Spring 2006
1
ME 201
Thermodynamics
Exam #2
Solution
Problem 1
Steam at 300 kPa with quality 0.96316 passes through a valve to convert it to
saturated vapor. Determine the exit pressure required.
Solution:
We begin with our template.
Substance Type: Compressible
System Type: Control Volume
Device: Valve
Process: Isenthalpic
Inlet State: Fixed
Outlet State: UNKNOWN
q = 0
w
sh
= 0
1st Law: h
2
 h
1
= 0
State 1
State 2
T
1
=
133.52ºC
T
2
=
81.31ºC
P
1
= 300 kPa
P
2
=
50 kPa
h
1
=
2645.2 kJ/kg
h
2
=
2645.2 kJ/kg
Phase: 2 phase with
x
1
= 0.96316
Phase: sat.vap
Italicized
values are from steam tables.
Bold
values are calculated.
Approach: Go to the steam tables and get h
f
and h
g
at 300 kPa. Use the quality to
calculate h
1
. Set h
2
= h
1
. Go to the saturation pressure table and find the pressure that
gives an h
g
= h
2
.
At 300 kPa, we have
h
f
= 561.43 kJ/kg and h
g
= 2724.9 kJ/kg
Then we can calculate
h
1
= (1x
1
)h
f
+x
1
h
g
= (10.96316)(561.43) + (0.96316)(2724.9)
= 2645.2 kJ/kg
Since our valve is isenthalpic
h
2
= h
1
= 2645.2 kJ/kg
Now going to the saturation pressure table we find
P
2
= 50 kPa
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentME 201 Thermodynamics
Spring 2006
2
Problem 2
A pistoncylinder device contains 0.001 kg of CO
2
at 750 kPa and 1340 K. The
device undergoes a polytropic process with polytropic exponent 0.3 or
PV
0.3
= constant
which decreases the pressure to 550.2 kPa. Determine the boundary work and heat
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 SOMERTON
 Thermodynamics, Wbnd, Constant Pressure Inlet

Click to edit the document details