thermodynamic and heat transfer examples

1. ASEN 3113 Thermodynamics and Heat Transfer
Homework #4
Assigned: September 28, 2006
Due: October 10, 2006 (class time)
(Total points noted in each section; must clearly show equations with values and units,
drawings, assumptions, etc.)
1. (20 points) A cold air standard Otto cycle has a compression ratio of 8. The cylinder state at
the end of the expansion process is T = 500 K and P = 620 kPa. The heat rejection per unit mass
kgkJ
m
Q
/16341
= Determine the following:
a) (10 points) Net work per unit mass of air.
b) (5 points) Thermal efficiency.
c) (5 points) Mean effective pressure.
Solution:
Using the given heat rejection, 14
41
uu
m
Q
−= . Thus, with Tcu v ∆=∆ and cv = 0.721 kJ/kg K:
K
kgKkJ
kgkJ
K
cm
Q
TT
v
93.273
/721.0
/163
50041
41 =−=
⋅
−=
For the isentropic compression .)/(/ 1
2112
−
= k
VVTT Therefore
KKT 35.629)93.273()8( 14.1
2 == −
Also for the isentropic expansion
KKTVVTVVT kk
7.1148)500()8()/()/( 14.1
4
1
214
1
343 ==⋅=⋅= −−−
a) The net work per unit mass of air is
kgkJkgkJKKkgKkJ
m
Q
TTc
m
Q
m
Q
m
W
v
cycle
/45.211/163)35.6297.1148(/721.0)( 41
23
4123
=−−=−−=−=
b) The thermal efficiency is
565.0
/45.374
/45.211
/
/
23
===
kgkJ
kgkJ
mQ
mWcycle
η
c) The mean effective pressure is given by

2. )/11(121 rV
W
VV
W
mep
cyclecycle
−
=
−
=
Evaluating the specific volume V1=V4
Where kgm
kPa
KkgKkJ
P
RT
V /2315.0
620
)500)(/2870.0( 3
4
4
4 ===
Thus kPa
kgm
kgkJ
mep 86.1043
)8/11)(/2315.0(
/45.211
3
=
−
=
2. (20 points) An air standard gas turbine engine has a compressor pressure ratio of 12. It
operates between 290 K and 1400 K. The turbine and compressor each have isentropic
efficiencies of 90%.Determine the following:
a) (10 points) Net work per unit mass of air flow.
b) (5 points) Heat rejected per unit mass flow of air.
c) (5 points) Thermal efficiency.
Solution:
Determine the unknown temperatures:
( ) KK
P
P
TT
kk
84.58912290
4.1/)14.1(
/)1(
1
2
12 ==





=
−
−
( ) KK
P
P
TT
kk
32.68812/11400
4.1/)14.1(
/)1(
3
4
34 ==





=
−
−
Determine the work input to the compressor and the work output of the turbine:
kgkJKKkgKkJTTcW pin /34.301)29084.589)(/005.1()( 12 =−=−=
kgkJKKkgKkJTTcW pout /24.715)32.6881400)(/005.1()( 43 =−=−=
a) Net work:
kgkJkgkJkgkJWWW inoutnet /9.413/34.301/24.715 =−=−=
b) Heat rejected:
kgkJKKkgkJTTcq pout /31.400)29032.688(/005.1)( 14 =−=−=
c) Thermal efficiency:

3. kgkJKKkgkJTTcq pin /21.814)84.5891400(/005.1)( 23 =−=−=
508.0
/21.814
/9.413
===
kgkJ
kgkJ
q
W
in
net
thη
3. (30 points) A four-cylinder, four-stroke internal combustion engine has a bore of 95.25 mm.
and a stroke of 87.63 mm. The clearance volume is 17% of the cylinder volume at bottom dead
center and the crankshaft rotates at 2600 RPM. The total volume of the cylinder is the volume of
bore and stroke volume plus the clearance volume. The processes within each cylinder are
modeled as an Otto cycle with a pressure of 1 atm and a temperature of 288 K at the beginning of
compression. The maximum temperature in the cycle is 2888 K. Use the following constants in
your analysis:
k = 1.4 for air/fuel mixture (compression) and k = 1.285 for combustion products (expansion)
Cv = 0.926 kJ/kg*K between steps 2 and 3, Cv = 0.825 kJ/kg*K between steps 4 and 1
(a) (4 points) Draw the P-v diagram; label Pressures, Temperatures, Qin, and Qout
(b) (2 points) Calculate the mass of air at the beginning of the cycle
(c) (12 points) Calculate the Pressure in kPa and Temperature in K at each step in the cycle
(d) (6 points) Calculate the net Work per cycle in Joules
(e) (3 points) Calculate the power developed by the engine in kW
(f) (3 points) Calculate the thermal efficiency of this engine
Solution
(a) See diagrams in Cengal pg 292-293.
( ) ( ) ( )
33
22
&
000752.017.0000624.0
17.008763.0
4
09525.0
17.0
4
mVVmV
Vm
m
Vh
d
VVV
TotalTotalTotal
totaltotalclearancestrokeboreTotal
=→+=
+







=+





=+=
ππ
( )( )
( )
g
K
Kg
J
m
m
N
RT
VP
m 92.0
288
*
2870.0
000752.0000,101 3
2
1
11
=





== (b solution)
1-2 Isentropic compression:

4. ( ) ( )
KT
kPa
kPa
K
P
P
TT
kPaP
m
m
kPa
v
v
PP
k
k
k
585
101
1205
288
1205
000128.0
000752.0
101
2
4.1
14.11
1
2
12
2
4.1
3
3
2
1
12
=→





→





=
=→





→





=
−−
2-3 Heat addition:
( ) ( ) ( )
( )( ) kPaP
K
K
kPa
T
T
V
V
PP
RT
VP
RT
VP
kJQ
KK
Kkg
kJkgTTmCvQKT
in
in
5950
585
2888
11205
96.1
5852888
*
926.000092.02888
3
2
3
3
2
23
3
33
2
22
233
=→





=











=→=
=
−



→−=→=
3-4 Isentropic Expansion:
( ) ( )
KT
kPa
kPa
K
P
P
TT
kPaP
m
m
kPa
v
v
PP
k
k
k
1743
5950
611
2888
611
000752.0
000128.0
5950
4
285.1
1285.11
3
4
34
4
285.1
3
3
4
3
34
=→





→





=
=→





→





=
−−
4-1 Heat rejection
( ) ( ) ( )
kJQ
KK
Kkg
kJkgTTmCvQ
out
out
10.1
1743288
*
825.000092.041
−=→
−



→−=
Net work per cycle:
JWkJkJQQW netoutinnet 85610.196.1 =→−=−=
Power developed by the engine:
( ) ( ) ( ) ( ) kWPowerJ
rev
cylindersW
RPM
cylindersPower net 2.74
sec60
min1
856
2
min
2600
4
sec60
min1
2
# =→













=











=
Thermal efficiency:
%6.43
1960
856
=→== th
in
net
th
J
J
Q
W
ηη
4. (30 points) The conditions at the beginning of compression in an air-standard Diesel cycle are
fixed by p1= 200 kPa, T1 = 380 K. The compression ratio is 20 and the heat addition per unit

5. mass is 900 kJ/kg. Use k = 1.4 for air/fuel mixture (compression) and k = 1.34 for combustion
products (expansion).
Hint: To determine T3, use the averaged temperature value between T3 and T2 to determine the
temperature corresponding to a value for average Cp and iterate until you calculate values for
average Cp and T3 that work in the equations. A first guess for the value of T3 will be between
1600 and 2000 K. Use three decimal places only for average Cp in kJ/kg*K and also round
temperature values to the nearest value in four significant digits for this problem (i.e. No decimal
places for temperatures). Table A-19 has values for air and the units for Cp should be J/kg *K.
(Total points noted in each section; must clearly show equations with values and units,
drawings, assumptions, etc.)
(a) (4 points) Draw the P-v diagram and label Pressures, Temperatures, Qin, and Qout
(b) (14 points) Calculate the Pressure in kPa and Temperature in K at each step in the cycle
(c) (4 points) Calculate the net Work per cycle in kJ/kg.
(d) (4 points) Calculate the cutoff ratio
(e) (4 points) Calculate the thermal efficiency of the cycle
Solution:
(a) See pg 298 in Cengal for diagrams.
1-2 Isentropic compression
( )
( ) kPaPkPa
v
v
PP
KT
v
v
TT
k
k
1330020200
125920380
2
4.1
1
2
12
2
4.0
2
1
12
=→→





=
=→→





=
2-3 Heat addition
( ) ( )KTC
kg
kJTTCq ppin 1259900 323 −=→−=
Choose T3 = 1983K, Calculate a Cp
( ) ( ) 243.11259198390023 =→−=→−= pppin CKC
kg
kJTTCq
Check average temperature of T3 and T2 in Cengal Table A-19
( )
( ) OKGuessTempCAirATableK
TT
p __243.1,191621
2
23
→≅→−→=
+
Calculate cutoff ratio

6. 575.1
1259
1983
_
2
3
2
3
3
33
2
22
==→





=





→→=
K
K
r
T
T
V
V
PConst
RT
VP
RT
VP
c
3-4 Isentropic expansion
kPaPkPar
v
v
PP
KTKr
v
v
TT
k
c
k
c
442575.1
20
1
13300
836575.1
20
1
1983
4
34.1
4
3
34
4
34.01
4
3
34
=→











→













=
=→











→













=
−
( ) ( )
kg
kJq
KK
Kkg
kJTTCvq
out
out
348
836380
*
764.041
−=→
−



→−=
Net work per cycle:
kJwkJkJqqw netoutinnet 552348900 =→−=−=
Thermal efficiency:
%3.61
900
552
=→== th
in
net
th
kJ
kJ
q
w
ηη

7. 575.1
1259
1983
_
2
3
2
3
3
33
2
22
==→





=





→→=
K
K
r
T
T
V
V
PConst
RT
VP
RT
VP
c
3-4 Isentropic expansion
kPaPkPar
v
v
PP
KTKr
v
v
TT
k
c
k
c
442575.1
20
1
13300
836575.1
20
1
1983
4
34.1
4
3
34
4
34.01
4
3
34
=→











→













=
=→











→













=
−
( ) ( )
kg
kJq
KK
Kkg
kJTTCvq
out
out
348
836380
*
764.041
−=→
−



→−=
Net work per cycle:
kJwkJkJqqw netoutinnet 552348900 =→−=−=
Thermal efficiency:
%3.61
900
552
=→== th
in
net
th
kJ
kJ
q
w
ηη