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
Q41
 163kJ / kg Determine the following:
m
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,
T1  T4 
Q41
 u 4  u1 . Thus, with u  cv T and cv = 0.721 kJ/kg K:
m
Q41
163kJ / kg
 500 K 
 273.93K
m  cv
0.721kJ / kgK
For the isentropic compression T2 / T1  (V1 / V2 ) k 1 . Therefore
T2  (8)1.41 (273.93K )  629.35K
Also for the isentropic expansion
T3  (V4 / V3 ) k 1  T4  (V1 / V2 ) k 1  T4  (8)1.41 (500 K )  1148.7 K
a) The net work per unit mass of air is
Wcycle
m

Q23 Q41
Q

 cv (T3  T2 )  41  0.721kJ / kgK (1148.7 K  629.35K )  163kJ / kg  211.45kJ / kg
m
m
m
b) The thermal efficiency is

Wcycle / m
Q23 / m

211.45kJ / kg
 0.565
374.45kJ / kg
c) The mean effective pressure is given by
mep 
Wcycle

Wcycle
V1  V2 V1 (1  1 / r )
Evaluating the specific volume V1=V4
Where V4 
RT4 (0.2870kJ / kgK )(500 K )

 0.2315m 3 / kg
P4
620kPa
Thus mep 
211.45kJ / kg
 1043.86kPa
(0.2315m 3 / kg)(1  1 / 8)
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:
P
T2  T1  2
 P1



( k 1) / k
 290 K 12
(1.4 1) / 1.4
 589.84 K
( k 1) / k
P 
(1.4 1) / 1.4
T4  T3  4 
 1400 K 1 / 12
 688.32 K
 P3 
Determine the work input to the compressor and the work output of the turbine:
Win  c p (T2  T1 )  (1.005kJ / kgK )(589.84 K  290 K )  301.34kJ / kg
Wout  c p (T3  T4 )  (1.005kJ / kgK )(1400 K  688.32 K )  715.24kJ / kg
a) Net work:
Wnet  Wout  Win  715.24kJ / kg  301.34kJ / kg  413.9kJ / kg
b) Heat rejected:
q out  c p (T4  T1 )  1.005kJ / kg(688.32 K  290 K )  400.31kJ / kg
c) Thermal efficiency:
qin  c p (T3  T2 )  1.005kJ / kg(1400 K  589.84 K )  814.21kJ / kg
 th 
Wnet
413.9kJ / kg

 0.508
qin
814.21kJ / kg
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.
VTotal  Vbore& stroke  Vclearance
VTotal  0.000624m 3  0.17VTotal

  0.09525m 2

h   0.17Vtotal  

4


 VTotal  0.000752m 3
 d 2
 
 4


3
N
P1V1 101,000 m 2 0.000752m
m

 0.92 g
RT1
 0.2870 J
288 K 
g*K

(b solution)

0.08763m   0.17Vtotal


1-2 Isentropic compression:
k
v
P2  P1  1
 v2

 0.000752m 3 
  101kPa

3 
0
.
000128
m



P
T2  T1  2
 P1



 k 1
k
 1205kPa 
 288K 

 101kPa 
1.4
 P2  1205kPa
1.4 1
1.4
 T2  585K
2-3 Heat addition:
2888 K  585 K 
T3  2888 K  Qin  mCvT3  T2   0.00092kg  0.926 kJ
kg * K 

Qin  1.96kJ
V
P2V 2 P3V3

 P3  P2  2
RT 2
RT 3
 V3
 T3

 T2

 2888K 
  1205kPa1
  P3  5950kPa
 585 K 

3-4 Isentropic Expansion:
k
v
P4  P3  3
 v4

 0.000128m 3 
  5950kPa

3 
 0.000752m 

P
T4  T3  4
 P3



 k 1
k
 611kPa 
 2888K 

 5950kPa 
1.285
 P4  611kPa
1.2851
1.285
 T4  1743K
4-1 Heat rejection
288K  1743K 
Qout  mCvT1  T4   0.00092kg  0.825 kJ
kg * K 

 Qout  1.10kJ
Net work per cycle:
Wnet  Qin  Qout  1.96kJ  1.10kJ  Wnet  856 J
Power developed by the engine:
 2600 rev
 RPM 
 1 min 

min
Power  # cylinders 
Wnet 
  4cylinders 
2
60
sec
2






Thermal efficiency:
W
856 J
 th  net 
  th  43.6%
Qin 1960 J

856 J  1 min   Power  74.2kW



 60 sec 

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
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
k
v
T2  T1  1
 v2

  38020 0.4  T2  1259 K

v
P2  P1  2
 v1

  200kPa20 1.4  P2  13300kPa

k
2-3 Heat addition
qin  C p T3  T2   900 kJ
kg
 C p T3  1259K 
Choose T3 = 1983K, Calculate a Cp
qin  C p T3  T2   900 kJ
kg
 C p 1983  1259K   C p  1.243
Check average temperature of T3 and T2 in Cengal Table A-19
T3  T2 
2
 1621K  Table A  19, Air   C p  1.243  Temp _ Guess _ OK
Calculate cutoff ratio
V
P2V2 P3V3

 Const _ P   3
RT 2
RT3
 V2
  T3
  
  T2

1983K
  rc 
 1.575
1259K

3-4 Isentropic expansion
k 1
 v
T4  T3   3
  v4
 
rc 

 
 v
P4  P3   3
  v4
 
 1 

rc   13300kPa  1.575 

  20 

 
 1 

 1983K   1.575 
  20 

k
0.34
 T4  836 K
1.34
 P4  442kPa
380 K  836 K 
q out  CvT1  T4    0.764 kJ
kg
*
K


 q out  348 kJ
kg
Net work per cycle:
wnet  qin  q out  900kJ  348kJ  wnet  552kJ
Thermal efficiency:
w
552kJ
 th  net 
  th  61.3%
qin 900kJ