Unformatted text preview: 31 2.3. RANKINE
The necessary heat transfer rate in the boiler is then
˙
Qin = m(h3 − h2 ),
˙
kg
=
104.697
s
= 269706 kW ,
= (2.123)
2758 kJ
kg − 181.94 kJ
kg , (2.124)
(2.125) 269.7 M W. (2.126) In the condenser, one ﬁnds
˙
Qout = m(h1 − h4 ),
˙
kg
=
104.697
s
= −169705 kW,
= (2.127)
173.88 kJ
kg − 1794.8 kJ
kg , (2.128)
(2.129) −169.7 M W. (2.130) Note also for the cycle that one should ﬁnd
˙
˙
˙
Wcycle = Qin + Qout = (269.7 M W ) − (169.7 M W ) = 100 M W. (2.131) For the condenser mass ﬂow rate now perform a mass balance:
dEcv
dt = ˙
˙
Qcv − Wcv +mc (hin − hout ) + m(h4 − h1 ),
˙
˙
=0 =0 =0 0=
mc
˙ = mc (hin − hout ) + m(h4 − h1 ),
˙
˙
m(h4 − h1 )
˙
−
,
hin − hout =
=
= − kg
s 1794.8 62.99 104.697 = (2.132) kJ
kg kg
,
s
kg
2027.79
s kJ
kg (2.133)
(2.134)
− 173.88 − 146.68 kJ
kg , (2.135)
(2.136) 2027.79 7.3 × 106 kJ
kg 3600 s
hr , (2.137) kg
.
hr (2.138) The enthalpy for the cooling water was found by assuming values at the saturated state at the respective
temperatures of 15 ◦ C and 35 ◦ C . Example 2.4
Compute the exergy at various points in the ﬂow of a Rankine cycle as considered in the previous
example problem. For that example, we had at state 1, the pump inlet that
h1 = 173.88 kJ
,
kg s1 = 0.5926 kJ
.
kg K (2.139) CC BYNCND. 18 November 2011, J. M. Powers. 32 CHAPTER 2. CYCLE ANALYSIS
After the pump, at state 2, we have
kJ
,
kg h2 = 181.94 s2 = 0.5926 kJ
.
kg K (2.140) After the boiler, at state 3, we have
h3 = 2758 kJ
,
kg s3 = 5.7432 kJ
.
kg K (2.141) After the turbine, at state 4, we have
h4 = 1794.8 kJ
,
kg s4 = 5.7432 kJ
.
kg K (2.142) Now for this example, kinetic and potential energy contributions to the exergy are negligible, so we
can say in general that
ψ = (h − To s) − (ho − To so ). (2.143) Now for this problem, we have To = 298.15 K , ho = 104.89 kJ/kg, so = 0.3674 kJ/kg/K . We have
estimated ho and so as the enthalpy and entropy of a saturated liquid at 25 ◦ C = 298.15 K .
So the exergies are as follows. At the pump entrance, we get
ψ1 = (h1 − To s1 ) − (ho − To so ) ,
kJ
kJ
− (298.15 K ) 0.5926
=
173.88
kg
kg K
kJ
kJ
− (298.15 K ) 0.3674
−
104.89
kg
kg K
= 1.84662 (2.144) , kJ
.
kg (2.145)
(2.146) After the pump, just before the boiler, we have
ψ2 = (h2 − To s2 ) − (ho − To so ) ,
kJ
kJ
=
181.94
− (298.15 K ) 0.5926
kg
kg K
kJ
kJ
− (298.15 K ) 0.3674
−
104.89
kg
kg K
= 9.90662 (2.147) , kJ
.
kg (2.148)
(2.149) So the exergy has gone up. Note here that ψ2 − ψ1 = h2 − h1 because the process is isentropic. Here
ψ2 − ψ1 = h2 = h1 = 181.94 − 173.88 = 8.06 kJ/kg .
After the boiler, just before the turbine, we have
ψ3 = (h3 − To s3 ) − (ho − To so ) ,
kJ
kJ
=
2758
− (298.15 K ) 5.7432
kg
kg K
kJ
kJ
− (298.15 K ) 0.3674
−
104.89
kg
kg K
= 1050.32 kJ
.
kg CC BYNCND. 18 November 2011, J. M. Powers. (2.150) , (2.151)
(2.152) ...
View
Full Document
 Fall '11
 ParkSou
 Dynamics, Heat Transfer, Trigraph, kg, HMS H3, HMS H4, hin − hout

Click to edit the document details