A Rankine steam cycle modified for reheat, a closed feedwater heater, and an open feedwater heater is shown below. The high-pressure turbine receives 100 kg/s of steam from the steam boiler. The feedwater heater exit states for the boiler feedwater and the condensed steam are the normally assumed ideal states. The following data tables give the saturation data for the pressures and data for h and s at selected states. ( a ) Sketch the T-s diagram for the ideal cycle. ( b ) Determine the net power output of the cycle, in MW. ( c ) If cooling water is available at 25°C, what is the minimum flow rate of the cooling water required for the ideal cycle, in kg/s? Take c p ,water = 4.18 kJ/kg·K. Process states and selected data State P , kPa T , °C h , kJ/kg s , kJ/kg·K 1 20 2 1400 3 1400 4 1400 5 5000 6 5000 700 3894 7.504 7 1400 3400 7.504 8 1200 3349 7.504 9 1200 600 3692 7.938 10 245 3154 7.938 11 20 2620 7.938

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Answer 1

Textbook Question

Chapter 10.9, Problem 111RP

A Rankine steam cycle modified for reheat, a closed feedwater heater, and an open feedwater heater is shown below. The high-pressure turbine receives 100 kg/s of steam from the steam boiler. The feedwater heater exit states for the boiler feedwater and the condensed steam are the normally assumed ideal states. The following data tables give the saturation data for the pressures and data for h and s at selected states. (a) Sketch the T-s diagram for the ideal cycle. (b) Determine the net power output of the cycle, in MW. (c) If cooling water is available at 25°C, what is the minimum flow rate of the cooling water required for the ideal cycle, in kg/s? Take c_p_,water = 4.18 kJ/kg·K.

Process states and selected data
State	P, kPa	T, °C	h, kJ/kg	s, kJ/kg·K
1	20
2	1400
3	1400
4	1400
5	5000
6	5000	700	3894	7.504
7	1400		3400	7.504
8	1200		3349	7.504
9	1200	600	3692	7.938
10	245		3154	7.938
11	20		2620	7.938

Chapter 10.9, Problem 111RP, A Rankine steam cycle modified for reheat, a closed feedwater heater, and an open feedwater heater

(a)

Expert Solution

To determine

Sketch the T-s diagram for the ideal cycle.

Answer to Problem 111RP

Sketch the T-s diagram for the ideal cycle is shown in Figure 1.

Explanation of Solution

Draw the T−s diagram of the given ideal regenerative Rankine cycle as shown in

Figure 1.

THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<, Chapter 10.9, Problem 111RP

(b)

Expert Solution

To determine

The net power output of the cycle.

Answer to Problem 111RP

The net power output of the cycle is 144.4 MW_.

Explanation of Solution

Write the formula for work done by the pump during process 1-2.

wpI,in=v1(P2−P1) (I)

Here, the specific volume is v, the pressure is P, and the subscripts 1 and 2 indicates the process states.

Write the formula for enthalpy (h) at state 2.

h2=h1+wpI,in (II)

Write the formula for work done by the pump during process 3-4.

wpII,in=v4(P5−P4) (III)

Here, the specific volume is v, the pressure is P, and the subscripts 4 and 5 indicates the process states.

Write the formula for enthalpy (h) at state 5.

h5=h4+wpII,in (IV)

Write the formula for an energy balance on the open feed water heater.

yh7(1−y)h3=1(h4) (V)

Here, the fraction of steam extracted is (y) from the high-pressure turbine.

Rewrite the Equation (V) to find out the value of (y).

y=h4−h3h7−h3 (VI)

Write the formula for an energy balance on the closed feed water heater.

zh10+(1−y)h2=(1−y)h3+zh12 (VII)

Here, the fraction of steam extracted is (z) from the low-pressure turbine.

Rewrite the Equation (V) to find out the value of (z).

z=(1−y)(h3−h2)(h10−h12) (VIII)

Write the formula for heat input in the boiler.

qin=(h6−h5)+(1−y)(h9−h8) (IX)

Write the formula for work output from the turbine.

wT,out=h6−yh7−(1−y)h8+(1−y)h9−zh10−(1−y−z)h11 (X)

Write the formula for net work output from the cycle.

wnet=wT,out−(1−y)wPI,in−wPII,in (XI)

Write the net power output of the cycle.

W˙net=m˙wnet (XII)

Conclusion:

From the Table A-5, “Saturated water-temperature Table” obtains the value of the enthalpy (h1) and specific volume (v1) at state 1 corresponding to the pressure of 20 kPa is 251.42 kJ/kg and 0.001017 m3/kg.

Substitute 0.001017 m3/kg for v1, 20 kPa for P1, and 1400 kPa for P2 in Equation (I).

wpI,in=(0.001017 m3/kg)(1400 kPa−20 kPa)=(0.001017 m3/kg)(1380 kPa)=1.40346 kPa⋅m3/kg×1 kJ1 kPa⋅m3=1.40346 kJ/kg

≅1.4035 kJ/kg

Substitute 251.42 kJ/kg for h1, and 1.4035 kJ/kg for wpI,in in Equation (II).

h2=251.42 kJ/kg+1.4035 kJ/kg=252.8 kJ/kg

From the Table A-5, “Saturated water-temperature Table” obtains the value of the enthalpy (h4) and specific volume (v4) at state 4 corresponding to the pressure of 1400 kPa is 829.96 kJ/kg and 0.001149 m3/kg.

Substitute 0.001149 m3/kg for v4, 1400 kPa for P4, and 5000 kPa for P5 in Equation (III).

wpII,in=(0.001149 m3/kg)(5000 kPa−1400 kPa )=(0.001149 m3/kg)(3600 kPa)=4.1364 kPa⋅m3/kg×1 kJ1 kPa⋅m3=4.1364 kJ/kg

≅4.14 kJ/kg

Substitute 829.96 kJ/kg for h3, and 4.14 kJ/kg for wpII,in in Equation (IV).

h5=829.96 kJ/kg+4.14 kJ/kg=834.1 kJ/kg

Refer Table A-5, “Saturated water-temperature Table”, and write the enthalpy at state 12 at pressure of 245 kPa using an interpolation method.

h3=h12=hf@245 kPa (XIII)

Here, enthalpy of saturation liquid at pressure of 245 kPa is hf@245 kPa.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XIV)

Here, the variables denote by x and y is pressure and specific enthalpy at state 12 respectively.

Show the specific enthalpy at state 12 corresponding to temperature as in Table (1).

Pressure at state 12 kPa	Specific enthalpy at state 12 h12(kJ/kg)
225 (x1)	520.71 (y1)
245 (x2)	(y2=?)
250 (x3)	535.35 (y3)

Substitute 225 kPa, 245 kPa, and 250 kPa for x1, x2 and x3 respectively, 520.71 kJ/kg for y1 and 535.35 kJ/kg for y3 in Equation (XIV).

y2=[(245−225)kPa×(535.35−520.71)kJ/kg(250−225)kPa+520.71 kJ/kg]=532.422 kJ/kg≅532 kJ/kg

Substitute 532 kJ/kg for hf@245 kPa in Equation (XIII).

h12=532 kJ/kg

Here, the throttle valve operation of specific enthalpy at the state 13 is equal to specific enthalpy at the state 12.

Substitute 830 kJ/kg for h4, 532kJ/kg for h3, and 3400 kJ/kg for h7 in Equation (VI).

y=(830 kJ/kg−532 kJ/kg)(3400 kJ/kg−532 kJ/kg)=(298 kJ/kg)(2868 kJ/kg)=0.1039

Substitute 252.8 kJ/kg for h2, 0.1039 for y, 532kJ/kg for h3, and 3154 kJ/kg for h10 in Equation (VIII).

z=(1−0.1039)(532 kJ/kg−252.8 kJ/kg)(3154 kJ/kg−532 kJ/kg)=(250.1911 kJ/kg)(2622 kJ/kg)=0.09542

Substitute 3894 kJ/kg for h6, 834.1 kJ/kg for h5, 0.1039 for y, 3692 kJ/kg for h9, and 3349 kJ/kg for h8 in Equation (IX).

qin=(3894−834.1)kJ/kg+(1−0.1039)(3692−3349)kJ/kg=3059.9 kJ/kg+(0.8961)×(343kJ/kg)=3059.9 kJ/kg+307.36 kJ/kg=3367 kJ/kg

Substitute 3894 kJ/kg for h6, 3400 kJ/kg for h7, 0.09542 for z, 0.1039 for y, 3692 kJ/kg for h9, 3349 kJ/kg for h8, 3154 kJ/kg for h10, and 2620 kJ/kg for h11 in Equation (X).

wT,out=[(3894 kJ/kg)−(0.1039)(3400 kJ/kg)−(1−0.1039)(3349 kJ/kg)+(1−0.1039)(3692 kJ/kg)−(0.09542)(3154 kJ/kg)−(1−0.1039−0.09542)(2620 kJ/kg)]=[(3894 kJ/kg)−(353.26 kJ/kg)−(0.8961)(3349 kJ/kg)+(0.8961)(3692 kJ/kg)−(0.09542)(3154 kJ/kg)−(0.80068)(2620 kJ/kg)]=[(3894 kJ/kg)−(353.26 kJ/kg)−(3001.039 kJ/kg)+(3308.401 kJ/kg)−(300.9547 kJ/kg)−(2097.782 kJ/kg)]=1449 kJ/kg

Substitute 1449 kJ/kg for wT,out, 4.14 kJ/kg for wpII,in, 1.41 kJ/kg for wpI,in, and 0.1039 for y in Equation (XI).

wnet=1449 kJ/kg−(1−0.1039)(1.41 kJ/kg)−(4.14 kJ/kg)=(1449 kJ/kg)−(1.2635 kJ/kg)−(4.14 kJ/kg)=1443.596 kJ/kg≅1444 kJ/kg

Substitute 100 kg/s for m˙ and 1444 kJ/kg for wnet in Equation (XII).

W˙net=(100 kg/s)×(1444 kJ/kg)=144400 kW=144400 kW×(1 MW1000 kW)=144.4 MW

Thus, the net power output of the cycle is 144.4 MW_.

(c)

Expert Solution

To determine

The minimum flow rate of the cooling water.

Answer to Problem 111RP

The minimum flow rate of the cooling water is 1311 kg/s_.

Explanation of Solution

Write the formula for heat rejected from the condenser.

qout=(1−y−z)h11+zh13−(1−y)h1 (XV).

The mass flow rate cooling water will be minimum when the cooling water exit temperature is a maximum as

Tw,2=T1=Tsat@20 kPa=60.1°C.

Write the formula for an energy balance on the condenser.

m˙qout=m˙wcp,w(Tw,2−Tw,1)m˙w=m˙qoutcp,w(Tw,2−Tw,1) (XVI)

Conclusion:

Substitute 0.1039 for y, 0.09542 for z, 2620 kJ/kg for h11, 532 kJ/kg for h13, and 251.4 kJ/kg for h1 in Equation (XV).

qout=[(1−0.1039−0.09542)(2620 kJ/kg)+(0.09542)(532 kJ/kg)−(1−0.1039)(251.4 kJ/kg)]=[(2097.782 kJ/kg)+(50.76344 kJ/kg)−(225.2795 kJ/kg)]=1923 kJ/kg

Substitute 100 kg/s for m˙, 1923 kJ/kg for qout, 4.18 kJ/kg⋅K for cp,w, 60.1 K for Tw,2, and 25 K for Tw,1 in Equation (XVI).

m˙w=(100 kg/s)(1923 kJ/kg)(4.18 kJ/kg⋅K)(60.1−25)K=(192300 kJ/s)(4.18 kJ/kg⋅K)(35.1)K=1310.67 kg/s≅1311 kg/s

Thus, the minimum flow rate of the cooling water is 1311 kg/s_.

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Answer 2

Textbook Question

Chapter 10.9, Problem 111RP

A Rankine steam cycle modified for reheat, a closed feedwater heater, and an open feedwater heater is shown below. The high-pressure turbine receives 100 kg/s of steam from the steam boiler. The feedwater heater exit states for the boiler feedwater and the condensed steam are the normally assumed ideal states. The following data tables give the saturation data for the pressures and data for h and s at selected states. (a) Sketch the T-s diagram for the ideal cycle. (b) Determine the net power output of the cycle, in MW. (c) If cooling water is available at 25°C, what is the minimum flow rate of the cooling water required for the ideal cycle, in kg/s? Take c_p_,water = 4.18 kJ/kg·K.

Process states and selected data
State	P, kPa	T, °C	h, kJ/kg	s, kJ/kg·K
1	20
2	1400
3	1400
4	1400
5	5000
6	5000	700	3894	7.504
7	1400		3400	7.504
8	1200		3349	7.504
9	1200	600	3692	7.938
10	245		3154	7.938
11	20		2620	7.938

Chapter 10.9, Problem 111RP, A Rankine steam cycle modified for reheat, a closed feedwater heater, and an open feedwater heater

(a)

Expert Solution

To determine

Sketch the T-s diagram for the ideal cycle.

Answer to Problem 111RP

Sketch the T-s diagram for the ideal cycle is shown in Figure 1.

Explanation of Solution

Draw the T−s diagram of the given ideal regenerative Rankine cycle as shown in

Figure 1.

THERMODYNAMICS (LL)-W/ACCESS >CUSTOM<, Chapter 10.9, Problem 111RP

(b)

Expert Solution

To determine

The net power output of the cycle.

Answer to Problem 111RP

The net power output of the cycle is 144.4 MW_.

Explanation of Solution

Write the formula for work done by the pump during process 1-2.

wpI,in=v1(P2−P1) (I)

Here, the specific volume is v, the pressure is P, and the subscripts 1 and 2 indicates the process states.

Write the formula for enthalpy (h) at state 2.

h2=h1+wpI,in (II)

Write the formula for work done by the pump during process 3-4.

wpII,in=v4(P5−P4) (III)

Here, the specific volume is v, the pressure is P, and the subscripts 4 and 5 indicates the process states.

Write the formula for enthalpy (h) at state 5.

h5=h4+wpII,in (IV)

Write the formula for an energy balance on the open feed water heater.

yh7(1−y)h3=1(h4) (V)

Here, the fraction of steam extracted is (y) from the high-pressure turbine.

Rewrite the Equation (V) to find out the value of (y).

y=h4−h3h7−h3 (VI)

Write the formula for an energy balance on the closed feed water heater.

zh10+(1−y)h2=(1−y)h3+zh12 (VII)

Here, the fraction of steam extracted is (z) from the low-pressure turbine.

Rewrite the Equation (V) to find out the value of (z).

z=(1−y)(h3−h2)(h10−h12) (VIII)

Write the formula for heat input in the boiler.

qin=(h6−h5)+(1−y)(h9−h8) (IX)

Write the formula for work output from the turbine.

wT,out=h6−yh7−(1−y)h8+(1−y)h9−zh10−(1−y−z)h11 (X)

Write the formula for net work output from the cycle.

wnet=wT,out−(1−y)wPI,in−wPII,in (XI)

Write the net power output of the cycle.

W˙net=m˙wnet (XII)

Conclusion:

From the Table A-5, “Saturated water-temperature Table” obtains the value of the enthalpy (h1) and specific volume (v1) at state 1 corresponding to the pressure of 20 kPa is 251.42 kJ/kg and 0.001017 m3/kg.

Substitute 0.001017 m3/kg for v1, 20 kPa for P1, and 1400 kPa for P2 in Equation (I).

wpI,in=(0.001017 m3/kg)(1400 kPa−20 kPa)=(0.001017 m3/kg)(1380 kPa)=1.40346 kPa⋅m3/kg×1 kJ1 kPa⋅m3=1.40346 kJ/kg

≅1.4035 kJ/kg

Substitute 251.42 kJ/kg for h1, and 1.4035 kJ/kg for wpI,in in Equation (II).

h2=251.42 kJ/kg+1.4035 kJ/kg=252.8 kJ/kg

From the Table A-5, “Saturated water-temperature Table” obtains the value of the enthalpy (h4) and specific volume (v4) at state 4 corresponding to the pressure of 1400 kPa is 829.96 kJ/kg and 0.001149 m3/kg.

Substitute 0.001149 m3/kg for v4, 1400 kPa for P4, and 5000 kPa for P5 in Equation (III).

wpII,in=(0.001149 m3/kg)(5000 kPa−1400 kPa )=(0.001149 m3/kg)(3600 kPa)=4.1364 kPa⋅m3/kg×1 kJ1 kPa⋅m3=4.1364 kJ/kg

≅4.14 kJ/kg

Substitute 829.96 kJ/kg for h3, and 4.14 kJ/kg for wpII,in in Equation (IV).

h5=829.96 kJ/kg+4.14 kJ/kg=834.1 kJ/kg

Refer Table A-5, “Saturated water-temperature Table”, and write the enthalpy at state 12 at pressure of 245 kPa using an interpolation method.

h3=h12=hf@245 kPa (XIII)

Here, enthalpy of saturation liquid at pressure of 245 kPa is hf@245 kPa.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XIV)

Here, the variables denote by x and y is pressure and specific enthalpy at state 12 respectively.

Show the specific enthalpy at state 12 corresponding to temperature as in Table (1).

Pressure at state 12 kPa	Specific enthalpy at state 12 h12(kJ/kg)
225 (x1)	520.71 (y1)
245 (x2)	(y2=?)
250 (x3)	535.35 (y3)