diesel / brayton cycles asen 3113. diesel cycle invented by rudolf christian karl diesel in 1893...
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Diesel Cycle• Invented by Rudolf
Christian Karl Diesel in 1893
• First engine was powered by powdered coal
• Achieved a compression ratio of almost 80
• Exploded, almost killed Diesel
• First working engine completed 1894 - generated 13 hp
Diesel Engine
• Also known as Compression Ignition Engine (CI)
• Can this engine “knock”?
• Difference from Otto Cycle?
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
Early CI Engine Cycle and the Thermodynamic Diesel Cycle
A I R
CombustionProducts
Fuel injectedat TC
IntakeStroke
Air
Air
BC
CompressionStroke
PowerStroke
ExhaustStroke
Qin Qout
CompressionProcess
Const pressure heat addition
Process
ExpansionProcess
Const volume heat rejection
Process
ActualCycle
DieselCycle
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
€
rc =vc
vb
=v3
v2
(BOOK)
Cut-off ratio:
( )m
VVP
m
Quu in 232
23 )()(−
−+=−
AIR23 Constant Pressure Heat Addition
now involves heat and work
)()( 222333 vPuvPum
Qin +−+=
)()( 2323 TTchhm
Qp
in −=−=
crv
v
T
T
v
RT
v
RTP ==→==
2
3
2
3
3
3
2
2
Qin
First Law Analysis of Diesel Cycle
Equations for processes 12, 41 are the same as those presented for the Otto cycle
)()( 34 m
W
m
Quu out+−=−
AIR
3 4 Isentropic Expansion
)()( 4343 TTcuum
Wv
out −=−=
note v4=v1 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
vP
T
vP c⋅=→=3
4
3
4
3
33
4
44
11
4
3
3
4−−
⎟⎠⎞
⎜⎝⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
kc
k
rr
vv
TT
23
1411hh
uu
mQ
mQ
in
out
cycleDiesel −
−−=−=η
€
ηDieselconst cV
=1−1
rk−1
1
k⋅
rck −1( )
rc −1( )
⎡
⎣ ⎢ ⎢
⎤
⎦ ⎥ ⎥
For cold air-standard the above reduces to:
Thermal Efficiency
1
11 −−= kOtto
rηrecall,
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 efficiencythan the Otto cycle
So why is a Diesel engine usually more efficient?
Typical CI Engines15 < r < 20
When rc (= v3/v2)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.
k = 1.3
k = 1.3
The cut-off ratio is not a natural choice for the independent variablea more suitable parameter is the heat input, the two are related by:
111
111 −⎟⎟
⎠
⎞⎜⎜⎝
⎛−−= kin
crVP
Qk
kr as Qin 0, rc1
€
MEP =Wnet
Vmax −Vmin
- compares performance of engines of the same size
Modern CI Engine Cycle and the Thermodynamic Dual Cycle
A I R
CombustionProducts
Fuel injectedat 15o before TDC
IntakeStroke
Air
AirTC
BC
CompressionStroke
PowerStroke
ExhaustStroke
Qin Qout
CompressionProcess
Const pressure heat addition
Process
ExpansionProcess
Const volume heat rejection
Process
ActualCycle
DualCycle
Qin
Const volume heat addition
Process
Process 1 2 Isentropic compressionProcess 2 2.5 Constant volume heat additionProcess 2.5 3 Constant pressure heat additionProcess 3 4 Isentropic expansionProcess 4 1 Constant volume heat rejection
Dual Cycle
Qin
Qin
Qout
11
2
2
2.5
2.5
33
44
)()()()( 5.2325.25.2325.2 TTcTTchhuum
Qpv
in −+−=−+−=
Thermal Efficiency
)()(11
5.2325.2
14
hhuu
uu
mQ
mQ
in
out
cycleDual −+−
−−=−=η
( )⎥⎦⎤
⎢⎣
⎡−+−
−−= − 1)1(
111 1
c
kc
kcconst
Dual rk
r
rv αα
αη
1
11 −−= kOtto
rη
( )( )⎥⎦
⎤⎢⎣
⎡
−−
⋅−= − 1
1111
1c
kc
kconst cDiesel
r
r
krV
η
Note, the Otto cycle (rc=1) and the Diesel cycle (=1) are special cases:
2
3
5.2
3 and where PP
vvrc ==
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), orii) maximum pressure P3.
Transformation of rc and into more natural variables yields
⎥⎦
⎤⎢⎣
⎡
−−−⎟⎟
⎠
⎞⎜⎜⎝
⎛−−= − 1
1111
111 krVP
Q
k
kr
kin
cα
α1
31
P
P
rk=
For the same initial conditions P1, V1 and the same compression ratio:
DieselDualOtto ηηη >>
For the same initial conditions P1, V1 and the same peak pressure P3 (actual design limitation in engines):
ottoDualDiesel ηηη >>
Brayton Cycle
• Introduced by George Brayton (an American) in 1872
• Used separate expansion and compression cylinder
• Constant Combustion process
Other applications of Brayton cycle
• Power generation - use gas turbines to generate electricity…very efficient
• Marine applications in large ships• Automobile racing - late 1960s Indy 500
STP sponsored cars
22
Brayton Cycle
•1 to 2--isentropic compression•2 to 3--constant pressure heat addition (replaces combustion process)•3 to 4--isentropic expansion in the turbine •4 to 1--constant pressure heat rejection to return air to original state
Brayton cycle analysis
in
net
q
w=η
compturbnet www −=
Efficiency:
Net work:
• Because the Brayton cycle operates between two constant pressure lines, or isobars, the pressure ratio is important.
•The pressure ratio is not a compression ratio.
24
€
wcomp = h2 − h1
1 to 2 (isentropic compression in compressor), apply first law
**When analyzing the cycle, we know that the compressor work is in (negative). It is standard convention to just drop the negative sign and deal with it later:
Brayton cycle analysis
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)
26
,hhq 41out −=
14out hhq −=
4 to 1 (constant pressure heat rejection)
We know this is heat transfer out of the system and therefore negative. In book, they’ll give it a positive sign and then subtract it when necessary.
Brayton cycle analysis
Thermal efficiency:
Brayton cycle analysis
in
net
q
w=η
)h(h
)h(h)h(h
23
1243
−−−−
=
)h(h
)h(h1
23
14
−−
−=η
Brayton cycle analysis
assume cold air conditions and manipulate the efficiency expression:
)T(Tc
)T(Tc1
23p
14p
−−
−=η
( )( )1TT
1TT
T
T1
23
14
2
1
−
−−=η
30
T
T
p
p
k
k2
1
2
1
1
=⎛
⎝⎜
⎞
⎠⎟
−
;T
T
p
p
p
p
k
k
k
k4
3
4
3
1
1
2
1
=⎛
⎝⎜
⎞
⎠⎟ =
⎛
⎝⎜
⎞
⎠⎟
− −
Using the isentropic relationships,
Define:
4
3
1
2p P
P
P
Pratiopressurer ===
Brayton cycle analysis
( )
4
3k1kp
1
2
T
Tr
T
T== −
Brayton cycle analysis
Then we can relate the temperature ratios to the pressure ratio:
Plug back into the efficiency expression and simplify:
( ) k1kpr
11
−−=η
Brayton cycle analysis
An important quantity for Brayton cycles is the Back Work Ratio (BWR).
turb
comp
w
wBWR=
The Back-Work Ratio is the Fraction of Turbine Work Used to Drive the
Compressor
The Back-Work Ratio is the Fraction of Turbine Work Used to Drive the
Compressor
EXAMPLE PROBLEM
The pressure ratio of an air standard Brayton cycle is 4.5 and the inlet conditions to the compressor are 100 kPa and 27C. The turbine is limited to a temperature of 827C and mass flow is 5 kg/s. Determine
a) the thermal efficiencyb) the net power output in kWc) the BWR
Assume constant specific heats.
Start analysis
Let’s get the efficiency:
( ) k1kpr
11
−−=η
From problem statement, we know rp = 4.5
( ) 349.05.4
11
4.114.1=−=
−η
Net power output:
Substituting for work terms:
€
˙ W net = ˙ m w net = ˙ m w turb − w comp( )
Net Power:
€
˙ W net = ˙ m (h3 −h4 )− (h2 −h1)( )
€
˙ W net = ˙ m cp (T3 −T4 )− (T2 −T1)( )
Applying constant specific heats:
Need to get T2 and T4
Use isentropic relationships:
T
T
p
p
k
k2
1
2
1
1
=⎛
⎝⎜
⎞
⎠⎟
−
;k
1k
3
4
3
4
p
p
T
T−
⎟⎟⎠
⎞⎜⎜⎝
⎛=
T1 and T3 are known along with the pressure ratios:
( ) K4614.5300T 1.40.42 ==T2:
T4: ( ) K7.7150.2221100T 1.40.44 ==
Net power is then: kW1120Wnet =&
€
˙ W net = ˙ m cp (T3 −T4 )− (T2 −T1)( )
Back Work Ratio
43
12
turb
comp
hh
hh
w
wBWR
−−
==
Applying constant specific heats:
42.07.7151100
300461
TT
TTBWR
43
12 =−−
=−−
=
Brayton Cycle
• In theory, as the pressure ratio goes up, the efficiency rises. The limiting factor is frequently 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
Brayton Cycle
• Irreversibilities– Compressor and turbine frictional effects -
cause increase in entropy– Also friction causes pressure drops through
heat exchangers– Stray heat transfers in components– Increase in entropy has most significance
• wc = h2 – h1 for the ideal cycle, which was isentropic
• wt = h3 – h4 for the ideal isentropic cycle