1
Applied Thermodynamics
2
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” ;
3
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.
4
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.
5
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.
6
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.
7
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.
8
Air standard assumptions 1. The working medium is a perfect gas with constant specific
heats and molecular weight corresponding to values at room temperature.
2. No chemical reactions occur during the cycle. The heat addition and heat rejection processes are merely heat transfer processes.
3. The processes are reversible.4. Losses by heat transfer from the apparatus to the atmosphere
are assumed to be zero in this analysis.5. 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.
Air standard Carnot CycleThe 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
10
Process 1 – 2: Reversible Adiabatic CompressionProcess 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 = =
1Q2 = 0 or
12211
VPVP
1
)( 21
TTmR
11
Process 2 -3: Isothermal Heat AdditionIn this air is heated isothermally so that volume increases and Temperature remains constant.Amount of heat supplied is equalto 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= p2V2 ln = mRT2 ln
2Q3 = p2V2 ln
12
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 = =
3Q4 = 0
14433
VPVP
1
)( 43
TTmR
13
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= p4V4 ln = mRT4 ln
4Q1 = p4V4 ln
14
15
16
max
minmax
max
minmax
ln
ln
T
TT
TrR
TTrRth
And also,
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 cycleWhen a confined body of gas (air, helium, whatever) isheated, its pressure rises. This increased pressure can push on a piston and do work. The body of gas is then cooled, pressure drops, and thepiston can return. The same cycle repeats over and over, using the samebody of gas. That is all there is to it. No ignition, no carburetion, no valvetrain, no explosions. Many people have a hard time understanding the Stirlingbecause it is so much simpler than conventional internalcombustion 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 Constantvolume 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
Where, r = compression OR expansion ratio and
.
129
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
35
The Efficiency of the cycle:
The efficiency of the cycle is given by,
Let, compression ratio, Cut-off ratio,
Expansion ratio,
Considering process 1-2,
36
Considering process 3-4,
Substituting for in eqn (1), we get
Considering process 2-3,
37
Mean effective pressure:
we know that work done per kg in Diesel cycle is given by,
And the mean effective pressure is given by:
38
Expression for cut-off ratio:Let ‘k’ be the cut-off in percentage of stroke (from
We know that,
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.
4
5
3
39
Dual combustion or Limited pressure or Mixedcycle:
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,
3 4
5
40
Dual combustion or Limited pressure or Mixed cycle:
Process 3-4: 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,
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,
3 4
5
41
42
The Efficiency of the cycle:
The efficiency of the cycle is given by,
Let, compression ratio,
Cut-off ratio,
Pressure ratio,
Expansion ratio,
5
43
Considering process 1-2,
Considering process 2-3,
Considering process 3-4,
Considering process 4-5,
Substituting for in (1)
44
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
45
Comparison between Otto, Diesel and Dual combustion cyclesThe 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.
46
Comparison with same compression ratio and heat supply:
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 1-2-3-4-1 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
50
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>q1We know that thermal efficiency is given by 1 – heat rejected/heat suppliedThermal efficiency of these engines under given circumstances is of the following orderDiesel>Dual>OttoHence in this case it is the diesel cycle which shows greater thermal efficiency.
51
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
25.1
1
3
TTrc
52
Solution: Process 1-2 is reversible adiabatic
....(1)..........
12
2
1
1
2
c
c
rTT
rVV
TT
Process 3-4 is reversible adiabatic
(2)..........
33
4
2
1
3
4
cc
c
rTrTT
rVV
TT
53
Work done = Heat added - Heat rejected
In the above equation T3, T1 and Cv are constants. Therefore for maximum work
11
313
1423
- C - - C
- C - - C
TrTrTT
TTTT
cc
0 drdW
54
25.1
1
3
14.1
1
31
1
1
3
TTr
TT
TTr
c
c
1
2
1
2
32
1
32
1
131
11
313
0 1C - C-
0 1C - C-
0 - C - - C
c
ccc
c
cc
cc
cc
cc
r
rrrr
TT
rTrT
rTrT
rTrT
TrTrTTdrd
55
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
3142 TTTT
56
Solution: For maximum work/kg in an Otto cycle•
31
21
1
31
11
1
3 112
11
1
3
1) problemin proved (as
TTTTT
TTTrTT
TTr
c
c
57
11
1
3
334
TT
TrTTc
3142
313
13
1
TTTT
TTTTT
Again
58
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
59
11
1
2
2
1
2
1
1
2
14
P
Prv
v
vv
PP
c
53% 58.6
11
11 efficiency Ideal
15.18%
1006.58
1 100 x
4.
1
2
1
2
1
c
cc
r
xvv
vv
vvr
60
bar 1.65 )158.6)(14.1(
)11)(6.58-1x6.58(1.5
1
1..
1.5 1421
ratio
1-1.4
1
2
3
c
c
r
rpPEM
pppressureExplosion
61
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 airSolution:Given:P2/P1 = 15 , T1 = 37ºC = 310 KT3 = 1950ºC = 2223 K
62
6.91
15
r
11
1
2
c2
1
1
2
c
c
r
P
Pr
vv
PP
54%
0.54 91.611
11
4.
cr
K 671.66 (31096.91) 1-1.412
crTT
63
Heat supplied = Cv(T3-T2) = 0.72(2223 - 671.66)
=1116.96 KJ/kg of air
Work done = 0.54 x 1116.96 = 603.16 KJ/kg of air
suppliedHeat doneWork
64
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
K 953.3 300(18) 1-1.4
12
crTT
65
Heat transferred = Cp(T3 – T2)
2000 = 1.005(T3 -953.3)]T3 = 2943.34 K
3.08 953.3
2943.34 ratio offcut 2
3 TT
58.6% 0.586 08
108.181
11
4.
cr
66
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.
67
Solution: Given:rc = 15,
P1 = 1 bar,
T1 = 27º C
d = 200 mm, L = 300 mm
3m .009424
.3 2.0 sV
V
2
s
x
dLvolumeSwept
68
3
23
23
2sc2
cc
s
c
s
c
sc
2
1c
c2
m 0.001427
009424.008.00006731.008.0 100
8 )(
stroke of 8%at place takesoff
m 0.0006731 14
0.009424
14
V V
14 1)-(15 )1(r V
V
V
V1
V
VV
V
V r
volumeclearance V
xVVV
VVV
Cut
V
V
s
s
69
kPaPEM
x
cr
r
rpPEM
c
c
cr
cut
2
4.114.1
4.11
2
3
10 x 7.41bar 7.14 ..
112.1512.4.115.
151
11
1..
59.8% 0.598
12.
112.15
1 1
1
2.12 0.00067310.001427
VV
ratio off
70
bar 44.3 3
2
3.441.4
1x15 1
2
25.8861-1.4300x15 1
2
kW 46.53 Power 60
400 x 6.98 Power
sec / cycles ofNumber x cycle / done Power Work
kJ/cycle 6.98 0.00942 10 x 7.41 cycle / done
meSwept volu
cycledonework ..
2
PP
barcrPP
KcrTT
xWork
PEM
71barP
rV
V
V
V
V
V
V
V
V
V
P
P
x
VceV
V
V
V
V
V
V
V
V
V
c
86.215
12.23.44
x x
K 858.99 15
12.285.1878
r
1
V sin x x
K 1878.85 2.12 x 886.25
4.1
14.1
1
14
111
72
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
73
62.18%
)107.1(1.4 367.1)1367.1(
11.07 x 367.1
14
11
)1()1(
111
367.123.40
55
P
P ratio pressureExplosion
bar 40.23
)14(1
4.1
14.1
1
2
3
4.112
x
r
rPP
c
c
74
112
2
323
3
33
2
22
112
2
32
3
43p
2334p
TP
PP
process lumecontant vo is 3-2 Pr
T T
1 1C
)()(C added
c
c
rTP
TT
T
V
T
V
ocess
r
T
TTC
T
TT
TTCTTHeat
75
kgm
c
cc
rP
RTSwept
xWork
T
rTCrT
/21
21
1
1
1
2121
1
1
1
14.11
14.11
11
11p
T 0.003091
14
11
10 x 1
T 0.287
11
V
V1VVV Volume
T 0.6439
6218.0T 1.0356 x addedHeat done
1.0356T
1367.11472.0107.114T 1.367 x 1.005
11C addedHeat
76
bar 083.2 ..
kPa 208.3
0.003091T
0.6439T
volume/kg
/kgdoneWork ..
1
1
PEM
SweptPEM
77
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
78
ration compressio where-(1)--- 8
1
8
788
78
18
1
300)(27
12732731000
5.3 ,4.1
2
21
211
211
1
3
cca
a
a
a
ba
rrV
V
VVV
VVVV
VVVV
Solution
KassumedT
KT
barPbarP
Given
79
85
83
81
87
have we2 and 1equation From
-(1)--- 8
5
8
38
58
5
2
211
211
c
c
b
a
cb
b
b
r
r
V
V
rV
V
VVVV
VVVVAgain
80
7
924.1
85
83
81
87
4 and 3 from
(4)----- 1.924
4.1
5.3
PP
4.1
11
ba
c
c
c
b
a
b
a
b
a
ba
r
r
r
V
V
P
P
V
V
VVBut
81
kJ/kg 240.6
445 x 0.5408
suppliedheat x output Network
kJ/kg 445
653.4)-0.718(1273
)(C addedHeat
K 653.4
7300
%08.547
11
11
23
14.1
1
2
112
14.11
TT
V
VTT
rc
82
Problem 9An 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 airii) Thermal efficiencyiii) Specific air consumption in kg/kWh.
83
(1)--- T x T
PP and PP
have we3-2 processFor
K 909.43
x16300T
TT have we2-1 process
300)(27
900273627
16
22
33
323
33
2
22
0.4
1
2
112
122
111
1
4
2
1
V
V
T
V
T
V
V
VTOr
VVFor
Solution
KassumedT
KT
rV
V
Given
c
84
1
2
11243
1
3
2
1
2
143
3
2
1
3
2
2
143
1
3
14
1
3
443
144
133
T
T
have we(1)Eqn from for
T
T
TT have we4-3 process
V
VTT
T
T
V
VT
V
VngSubstituti
V
V
V
VT
V
VT
V
VTOr
VVFor
85
kwhkgW
Specific
W
W
TTCHeat
k
V
VTT
p
/57.55.658
36003600 n consumptioair
%45.606045.03.1089
5.658
q efficiency Thermal
658.5kJ/kg
8.4303.1089 doneWork
g1089.3kJ/k
909.43]-[1993.3 x 1.005
)(q massunit per supplied
3.199361 x 09.439 x 900
T
3-2
233-2
4.11
4.04.0
11
2
11243
86
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 ratioii) expansion ratio and iii) the cycle efficiency
87
1
3
2c
4
3
1
3
2
2
1
4
3
1
3
2
2
4
1
3
4
4
3
14.1112
1
4
x T
T
x T
T
x T
T
42.90961 x300T
300)(27
900273627
16
V
Vr
V
V
V
V
V
V
V
V
V
V
KrT
Solution
KassumedT
KT
r
Given
c
881993.28K
)909.42 x 61 x 009(
)T(
T
T x T
x T
T
T
T
PP and PP
4.1114.114.1
112
143
12
143
12
14
133
13
121
1
3
2c
4
3
3
2
3
2
323
33
2
22
TrT
TrT
TrT
T
Tr
T
Tr
V
V
T
V
T
V
c
c
c
c
89
29.7900
28.1993
r ratioExpansion
60.46%
119.2
119.2
4.1
61-1
1
1r-1
19.2
42.909
28.1993
T
T ratio offcut
14.1
1
1
1
4
3
3
4E
4.11.4-1
1c
2
3
T
T
V
V
90
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
91
354.12.4431
6000
P
26.443151 x100P
1675kJ/kgMJ/kg 1.675 added
60006
40
1001.0
15
2
3
14.112
43
1
1
P
KParP
Solution
Heat
KPaMPaPP
CT
KPaMPaP
r
Given
c
c
92
/kg1439.286kJ235.71-1675
olumeconstant vat addedheat -
addedheat Total pressureconstant at addedHeat
g235.71kJ/k
)65.924251.9910.72(
)0.72( olumeconstant vat addedHeat
K 1251/99
1.354 x 6.924
924.65k
)15(313
23
23
14.1112
TT
JTT
rTT C
93
60.56%
11438.21.4 x 354.11354.1
12.1438 x 354.1
15
1-1
11
1
r
1-1 effeciency standard
11.2684T
)99.12511.005(T1439.286
C pressureconstant at added
4.1
14.1
1-c
4
4
34p
Air
K
TTHeat
KPa 2000.13
1.14382 x 354.11511438.2.3541 x 4.11354.111514.1
15 x 100
11111
..
4.14.114.1
11
c
c
rr
rPPEM