2. 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
3. Diesel Engine
• Also known as
Compression
Ignition Engine
(CI)
• Can this engine
“knock”?
• Difference from
Otto Cycle?
4. 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 starts
Fuel injection starts
Early CI engine Modern CI engine
5. Early CI Engine Cycle and the Thermodynamic Diesel Cycle
A
I
R
Combustion
Products
Fuel injected
at 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
Diesel
Cycle
6. 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:
7. ( )
m
VVP
m
Q
uu in 232
23 )()(
−
−+=−
AIR
23 Constant Pressure Heat Addition
now involves heat and work
)()( 222333 vPuvPu
m
Qin
+−+=
)()( 2323 TTchh
m
Q
p
in
−=−=
cr
v
v
T
T
v
RT
v
RT
P ==→==
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
8. )()( 34
m
W
m
Q
uu out
+−=−
AIR
3 4 Isentropic Expansion
)()( 4343 TTcuu
m
W
v
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
−−
=
=
κ
χ
κ
ρ
ρ
ϖ
ϖ
Τ
Τ
9. 23
1411
hh
uu
mQ
mQ
in
out
cycle
Diesel
−
−
−=−=η
ηDiesel
const cV
=1−
1
rk−1
1
k
⋅
rc
k
−1( )
rc −1( )
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
For cold air-standard the above reduces to:
Thermal Efficiency
1
1
1 −
−= κΟττο
ρ
ηrecall,
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?
10. Typical CI Engines
15 < 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.
11. 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:
1
11
11
1 −
−
−= κ
ιν
χ
ρςΠ
Θ
κ
κ
ρ as Qin 0, rc1
MEP =
Wnet
Vmax −Vmin
- compares performance
of engines of the same
size
12. Modern CI Engine Cycle and the Thermodynamic Dual Cycle
A
I
R
Combustion
Products
Fuel injected
at 15o
before
TDC
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
Dual
Cycle
Qin
Const volume
heat addition
Process
13. Process 1 2 Isentropic compression
Process 2 2.5 Constant volume heat addition
Process 2.5 3 Constant pressure heat addition
Process 3 4 Isentropic expansion
Process 4 1 Constant volume heat rejection
Dual Cycle
Qin
Qin
Qout
1
1
2
2
2.5
2.5
3
3
4
4
)()()()( 5.2325.25.2325.2 TTcTTchhuu
m
Q
pv
in
−+−=−+−=
15. 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 rc and α into more natural variables yields
−
−
−
−
−= − 1
111
1 1
11 krVP
Q
k
k
r k
in
c
α
α
1
31
P
P
rk
=α
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 >>
16.
17. Brayton Cycle
• Introduced by George
Brayton (an
American) in 1872
• Used separate
expansion and
compression cylinder
• Constant Combustion
process
19. 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. 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
23. 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. 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. 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. 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
31. ( )
4
3k1k
p
1
2
T
T
r
T
T
== −
Brayton cycle analysis
Then we can relate the temperature ratios to
the pressure ratio:
Plug back into the efficiency
expression and simplify:
( ) k1k
pr
1
1 −
−=η
33. Brayton cycle analysis
An important quantity for Brayton cycles is
the Back Work Ratio (BWR).
turb
comp
w
w
BWR =
34. The Back-Work Ratio is the Fraction
of Turbine Work Used to Drive the
Compressor
35. 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 27°C. The
turbine is limited to a temperature of 827°C
and mass flow is 5 kg/s. Determine
a) the thermal efficiency
b) the net power output in kW
c) the BWR
Assume constant specific heats.
37. Start analysis
Let’s get the efficiency:
( ) k1k
pr
1
1 −
−=η
From problem statement, we know rp = 4.5
( ) 349.0
5.4
1
1 4.114.1
=−= −
η
38. Net power output:
Substituting for work terms:
˙Wnet = ˙mwnet = ˙m wturb − wcomp( )
Net Power:
˙Wnet = ˙m (h3 −h4 )− (h2 −h1)( )
˙Wnet = ˙mcp (T3 −T4 )− (T2 −T1)( )
Applying constant specific heats:
39. Need to get T2 and T4
Use isentropic relationships:
T
T
p
p
k
k
2
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
1.40.4
2 ==T2:
T4: ( ) K7.7150.2221100T
1.40.4
4 ==
Net power is then: kW1120Wnet =&
˙Wnet = ˙mcp (T3 −T4 )− (T2 −T1)( )
41. 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
42. 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
43. Brayton Cycle
• In order to deal with irreversibilities, we
need to write the values of h2 and h4 as h2,s
and h4,s.
• Then
s,43
act,43
s,t
a,t
t
hh
hh
w
w
−
−
==η
act,21
s,21
a,c
s,c
c
hh
hh
w
w
−
−
==η
