ICE CYCLES

1. ME 14 - THERMODYNAMICS 2
Instructor: ENGR. YURI G. MELLIZA
AIR STANDARD CYCLE
INTERNAL COMBUSTION ENGINE CYCLE
Air Standard Cycle
o Otto Cycle
o Diesel Cycle
o Dual Cycle
Working Substance: Air
For Cold Air Standard: k = 1.4
For Hot Air Standard: k = 1.3
AIR STANDARD OTTO CYCLE
Otto, Nikolaus August
Born:June 10, 1832, Holzhausen, Nassau
Died: Jan. 26, 1891, Cologne
German engineer who developed the four-stroke internal-combustion engine, which offered the first practical alternative to the
steam engine as a power source.
Otto built his first gasoline-powered engine in 1861. Three years later he formed a partnership with the German industrialist Eugen
Langen, and together they developed an improved engine that won a gold medal at the Paris Exposition of 1867.
In 1876 Otto built an internal-combustion engine utilizing the four-stroke cycle (four strokes of the piston for each ignition). The
four-stroke cycle was patented in 1862 by the French engineer Alphonse Beau de Rochas, but since Otto was the first to build an
engine based upon this principle, it is commonly known as the Otto cycle. Because of its reliability, its efficiency, and its relative
quietness, Otto's engine was an immediate success. More than 30,000 of them were built during the next 10 years, but in 1886 Otto's
patent was revoked when Beau de Rochas' earlier patent was brought to light.
Processes
1 to 2 – Isentropic Compression (S = C)
2 to 3 – Constant Volume Heat Addition (V = C)
3 to 4 – Isentropic Expansion (S = C)
4 to 1 – Constant volume Heat Rejection (V = C)

2. 1
5
Eq.
)
T
-
(T
mC
=
Q 2
3
v
A →
6
Eq.
)
T
-
(T
mC
=
Q 1
4
v
R →
  8
Eq.
)
T
-
(T
-
)
T
-
(T
mC
W
7
Eq.
Q
-
Q
=
W
Q
W
1
4
2
3
v
R
A
→
=
→

=
Compression Ratio (r)
1
V
V
V
V
r
3
4
2
1
→
=
=
V1 = V4 and V2 = V3
V1 - volume at bottom dead center (BDC)
V2 – volume at top dead center (TDC)(clearance volume)
Displacement Volume (VD)
VD =V1 – V2 → Eq. 2
Percent Clearance (C)
3
Eq.
V
V
C
D
2
→
=
4
Eq.
C
C
1
r
CV
CV
V
r
D
D
D
→
+
=
+
=
Heat Added (QA)
At V = C ;
Q = mCv(T)
Heat Rejected (QR)
Net Work (W)
P,V and T Relations
At point 1 to 2 (S = C)
( )
9
Eq.
)
r
(
T
T
P
P
r
V
V
T
T
1
k
1
2
k
1
k
1
2
1
k
1
k
2
1
1
2
→
=








=
=








=
−
−
−
−
At point 2 to 3 (V = C)
10
Eq.
P
P
T
T
2
3
2
3
→
=
At point 3 to 4 (S = C)
1
1
.
Eq
)
r
(
T
T
)
r
(
P
P
V
V
T
T
1
k
4
3
1
k
k
1
k
4
3
1
k
3
4
4
3
→
=
=








=








=
−
−
−
−

3. 2
23
Eq.
kg
m
-
V
22
Eq.
m
V
V
V
3
2
1
D
3
2
1
D
→


=
→
−
=
At point 4 to 1 (V = C)
12
Eq.
P
P
T
T
1
4
1
4
→
=
Entropy Change
a) S during Heat Addition
13
Eq.
T
T
ln
mC
S
S
2
3
V
2
3 →
=
−
b) S during Heat Rejection
15
Eq.
)
S
S
(
S
S
14
Eq.
T
T
ln
mC
S
S
2
3
4
1
4
1
V
4
1
→
−
−
=
−
→
=
−
Thermal Efficiency
Eq.16
100%
x
Q
W
e
A
→
=
17
Eq.
100%
x
Q
Q
Q
e
A
R
A
→
−
=
18
Eq.
100%
x
Q
Q
1
e
A
R
→






−
=
)
T
T
(
)
r
(
)
T
T
(
11
Eq.
and
9
Eq.
From
19
Eq.
100%
x
)
T
T
(
)
T
T
(
1
e
1
4
1
k
2
3
2
3
1
4
−
=
−
→






−
−
−
=
−
20
Eq.
100%
x
)
r
(
1
1
e 1
k
→






−
= −
Mean Effective Pressure
21
Eq.
KPa
V
W
P
D
m →
=
where:
Pm – mean effective pressure, KPa
W – Net Work KJ, KJ/kg, KW
VD – Displacement Volume m3
, m3
/kg, m3
/sec
Displacement Volume

4. 3
AIR STANDARD DIESEL CYCLE
Diesel, Rudolf (Christian Karl)
Born: March 18, 1858, Paris, France
Died: September 29, 1913, at sea in the English Channel
German thermal engineer who invented the internal-combustion engine that bears his name. He was also a distinguished connoisseur
of the arts, a linguist, and a social theorist.
Diesel, the son of German-born parents, grew up in Paris until the family was deported to England in 1870 following the outbreak
of the Franco-German War. From London Diesel was sent to Augsburg, his father's native town, to continue his schooling. There
and later at the Technische Hochschule (Technical High School) in Munich he established a brilliant scholastic record in fields of
engineering. At Munich he was a protégé of the refrigeration engineer Carl von Linde, whose Paris firm he joined in 1880.
Diesel devoted much of his time to the self-imposed task of developing an internal combustion engine that would approach the
theoretical efficiency of the Carnot cycle. For a time he experimented with an expansion engine using ammonia. About 1890, in
which year he moved to a new post with the Linde firm in Berlin, he conceived the idea for the diesel engine . He obtained a German
development patent in 1892 and the following year published a description of his engine under the title Theorie und Konstruktion
eines rationellen Wäremotors (Theory and Construction of a Rational Heat Motor). With support from the Maschinenfabrik
Augsburg and the Krupp firms, he produced a series of increasingly successful models, culminating in his demonstration in 1897 of
a 25-horsepower, four-stroke, single vertical cylinder compression engine. The high efficiency of Diesel's engine, together with its
comparative simplicity of design, made it an immediate commercial success, and royalty fees brought great wealth to its inventor.
Diesel disappeared from the deck of the mail steamer Dresden en route to London and was assumed to have drowned.
Processes:
1 to 2 – Isentropic Compression (S = C)
2 to 3 – Constant Pressure Heat Addition (P = C)
3 to 4 – Isentropic Expansion (S = C)
4 to 1 – Constant Volume Heat Rejection (V = C)

5. 4
5
Eq.
V
-
V
=
V 2
1
D →
7
Eq.
)
T
-
(T
mkC
Q
kC
C
;
C
C
k
6
Eq.
)
T
-
(T
mC
Q
T
mC
Q
C
P
at
2
3
V
A
V
p
v
p
2
3
p
A
p
→
=
=
=
→
=

=
=
8
Eq.
)
T
T
(
mC
Q 1
4
V
R →
−
=
  10
Eq.
)
T
T
(
T
T
(
k
mC
W
9
Eq.
Q
Q
W
1
4
)
2
3
v
R
A
→
−
−
−
=
→
−
=
Compression Ratio
1
Eq.
V
V
V
V
r
2
4
2
1
→
=
=
3
4
2
1
V
V
V
V

Cut-Off Ratio(rc)
2
Eq.
V
V
r
2
3
c →
=
Percent Clearance
3
Eq.
V
V
C
D
2
→
=
4
Eq.
C
C
1
r →
+
=
Displacement Volume (VD)
Heat Added (QA)
Heat Rejected (QR)
Net Work (W)
P,V and T Relations
At point 1 to 2 (S = C)
( )
11
Eq.
)
r
(
T
T
P
P
r
V
V
T
T
1
k
1
2
k
1
k
1
2
1
k
1
k
2
1
1
2
→
=








=
=








=
−
−
−
−

6. 5
16
Eq.
T
T
ln
mC
S
S
2
3
P
2
3 →
=
−
18
Eq.
)
S
S
(
S
S
17
Eq.
T
T
ln
mC
S
S
2
3
4
1
4
1
v
4
1
→
−
−
=
−
→
=
−
At point 2 to 3 (P = C)
c
2
3
2
3
r
V
V
T
T
=
= → Eq. 12
)
r
(
)
r
(
T
T c
1
k
1
3
−
= → Eq. 13
At point 3 to 4 (S = C)
14
.
Eq
)
r
(
T
T
V
V
V
V
V
V
T
T
V
V
)
r
(
)
r
(
T
T
P
P
V
V
T
T
k
c
1
4
1
k
4
1
k
3
2
3
1
k
2
1
k
1
1
4
1
k
4
3
c
1
k
1
4
k
1
k
3
4
1
k
4
3
3
4
→
=
=








=








=








=
−
−
−
−
−
−
−
−
At point 4 to 1 (V = C)
1
4
1
4
P
P
T
T
= → Eq. 15
Entropy Change
a) S during Heat Addition
b) S during Heat Rejection
Thermal Efficiency
19
Eq.
%
100
x
Q
W
e
A
→
=
20
Eq.
%
100
x
Q
Q
Q
e
A
R
A
→
−
=
21
Eq.
100%
x
Q
Q
1
e
A
R
→






−
=
22
Eq.
100%
x
)
T
T
(
k
)
T
T
(
1
e
2
3
1
4
→






−
−
−
=

7. 6
Substituting Eq. 11, Eq. 13 and Eq. 14 to Eq. 22
23
Eq.
%
100
x
)
1
r
(
k
1
r
)
r
(
1
1
e
c
k
c
1
k
→
















−
−
−
= −
Mean Effective Pressure
24
Eq.
KPa
V
W
P
D
m →
=
where:
Pm – mean effective pressure, KPa
W – Net Work KJ, KJ/kg, KW
VD – Displacement Volume m3
, m3
/kg, m3
/sec
Displacement Volume
VD = V1 – V2 m3
→ Eq. 25
VD = 1 - 2 m3
/kg → Eq.26
AIR STANDARD DUAL CYCLE
Processes:
1 to 2 – Isentropic Compression (S = C)
2 to 3 – Constant Volume Heat Addition Q23 ( V = C)
3 to 4 – Constant Pressure Heat Addition Q34 (P = C)
4 to 5 – Isentropic Expansion (S = C)
5 to 1 - Constant Volume Heat Rejection (V = C)
Compression Ratio
1
Eq.
V
V
r
2
1
→
=
V1 = V5 and V2 = V3
Cut-Off Ratio
2
Eq.
V
V
r
3
4
c →
=

8. 7
Pressure Ratio
3
Eq.
P
P
r
2
3
p →
=
P3 = P4
Percent Clearance
4
Eq.
V
V
C
D
2
→
=
5
Eq.
C
C
1
r →
+
=
Displacement Volume
VD = V1 – V2 m3
→ Eq. 6
VD = 1 - 2 → Eq. 7
P, V, and T Relations
At point 1 to 2 (S = C)
( )
8
Eq.
)
r
(
T
T
P
P
r
V
V
T
T
1
k
1
2
k
1
k
1
2
1
k
1
k
2
1
1
2
→
=








=
=








=
−
−
−
−
At point 2 to 3 (V = C)
9
Eq.
r
P
P
T
T
P
2
3
2
3
→
=
=
10
Eq.
)
r
(
)
r
(
T
T P
1
k
1
3 →
= −
At point 3 to 4 (P = C)
11
Eq.
)
r
)(
r
(
)
r
(
T
T
r
V
V
T
T
c
P
1
k
1
4
c
3
4
3
4
→
=
=








=
−
At point 4 to 5 (S = C)
( )
13
Eq.
)
r
(
)
r
(
T
T
12
Eq.
V
V
V
V
r
V
V
T
T
V
V
T
T
P
k
c
1
5
1
k
1
1
k
4
2
4
P
1
k
2
1
k
1
1
5
1
k
5
4
4
5
→
=
→








=








=
−
−
−
−
−
At 5 to 1 (V = C)
14
Eq.
P
P
T
T
1
5
1
5
→
=

9. 8
  22
Eq.
)
T
T
(
k
)
T
T
(
mC
Q
21
Eq.
)
T
T
(
mkC
)
T
T
(
mC
Q
20
Eq.
)
T
T
(
mC
Q
19
Eq.
Q
Q
Q
3
4
2
3
V
A
3
4
v
3
4
p
34
2
3
V
23
34
23
A
→
−
+
−
=
→
−
=
−
=
→
−
=
→
+
=
23
Eq.
)
T
T
(
mC
Q 1
5
v
R →
−
=
25
Eq.
)]
T
-
(T
-
)
T
-
k(T
+
)
T
-
[(T
mC
=
W
24
Eq.
)
Q
-
(Q
=
W
1
5
3
4
2
3
v
R
A
→
→
Entropy change
a) At 2 to 3 (V = C)
15
Eq.
T
T
ln
mC
S
S
2
3
V
2
3 →
=
−
b) At 3 to 4 (P = C)
16
Eq.
T
T
ln
mC
S
S
3
4
p
3
4 →
=
−
S4 – S2 = (S3 – S2) + (S4 – S3) → Eq. 17
c) At 5 to 1 (V = C)
18
Eq.
T
T
n
mC
S
S
5
1
vl
5
1 →
=
−
Heat Added
Heat Rejected
Net Work
Thermal Efficiency
26
Eq.
%
100
x
Q
W
e
A
→
=
27
Eq.
%
100
x
Q
Q
Q
e
A
R
A
→
−
=
28
Eq.
100%
x
Q
Q
1
e
A
R
→






−
=
29
Eq.
100%
x
)
T
T
(
k
)
T
T
(
)
T
T
(
1
e
3
4
2
3
1
5
→






−
+
−
−
−
=
Substituting Eq. 8, Eq. 10 and Eq. 11and Eq. 13 to Eq. 29
30
Eq.
100%
x
)
1
r
(
kr
)
1
r
(
1
r
r
)
r
(
1
1
e
c
P
P
k
c
P
1
k
→
















−
+
−
−
−
= −

10. 9
( )
( )
K
-
kg
KJ
355
.
0
T
T
ln
C
)
S
-
(S
K
072
.
505
)
r
(
T
T
)
r
(
V
V
T
T
KPa
984
.
163
r
P
P
V
P
V
P
KPa
2500
T
T
P
P
K
308
273)
(35
T
KPa
53
.
1524
)
7
(
100
r
P
P
V
P
V
P
K
8
.
670
)
7
(
308
r
T
T
kg
m
758
.
0
V
kg
m
126
.
0
kg
m
126
.
0
r
kg
m
884
.
0
kg
m
884
.
0
P
RT
4
1
v
4
1
1
k
3
4
1
k
1
k
3
4
4
3
k
3
4
k
4
4
k
3
3
2
3
2
3
3
4
.
1
k
1
2
k
2
2
k
1
1
1
4
.
1
1
k
1
2
3
2
1
D
3
3
3
2
2
1
3
4
3
1
1
1
−
=
=
=
=
=








=
=
=
=
=








=
=
+
=
=
=
=
=
=
=
=
=

−

=
=

=



=
=

=
=

−
−
−
−
−
KPa
82
.
219
V
W
P
%
1
.
54
%
100
x
Q
W
e
kg
KJ
56
.
166
Q
Q
W
kg
KJ
4
.
141
)
1
T
4
T
(
Cv
Q
kg
KJ
954
.
307
Q
K
kg
KJ
7175
.
0
1
4
.
1
287
.
0
1
k
R
C
)
T
T
(
C
Q
D
m
A
R
A
R
A
v
2
3
v
A
=
=
=
=
=
−
=
=
−
=
=
−
=
−
=
−
=
−
=
SAMPLE PROBLEMS
1. In an air standard Otto cycle, the compression ratio is 7 and the compression begins at 35C and 100 KPa. The maximum
temperature of the cycle is 1100C. Find
a. The temperature and pressure at each pont in the cycle
b. The heat supplied in KJ/kg of air
c. The work done
d. The mean effective pressure in KPa
e. The entropy change during heat rejection

11. 10
( )
( )
K
25
.
1325
T
V
V
T
T
kg
m
898
.
0
KPa
4
.
423
V
V
P
P
V
P
V
P
KPa
27
.
4431
P
P
K
15
.
2592
T
C
Q
T
)
T
T
(
C
Q
KPa
27
.
4431
r
P
P
V
P
V
P
K
66
.
924
r
T
T
kg
m
838
.
0
V
kg
m
0599
.
0
r
kg
m
898
.
0
kg
m
898
.
0
P
RT
4
1
k
4
3
3
4
3
1
4
k
4
3
3
4
k
4
4
k
3
3
2
3
2
p
A
3
2
3
p
A
k
1
2
k
2
2
k
1
1
1
k
1
2
3
2
1
D
3
2
2
1
3
4
3
1
1
1
=








=
=

=

=








=
=
=
=
=
+
=
−
=
=
=
=
=
=
=

−

=
=



=
=

=
=

−
−
K
-
kg
KJ
7175
.
0
C
K
-
kg
KJ
0045
.
1
C
K
-
kg
KJ
03546
.
1
T
T
Cvln
)
S
-
(S
K
-
kg
KJ
03546
.
1
T
T
ln
C
)
S
-
(S
KPa
55
.
1131
838
.
0
71
.
948
VD
W
Pm
%
6
.
56
%
100
x
Q
W
e
kg
KJ
71
.
948
Q
Q
W
kg
/
KJ
27
.
726
)
T
T
(
C
Q
v
p
4
1
4
1
2
3
p
2
3
A
R
A
1
4
v
R
=
=
−
=
=
=
=
=
=
=
=
=
=
−
=
=
−
=
2. In a diesel cycle, the compression ratio is 15. Compression begins at 100 KPa and 40C. the heat added is 1,675 KJ/kg.
Find
a. The maximum temperature in the cycle
b. The work done in KJ/kg
c. The cycle thermal efficiency
d. The temperature at the end of isentropic expansion
e. The cut – off ratio
f. The MEP of the cycle
g. The entropy change during heat addition and heat rejection process

12. 11
kg
m
0.786
V
22
.
1
P
P
r
02
.
2
r
K
71
.
127
,
1
5720
7000
5
.
921
T
P
P
T
T
kg
m
046
.
0
kg
m
046
.
0
5720
5)
0.287(921.
KPa
5720
)
18
(
100
(r)
P
P
K
5
.
921
)
18
(
290
T
)
r
(
T
T
kg
m
093
.
0
7000
)
2273
)(
287
.
0
(
kg
m
832
.
0
100
)
290
(
287
.
0
K
290
273
17
T
;
KPa
100
P
KPa
7,000
P
P
K
2273
273
2000
T
3
2
1
D
2
3
p
3
4
c
3
2
3
2
3
3
3
2
3
2
4
.
1
k
1
2
1
4
.
1
2
1
k
1
2
3
4
5
1
3
1
1
1
4
3
4
=

−

=
=
=
=


=
=






=
=
=

=

=
=

=
=
=
=
=
=
=
=


=

=
=

=
+
=
=
=
=
=
+
=
−
−
KPa
053
,
1
786
.
0
69
.
827
V
W
P
7455
.
63
%
100
x
Q
W
e
kg
KJ
827.69
Q
Q
W
kg
KJ
470.75
)
290
1
.
946
(
7175
.
0
)
T
T
(
C
Q
kg
KJ
1298.44
1150.44
148
Q
kg
KJ
1150.44
1127.71)
-
3
1.0045(227
)
T
-
(T
C
Q
kg
KJ
148
921.5)
-
7.71
0.7175(112
)
T
-
(T
C
Q
KPa
7
.
325
P
832
.
0
093
.
0
7000
P
P
K
1
.
946
832
.
0
093
.
0
2273
T
T
T
D
m
A
R
A
1
5
v
R
A
3
4
p
34
2
3
v
23
5
4
.
1
k
5
4
4
5
1
4
.
1
5
1
k
5
4
4
5
=
=
=
=
=
=
−
=
=
−
=
−
=
=
+
=
=
=
=
=
=
=
=






=










=
=






=










=
−
−
3. In a dual-combustion cycle the maximum temperature is 2000C and the maximum pressure is 70 Bar. Calculate the cycle
thermal efficiency and the mean effective pressure when the pressure and temperature at the start of compression are 1 Bar
and 17C respectively. The compression ratio is 18. (63.6% 10.53 Bar)

13. 12