Reheat Rankine Cycle with Open and Closed Feedwater heater

1. A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The
thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water
are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),
h1 = h f @ 10 kPa = 191.81 kJ/kg
v 1 = v f @ 10 kPa = 0.00101 m 3 /kg
w p ,in = v 1 (P2 − P1 )
3
( ) ⎛ 1 kJ
= 0.00101 m 3 /kg (7,000 − 10 kPa )⎜
⎜ 1 kPa ⋅ m 3
⎞
⎟
⎟
7 MPa
⎝ ⎠ qin
= 7.06 kJ/kg 2
10 kPa
h2 = h1 + w p ,in = 191.81 + 7.06 = 198.87 kJ/kg 1 4
qout
P3 = 7 MPa ⎫ h3 = 3411.4 kJ/kg s
⎬
T3 = 500°C ⎭ s 3 = 6.8000 kJ/kg ⋅ K
P4 = 10 kPa ⎫ s 4 − s f 6.8000 − 0.6492
⎬ x4 = = = 0.8201
s 4 = s3 ⎭ s fg 7.4996
h4 = h f + x 4 h fg = 191.81 + (0.8201)(2392.1) = 2153.6 kJ/kg
Thus,
q in = h3 − h2 = 3411.4 − 198.87 = 3212.5 kJ/kg
q out = h4 − h1 = 2153.6 − 191.81 = 1961.8 kJ/kg
wnet = q in − q out = 3212.5 − 1961.8 = 1250.7 kJ/kg
and
wnet 1250.7 kJ/kg
η th = = = 38.9%
q in 3212.5 kJ/kg
W&net 45,000 kJ/s
(b) m& = = = 36.0 kg/s
wnet 1250.7 kJ/kg
(c) The rate of heat rejection to the cooling water and its temperature rise are
Q& = m& q = (35.98 kg/s )(1961.8 kJ/kg ) = 70,586 kJ/s
out out

Q& out 70,586 kJ/s
ΔTcooling water = = = 8.4°C
(m& c) cooling water (2000 kg/s )(4.18 kJ/kg ⋅ °C)

, 2. A steam power plant that operates on the ideal reheat Rankine cycle is considered. The turbine work output and
the thermal efficiency of the cycle are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.
Analysis From the steam tables (Tables A-4, A-5, and A-6),
h1 = h f @ 20 kPa = 251.42 kJ/kg
T
v 1 = v f @ 20 kPa = 0.001017 m 3 /kg
w p ,in = v 1 (P2 − P1 )
3 5

( ) ⎛ 1 kJ ⎞
= 0.001017 m 3 /kg (6000 − 20 kPa )⎜ ⎟
⎜ 1 kPa ⋅ m 3 ⎟
6 MPa
= 6.08 kJ/kg ⎝ ⎠ 4

h2 = h1 + w p ,in = 251.42 + 6.08 = 257.50 kJ/kg 2

P3 = 6 MPa ⎫ h3 = 3178.3 kJ/kg 20 kPa
⎬
T3 = 400°C ⎭ s 3 = 6.5432 kJ/kg ⋅ K 1 6
s
P4 = 2 MPa ⎫
⎬ h4 = 2901.0 kJ/kg
s 4 = s3 ⎭
P5 = 2 MPa ⎫ h5 = 3248.4 kJ/kg
⎬
T5 = 400°C ⎭ s 5 = 7.1292 kJ/kg ⋅ K
s 6 − s f 7.1292 − 0.8320
P6 = 20 kPa ⎫ x 6 = = = 0.8900
⎬ s fg 7.0752
s 6 = s5 ⎭
h6 = h f + x 6 h fg = 251.42 + (0.8900 )(2357.5) = 2349.7 kJ/kg
The turbine work output and the thermal efficiency are determined from
wT,out = (h3 − h4 ) + (h5 − h6 ) = 3178.3 − 2901.0 + 3248.4 − 2349.7 = 1176 kJ/kg
and
q in = (h3 − h2 ) + (h5 − h4 ) = 3178.3 − 257.50 + 3248.4 − 2901.0 = 3268 kJ/kg
wnet = wT ,out − w p ,in = 1176 − 6.08 = 1170 kJ/kg
Thus,
wnet 1170 kJ/kg
η th = = = 0.358 = 35.8%
q in 3268 kJ/kg