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Ideal Regenerative Rankine Cycle with one Closed FWH

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 π‘˜π‘ƒπ‘Ž
π‘†π‘Žπ‘‘. πΏπ‘–π‘žπ‘’π‘–π‘‘ }
β„Ž1 = β„Žπ‘“@10π‘˜π‘ƒπ‘Ž = 191.81 π‘˜π½β„π‘˜π‘”
𝑣1 = 𝑣𝑓@10π‘˜π‘ƒπ‘Ž = 0.001010 π‘š3β„π‘˜π‘”
Stage 2
𝑃2 = 10 π‘€π‘ƒπ‘Ž
𝑠2 = 𝑠1
}
𝑀𝑃𝐼,𝑖𝑛 = 𝑣1
(𝑃2 βˆ’ 𝑃1
)
=(0.001010π‘š3β„π‘˜π‘”)(10000βˆ’10 π‘˜π‘ƒπ‘Ž)(1π‘˜π½ 1π‘˜π‘ƒπ‘Ž.π‘š3 ⁄ )
=10.1π‘˜π½β„π‘˜π‘”
β„Ž2 = β„Ž1 + 𝑀𝑃𝐼,𝑖𝑛 = (191.81 + 10.1) π‘˜π½β„π‘˜π‘” = 201.91 π‘˜π½β„π‘˜π‘”
Stage 4
𝑃4 = 10 π‘€π‘ƒπ‘Ž
𝑇4 = 500π‘œπΆ
}
β„Ž5 = 3375.1 π‘˜π½β„π‘˜π‘”
𝑠5 = 6.5995 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾
Stage 5
𝑃5 = 1 π‘€π‘ƒπ‘Ž
𝑠5 = 𝑠4
} β„Ž5 = 2784 π‘˜π½β„π‘˜π‘” 𝑏𝑦 π‘–π‘›π‘‘π‘’π‘Ÿπ‘π‘œπ‘™π‘Žπ‘‘π‘–π‘œπ‘›
π‘“π‘Ÿπ‘œπ‘š π‘‘β„Žπ‘’ π‘ π‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘‘ π‘ π‘‘π‘’π‘Žπ‘š π‘‘π‘Žπ‘π‘™π‘’ π‘‘β„Žπ‘’ 𝑠𝑔@1.0 π‘€π‘ƒπ‘Ž = 6.5850 π‘˜π½β„π‘˜π‘” βˆ’ 𝐾
π‘€β„Žπ‘–π‘β„Ž 𝑖𝑠 𝑙𝑒𝑠𝑠 π‘‘β„Žπ‘Žπ‘› 𝑠5,π‘‘β„Žπ‘’π‘Ÿπ‘“π‘œπ‘Ÿπ‘’ 𝑖𝑑 𝑖𝑠 π‘ π‘’π‘π‘’π‘Ÿβ„Žπ‘’π‘Žπ‘‘π‘’π‘‘
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 βˆ’ 6.5995
6.6956 βˆ’ 6.580 =
200 βˆ’ 𝑇
200 βˆ’ 179.88
0.8689 βˆ— (200 βˆ’ 179.88) = 200 βˆ’ 𝑇
𝑇 = 200 βˆ’ (0.8689 βˆ— (200 βˆ’ 179.88)) = 182.5 𝐢
Similarly
6.6956 βˆ’ 6.5995
6.6956 βˆ’ 6.580 =
2828.3 βˆ’ β„Ž
2828.3 βˆ’ 2777.1
0.8689 βˆ— (2828.3 βˆ’ 2777.1) = 2828.3 βˆ’ β„Ž
β„Ž = 2828.3 βˆ’ (0.8689 βˆ— (2828.3 βˆ’ 2777.1)) = 2784 π‘˜π½β„π‘˜π‘”
Stage 6
𝑃6 = 10 π‘˜π‘ƒπ‘Ž
𝑠6 = 𝑠4
} π‘₯6 =
𝑠6 βˆ’ 𝑠𝑓

𝑠𝑓𝑔

6.5995 βˆ’ 0.6492
7.4996 = 0.793
β„Ž6 = β„Žπ‘“ + β„Žπ‘“π‘” = 191.81 + (0.793)(2392.1) = 2088.75 π‘˜π½β„π‘˜π‘”
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 π‘€π‘ƒπ‘Ž
π‘₯7 = 0
}
β„Ž7 = 762.51 π‘˜π½β„π‘˜π‘”
𝑇7 = 179.88π‘œπΆ
β„Ž8 = β„Ž7 = 762.51 π‘˜π½β„π‘˜π‘”
𝑃3 = 10 π‘€π‘ƒπ‘Ž
𝑇3 = 𝑇7 = 179.88π‘œπΆ
} β„Žπ‘“ = 771.51 π‘˜π½β„π‘˜π‘”
At T7 the saturation pressure is 1.0 MPa, therefore the h7 can be worked from the following formula
β„Ž3 = β„Ž7 + 𝑣7
(𝑃7 βˆ’ π‘ƒπ‘ π‘Žπ‘‘@𝑇7
)
= 762.51 π‘˜π½β„π‘˜π‘” + ((0.001 π‘š3β„π‘˜π‘”)(10000 βˆ’ 1000 π‘˜π‘ƒπ‘Ž) (1π‘˜π½ 1π‘˜π‘ƒπ‘Ž. π‘š3 ⁄ )) = 771.51 π‘˜π½β„π‘˜π‘”
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

  • = +
  • = +
    οƒ₯ =οƒ₯
       
     
    Rearranging,
    𝑦 =
    β„Ž3 βˆ’ β„Ž2

β„Ž5 βˆ’ β„Ž7

771.51 βˆ’ 201.91
2784 βˆ’ 762.51 = 0.282
Then,
π‘žπ‘–π‘› = β„Ž4 βˆ’ β„Ž3 = (3375.1 βˆ’ 763.05) π‘˜π½β„π‘˜π‘” = 2612.05 π‘˜π½β„π‘˜π‘”
𝑀𝑑,π‘œπ‘’π‘‘ = (β„Ž4 βˆ’ β„Ž5
) + (1 βˆ’ 𝑦)(β„Ž5 βˆ’ β„Ž6)
= (3375.1 βˆ’ 2784) + (1 βˆ’ 0.282)(2784 βˆ’ 2088.75) π‘˜π½β„π‘˜π‘” = 1090.3 π‘˜π½β„π‘˜π‘”
𝑀𝑛𝑒𝑑 = 𝑀𝑑,π‘œπ‘’π‘‘ βˆ’ 𝑀𝑝,𝑖𝑛 = (1090.3 βˆ’ 10.1) π‘˜π½β„π‘˜π‘” = 1070.2 π‘˜π½β„π‘˜π‘”
πœ‚π‘‘β„Ž =
𝑀𝑛𝑒𝑑

π‘žπ‘–π‘›

1070.2
2612.05 = 0.4097 = 41.0%
The EES code for the above problem is presented here
P[4] = 10convert(MPa,kPa) T[4] = 500 [C] P[5] =1convert(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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