Polytropic Efficiency
kJ := 1000J
consider a two stage compressor with a stage efficiency = 0.9
p 1 := 1bar
p 2 := 5bar
kJ
s1 := 1
kg⋅ K
p 3 := 20bar
η stage := 0.9
calculations
the states resulting are plotted ...
Temperature
800
600
400
200
0.95
1
1.05
1.1
1.15
entropy
using the gas laws and stage effiiciency = ηstage , after the first stage the states will be
kJ
s2 = 1.04
kg⋅ K
T2 = 491.556 K
p 2 = 5 bar
after the second stage the states will be
T3 = 756.994 K
kJ
s3 = 1.076
kg⋅ K
p 3 = 20 bar
now if we calculate efficiency of the compressor
η comp =
∆Ts
∆T
=
T3s − T1
T3s − T1
T3 − T1
T3 − T1
= 0.88
as a check ...
T33s − T2
T3 − T2
= 0.9
This effect can be accounted for by using polytropic efficiency - or small stage efficiency.
reset variables T, s, p
ηc =
h 2s − h 1
h2 − h1
=
T2s − T1
T2 − T1
=
∆Ts
∆T
compressor
11/9/2005
1
T2s − T1
T2 − T1
= 0.9
as the pressure ratio approaches unity
∆Ts
∆T
=>
ds =
dh
T
=
d
Ts
dT
for simple compressible substance
dT
dp
⋅dp = cpo⋅
− R⋅
T
T
p
ideal gas and definition of c po
isentropic ds = 0
substitute
γ=
d
Ts
dT
dT s = η pc⋅dT
v
−
η pc =
define ...
cpo⋅
dT s = η pc⋅dT
cpo
cvo :=
cvo
dT s
T
= R⋅
cpo⋅ η pc⋅
cpo
dp
p
dT
T
= R⋅
dp
 T2 
rearrange and
integrate .....
p
R := cpo − cvo
γ
R
ln
 T1 
simplify →
η pc⋅cpo
T2
T1
=
 p2 
p
 1
=
use in definition
of ηc
ηc =
cpo⋅η pc
⋅ln
 p1 
γ ⋅ η pc
T2s
γ ⋅ η pc
 p2 
R
γ−1
γ−1
raise to exponents
(7.7)
T⋅ ds = dh − v ⋅ dp
T2s − T1
T2 − T1
=
T1
T2
T1
γ−1
−1
γ
r
=
−1
γ−1
−1
r
γ ⋅ η pc
−1
example polytropic efficiency = 0.9; calculate isentropic efficiency for p 2 /p1 = 2, 16, 30; use air as working fluid
γ−1
i := 0 .. 2
η pc := 0.9
if T1 were 25 deg C
γ := 1.4
 2 
r :=  16

 30 
 p2 
=
T1
 p1 
T2 := T1 ⋅r
11/9/2005
T2 := T1 +
ηc =
T2s − T1
γ ⋅ η pc
T2 = T1 +
T2 − T1
T2s − T1
η pc
−1
 0.89 
η c =  0.856

 0.845 
−1
 371.535 
T2 =  718.944

 877.733 
γ ⋅ η pc
whereas T2 calculated using
ηpc as ηc
 363.449 
T2s =  658.369

 787.897
(ri)
γ − 1
γ ⋅ η pc
γ
γ−1
i
T1 := 25 + 273.15
γ−1
T2
η c :=
(ri)
 370.704 
T2 =  698.393

 842.313 
2
T2s − T1
ηc
γ−1
T2s := T1 ⋅r
γ
temperature above is
higher indicating more
energy required for
compressor
consistent with lower
efficiency
another observation ... if we say the two stages have a polytropic efficiency of 0.9 then using ...
γ−1
 p2 
=
T1
 p1 
T2
γ−1
γ ⋅ η pc
and ...
γ−1
T3
T1
=
 p3 
 p3 
=
T2
 p2 
T3
γ−1
γ ⋅ η pc
T2 T3
T3
=
=
⋅
T1 T2
T1
=>
 p2 

 p1 
γ−1
γ ⋅ η pc
 p3 
⋅
 p2 
γ−1
γ ⋅ η pc
γ ⋅ η pc
 p3 
= 
 p1 
the polytropic efficiency of the compressor is identical, i.e T 3 is determined from
γ ⋅ η pc
the polytropic efficiency that is the same as the two stages
since polytropic efficiency approaches isentropic efficiency for pressure ratio ~
1, this is the same as saying that for a compressor with a large number of
stages each with pressure ration near 1, the polytropic efficiency of the
compressosor is isentropic efficiency of the individual stages

 p1 
turbine would be similar with exception of inversion of relationship
ηt =
T3 − T4
η pt =
=>
T3 − T4s
dT
dT s =
dT s
dT
η pt
turbine
cpo⋅
dT s
T
= R⋅
dp
p
cpo dT
dp
⋅
= R⋅
p
η pt T
=>
=>
R⋅η pt dp
=
⋅
T
cpo p
dT
η pc⋅
 p3 
= 
T4
 p4 
T3
could do by direct analogy if write
ηt
=
4=2
T4s − T3
=>
T4 − T3
and ...
T4
T3
3=1
η pt⋅
=
 1 

r
 p2 
=
T1
 p1 
T2
γ ⋅ η pc
γ
γ−1
γ−1
1
γ−1

1


η pt

γ⋅ 
morphs to ...
T4
T3
γ−1
=
 p4 

 p3 
γ
same example polytropic efficiency = 0.9; calculate isentropic efficiency for p 3 /p4 = 2, 16, 30; use air as
working fluid
i := 0 ..
2
η pt := 0.9
if T3 were 700 deg C
11/9/2005
γ := 1.4
1
1 −  
r
 i
η t :=
i
2
r :=  16

 30 
1−
T3 := 700 + 273.15
3
η pt⋅
γ−1
γ
γ−1
 1 
r
 i 
γ
 0.909 
η t =  0.932

 0.938 
T4
1
= 

T3  r 
η pt⋅
γ−1
γ
1
T4 := T3 ⋅  
r
whereas T4 calculated using
ηpt as ηt
 798.309 
T4s =  440.702

 368.252 
ηt =
η pt⋅
γ−1
 814.277 
T4 =  477.034

 405.834 
γ
T3 − T4
T3 − T4s
(
T4 − T3
=
T4 := T3 + η pt⋅ T4s − T3
(
T4 = T3 + η pt⋅ T4s − T3
T4s − T3
 815.793 
T4 =  493.947

 428.742 
)
γ−1
)
1
T4s := T3 ⋅  
r
γ
 
above temperature is lower indicating
more energy extracted from fluid,
consistent with higher efficiency
direct approach for calculating T 2 modeling as discrete multiple stages. increasing number_of_stages should
make ηc_1 approach ηc (back to compressor for calculations)
number_of_stages := 4
j := 0 .. 2
TT
1
r_per_stage := r
number_of_stages
0, j
:= 25 + 273.15
power :=
T
r_per_stage = ( 1.189 2 2.34 )
2
r =  16

 30 
γ := 1.4
γ−1
γ
temperature after each stage
TT
1 , ns
= TT
0 , ns
(r_per_stage j)
+
n := 0 .. number_of_stages
power
⋅TT
0, j
− TT
power
T2s = T ⋅r
TT
1
η c_1 :=
j
 0.892 
η c_1 =  0.869

 0.862 
ηc =
η pc
TT
number_of_stages = 4
0, j
TT
isentropic
continuous model
 0.89 
η c =  0.856

 0.845 
( j)
⋅ r
0, j
power
n+ 1, j
− TT
number_of_stages , j
 0.891 
η c_1 =  0.862

 0.852 
T2 − T1
:= TT
n, j
T2 = T1 +
(r_per_stage j)
+
T2s − T1
ηc
power
⋅TT
n, j
η pc
0, j
20 stages
 0.89 

η c_1 = 0.859

 0.849 
50 stages
 0.89 

η c_1 = 0.857

 0.846 
further evidence that for a compressor with a large number of stages each with pressure ration
near 1, the polytropic efficiency of the compressosor is isentropic efficiency of the individual
stages. this follows through to determine isentropic efficiency for the compressor based on
equating polytropic efficiency of small stages to the isentropic efficiency of the (small) stage.
11/9/2005
4
− TT
0, j
− TT
10 stages
T2s − T1
n, j