2. INTERNAL
COMBUSTION ENGINE
❑ Internal Combustion Engine or I.C.E.
-is an engine where the generation of heat is effected
inside the work producing unit.
-the combustor and the work producing unit are the
same and the products of combustion eventually become
the working fluid.
EXAMPLES:
1. Gasoline Engine
2. Diesel Engine
4. OTTO CYCLE
(S-V-S-V)
❑ It is an idealized thermodynamic
cycle which describes the function of
a typical spark ignition
reciprocating piston engine, the
thermodynamic cycle most found in
automobile engines.
❑ The first person to build a working
four-stroke engine, a stationary
engine using a coal gas-air mixture
for fuel (a gas engine), was German
engineer Nicholas Otto.
6. PROCESSES OF
OTTO CYCLE
1. Isentropic compression (compression stroke)
The gas (fuel-air mixture) is compressed adiabatically from
state 1 to state 2. The surroundings do work on the gas,
increasing its internal energy (temperature) and compressing it.
On the other hand, the entropy remains unchanged. The changes
in volumes and it’s the ratio (V1/V2) is known as the compression
ratio.
2. Isometric compression (ignition phase)
In this phase (between state 2 and state 3) there is a
constant-volume (the piston is at rest) heat transfer to the air from
an external source. This process is intended to represent the ignition
of the fuel–air mixture injected into the chamber and the subsequent
rapid burning.
7. PROCESSES OF
OTTO CYCLE
3. Isentropic expansion (power stroke)
The gas expands adiabatically from state 3 to state 4, as the
piston moves from top dead center to bottom dead center. The gas
does work on the surroundings (piston) and loses an amount of
internal energy equal to the work that leaves the system. Again, the
entropy remains unchanged.
4. Isometric expansion (exhaust stroke)
In this phase the cycle completes by a constant-
volume process in which heat is rejected from the air while the
piston is at bottom dead center. The working gas pressure drops
instantaneously from point 4 to point 1. The exhaust valve opens at
point 4. The exhaust stroke is directly after this decompression.
8. PROCESSES OF
OTTO CYCLE
During the Otto cycle, work is done on the gas by
the piston between states 1 and 2 (isentropic
compression).
Work is done by the gas on the piston
between stages 3 and 4 (isentropic expansion).
The difference between the work done by the gas
and the work done on the gas is the network produced
by the cycle and it corresponds to the area enclosed by
the cycle curve. The work produced by the cycle times
the rate of the cycle (cycles per second) is equal to the
power produced by the Otto engine.
10. ANALYSIS OF
OTTO CYCLE
1. Heat Added, Qa
Qa= m Cv (T3 – T2)
2. Heat Rejected, QR
QR = m *Cv*(T4 – T1)
3. Net Work, Wnet
Wnet = Qa – QR, kW
4. Cycle Efficiency, e
e = WNET / QA
e = 1 −
𝑇4−𝑇1
𝑇3−𝑇2
e = 1 −
1
𝑟𝑘
𝑘−1
9. Mean Effective Pressure, PMEP
PMEP =
WNET
VD
=
WNET
(V1 –V2 )
5. Compression Ratio, rk
rk =
V1
V2
=
V4
V3
rk =
(1 + c)
c
6. Percent Clearance, c
c =
1
(rk− 1)
7. Volume Displacement, VD
VD = (V1 –V2 ) = (V4 –V3)
8. Clearance Volume, VC
VC = c*VD
11. ANALYSIS OF
OTTO CYCLE
The compression ratio - is a value that represents the ratio of
the volume of its combustion chamber from its largest capacity to
its smallest capacity. It is a fundamental specification for many
common combustion engines.
Clearance volume - It is the volume of the combustion chamber
(including head gasket).
MEP = valuable measure of an engine's capacity to do work that
is independent of engine displacement.
14. SAMPLE PROBLEM #1
1. The conditions at the beginning of compression in
an Otto engine operating on hot-air standard with
k=1.34 are 101.3 kPa, 0.039m3, and 32°C. The
clearance is 10% and 12.6 kJ are added per cycle.
Determine (a) 𝑽𝟐, 𝑻𝟐, 𝑷𝟐, 𝑻𝟑, 𝑷𝟑, 𝑻𝟒 &𝑷𝟒 (b) W, (c) e,
and (d) MEP. Use R=0.28708
𝒌𝑱
𝒌𝒈𝒎∙𝑲
15. SAMPLE PROBLEM #1
1. The conditions at the beginning of compression in an Otto
engine operating on hot-air standard with k=1.34 are 101.3
kPa, 0.038m3, and 32°C. The clearance is 10% and 12.6 kJ are
added per cycle. Determine (a) 𝑽𝟐, 𝑻𝟐, 𝑷𝟐, 𝑻𝟑, 𝑷𝟑, 𝑻𝟒 &𝑷𝟒 (b) W,
(c) e, and (d) MEP. Use R=0.28708
𝒌𝑱
𝒌𝒈𝒎∙𝑲
GIVEN:
𝑘 = 1.34
𝑃1 = 101.3 𝑘𝑃𝑎
𝑇1 = 30°𝐶 + 273 = 305𝐾
𝑉1 = 0.038𝑚3
𝑄𝐴 = 12.6 𝑘𝐽
𝑅 = 0.28708
𝑘𝐽
𝑘𝑔𝑚 ∙ 𝐾
16. SOLUTION:
FIRST, WE NEED TO COMPUTE FOR THE MASS & COMPRESSION RATIO.
TO GET T2 WE WILL GET THE RELATION OF TEMPERATURE AND VOLUME @ISENTROPIC PROCESS
(a) @POINT 1:
𝑉2 =?
𝑚 =
𝑃1𝑉1
𝑅𝑇1
=
(101.3 𝑘𝑃𝑎 )(0.038𝑚3)
0.28708
𝑘𝐽
𝑘𝑔𝑚 ∙ 𝐾
(305𝐾)
= 0.04396 𝑘𝑔𝑚
𝑘 = 1.34
𝑃1 = 101.3 𝑘𝑃𝑎
𝑇1 = 30°𝐶 + 273 = 305𝐾
𝑉1 = 0.038𝑚3
𝑅 = 0.28708
𝑘𝐽
𝑘𝑔𝑚 ∙ 𝐾
𝑟𝑘 =
1 + 𝑐
𝑐
=
1 + 0.10
0.10
= 11
𝑟𝑘 =
𝑉1
𝑉2
⇒ 𝑉2 =
𝑉1
𝑟𝑘
=
0.038𝑚3
11
= 0.003455𝑚3
𝑇2
𝑇1
=
𝑉1
𝑉2
𝑘−1
⇒ SINCE 𝑟𝑘 =
𝑉1
𝑉2
⇒ 𝑇2 = 𝑇1 𝑟𝑘
𝑘−1
𝑇2 = 305𝐾 11 1.34−1
= 689𝐾
17. SOLUTION:
TO GET P2 WE WILL GET THE RELATION OF PRESSUREAND VOLUME @ISENTROPIC PROCESS
@POINT 3:
𝑄𝐴 = 𝑚 𝑐𝑣 (𝑇3 − 𝑇2) to get cv 𝑐𝑣 =
𝑅
𝑘 − 1
=
0.28708
1.34 − 1
= 0.8444
𝑘𝐽
𝑘𝑔𝑚 ∙ 𝐾
𝑃2
𝑃1
=
𝑉1
𝑉2
𝑘
⇒ SINCE 𝑟𝑘 =
𝑉1
𝑉2
⇒ 𝑃2= 𝑃1 𝑟𝑘
𝑘 = 101.3𝑘𝑃𝑎 11 1.34 = 2518 𝑘𝑃𝑎
12.6 𝑘𝐽 = 0.04396 𝑘𝑔𝑚 0.8444
𝑘𝐽
𝑘𝑔𝑚 ∙ 𝐾
(𝑇3 − 689𝐾)
𝑇3 = 1028 𝐾
TO GET P3 WE WILL GET THE RELATION OF PRESSUREAND VOLUME @ISOCHORIC PROCESS
𝑃3 = 𝑃2×
𝑇3
𝑇2
= 2518 𝑘𝑃𝑎
1038 𝐾
689𝐾
= 3757𝑘𝑃𝑎
𝑃1
𝑃2
=
𝑇1
𝑇2
18. SOLUTION:
TO GET T4 & P4 WE WILL GET THE RELATION OF PRESSUREAND VOLUME @ISENTROPIC PROCESS
@POINT 4:
𝑇4
𝑇3
=
𝑉3
𝑉4
𝑘−1
⇒ SINCE 𝑟𝑘 =
𝑉4
𝑉3
⇒ 𝑇4 = 𝑇3
1
𝑟𝑘
𝑘−1
𝑇4 = 1028𝑘
1
11
1.34−1
= 455𝐾
𝑃4
𝑃3
=
𝑉3
𝑉4
𝑘
⇒ SINCE 𝑟𝑘 =
𝑉4
𝑉3
⇒ 𝑃4= 𝑃3
1
𝑟𝑘
𝑘
𝑃2= (3757𝑘𝑃𝑎)
1
11
1.34
𝑃2= 151 𝑘𝑃𝑎
