Post on 22-Dec-2015
Shaft Power Cycles Ideal cycles
Assumptions:• Both compressor and turbine are isentropic
(reversible adiabatic, η= 100 %)• Pressure losses in both of combustion chamber and
heat exchanger are negligible.• Working fluid is air. It behaves as an ideal gas• Mass flow rate is constant and combustion process
is replaced by heat addition process.• Changes of K.E. of the working fluid between inlet
and outlet of each component are negligible.• Heat transfer in heat exchanger is complete
(ε=100%)
Shaft Power Cycles
• Description of simple cycle
• Specific work, w• Thermal efficiency, η=
w / q
Shaft Power Cycles
• First law of thermodynamics (conservation of energy)
• q – w = Δh• Compressor
(adiabatic, q=0.)
• Turbine
)( 2121 TTchhw
hw
pc
c
)( 4334 TTchhw
hw
pt
t
Shaft Power Cycles• Combustion chamber
)(
.0,
2323 TTchhq
whq
ph
h
ct
h
netth q
ww
q
w
/)1(/)1(
1
2
1
2 )()( thus4,-3 and 2-1 processes Isentropic rp
p
T
T
/)1(/)1(
4
3
4
3 )()( Also rp
p
T
T
23
14
23
2143 1)(
)()(
TT
TT
TTc
TTcTTc
p
ppth
Shaft Power Cycles
)1
(1)1( /)1(
3
4 rT
T
Shaft Power Cycles
)]()[(
)()( ww
by given is work specific
1243
1243t
TTTTcp
hhhhw
The
c
13/TT of ratio theis t where
)r1())r
1(1(t
)r1())r
1(1(
T
T
Tc
w
/)1(/)1(
/)1(/)1(
1
3
1p
,,cby dividing 1p thusT
ration compressio optimum thegivesequation theating,Differenti
Shaft Power Cycles
42
1
3
4
3
1
2
4
3
1
2/)1(
/)1(
2/)1(/)22(/)1(/)21(
/)1(/)21(
1p
TT
T
Tt
T
T
T
T and
T
T
T
Tr since
tr
)r(rt ,r)r(t
0r1
))r(1
(t : 0Tc
w
r
nsapplicatiocraft air 5.55
45.3
fort
turbinesgasindustrialfort
Simple cycle
Regeneration cycle (heat exchange cycle)
thus,T T cycle ideal
)(
)()(
q
w
54
53
1243
53h
net
for
TTc
TTcTTc
q
ww
p
ppct
To benefit from the exhaust, a regenerator is used to reduce the heat addition, the efficiency is improved since heat rejection is reduced.
/)1(/)1(/)1(
/)1(
/)1(
/)1(
3
4
1
2
3
1
1
1)1)(/1(
1
11
/11
1
11
)1(
)1(
1
rtrr
r
t
r
r
tT
TT
T
T
T
Regeneration cycle (heat exchange cycle)
Recall that .
cycle
cycle
simpleExcjangerHeat
simpleExcjangerHeat ww
Regeneration cycle (heat exchange cycle)
In the reheat cycle, both of the heat addition and work done are increased. The thermal efficiency may increase or decrease depending on the values of T4 in comparison to T3.
Reheat Cycle
Reheat Cycle
)1()()(/
)()()(
1
2
1
6
1
5
1
4
1
31
126543
T
T
T
T
T
T
T
T
T
TTcw
TTcTTcTTcwwww
p
pppCLPTHPT
16235461
5
4
3 , , ),( PPPPPPP
P
P
P
For optimum conditions, i.e. maximum work, thus.
This gives
Reheat Cycle
c
tt
Tc
w
p
2 - 1 c - 2
1
Thus gives
11
1
2
1
4
3
4
3
3
4
1
3
3
4
1
4
1
3
6453 HPT
)(r )P
P( C
where )P
P(
T
T
T
T
T
T
T
T
T
T and
T
T t
TThen , T IF , T
CBut
tAlso
TTT LPT
Reheat Cycle
ctct
ctct
Tcq
Tcw
c
tct
T
T
T
T
T
T
T
T
Tc
q
TTcTTcq
ctctTcw
pin
p
p
in
ppin
p
/2
/212
)/(
)/(
2
)]()[(
)()(
/212)/(
but ,q
w
1
1
1
4
1
3
3
2
1
3
1
4523
1
in
net
The efficiency is calculated as
Cycle with reheat and heat exchange
8276 & TTTT
For ideal regenerator;
/)1(
1
1
1
21
in
net
where
)1()1
1(2
and /22
/212
)/(
)/(
tproven tha becan It
q
w
rc
cc
tTc
wctt
ctct
Tcq
Tcw
www
p
pin
pnet
Rcc
cTT
Methods for accounting for component losses
• Performance of real cycles differ from ideal cycles for the following reasons:
• Velocities in turbomachinery are high thus Δ K.E. is not equal to zero.
• Compression and expansion are irreversible.• Friction losses in H.E. and CC.• H.E. effectiveness is less than 100%• Specific heat specific heat ratio are not constant through the cycle
processes• More work is required for compression process to overcome
bearings and friction in transmission between compressor and turbine
• Cycle efficiency is different from unit efficiency• Some compressed air is bled for turbine blade cooling.
Stagnation properties
The energy equation for a flow coming to rest is
Where
h is the static enthalpyO refers to the stagnation conditionC is the velocity
2
2chvelocityzeroho
Stagnation properties
• Thus:
2
2chho
po c
cTT
2
2
re temperatudynamic theis 2
2
pc
c
and re temperatustatic theis T
re, temperatu(or total) stagnation theis oTwhere
Stagnation properties• Thus the first law equation is
)()(
)2
()2
(
1212
21
1
22
2
oopoo TTchh
ch
chwq
)1/(12
)1/(2
2
)2
11(
,)2
11(
,2
11
M
Mp
p
MT
T
o
o
o
bygiven are process adiabatican in properties stagnation the
then,RT aby given sound of speed theawith
numberMach theis a
cM and )1/(c p
RgConsiderin
1
o1
o2
o1
o2 )T
T(
P
P
process isentropican
For
Compressor and turbine efficiency
workactual
workideal = efficiency
1
1)(
1
1
01
02
/)1(
01
02
01
02
01
02
0102
0102
0102
0102
'
'
'''
T
Tp
p
T
TT
T
TT
TT
TT
TT
Tc
Tc
h
h
w
w
c
op
op
o
o
actual
idealc
]1)[( /)1(
01
0201
01020102
'
p
pT
TTTT
c
c
Compressor efficiency
ediscalculatNormally o2T andgiven is
Compressor and turbine efficiency
workideal
workactual = efficiency
] ) ( / 11[
/)1(
04
03030403
p
pTTT
or
t
Turbine efficiency
/)1(
03
04
03
04
03
04
03
04
0403
0403
0403
0403
,
)(1
1
1
1
''
''
p
pT
T
T
TT
T
TT
TT
TT
TT
Tc
Tc
h
h
w
w
t
idealop
op
o
o
ideal
actualt
Polytropic efficiency
T
T
T
T
s
idealscs
',
,
Consider a compressor of different stages. The Polytropic efficiency (defined as the isentropic efficiency of an elemental stage in the process such that it is constant throughout the whole process) given by
c
cc
c
n
nor
TTpp
p
p
T
T
p
dp
T
dT
p
dp
T
dT
dppp
TdppCdT
consp
T
11
ln
ln1
or ln1
ln
1 and
1
11
tan
01
02
01
02
01
02
01
02
'
/1/)1(
/1'
/)1(
Polytropic efficiency
• For an expansion (such as a turbine)
idealideal,s
st,s dT
dT
T
T
ttancons
p
T/)1(
11
ln1
ln
ln1
ln
1 thus
1
03
04
03
04
03
04
03
04
t
t
t
tideal
n
nor
p
pT
T
p
p
T
T
p
dp
T
dT
p
dp
T
dT
Polytropic efficiency• For a compressor
]1)[(
]1)[(]1[
/)1(
01
0201
/)1(
01
0201
01
02010102
c
p
pT
P
PT
T
TTTT nn
])/(11[
])/(11[
/)1(
04
0303
/)1(
04
03030403
0403
0403
, ''
t
p
pT
p
pTTT
TT
TT
Tc
Tc
h
h
w
w
nn
idealop
op
o
o
ideal
actualt
• For a turbine
Pressure losses
hga ppp 04
]1[0202
02
0203
p
p
p
pp
pppp
ab
ab
Heat exchanger effectiveness
0204
0205
energy ideal
,
TT
TT
airbyreceivedenergyactualesseffectiven
Mechanical Losses and Variations in specific heats
99.0
/lossesh work wit
losseshout work wit
m
mct
m
ww
5.31
,4.1
)/(005.1
;
a
pa kgKkJc
airFor
Variations in specific heats
Mechanical Losses
41
,333.1
)/(148.1
;
a
pg kgKkJc
gasescombustionFor
Fuel Air ratio
C) 86.08% and H (13.92% nshydrocarbo liquid ofheat specific : )/(2
nshydrocarbo liquid ofreaction ofenthalpy : )/(43100
)298()1()298( )298( 0302
kgKkJc
kgkJH
TcfTfcHfTc
pf
c
pgfpfcpa
T given for f actual
T given for f ltheoreticab
K 298 of ture temperareferencefor ; .
.
a
f
m
mf
fm
af mm..
am
.
f11
f
Fuel Air ratio
Specific fuel consumption
net
a
net
a
fnetf wf
m
W
m
mWmsfc /3600 / .
..
.
....
HHVsfcHHVxfwnet /3600) (/
..
Example 2.1• A heat exchange cycle having the following data:
– turbine inlet temperature 1100(K)– compressor pressure ratio 4.0 – isentropic efficiency of compressor 0.85– isentropic efficiency of turbine 0.87– combustion efficiency 0.98 – mechanical transmission efficiency 0.99– heat exchange effectiveness 0.80– ambient conditions 1(bar), 288 (k)– pressure losses
• combustion chamber 2% of compressor delivery pressure
• heat exchanger air side 3% of compressor delivery pressure
• heat exchanger gas side 0.04 (bar)
Example 2.102 012 1
12,
02, ( 1) / ( 1) /02 02
02, 01 01 01 0102 01 01
02 0202 01
01 01
(1.4 1) /1.4
02 01
( )( )
1 ( ) 1 ( ) 1
and 1 1
(4) 1288 164.7( )
0.851.005 164.7
0.
pactual
m m
ideal
idealc
c
c
c T Th hw
T p pT T T p p
T T TT TT TT T
T T K
Xw
167.2( / )99
kJ kg
Example 2.1
03 04
04 04
( 1) /03 04 03 03 0403 04 03
( 1) /04, 0403 04, 03
0303
03 02 04
( )
1 1
and (1 ( ) )1 ( )1
4(1 0.03 0.02) 3.8( ) and
1 0.04 1.
t pg
t tidealideal
ha b a hg
w c T T
T T
T T T T pT T T
T pT T ppT
p p p p bar p p p
( 1) /0403 04 03
03
03 04
04( )
1 1.148( / ) and 1.333, 0.25
(1 ( ) ) 264.8( )
( ) 1.148 264.8 304.0( / )
136.8( / )
pg
t
t pg
net t c
bar
c kJ kgK
pT T T K
p
w c T T X kJ kg
w w w kJ kg
Example 2.1
02 01 02 01
05 02
04 02
04 03 03 04
05 04 02 02
( ) 288 164.7 452.7( )
H.E. effectiveness
( ) 1100 264.8 835.2( )
effectiveness ( ) 758.7
T T T T K
T T
T T
T T T T K
T X T T T K
b
theoretical (from chart) 0.0094
theoretical f0.0096
f
f
Example 2.1
0.00963600 0.253( / )
136.8t c
Specific fuel consumption is given by
fSFC kg kWh
w w
43100 /
3600 36000.331
0.253 43100
fuelnet
net
Q kJ kg
SFC Q X
Example 2.1: Using EESKnown Information
Ta = 288 [K] Pa = 100 [kPa] To3 = 1100 [K] Pr = 4
Ettab = 0.98 Ettac = 0.85 Etta t = 0.87 Ettam = 0.99 Effecthe = 0.8
Ploss b = 0.02 Ploss ha = 0.03 Ploss hg = 4
cpa = 1.005 cpg = 1.148 R = 0.287 Gam a = 1.4 Gam g = 1.33333
HHV = 43100 Tf = 298 cpf = 1.2
To1 = Ta
Po1 = Pa
Po2 = Po1 · Pr
Gamra = Gam a – 1
Gam a
Gamrg = Gam g – 1
Gam g
To2d
To1 =
Po2
Po1
Gamra
Ettac = To2d – To1
To2 – To1
Wc = cpa · ( To2 – To1 )
Example 2.1: Using EES
Wtc = Wc
Ettam
Po3 = Po2 · ( 1 – Ploss b – Ploss ha )
Po4 = Po1 + Ploss hg
Etta t = To3 – To4
To3 – To4d
To3
To4d =
Po3
Po4
Gamrg
Wt = cpg · ( To3 – To4 )
Wnet = Wt – Wtc
Effecthe = To5 – To2
To4 – To2
( 1 + f ) · cpg · ( To3 – 288 ) = f · HHV + cpa · ( To5 – 288 ) + f · cpf · ( Tf – 288 )
fact = f
Ettab
sfc = fact
Wnet · 3600
Etta = Wnet
fact · HHV
Example 2.1: Using EES
cpa = 1.005 cpf = 1.2 cpg = 1.148 Effecthe = 0.8 Etta = 0.2858 Ettab = 0.98 Ettac = 0.85
Ettam = 0.99 Ettat = 0.87 f = 0.01089 fact = 0.01111 Gamra = 0.2857 Gamrg = 0.25 Gama = 1.4
Gamg = 1.333 HHV = 43100 Pa = 100 [kPa] Plossb = 0.02 Plossha = 0.03 Plosshg = 4 Po1 = 100 [kPa]
Po2 = 400 Po3 = 380 Po4 = 104 Pr = 4 R = 0.287 [kJ/kgK] sfc = 0.2922 Ta = 288 [K]
Tf = 298 To1 = 288 [K] To2 = 452.7 To2d = 428 To3 = 1100 [K] To4 = 835.2 To4d = 795.6
To5 = 758.7 Wc = 165.5 Wnet = 136.8 Wt = 304 Wtc = 167.2
Example 2.2
• A simple gas turbine with free turbine power having the following data:– turbine inlet temperature 1350(K)– compressor pressure ratio 12.0– isentropic efficiency of compressor 0.86– isentropic efficiency of each turbine 0.89– combustion efficiency 0.99– mechanical efficiency of each shaft 0.99– ambient conditions 1(bar), 288 (k)– combustion chamber pressure loss 6%
comp. deliv. pres.– exhaust pressure loss 0.03(bar)
Example 2.2
02, ( 1) / ( 1) /02 02
02, 01 01 01 0102 01 01
02 0202 01
01 01
(1.4 1) /1.402 01
02 01
1 ( ) 1 ( ) 1
and 1 1
( )(12) 1 1.005 346.3288 346.3( ),
0.86 0.99
351.5( / )
ideal
idealc
c
ptc
m
T p pT T T p p
T T TT TT TT T
c T T XT T K w
kJ kg
Example 2.202 01 01 02
03 04
03 04 04 03 03 04
04 04
03 04 03 03
04,03 04,
03
288 346.3 634.3( )
( ), 1.148( / ), and 1.333
306.2( ) and 793.8( )
1 1
1 (1
tc pg pg
tc
pg
tidealideal
T T T K
w c T T c kJ kgK
wT T K T T T K
c
T T
T T T TTT T
T
( 1) /04
03
04
/( 1)04 03
03
03 02 04
)
1
(1 ) 3.243
12(1 0.06) 11.28( ), thus 3.478( )t
b
p
p
T
p T
p
p p p bar p bar
Example 2.2
05
04 05
04 05,
( 1) /0504 05 04
04
0.25
04 05
1 0.03 1.03( )
, thus,
(1 ( ) )
1.030.89 1043.8 (1 ( ) )
3.478243.7( )
( ) 277( / )
a exhaust
tideal
t
tp m pg
p p p bar
T T
T T
pT T T
p
X X
K
w c T T kJ kg
Example 2.2
0.02043600 0.265 /
277.9tp
fSFC kg kWh
w
3600 360031.5 %
0.265 43100netSFCXQ X
th
b
f 0.0202 and 0.0204thf f
Example 2.3• Reheat Cycle
– net work output 240 MW – turbine inlet temperature 1525(K)– compressor pressure ratio 30– polytropic efficiency for compressor and turbines
»0.89
– Pressure loss in first combustor is 2%– Pressure loss in second combustor 4%– exhaust pressure 1.02 (bar)– ambient conditions 288(K), 1.01(bar)
Assume that pressure ratios in HP and LP turbines are equal
Example 2.3
( 1) /02 0202
01 01
02 01
( 1) / ( 1) / 0.321
( ) /( 1) ( 1) / 0.2223
( ) 2.98, 858.1( )
( ) 573( / )
c
c
n n
c p
n n
n n
T pT K
T p
w c T T kJ kg
Example 2.3
06
0202 01 03 02 1
01
03 03 03
04 06 04
loss 04 0303 04
assume 1.02( )
30.3( ) and 29.69( )
29.695.395. Take 5.3 to consider
1.02
1 4 % P in the , , 5.602( )
/
exhaust
b
p p bar
pp p bar p p p bar
p
p p p
p p p
reheater thus p p barp p
T
( 1) /( 1) / 0.89(1.333 1) /1.33303 03 03
04 04 04
04
05 04 2
( ) ( ) 5.3 1.449
1052.6( )
5.602(1 0.04) 5.378( )
tn n
b
p p
T p p
T K
p p p bar
Example 2.303 04
05 06
( ) 536.9( / )
( ) 535.5( / )
499.4( / )
HP m pg
LP m pg
net HP LP c
w c T T kJ kg
w c T T kJ kg
w w w w kJ kg
.
480.6( / )net
net
Wm kg s
w
, 1theoretical 0.0197th bf
, 2
, 1 , 2
b
theoretical 0.0142
0.0342
th b
th b th b
f
f ff
33.9 %
netw
f X HHV