Diesel / Brayton Cycles
ASEN 3113
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)
Fuel injection starts
Early CI engine
Fuel injection starts
Modern CI engine
• 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
Early CI Engine Cycle and the Thermodynamic Diesel Cycle
Fuel injected
at TC
A
I
R
Combustion
Products
Air
Actual
Cycle
Intake
Stroke
Compression
Stroke
Power
Stroke
Qin
Diesel
Cycle
Exhaust
Stroke
Qout
Air
BC
Compression
Process
Const pressure
heat addition
Process
Expansion
Process
Const volume
heat rejection
Process
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
Cut-off ratio:
Air-Standard Diesel cycle
rc 

vc
v
 3 (BOOK )
vb
v2
First Law Analysis of Diesel Cycle
Equations for processes 12, 41 are the same as those presented
for the Otto cycle
23 Constant Pressure Heat Addition
now involves heat and work
Q
P V  V2 
(u3  u 2 )  ( in )  2 3
m
m
Qin
 (u3  P3v3 )  (u 2  P2 v2 )
m
Qin
 (h3  h2 )  c p (T3  T2 )
m
P
T
v
RT2 RT3

 3  3  rc
v2
v3
T2 v2
AIR
Qin
3  4 Isentropic Expansion
(u 4  u3 ) 
W
Q
 ( out )
m
m
Wout
 (u3  u4 )  cv (T3  T4 )
m
note v4=v1 so
v4 v4 v2 v1 v2 r
    
v3 v2 v3 v2 v3 rc
P4v4 P3v3
P T r

 4  4 c
T4
T3
P3 T3 r
T4  v3 
  
T3  v4 
k 1
r 
 c 
r
k 1
AIR
Thermal Efficiency
 Diesel
cycle
Qout m
u4  u1
 1
 1
Qin m
h3  h2
For cold air-standard the above reduces to:
Diesel
const cV
k
1 1 rc 1

 1 k1  
r 
k
 rc 1

recall,
Otto  1 
1
r k 1
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?
Thermal Efficiency
Typical CI Engines
15 < r < 20
When rc (= v3/v2)1 the Diesel cycle efficiency approaches the
efficiency of the Otto cycle
Higher efficiency is obtained by adding less heat per cycle, Qin,
 run engine at higher speed to get the same power.
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:
k  1  Qin  1

 k 1
rc  1 
k  P1V1  r
k = 1.3
W net
MEP 
Vmax  Vmin
- compares performance
of engines of the same
size
k = 1.3
as Qin 0, rc1
Modern CI Engine Cycle and the Thermodynamic Dual Cycle
Fuel injected
at 15o before
TDC
A
I
R
Air
Combustion
Products
Actual
Cycle
Intake
Stroke
Compression
Stroke
Power
Stroke
Qin
Dual
Cycle
Air
Exhaust
Stroke
Qin
Qout
TC
BC
Compression
Process
Const volume
heat addition
Process
Const pressure
heat addition
Process
Expansion
Process
Const volume
heat rejection
Process
Dual Cycle
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
2.5
Qin
3
3
2
Qin
4
2.5
4
2
1
1
Qout
Qin
 (u2.5  u2 )  (h3  h2.5 )  cv (T2.5  T2 )  c p (T3  T2.5 )
m
Thermal Efficiency
 Dual  1 
cycle
Qout m
u4  u1
 1
Qin m
(u2.5  u2 )  (h3  h2.5 )
 Dual
const cv
where rc 

rck  1
1 
 1  k 1 
r  (  1)  k rc  1
v3
v2.5
and  
P3
P2
Note, the Otto cycle (rc=1) and the Diesel cycle (=1) are special cases:
Otto  1 
1
r k 1
 Diesel
const cV




1  1 rck  1 
 1  k 1  

r  k rc  1 
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  into more natural variables yields
rc  1 
k  1  Qin  1
  1



k  P1V1  r k 1 k  1 

1 P3
r k P1
For the same initial conditions P1, V1 and the same compression ratio:
Otto   Dual   Diesel
For the same initial conditions P1, V1 and the same peak pressure P3
(actual design limitation in engines):
 Diesel   Dual  otto
Brayton Cycle
• Introduced by George
Brayton (an
American) in 1872
• Used separate
expansion and
compression cylinder
• Constant Combustion
process
Brayton Cycle
18
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
Schematic of simple cycle
Idealized Brayton Cycle
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
22
Brayton cycle analysis
• 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.
Efficiency:
Net work:
w net

q in
w net  w turb  w comp
Brayton cycle analysis
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:
wcomp  h2  h1
24
Brayton cycle analysis
2 to 3 (constant pressure heat addition treated as a heat exchanger)
q in  q 23  h 3  h 2
3 to 4 (isentropic expansion in turbine)
 w turb  h 4  h 3 , or
w turb  h 3  h 4
25
Brayton cycle analysis
4 to 1 (constant pressure heat rejection)
q out  h1  h 4 ,
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.
q out  h 4  h1
26
Brayton cycle analysis
net work:
w net  w turb  w comp
Substituting:
w net  (h 3  h 4 )  (h 2  h1 )
Brayton cycle analysis
Thermal efficiency:
w net (h 3  h 4 )  (h 2  h1 )


q in
(h 3  h 2 )
(h 4  h1 )
 1 
(h 3  h 2 )
Brayton cycle analysis
assume cold air conditions and manipulate
the efficiency expression:
 1 
c p (T4  T1 )
c p (T3  T2 )
T1 T4 T1  1
  1
T2 T3 T2  1
Brayton cycle analysis
Using the isentropic relationships,
T2  p 2 
 
T1  p1 
k 1
k
;
T4  p4 
 
T3  p3 
k 1
k
 p1 
 
 p2 
k 1
k
Define:
P2 P3
rp  pressure ratio 

P1 P4
30
Brayton cycle analysis
Then we can relate the temperature ratios to
the pressure ratio:
T3
T2
 k 1 k
 rp

T1
T4
Plug back into the efficiency
expression and simplify:
  1
1
k 1 k
rp
Brayton cycle analysis
32
Brayton cycle analysis
An important quantity for Brayton cycles is
the Back Work Ratio (BWR).
BWR 
w comp
w turb
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 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.
Draw diagram
P
2
3
1
4
v
Start analysis
Let’s get the efficiency:
  1
1
k 1 k
rp
From problem statement, we know rp = 4.5
  1
1
1.41 1.4
4.5
 0.349

Net power output:
Net Power:

Ýnet  m
Ýw net  m
Ý w turb  w comp
W

Substituting for work terms:
Ý m
Ý(h3 h4 ) (h2 h1)
W
net
Applying constant specific heats:
Ýnet  m
Ýc p (T3 T4 ) (T2 T1)
W
Need to get T2 and T4
Use isentropic relationships:
 p2 
T2


T1
 p1 
k 1
k
 p4 
T4




T3
p
 3
;
k 1
k
T1 and T3 are known along with the
pressure ratios:
T2: T2  3004.5
 461 K
0.4 1.4
T4:
T4  11000.222
0.4 1.4
 715.7 K
Ýnet  m
Ýc p (T3 T4 ) (T2 T1)
W

Net power is then: W
net
 1120 kW
Back Work Ratio
BWR 
w comp
w turb

h 2  h1
h3  h4
Applying constant specific heats:
T2  T1
461  300
BWR 

 0.42
T3  T4 1100  715.7
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
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
w t ,a h 3  h 4,act
t 

w t ,s
h 3  h 4,s
w c ,s
h1  h 2,s
c 

w c ,a h1  h 2,act