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Water is the working fluid in an ideal regenerative Rankine cycle with one closed feedwater

heater. Superheated vapor enters the turbine at 10 MPa, 500

oC, and the condenser pressure is

10 kPa. Steam expands through the first-stage turbine where some is extracted and diverted to

a closed feedwater heater at 1.0 MPa. Condensate drains from the feedwater heater as

saturated liquid at 1.0 MPa and is trapped into the condenser. The feedwater leaves the heater

at 10 MPa and a temperature equal to the saturation temperature at 1.0 MPa. Determine for

the cycle

(a) the heat transfer to the working fluid passing through the steam generator, in kJ per kg of

steam entering the first stage turbine.

(b) the thermal efficiency.

(c) the heat transfer from the working fluid passing through the condenser to the cooling

water, in kJ per kg of steam entering the first-stage turbine

An ideal regenerative Rankine cycle with a closed feedwater heater is considered. The work

produced by the turbine, the work consumed by the pumps, and the heat added in the boiler

are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are

negligible.

Analysis From the steam tables,

Stage 1

𝑃1 = 10 𝑘𝑃𝑎

𝑆𝑎𝑡. 𝐿𝑖𝑞𝑢𝑖𝑑 }

ℎ1 = ℎ𝑓@10𝑘𝑃𝑎 = 191.81 𝑘𝐽⁄𝑘𝑔

𝑣1 = 𝑣𝑓@10𝑘𝑃𝑎 = 0.001010 𝑚3⁄𝑘𝑔

Stage 2

𝑃2 = 10 𝑀𝑃𝑎

𝑠2 = 𝑠1

}

𝑤𝑃𝐼,𝑖𝑛 = 𝑣1

(𝑃2 − 𝑃1

)

=(0.001010𝑚3⁄𝑘𝑔)(10000−10 𝑘𝑃𝑎)(1𝑘𝐽 1𝑘𝑃𝑎.𝑚3 ⁄ )

=10.1𝑘𝐽⁄𝑘𝑔

ℎ2 = ℎ1 + 𝑤𝑃𝐼,𝑖𝑛 = (191.81 + 10.1) 𝑘𝐽⁄𝑘𝑔 = 201.91 𝑘𝐽⁄𝑘𝑔

Stage 4

𝑃4 = 10 𝑀𝑃𝑎

𝑇4 = 500𝑜𝐶

}

ℎ5 = 3375.1 𝑘𝐽⁄𝑘𝑔

𝑠5 = 6.5995 𝑘𝐽⁄𝑘𝑔 − 𝐾

Stage 5

𝑃5 = 1 𝑀𝑃𝑎

𝑠5 = 𝑠4

} ℎ5 = 2784 𝑘𝐽⁄𝑘𝑔 𝑏𝑦 𝑖𝑛𝑡𝑒𝑟𝑝𝑜𝑙𝑎𝑡𝑖𝑜𝑛

𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑒𝑑 𝑠𝑡𝑒𝑎𝑚 𝑡𝑎𝑏𝑙𝑒 𝑡ℎ𝑒 𝑠𝑔@1.0 𝑀𝑃𝑎 = 6.5850 𝑘𝐽⁄𝑘𝑔 − 𝐾

𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 𝑠5,𝑡ℎ𝑒𝑟𝑓𝑜𝑟𝑒 𝑖𝑡 𝑖𝑠 𝑠𝑢𝑝𝑒𝑟ℎ𝑒𝑎𝑡𝑒𝑑

𝑏y interpolation at 1 MPa we need to determine the enthalpy.

T (C) s (kJ/kg.K) h (kJ/kg)

179.88 6.5850 2777.1

182.5 6.5995 2784

200 6.6956 2828.3

6.6956 − 6.5995

6.6956 − 6.580 =

200 − 𝑇

200 − 179.88

0.8689 ∗ (200 − 179.88) = 200 − 𝑇

𝑇 = 200 − (0.8689 ∗ (200 − 179.88)) = 182.5 𝐶

Similarly

6.6956 − 6.5995

6.6956 − 6.580 =

2828.3 − ℎ

2828.3 − 2777.1

0.8689 ∗ (2828.3 − 2777.1) = 2828.3 − ℎ

ℎ = 2828.3 − (0.8689 ∗ (2828.3 − 2777.1)) = 2784 𝑘𝐽⁄𝑘𝑔

Stage 6

𝑃6 = 10 𝑘𝑃𝑎

𝑠6 = 𝑠4

} 𝑥6 =

𝑠6 − 𝑠𝑓

6.5995 − 0.6492

7.4996 = 0.793

ℎ6 = ℎ𝑓 + ℎ𝑓𝑔 = 191.81 + (0.793)(2392.1) = 2088.75 𝑘𝐽⁄𝑘𝑔

For an ideal closed feedwater heater, the feedwater is heated to the exit temperature of the

extracted steam, which ideally leaves the heater as a saturated liquid at the extraction pressure.

Stage 7

𝑃7 = 1.0 𝑀𝑃𝑎

𝑥7 = 0

}

ℎ7 = 762.51 𝑘𝐽⁄𝑘𝑔

𝑇7 = 179.88𝑜𝐶

ℎ8 = ℎ7 = 762.51 𝑘𝐽⁄𝑘𝑔

𝑃3 = 10 𝑀𝑃𝑎

𝑇3 = 𝑇7 = 179.88𝑜𝐶

} ℎ𝑓 = 771.51 𝑘𝐽⁄𝑘𝑔

At T7 the saturation pressure is 1.0 MPa, therefore the h7 can be worked from the following formula

ℎ3 = ℎ7 + 𝑣7

(𝑃7 − 𝑃𝑠𝑎𝑡@𝑇7

)

= 762.51 𝑘𝐽⁄𝑘𝑔 + ((0.001 𝑚3⁄𝑘𝑔)(10000 − 1000 𝑘𝑃𝑎) (1𝑘𝐽 1𝑘𝑃𝑎. 𝑚3 ⁄ )) = 771.51 𝑘𝐽⁄𝑘𝑔

An energy balance on the heat exchanger gives the fraction of steam extracted from the

turbine (

5 4 = m

/ m

) for closed feedwater heater:

5 2 3 7

5 5 2 2 3 3 7 7

yh 1h 1h yh

m h m h m h m h

m h m h i i e e

- = +
- = +

=

Rearranging,

𝑦 =

ℎ3 − ℎ2

771.51 − 201.91

2784 − 762.51 = 0.282

Then,

𝑞𝑖𝑛 = ℎ4 − ℎ3 = (3375.1 − 763.05) 𝑘𝐽⁄𝑘𝑔 = 2612.05 𝑘𝐽⁄𝑘𝑔

𝑤𝑡,𝑜𝑢𝑡 = (ℎ4 − ℎ5

) + (1 − 𝑦)(ℎ5 − ℎ6)

= (3375.1 − 2784) + (1 − 0.282)(2784 − 2088.75) 𝑘𝐽⁄𝑘𝑔 = 1090.3 𝑘𝐽⁄𝑘𝑔

𝑤𝑛𝑒𝑡 = 𝑤𝑡,𝑜𝑢𝑡 − 𝑤𝑝,𝑖𝑛 = (1090.3 − 10.1) 𝑘𝐽⁄𝑘𝑔 = 1070.2 𝑘𝐽⁄𝑘𝑔

𝜂𝑡ℎ =

𝑤𝑛𝑒𝑡

1070.2

2612.05 = 0.4097 = 41.0%

The EES code for the above problem is presented here

P[4] = 10*convert(MPa,kPa) T[4] = 500 [C] P[5] =1*convert(MPa,kPa)

P[1] = 10 [kPa]

P[6]=P[1];P[2]=P[4];P[3]=P[4];P[7]=P[5];P[8]=P[6]

“Analysis”

“pump I”

x[1]=0

h[1]=enthalpy(Water, P=P[1], x=x[1])

v[1]=volume(Water, P=P[1], x=x[1])

w_pumpI_in=v[1]*(P[2]-P[1]) h[2]=h[1]+w_pumpI_in{*(1-y_1)}

“turbine High Pressure”

h[4]=enthalpy(Steam, P=P[4], T=T[4])

s[4]=entropy(Steam, P=P[4], T=T[4])

s[5]=s[4]

“turbine Low Pressure”

h[5]=enthalpy(Steam, P=P[5], s=s[5])

x[5]=quality(Steam, P=P[5], s=s[5])

s[6]=s[4]

h[6]=enthalpy(Steam, P=P[6], s=s[6])

x[6]=quality(Steam, P=P[6], s=s[6])

h[7] = enthalpy(Water, P=P[7], x=x[1])

T[7] = temperature(Water, P=P[7], x=x[1])

v[7]=volume(Water, P=P[7], x=x[1])

h[8] = h[7] “throttling”

P_sat[1]=P_sat(Water,T=T[7])

h[3] = h[7] + (v[7]*(P[3]-P_sat[1])) “Steam extracted from the turbine Closed FWH ” y = (h[3] – h[2])/(h[5] -h[7]) “y = m_extract/m_feed)” “cycle” q_in=h[4]-h[3] w_turbine=h[4]-h[5]+(1-y)*(h[5]-h[6])

w_net = w_turbine- w_pumpI_in

Eta_th = w_net/q_in

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