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Diesel / Brayton Cycles
ASEN 3113
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Diesel Cycle
Invented by Rudolf Christian Karl Diesel in1893
First engine was powered bypowdered coal
Achieved a compression ratioof almost 80
Exploded, almost killedDiesel
First working enginecompleted 1894 - generated13 hp
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Diesel Engine
Also known asCompressionIgnition Engine
(CI) Can this engine
³knock´?
Difference fromOtto Cycle?
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Thermodynamic Cycles for CI engines
In early CI engines the fuel was injected when the piston reached TDC
and thus combustion lasted well into the expansion stroke.
In modern engines the fuel is injected before TDC (about 15o)
The combustion process in the early CI engines is best approximated by
a constant pressure heat addition process Diesel Cycle
The combustion process in the modern CI engines is best approximated
by a combination of constant volume and constant pressure Dual Cycle
Fuel injection startsFuel injection starts
Early CI engine Modern CI engine
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Early CI Engine Cycle and the Thermodynamic Diesel Cycle
AI
R
CombustionProducts
Fuel injectedat TC
Intake
Stroke
Air
Air
BC
Compression
Stroke
Power
Stroke
Exhaust
Stroke
Qin Qout
Compression
Process
Const pressure
heat addition
Process
Expansion
Process
Const volume
heat rejection
Process
Actual
Cycle
DieselCycle
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Process a b
Isentropic
compression
Process b
c Constant
pressure heat
addition
Process c d
Isentropic
expansion
Process d a
Constant volume
heat rejection
- a=1,b=2,etc«for
book
Air-Standard Diesel cycle
r c!
vc
vb
!
v3
v2
( BOOK )
Cut-off ratio:
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m
V V P
m
Q
uu
in 232
23 )()(
!
AIR
23 Constant Pressure Heat Addition
now involves heat and work
)()( 222333 v P uv P um
Qin!
)()( 2323 T T chhm
Q p
in !!
cr v
v
T
T
v
RT
v
RT P !!p!!
2
3
2
3
3
3
2
2
Qin
First LawAnalysis of Diesel Cycle
Equations for processes 12, 41 are the same as those presented
for the Otto cycle
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)()( 34m
W
m
Quu out ! AIR
3 4 Isentropic Expansion
)()( 4343 T T cuum
W v
out !!
notev 4=v
1 so cr
r
v
v
v
v
v
v
v
v
v
v!!!
3
2
2
1
3
2
2
4
3
4
r
r
T
T
P
P
T
v P
T
v P c!p!
3
4
3
4
3
33
4
44
11
4
3
3
4
¹ º ¸©
ª¨!¹¹
º ¸©©
ª¨!
k
c
k
r
r
v
v
T
T
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23
1411hhuu
mQmQ
in
out
cycle Diesel
!!L
L Diesel con st cV !1
1
r k 1
1
k
r c
k 1 r c 1
«
-¬¬
»
½¼¼
For cold air-standard the above reduces to:
Thermal Efficiency
1
11
!k
Otto r
Lrecall,
Note the term in the square bracket is always larger than one so for the
same compression ratio, r , the Diesel cycle has a lower thermal efficiency
than the Otto cycle
So why is a Diesel engine usually more efficient?
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Typical CI Engines
15 < r < 20
When r c
(= v3/v
2)1 the Diesel cycle efficiency approaches the
efficiency of the Otto cycle
Thermal Efficiency
Higher efficiency is obtained by adding less heat per cycle, Qin,
run engine at higher speed to get the same power.
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k = 1.3
k = 1.3
The cut-off ratio is not a natural choice for the independent variable
a more suitable parameter is the heat input, the two are related by:
111
111¹¹
º ¸©©
ª¨! k
inc
r V P Q
k k r as Qin 0, r c 1
ME P ! W net V max V min
- compares performance
of engines of the samesize
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Modern CI Engine Cycle and the Thermodynamic Dual Cycle
AI
R
CombustionProducts
Fuel injectedat 15o beforeTDC
Intake
Stroke
Air
Air
TC
BC
Compression
Stroke
Power
Stroke
Exhaust
Stroke
Qin Qout
Compression
Process
Const pressure
heat addition
Process
Expansion
Process
Const volume
heat rejection
Process
Actual
Cycle
DualCycle
Qin
Const volume
heat addition
Process
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Process 1 2 Isentropic compression
Process 2 2.5 Constant volume heat additionProcess 2.5 3 Constant pressure heat addition
Process 3 4 Isentropic expansion
Process 4 1 Constant volume heat rejection
Dual Cycle
Q in
Q in
Q out
11
2
2
2.5
2.5
33
44
)()()()( 5.2325.25.2325.2 T T cT T chhuu
m
Q pv
in !!
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Thermal Efficiency
)()(115.2325.2
14hhuu
uu
mQ
mQ
in
out
cycle Dual
!!L
¼½»
¬-
«
!
1)1(
111
1 c
k
c
k
ccon st Dual
r k
r
r v EE
E
L
1
11
!
k Ottor
L ¼½
»¬-
«
! 1
1111
1c
k c
k con st c Diesel
r
r
k r V
L
Note, the Otto cycle (r c=1) and the Diesel cycle (E=1) are special cases:
2
3
5.2
3 andwhere P
P v
vr c !! E
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The use of the Dual cycle requires information about either:
i) the fractions of constant volume and constant pressure heat addition(common assumption is to equally split the heat addition), or
ii) maximum pressure P3.
Transformation of r c and E into more natural variables yields
¼½»¬
-«
¹¹
º ¸©©
ª¨!
11111
111 k r V P
Q
k
k r k
inc
E
E1
31
P
P
r k
!E
For the same initial conditions P1, V1 and the same compression ratio:
Diesel Dual Otto LLL ""For the same initial conditions P1, V1 and the same peak pressure P3
(actual design limitation in engines):
otto Dual Diesel LLL ""
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Brayton Cycle
Introduced by George
Brayton (an
American) in 1872
Used separateexpansion and
compression cylinder
Constant Combustion
process
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18
Brayton Cycle
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Other applications of Braytoncycle
Power generation - use gas turbines togenerate electricity«very efficient
Marine applications in large ships
Automobile racing - late 1960s Indy 500STP sponsored cars
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Schematic of simple cycle
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Idealized Brayton Cycle
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22
Brayton Cycle
1 to 2--isentropiccompression
2 to 3--constant pressure
heat addition (replacescombustion process)
3 to 4--isentropicexpansion in the turbine
4 to 1--constant pressureheat rejection to return airto original state
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Brayton cycle analysis
in
net
q
w!L
com pturbnet www !
Efficiency:
Net work:
Because the Brayton cycle operates between two constantpressure lines, or isobars, the pressure ratio is important.
The pressure ratio is not a compression ratio.
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24
wcomp
! h2 h
1
1 to 2 (isentropic compression incompressor), apply first law
**When analyzing the cycle, we know thatthe compressor work is in (negative). It isstandard convention to just drop the negativesign and deal with it later:
Brayton cycle analysis
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25
2323in hhqq !!
2 to 3 (constant pressure heat addition -treated as a heat exchanger)
Brayton cycle analysis
or ,hhw 34turb !
43turb hhw !
3 to 4 (isentropic expansion in turbine)
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26
,hhq 41out !
14out hhq !
4 to 1 (constant pressure heat rejection)
We know this is heat transfer out of thesystem and therefore negative. In book,they¶ll give it a positive sign and then
subtract it when necessary.
Brayton cycle analysis
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Brayton cycle analysis
Substituting:
com pturbnet www
!
net work:
)h(h)h(hw 1243net !
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Thermal efficiency:
Brayton cycle analysis
in
net
qw!L
)h(h)h(h)h(h
23
1243
!
)h(h
)h(h1
23
14
!L
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Brayton cycle analysis
assume cold air conditions and manipulatethe efficiency expression:
)T(Tc
)T(Tc1
23 p
14 p
!L
1TT
1TT
T
T1
23
14
2
1
!L
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30
TT
p p
k
k 2
1
2
1
1
! ¨ª© ¸
º¹
; TT
p p
p p
k
k
k
k 4
3
4
3
1
1
2
1
! ¨ª© ¸
º¹ ! ¨
ª© ¸
º¹
Using the isentropic relationships,
Define:
4
3
1
2 p
P
P
P
Pratio pressurer !!!
Brayton cycle analysis
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4
3k 1k p
1
2
TTr
TT !!
Brayton cycle analysis
Then we can relate the temperature ratios tothe pressure ratio:
Plug back into the efficiency
expression and simplify:
k 1k
pr
11
!L
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32
Brayton cycle analysis
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Brayton cycle analysis
An important quantity for Brayton cycles isthe Back Work Ratio (BWR).
turb
com p
w
wBWR !
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The Back-Work Ratio is the Fractionof Turbine Work Used to Drive the
Compressor
The Back-Work Ratio is the Fractionof Turbine Work Used to Drive the
Compressor
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EXAMPLE PROBLEM
The pressure ratio of an air standard Braytoncycle is 4.5 and the inlet conditions to thecompressor are 100 kPa and 27rC. The
turbine is limited to a temperature of 827rCand mass flow is 5 kg/s. Determine
a) the thermal efficiency
b) the net power output in kWc) the BWR
Assume constant specific heats.
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Draw diagram
P
v
1
2 3
4
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Start analysis
Let¶s get the efficiency:
k 1k pr
1
1
!L
From problem statement, we know rp = 4.5
349.05.4
114.114.1
!!
L
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Net power output:
Substituting for work terms:
Ý Wnet ! Ý m w net ! Ý m w turb w com p Net Power:
Ý Wnet ! Ý m (h3h4 ) (h2 h1)
Ý Wnet ! Ý mc p (T3T4 ) (T2 T1)
Applying constant specific heats:
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Need to get T2 and T4
Use isentropic relationships:
T
T
p
p
k
k
2
1
2
1
1
!
¨
ª©
¸
º¹
;
k
1k
3
4
3
4
p
p
T
T
¹¹ º
¸©©ª
¨!
T1 and T3 are known along with thepressure ratios:
K 4614.5300T1.40.4
2 !!T2:
T4: K 7.7150.2221100T1.40.4
4 !!
Net power is then: kW1120Wnet
!
Ý Wnet ! Ý mc p (T3T4 ) (T2 T1)
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Back Work Ratio
43
12
turb
com p
hh
hh
w
wBWR
!!
Applying constant specific heats:
42.0
7.7151100
300461
TT
TTBWR
43
12!
!
!
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Brayton Cycle
In theory, as the pressure ratio goes up,the efficiency rises. The limiting factor isfrequently the turbine inlet
temperature. The turbine inlet temp is restricted to
about 1,700 K or 2,600 F.
Consider a fixed turbine inlet temp., T3
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Brayton Cycle
Irreversibilities
± Compressor and turbine frictional effects -cause increase in entropy
± Also friction causes pressure drops throughheat exchangers
± Stray heat transfers in components
± Increase in entropy has most significance
wc = h2 ± h1 for the ideal cycle, which wasisentropic
wt = h3 ± h4 for the ideal isentropic cycle
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Brayton Cycle
In order to deal with irreversibilities, weneed to write the values of h2 and h4 ash2,s and h4,s.
Then
s,43
act,43
s,t
a,tt
hh
hh
w
w
!!L
act,21
s,21
a,c
s,c
chh
hh
w
w
!!L