1
Chapter 1
Gas Power Cycles – Reciprocating Engines
A power cycle refers to power generation which is accomplished by a system that
operates on a thermodynamic cycle. The system or device is often called an engine. In
gas cycles, the working fluid remains gaseous throughout the entire cycle. Examples
of gas power cycles are : internal combustion engine (ICE) and gas-turbine engine.
1.1 The Carnot cycle
The Carnot cycle is the most efficient power cycle. It is an ideal cycle with thermal
efficiency expressed as
T
th,Carnot  1  L
TH
when operating between a heat source at temperature TH and a sink at temperature TL.
Thermal efficiency increases with an increase in TH or with a decrease in TL, which
are also true for actual cycle. In practice, the highest temperature in the cycle is
limited by the metallurgical limits of the components (piston, cylinder wall, turbine
blade, etc.). In an actual cycle, the lowest temperature is limited by the temperature of
the cooling medium used in the cycle such as a lake, a river, or the atmospheric air.
1.2 Air-standard assumptions
In internal combustion engines
 energy (heat) is obtained by burning fuel within the system boundaries
 the combustion process changes the composition of the working fluid from air
and fuel to combustion products
All internal combustion engines
 operate on a mechanical cycle; the working fluid does not undergo a complete
thermodynamic cycle
 work on an open cycle
The actual gas power cycles are complex. To simplify the analysis it is common to
utilize the air-standard assumptions:




the working fluid is air, which continuously circulates in a closed loop and
always behaves as an ideal gas
all the processes that make up the cycle are internally reversible
the combustion process is replaced by a heat-addition process from an external
source
the exhaust process is replaced by a heat-rejection process that restores the
working fluid to its initial state
When assuming the air has constant specific heats determined at room temperature of
25 oC, the air-standard assumptions become cold-air-standard assumptions.
A cycle using the air-standard assumptions is known as an air-standard cycle.
2
1.3 Overview of reciprocating engines
A reciprocating engine is a piston-cylinder device which is simple with a wide range
of applications. It is used in automobiles, light aircraft, ships and many other devices.
Intake valve
Exhaust valve
Bore
TDC
piston
stroke
connecting rod
BDC
crankshaft
In a reciprocating engine, the piston reciprocates in the cylinder between top dead
center (TDC) and bottom dead center (BDC). The distance between TDC and BDC is
the stroke of the engine. The diameter of the piston is known as the bore. The air or
air-fuel mixture is drawn into the cylinder through the intake valve, and the
combustion gases are expelled from the cylinder through the exhaust valve.
When the piston is at TDC, the minimum volume formed in the cylinder is called the
clearance volume. The volume displaced by the piston as it moves between TDC and
BDC is known as the displacement volume. The ratio of the maximum volume
formed in the cylinder to the minimum volume is the compression ratio of the engine:
r
Vmax VBDC

Vmin VTDC
(Note : this is not to be confused with the pressure ratio)
The mean effective pressure (MEP) is a fictitious pressure, that if it acted on the
piston during the entire power stroke, would produce the same amount of net work as
that produced during the actual cycle.
Wnet = MEP x Piston area x Stroke = MEP x Displacement volume
or
MEP =
Wnet
wnet

Vmax  Vmin vmax  vmin
………. (kPa)
The engine with a larger value of MEP delivers more net work per cycle and therefore
is a better performing engine.
3
P
Wnet
MEP
V
Vmin
TDC
Vmax
BDC
Fig. 1.1 Graphical representation of the mean effective pressure (MEP)
Reciprocating engines can be classified as:
 Spark-ignition (SI) engines – combustion is initiated by a spark plug
 Compression-ignition (CI) engines – air-fuel mixture is self-ignited as a result
of compressing the mixture above its self-ignition temperature.
The Otto cycle is an ideal cycle for SI engines, whereas the Diesel cycle is an ideal
cycle for CI engines.
1.3.1 Four-stroke engines
Figure 1.2 shows the actual and ideal cycles in four-stroke SI engines and their P-v
diagrams.
1 – 2 Induction stroke: The piston travels from TDC to BDC and air-fuel mixture is
drawn into the cylinder through the intake valve. Note that pressure in the cylinder is
slightly below atmospheric.
2 – 3 Compression stroke: The piston moves from BDC to TDC, compressing the airfuel mixture. Just before reaching TDC, the spark plug fires and the mixture ignites,
increasing the pressure and temperature of the system.
3 – 4 Expansion stroke: The high-pressure gases force the piston down from TDC to
BDC, forcing the crankshaft to rotate, producing useful work output during this stroke
which is also known as power stroke.
4 – 1 Exhaust stroke: The piston moves from DBC to TDC, purging the exhaust gases
through the exhaust valve. During this stroke, the pressure in the cylinder is slightly
above atmospheric.
It takes 4 strokes (and 2 crankshaft revolutions) to complete 1 “thermodynamic” cycle
(actually mechanical cycle).
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1.3.2 Two-stroke engines
In two-stroke engines, the induction and exhaust strokes are part of the compression
stroke, as shown by the timing diagram. On a timing diagram, the angular position is
in terms of crank angle position in relation to TDC and BDC positions.
For clockwise rotation, moving from BDC to TDC (compression stroke), induction,
exhaust and transfer (air-fuel mixture from crankcase to cylinder) processes are part
of the compression stroke.
Similarly, during the expansion stroke (moving from TDC to BDC), induction,
exhaust and transfer (air-fuel mixture from crankcase to cylinder) processes are also
part of the expansion stroke.

Two-stroke engines are less efficient than four-stroke engines due to
incomplete expulsion of exhaust gases and partial expulsion of fresh air-fuel
mixture with exhaust gases
Two-stroke engines are relatively simple and inexpensive
Two-stroke engines have high power-to-weight and power-to-volume ratios,
which make them suitable for applications that require small size and weight
(e.g. motorcycles, chain saws, lawn mowers etc.)


1.4 Otto cycle – ideal cycle for SI engines
The Otto cycle is the ideal cycle for SI reciprocating engines. Most SI engines are
four-stroke internal combustion engines. Figure 1.3 shows the Otto cycle on a P-v and
T-s diagrams. This cycle closely resembles the actual four-stroke or two-stroke cycles
of SI engines.
P
T
3
3
.
qin
isentropic expansion
qin
v=C
2
4
2
4
qout
isentropic compression
qout
1
v=C
1
v
TDC
BDC
Figure 1.3 P-v and T-s diagrams for Otto cycle
The (air-standard) Otto cycle consists of 4 internally reversible processes:
1–2
2–3
3–4
4–1
Isentropic compression
Constant-volume heat addition
Isentropic expansion
Constant-volume heat rejection
s
5
The cycle is executed in a closed system, and ignoring changes in kinetic and
potential energies, the energy balance for any of the processes, on a unit-mass basis is
qin  qout  ( win  wout )  u ……… (kJ/kg)
Processes 2 – 3 and 4 – 1 take place at constant volume, thus the work terms are zero.
Therefore,
qin  u3  u 2  cv T3  T2 
and
qout  u 4  u1  cv T4  T1 
Using cold air standard assumptions, the thermal efficiency of the ideal Otto cycle is
th,Otto 
wnet qin  qout
q
T T

 1  out  1  4 1
qin
qin
qin
T3  T2
Processes 1 – 2 and 3 – 4 are isentropic, and v2 = v3 and v4 = v1. Therefore,
T1  v 2 
 
T2  v1 
k 1
v 
 3
 v4 
k 1

T4
T3
i.e.,
T2  T1 r k 1
where r is the compression ratio ( r 
 th,Otto  1 
and
T3  T4 r k 1
cp
vmax v1
 ) and k 
. Thus,
cv
vmin v2
T4  T1
T4  T1
1
1
 1  k 1
k 1
k 1
T3  T2
T4 r  T1 r
r
The thermal efficiency increases with both r and k. This is also true for actual SI
internal combustion engines. For a given r, the thermal efficiency of an actual cycle is
less than that of the ideal Otto cycle due to irreversibilities (friction) and incomplete
combustion in real engines. Typical values of r for real SI engines range from 7
through 10.
Example 1a (constant sepcific heats)
The compression ratio of an ideal Otto cycle is 8. At the start of compression, air is at
100 kPa, 17 C, and 800 kJ/kg of heat is transferred to the air during the constantvolume heat addition process. Assuming cv = 0.718 k and k = 1.4, determine (a) the
maximum temperature and pressure in the cycle, (b) the net work output, (c) the
thermal efficiency, and (d) the mean effective pressure for the cycle.
k 1
v 
(a) T2  T1  1   (17  273)8 0.4 = 666.3 K
 v2 
qin  u3  u 2  cv T3  T2 
800 = 0.718 (T3  666.3)
T3 = 1780.5 K
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k
v 
P2  P1  1   100(8)1.4 = 1837.9 kPa
 v2 
T 
1780.5 
= 4911.3 kPa
P3  P2  3   1837.9
 666.3 
T2 
k
Pmaz
v 
1 
P4  P3  3   4911.3  = 267.2 kPa
8 
 v4 
P 
 267.2 
T4  T1  4   290
 = 774.9 K
P
100

 1
1.4
(b) qin  = 800 kJ/kg (given)
qout  u 4  u1  cv T4  T1  = 0.718(774.9 – 290) = 348.2 kJ/kg
wnet = qin – qout = 800 – 348.2 = 451.8 kJ/kg
w
451.8
(c) Otto  net 
= 0.565
qin
800
RT
0.287(290)
(d) v1  1 
= 0.8323 m3/kg
P1
100
v
0.8323
= 0.1040 m3/kg
v2  1 
r
8
wnet
451.8
= 620.3 kPa
MEP 

v1  v2 0.8323  0.1040
Example 1b (variable specific heats)
The compression ratio of an ideal Otto cycle is 8. At the start of compression, air is at
100 kPa, 17 C, and 800 kJ/kg of heat is transferred to the air during the constantvolume heat addition process. Accounting for the variation of specific heats of air
with temperature, determine (a) the maximum temperature and pressure in the cycle,
(b) the net work output, (c) the thermal efficiency, and (d) the mean effective pressure
for the cycle.



Assume air standard assumptions
Kinetic and potential energy changes are negligible
Specific heats vary with temperature (use table for air).
P
3
.
isentropic expansion
qin
qin = 800 kJ/kg
r =8
2
4
qout
isentropic compression
1
v
TDC
BDC
P1 = 100 kPa
T1 = 290 K
7
T1 = 290 K
(using Table A-17)
Process 1 – 2 is isentropic,
u1 = 206.91 kJ/kg, vr1 = 676.1
v2 vr 2 1


v1 v r1 r
v r1 676 .1
= 84.51 (Table A-17)
T2 = 652.4 K, u2 = 475.11 kJ/kg

r
8
 T  v 
p v
pv
 652.4 
P2  P1  2  1  = 100
R 2 2  1 1
8 = 1799.7 kPa
T2
T1
 290 
 T1  v 2 
vr 2 
or
Process 2 – 3 is constant-volume heat addition:
qin = u3 – u2
800 = u3 – 475.11
u3 = 1275.11
(Table A-17) T3 = 1575.1 K
R
T
P3  P2  3
 T2
p 3 v3 p 2 v 2

T3
T2
, vr3 = 6.108
 v 2 
 1575 .1 
  = 1799 .7
1 = 4345 kPa
 652 .4 
 v3 
(b) Process 3 – 4 is isentropic expansion,
v4 vr 4

r
v3 v r 3
v r 4  rv 3  8 (6.108) = 48.864
Table A-17 gives
T4 = 795.6 K, u4 = 588.74 kJ/kg
qout = u4 – u1 = 588.74 – 206.91 = 381.83 kJ/kg
wnet,out = qnet,in = qin – qout = 800 – 381.83 = 418.17 kJ/kg
(c) th 
wnet ,out
(d) v1 
RT1 0.287(290)
= 0.832 m3/kg

P1
100
qin
MEP 

418 .17
= 0.523 @ 52.3 %
800
wnet
wnet
wnet
418 .17
= 574 kPa



v
v1  v 2
 1
 1
1
v1 
v1 1   0.8321  
r
 r
 8
i.e A constant pressure of 574 kPa during the power stroke would produce the
same amount of work output as the entire cycle.
1.5 Diesel Cycle – ideal cycle for compression-ignition engines
In SI (gasoline) engines, air-fuel mixture is compressed to a temperature below the
autoignition temperature of the fuel, and the combustion process is initiated by firing
a spark plug. In CI (i.e. diesel) engines, the air is compressed to a temperature above
the autoignition temperature of the fuel, and combustion takes place as the fuel is
injected into this hot air. The spark plug and carburetor (in SI engines) is replaced
with a fuel injector in diesel engines.
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The compression ratios in SI engines are limited by the onset of autoignition or
engine knock. However, diesel engines can operate at higher compression ratios
(typically between 12 and 24), because only air is compressed during the compression
stroke.
Fuel is injected in diesel engines when the piston approaches TDC and continues
during the first part of the power stroke. That is, the combustion takes place over a
longer interval. Thus, the combustion process in ideal Diesel cycle is modelled as a
constant-pressure heat-addition process. (note that in Otto cycle heat addition is at
constant volume).
The Otto and Diesel cycles differ only on the heat-addition process. The other
processes are the same for both ideal cycles.
P
qin
2
T
P=C
3
.
3
qin
2
isentropic expansion
V=C
4
qout
isentropic compression
4
qout
1
1
TDC
v
BDC
P-v and T-s diagrams for ideal Diesel cycle
s
The Diesel cycle is executed in a piston-cylinder device, which forms a closed system.
Th cycle consists of 4 internally reversible processes:
1–2
2–3
3–4
4–1
Isentropic compression
Constant-pressure heat addition
Isentropic expansion
Constant-volume heat rejection
The energy balance for process 2 – 3 is written as
qin  wb ,out  u 3  u 2
qin  P2 v3  v2   u3  u 2 
= h3  h2  c p T3  T2 
The heat rejection (4 – 1) is similar to the ideal Otto cycle
qout  u 4  u1  cv T4  T1 
The thermal efficiency of the ideal Diesel cycle under the cold-air-standard
assumptions is
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th, Diesel 
wnet qin  qout
q
c T  T 

 1  out  1  v 4 1
qin
qin
qin
c p T3  T2 
= 1
T4  T1 
k T3  T2 
The cutoff ratio rc is defined as
rc 
T2  T1 r k 1
P3 v3 RT3

P2 v 2 RT 2
process 1 – 2 :
process 2 – 3 :
V3 v3

V2 v 2
T3 v3

 rc
T2 v2
gives
T3  rc T2  rc T1 r k 1
process 3 – 4 : Tv
k 1
v 
T
 C …… 4   3 
T3  v 4 
r 
T4  T3  c 
r
Therefore,
 th, Diesel = 1 
When rc
1,
k 1
 rcT1r
T4  T1 
k T3  T2 
 th, Diesel
k 1
k 1
 rc 
 
r
v 
  3 
 v1 
k 1
  v3
 
v
  2
 v
  1
  v 2

 


 
 
k 1
r 
 c 
r
k 1
k 1
 rck T1
1  r k 1 
1  rck  1 



= 1   k c1

1

k  rc r  r k 1 
r k 1  k rc  1 
1
1  k 1 (see Richard Stone, “Introduction to
r
internal combustion engines”)
Example 2
An ideal Diesel cycle has a compression ratio of 18 and a cutoff ratio of 2. At the start
of the compression process, the working fluid is at 0.1 Mpa and 27 oC. Utilizing the
cold-air-standard assumptions, determine (a) the thermal efficiency, and (b) the mean
effective pressure.
Data: P1 = 100 kPa, T1 = 300 K, r 
v
v1
 18 , rc  3  2
v2
v2
k 1
v 
(a) T2  T1  1   300(18) 0.4 = 953.3 K
 v2 
v
T3  T2 3  953.3(2)  1906.6 K
v2
v 
T4  T3  3 
 v4 
k 1
 v3 
 v 
2

 T3 
v
 4 
 v2 
k 1
2
 1906.6 
 18 
0.4
= 791.4 K
10
qin = cp (T3 – T2) = 1.005 (1906.6 – 953.3) = 958.1 kJ/kg
qout = cv (T4 – T1) = 0.718 (791.4 – 300) = 352.5 kJ/kg
qin  qout 958.1  352.8
= 0.632 @ 63.2 %

qin
958.1
RT
0.287(300)
(b) v1  1 
= 0.8610 m3/kg
P1
100
v
0.8610
= 0.0478 m3/kg
v2  1 
r
18
wnet
958.1  352.8
= 744.3 kPa
MEP 

v1  v2 0.8610  0.0478
 th, Diesel =