Brayton Cycle Summary
define some units
3
kJ := 10 ⋅J
QH_dot
3
2
Gas Turbine represented by air standard
Brayton cycle
ma_dot air
compressor
turbine
1
4
W_dotnet=
Wt_dot+Wc_dot
QL_dot
Brayton cycle consists of:
1-2 adiabatic compression
2-3 heat addition ~ constant pressure
3-4 adiabatic expansion in turbine
4-1 heat rejection ~ constant pressure
p-v and T - s plots for Brayton cycle shown below for reversible cycle. in
irreversible cycle, p 2 > p3 and p4 > p1 , s2 > s1 , s4 > s3
starting conditions
p 1_plot := 1
T1_plot := 25 + 273.15
after compression
p 2_plot := 10
max temperature after heat addition
T3_plot := 1000 + 273.15
calculations
11/14/2005
1
s1_plot := 1
p-v plot of Brayton cycle
adiabatic compression
heat addition
adiabatic expansion in turbine
heat rejection
{3}
10
pressure
{2}
5
{1}
0
0
0.2
0.4
0.6
{4}
0.8
1
1.2
1.4
1.6
1.8
volume
T-s plot of Brayton cycle (reversible)
1400
adiabatic compression
heat addition
adiabatic expansion in turbine
heat rejection
1200
{3}
temperature
1000
800
{4}
{2}
600
400
{1}
200
0.9
1
1.1
1.2
1.3
1.4
1.5
entropy
11/14/2005
2
1.6
1.7
1.8
1.9
Ideal (reversible) basic Brayton cycle
(
compressor work
wc = − h 2 − h 1
turbine work
wt = h 3 − h 4
η th =
qH + qL
qH
qL
=1+
(
qH
=
(
q H = h 3 − h 2
heat addition
(
qL = − h4 − h1
heat rejection
)
wt + wc
q H
)
h 4s := Cp ⋅ T4s − T1 + h 1
(2)
)
assuming perfect gas, constant specific heat.
h only a function of temperature; (5.23) VW &S, Joule's
experiment shows u is f(T) only, pv = RT => h=f(T).
)
h 3 := Cp ⋅ T3 − T2s + h 2s
factor out T1 / T2s
η th := 1 −
(1)
h 4s − h 1
T4s − T1
η th → 1 −
h 3 − h 2s
T4s
η th := 1 −
T3 − T2s
T1
⋅
T2s T3
T2s
γ−1
isentropic compression (and expansion) ...
 p 2s 
=
T1
 p1 
T2s
γ−1
p 2s
since
p1
=
 p 2s 
=
T1
 p1 
p3
T2s
p 4s
γ
T1
−1
−1
this is reversible adiabatic
process with ideal gas and
constant specific heat
γ
γ−1
 p3 
=
 p 4s 
γ
=
T3
T4s
=>
T4s
T1
=
T3
T2s
γ−1
 p1 
η th = 1 −
=1−
T2s
 p2s 
T1
γ
=1−
1
γ−1
 p 2s 

 p1 
1
=1−
r = pressure_ratio
γ−1
γ
r
γ
example; for 50 % efficiency, and some typical gas constants ...
 1.29 
γ :=
 1.4

 1.67 
η th := 0.5
η th = 1 −
CO2
air
r
1
γ −1
γ
γ
r
= 1 − η th
−γ
(
) γ −1
r = 1 − η th
monotonic gasses, He, Ar, Ne, He
i := 0 ..2 − γi
(
γ i−1
)
r := 1 − η th
i
11/14/2005
1
γ −1
3
 21.83 
r =  11.31

 5.63 
so for air as the working fluid,
a pressure ratio of 11.3 will
provide 0.5 isentropic
efficiency
effect of pressure ratio on isentropic efficiency
1
η th( r, γ ) := 1 −
r
 1.29 
γ =  1.4

 1.67 
r := 0 .. 25
γ −1
γ
efficiency vs pressure ratio
0.8
efficiency
0.6
0.5
0.4
CO2 - gamma = 1.29
air - gamma = 1.4
monotonic - gamma = 1.67
0.2
0
0
5
10
15
pressure ratio
11/14/2005
4
20
25
regeneration ...
T-s plot of Brayton cycle (reversible)
1400
QH_dot
2
3
1200
ma_dot air
temperature
1000
compressor
turbine
W_dotnet=
Wt_dot+Wc_dot
1
4
QL_dot
regenerator
800
600
η th =
wnet
qH
=
400
wt + wc
q H
(
)
(
)
q H = c p ⋅ T3 − Tx
wt = cp ⋅ T3 − T4
200
η th = 1 +
wc
wt
1.2
1.6
1.8
adiabatic compression
heat addition
adiabatic expansion in turbine
heat rejection
T2
T4
wt = q H
γ−1




γ


 T2
 p2 
T1 ⋅ 
T1 ⋅ 
−1
− 1
c p ⋅ ( T2 − T1 )
 T1
 =
 p1 

=1−
=1−
γ− 1
c p ⋅ ( T3 − T4 )
T4 


T3 ⋅  1 −

γ 
T3
  p1  


T3 ⋅ 1 − 

  p2  
b
a −1
form is ...
1.4
entropy
Tx = temperature into regenerator
out of regenerator = T2
max when Tx = T4 then
1
1−
1
a
b
=
b
a −1
b
a −1
b
a
=a
as ... p1/p2 = p4/p3
γ−1
b
 p2 
η th = 1 −
⋅
T3 p 1
 
T1
γ
Q.E.D.
for example, plot ηth vs pr for γ = 1.4 (air) with regeneration and T1/T3 = 0.25 figure 9.27
γ−1
γ := 1.4
11/14/2005
r := 1 .. 14
T1_over_T3 := 0.25
η th_reg( r, γ , T1_over_T3) := 1 − T1_over_T3⋅ r
5
γ
r := 2
efficiency vs pressure ratio
Given
1
efficiency
solve for pressure ratio
at intersection
η th_reg( r, γ , T1_over_T3) = η th( r, γ )
r_intersect := Find( r)
0.5
r_intersect = 11.314
T1 := 300
say ...
T3 := 1200
at this pressure ratio
0
0
5
10
15
γ−1
T2_intersect := T1 ⋅r_intersect
pressure ratio
air - gamma = 1.4
air - gamma = 1.4 with regen. T3/T1=4
γ
T2_intersect = 600
γ−1
1

T4_intersect := T3 ⋅ 
 r_intersect 
γ
T4_intersect = 600
at the r_intersect the temperature out of the turbine matches the temperature out of the compressor,
hence regeneration is infeasible
air-standard cycles ...
1. air as ideal gas is working fluid throughout cycle -no inlet or exhaust process
2. combustion process replaced by heat transfer process
3. cycle is completed by heat transfer to surroundings
4. all processes internally reversible
5. usually constant specific heat
(page 311)
these are our
assumptions for this
analysis
reset variables
Intercooled Brayton cycle
QH_dot
figure later
T3 = low temperature from first
intercooler, T4 second
compressor. additional stages
replicated at T3 and T4 which =
T1 and T2 respectively. T5 is
turbine inlet
4
QL_dot
compressor
5
ma_dot air
3
compressor
2
1
turbine
6
W_dotnet=
Wt_dot+Wc_dot
QL_dot
example plot of intercooled Brayton cycle
parameters for plot. to retain states 2, 3 & 4 as previously defined two points 1a and 1b are inserted rather than
renumbering. for intercooling, T 1 => T1a => T1b =>T2
11/14/2005
6
p1 => p1a => p1b =>p2
s 1 => s1a => s1b =>s2
starting conditions
p 1_plot := 1
after first stage compression
p 1a_plot :=
intercooler final temperature
T1b_plot := T1_plot
after second stage compression
p 2_plot := 10
max temperature after heat addition
T3_plot := 1000 + 273.15
T1_plot := 25 + 273.15
10
calculations
p-v Brayton cycle (rev.) 1 stg interclg
adiabatic compression 1st stage
intercooling
adiabatic compression 2nd stage
heat addition
adiabatic expansion in turbine
heat rejection
pressure
10
5
0
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
volume
T-s Brayton cycle (rev.) 1 stg interclg
1400
adiabatic compression
intercooling heat rejection
adiabatic compression second stage
heat addition
adiabatic expansion in turbine
heat rejection
temperature
1200
1000
800
600
400
200
0.6
0.8
1
1.2
1.4
entropy
11/14/2005
7
1.6
1.8
s1_plot := 1
for these calculations
γ := 1.667
η th_ic = 1 +
T1 := 300
QL
assume ...
observe ..
T3 := T1
for all intercooled stages
taking advantage of constant c po
η th_ic = 1 −
(
T6 − T1 + N⋅ T2 − T1
power :=
)
N := 1
γ−1
1
T6 ( pr) := T5 ⋅  
 pr 
γ
power
⋅ T1
η th_ic( pr , N) := 1 −
stages of intercooling
1
power
rc(pr , N)
:= pr
(
T6 ( pr) − T1 + N⋅ T2 (pr , N) −
T1
0.6
efficiency
0.4
1 stage intercooling
basic Brayton cycle
4 stages intercooling
0.2
0.2
0
2
3
4
γ = 1.667
intercooled and basic Brayton cycle
0.4
1
)
1 and 4 stages of intercooling
γ = 1.667
0.6
0
N+ 1
T5 − T2 (pr )
efficiency with intercooling
thermal efficiency (ideal)
pr := 1 .. 5
T5 − T2
T2 (pr , N) := rc(pr , N)
N= 1
T4 := T2
2
as ...
QH
maximum
T5 := 1200
5
1
2
3
4
pressure ratio (overall)
pressure ratio
as we observed in class both T H and TL are lowered by intercooling. Intercooling (by itself) slightly reduces
ideal efficiency. Increased number of stages doesn't reduce efficiency significantly further.
reset variables
11/14/2005
8
5
Intercooled Regenerative Brayton cycle
QH_dot
5
T3 = low temperature from first
intercooler, T4 second
compressor. additional stages
replicated at T3 and T4 which =
T1 and T2 respectively. T5 is
turbine inlet
ma_dot air
QL_dot
compressor
3
compressor
W_dotnet=
Wt_dot+Wc_dot
1
T1 := 300
for these calculations
η th_ic = 1 +
QL
as ...
QH
(
)
γ−1
T2 (pr , N) := rc( pr , N)
power
⋅ T1
maximum
(
η th_ic_reg = 1 −
1
T6 ( pr) := T5 ⋅  
pr
start with 1+ as
pr := 1.01 .. 5.01 η = 1
mathematically
)
and ...initial
stage of qL is ...
(
T2 − T1 + N⋅ T2 − T1
T5 − T6
)
= 1 −
T2 − T1
(
(N + 1)⋅ T2 − T1
T5 − T6
1
power
rc(pr , N)
:= pr
 
η th_ic_reg( pr , N) := 1 −
T3 := T1
observe ..
QH = T5 − T6
with
regeneration
N := 2
γ
regenerator
for all intercooled stages
so thermal efficiency becomes
power :=
T4 := T2
7
8
T5 := 1200
assume ...
taking advantage of constant c po
intercooled only from above
T6 − T1 + N⋅ T2 − T1
η th_ic = 1 −
T5 − T2
turbine
4
2
QL_dot
γ := 1.667
6
(
N+ 1
( N + 1)⋅ T2 ( pr , ) −
T1
)
T5 − T6 ( pr)
ideal efficiency Brayton cycles
thermal efficiency (ideal)
0.8
0.6
regeration was derived
above leaving T1/T3 now
renumbered to T1/T5
explicit. so variable T1/T5
inserted in arguments
0.4
intercooled with regeneration
regeration only
basic Brauton cycle
intercooling only
0.2
0
1
1.5
2
2.5
3
3.5
4
4.5
pressure ratio
11/14/2005
9
5
5.5
)
reset variables
intercooling, reheating and regenerative
QH_dot
5
6
ma_dot air
QL_dot
compressor
QH_dot
7
3
compressor
2
reheater
turbine
4
8
W_dotnet=
Wt_dot+Wc_dot
turbine
1
9
QL_dot 10
regenerator
example plot of intercooled Brayton cycle with reheat (and regeneration)
parameters for plot. to retain states 2, 3 & 4 as previously defined two points 1a and 1b are inserted rather than
renumbering. for intercooling, T 1 => T1a => T1b =>T2
p1 => p1a => p1b =>p2
s 1 => s1a => s1b =>s2
for reheat return to T 3; T3 => T3a => T3b =>T4
p3 => p3a => p3b =>p4
s 3 => s3a => s3b =>s4
starting conditions
p 1_plot := 1
after first stage compression
p 1a_plot :=
intercooler final temperature
T1b_plot := T1_plot
after second stage compression
p 2_plot := 10
max temperature after heat addition
T3_plot := 1000 + 273.15
after first turbine expansion
p 3a_plot :=
max temperature after reheat addition
T3b_plot := 1000 + 273.15
T1_plot := 25 + 273.15
10
10
calculations
11/14/2005
10
s1_plot := 1
p-v Brayton cycle (rev.) interclg & rht
adiabatic compression 1st stage
intercooling
adiabatic compression 2nd stage
heat addition
adiabatic expansion in 1st turbine
reheat
adiabatic expansion in 2nd turbine
heat rejection
10
pressure
8
6
4
2
0
0
0.5
1
1.5
2
2.5
3
volume
T-s Brayton cycle (rev.) interclg & rht
1400
adiabatic compression 1st stage
intercooling heat rejection
adiabatic compression 2nd stage
heat addition
adiabatic expansion in 1st turbine
reheat
adiabatic expansion in 2nd turbine
heat rejection
1200
temperature
1000
800
600
400
200
0.6
0.8
1
1.2
1.4
entropy
11/14/2005
11
1.6
1.8
2
2.2
η th_ic_reh_reg = 1 +
QL
γ := 1.667
for these calculations
QH
T1 := 300
T5 := 1200
maximum
assume ...
as ...
taking advantage of constant c po
T4 := T2
T3 := T1
for all intercooled stages
observe ..
figure later
T5 inlet to turbine, stages of
turbine are at T5 - T6 for all,
for ease of calculations
number of reheat and
intercooling are the same so
pressure ratios are identical
and upper and lower temperature for reheat are at T5 and T6
η th_ic_reh_reg = 1 −
(
)
(N + 1)⋅ ( T5 − T6 )
(N + 1)⋅ T2 − T1
N := 2
pr := 1.01 .. 5.01
1
power :=
γ−1
rc(pr ,N) := pr
γ
η th_ic_reh_reg( pr , N) := 1 −
N+ 1
T2 (pr , N) := rc(pr , N)
power
1

T6 (pr , N) := T5 ⋅ 
rc(pr , )


⋅ T1
(
)
(N + 1 )⋅ ( T5 − T6 (pr , N) )
(N + 1)⋅ T2 (pr , N) − T1
Brayton cycle efficiency
thermal efficiency (ideal)
0.8
0.6
0.4
intercld - reheat - regen
intercld - regen
regen
basic Brayton cycle
intercld
0.2
0
1
1.5
2
2.5
3
3.5
4
pressure ratio overall
11/14/2005
12
4.5
5
5.5
power
example plot of multiple intercooled Brayton cycle with multiple reheat (and regeneration
parameters for plot. to retain states 2, 3 & 4 as previously defined two points 1a and 1b are inserted rather than
renumbering. for intercooling, T 1 => T1a => T1b =>T2
p1 => p1a => p1b =>p2
s 1 => s1a => s1b =>s2
for reheat return to T 3; T3 => T3a => T3b =>T4
p3 => p3a => p3b =>p4
s 3 => s3a => s3b =>s4
starting conditions
p 1_plot := 1
pressure ratio
pr_plot := 20
number of compression stages ...
T1_plot := 25 + 273.15
n_comp := 4
intercooler final temperature
T1_plot
max temperature after heat addition
T3_plot := 1000 + 273.15
number of expansion stages ...
max temperature after reheat addition
n_exp := 4
T3_plot
calculations
11/14/2005
13
s1_plot := 1
p-v Brayton cycle (rev.) interclg & rht
20
adiabatic compression 1st stage
intercooling and compression stages
heat addition first stage
adiabatic expansion and reheat
adiabatic expansion in last turbine
heat rejection
pressure
15
10
5
0
0
0.5
1
1.5
2
2.5
3
volume
T-s Brayton cycle (rev.) interclg & rht
adiabatic compression 1st stage
intercooling and compression stages
heat addition first stage
adiabatic expansion and reheat
adiabatic expansion in last turbine
heat rejection
1200
temperature
1000
800
600
400
0.2
0.4
0.6
0.8
1
1.2
entropy
11/14/2005
14
1.4
1.6
1.8
2
2.2
as number of reheat and intercooled stages increases, ideal efficiency should approach Carnot
η th_carnot := 1 −
T1
T5
N := 1 .. 20
pr := 5
this calculation fixes pressure ratio overall = 5 and
looks at variation with number of stages of
intercooling and reheat (same)
Intercooled, Reheat, Regen Brayton cycle
thermal efficiency (ideal)
0.85
0.8
η th_carnot
0.75
0.7
0.65
0
5
10
15
number of intercool and reheat stages
11/14/2005
15
20