Applied
Thermodynamics
1
1. Air Standard Power Cycles
Introduction
Two important applications of thermodynamics are
power generation and refrigeration.
Both are usually accomplished by systems that operate
on thermodynamic cycles.
Hence the thermodynamic cycles are usually divided into
two general categories, viz., “power cycles” and “
refrigeration cycles”;
Power or refrigeration cycles are further classified as “
gas cycles” and “ vapour cycles” ;
2
In case of gas cycles, the working substance will be in
gaseous phase throughout the cycle, where as in
vapour cycles, the working substance will be in liquid
phase in one part of the cyclic process and will be in
vapour phase in some other part of the cycle;
Thermodynamic cycles are also classified as
“
closed cycles” and “ open cycles”.
In closed cycles, the working fluid is returned to its
original state at the end of each cycle of operation
and is recirculated.
3
In an open cycle, the working substance is renewed at
the end of each cycle instead of being re-circulated.
In automobile engines, the combustion gases are
exhausted and replaced by fresh air-fuel mixture at the
end of each cycle.
Though the engine operates in a mechanical cycle, the
working substance does not go through a complete
thermodynamic cycle.
4
Basic Considerations in the Analysis of Power Cycles
The cycles encountered in actual devices are difficult to
analyze because of the presence of friction, and the
absence of sufficient time for establishment of
equilibrium conditions during the cycle.
In order to make an analytical study of a cycle feasible, we
have to make some idealizations by neglecting internal
Irreversibilities and complexities.
Such cycles resemble the actual cycles closely but are
made up of internal reversible processes.
These cycles are called ideal cycles.
5
Air Standard Cycles
In gas power cycles, the working fluid will be in gaseous
phase throughout the cycle.
Petrol engines (gasoline engines), diesel engines and gas
turbines are familiar examples of devices that operate on
gas cycles.
All these devices are called “ Internal combustion
engines” as the fuel is burnt within the boundaries of the
system.
Because of the combustion of the fuel, the composition of
the working fluid changes from a mixture of air and fuel
to products of combustion during the course of the cycle.
6
However, considering that air is predominantly
nitrogen which hardly undergoes any chemical
reaction during combustion, the working fluid closely
resembles air at all times.
The actual gas power cycles are complex.
In order that the analysis is made as simple as
possible, certain assumptions have to be made.
These assumptions result in an analysis that is far from
correct for most actual combustion engine processes,
but the analysis is of considerable value for indicating
the upper limit of performance.
7
Air standard assumptions
1.
2.
3.
4.
5.
The working medium is a perfect gas with constant specific
heats and molecular weight corresponding to values at room
temperature.
No chemical reactions occur during the cycle.
The
heat addition and heat rejection processes are merely heat
transfer processes.
The processes are reversible.
Losses by heat transfer from the apparatus to the atmosphere
are assumed to be zero in this analysis.
The working medium at the end of the process (cycle) is
unchanged and is at the same condition as at the beginning of
the process (cycle).
i.e Changes in kinetic and potential energies of the working
substance are very small and hence negligible.
8
Air standard Carnot Cycle
The Carnot cycle is represented on P-v
and T-s diagrams as in Fig.
The Carnot cycle is composed of four
totally reversible processes: isothermal
heat addition, isentropic expansion,
isothermal heat rejection, and isentropic
compression.
The Carnot cycle is the most efficient
cycle that can be executed between a
heat source at temperature and a sink
at temperature , and its thermal
efficiency is expressed as
9
Process 1 – 2: Reversible Adiabatic Compression
Process 1-2: In this, air is compressed isentropically from volume
During this process heat rejected is zero. i.e.,
P: Increases from p1 to p2
V: Decreases from V1 to V2
T: Increases from T1 to T2
S: Remains same.
1W2
= P V  P V=
1 1
2 2
 1
1Q2
mR (T1  T 2 )
 1
= 0 or
10
Process 2 -3: Isothermal Heat Addition
In this air is heated isothermally
so that volume increases and
Temperature remains constant.
Amount of heat supplied is equal
to the work done by the gas.
P: Decreases from p2 to p3
V: Increases from V2 to V3
T: Remains same.
S: Increases from S2 to S3
2W3=
2Q3
p2V2 ln
= mRT2 ln
= p2V2 ln
11
Process 3 – 4:
Reversible Adiabatic Expansion
This is isentropic(Adiabatic) expansion
process.
Heat supplied during the process is zero. i.e.,
P: Decreases from p3 to p4
V: Increases from V3 to V4
T: Decreases from T3 to T4
S: Remains same.
3W4
= P3V 3  P4 V=4
mR (T3  T 4 )
=0 1
 1
3Q4
12
Process 4 – 1:
Isothermal Heat Rejection
P: Increases from p4 to p1
V: Decreases from V4 to V1
T: Remains same.
S: Decreases from S4 to S1
4W1=
4Q1
p4V4 ln
= mRT4 ln
= p4V4 ln
13
14
15
And also,
 th 

R ln
 r T max  T min 
R ln  r  T max
T max  T min
T max
16
17
Mean Effective Pressure
Mean effective pressure may
be defined as the theoretical
pressure which, if it is maintained
constant throughout the volume
change of the cycle, would give the
same work output as that obtained from the cycle.
Or it is the constant pressure which produces the same work output
while causing the piston to move through the same swept volume
as in the actual cycle.
18
Mean effective Pressure:
When the piston moves from TDC to BDC, the air inside expands
resulting in work output. If Pm1 is the average pressure on the piston during
this stroke, the average force on the piston is
Where d = diameter of piston or cylinder bore
Work output = average force on piston X stroke length
During the return stroke, as the piston moves from BDC to TDC, air is
compressed requiring work input of the average pressure on the piston
during this stroke is Pm2, the work input is given by;
Where Pm is known as mean effective pressure and
is the swept volume.
Usually the net work output is in kJ, volume in m3 and mean effective
pressure in bar.
19
Stirling cycle
When a confined body of gas (air, helium, whatever) is
heated, its pressure rises.
This increased pressure can push on a piston and do work.
The body of gas is then cooled, pressure drops, and the
piston can return.
The same cycle repeats over and over, using the same
body of gas.
That is all there is to it. No ignition, no carburetion, no valve
train, no explosions.
Many people have a hard time understanding the Stirling
because it is so much simpler than conventional internal
combustion engines.
20
Stirling Cycle:
The Stirling cycle is represented
on P-v and T-s diagrams as in Fig.
It consists of two isothermal
processes and two isochors.
Process 1-2: In this air is heated
isothermally so that volume
increases from Temperature
remains constant.
Amount of heat supplied is equal
to the work done by the gas.
21
Stirling Cycle:
Process 2-3: This is constant volume
heat rejection process. Temperature
decreases from pressure decreases
from
the heat rejected during the
process is given by,
Process 3-4: In this air is compressed
isothermally from volume
During this process heat rejected is
equal to the work done by the gas.
22
Stirling Cycle:
Process 4-1:
This is constant volume heat
addition process.
Temperature increase from
The heat added during the process
is given by,
23
24
The Efficiency of the cycle:
Due to heat transfers at constant volume processes, the
efficiency of the Stirling cycle is less than that of the
Carnot cycle.
However if a regenerative arrangement is used such that,
i.e., the area under 2-3 is equal to the area
under 4 -1 on T-s diagram, then the efficiency,
25
Otto cycle OR Constant
volume cycle:
The Otto cycle is the ideal
cycle for spark-ignition reciprocating
engines.
It is named after Nikolaus A. Otto, who
built a successful four-stroke engine in
1876.
This cycle is also known as constant
volume cycle as the heat is received and
rejected at constant volume.
The cycle consists of two adiabatic
processes and two constant volume
processes as shown in P-v and T-s
diagrams.
26
Otto cycle OR Constant
volume cycle:
Process 1-2:
In this air is compressed
isentropically from V1 to V2 Temperature
increases from T1 to T2.
Since this is an adiabatic process heat
rejected is zero. i.e.
Process 2-3:
In this air is heated at constant volume
and temperature increases from T2 to T3.
Heat supplied during this process is given
by,
27
Otto cycle OR Constant
volume cycle:
Process 3-4:
In this air is expanded isentropically from
V3 to V4 and temperature decreases from
T3 to T4.
Since this is an adiabatic
process, the heat supplied is zero. i.e.,
Process 4-1:
In this air is cooled at constant volume
and temperature decreases from T4 to
T1. Heat rejected during this process is
equal to change in internal energy and is
given by,
28
The Efficiency of the cycle:
.
Efficiency of the cycle is given by,
Considering isentropic expansion process 3-4,
Or
Considering isentropic compression process 1-2,
Or
Substituting for
in eqn (1)
Or
1
Where, r = compression OR expansion ratio and
29
Mean effective pressure:
We know that for Otto cycle,
the pressure ratio
30
31
Diesel cycle OR Constant pressure cycle:
The Diesel cycle is the ideal cycle for
Compression Ignition reciprocating
engines.
The CI engine was first proposed by
Rudolph Diesel.
The Diesel cycle consists of one constant
pressure heating process, one constant
volume cooling process and two
adiabatic processes as shown in P-v and
T-s diagrams.
This cycle is also known as constant
pressure cycle because heat is added at
constant pressure.
32
Diesel cycle OR Constant pressure cycle:
Process 1-2:
During this process air is compressed
adiabatically and volume decreases
from V1 to V2 Heat rejected during
this process is zero. i.e.,
Process 2-3:
During this process air is heated at
constant pressure and temperature
rises from T2 to T3 Heat supplied
during this process is given by,
33
Diesel cycle OR Constant pressure cycle:
Process 3-4:
During this process air is expanded
adiabatically and volume increases from
V3 to V4.
Heat supplied during the process is zero.
i.e.,
Process 4-1:
In this air is cooled at constant volume
and temperature decreases from T4 to
T1 .
Heat rejected during this process is given
by,
34
The Efficiency of the cycle:
The efficiency of the cycle is given by,
Considering process 1-2,
Let, compression ratio,
Cut-off ratio,
Expansion ratio,
35
Considering process 2-3,
Considering process 3-4,
Substituting for
in eqn (1), we get
36
Mean effective pressure:
we know that work done per kg in Diesel cycle is given by,
And the mean effective pressure is given by:
37
Expression for cut-off ratio:
Let ‘k’ be the cut-off in percentage of stroke (from
We know that,
38
Dual combustion or Limited pressure or Mixed cycle:
This cycle is a combination of Otto
and Diesel cycles.
It is also called semi-diesel cycle
because semi-diesel engines work
on this cycle.
In this cycle heat is absorbed partly
at constant volume and partly at
constant pressure.
It consists of two reversible
adiabatic or isentropic, two
constant volume and a constant
pressure processes as shown in P-v
and T-s diagrams.
3
4
5
39
Dual combustion or Limited pressure or Mixed
cycle:
4
3
Process 1-2:
The air is compressed
reversibly and adiabatically from
temperature T1 to T2 .
No heat is rejected or absorbed
by the air.
Process 2-3:
The air is heated at constant
volume from T2 to T3.
Heat absorbed by the air is given
by,
5
40
Dual combustion or Limited pressure or Mixed cycle:
Process 3-4:
4
3
The air heated at constant pressure from
temperature T3 to T4.
The heat supplied by the fuel or heat
absorbed by the air is given by,
5
Process 4-5:
The air is expanded reversibly and
adiabatically from temperature T4 to T5 .
No heat is absorbed or rejected during
the process.
Process 5-1: The air is now cooled at
constant volume from temperature T5 to
T1 . Heat rejected by the air is given by,
41
The Efficiency of the cycle:
The efficiency of the cycle is given by,
5
Let, compression ratio,
Cut-off ratio,
Pressure ratio,
Expansion ratio,
42
Considering process 1-2,
Considering process 2-3,
Considering process 3-4,
Considering process 4-5,
Substituting for
in (1)
43
Mean effective pressure:
We know that work done per kg in dual cycle is given by,
And the mean effective pressure is given by:
Note:
1) For Otto cycle
2) For Diesel cycle
44
Comparison between Otto, Diesel and Dual
combustion cycles
The important variables which are used as the basis
for comparison of the cycles are compression ratio,
peak pressure, heat supplied, heat rejected and the
net work output.
In order to compare the performance of the Otto,
Diesel and Dual combustion cycles some of these
variables have to be fixed.
45
Comparison with same compression ratio and
heat supply:
46
The comparison of these cycles
for the same compression ratio
and same heat supply are
shown in on both p – V and T –
S diagrams.
In these diagrams, cycle 1-2-34-1 represents Otto Cycle,
cycle 1-2-3’-4’-1 represents
diesel cycle and cycle 1-2”-3”4”-1 represents the dual
combustion cycle for the same
compression ratio and heat
supply.
47
From the T-S diagram, it
can be seen that area
5236 = area 522”3”6” =
area 523’6’ as this area
represents the heat
supply which is same for
all the cycles.
All the cycles start from
the same initial point 1
and the air is compressed
from state 1 to state 2 as
the compression ratio is
same.
48
It is seen from the T-s
diagram, that for the same
heat supply, the heat
rejection in Otto cycle (area
5146) is minimum and heat
rejection in Diesel cycle
(area 514’6’) is maximum.
Consequently Otto cycle has
the highest work output and
efficiency. Diesel cycle has
the least efficiency and dual
cycle has the efficiency
between the two.
49
Therefore for the same compression ratio and same heat
rejection, Otto cycle is the most efficient while the Diesel cycle is
the least efficient.
It can also be seen from the same diagram that q3>q2>q1
We know that thermal efficiency is given by
1 – heat rejected/heat supplied
Thermal efficiency of these engines under given circumstances is
of the following order
Diesel>Dual>Otto
Hence in this case it is the diesel cycle which shows greater
thermal efficiency.
50
Problem 1
In an Otto cycle, the upper and lower limits for
the absolute temperature respectively are T1
and T2.
Show that for the maximum work, the ratio of
compression should have the value
rc 





T
T
1 . 25

3


1
51
Solution:
Process 1-2 is reversible adiabatic





T2
V

T1 V

1 


2 
 
 rc  
T 2  T1rc  
.......... ....(1)
Process 3-4 is reversible adiabatic





T4
V

T3 V
T4 

1 


2 
 
 rc  
T3
 

T
r
3 c
rc  
.......... (2)
52
Work done = Heat added - Heat rejected

C   T 3 - T 2  - C   T 4 - T1 

C   T 3 - T1rc    - C   T 3 rc1   - T1 








In the above equation T3, T1 and Cv are constants.
Therefore for maximum work
dW
dr
 0
53
d C T - T r     - C T r 1   - T    0

 3
 3 c

1 c
1  




dr 
- C      T1rc     - C  1     T 3 rc1       0





- C      T1rc  2 - C  1     T 3 rc 







 0
T1rc  2  T 3 rc 





T2
T1





rc  2

rc 
 rc      rc  
 rc    1
rc
rc







T3
T1





1
   1 




T3
T1
1 . 25











T3
T1
1
   . 4 1 




54
Problem 2
An engine working on Otto cycle in which
salient points are 1,2,3 and 4 has upper and
lower temperature limits T3 and T1.
If the maximum work per kg of air is to be
done,
show
that
the
intermediate
temperatures are given by
T2
 T4

T1T 3
55
Solution: For maximum work/kg in an Otto cycle
•
rc 





T 3 

T1 
1
  1 
T2  T r
 
1 c


1


T
 T
T
12

3


1
(as proved in problem 1)
 T





1








T 3 

T1 
1 
  1  






 
T1T 3
56
T4

T3
rc  
T3















T3
T1





1

  1  







 
Again


3


T1 

T 3 
 T4

 T
T2
1


T1T 3
T1T 3
57
Problem 3
An engine working on the otto cycle has a suction
pressure of 1 bar and a pressure of 14 bar at the end
of compression.
Find Compression ratio, Clearance volume as a
percentage of cylinder volume
The ideal efficiency and MEP if the pressure at the end
of combustion is 21 bar.
Solution:
Given: P1 = 1 bar, P2 = 14 bar, P3 = 21 bars
58


v
P2
 1 
v2 
P1












2 


1 
P
v1
 rc 
v2
P
1

 14 

1


v1
v1
rc 

v2 vc
v2
1
x 100 
x100
v1
6.58
 15.18%
Ideal efficiency
  1   1
rc  
 1
1
6 . 58  . 4  
 53%
59
   Explosion
M .E .P 

pressure
p3
ratio  
p2
 21  1.5
14
p1        rc    1 






     rc  1 

1x6.58(1.5

-1)(6.58
1.4 - 1
 1)
(1 . 4  1)( 6 . 58  1)
 1.65 bar
60
Problem 4
In a constant volume cycle the pressure at the end of compression is 15
times that at the start, the temperature of air at the beginning of
compression is 37° C and the maximum temperature attained in the
cycle is 1950°C. Find,
(i) the compression ratio
(ii) thermal efficiency of the cycle
(iii) heat supplied per kg of air
(iv) the work done per kg of air
Solution:
Given:
P2/P1 = 15 , T1 = 37ºC = 310 K
T3 = 1950ºC = 2223 K
61
P2

P1




rc 






v1
v2











P2
P1
 r c
1





15



1

rc  6.91
 1  
1
rc
 1

1
 0.54
 .4  
6 . 91
 54%
T 2  T1rc   (31096.91)
1.4 -1
 671.66
K
62
Heat supplied = Cv(T3-T2) = 0.72(2223 - 671.66)
=1116.96 KJ/kg of air

Work done
Heat supplied
Work done = 0.54 x 1116.96
= 603.16 KJ/kg of air
63
Problem 5
An air standard Diesel cycle has a compression ratio
of 18 and the heat transferred to the working fluid
per cycle is 2000 kJ/kg.
At the beginning of the compression stroke, the
pressure is 1 bar and the temperature is 300 K.
Calculate the thermal efficiency.
Given:
rc = 18
P1 = 1 bar
T1 = 300 K
T 2  T1rc   300(18)
 953.3 K
1.4 -1
64
Heat transferred = Cp(T3 – T2)
2000 = 1.005(T3 -953.3)]
T3 = 2943.34 K
T3
2943.34
cut off ratio   

T2
953.3
 3.08
rc       1 
 1  


   

18    . 4   . 08   1 


 1  


 08   

 0.586  58.6%
65
Problem 6
An engine with 200 mm cylinder diameter and 300
mm stroke length, works on the theoretical Diesel
cycle. The initial pressure and temperature of air are
1 bar and 27° C. The cut off is at 8% of the stroke and
compression ratio is 15. Determine
(i) Pressure and temperatures at all salient points of the
cycle.
(ii) theoretical air standard efficiency.
(iii) mean effective pressure.
(iv) power developed if there are 400 working strokes
per minute.
66
Solution:
Given:
rc = 15,
P1 = 1 bar,
T1 = 27º C
d = 200 mm, L = 300 mm
Swept volume V s   dL

V s   0 . 2 2 x .3

 .009424 m 3
67
V 2  V c  clearance
rc 
Vs
Vc
V1

Vc  Vs
V2
volume
1
Vc
Vs
Vc
 (r c  1)  (15 - 1)  14
V 2  Vc 
Vs

0.009424
14
 0.0006731
m
2
14
Cut off takes place at 8% of stroke
(V 3  V 2 ) 
8
100
Vs
V 3  V 2  0 . 08 V s  0 . 0006731  0 . 08 x 0 . 009424
 0.001427
m
3
68
cut off ratio     
V3
V2
r
 1  
1  
c


0.001427
 2.12
0.0006731
1   .4
15
    1

  1 

    
  . 12   1 


  . 12   
 0.598  59.8%

p r
  




1
c
M .E .P 
      r
   1  
  
 c 
     r  1 


c

1 .4
 . 4   . 12    15 1  1 .4   . 12   1   




  

 .    15  1  
1 x15
M . E . P  7.14 bar  7.41 x 10 kPa
2
69
work done cycle
M .E .P 
Swept volu
me
Work done / cycle  7.41 x 10 x 0.00942  6.98 kJ/cycle
2
Power Work
done / cycle x Number
Power  6.98 x
of cycles / sec
400
60
Power  46.53 kW
 
T T r
2
1c

 300x15 1.4 -1  886 . 25 K
P  P r  1x15
2
1c
1.4
P  P  44.3 bar
2
3
 44 . 3 bar
70
  


       886.25 x 2.12  1878.85 K


 V
 
 V




 1
1

  x 
r

 V V
 
x
 V V
 V
 
P  V 
P




 V V
 
x
 V  V




 1
sin ce V 4  V1
 1
 2 . 12 
   1878 . 85 

 15 





 1
1 . 4 1
 V V
 
x
 V V
 2 . 12 
P  44 . 3 

 15 
 858.99 K





 V V
 
x
 V  V





 
  
 rc 

1 .4
 2 . 86 bar
71
Problem 7
In a dual combustion cycle the compression ratio is
14, maximum pressure is limited to 55 bar.
The cut-off ratio is 1.07. Air is admitted at a pressure
of 1 bar. Find the thermal efficiency and M.E.P of
the cycle.
Solution: (Given):
rc = 14
P1 = 1 bar
P3 = 55 bar
Cut off ratio =  =1.07
72

P2  P1 rc  1(14 )
1 .4
 40.23 bar
Explosion
pressure
ratio
 
P3
P2
  1
1
 1
rc
14
 1 . 367
40 . 23



  1


 (  1)   (   1) 
1
 1

55
1 . 4 1
1 .4


1 . 367 x 1.07
1


 (1 . 367  1)  1 . 367 x 1.4 (1 . 07  1) 
 62.18%
73
Heat added  C p (T 4  T 3 )  C  (T 3  T 2 )
 T4

 T3

 C p T 3 
 1   C  T 2 
 1 
 T2

 T3

 1
T 2  T1 rc
Pr ocess 2 - 3 is contant vo
P 2V 2

T2
T3  T 2
lume process
P3V 3
T3
P3
P2
 1
 T 2   T1 rc
74
 1
 Heat added  C p  T1 rc

 1   C  T1 rc
 1
 1.005 x 1.367 T1 14 
1 .4 1
 1.0356T

1 . 07
 1
 1   0 . 72 T114
1 .4 1
1 . 367
1
Work done  Heat added x   1.0356 T1 x 0 . 6218
 0.6439 T1
Swept Volume

V2 
RT 1 
1 




 V1  V 2  V1  1 

1



V
P
r
1 
1 
c 

0.287 T1 
1 

1 

2
1 x 10 
14 
 0.003091 T1 m
2
/ kg
75
 1
M .E .P 
Work done /kg
Swept volume/kg

0.6439T
0.003091T
1
1
 208.3 kPa
M . E . P  2 . 083 bar
76
Problem 8
From the PV diagram of an engine working on the Otto
cycle, it is found that the pressure in the cylinder after
1/8th of the compression stroke is executed is 1.4 bar.
After 5/8th of the compression stroke, the pressure is
3.5bar. Compute the compression ratio and the air
standard efficiency. Also if the maximum cycle
temperature is limited to 1000.C, find the net work out
put
77
Given
Pa  1 . 4 bar , Pb  3 . 5 bar
T 3  1000  273  1273 K
T1  27  ( assumed )  300 K
Solution
V1  V a 
1
V a  V1 
1
Va 
7
Va
7
V2

8
8
8
8
V1  V 2 
V1  V 2 
V1 
V2
rc 
1
8
8
- - - -(1) where rc  compressio
n ratio
78
V1  V b 
Again
V b  V1 
Vb

V2
3
8
5
8
5
rc 
Vb
 V2 
- - - -(1)
8
 8 r

3 8 r

7

8
V1
V1  V 2 
From equation
Va
5
c
c
1 and 2 we have
1
8
5
8
79


Pa V a  P bV b
But
1
Va
Vb
 Pa
 
 P
 b
1

 3 . 5  1 .4
  


 1 .4 

Va
 1.924
- - - - - (4)
Vb
from 3 and 4
 8 r

3 8 r

7
c
c
rc  7
1
8  1 . 924
5
8
80
1
  1
 1
 1
rc
T2
 V1
 T1 
V
 2
1
7




1 . 4 1
 54 . 08 %
 1
 300  7 
1 . 4 1
 653.4 K
Heat added
 C  (T 3  T 2 )
 0.718(1273
- 653.4)
 445 kJ/kg
Network
output
  x heat supplied
 0.5408
x 445
 240.6 kJ/kg
81
Problem 9
An air standard diesel cycle has a compression ratio of 16. The
temperature before compression is 27°C and the temperature
after expansion is 627°C. Determine:
i) The net work output per unit mass of air
ii) Thermal efficiency
iii) Specific air consumption in kg/kWh.
82
Given
V1
V2
rc  16
T 4  627  273  900 K
T1  27  ( assumed
)  300 K
Solution
For process
Or
T2
1 - 2 we have T1V 1
 V1
 T1 
V
 2




 1
 1
 T 2V 2
 1
 300 x16
0.4
 909.43 K
For process
2 - 3 we have
P 2V 2
T2

T3 
V3
V2
x T2

P 3V 3
T3
and P 2  P 3
- - - (1)
83
 1
For process
Or
T3
T3
3 - 4 we have T 3V 3
 V4
 T4 
V
 3




 V1
 T4 
V
 2
Substituti ng for
 1
 V1
 T4 
V
 3
 V 2

 V
 3
V2








 1
 T 4V 4
 1
 1
from Eqn (1) we have
V3
 V1
 T4 
V
 2
T3
T3

 T4T2




 1
 1 
 T2

T
 3
V1 


V 
 2 




 1
 1
84

  1  V1
T3   T 4 T 2 
V2




 900 x 9 09.43
Heat supplied




 1
0 .4




1

x 16
0 .4

1
1 .4
 1993 . 3 k
per unit mass  q 2 - 3  C p (T 3  T 2 )
 1.005 x [1993.3 - 909.43]
 1089.3kJ/k g
Work done  W  1089 . 3  430 . 8
 658.5kJ/kg
 Thermal
efficiency
W
 

q 2 -3
Specific air consumptio
n
3600
W
658 . 5
 0 . 6045  60 . 45 %
1089 . 3

3600
658 . 5
 5 . 57 kg / kwh
85
Problem 10
The compression ratio of a compression ignition
engine working on the ideal Diesel cycle is 16. The
temperature of air at the beginning of compression is
300K and the temperature of air at the end of
expansion is 900K. Determine
i) cut off ratio
ii) expansion ratio and
iii) the cycle efficiency
86
Given
rc  16
T 4  627  273  900 K
T1  27  ( assumed
)  300 K
Solution
T 2  T1 r
T3
T4
T3
T4
T3
T4
 1
 V4
 
V
 3




 300 x 1 6
 1
1 .4 1
 909 . 42 K
 V4
V2 

 
V x V 
3 
 2
 V1
V2 

 
V x V 
3 
 2

V2 

 
 rc x V 
3 

 1
 1
 1
87
P 2V 2

P3V 3
T2
T3
V2


V3
T3
T4

T2
T3

T2 

  rc x

T
3 

 T3 x T 3
T3
and P 2  P3
 1
 1
 1
 rc
 1
 T 4 rc
T2
 1
 1
 1
 1 1 
 T 4 rc
T 3  ( T 4 rc
T2
T2
 ( 9 00 x 1 6
 1993.28K
T2
T3
 1
 1
 1
)
1 . 4 1
x 909.42
1 . 4 1 1 1 . 4
)
88
 
 cut off ratio
T3

T2
1993 . 28
909 . 42
  2 . 19
1 

  1-
 1-
rc

16
  1


  1 


1-1.4
1 .4
 2 . 19 1 . 4  1 


 2 . 19  1 


 60.46%
1
Expansion
ratio
rE 
V4
V3
 T3
 
 T4
  1



1
 1993 . 28  1 . 4 1
 7 . 29


 900

89
Problem 11
An air standard limited pressure cycle has a compression
ratio of 15 and compression begins at 0.1 MPa, 40°C. The
maximum pressure is limited to 6 MPa and heat added is
1.675 MJ/kg. Compute
(i) the heat supplied at constant volume per kg of air
(ii) the heat supplied at constant pressure per kg of air
(iii) the work done per kg of air
(iv) the cycle efficiency
(v) cut off ratio and
(vi) the m.e.p of the cycle
90
Given
rc  15
P1  0 . 1 MPa  100 KPa
T1  40  C
P3  P4  6 MPa  6000 KPa
Heat added  1.675 MJ/kg  1675kJ/kg
Solution

P 2  P1 rc  100 x 1 5
 
P3
P2

6000
1 . 4 1
 4431 . 26 KPa
 1 . 354
4431 . 2
91
 1
T 2  T1 rC
 313 (15 )
1 . 4 1
 924.65k
T 3  T 2  924 . 6 J x 1.354
 1251/99 K
Heat added at constant v olume  0.72( T 3  T 2 )
 0.72( 1 251.99  924 . 65 )
 235.71kJ/k g
Heat added at constant
pressure
 Total heat added
- heat added at constant v olume
 1675 - 235.71  1439.286kJ /kg
92
Heat added at constant
 C p T 4  T 3 
pressure
1439.286  1.005(T
4
 1251 . 99 )
T 4  2684 . 11 K
Air standard
effeciency


1 
  1
 1 -  -1 

rc    1       1  
1
 115
1 . 4 1
1 .4


1 . 354 x 2.1438
1


 1 . 354  1   1 . 354 x 1.4  2 . 1438  1  
 60.56%
M .E .P 


P1 r

 1  rc  1 
100 x 15
1 . 4  1 15
  1      1  r 
1 .4
 1
1 
c
1 . 354

1

 1   1 . 4 x 1 .354  2 . 1438  1   15
1 1 .4
1 . 354
x 2 .1438
1 .4
 2000.13 KPa
93
1
