Water is the working fluid in a regenerative Rankine cycle with one open feedwat
ID: 1717956 • Letter: W
Question
Water is the working fluid in a regenerative Rankine
cycle with one open feedwater heater. Steam enters the
turbine at 1400 lbf/in.2 and 1000 oF and expands to 120 lbf/in.2,
where some of the steam is extracted and diverted to the
open feedwater heater operating at 120 lbf/in.2 The remaining
steam expands through the second-stage turbine to the
condenser pressure of 2 lbf/in.2 Each turbine stage and both
pumps have isentropic efficiencies of 85%. Flow through the
condenser, open feedwater heater, and steam generator is at
constant pressure. Saturated liquid exits the open feedwater
heater at 120 lbf/in.2 The net power output of the cycle is
1x109 Btu/h. Determine for the cycle
(a) the mass flow rate of steam entering the first stage of
the turbine, in lb/h.
(b) the rate of heat transfer, in Btu/h, to the working fluid
passing through the steam generator.
(c) the thermal efficiency.
Explanation / Answer
Let us assume the mass flowing through steam generator be 1 lb/h
At 1400 psia & 1000 °F
h1 = 1493.42 Btu/lbm
S1 = 1.61068 Btu/lbm.R
Since S1 = S2 = S3 = 1.61068 Btu/lbm.R
At 120 psia
Sg = 1.5900 < S1
From superheated steam tables
h2 = 1192.95 Btu/lbm
h21 = 1493.42 - 0.85 (1493.42 - 1192.95)
= 1238.02 Btu/lbm
At 2 psia
Sg = 1.9205 Btu/lbm.R
1.61068 = 0.1817 + x (1.7292)
x = 0.8264
h3 = 97.97 + 0.8264 (1019.78)
= 940.716 Btu/lbm
h31 = 1192.95 - 0.85 (1192.95 - 940.716)
= 978.55 Btu/lbm
h4 = 97.97 Btu/lbm
h5 = 97.97 + 0.01625 (120 - 2)*144
= 97.97 + 0.3548
= 98.324 Btu/lbm
h51 = 97.97 + (98.325 - 97.97)/0.85
= 98.38 Btu/lbm
h6 = 311.29 btu/lbm
h7 = 311.29 + 0.01787 (1000 - 120) * 144
= 311.29 + 2.91
= 314.2 Btu/lbm
h71 = 311.29 + (314.2 - 311.29)/0.85
= 314.71 Btu/lbm
Given network output = 109 Btu/h
109 = (1493.42 - 1238.02) + (1 - m) (1192.95 - 978.55) - (1 - m)(98.38 - 97.97) - (314.71 - 311.29)
Mass flow rate
m * 1192.95 = 311.29 - (1 - m) * 98.324
m = 0.1945
109 = m (255.4) + 0.8055m(214.4) - 0.8055m (0.41) - m (3.42)
m = 109/424.34
= 2356551.163 lbm/hr
(a) Mass flow rate entering 1st turbine = 2356551.16 lbm/hr
(b) Heat transfer = 2356551.163 (h1 - h71)
= 2356551.163 (1493.42 - 314.71)
= 2.7776 * 109 Btu/h
(c) Thermal efficiency = Net work output / Heat transfer
= 109 / (2.7776 * 109)
= 0.36 = 36%
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.