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Unformatted text preview: 29 2.3. RANKINE T
3 2
1 4
s Figure 2.4: T − s plane for Rankine cycle.
• thermal eﬃciency
• back work ratio
• mass ﬂow rate of steam
˙
• rate of heat transfer Qin to the ﬂuid in the boiler
˙
• rate of heat transfer Qout in the condenser
• mass ﬂow rate of condenser cooling water if the cooling water enters at 15 ◦ C and exits at 35 ◦ C .
Use the steam tables to ﬁx the state. At the turbine inlet, one has P3 = 8.0 M P a, and x3 = 1 (saturated
steam). This gives two properties to ﬁx the state, so that
h3 = 2758 kJ
,
kg s3 = 5.7432 kJ
.
kg K (2.104) State 4 has P4 = 0.008 M P a and s4 = s3 = 5.7432 kJ/kg/K , so the state is ﬁxed. From the saturation
tables, it is found then that
5.7432
s4 − sf
=
x4 =
sg − sf kJ
kg K − 0.5926 7.6361 kJ
kg K kJ
kg K = 0.6745. (2.105) Note the quality is 0 ≤ x4 ≤ 1, as it must be. The enthalpy is then
h4 = hf + x4 hf g = 173.88 kJ
kg + (0.6745) 2403.1 kJ
kg = 1794.8 kJ
.
kg (2.106) State 1 is saturated liquid at 0.008 M P a, so x1 = 0, P1 = 0.008 M P a. One then gets h1 = hf =
173.88 kJ/kg, v1 = vf = 0.0010084 m3 /kg .
Now state 2 is ﬁxed by the boiler pressure and s2 = s1 . But this requires use of the sparse
compressed liquid tables. Alternatively, the pump work is easily approximated by assuming an incomCC BYNCND. 18 November 2011, J. M. Powers. 30 CHAPTER 2. CYCLE ANALYSIS
pressible ﬂuid so that
h2 = h2 ˙
W
= h1 + v1 (P2 − P1 ),
m
˙
m3
kJ
+ 0.0010084
173.88
kg
kg = (2.107) h1 + The net power is (8000 kP a − 8 kP a) = 181.94 kJ
.
kg ˙
˙
˙
Wcycle = Wt + Wp . (2.108) (2.109) Now the ﬁrst law for the turbine and pump give
˙
Wt
= h3 − h4 ,
m
˙ ˙
Wp
= h1 − h2 .
m
˙ (2.110) The energy input that is paid for is
˙
Qin
= h3 − h2 .
m
˙ (2.111) ˙
˙
(h3 − h4 ) + (h1 − h2 )
Wt + Wp
=
,
˙ in
h3 − h2
Q (2.112) The thermal eﬃciency is then found by
η = = − 1794.8 kJ
kg kJ
kg + 173.88 kJ
kg 2758 2758
= kJ
kg − 181.94 kJ
kg − 181.94 kJ
kg , 0.371. (2.113)
(2.114) By deﬁnition the back work ratio bwr is the ratio of pump work to turbine work:
= ˙
Wp
,
˙
Wt (2.115) = bwr h1 − h2
,
h3 − h4 (2.116) 173.88
= kJ
kg 2758
= kJ
kg − 181.94
− 1794.8 kJ
kg , kJ
kg (2.117) 0.00837. (2.118) The desired mass ﬂow can be determined since we know the desired net power. Thus
m
˙ = ˙
Wcycle
,
(h3 − h4 ) + (h1 − h2 ) =
2758 kJ
kg kg
,
s
kg
104.697
s − 1794.8 (2.119)
100 × 103 kW
kJ
kg + 173.88 kJ
kg − 181.94 = 104.697
= kJ
kg , (2.120)
(2.121) 3600 s
hr CC BYNCND. 18 November 2011, J. M. Powers. = 3.769 × 105 kg
.
hr (2.122) ...
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 Fall '11
 ParkSou
 Dynamics, Heat Transfer

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