Assignment #7
PROBLEM 6.20
PROBLEM STATEMENT:
An adiabatic steam turbine operates with inlet steam
conditions of 500°C, 2 MPa, and a measured outlet condition of 10 kPa, x
=
0.95.
(a) How much work per unit mass (kJ/kg) will this turbine produce?
(b) If the turbine were adiabatic and reversible, what would the outlet quality be,
and how much work per unit mass could it produce for the same outlet
pressure?
DIAGRAM DEFINING SYSTEM AND PROCESS:
GIVEN:
H
2
O, adiabatic turbine
State 1: P
1
=
2 MPa, T
1
=
500°C
State 2: P
2
=
10 kPa, x
2
=
0.95
FIND:
w
OUT,1-2
(kJ/kg), x
2,s
, w
OUT,1-2s
(kJ/kg)
ASSUMPTIONS:
No change in kinetic and potential energy.
GOVERNING RELATIONS:
1.
First Law for control volume:
IN,1-2
1
OUT,1-2
2
q
+h = w
+h
2.
Second law: Adiabatic and reversible process is isentropic, s
2
=s
1
QUANTITATIVE SOLUTION:
a). From First Law equation:
NET,1-2
q
(
29
0
1
2
OUT,1-2
OUT,1-2
1
2
+ h -h
= w
w
= h -h
⇒
From the property tables for H
2
O, the following values can be obtained:
(
29
(
29
(
29
(
29
1
2
L,2
2 LV,2
h = h P = 2 MPa, T = 500°C = 3467.70 kJ/kg
Table 12s
h = h P =10 kPa, x = 0.95 = h
+ x h
=191.83+0.95 2392.0 = 2464.23 kJ/kg
Table 11s
×
Thus, work produced by the turbine is
OUT,1-2
1
2
w
= h -h = 3467.70 - 2464.23
=1003.47
kJ/kg
b). The turbine is adiabatic and reversible, i.e. isentropic. From Table 12s:
1
2,s
s = 7.4318 kJ/kg K = s
( isentropic)
g
Q
P
1
=
2 MPa
T =500°C
Comp.
P
2
=
10 kPa
x
2
=
0.95
2
1
Turbine
w
OUT,1-2