ks =

State 8 is two-phase, so to find ( )8 8 8,h P s 8 ,88

,8

7.7938 0.4764 0.9247.9187

f

fg

s sx

s− −

= = =

( )8 137.82 0.924 2423.7 2377.5kJ / kgh = + =

( )6,

kg kJ 1hr 1kW2.5 10 3275.4 2827.3 311,180kWhr kg 3600s 1kJ/sT AW

⎛ ⎞⎛ ⎞⎛ ⎞= × − =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

( )( )6,

12.5 10 3433.8 2818.0 427,640kW3600T BW ⎛ ⎞= × − =⎜ ⎟

⎝ ⎠

( )( )6,

12.5 10 3271.9 2377.5 621,110kW3600T CW ⎛ ⎞= × − =⎜ ⎟

⎝ ⎠

( )( )6 12.5 10 167.9 137.82 20,900kW3600PW ⎛ ⎞= × − =⎜ ⎟

⎝ ⎠

8- 26

311,180 427,640 621,110 20,900 1,339,000kWnetW = + + − = Answer

b) The cycle thermal efficiency is cyclenet

in

WQ

η =

,1 ,2in boiler reheater reheaterQ Q +Q Q= +To find the heat inputs, apply the mass and energy equations to the water flows through the boiler and both reheaters, and assume steady, no work, and negligible potential and kinetic energy effects:

( ) ( )( )63 2

12.5 10 3275.4 167.9 2,158,000 kW3600boilerQ m h h ⎛ ⎞= − = × − =⎜ ⎟

⎝ ⎠

( )( )( )6, 1 5 4

12.5 10 3433.8 2827.3 421, 200kW3600reheaterQ m h h ⎛ ⎞= − × − =⎜ ⎟

⎝ ⎠

( ) ( )( )6,2 7 6

12.5 10 3271.9 2818.0 315,200kW3600reheaterQ m h h ⎛ ⎞= − = × − =⎜ ⎟

⎝ ⎠

kW 2,158,000 421,200 315, 200 2,894,000kWinQ = + + =

1,339,000kW 0.4632,894,000kWcycleη = = Answer

8- 27

8-23 For Problem P 8-22, assume the three turbine stages have isentropic efficiencies of 91%, 87%, and 83%, respectively, and the pump isentropic efficiency is 80%. Determine:

a. the net power (in kW) b. the cycle thermal efficiency.

Approach:

The same approach is used as in P 8-22. The only difference is the definition of isentropic efficiency is used.

Assumptions:


Solution:

a) Using information given in P 8-22 and incorporating the effects of isentropic efficiencies:

Turbine A: ( )3 4, 4

3 4

kJ3275.4 0.91 3275.4 2827.3 2867.6kgT A

s

h hh

h hη

−= → = − − =

−

( ) ( )6, 3 4

kg kJ 1hr 1kW2.5 10 3275.4 2867.6 283,200kWhr kg 3600 s 1 kJ/sT AW m h h ⎛ ⎞ ⎛ ⎞⎛ ⎞= − = × − =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

Turbine B: ( )5 6, 6

5 6

kJ3433.8 0.87 3433.8 2818.0 2898.1kgT B

s

h hh

h hη

−= → = − − =

−

( ) ( )( )( )6, 5 6 2.5 10 3433.8 2898.1 1/ 3600 372,000kWT BW m h h= − = × − =

Turbine C: ( )7 8, 8

7 8

kJ3271.9 0.83 3271.9 2377.5 2529.5kgT C

s

h hh

h hη

−= → = − − =

−

( ) ( )6, 7 8

kg kJ 1hr 1kW2.5 10 3271.9 2529.5 515,600kWhr kg 3600 s 1 kJ/sT CW m h h ⎛ ⎞ ⎛ ⎞⎛ ⎞= − = × − =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

Pump: 1 22

1 2

137.82 167.9 kJ137.82 175.50.80 kg

sP

h hh

h hη

− −= → = − =

−

( )( ) ( )62.5 10 175.5 137.82 1/ 3600 26, 200 kWPW = × − =

283,200 372,000 515,600 26, 200 1,145,000kWnetW = + + − = b) Using information from P 8-22 and Part (a) above ( ) ( )( )( )6

3 2 2.5 10 3275.4 175.5 1/ 3600 2,153,000 kWboilerQ m h h= − = × − =

( ) ( )( )( )6 1 5 4 2.5 10 3433.8 2867.6 1/ 3600 393,200kWreheater,Q m h h= − = × − =

( ) ( )( )( )6, 2 7 6 2.5 10 3271.9 2898.1 1/ 3600 259,600kWreheaterQ m h h= − = × − =

2,153,000 393, 200 259,600 2,806,000kWinQ = + + =

1,145,000kW 0.4082,806,000 kWcycleη = = Answer

Comments:

As expected, when non-isentropic turbines and pump are used, both the net work and the cycle thermal efficiency decrease. The turbine inefficiencies contribute the vast majority of the reduced performance.

8- 28

8-24 For the Rankine cycle power plant shown in the figure below, determine: a. the mass flow rate (in kg/s) b. the total heat addition (in kW) c. the net power output (in kW) d. the cycle thermal efficiency.

Approach:

We are given the heat transfer rate rejected from the condenser and have enough information to apply conservation of energy to it to find the steam flow rate. Likewise, sufficient information is given to evaluate the properties at the entrance and exit of the other devices. Hence, by applying the mass and energy equations to the other devices, the quantities sought can be determined.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the condenser, boiler, or

reheater. 4. The subcooled liquid approximation is valid. 5. Liquid water is incompressible.

Solution: a) We can determine the mass flow rate from an energy and mass balance on the steam in the condenser. Assume steady, no work, and negligible potential and kinetic energy effects: and 6 1m m m= = ( ) ( )6 1 out 6 1/outQ m h h m Q h h= − → = − Evaluating the fluid properties using Appendices A10—A-12: ( ) ( )6 6 6 6 6, 289.23 0.99 2336.1 2602.0kJ / kgf fgh P x h x h= + = + =

( )1 1 209.33kJ / kgfh h T =Therefore,

( )

1kJ/s1500kW1 kW 0.627 kg / s

2602.0-209.33 kJ / kgm

⎛ ⎞⎜ ⎟⎝ ⎠= = Answer

b) Total heat input is . Apply the energy and mass equations to the water flowing through the boiler and reheater, and assume steady, no work, and negligible potential and kinetic energy effects:

in boiler reheaterQ = Q +Q

( )3 2boilerQ m h h= −Evaluating the properties: ( )3 3 3, 3625.3kJ / kgh P T =

assuming incompressible and reversible flow 2 1 Ph h h= + ∆ → ( )1 2 1P fh v P P∆ = −From the definition of isentropic efficiency

( ) ( )3

21 2 1

m kN kJ0.001012 10000 30kg 1kN-mm

11.2kJ / kg0.90

fP PP

P P

v P Ph hh

hη

η η

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟−∆ ∆ ⎝ ⎠⎝ ⎠= → ∆ = = = =

∆

2 209.33 11.2 220.5kJ / kgh = + =

( )( )0.627 kg s 3625.3 220.5 kJ kg 2135kWboilerQ = − = Likewise, for the reheater ( )5 4reheaterQ m h= − h

8- 29

( )5 5 5, 3066.8kJ / kgh P T =

For , we need to use the isentropic turbine efficiency definition4h 3 4

3 4T

s

h hh h

η−

=−

. With 4 3s s= ,

( )3 3 3, 6.902kJ kg Ks P T = i

By interpolation in the superheated vapor table ( )4 4 4, 2741kJ ksh P s = g

( ) ( )4 3 3 4 3625.3 0.88 3625.3 2741 2847 kJ kgT sh h h hη= − − = − − =

( )( )0.627 kg s 3066.8 2847 kJ kg 137.7 kWreheaterQ = − =

2135 137.7 2273kWinQ = + = Answer c) Net power output is 2273kW 1500kW 772.7 kWnet T P in outW W W Q Q= − = − = − = d) The cycle thermal efficiency is

772.7 kW 0.3402273kWQ

netcycle

in

Wη = = = Answer

8- 30

8-25 In a Rankine cycle power plant, superheated steam leaves the boiler at 1250 psia, 1000 ºF. Saturated liquid exits the condenser that operates at a pressure of 2 psia. Pump isentropic efficiency is 90%. Determine:

a. the cycle thermal efficiency if expansion is through a single turbine with an isentropic efficiency of 90%

b. the cycle thermal efficiency if reheat is used in which the steam is extracted at 100 psia, is reheated to 800 ºF, and is expanded to the same condenser pressure; both low- and high-pressure turbines have isentropic efficiencies of 90%.

Approach:

From the definition of cycle efficiency, we need to evaluate net work and heat input. We are given sufficient to evaluate the properties at the entrance and exit of the turbines, pump, boiler, and reheater. By applying the mass and energy equations to the other devices, the quantities sought can be determined.

Assumptions:


reheater. 4. The subcooled liquid approximation is valid. 5. Liquid water is incompressible.

Solution: a) The cycle thermal efficiency is

Q

netcycle

in

Wη = and net T PW W W= −

For one stage expansion, apply the mass and energy equations, from Point 3 to Point 6 with no extraction, and assume steady, adiabatic, and negligible potential and kinetic energy effects: and 3 6m m m= = ( )3 6TW m h h= −Evaluating the properties using Appendices A-10—A-12: ( )3 3 3, 1498.2Btu / lbmh P T = 3 1.6244Btu / lbm Rs =

With no reheat and for an isentropic process, 6 3s s= , and ( )6 6 6,sh P s is in the two-phase region, so:

6 ,66

,6

1.6244 0.17499 0.8311.7448

fs

fg

s sx

s− −

= = =

( )6 ,6 6 ,6 94.02 0.831 1022.1 943.1Btu / lbms f s fgh h x h= + = + =

( )( )3 66

3 6

1498.2 0.90 1498.2 943.1 998.6Btu / lbmTs

h hh

h hη

−= → = − − =

−

3 6 1498.2 998.6 499.6Btu / lbmTW h hm

= − = − =

Likewise, for the pump and 2 1m m m= = ( )2 1PW m h h= −

State 1 is a subcooled liquid, so using the subcooled liquid approximation ( )1 1 1 94.02kJ kgfh h P= =

For h2, which is a subcooled, incompressible liquid, ( )2 1 1 2 1s fh h v P P− = −

( )3 2

2 1 2 2

ft lbf 144in. 1 Btu Btu0.01623 1250 2 3.75lbm 778.2 ft-lbf lbmin. ftsh h

⎛ ⎞ ⎛ ⎞⎛ ⎞− = − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

1 22

1 2

3.75 Btu94.02 98.20.90 lbm

sP

h hh

h hη

−= → = + =

−

8- 31

98.2 94.02 4.2Btu / lbmPWm

= − =

499.6 4.2 495.4Btu / lbmnetW m = − = For the heat input, apply the mass and energy equations to the water flowing through the boiler (from Point 1 to Point 2), and assume steady, no work, and negligible potential and kinetic energy effects: ( )3 2 3 2 1498.2 98.2 1400Btu lbmin inQ m h h Q m h h= − → = − = − = Finally,

495.4Btu lbm 0.3541400Btu lbmcycleη = = Answer

b) This problem is similar to part (a), but now we take into account the reheat. Applying the mass and energy equations to the turbines using the same assumptions as before: ( )3 4T,AW m h h= − ( ), 5T BW m h h= − 6

3

For the isentropic turbines, 4s s= and 6 5s s= . For ( )4 4 4,sh P s , this is in the superheated vapor region, so by interpolation 4 1204.7 Btu lbmsh = . Using isentropic efficiency ( )( )4 1498.2 0.9 1498.2 1204.7 1234.0Btu / lbmh = − − =For the second turbine: ( )5 5 5, 1429.6Btu / lbmh P T = 5 1.8449Btu / lbm Rs =

State 6 is in the two-phase region, so to determine ( )6 6, 6sh P s

6 ,66

,6

1.8449 0.17499 0.9571.7448

fs

fg

s sx

s− −

= = =

( )6 94.02 0.957 1022.1 1072.2Btu / lbmsh = + =

Using isentropic efficiency ( )6 1429.6 0.9 1429.6 1072.2 1108.0Btu / lbmh = − − = Therefore,

( ) ( )3 4 5 6net PW W

h h h hm m

= − + − −

( ) ( ) Btu1498.2 1234.0 1429.6 1108.0 4.2 581.6lbm

= − + − − =

For the heat transfer input, both the boiler and reheater need to be taken into account. Analyzing the reheater similar to the boiler in part (a):

( ) (3 2 5 4in boiler reheaterQ Q Q

+ h h hm m m

= = − + )h−

( ) ( )1498.2 98.2 1429.6 1234.0 1595.6Btu / lbm= − + − =Therefore, the cycle efficiency using reheat is:

581.6Btu/lbm 0.3651595.6 Btu/lbmcycleη = = Answer

Comments:

As expected the addition of a reheater increased the cycle thermal efficiency. Note also that the net work per unit mass flow increased, and the exit quality at the lowest pressure increased, too.

8- 32

8-26 In an ideal reheat Rankine cycle, steam enters the high-pressure turbine at 9 MPa, 500 ºC and is extracted at a lower pressure, reheated to 500 ºC, and then, in the low-pressure turbine, expanded to 10 kPa. To minimize possible damage in the low-pressure turbine, the minimum quality at the turbine outlet is specified to be 90%. Determine:

a. the pressure at which reheating takes place (in kPa) b. the total heat addition per unit mass (in kJ/kg) c. the total heat rejection per unit mass (in kJ/kg) d. the cycle thermal efficiency.

Approach:

Part (a) is determined solely using information about thermodynamic states. The heat rejection and addition are determined by application of conservation of mass and energy to the condenser and boiler, respectively.

Assumptions:


reheater. 4. The subcooled liquid approximation is valid. 5. Pump and turbine are isentropic.

Solution:

a) For an isentropic turbine, 6 5s s= . Using data from Appendices A-10—A-12, and based on the given minimum outlet quality: ( ) ( ) ( )( )6 6 6 6 6 6 0.6493 0.90 7.5009 7.400kJ / kg Kf fgs s P x s P= + = + = Likewise, the outlet quality can be determined: ( )( )6 191.83 0.90 2392.8 2345.4kJ / kgh = + =We know that the steam is reheated to 500 °C upstream of the turbine, and we know its entropy. Because this is superheated vapor, by interpolation and ( )5 5 5, 2.15MPP T s = a 5 3466kJ / kgh = Answer

b) The total heat input is: . Apply the mass and energy equations to the water flowing through the boiler and reheater, and assuming steady, no work, and negligible potential and kinetic energy effects:

in boiler reheaterQ Q +Q=

( ) (3 2 5 4inQ m h h h h= − + − ) To obtain h2, across the pump we know that

( ) ( )3

2 1 1 2 1 2

m kN kJ0.00101 9000 10 9.08kJ / kgkg 1 kN-mmP P fh h h h v P P

⎛ ⎞ ⎛ ⎞= + ∆ → ∆ = − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

The pump inlet is saturated liquid: ( )1 1 191.83kJ / kgfh h P= = Therefore, 2 191.83 9.08 200.9kJ / kgh = + =

States 3 and 4 are superheated vapor: ( )3 3 3, 3386.1kJ / kgh P T = 3 6.6576kJ / kg Ks =

With 4 3s s= , and noting state 4 is superheated vapor, by interpolation ( )4 4 4, 2979.5kJ / kgh P s = , so that

( ) ( )3386.1 200.9 3466.0 2979.5 3672kJ/ kginQ m = − + − = Answer c) The heat rejection is obtained by a mass and energy balance on the water flowing through the condenser with the same assumptions as used above for the boiler; ( )5 1 2345.4 191.83 2154 kJ / kgoutQ m h h= − = − = Answer d) The cycle thermal efficiency is:

2154kJ / kg1 1 0.4143672kJ / kg

net outcycle

in in

W QQ Q

η = = − = − = Answer

8- 33

8-27 In an ideal reheat Rankine cycle, steam enters the high-pressure turbine at 800 psia, 900 ºF and exits as a saturated vapor. The steam then enters the reheater where its temperature is raised to 800 ºF. The steam then expands in the low-pressure turbine to 1 psia. Total heat addition is 2.2×108 Btu/h. Determine:

a. the pressure at which reheat takes place (in psia) b. the mass flow rate (in lbm/s) c. the heat rejection (in Btu/hr) d. the cycle thermal efficiency e. the net power output (in Btu/hr, hp, and kW).

Approach:

Part (a) is determined solely from thermodynamic properties. Mass flow is obtained from the total heat addition and conservation of energy and mass applied to the boiler. The other quantities are obtained using their definitions and conservation of mass and energy.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in condenser, boiler, or reheater. 4. The subcooled liquid approximation is valid. 5. Pump and turbine are isentropic.

Solution: a) For an isentropic turbine, 4 3s s= . Using Appendices A-10—A-12, we obtain ( )3 3 3, 1.6408Btu / lbm Rs P T = and 3 1455.6Btu / lbmh =Based on the statement that the outlet is saturated vapor, by interpolation:

( ) 24 4 ,saturated vapor 62.8lbf in.P s = 4 1178.9Btu / lbmh = Answer

b) The mass flow rate is obtained by a mass and energy balance on the water flowing through the boiler and reheater. Assuming steady, no work, and negligible potential and kinetic energy effects:

( ) ( ) ( ) ( )3 2 5 43 2 5 4h

inin

QQ m h h m h h m

h h h= − + − → =

− + −

Evaluating properties, , and noting that across the pump we assume incompressible and reversible:

( )1 1 69.74Btu / lbmfh h P= =

( ) ( )3 2

2 1 1 1 2 1 2 2

Btu ft lbf 144in. 1Btu69.74 0.01614 800 1 72.1Btu / lbmlbm lbm 778.2 ft-lbfin. ftP fh h h h v P P

⎛ ⎞ ⎛ ⎞⎛ ⎞= + ∆ = + − = + − =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

( )5 5 5, 1431.1Btu / lbmh P T =

( ) ( )

82.2 10 Btu / hr lbm lbm134,500 37.4Btu hr s1455.6 72.1 1431.1 1178.9lbm

m ×= = =

− + −⎡ ⎤⎣ ⎦

Answer

c) The heat output is obtained by applying the energy and mass equations to the condenser, and assuming steady, no work, and negligible potential and kinetic energy effects: ( )6 1outQ m h h= −

For h6, we note that 6 5s s= and that state 5 is superheated vapor, so by interpolation ( )5 5, 5 1.8977 Btu lbmRs P T =

6 ,66

,6

1.8977 0.13266 0.9551.8453

f

fg

s sx

s− −

= = = → ( )( )6 6 69.74 0.955 1036 1060.7 Btu / lbmgf fh h x h= + = + =

( )( ) 8134500lbm hr 1060.7 69.24 Btu lbm 1.333 10 Btu / hroutQ = − = × Answer

d) Cycle thermal efficiency is: ( ) ( )8 81 1 1.333×10 Btu hr 2.2×10 Btu hr 0.394cycle out inQ Qη = − = − = Answer e) For the net work,

( ) 6 7Btu Btu0.394 2.2 10 8.67 10 25, 413kWhr hrcycle net in net cycle inW Q W Qη η ⎛ ⎞= → = = × = × =⎜ ⎟

⎝ ⎠ Answer

8- 34

8-28 An ideal Rankine cycle uses one stage of reheat. Vapor leaves the boiler at 2000 psia, 1000 ºF, expands in the high-pressure turbine to 500 psia, at which point the steam is extracted and routed back through the reheater where the steam temperature is raised to 1000 ºF. The steam then expands through the low-pressure turbine to the condenser pressure of 5 psia. The condensate leaves the condenser as a saturated liquid. The steam mass flow rate is 5 lbm/s. Determine:

a. the net power produced (in kW) b. the heat input to the boiler (without reheat) (in kW) c. the heat input in the reheater (in kW) d. the cycle thermal efficiency.

Approach:

Basic definitions and conservation of mass and energy applied to the devices are used to determine the quantities sought.

Assumptions:



Solution: a) Net power produced is . To find power, apply the mass and energy equations to the turbines and pump, and assume steady, isentropic, and negligible potential and kinetic energy effects:

, ,net T A T B PW W W W= + −

and 3 4 5 6m m m m m= = = = 1 2m m m= =

)4

(, 3T AW m h h= − ( ), 5T BW m h h= − 6 ( )2 1PW m h h= −

The inlet to the pump is saturated liquid, so using Appendices B-10—B-12, ( )1 1 1 130.17 Btu / lbmfh h P= = . For

an incompressible, reversible process across the pump ( )2 1 1 2 1fh h v P P− = −

( )3 2

2 2 2

Btu ft lbf 144 in. 1Btu Btu130.17 0.016407 2000 5 130.17 6.06 136.2lbm lbm 778.2ft-lbf lbmin. 1ft

h⎛ ⎞ ⎛ ⎞⎛ ⎞

= + − = + =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

( )3 3 3, 1474.1Btu / lbmh P T = 3 1.5598Btu / lbm Rs = For the isentropic turbine, 4 3s s= , and state 4 is superheated vapor, so by interpolation ( )4 4 4, 1299.7 Btu lbmh P s = . ( )5 5 5, 1520.7 Btu / lbmh P T = 5 1.7371Btu lbmRs = Note that 6 5s s= , and state 6 is in the two-phase region, so:

( ) ( )6 6 6 ,6 1.7371 0.23486 1.6093 0.933f fgx s s s= − = − = ( )( )6 130.17 0.933 1000.9 1064.5Btu / lbmh = + =

( )( ),Btu 1.055kW5lbm s 1474.1 1299.7 920.0kWlbm 1Btu/sT AW

⎛ ⎞= − =⎜ ⎟

⎝ ⎠

( )( )( ), 5 1520.7 1064.5 1.055 2407 kWT BW = − = ( )( )( )5 6.06 1.055 32.0kWPW = =

920.0 2407 32.0 3300kWnetW = + − = Answer b) To obtain the heat input, apply the mass and energy equation to the water flowing through the boiler, and assume steady, no work, and negligible potential and kinetic energy effects and 3 2m m m= = ( ) ( )( )( )3 2 5 1474.1 136.2 1.055 7060kWboilerQ m h h= − = − = Answer c) The reheater heat transfer rate calculation is similar to part (b): ( ) ( )( )5 4 5 1520.7 1299.7 1.055 1170 kWreheaterQ m h h= − = − = Answer d) The cycle thermal efficiency is

( )

( )920.0 2407 32.0 kW

0.4017060 1170 kW

T,A T,B Pnetcycle

in boiler reheater

W -W -WWQ Q +Q

η+ −

= = = =+

Answer

8- 35

8-29 A closed feedwater heater is used in a Rankine cycle. Steam leaves the boiler at 20 MPa, 600 ºC. Between the high- and low-pressure turbines, steam at 1 MPa is extracted and delivered to the closed feedwater heater. Feedwater exits the feedwater heater at 20 MPa and the saturation temperature of the 1 MPa steam; saturated liquid condensate is fed through a steam trap back to the condenser. Steam from the second stage turbine enters the condenser at 10 kPa, and saturated liquid exits the condenser. Both stages of the turbine have isentropic efficiencies of 90%, and the pump isentropic efficiency is 85%. Determine:

a. the fraction of steam entering the turbine that must be extracted b. the net work per unit mass of steam entering the first turbine stage (in kJ/kg) c. the heat added per unit mass of steam entering the first turbine stage (in kJ/kg) d. the cycle thermal efficiency.

Approach:

An energy balance is used on the feedwater heater to determine the steam fraction extracted from the turbine. Conservation of mass and energy applied to the two turbine states are used to determine the work.

Assumptions:



Solution: a) To determine the steam fraction, y, extracted from the turbine perform a mass and energy balance around the closed feedwater heater, and assume steady, adiabatic, no work, and negligible potential and kinetic energy effects: 4 5 2 2 7 7 3 30 0

in out

m h m h y m h m h m h m h− = ⇒ + − − =∑ ∑ But and 2 3m m m= = 5 57m y m= so that substituting in these expressions and simplifying:

3 2

5 7

h hyh h−

=−

To evaluate the properties, for subcooled, incompressible liquid ( )2 1 1 2 1s fh h v P P− = − and ( )1 1 1fh h P=

And ( )1 2 11 2 1 2

2 1 11 2

fs sP

P P

v P Ph h h hh h h

h hη

η η−− −

= → = − = −−

Using Appendices A-10—A-12 (by interpolation as needed)

( )

3

2

2

m kN 1kJ0.001010 20000 10kg 1 kN-mmkJ kJ191.83 191.83 23.8 215.6

kg 0.85 kgh

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠= + = + =

For the subcooled liquid at point 3, using the same approach as for point 2: ( ) ( ) ( )( )3 3 3 3 3 762.81 0.001127 20000 1000 784.2kJ kgf f sath h T v P P T= + − = + − =⎡ ⎤⎣ ⎦

( )7 7 762.87 kJ / kgfh h P= =

For state 5, we recognize that 5 4s s= , so ( )4 4 4, 3537.6kJ kgh P T = and 4 6.5048kJ kgKs =

Now for , we recognize that state 5 is in the two phase region (5 5 5,sh P s )5 5

55

6.5048 2.1387 0.9824.4478

fs

fg

s sx

s− −

= = = and ( )( )5 5 5 5 762.81 0.982 2015.3 2741.1kJ kgs f s fgh h x h= + = + =

Using the isentropic efficiency ( ) ( )( )5 4 4 5 3537.6 0.90 3537.6 2741.1 2820.7 kJ kgT sh h h hη= − − = − − =

8- 36

784.2 215.6 0.2762820.7 762.81

y −= =

− Answer

b) The net work is . Applying the mass and energy equations to the turbines and pump, and assuming steady, adiabatic, and negligible potential and kinetic energy effects

, ,net T A T B PW W W W= + −

( ), 4 5 , 4T A T AW m h h W m h h= − → = − 5

( )( ), 5 6 1T BW m h h y= − −

2 1PW m h h= − State 6 is in the two-phase region, and 6 5s s= , so in the superheated vapor region, by interpolation

( )5 5 5, 6.6785kJ / kg Ks P h =

( ) ( )( )6 6 6 66.6785 0.6493 0.804 , 191.83 0.804 2392.8 2115.1kJ kg

7.5009s sx h P s−= = → = + =

Using the isentropic efficiency ( ) ( )( )6 5 5 6 2820.7 0.90 2820.7 2115.1 2185.6kJ kgT sh h h hη= − − = − − = Therefore, , 3537.6 2820.7 716.9kJ / kgT AW m = − =

( )( ), 1 0.276 2820.7 2185.6 459.7 kJ / kgT BW m = − − =

23.8kJ / kgPW m = 716.9 459.7 23.8 1152.8kJ / kgnetW m = + − = Answer c) For the heat added, apply the mass and energy equations to the water flowing through the boiler, and assume steady, no work, and negligible potential and kinetic energy effects 4 3 3537.6 784.2 2753.4kJ / kginQ m h h= − = − = d) The cycle thermal efficiency is:

1152.8kJ / kg 0.4192753.4kJ / kg

netcycle

in

W m=

Q mη = = Answer

8- 37

8-30 For Problem P 8-29 the closed feedwater heater is changed to an open feedwater heater, from which saturated liquid at 1 MPa exits. With this change, a second pump must be added to the system. Determine:

a. the fraction of steam entering the turbine that must be extracted b. the net work per unit mass of steam entering the high-pressure turbine (in kJ/kg) c. the heat added per unit mass of steam entering the high-pressure turbine (in kJ/kg) d. the cycle thermal efficiency.

Approach:

The same approach as used in P 8-29 is used, except an open feedwater heater is substituted for the closed feedwater heater, and a second pump is now used.

Assumptions:



Solution: a) To determine y, perform a mass and energy balance on the open feed water heater, and assume steady, adiabatic, no work and negligible potential and kinetic energy effects

→ ( )5 6 5 2 5 31 0y m h y m h m h+ − − = 3 2

6 2

h hyh h−

=−

Evaluating the enthalpies, for subcooled, incompressible liquid ( )2 1 1 2 1s fh h v P P− = − and ( )1 1fh h P=

( )1 2 11 2 1 2

, 2 1 11 2 , ,

fs sP A

P A P A

v P Ph h h hh h h

h hη

η η−− −

= → = + = +−

From Appendices A-10—A-12 as needed: 1 191.83kJ / kgh =

( )[ ]3

2

2

kN 1kJ0.001010m kg 1000 101 KN-mm191.83 191.83 1.17 193.00kJ / kg

0.85h

⎛ ⎞− ⎜ ⎟⎝ ⎠= + = + =

( )3 3 762.81kJ / kgfh h P= =

For the isentropic turbine 6 5s s= , so first evaluating ( )5 5 5, 3537.6kJ / kgh P T = and 5 6.5048kJ kg Ks =

To find , we recognize that it is in the two-phase region, so (6 6 6,sh P s )

6 ,66

,6

6.5048 2.1387 0.9824.4478

fs

fg

s sx

s− −

= = =

( )( )6 ,6 6 ,6 762.81 0.982 2015.3 2741.1kJ / kgs f s fgh h x h= + = + = Using isentropic efficiency ( ) ( )( )6 5 5 6 3537.6 0.90 3537.6 2741.1 2820.7 kJ kgT sh h h hη= − − = − − =

762.81 193.0 0.2172820.7 193.0

y −= =

− Answer

b) The net work is , , , ,net T A T B P A P BW W W W W= + − − Apply the mass and energy equations to the turbines and pumps, and assume steady, adiabatic, and negligible potential and kinetic energy effects: ( ), 5 6 , 5T A T AW m h h W m h= − → = − 6h

( )(, 6 7 1T BW m h h y= − − )

, 2P AW m h h= − 1

8- 38

, 4P BW m h h= − 3 Similar to the analysis of the first pump

( ) ( )[ ],3 4 3

4 3,

0.001127 20000 1000 kJ762.83 762.83 25.2 788.00.85 kg

f

P B

v P Ph h

η− −

= + = + = + =

For state 7, 7 6s s= , so first evaluating state 6, we recognize it is superheated vapor, so by interpolation . State 7 is in the two-phase region, so: ( )6 6 6, 6.6785kJ / kg Ks P h =

( )( )7 77 7

,7

6.6785 0.6453 0.804 191.83 0.804 2352.8 2115.1kJ / kg7.5009

fs s

fg

s sx h

s− −

= = = → = + =

using isentropic efficiency ( ) ( )( )7 6 6 7 2820.7 0.90 2820.7 2115.1 2185kJ / kgT sh h h hη= − − = − − =

, 3537.6 2820.7 716.9kJ / kgT AW m = − =

( )( ), 1 0.217 2820.7 2185.6 497.3kJ / kgT BW m = − − =

, 1.17 kJ / kgP AW m =

, 25.2 kJ / kgP BW m =

716 497.3 1.17 25.2 1187 kJ / kgnetW m = + − − = Answer

c) For the heat added, apply the mass and energy equations to the water flowing through the boiler, and assume steady, no work, and negligible potential and kinetic energy effects: 5 4 3537.6 788.0 2749.6kJ / kginQ m h h= − = − = Answer d) The cycle thermal efficiency is:

1187 kJ / kg 0.4322749.6kJ / kg

netcycle

in

W mQ m

η = = = Answer

8- 39

8-31 A Rankine cycle power plant uses one open feedwater heater. Steam leaves the boiler at 1000 psia, 800 ºF and enters the turbine. At 100 psia, steam is extracted and routed to the open feedwater heater; the feedwater exits the feedwater heater as a saturated liquid. The condenser pressure is 2 psia. The turbine and pump are isentropic. Determine:

a. the fraction of steam entering the turbine that is extracted b. the cycle thermal efficiency c. the cycle thermal efficiency if there were no feedwater heater.

Approach:

Conservation of mass and energy are applied to the feedwater heater to determine the steam fraction extracted from the turbine. To determine cycle thermal efficiency, basic definitions are used in conjunction with the conservation of mass and energy applied to other devices.

Assumptions:



Solution: a) To determine the fraction, y, of steam extracted from the turbine, apply the mass and energy equations to the open feedwater heater, and assume steady, adiabatic, no work, and negligible potential and kinetic energy effects

→ ( )5 6 5 2 5 31y m h y m h m h+ − − = 0 3 2

6 2

h hyh h−

=−

Evaluating the enthalpies using Appendices B-10—B-12: ( )1 1 94.02Btu / lbmfh h P= =

For state 2, assuming incompressible liquid and reversible process 2 1 Ph h h= + ∆

( ) ( )3 2

,1 2 1 2 2

ft lbf 144in. 1Btu Btu0.01623 100 1 0.3lbm 778.2ft-lbf lbmin. ftP fh v P P

⎛ ⎞ ⎛ ⎞⎛ ⎞∆ = − = − =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

2 94.02 0.3 94.3Btu / lbmh = + =

( )3 3 298.61Btu / lbmfh h P= =

For state 6, we note that 6 5s s= , so ( )5 5 5, 1.5664Btu lbm s P T = R and 5 1388.5Btu lbmh =

State 6 is in the two-phase region, so 6 66

6

1.5664 0.47439 0.9671.1290

f

fg

s sx

s− −

= = =

( ) ( )( )6 6 6 6 6 6, 298.61 0.967 889.2 1158.7 Btu / lbmf fgh P s h s h= + = + =

298.61 94.3 0.1921158.7 94.3

y −= =

− Answer

b) The cycle thermal efficiency is 1Q

net outcycle

in in

W QQ

η = = −

Apply the mass and energy equations to the water flowing through the boiler and condenser, and assume steady, no work, and negligible potential and kinetic energy effects ( )5 4inQ m h h= − ( ) ( )7 11outQ y m h= − − h Similar to pump 1

( ) ( ) ( )( )4 3 3 3 4 3144 Btu298.61 0.017736 10000 100 301.6

778.2 lbmP f fh h h h P v P P ⎛ ⎞= + ∆ = + − = + − =⎜ ⎟⎝ ⎠

8- 40

For state 7, we note that 7 6s s= and state 7 is in the two phase region, so

7 77

7

1.5664 0.17499 0.7971.7448

f

fg

s sx

s− −

= = =

( )( )7 7 7 7 94.02 0.797 1022.1 909.1Btu / lbmf fgh h x h= + = + =

1388.5 301.6 1086.9Btu / lbminQ m = − =

( )( )1 0.192 909.1 94.02 658.6Btu / lbmoutQ m = − − =

658.6Btu / lbm1 0.3941086.9Btu / lbmcycleη = − = Answer

c) If there is no feedwater heater, then both and change. Ignoring the reheat portion of the cycle inQ outQ

5 4 1388.5 97.0 1291.5Btu / lbminQh h

m= − = − =

4h changes, so that

( ) ( ) ( )( )4 1 1 4 1144 Btu94.02 0.01623 1000 1 97.0

778.2 lbmf fh h P v P P ⎛ ⎞= + − = + − =⎜ ⎟⎝ ⎠

7 1 909.1 94.02 815.1Btu / lbmoutQh h

m= − = − =

815.1Btu / lbm1 0.3691291.5Btu / lbmcycleη = − = Answer

Comments:

With no feedwater heater, additional heat must be added to the system from an external source, and this heat is used to raise the liquid feedwater temperature. This results in a lower cycle thermal efficiency.

8- 41

8-32 A Rankine cycle power plant uses one closed feedwater heater. Steam leaves the boiler at 6 MPa, 400 ºC. At 400 kPa, 15 % of the steam entering the turbine is extracted and routed to the closed feedwater heater; the condensate exits the feedwater heater as a saturated liquid and is routed back to the condenser. The condenser pressure is 7.5 kPa. The turbine and pump are isentropic. Determine:

a. the cycle thermal efficiency b. the cycle thermal efficiency if there were no feedwater heater.

Approach:

Begin with the definition of cycle thermal efficiency. Each term is evaluated with conservation of energy and mass. Information is given about all fluid states upstream and downstream of each device.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. No work occurs in the heat exchangers. 4. The subcooled liquid approximation is valid. 5. The pump and turbines are isentropic.

Solution:

a) Cycle thermal efficiency is defined as: ( ) ( ), ,cycle net in in out in T A T B P inW Q Q Q Q W W W Qη = = − = + −

4

5

We need to evaluate each term on the right hand side of the equation. Apply the mass and energy equations to the turbines, and assume steady, isentropic, and negligible potential and kinetic energy effects: and 5m m= ( )5 61m y m− =

( ), 4 4T AW m h h= − Evaluating the properties using Appendices A-10—A-12: ( )4 4 4, 3177.2kJ kgh P T = 4 6.5408kJ kg Ks =

With 5 4s s= , we recognize that ( )5 5 5,h P s is in the two-phase region, so

5 ,5 55

,5

604.24kJ kg6.5408 1.7766 0.9315.1193 2133.8kJ kg

f

fg

s s hx

s− −−

= = = = 5kJ2590.0kg

h =

Likewise for turbine 2, ( )(,2 5 5 61TW m y h h= − − )

With 6 5s s= , is in the two-phase region, so (6 6 6,h P s ) 6 ,66

,6

6.5488 0.5764 0.7787.6750

f

fg

s sx

s− −

= = =

( )6 ,6 6 168.79 0.778 (2406.0) 2040.7 kJ / kgf fgh h x h= + = + = To calculate the heat input rate, apply the mass and energy equations to the boiler, and assuming steady, no work, and negligible potential and kinetic energy effects: and 4m m= 3

= 4

( )4 4 3inQ m h h= − To obtain h3, apply the mass and energy equations to the closed feedwater heater, and assuming steady, no work, negligible potential and kinetic energy effects, and adiabatic: but ( ) ( )2 2 4 5 4 7 3 3 0m h y m h y m h m h+ − − 2 3m m m= = so Evaluating the properties:

3 2 5h h y h y h= + − 7

( ) ( )7 7 7 7, 604.74 kJ / kgfh P T h P= = Because 2 is a subcooled liquid

( ) ( ) ( )3

2 ,1 1 ,1 2 sat, 1 2

kJ m kN kJ168.79 0.001008 6000 7.5 168.79 6.0 174.8kg kg kgmf fh h P v P P

⎛ ⎞= + − = + − = + =⎜ ⎟

⎝ ⎠

( ) ( )3 174.8 0.15 2590.0 0.15 604.74 472.7 kJ / kgh = + − =

8- 42

( ) ( ) ( ) ( )

( ), , 4 4 5 4 5 6 4 2 1

4 4 3

1T A T B Pcycle

in

W W W m h h y m h h m h hh

m h hQ+ − − + − − − −

= =−

( ) ( )( ) ( )3177.2 2590.0 1 0.15 2590.0 2040.7 6.0

0.3883177.2 472.7

− + − − −= =

− Answer

b) Without the feedwater heater, the cycle thermal efficiency is defined as:

( ) ( )

( )4 4 6 4 2 1

4 4 2

T Pcycle

in

m h h m h hW -W= =

m h hQη

− − −−

( ) ( )3177.2 2040.7 6

0.3773177.2 174.8

− −= =

− Answer

8- 43

8-33 An ideal Rankine cycle uses two open feedwater heaters and three pumps. Steam at a flow rate of 12 lbm/s leaves the boiler at 1500 psia, 1600 ºF, enters the high-pressure turbine and expands to 250 psia, where steam is extracted for the first open feedwater heater. The steam expands in the intermediate-pressure turbine to a pressure of 100 psia, where steam is extracted for the second open feedwater heater. The steam expands in the low-pressure turbine to the condenser pressure of 4 psia. Water leaves the condenser and the feedwater heaters as saturated liquid. All pumps and turbines are isentropic. Determine:

a. the fraction of steam entering the high-pressure turbine that is extracted for each of the feedwater heaters

b. the power required to operate each pump (in kW) c. the power produced by each turbine stage (in kW) d. the heat input in the boiler (in kW) e. the cycle thermal efficiency.

Approach:

Apply conservation of energy to each of the feedwater heaters to determine the steam fractions extracted from the turbines. Apply mass and energy equations to all other devices to determine power and heat input.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy

effects. 3. No work occurs in the condenser, boiler, or

reheater.

4. The subcooled liquid approximation is valid. 5. Pump and turbine are isentropic. 6. Feedwater heaters are adiabatic.

Solution:

a) To determine the fraction of flow, y, extracted from the turbine, apply the mass and energy equations to the feed water heater A, and assume steady, adiabatic, no work and negligible potential and kinetic energy effects: 4 4 7 8 5 5 0m h y m h m h+ − =

but and ( )4 1m y= − 7m 5 45 7

8 4

h hm m y

h h−

= → =−

A similar analysis is used on feed water heater B to find y′ '

2 2 9 9 3 3 0m h y m h m h+ − =

but and 3m m= 4 ( ) ( )( )3 22 7

9 2

11

y h hm y y m y

h h− −

′ ′= − − → =−

Now, evaluating properties using Appendices B-10—B-12 ( )1 1 120.89 Btu / lbmfh h P= = 2 1 ,P Ah h h= + ∆

For the pump, assume incompressible and reversible, so that ( ), ,1 2P A fh v P P∆ = − 1

( )3 2

, 2 2

ft lbf 144in. 1Btu0.016358 100 4 0.3Btu / lbmlbm 778.2ft-lbfin. ftP Ah

⎛ ⎞ ⎛ ⎞⎛ ⎞∆ = − =⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

2 120.89 0.3 121.2Btu / lbmh = + =

( )3 3 298.61Btu / lbmfh h P= =

Similar to pump A 4 3 ,P Bh h h= + ∆ ( )( ),1440.017736 250 100 0.5Btu / lbm

788.2P Bh ⎛ ⎞∆ = − =⎜ ⎟⎝ ⎠

4 298.61 0.5 299.1Btu / lbmh = + =

8- 44

( )5 5 376.2Btu / lbmfh h P= =

Similar to pump A 6 5 ,P Ch h h= + ∆ ( )( ),144 Btu0.01865 1500 250 4.3

778.2 lbmP Ch ⎛ ⎞∆ = − =⎜ ⎟⎝ ⎠

6 376.2 4.3 380.5Btu lbmh = + =

( )7 7 7, 1.8031Btu lbm Rs P T = 7 1843.7 Btu lbmh =

Point 8 is superheated vapor, and noting that 8 7s s= , so by interpolation ( )8 8 8, 1508.0 Btu / lbmh P s =

Point 9 is superheated vapor, and noting that 9 8s s= , so by interpolation ( )9 9 9, 1379.2Btu lbmh P s = Point 10 is in the two-phase region and 10 9s s= , so

10 ,1010

,10

1.8031 0.21983 0.9641.6426

f

fg

s sx

s− −

= = =

( ) ( )( )910 10 10 ,10 10, 120.89 0.964 1006.4 1090.9Btu / lbmf fh P s h x h= + = + =

376.2 299.1 0.06381508.0 299.1

y −= =

− Answer

( )( )1 0.0638 298.61 121.2

0.1321379.2 121.2

y− −

′ = =−

Answer

b) Power for each pump is P PW m h= ∆

( ), 1 , 71 ,P A P AW m h y y m h′= ∆ = − − ∆ P A

( ) lbm Btu 1.055kW1 0.0638 0.132 12 0.3 3.05kW 4.1hps lbm 1Btu/s

⎛ ⎞⎛ ⎞⎛ ⎞= − − = =⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

Answer

( ) ( )( )( )( ), 3 , 7 ,1 1 0.0638 12 0.5 1.055 5.9kW 7.9hpP B P B P BW m h y m h= ∆ = − ∆ = − = = Answer

Answer ( )( )( ), 5 , 12 4.3 1.0551 54.4kW=73.0hpP C P CW m h= ∆ = = c) For each turbine, assuming steady, adiabatic, and negligible potential and kinetic energy effects so TW m h= ∆

( ) ( ), 7 7 8lbm Btu 1.055kW12 1843.7 1508.0 4250kW 5700hp

s lbm 1Btu/sT AW m h h⎛ ⎞⎛ ⎞= − = − = =⎜ ⎟⎜ ⎟


( ) ( ) ( )( )( )( ), 7 8 91 1 0.0638 12 1508.0 1379.2 1.055 1527 kW 2047 hpT BW y m h h= − − = − − = = Answer

( ) ( ) ( )( )( )( ), 7 9 101 1 0.0638 0.132 12 1379.2 1090.9 1.055 2936kW 3937 hpT CW y y m h h′= − − − = − − − = =

Answer d) To obtain the input heat transfer rate, apply the energy and mass equations to the water flowing through the boiler, and assume steady, no work, and negligible potential and kinetic energy effects

( ) ( )6 7 6lbm Btu 1.055kW12 1843.7 380.5 18,520kW 24,840hp

s lbm 1Btu/sinQ m h h⎛ ⎞⎛ ⎞= − = − = =⎜ ⎟⎜ ⎟


e) The cycle thermal efficiency is:

, , , , , ,T A T B T C P A P B P Cnetcycle

in in

W W W W W WWQ Q

η+ + − − −

= =

( )4250 1527 2936 3.05 5.9 54.4 kW

0.46718520kW

+ + − − −= = Answer

8- 45

8-34 A Rankine cycle uses one stage of reheat and one open feedwater heater. Steam leaves the boiler at 5 MPa, 550 ºC and expands in the high-pressure turbine to a pressure of 1 MPa. The steam is extracted, some of which flows through the reheater where the steam temperature is raised to 500 ºC and the remainder flows to the open feedwater heater. The steam from the reheater enters the low-pressure turbine and expands to the condenser pressure of 20 kPa; saturated liquid exits the condenser. Feedwater from the open feedwater heater leaves as saturated liquid at 1 MPa. Each pump has an isentropic efficiency of 88% and each turbine stage has an isentropic efficiency of 92%. Determine:

a. the fraction of the mass flow entering the high-pressure turbine that is extracted to flow to the open feedwater heater

b. the work output of the high- and low-pressure turbines per unit of mass flowing into the high-pressure turbine (in kJ/kg)

c. the work input of the low pressure and high pressure pumps per unit of mass flowing into the high-pressure turbine (in kJ/kg)

d. heat input in the boiler and reheater per unit of mass entering the high-pressure turbine e. the cycle thermal efficiency.

Approach: We are given information about states upstream and downstream of each device. Application of conservation of mass and energy is needed to evaluate the quantities sought.

Assumptions:

1. The system is steady. 2. Potential and kinetic energy effects are

negligible. 3. The pump, turbine, and feedwater heater

are adiabatic. 4. No work is done in the feedwater heater. 5. The liquid water is incompressible.

Solution: a) To determine the fraction of flow, y, extracted from the turbine, perform an energy balance on the open freshwater heater. Assume steady, adiabatic, and negligible potential and kinetic energy effects: 1y m ( )2 11h y m+ − 6 1h m− 7 0h =

7 6

2 6

h hyh h−

=−

To obtain , use the definition of isentropic efficiency: 2h

( )1 22 1 1 2

1 2T T

s

h hh h h h

h hη η

−= → = − −

− s

From Appendices A-10—A-12 at the given conditions (with interpolation as needed): ( ) ( )1 15MPa,550 C 3550 kJ / kg, 5MPa,550 C 7.1174 kJ / kgKh s= = ,

with 2 1s s= , ( )2 21MPa, 3048.2kJ / kgsh s =

( )2 3550 0.92 3550 3048.2 3088.3kJ / kgh = − − =

( ) ( )7 1MPa, 0 1MPa 762.8kJ / kgfh x h= = =

from an energy balance on the pump with an incompressible liquid

, using the definition of pump isentropic efficiency 6h → 6 5dP sw v P v P h= − = ∆ = −∫ h

)(6 5 5 6 5s fh h v P P= + − ( )6 5 6 5s Ph h h h η= + −

( ) ( )5 20kPa, 0 20kPa 251.4kJ / kgfh x h= = =

( ) ( ) 320kPa 20kPa, 0 0.001017 m / kgf fv v x= = =

8- 46

( )( )36 2

kN kJ251.4kJ kg 0.001017 m kg 1000 20 252.4kJ kgkN-mmsh = + − =

( )6 251.4 252.4 251.4 0.88 252.5kJ / kgh = + − =

762.8 252.5 0.1803088.3 252.5

y −= =

− Answer

b) With mass and energy balances on the turbines and pumps, assuming steady, negligible potential and kinetic energy effects, and adiabatic, and dividing the energy equation through by the mass flow rate: ( ),1 1 2 3550 3088.3 kJ kg 461.7 kJ / kgTw h h= − = − = Answer

( ),2 3 4 T 3 4T sw h h h hη′ = − = − This is work per unit mass of flow entering turbine 2. To express in terms of work per unit mass of flow entering turbine 1, multiply by (1-y). Like wise, for ,1Pw the same is true.

( ),2 ,21T Tw y ′= − wEvaluating the properties

( )3 1MPa, 500 C 3478.5kJ / kgh = , → ( )3 1MPa, 500 C 7.7622 kJ / kgKs = 47.7622 0.8320 0.979

7.0766sx −= =

with 4 3s s= → ( ) ( )( )4 420kPa, 251.4 0.979 2358.3 2560.9kJ / kgsh s = + =

( )( ),2 (1 0.180) 0.92 3478.5 2560.9 692.2kJ / kgTw = − − = Answer c) For the pumps, the same approach is used as for the turbines:

1 6 5 252.5 251.4 1.1kJ / kgPw h h= − = − = ( )( ) ( )( ),1 6 51 1 0.180 252.5 251.4 kJ / kg 0.90kJ / kgPw y h h= − − = − − = Answer

where ,2 8 7Pw h h= − ( )8 7 7 8 7s fh h v P P= + − and ( )8 7 8 7s Ph h h h η= + − Evaluating the properties , ( ) 3

7 1000kPa, 0 0.001127 m / kgv x = = ( )( )8 762.8 0.001127 5000 1000 767.3kJ / kgsh = + − =

( )8 762.8 767.3 762.8 0.88 767.9kJ / kgh = + − =

,2 767.9 762.8 5.1kJ / kgPw = − = Answer d) Applying the mass and energy equations to the steam flowing through the boiler and reheater, assuming steady, negligible potential and kinetic energy effects, no work, and dividing the energy equation by the mass flow rate: 1 8in, boilerq h h 3550 767.9 2782kJ / kg= − == − Answer

( )( ) ( )( )3 21 1 0.180 3478.5 3088.3 320.0 kJ / kgin, reheaterq y h h= − − = − − = Answer e) Using the definition of cycle thermal efficiency:

net net

inin

W wqQ

η = =

,1 ,2 ,1 ,2 461.7 692.2 0.90 5.1 0.3702782 320.0

T T P P

in

w w w wq

+ − − + − −= = =

+ Answer

8- 47

8-35 A regenerative Rankine cycle that produces 500 MW uses one closed and one open feedwater heater. Steam exits the boiler at 10 MPa, 550 ºC. Steam at 1 MPa is extracted between the high- and intermediate-pressure turbines and is sent to the closed feedwater heater, from which the feedwater leaves at 10 MPa, 150 ºC; saturated liquid condensate is pumped forward and injected into the boiler feedwater line. Steam at 0.15 MPa is extracted between the intermediate- and low-pressure turbines and is sent to an open feedwater heater, from which saturated liquid water leaves at 0.15 MPa. Steam leaves the low-pressure turbine at 6 kPa. The turbines and pumps are isentropic. Determine:

a. the cycle thermal efficiency b. the required mass flow rate (in kg/s).

Approach:

Begin with the definition of cycle thermal efficiency. Each quantity is evaluated with application of conservation of mass and/or energy, since information on states upstream and downstream of each device is given.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy

effects. 3. No work occurs in the heat exchangers. 4. The subcooled liquid approximation is

valid. 5. The pumps and turbine are isentropic.

Solution:

The cycle thermal efficiency is defined as: ( )cycle net in in out inW Q Q Q Qη = = − (1) The heat input is obtained with an energy balance around the water flowing through the boiler. Assuming steady, negligible potential and kinetic energy effects, and no work: (2) (10 1 10inQ m h h= − )

0

where 10 1 unknownm m= = Sufficient information is given to evaluate h1, but we will wait to evaluate all the properties at once. To find , we perform a mass and energy balance on the location where streams 9 and 12 mix. Assuming steady, negligible potential and kinetic energy effects, adiabatic, and no work:

10h

and 9 9 12 12 10 10 0m h m h m h+ − = 9 12 1m m m+ = Note that 12 10m m y= and 9 10 1m m y= − . Substituting these two expressions into the main energy equation and dividing by : 10m ( ) ( )9 12 10 10 91 0 1 12y h y h h h y h y h− + − = ⇒ = − + We need to determine y by performing a mass and energy energy balance on the closed feedwater heater. With the same assumptions as for the mixer and bringing in the definition of y:

→ ( ) ( )1 2 1 8 1 11 1 91 1 0ym h y m h ym h y m h+ − − − − = 9 8

2 8 11 9

h hy

h h h h−

=− − +

)h

0

To find the output heat transfer rate, apply conservation of mass and energy to the steam flowing through the condenser with the same assumptions as for the boiler. (3) ( ) (1 4 51outQ y y m h′= − − −

We need to find , and to do so we perform a mass and energy balance on the open feedwater heater with the same assumptions as for the mixer:

y′

→ ( ) ( )1 3 1 6 1 71 1ym h y y m h y m h′+ − − − − =( )( )7 6

3 6

1 y h hy

h h− −

′ =−

Now, substituting Eqs. (2) and (3) into Eq. (1) and noting that 10 1m m= :

8- 48

( ) ( ) ( )

( )( ) ( )( )

( )10 1 10 1 4 5 1 10 4 5

10 1 10 1 10

1 1cycle

m h h y y m h h h h y y h hm h h h h

η′ ′− − − − − − − − − −

= =− −

Now evaluating all properties from Appendices A-10—A-12 (with interpolation as required), and then and : y y′

( )1 10 MPa,500 C 3501kJ / kgh = ( )1 10 MPa,500 C 6.75611kJ / kgKs =

with 2 1s s= , ( )2 21MPa, 2859 kJ / kgh s =

with 3 2s s= , state 3 is in the two-phase region, so 36.7561 1.4336 0.919

5.7897x −= =

( ) ( )( )3 30.15MPa, 467.11 0.919 2226.5 2514kJ / kgh s = + =

Likewise, with 4 3s s= , 36.7561 0.5164 0.800

7.8212x −= = ( ) ( )( )4 46kPa, 150.2 0.800 2416.6 2084kJ / kgh s = + =

( ) ( )5 5 ,56kPa, 0 6kPa 150.2kJ / kgfh x h= = =

For apply an energy balance to the liquid pump 6h

( )6 5 6 5 ,5 6 5d fw v P v P h h h h v P P= − = ∆ = − → = + −∫

( )( )3 26 150.2kJ kg 0.001006m kg 150 6 kN m 150.3kJ / kgh = + − =

( ) ( )7 9 ,70.15MPa, 0 0.15MPa 467.1kJ / kgfh x h= = =

is determined similar to 8h ( )6 8 7 ,7 8fh h h v P→ = + − 7P

( )( )3 28 467.1kJ kg 0.001053m kg 10000 150 kN m 477.4kJ / kgh = + − =

( ) ( ) ( ) ( )( )9 9 910 MPa, 150 C 632.2 0.001091 10000 475.8 642.6 kJ / kgfg fg sath h T v P P= + − = + − =

( ) ( )11 11 ,111MPa, 0 1MPa 762.8 kJ / kgfh x h= = =

is determined similar to12h ( )6 12 11 ,11 12 11fh h h v P→ = + − P

( )( )3 212 762.8kJ kg 0.001127 m kg 10000 1000 kN m 772.9 kJ / kgh = + − =

642.6 477.4 0.07312857 477.4 762.8 642.6

y −= =

− − +

( )( ) ( )( )10 1 0.0731 642.6kJ / kg 0.0731 772.9kJ / kg 652.1kJ / kgh = − + =

( )( )1 0.0731 467.1 150.2

0.1242514 150.2

y− −

′ = =−

( ) ( ) ( )3501 652.1 1 0.0731 0.124 2084 150.2

0.4553501 652.1cycleη

− − − − −= =

− Answer

For the mass flow rate, we use the net power output: ( ) ( ) ( )10 1 10 1 4 5500MW 1net in outW Q - Q m h h y y m h′= = = − − − − − h

)

With , and solving for the mass flow rate: 10 1m m=

( ) ( )(1

1 10 4 51netW

mh h y y h h

=′− − − − −

( )

( ) ( )( )

1kJ / sec500,000 kW1kW 385.7 kg / sec

3501-652.1 1 0.0731 0.124 2084 150.2

⎛ ⎞⎜ ⎟⎝ ⎠= =

− − − − Answer

8- 49

8-36 A Rankine cycle power plant uses reheat and one closed feedwater heater. Steam leaves the boiler at 6 MPa, 400 ºC and enters the high-pressure turbine. At 400 kPa all the steam is extracted; 85% of the flow is routed through a reheater where the steam temperature is raised to 400 ºC, and then is returned to the low-pressure turbine. The other 15% of the extracted steam is routed to the closed feedwater heater; the condensate exits the feedwater heater as a saturated liquid and is routed back to the condenser. The condenser pressure is 7.5 kPa. The turbine and pump are isentropic. Determine:

a. the cycle thermal efficiency b. compare this result to that obtained in Problem P8-32.

Approach:

Begin with basic definitions, and use conservation of mass and energy applied to individual devices to evaluate terms in the definitions.

Assumptions:



Solution: a) The cycle thermal efficiency is

1 out netcycle

in in

Q WQ Q

η = − =

net T PW W W= −

)

To determine the turbine and pump work, apply the mass and energy equations to the turbines and pump, and assume steady, isentropic, and negligible potential and kinetic energy effects ( ) ( ) (4 4 5 4 6 71TW m h h y m h h= − + − −

Evaluating the properties using Appendices A-10—A-12: ( )4 4 4, 3177.2kJ kgh P T = 4 6.5408kJ / kg Ks =

State 5 is in the two-phase region, and noting 5 4s s= ,

5 55

5

6.5408 1.7766 0.9315.1193

f

fg

s sx

s− −

= = =

( )( )5 5 5 604.74 0.931 2133.8 2590.5kJ / kgf fh h x h= + = + =

( )6 6 6, 3273.4kJ / kgh P T = 6 7.8985kJ / kg Ks = State 7 is in the two-phase region, and noting 7 6s s=

77.8985 0.5764 0.954

7.6750x −= =

( )7 168.79 0.954 2406.0 2464.2kJ / kgh = + =

( ) ( )( )3177.2 2590.5 1 0.15 3273.4 2464.2 1274.5kJ / kgTW m = − + − − =

For the pump ( )( 2 11PW m y h h= − − ) . Assuming incompressible liquid and reversible,

( )2 1 1 2 1P fh h h v P P∆ = − = −

( )3

2

m kN 1kJ0.001008 6000 7.5 6.0kJ / kgkg 1 kN-mmPh

⎛ ⎞ ⎛ ⎞∆ = − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Therefore ( )( )1 0.15 6.0 kJ/kg 5.1kJ / kgPW m = − = For the heat input, apply the mass and energy equations to the water flowing through the boiler and reheater, and assume steady, no work, and negligible potential and kinetic energy effects:

8- 50

in boiler reheaterQ Q Q= +

and ( 4 3boilerQ m h h= − ) ( ) ( )6 51reheaterQ y m h= − − h

=

To evaluate , apply the mass and energy equations to the closed feedwater heater, and assume steady, adiabatic, no work, and negligible potential and kinetic energy effects:

3h

4 2 4 5 4 8 4 3 0m h y m h y m h m h+ − −

( )3 2 5 8h h y h h= + −Evaluating properties ( )8 8 604.74kJ / kgfh h P= =

Across the pump 2 1 2 1P Ph h h h h h− = ∆ → = + ∆ ( )1 1 2168.79kJ / kg 168.79 5.1 173.9kJ / kgfh h P h= = → = + =

( )( )3 173.9 0.15 2590.5 604.74 471.8kJ / kgh = + − = ( ) ( )(4 3 6 51inQ m h h y h h= − + − − )

( ) ( )( )3177.2 471.8 1 0.15 3273.4 2590.5 3285.9 kJ / kg= − + − − =

( )1274.5 5.1 kJ / kg

0.3863285.9 kJ / kgcycleη

−= = Answer

8- 51

8-37 Air at 1 atm, 40 ºF enters the compressor of an ideal Brayton cycle. The pressure ratio is 10. The maximum temperature in the cycle is 1500 ºF. Using an air-standard analysis, determine:

a. the cycle thermal efficiency b. the back work ratio c. the temperature of the air at the turbine exit (in ºC).

Approach:

Begin with the definition of cycle thermal efficiency, and apply conservation of mass and energy to individual devices to evaluate the terms in the definition. The same terms are used to determine the backwork ratio.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. Compressor and turbine are isentropic. 4. An-air-standard analysis is used.

Solution: a) Cycle thermal efficiency is cycle net inW Qη = where . Apply the energy and mass equations to the turbine and compressor, and assume steady, isentropic, and negligible potential and kinetic energy effects:

net T CW W W= −

and ( 3 4TW m h h= − ) ( )2 1CW m h h= −

From the air table Appendix B-9 ( )1 500R 119.48Btu / lbmh = ,1 1.0590rP =

( )3 1960R 493.6 Btu / lbmh = ,3 160.5rP = Using relative pressures by interpolation ( ) ( ),2 ,1 2 1P / 1.059 10 10.59r rP P P= = = 2 230.9Btu / lbmh =

by interpolation ( ) ( ),4 ,3 4 3/ 160.5 1/10 16.05r rP P P P= = = 4 259.9Btu / lbmh =

( ) ( ) ( ) ( )3 4 2 1 493.6 259.9 230.9 119.48 233.7 111.5 122.2Btu / lbmnetW m h h h h= − − − = − − − = − = Apply the energy and mass equations to the air flowing through the combustor, and assume steady, no work, and negligible potential and kinetic energy effects: ( )3 2 3 2 493.6 230.9 262.7 Btu lbmin inQ m h h Q m h h= − → = − = − =

122.2 Btu / lbm 0.465267.7 Btu / lbmcycleη = = Answer

b) The back work ratio is: ( )( )

111.5Btu / lbm 0.477233.7 Btu / lbm

CC

T T

W mWBWRW W m

= = = = Answer

c) The turbine outlet temperature is: by interpolation ( )4 4T h → 4 1075.7 R 615.7 FT = = Answer

8- 52

8-38 For Problem P 8-37, rework the problem using an air-standard analysis, with a turbine efficiency of 89% and a compressor efficiency of 83%. Determine:

a. the cycle thermal efficiency b. the back work ratio c. the temperature of the air at the turbine exit (in ºC).

Compare with the results from P 8-37.

Approach: Basic definitions, now including isentropic efficiency, are used with conservation of mass and energy applied to individual components to evaluate terms in the definitions.

Assumptions:


Solution: a) The operating conditions are the same except now the isentropic efficiencies ( )0.89, 0.83T Cη η= = must be incorporated into the solution. Following the solution to P 8-39: 3TW m h h= − 4 and 2 1CW m h h= − Again using Appendix B-9 ( )1 500R 199.48Btu / lbmh = ,1 1.0590rP =

( )3 1960 R 493.6 Btu / lbmh = ,3 160.5rP = For an isentropic process (from previous solution) 2 230.9Btu / lbmsh = 4 259.9Btu / lbmsh = Using the definition of isentropic efficiency

( ) ( )( )3 44 3 3 4

3 4

493.6 0.89 493.6 259.9 285.6Btu / lbmT T ss

h hh h h h

h hη η

−= → = − − = − − =

−

2 1 2 12 1

2 1

230.9 119.48119.48 253.7 Btu / lbm0.83

s sC

C

h h h hh h

h hη

η− − −

= → = + = + =−

493.6 285.6 208.0Btu / lbmTW m = − = 253.7 119.48 134.2Btu / lbmCW m = − =

208.0 134.2 73.8Btu / lbmnetW = − =

Likewise, for 3 2 493.6 253.7 239.9Btu / lbminQ m h h= − = − =

73.8Btu / lbm 0.308239.9Btu / lbmcycleη = = Answer

b) The back work ratio is: 134.2Btu / lbm 0.645208.0Btu / lbm

C

T

W mBWRW m

= = = Answer

c) The turbine exit temperature is: by interpolation ( )4 4T h → 4 1177.6R 717.6 FT == Answer

Comments:

The inefficiencies in the turbine and compressor have a dramatic detrimental effect on the cycle thermal efficiency. Note that because less energy is removed from the turbine, its outlet temperature is much higher.

8- 53

8-39 Air at 100 kPa, 27 ºC at a volumetric flowrate of 10 m3/s enters an ideal Brayton cycle and is compressed to 1750 kPa. The air temperature at the entrance to the turbine is 1073 ºC. Using a cold-air-standard analysis, determine:

a. the net power (in kW) b. the heat addition (in kW) c. the cycle thermal efficiency d. back work ratio.

Approach:

Basic definitions are used with conservation of mass and energy applied to individual devices.

Assumptions:

1. The processes are steady. 2. Neglect potential and kinetic energy effects. 3. Compressor and turbine are isentropic. 4. A cold-air-standard analysis is used.

Solution: a) Net power is . Apply the mass and energy equations to the turbine and compressor, and assume steady, isentropic, and negligible potential and kinetic energy effects

net T CW W W= −

and ( 3 4TW m h h= − ) ( )2 1CW m h h= − Using a cold-air-standard analysis where Ph c T∆ = ∆ , and from Appendix A-8 for air ,

: 1.005kJ / kg KPc =

1.4k = and ( 3 4T PW mc T T= − ) ( )2 1C PW mc T T= − For an isentropic process of an ideal gas with constant specific heats

1 1.4 1

1.42

2 11

1750300K 679.6K100

kkP

T TP

− −⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1 1.4 1

1.44

4 33

1001300K 573.8K1750

kkP

T TP

− −⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

1 1 1 1m V V vρ= = ( )( )( )

( )( )3

11 2

1

8.314kJ kmol K 300K kN-m 1kJ m0.861kg28.97 kg kmol 100kN m

RTv

MP= = =

( )( )

3

3

10m /min11.6kg / min 0.194kg / s

0.861m /kgm == =

( )( )( ) ( )0.194kg s 1.005kJ kg K 1300 573.8 K 1kW 1 kJ/s 141.6kWTW = − =

=

( )( )( )0.194 1.005 679.6 300 74.0kWCW = −

141.6 74.0 67.6 kWnetW = − = Answer b) To obtain the input heat transfer rate, apply the energy and mass equations to the combustor, and assume steady, no work and negligible potential and kinetic energy effects: ( ) (3 2 3 2in PQ m h h mc T T= − = − ) ( )( )( )0.194 1.005 1300 679.6 121.0inQ = − = Answer

c) The cycle thermal efficiency is: 67.6kW 0.559121.0kW

netcycle

in

WQ

η = = = Answer

d) The back work ratio is: C 74.0kW 0.523141.6kWT

WBWRW

= = = Answer

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8-40 Redo Problem P 8-39 with an air-standard analysis. Approach:

Basic definitions are used with conservation of mass and energy applied to each device. Variable property effects are taken into account using the air tables.

Assumptions:


Solution: a) All conditions are the same as in P 8-39, so for the net power: and ( 3 4TW m h h= − ) ( )2 1CW m h h= − (remains unchanged) 0.194 kg / sm =From the air tables, Appendix A-9 ( )1 300K 300.19 kJ / kgh = ,1 1.3860rP =

( )3 1300 K 1395.97 kJ / kgh = ,3 330.9rP =

For , use relative pressures 2h ( ),2 ,1 2 1r rP P P P=

( ),2 1.386 1750 100 24.26rP = = so by interpolation 2 679.5kJ kgh = Likewise for state 4: ( ) ( ),4 ,3 4 3 330.9 100 1750 18.91r rP P P P= = = so by interpolation 4 632.0kJ / kgh =

( )( ) kJ 1kW0.194kg s 1395.97 632.0 148.2kWkg 1 kJ/sTW ⎛ ⎞= − =⎜ ⎟

⎝ ⎠

( )( )0.194 679.5 300.19 73.6kWCW = − =

142.2 73.6 74.6 kWnetW = − = Answer b) The heat input rate is: ( ) ( )( )3 2 0.194 1395.97 679.5 139.0kWinQ m h h= − = − = Answer

c) The cycle thermal efficiency is: 74.6kW 0.537139.0kW

netcycle

in

WQ

η = = = Answer

d) The back work ratio is: 73.6kW 0.497148.2kW

C

T

WBWRW

= = = Answer

Comments:

Taking into account the variability of specific heat with temperature resulted in the cycle thermal efficiency and the back work ratio decreasing.

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ch08_1_40

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