Spring 2006
1
ME 201
Thermodynamics
Final Exam Solutions
Directions:
Work all three problems.
The exam is open notes and open text book.
All problems have equal weight.
Note that you may round where appropriate to avoid
interpolation.
Problem 1
A more realistic model for the compression process in an internal combustion uses a
polytropic process followed by an isentropic process instead of a single isentropic
process.
The engine operates with the following conditions:
Six cylinder, four stroke engine with displacement of 2.8 liters
Compression ratio: 9.065
Compression ratio for polytropic process: 3
Polytropic exponent for first compression process: 1.203
Initial air conditions: 280 K and 120 kPa
Engine speed of 1300 rpm
Determine
a.) the air temperature at the end of both compression processes
b.
the heat transfer for the polytropic process
b.) the engine power in kW for the two compression processes
Solution:
We begin with a block diagram of the system nodded out for the two processes of
interest.
Next we need to
obtain our
volumes.
We are given
V
disp
= 2.8 x 10
3
m
3
and r = 9.065
then
TDC
BDC
disp
V
V
V

=
TDC
BDC
rV
V
=
3
V
/
V
2
BDC
=
Polytropic
Compression
Isentropic
Compression
1
2
3
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ME 201 Thermodynamics
Spring 2006
2
Substituting gives
3
4

3
disp
TDC
m
10
x
3.4718
1
065
.
9
10
x
8
.
2
1
r
V
V
=

=

=

and
3
3
4
BDC
m
10
x
1472
.
3
)
10
x
4718
.
3
)(
065
.
9
(
V


=
=
Following the polytropic process we have
3
3
3
BDC
2
m
10
x
1.0491
3
/
)
10
x
14724
.
3
(
3
/
V
V
=
=
=

Now setting up our table and entering our operating information
Node
T(K)
P(kPa)
V(m
3
)
v(m
3
/kg) u(kJ/kg)
v
r
1
280
120
3.1472 x 10
3
0.6697
199.75
2
350
449.94
1.0491 x 10
3
0.2232
250.02
422.2
3
540
2098
3.4718 x 10
4
0.0739
389.34
139.72
Italicized
values from ideal gas relations.
Bold
values calculated.
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 Spring '06
 SOMERTON
 Thermodynamics, Heat, Heat Transfer, high temperature reservoir, microwave heating

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