Gas Power Cycles
Deal with systems that produce power in which the
working fluid remains a gas throughout the cycle, i.e., no
change in phase.
Internal Combustion (IC) Engine
There are two types of reciprocating engines:
1) Spark ignition – Otto cycle
2) Compression – Diesel cycle
Terminology:
Engine displacement =
(# of cylinders)
x (stroke length)
x (bore area)
(TDC)
(usually given in cc or liters)
(BDC)
Compression ratio =
cylinder volume (BDC)
cylinder volume (TDC
Crank shaft
1 stroke = ½ revolution
of the crank shaft
167
Four stroke Spark Ignition Engine: Cycle consists of
four distinct strokes (processes)
A
FUEL
Intake
I
valve
R
opens
Ignition
Fuel/Air
Mixture
Intake
Stroke
Compression
Stroke
Exhaust
valve
opens
Combustion
Products
Power
Stroke
Exhaust
Stroke
A complete cycle (4 strokes) requires two revolutions of
the crank shaft
x
Ignition
1 atm
168
A comprehensive study of IC engines requires
consideration of the details of the air intake, combustion,
and exhaust processes (MECH 435 – 4th year elective)
Thermodynamic Air-Standard Analysis
Used to perform elementary analyses of IC engines.
Simplifications to the real cycle include:
1) Fixed amount of air (ideal gas) for working fluid
2) Combustion process replaced by constant volume
heat addition with piston at TDC
3) Intake and exhaust not considered, cycle completed
with constant volume heat removal with piston at
BDC
4) All processes considered internally reversible
Air-Standard Otto cycle
Process 1 2
Isentropic compression
Process 2  3
Constant volume heat addition
Process 3  4
Isentropic expansion
Process 4  1
Constant volume heat removal
169
Qin
AIR
Qout
AIR
AIR
TDC
AIR
BDC
(1)  (2)
(2)  (3)
(3)  (4)
(4)  (1)
Qin
Qout
v2
v1
TDC
BDC
TDC
compression ratio
r
V1 V4

V2 V3
since fixed mass
r
v1 v4

v2 v3
BDC
170
Apply First Law ( u 
Q W
 ) to each process:
m m
12 Isentropic Compression
(u 2  u1 ) 
W
Q
 ( in )
m
m
AIR
Win
 (u 2  u1 )
m
vr2
vr1

v2 1

v1 r
R
P2 v2 P1v1
P T v

 2  2 1
T2
T1
P1 T1 v2
23 Constant Volume Heat Addition
(u3  u 2 )  (
Qin W
)
m
m
AIR
Qin
Qin
 (u3  u 2 )
m
v
P
P T
P2
 3  3  3
RT2 RT3
P2 T2
171
3  4 Isentropic Expansion
(u 4  u3 ) 
W
Q
 ( out )
m
m
AIR
Wout
 (u3  u 4 )
m
vr4
v4

r
vr3 v3
P4 v4 P3v3
P T v

 4  4 3
T4
T3
P3 T3 v4
4  1 Constant Volume Heat Removal
(u1  u 4 )  (
Qout W
)
m
m
AIR
Qout
Qout
 (u 4  u1 )
m
P4
P1
P4 P1
v



RT4 RT1
T4 T1
172
Otto cycle thermal efficiency is
cycle 
Wcycle m
Qin m
otto 

Wout m  Win m u3  u 4   u 2  u1 

u3  u2 
Qin m
u3  u2   u4  u1 
u3  u 2
cycle
u 4  u1
 1
u3  u 2
Cold Air-Standard Analysis
The specific heats are assumed to be constant at their
ambient temperature values.
For the two isentropic processes in the cycle, assuming
constant specific heat yields:
12:
34:
T2  v1 
  
T1  v2 
k 1
T4  v3 
  
T3  v4 
k 1
 r k 1
1
 
r
and
k 1
and
k 1
 k
T2  P2
  
T1  P1 
k 1
 k
T4  P4
  
T3  P3 
where k= 1.4 for air at ambient temperature.
173
Furthermore, for constant specific heats u  cV T so
otto
cycle
but
T1  T4  1
T1 
c T  T 
u u
 1 4 1  1 V 4 1  1 
T
u3  u 2
cV T3  T2 
T2  3  1
 T2 
r k 1 
T
T2 T3
T

 4  3
T1 T4
T1 T2
otto
const cV
 1
so
T1
1
 1  k 1
T2
r
For high thermal efficiency want a high compression ratio
Current
spark ICE
7 < r < 10
174
Spark ignition engine compression ratio limited due to a
phenomenon known as “knock”
Knock refers to the noise produced when combustion
occurs explosively  damaging to engine.
In order to reduce knock between 1920 and 1975
tetraethyl lead added to gasoline  r = 12 possible
Also, larger r  higher T2  higher T3  more emissions
such as NOx and Sox
175
Engine Power W (“horse power”)
work work cycles work
cycles crank revs





time cycle sec
cycle crank revs
sec
For a 4 stroke engine, 2 crank shaft revs per cycle
N
W  Wcycle 
2
where N  crank shaft speed (rev/s)
Increase power by increasing the engine speed N
Another way to increase power is to increase the net work
output per cycle Wcycle by either:
(i)
Increasing the compression ratio, or
(ii) Increase Qin (increase the engine displacement).
P
3
(ii)
Qin
Wcycle
4
2
(i)
1
V2
V1
V
176
Another Parameter used to indicate engine performance is
the mean effective pressure (mep) – the theoretical
constant pressure that, if it acted on the piston during the
power stroke, would produce the same net work as
actually developed in one cycle, i.e.,
Wcycle
net work per cycle
mep 

displacement volume V1  V2
For two engines with same displacement volume, the one
with a higher mep produces more net work, and at the
same engine speed would produce more power.
177
Four stroke Compression Engine
No spark plug for ignition, rely on autoignition at high
temperature for initiating combustion
A
I
R
Intake
valve
opens
Fuel Injector
Combustion
Products
Air
Intake
Stroke
Compression
Stroke
Exhaust
valve
opens
Power
Stroke
Exhaust
Stroke
When piston approaches TDC when air is near its
maximum temperature fuel injection is started 
combustion initiated.
Continue to inject fuel as the piston is moving down
(mixture volume is increasing) at the same time liquid
fuel evaporates and reacts as it passes through the
standing flame  combustion occurs at roughly constant
pressure
178
Air-Standard Diesel cycle
Process 1 2
Isentropic compression
Process 2  3
Constant pressure heat addition
Process 3  4
Isentropic expansion
Process 4  1
Constant volume heat removal
Qin
AIR
Qout
AIR
AIR
TDC
AIR
BDC
(1)  (2)
(2)  (3)
(3)  (4)
(4)  (1)
Qin
Qout
v2
v1
TDC
BDC
179
Recall compression ratio r 
v1
v2
Define the cut-off ratio rc 
V3 v3

V2 v2
Apply First Law ( u 
Q W
 ) to each process:
m m
12 Isentropic Compression
(u 2  u1 ) 
W
Q
 ( in )
m
m
AIR
Win
 (u 2  u1 )
m
vr2
v2 1


vr1 v1 r
P2 v2 P1v1
P2 T2 v1
R


 
T2
T1
P1 T1 v2
180
23 Constant Pressure Heat Addition
Qin
P2 V3  V2 
(u3  u 2 )  (
)
m
m
AIR
Qin
Qin
 (u3  P3v3 )  (u 2  P2 v2 )
m
Qin
 ( h3  h2 )
m
T
v
RT2 RT3

 3  3  rc
v2
v3
T2 v2
P
3  4 Isentropic Expansion
(u 4  u3 ) 
W
Q
 ( out )
m
m
AIR
Wout
 (u3  u 4 )
m
vr4
vr3
vr4
vr3

v4
v3

v4 r

v3 rc
note v4=v1 
and
v4 v4 v2 v1 v2 r
    
v3 v2 v3 v2 v3 rc
P4 v4 P3v3
P T r

 4  4
T4
T3
P3 T3 rc
181
4  1 Constant Volume Heat Removal
(u1  u 4 )  (
Qout W
)
m
m
AIR
Qout
Qout
 (u 4  u1 )
m
P4
P1
P4 P1
v



RT4 RT1
T4 T1
Diesel cycle thermal efficiency is
cycle 
Wout m  Win m Qin m  Qout m


m
Qin m
Qin m
Wcycle m
Qin
 Diesel  1 
cycle
Qout m
u u
 1 4 1
Qin m
h3  h2
182
Cold Air-Standard Analysis
The specific heats are assumed to be constant at their
ambient temperature values.
For the two isentropic processes in the cycle, assuming
constant specific heat yields:
T2  v1 
  
12:
T1  v2 
k 1
T4  v3 
  
T3  v4 
k 1
34:
r
k 1
r 
 c
r
and
k 1
and
k 1
 k
T2  P2
  
T1  P1 
k 1
 k
T4  P4
  
T3  P3 
where k= 1.4 for air at ambient temperature.
For constant specific heats u  cV T and h  c P T so
 Diesel
cycle
T1  T4  1
T1 
c T  T 
u u
1
 1 4 1  1 V 4 1  1  
h3  h2
c P T3  T2 
k T  T3  1

2
 T2 
183
recall:
T1
1
 k 1
T2 r
T4  rc 
 
T3  r 
T3
 rc
T2
T4 T4 T3 T2  rc 
    
T1 T3 T2 T1  r 
k 1
k 1
 rc  r k 1  rck
substituting yields
 Diesel
const cV




1  1 rck  1 
 1  k 1  

r  k rc  1 
For given rc higher thermal efficiency is obtained via
higher compression ratio r
For a given r higher thermal efficiency is obtained via
lower cut-off ratio rc
However a smaller rc yields less net work per cycle, so to
achieve the same power need higher engine speeds
184
Compare Efficiency of Otto and Diesel Cycles
Otto
const k
 1
1
r k 1
 Diesel
const k




1  1 rck  1 
 1  k 1  

r  k rc  1 
Current Diesel ICE
12 < r < 23
Note: for a given compression ratio the diesel engine is
less efficient than the spark ignition engine
Otto
const cV
 1
1
r
k 1
 Diesel  1 
const cV
1
r
k 1
 1
185
Compare the Otto cycle (1-2-3O-4O) and the Diesel cycle
(1-2-3D-4D) for:
- the same compression ratio, v1 v2 Otto  v1 v2 Diesel
- the same heat input, 23D Tds  23O Tds (same area
under the curve 2-3 for both cycles)
3O
P= const
V= const
3D
2
4D
4O
V= const
1
186
The heat rejected by the Otto cycle Qout is less than that
for the Diesel cycle:
4O Tds  4 D Tds
1
1
Since Wcycle = Qin – Qout the net cycle work for the Otto
cycle is greater than that for the Diesel since Qout is
smaller for the Otto cycle
Since cycle 
Wcycle
Qin

Otto   Diesel
cycle
cycle
However, since one can use a higher compression ratio
for the Diesel one can achieve obtain a higher cycle
efficiency!
187