chapter 7 thermo
TRANSCRIPT
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Chapter 7
Air Standard Cycles
The power cycles can be classified into two important fields. The first is the
power generation which the work done output of the system such as Heat Engine. The
second is the refrigeration and air conditioning which the work done input to the system
such as Heat Pump. Both of it are usually operating on a thermodynamic cycle. In the
power generation systems, we will use the air standard cycles which the working fluid is
returned to the initial state at the end of the cycle. Some assumptions were made to
study the air standard cycles such as,
(a)The working fluid is air and always ideal gas.(b)All processes are internally reversible.(c)Heat added to the cycle is from external heat source.(d)Heat rejected from the cycle is to the surrounding.
The simplified model of heat engine is a piston cylinder device which called
reciprocating engine and the basic components are shown in Fig. 7-1.
Fig. 7-1 Basic components of reciprocating engine
The piston reciprocates in the cylinder between two fixed positions called the top dead
center (TDC)and the bottom dead center (BDC). The distance between the TDC and
the BDC is called the stroke, Lst. The diameter of the piston is called the bore, d. The
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thermodynamic cycle of reciprocating engine is four processes in two strokes as shown
in Fig. 7-2. The maximum volume when the piston is at BDC. The minimum volume
when the piston is at TDC which is called the clearance volume. The volume between
TDC and BDC is called the displacement volume or swept volume. The ratio between
maximum to minimum volume is called the compression ratio, r.
Fig. 7.2 Thermodynamic cycle of reciprocating engine
min
max
V
Vr=
The mean effective pressure, MEP, is the exerted pressure on the piston during the
power stroke to produce the same net work done.
minmax VV
W
V
WMEP net
stroke
net
==
The mean effective pressure can be used as a parameter to compare the performance of
reciprocating engines of equal size.
Reciprocating engines are classified as spark-ignition (SI) engines and
compression-ignition (CI)engines. In spark-ignition engines, the combustion of the air
fuel mixture is initiated by a spark plug. But in CI engines, the air fuel mixture is self
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ignited as a result of the compression temperature. The Otto and Diesel cycles are the
ideal cycles for SI and CI reciprocating engines.
Otto Cycle
The Otto cycle is the ideal cycle for spark ignition, (SI), reciprocating engines.
The thermodynamic analysis of four-stork cycle can be simplified if the air standard
assumptions are used. The Otto cycle is consists of four internally reversible processes
as shown in Fig. 7-3.
Fig. 7-3 Otto cycle and P-v, T-s diagrams
The Otto cycle is executed in a closed system and the change of kinetic and potential
energies are disregarded. The energy balance for any of the processes is expressed in a
unit mass basis as,
Process 12, isentropic compression and ris the compression ratio,
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1
12
1
1
2
1
1
2
2
1
.,
==
=
=
kk
k
rTTrv
v
T
T
v
vr
Process 23, constant volume heat addition, v2= v3,
)( 23 TTCq vin =
Process 34, isentropic expansion,
1
43
1
1
3
4
4
3
3
4
2
1
.,
==
=
==
kk
k
rTTrv
v
T
T
v
v
v
vr
Process 41, constant volume heat rejection, v4= v1,
)( 14 TTCq vout =
Otto cycle net work done is,
outinnet qqw =
The thermal efficiency of Otto cycle is,
)(
)(11
23
14,
TTC
TTC
q
q
q
qq
q
w
v
v
in
out
in
outin
in
net
Ottoth
==
==
1.
14
14
11
1
1
4
14
23
14.
11
11
..11
=
=
=
=
kOttoth
kkkOttoth
r
TT
TT
rrTrT
TT
TT
TT
From the relation of thermal efficiency for Otto cycle, it is clear that the thermal
efficiency is very dependent on the compression ratio, r, and the specific heat ratio, k =
Cp/Cvas shown in Fig. 7-4. The Otto cycle thermal efficiency increases with increasing
the compression ratio rand the specific heat ratio k.
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Fig. 7-4 Dependent of Otto thermal efficiency on rand k
Diesel Cycle
The diesel cycle is the ideal cycle for compression ignition, (CI), reciprocating
engines. The thermodynamic analysis of four-stork cycle can be simplified if the air
standard assumptions are used. The Diesel cycle is consists of four internally reversible
processes as shown in Fig. 7-5.
Fig. 7-5 Diesel cycle P-v and T-s diagrams
Process 12, isentropic compression and ris the compression ratio,
1
12
1
1
2
1
1
2
2
1
.,
==
=
=
kk
k
rTTrv
v
T
T
v
vr
Process 23, constant pressure heat addition, P2= P3, and the cut-off ratio, rc,
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)(
...,//,
23
1
1232323
2
3
TTCq
rrTrTTrTTvvv
vr
pin
c
k
ccc
=
=====
Process 34, isentropic expansion,
k
c
k
c
c
k
k
c
k
c
k
c
k
rTr
rrrTT
r
rTT
r
r
vr
v
v
v
T
T
....
,
1
1
1
14
1
34
11
2
1
1
3
4
4
3
=
=
=
=
=
=
Process 41, constant volume heat rejection, v4= v1,
)( 14 TTCq vout =
Diesel cycle net work done is,
outinnet qqw =
The thermal efficiency of Diesel cycle is,
)(
)(1
)(
)(11
23
14
23
14
,
TTk
TT
TTC
TTC
q
q
q
qq
q
w
p
v
in
out
in
outin
in
net
Dieselth
=
==
==
[ ]11
1
1
11
23
14.
...
.1
)(1
=
=
k
c
k
k
c
DieselthrTrrTk
TrT
TTk
TT
=
)1(
111
1.
c
k
c
kDieselth rk
r
r
From the relation of thermal efficiency for Diesel cycle, it is clear that the thermal
efficiency is very dependent on the compression ratio, r, cut-off ratio, rc, and the
specific heat ratio, k = Cp/Cvas shown in Fig. 7-6. The Diesel cycle thermal efficiency
increases with increasing the compression ratio rand the specific heat ratio k. But the
Diesel cycle thermal efficiency increases with decreasing the cut-off ratio rcand
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Fig. 7-6 Dependent of Diesel thermal efficiency on r, rc, and k
From Fig. 7-6, when the cut-off ratio equal 1, it becomes the thermal efficiency of Otto
cycle. Therefore,
DieselthOttoth ,, >
When both cycles operate on the same compression ratio r. Also, as the cut-off ratio
decreases the thermal efficiency of diesel cycle increased as shown in Fig. 7-6.
Dual cycle
Fig. 7-7 Dual cycle P-v and T-s diagrams
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The ideal power cycle which the heat addition partially at constant volume and constant
pressure processes is called Dual cycle as shown in Fig. 7-7. Compression ratio, r =
v1/v2, cut-off ratio, rc= v4/v3, pressure ratio, rp= P3/P2.
Process 12, isentropic compression,
1
12
1
2
1
1
2 .,
=
= k
k
rTTV
V
T
T
Process 23, heat added at constant volume, v3= v2, or P/T=C
)(
...,
23,
1
12
2
3
23
2
2
3
3
TTCq
rrTrTP
PTT
T
P
T
P
vcvin
k
pp
=
====
=
Process 34, heat added at constant pressure, P = C, or V/T=C
)(
...,.,
34,
1
143
3
434
3
3
4
4
TTCq
rrrTTrTV
VTT
T
V
T
V
pcpin
cp
k
c
=
====
=
Process 45, isentropic expansion,
k
cp
k
c
cp
k
k
c
k
c
k
c
k
c
k
rrTr
rrrrTT
r
rTT
r
r
V
Vr
V
Vr
V
V
T
T
.....
,
1
1
1
15
1
45
11
1
2
1
5
3
1
5
4
4
5
=
=
=
=
=
=
=
Process 51, heat rejection at constant volume,
)( 15, TTCq vcvout ==
The thermal efficiency of Dual cycle is,
)()(
1
3423,, TTCTTCqqq
q
q
q
qq
q
w
pvcpincvinin
in
out
in
outin
in
net
th
+=+=
=
==
==
)()(
)(1
)()(
)(1
3423
15
3423
15
,TTkTT
TT
TTCTTC
TTC
pv
v
Dualth+
=
+
=
+
=
+
=
+
=
)1(.)1(
111
)1(..)1(
)1.(1
)()(
)(1
1,
1
1
1
1
1
3423
15,
cpp
k
cp
kDualth
cp
k
p
k
k
cp
Dualth
rrkr
rr
r
rrrkTrrT
rrT
TTkTT
TT
For Otto cycle, the cut-off ratio, rc= 1, the thermal efficiency yields,
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111,1
11
111
)1(.)1(
111
=
=
+
=
k
p
p
k
cpp
k
cp
kOttothrr
r
rrrkr
rr
r
For Diesel cycle, the pressure ratio, rp= 1, the thermal efficiency yields,
=
+
=
)1(
111
)1(.)1(
111
11,
c
k
c
k
cpp
k
cp
kdieselth rk
r
rrrkr
rr
r
Brayton Cycle
The Brayton cycle is used for gas turbines only where both the compression and
expansion processes take place in a rotating machinery and can be modeled as a closedcycle as shown in Fig. 7-8.
Fig. 7-8 A closed cycle of gas turbine engine
The air standard assumptions are used for ideal cycle and the compression and
expansion processes are internally reversible adiabatic. The heat addition and heat
rejection processes are at constant pressure and occurred in heat exchangers. The four
processes are shown in Fig. 7-9 as follows,
Process 12, isentropic compression in a compressor and the pressure ratiorp= P2/P1,
kk
p
kk
rP
P
T
T /)1(/)1(
1
2
1
2
=
=
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Fig. 7-9 Brayton cycle P-v and T-s diagrams
Process 23, heat addition at constant pressure, P2= P3,
)( 23 TTCq pin = Process 34, isentropic expansion in a turbine, and rp= P3/P4,
kk
p
kk
rP
P
T
T /)1(/)1(
4
3
4
3
=
=
4
3
1
23241 ,,
T
T
T
TPPPP ===Q
Process 41, heat rejection at constant pressure, P4= P1,
)( 14 TTCq pout = The Brayton thermal efficiency is,
kk
p
kk
p
kk
p
th
in
out
in
outin
in
net
th
rrTrT
TT
TT
TT
q
q
q
qq
q
w
/)1(/)1(
1
/)1(
4
14
23
14
11
..1
11
=
=
==
==
Examples of Air Standard Cycles
1. The compression ratio of air-standard Otto cycle is 8. At the beginning of thecompression stroke the pressure is 1 bar and the temperature is 17 C. The heat of
800 kJ/kg is added during the constant volume process. Determine:
a. The pressure and temperature at each corner of the cycle,b. The thermal efficiency, andc. The mean effective pressure.
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Solution
Process 12, isentropic compression, r = 8.
barrPPrv
v
P
P
KrTTrv
v
T
T
kgmvv
vr
kgmP
RTv
kk
k
kk
k
38.1881.,
245.6668290.,
/10404.08/8323.0,
/8323.01001
290287.0
4.1
12
2
1
1
2
4.01
12
1
1
2
1
1
2
3
2
2
1
3
1
11
====
=
====
=
===
=
==
Process 23, constant volume heat addition, v2= v3,
barT
TPP
KT
TTTCq vin
118.49245.666
45.178038.18
45.1780
)245.666(718.0800),(
2
323
3
323
===
=
==
Process 34, isentropic expansion,
barrPPrv
v
P
P
KrTTrv
v
T
T
k
k
k
k
k
k
672.28/118.49/,1
986.7748/45.1780/,1
4.1
34
4
3
3
4
4.01
341
1
4
3
3
4
====
=
====
=
Process 41, constant volume heat rejection, v4= v1,
kgkJTTCq vout /22.348)290986.774(718.0)( 14 ===
The work done net is,
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kgkJqqw outinnet /78.45122.348800 ===
%47.565647.0800
78.451, ====in
netOttoth
q
w
barKPavv
wMEP net 204.6369.620
10404.08323.0
78.451
21
==
=
=
2- An engine operates on the theoretical diesel cycle with a compression ratio of 15. The
heat of 1300 kJ/kg is added at constant pressure for 10 % of the stroke volume. The
pressure and temperature of the air at the beginning of compression are 100 kPa and
27 C. Determine, (a) The cut-off ratio, (b) The pressure and temperature at the end
of each process, (c) The thermal efficiency of the cycle, (d) The mean effective
pressure, and the output power from the engine if the mass flow rate equal 0.15 kg/s.
Solution
Process 12, isentropic compression, r = 15.
kgmvvvr
kgmP
RTv
/0574.015/8323.0,
/861.0100
300287.0
32
2
1
3
1
11
===
=
==
KrTTrv
v
T
T kkk
253.88615300.,4.01
12
1
1
2
1
1
2 ====
=
KParPPrv
v
P
P kkk
265.443115100., 4.1122
1
1
2 ====
=
Process 2
3, constant pressure heat addition, P2= P3,
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==
=
==
11.01),(1.0
785.2179
)253.886(005.11300),(
2
1
2
3
2123
3
323
v
v
v
vvvvv
KT
TTTCq pin
( ) ( )
kgmvrv
rr
c
c
/13776.00574.04.2
4.21151.0111.01
3
23 ===
=+=+=
Process 34, isentropic expansion,
kPaP
r
r
v
rv
v
v
P
P
KT
r
r
v
rv
v
v
T
T
k
c
k
c
k
k
c
k
c
k
64.34015
4.2265.4431
.
277.104715
4.2785.2179
.
4.1
4
1
2
4
3
3
4
14.1
4
11
1
2
1
4
3
3
4
=
=
=
=
=
=
=
=
=
=
Process 41, constant volume heat rejection, v4= v1,
kgkJTTCq vout /346.536)300277.1047(718.0)( 14 ===
The work done net is,
kgkJqqw outinnet /654.763346.5361300 ===
%74.585874.01300
654.763, ====
in
net
Dieselthq
w
kWmwPower
barKPavv
wMEP
airnet
net
548.11415.0654.763
5029.929.9500574.0861.0
654.763
21
===
==
=
=
&
3. An air-standard Dual cycle operates with a compression ratio of 14. The conditionsat the beginning of compression are 100 kPa and 300 K. The maximum temperature
in the cycle is 2200 K and the heat added at constant volume is twice the heat added
at constant pressure. Determined, (a) The pressure, temperature, and specific volume
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at each corner of the cycle, (b) The thermal efficiency of the cycle, and (c) The mean
effective pressure.
Solution
Process 12, isentropic compression, r = 14.
kgmP
RTv
kgmP
RTv
kParPPv
v
P
P
KrTTv
v
T
T
k
k
k
k
/0615.0271.4023
129.862287.0
/861.0100
300287.0
271.402314100.,
129.86214300.,
3
2
22
3
1
1
1
4.1
12
2
1
1
2
4.01
12
1
2
1
1
2
=
==
=
==
===
=
===
=
Process 23, heat added at constant volume, v3= v2, or P/T=C
KT
TT
TTCTTCqq pvvpincvin
877.1847
)2200(005.12)129.862(718.0
)(2)(,2
3
33
3423,,
=
=
== ==
( ) kgkJTTCq
kgmvv
kPaT
TPP
T
P
T
P
vcvin /767.707129.862877.1847718.0)(
/0615.0
43.8623129.862
877.1847271.4023,
23,
3
23
2
3
23
2
2
3
3
===
==
====
=
Process 34, heat added at constant pressure, P = C, or V/T=C
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( ) kgkJTTCq
v
vr
kgmP
RT
v
KTkPaPP
pcpin
c
/884.353877.18472200005.1)(
19.10615.0
0732.0
/0732.043.8623
2200287.0
2200,43.8623
34,
3
4
3
4
4
4
434
===
===
=
==
===
=
Process 45, isentropic expansion,
kgmvv
kPaPv
v
v
v
P
P
KTv
v
v
v
T
T
kk
kk
/861.0
139.731861.0
0732.043.8623,
88.819861.0
0732.02200,
3
15
4.1
5
1
4
5
4
4
5
14.1
5
1
1
4
1
5
4
4
5
==
=
=
=
=
=
=
=
=
Process 51, heat rejection at constant volume,
( ) kgkJTTCq vcvout /273.37330088.819718.0)( 15, ====
The thermal efficiency of Dual cycle is,
kgkJqqw
kgkJqqq
outinnet
cpincvinin
/377.688273.37365.1061
/651.1061884.353767.707,,
===
=+=+= ==
barkPavv
wMEP
q
q
q
qq
q
w
net
in
out
in
outin
in
net
Dualth
61.801.8610615.0.861.0
377.688
%84.646484.0651.1061
273.37311
21
,
==
=
=
====
==
4. A gas turbine power plant operates on a simple ideal Brayton cycle that has apressure ratio of 12. The compressor inlet pressure and temperature are 1 bar and 300
K. The inlet temperature to the turbine is 1000 K. Determine the thermal efficiency
and the air mass flow rate required for a net power of 3 MW.
Solution
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Process 12, isentropic compression,rp= P2/P1=12,
barPrP
KTrP
P
T
T
p
kk
p
kk
12112
181.61012300,
12
4.1/)14.1(
2
/)1(
/)1(
1
2
1
2
===
===
=
Process 23, heat addition at constant pressure, P2= P3, T3= 1000 K.
kgkJTTCq pin /768.391)181.6101000(005.1)( 23 ===
Process 34, isentropic expansion, rp= P3/P4= 12.
KrTT
rP
P
T
T
kk
p
kk
p
kk
657.49112/1000/ 4.1/)14.1(/)1(
34
/)1(
/)1(
4
3
4
3
===
=
=
Process 41, heat rejection at constant pressure, P4= P1,
kgkJqqw
kgkJTTCq
outinnet
pout
/153.199615.192768.391
/615.192)300657.491(005.1)( 14
===
===
skgmmwPower
r
q
w
airairnet
kk
p
th
in
net
th
/064.15153.199/103,
%83.5012
11
11
%83.505083.0
768.391
153.199
3
4.1/)14.1(/)1(
===
===
====
&&
Problems
1- The compression ratio in air-standard Otto cycle is 8. At the beginning of the
compression stroke the pressure is 1 bar and the temperature is 25 C. The heat
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transfer to the air per cycle is 1800 kJ/kg. Determine; the pressure and temperature at
the end of each process of the cycle, the thermal efficiency, and the mean effective
pressure.
2- 1 kg of air undergoes an Otto cycle with a compression ratio of 8.7. Initial state of air
is 1 bar and 15 C. The heat added is 1500 kJ. Determine; the pressure, volume, and
temperature at the beginning of each process, the expansion work, compression
work, and net work, the thermal efficiency of the cycle, and the mean effective
pressure.
3- An air-standard diesel cycle has compression ratio of 15 and the heat transferred to
the working fluid per cycle is 1800 kJ/kg. At the beginning of the compression
process the pressure is 1 bar and the temperature is 25 C. Determine; the pressure
and temperature at each point of the cycle, the thermal efficiency of the cycle, and
the mean effective pressure
4- An engine operates on the theoretical diesel cycle with a compression ratio of 15. The
heat is added for 10% of the stroke. The pressure and temperature of the air at thebeginning of compression are 98 kPa and 17 C. Determine; the cut-off ratio, the
pressure and temperature at the end of each process, the amount of heat added, the
thermal efficiency of the cycle, and the volume flow rate of air, measured at the
beginning of compression, needed to produce 200 kW.
5- The intake conditions for an air-standard Dual cycle operating with a compression
ratio of 15 are 1bar and 300 K. The pressure ratio during constant volume heating is
1.5 and the volume ratio during the constant pressure part of the heating process is
1.8. Calculate; the temperatures and pressures around the cycle, the heat input and
the heat rejection, and the thermal efficiency of the cycle.
6- An air-standard Dual cycle operates with a compression ratio of 18. The conditions at
the beginning of compression are 1bar and 300 K. the maximum temperature in the
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cycle is 2500 K. The heat added at constant volume is twice the heat added at
constant pressure. Plot the cycle on P-V and T-S diagrams and find; the pressure,
temperature, specific volume at each corner of the cycle, the thermal efficiency of
the cycle, and the mean effective pressure.
7- Air is used as the working fluid in the simple Brayton cycle that has a pressure ratio
of 12. The compressor inlet pressure and temperature are 0.96 bar and 300 K. The
inlet temperature to the turbine is 1000 K. Determine the thermal efficiency and the
mass flow rate of air required for a net power of 30 MW.
8- Air is used as the working fluid in the simple Brayton cycle that has a pressure ratio
of 14. The compressor inlet temperature is 300 K, and a turbine inlet temperature of
1100 K. Determine the mass flow rate of air for a net power output of 100 MW
assuming that the adiabatic efficiency of the compressor is 90 % and the adiabatic
efficiency of the turbine is 85 %. Calculate also the work ratio, back work ratio.