Spring 2006
1
ME 201
Thermodynamics
Homework 18
Solutions
1. Determine the work per mass output of an adiabatic turbine with isentropic
efficiency 0.83 that has a steam input of 15 MPa and 650
°
C and an outlet pressure
of 50 kPa.
Solution:
The actual work will be calculated from the isentropic efficiency by
w
act
=
η
s
w
ideal
Hence we need to calculate the ideal work. We begin by setting up our table and
working the 1
st
law problem
System Type: Control Volume (Turbine)
Working Fluid: Steam (compressible Substance)
Process: Isentropic (ideal) Adiabatic (actual) with
η
s
= 0.83
State 1
State 2s
State 2a
T
1
= 650
°
C
T
2s
=
81.32
°
C
T
2a
=
81.32
°
C
P
1
= 15 MPa
P
2s
= 50 kPa
P
2a
= 50 kPa
h
1
=
3712.1 kJ/kg
h
2s
=
2372.4 kJ/kg
h
2a
=
2600 kJ/kg
s
1
=
6.8233
kJ/(kg
⋅
K)
s
2s
=
6.8233 kJ/(kg
⋅
K)
s
2a
=
7.4649 kJ/(kg
⋅
K
)
phase: sup.vap.
phase:
2 phase with
x
2s
= 0.882
phase:
2 phase with
x
2a
= 0.98
Italicized
values are from steam tables.
Bold
values are calculated.
State1: Fixed
State2s: UNKNOWN
State2a: UNKNOWN
°
Q
= 0
°
W
sh
= UNKNOWN
1st Law:
[
]
°
°
m h
h
= W
1
2
sh

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Spring 1999
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 Spring '06
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 Thermodynamics, Isentropic efficiency, steam tables, 650°C

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