Lec 22: Diesel cycle, Brayton
cycle
• For next time:
– Read: § 8-8 to 8-10.
– Methods section due Monday, November 17,
2003
• Outline:
– Diesel cycle
– Brayton cycle
– Example problem
• Important points:
– Understand differences between the Otto,
Diesel and Brayton cycles
– Know how to compute the cycles using
variable or constant specific heats
– Know what components make up each of the
cycles.
Brayton Cycle
Proposed by
George
Brayton in
1870!
http://www.pwc.ca/en_markets/demonstration.html
3
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
TEAMPLAY
• What expression of the first law applies?
u or h?
Why?
6
Idealized Brayton Cycle
Brayton Cycle
8
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
9
Brayton Cycle
• 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.
10
Brayton cycle analysis
As with any cycle, we’re going to concern
ourselves with the efficiency and net work
output:
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:
wcomp  h 2  h 1
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:
w comp  h 2  h1
12
Brayton cycle analysis
2 to 3 (constant pressure heat addition treated as a heat exchanger)
q in  q 23  h 3  h 2
13
Brayton cycle analysis
3 to 4 (isentropic expansion in turbine)
 w turb  h 4  h 3 , or
w turb  h 3  h 4
14
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
15
Brayton cycle analysis
Let’s get the net work:
w net  w turb  w comp
Substituting:
w net  (h 3  h 4 )  (h 2  h1 )
Brayton cycle analysis
Let’s get the 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
Let’s 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
Let’s define:
P2 P3
rp  pressure ratio 

P1 P4
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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
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What does this expression
assume?
  1
1
k 1 k
rp
Brayton cycle analysis
An important quantity for Brayton cycles is
the Back Work Ratio (BWR).
BWR 
w comp
w turb
Why might this be important?
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:

 m
 w net  m
 w turb  w comp
W
net

Substituting for work terms:
 m

(h 3  h 4 )  (h 2  h1 )
W
net
Applying constant specific heats:
 m
 cp (T3  T4 )  (T2  T1 )
W
net
Need to get T2 and T4
Use isentropic relationships:
T2  p 2 
 
T1  p1 
k 1
k
T4  p 4 
  
;
T3  p 3 
k 1
k
T1 and T3 are known along with the
pressure ratios:
Solving for temperatures:
T2:
T2  3004.5
T4:
T4  11000.222
 461 K
0.4 1.4
0.4 1.4
 715.7 K
Net power is then:
Wnet  (5 kg/ s)(1.0035 kJ / (kg K )) 
(1100
 715.7) 
(461 300)
  1120 kW
W
net

K
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
We can also do the analysis with variable
specific heats….we’ll use relative pressures.
p r2
 p2 
p2

 
p1
p r1
 p1  s
p r3 p 2
p3  p3 
  

p 4  p 4  s p r4
p1
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TEAMPLAY
Work the same problem with EES using
variable specific heats. How do the
numbers change?
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
For fixed values of Tmin and Tmax, the net
work of the Brayton cycle first increases
with the pressure ratio, then reaches a
maximum at rp=(Tmax/Tmin)k/[2(k-1)], and
finally decreases
Brayton Cycle
• Irreversibilities
• We had wc = h2 – h1 for the ideal cycle,
which was isentropic
• And we had 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
Effect of irreversibilities on the
cycle
TEAMPLAY
Problem 8-69