EDUSAT PROGRAMME 15 Turbomachines (Unit 3) Axial … · EDUSAT PROGRAMME 15 Turbomachines (Unit 3)...
Transcript of EDUSAT PROGRAMME 15 Turbomachines (Unit 3) Axial … · EDUSAT PROGRAMME 15 Turbomachines (Unit 3)...
EDUSAT PROGRAMME 15
Turbomachines (Unit 3)
Axial flow compressors and pumps
Axial flow compressors and pumps are power absorbing turbomachines. These
machines absorb external power and thereby increase the enthalpy of the flowing
fluid. Axial flow turbomachines use large quantity of fluid compared to mixed and
centrifugal type of turbomachines. However, the pressure rise per stage is lower in
case of axial flow turbomachines than mixed and centrifugal flow turbomachines.
Axial flow compressors are used in aircraft engines, stand alone power generation
units, marine engines etc.
Design of axial flow compressors is more critical than the design of turbines
(power generating turbomachines). The reason is that, in compressors the flow
moves in the direction of increasing pressure (adverse pressure gradient). If the
flow is pressurized by supplying more power, the boundary layers attached to the
blades and casing get detached and reverse flow starts which leads to flow
instability leading to possible failure of the machine. However, in turbines the fluid
moves in the decreasing pressure (favourable pressure gradient). To transfer a
given amount of energy more number of stages are required in compressors than in
turbines. Generally, the fluid turning angles are limited to 20˚ in compressors and
is 150˚ to 165˚ in case of turbines. Thus the pressure rise per stage is limited in
case of compressors. Fluid flows in direction of increasing pressure, this increases
the density of the fluid, thus the height of the blade decreases from the entrance
to the exit. Axial flow compressors have inlet guide vanes at the entrance and
diffuser at the exit.
Figure 1 shows the flow through the compressor and fig. 2 shows the
corresponding inlet and exit velocity triangles. Generally for a compressor the
angles are defined with respect to axial direction (known as air angles). It can be
seen that the fluid turning angle is low in case of compressors. Figure 3 shows the
flow through the turbine blade and fig. 4 the corresponding inlet and exit velocity
triangles for a turbine.
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Fig. 1 Flow across a compressor blade
Fig. 2 Inlet and exit velocity triangles
Fig. 1 Flow across a compressor blade
Fig. 2 Inlet and exit velocity triangles
ADARSHA H G
Fig. 3 Flow across a turbine blade
Fig. 4 Inlet and exit velocity triangles for turbine
Fig. 3 Flow across a turbine blade
Fig. 4 Inlet and exit velocity triangles for turbine
Fig. 4 Inlet and exit velocity triangles for turbine
ADARSHA H G
Expression for degree of reaction for axial flow compressor
Degree of reaction is defined as a measure of static
rotor expressed as a percentage of the total static enthalpy
Assumptions:
1. The analysis is for 2-dimensional flow
2. The flow is assumed to take
peripheral velocities at inlet and outlet are same and there is no flow in the
radial direction
3. It is common to express blade angles w.r.t. axial direction in case of axial
flow compressor (air angles)
4. Axial velocity is assumed to remain constant
Figure 2 shows the inlet and exit velocity triangles for axial flow compressors
Static
StaticR =
o∆h =
V1
tanγa
VU +=
(tanVU a=
tantanγtanγ 21 =−
2(h
VR
02
2
r1 −=
Expression for degree of reaction for axial flow compressor
s defined as a measure of static enthalpy rise that occurs in the
rotor expressed as a percentage of the total static enthalpy across the rotor.
dimensional flow
The flow is assumed to take place at a mean blade height where blade
peripheral velocities at inlet and outlet are same and there is no flow in the
It is common to express blade angles w.r.t. axial direction in case of axial
flow compressor (air angles)
y is assumed to remain constant
Figure 2 shows the inlet and exit velocity triangles for axial flow compressors
stageinriseenthalpyStatic
rotorinriseenthalpyStatic
1)u1
Vu2
U(V −−−−=
-2-)0
tanγ1
(tanγa
V0
tanγa
V +=
3)tanγ(tanγ 32 −−+
4tanγtanγ 03 −−−−
5h
V
01
2
r2 −−−−
−
)
Expression for degree of reaction for axial flow compressor
enthalpy rise that occurs in the
across the rotor.
place at a mean blade height where blade
peripheral velocities at inlet and outlet are same and there is no flow in the
It is common to express blade angles w.r.t. axial direction in case of axial
Figure 2 shows the inlet and exit velocity triangles for axial flow compressors.
ADARSHA H G
Substituting eq. 6 and 7 in eq.5
6)γtanγ(tanVVV
γtanVVV
γtanVVV
22
122
a
2
r2
2
r1
222
a
2
a
2
r2
122
a
2
a
2
r1
−−−−=−
+=
+=
7)tanγ(tanγUV∆h
tanγVVandtanγVV
)VU(Vhh
03a0
0au13au2
U1u20102
−−−−=
==
−=−
)tanγ(tanγ2UV
)γtanγ(tanVR
03a
22
122
a
−
−=
4tanγtanγtanγtanγ 0321 −−−−=−
10)tanβtanβ
tanβtanβ(
2U
VR
)tanβ
1
tanβ
1(
2U
VR
9)cotβ(cotβ2U
VR
82U
)tanγ(tanγVR
)tanγ(tanγ2UV
)γtanγ(tanVR
21
21a
21
a
21a
21a
21a
22
122
a
−−−+
=
+=
−−−+=
−−−−+
=
−
−=
ADARSHA H G
( ) ((
−==
+−
−=
VVWDE
VVV
VVR
2
2
1
r2
2
2
2
1
2
r1
2
r2
)) ( )
−+
−
2
VV
V
2
r1
2
r2
2
2
r1
2
2
ADARSHA H G
ADARSHA H G
Problem 1
In a mixed flow pump absolute fluid velocity at the inlet is
radial velocity at the exit. Inlet hub
250 mm. Pump speed is 3000 rpm. Find the degree of reaction and
input to the fluid, if the relative velocity at the
speed.The fluid leaves the rotor in the radial direction
D1= 80 mm, D2= 250 mm, N=3000 rpm
The next step is to draw the velocity triangles
Problem 2
The total power input at a stage in an axial flow compressor with symmetric inlet
and outlet velocity triangles (R=0.5) is 27.85 kJ/kg of air flow. If the blade speed
is 180m/s throughout the rotor,
blade angles. Do you recommend the use of such compressor?. Assume the axial
velocity component to be 120m/s.
V12.57m/s;UV 21r2 ==
39.239.2V-UWDE
12.57m/s;VUV
u22
11rd2
×===
===
( )
0.5R
E2
V(VUUR
2
r1
2
2
2
1
=
×
−−−=
In a mixed flow pump absolute fluid velocity at the inlet is axial and equal to
radial velocity at the exit. Inlet hub diameter is 80 mm and impeller tip diameter is
speed is 3000 rpm. Find the degree of reaction and
input to the fluid, if the relative velocity at the exit equals the inlet tangential blade
the rotor in the radial direction Given: V1 = Va1= Vrd2 ;
mm, N=3000 rpm, R = ?, WD=?, Vr2=U1
The next step is to draw the velocity triangles
The total power input at a stage in an axial flow compressor with symmetric inlet
and outlet velocity triangles (R=0.5) is 27.85 kJ/kg of air flow. If the blade speed
is 180m/s throughout the rotor, draw the velocity triangles and compute the rotor
Do you recommend the use of such compressor?. Assume the axial
velocity component to be 120m/s.
41.23m/sVU2
r2
2
2 =+=
1.543KJ/kg39.2
17.78m/sUVV2
1
2
1r1
=
=+=
)V2
r2
axial and equal to
diameter is 80 mm and impeller tip diameter is
speed is 3000 rpm. Find the degree of reaction and the energy
exit equals the inlet tangential blade
Given: V1 = Va1= Vrd2 ;
The total power input at a stage in an axial flow compressor with symmetric inlet
and outlet velocity triangles (R=0.5) is 27.85 kJ/kg of air flow. If the blade speed
draw the velocity triangles and compute the rotor
Do you recommend the use of such compressor?. Assume the axial
ADARSHA H G
Given: Symmetric velocity triangles with R = 0.5; P= 27.85 kJ/kg of air flow;
U=180m/s; Va=120m/s; β1=? and
Draw the velocity triangles.
Since the R=0.5 and velocity triangles are symmetrical
α1=β2; α2=β1; V1=Vr2 and V2=Vr1
Problem 3
The axial component of air velocity at the exit of the nozzle of an axial flow
reaction stage is 180m/s, the nozzle inclination to the direction of rotation is 27
Find the rotor blades angles if the degree of reaction is 50% and the blade speed is
180m/s. Also, for the same blade speed, axial velocity and nozzle angle find degree
of reaction, if the absolute velocity at the rotor outlet should be axial and equal to
axial velocity at the inlet.
Given: 1) Va1=180m/s, α1=27
2) U=180m/s, Va1=180m/s,
Sol: 1) R = 50%, V1=Vr2, V2=Vr2,
Inlet velocity triangle is shown in the figure
154.72m/s∆V
27850kJ/Kg)VU(∆WD
u
u
=
==
12.64X
(UX;∆VX2U u
=
==×−
12.64180
120tanβ
83.9812.64
120tanβ
1
1
12
=
−=
=
=
−
−
o48.34ββ 12 =−
Given: Symmetric velocity triangles with R = 0.5; P= 27.85 kJ/kg of air flow;
1=? and β2=?; Do you recommend such compressor?
Draw the velocity triangles.
Since the R=0.5 and velocity triangles are symmetrical
1; V1=Vr2 and V2=Vr1
The axial component of air velocity at the exit of the nozzle of an axial flow
reaction stage is 180m/s, the nozzle inclination to the direction of rotation is 27
Find the rotor blades angles if the degree of reaction is 50% and the blade speed is
. Also, for the same blade speed, axial velocity and nozzle angle find degree
of reaction, if the absolute velocity at the rotor outlet should be axial and equal to
1=27˚, R = 50%, U=180m/s, β1=?, β2=?
) U=180m/s, Va1=180m/s, α1=27˚, R=?, if V2=Va2=Va1
Sol: 1) R = 50%, V1=Vr2, V2=Vr2, α1=β2 and α2=β1
Inlet velocity triangle is shown in the figure
27850kJ/Kg
)/2∆Vu−
o
o
35.64
83.98
=
Given: Symmetric velocity triangles with R = 0.5; P= 27.85 kJ/kg of air flow;
2=?; Do you recommend such compressor?
The axial component of air velocity at the exit of the nozzle of an axial flow
reaction stage is 180m/s, the nozzle inclination to the direction of rotation is 27˚.
Find the rotor blades angles if the degree of reaction is 50% and the blade speed is
. Also, for the same blade speed, axial velocity and nozzle angle find degree
of reaction, if the absolute velocity at the rotor outlet should be axial and equal to
ADARSHA H G
From the exit velocity triangles
Turbines
1.Utilization factor is
2.Adiabatic efficiency is the
3.Overall efficiency is product of adiabatic efficiency and
efficiency
4.Mechanical efficiency of majority of TM’s is nearly 100%
5.Therefore, overall efficiency is almost equal to
6. However, adiabatic efficiency is product of utilization factor
efficiency) and efficiency associated with various losses
7.Utilization factor deals with what is maximum
obtained from a turbine
8.Utilization factor is the ratio of ideal work
for conversion to work
9.Under ideal conditions it should be possible to utilize all the K.E. at inlet
and increase the K.E. due to reaction effect
o27ββα
173.26
180tanβ
173.26m/sUcosαV
V180;sinαV
221
11
11
111
==
==
=−
==
−
( ) (VVV
VVR
V396.48m/sV
180173.26V
254.55m/sV,45β
2
r2
2
212
2
r1
2
r2
21
2
r1
r22
+−
−=
==
+=
== o
From the exit velocity triangles
Turbines- Utilization factor
is defined only for PGTM- Turbines
Adiabatic efficiency is the quantity of interest in turbines
Overall efficiency is product of adiabatic efficiency and mechanical
Mechanical efficiency of majority of TM’s is nearly 100%
Therefore, overall efficiency is almost equal to adiabatic efficiency
However, adiabatic efficiency is product of utilization factor
efficiency) and efficiency associated with various losses.
Utilization factor deals with what is maximum energy that can be
obtained from a turbine without considering the losses in the turbine
Utilization factor is the ratio of ideal work output to the energy available
to work
Under ideal conditions it should be possible to utilize all the K.E. at inlet
ncrease the K.E. due to reaction effect
o46.09
173.26m/s
396.48m/s=
)0.0198
V
180m/s
249.83m/s180
254.55m/s
2
r1
2
2
=−
=
=
mechanical
efficiency
However, adiabatic efficiency is product of utilization factor (diagram
energy that can be
ut considering the losses in the turbine
output to the energy available
Under ideal conditions it should be possible to utilize all the K.E. at inlet
ADARSHA H G
10. The ideal energy available for conversion into work
11.The work output given by Euler’s Turbine Equation is
11. Utilization factor is given by
12. Utilization factor for modern TM’s is between 90% to 95%
Relation between utilization factor and degree of reaction
Utilization factor is given by
The degree of reaction is given by
Substituting the value of X in the expression for utilization factor and
simplifying
( ) ( )[ ]2
VVUUVw
2
r2
2
r1
2
2
2
1
2
1a
−−−+=
( ) ( ) ( )[ ]
( ) ( ) ( )[ ]( ) ( )[ ]2
r2
2
r1
2
2
2
1
2
1
2
r2
2
r1
2
2
2
1
2
2
2
1
a
2
r2
2
r1
2
2
2
1
2
2
2
1
VVUUV
VVUUVV
w
wε
2
VVUUVVw
−−−+
−−−+−==
−−−+−=
( ) ( ) ( )[ ]( ) ( )[ ]2
r2
2
r1
2
2
2
1
2
1
2
r2
2
r1
2
2
2
1
2
2
2
1
a VVUUV
VVUUVV
w
wε
−−−+
−−−+−==
( ) ( )( ) ( ) ( )
( ) ( )E2
VVUU
VVUUVV
VVUUR
2
r2
2
r1
2
2
2
12
r2
2
r1
2
2
2
1
2
2
2
1
2
r2
2
r1
2
2
2
1
×
−−−=
−−−+−
−−−=
( ) ( )
( )( )
R1
VVRX
XVV
XR
VVUUX
2
2
2
1
2
2
2
1
2
r2
2
r1
2
2
2
1
−
−=
+−=
−−−=
( )( )2
2
2
1
2
2
2
1
RVV
VVε
−
−=
ADARSHA H G
The above equation is valid for single rotor under the conditions where Euler’s
turbine equations are valid.
equation is valid in the following range of
Maximum Utilization factor
Utilization factor is given by
Utilization factor maximum if the exit absolute velocity is minimum. This is
possible when the exit absolute velocity is in axial direction
Maximum utilization factor is given by
(( 1V
Vε =
m
m
12
1ε
V
Vε
VV
=
=
=
m1
cosε
−=
The above equation is valid for single rotor under the conditions where Euler’s
The above equation is invalid when R = 1
equation is valid in the following range of R 0 ≤ R < 1
Maximum Utilization factor
Utilization factor is given by
Utilization factor maximum if the exit absolute velocity is minimum. This is
possible when the exit absolute velocity is in axial direction
Maximum utilization factor is given by
))2
2
2
1
2
2
2
1
RV
VV
−
−
12
12
122
1
2
1
122
1
2
1
11
αRsin
αcos
αsinRVV
αsinVV
sinα
−
−
−
1
2
12
αRsin
αcos
−0αwhen1ε 1m ==
The above equation is valid for single rotor under the conditions where Euler’s
= 1. The above
Utilization factor maximum if the exit absolute velocity is minimum. This is
ADARSHA H G
Maximum Utilization factor for impulse turbine
For an impulse turbine R = 0, thus
From the velocity triangles OAB and OBD are
similar. Thus AB = U,
α1 is made as small as possible(15
Maximum Utilization factor for 50% reaction turbines
50% reaction turbines have V1=Vr2, V2=Vr1,
maximum utilization V2 must be in axial direction. The corresponding
velocity triangles are
Speed2
cosα
V
U
UUcosαV
1
1
11
==
=+=
φcosαV
U
cosαVAlso,
1ε0.5,R
αRsin1
αcosε
1
1
11
m
1
2
1
2
m
===
=
==
−=
Maximum Utilization factor for impulse turbine
For an impulse turbine R = 0, thus
From the velocity triangles OAB and OBD are
similar. Thus AB = U,
is made as small as possible(15˚-20˚)
Maximum Utilization factor for 50% reaction turbines
50% reaction turbines have V1=Vr2, V2=Vr1, α1=β2 and α2=
maximum utilization V2 must be in axial direction. The corresponding
12
m αcosε =
φRatioSpeed
2U
=
=
ratiospeed
U
α0.5sin
αcos
12
1
2
1
=
=
−
Maximum Utilization factor for impulse turbine
Maximum Utilization factor for 50% reaction turbines
2=β1 and for
maximum utilization V2 must be in axial direction. The corresponding
ADARSHA H G
Comparison between Impulse and 50% reaction
A) When both have same blade speed
Let UI and UR be the blade speeds of impulse and 50% reaction
turbines. The velocity triangles for maximum utilization are
Comparing eq. 1 and 2, it is clear that impulse turb
of energy per unit mass than 50% reaction turbin
utilization is maximum.
• However, 50% reaction turbines are more efficient than impulse turbines
• But 50% reaction turbines transfer half the en
turbines.
• If only 50% reaction turbines are used more stages are required or if only
impulse turbines are used stages are less but efficiency is low
• In steam turbines where large pressure ratio is available it is common to use
one or two impulse stages followed by reaction stages
UE
VBut
UE
50%For
2UE
From
UE
R
R
I
1I
=
=
=
=
Comparison between Impulse and 50% reaction turbine at
maximum utilization
When both have same blade speed
Let UI and UR be the blade speeds of impulse and 50% reaction
turbines. The velocity triangles for maximum utilization are
Comparing eq. 1 and 2, it is clear that impulse turbine transfers twice the amount
of energy per unit mass than 50% reaction turbine for the same blade speed when
However, 50% reaction turbines are more efficient than impulse turbines
But 50% reaction turbines transfer half the energy compared to impulse
If only 50% reaction turbines are used more stages are required or if only
impulse turbines are used stages are less but efficiency is low.
In steam turbines where large pressure ratio is available it is common to use
ne or two impulse stages followed by reaction stages.
2U
UV
VU
turbinereaction50%
12U
2UVtrianglevelocity
V
2
R
Ru1
u1R
2
I
Iu1
u11
−−−
=
−−−
=
turbine at
Let UI and UR be the blade speeds of impulse and 50% reaction
turbines. The velocity triangles for maximum utilization are
ine transfers twice the amount
e for the same blade speed when
However, 50% reaction turbines are more efficient than impulse turbines.
ergy compared to impulse
If only 50% reaction turbines are used more stages are required or if only
In steam turbines where large pressure ratio is available it is common to use
ADARSHA H G
Comparison between Impulse and 50% reaction turbine at
b) When both have same energy transfer
c) When V1 and α1 are same in both TM’s
Speed ratio for impulse and 50% reaction stage for maximum utilization
Problem 1
Find an expression for the utilization factor for an axial flow impulse turbine stage
which has equiangular rotor blades, in terms fixed inlet blade angle and speed ratio
ф.
Given: Equiangular rotor blades
turbine R = 0, Vr1=Vr2, find expression
for utilization factor in terms of
From the velocity triangles
1.414U2UU
2UU
EE
I
2
IR
2
I
2
R
IR
==
=
=
62UU
U;cosαφV
U
2U;2
cosαφ
V
U
IR
R1
I
R
I1
I
I
−−=
==
==
r222
r1
2
1
r222
r2
2
2
cos2UVUVV
cos2UVUVV
−+=
−+=
Comparison between Impulse and 50% reaction turbine at
maximum utilization
b) When both have same energy transfer
1 are same in both TM’s
Speed ratio for impulse and 50% reaction stage for maximum utilization
Find an expression for the utilization factor for an axial flow impulse turbine stage
which has equiangular rotor blades, in terms fixed inlet blade angle and speed ratio
Given: Equiangular rotor blades β1=β2, axial flow turbine U1=U2=U, Impulse
turbine R = 0, Vr1=Vr2, find expression
for utilization factor in terms of α1 and ф.
3I −−−
5cosαV
4cosα
1I
1I
−−=
−−=
( )2
2
β180cos
cosβ
−
Comparison between Impulse and 50% reaction turbine at
Speed ratio for impulse and 50% reaction stage for maximum utilization
Find an expression for the utilization factor for an axial flow impulse turbine stage
which has equiangular rotor blades, in terms fixed inlet blade angle and speed ratio
2, axial flow turbine U1=U2=U, Impulse
ADARSHA H G
Problem 2 For a 50% degree of reaction axial flow turbomachine, the inlet fluid velocity is 230m/s; outlet angle of inlet guide blade 3060˚ and the outlet rotor angle is 25˚. Find the utilization factor, axial thrust and the power output/unit maGiven: R=50%; V1=230m/s;
ε = ?; Fax=?; P = ?
Although R = 50%, α1≠
Inlet and exit velocity triangles are
drawn
cosβ4UVVV 1r1
2
2
2
1 =−
(
1
1
1
2
1
2
2
2
1
2
2
2
1
2
2
2
1
V
Uφ where,
φ)(cosα4φε
cosV4U
V
VVε
RVV
VVε
=
−=
=−
=
−
−=
132.8m/scosβVcosαVU
132.8m/ssinβ
VV
115m/ssinαVV
1r111
1
a1r1
11a1
=−=
==
==
( ) (
233.6m/sV
26450120.36VV
17635.8240.72V2V
cos2UVUVV
132.8VVVV
VVVV
VVR
r2
r2
2
r2
r2
2
r2
r2
22
r2
2
2
2
2
2
r2
2
1
2
r1
r1
2
r2
2
2
2
1
2
r1
2
r2
=
−−
+−
−+=
=+=+
−+−
−=
For a 50% degree of reaction axial flow turbomachine, the inlet fluid velocity is 230m/s; outlet angle of inlet guide blade 30˚; inlet rotor angle
˚ and the outlet rotor angle is 25˚. Find the utilization factor, axial thrust and the power output/unit mass flow. Given: R=50%; V1=230m/s; α1=30˚; β1=60˚; β2=25˚
≠ β2; as Va1≠Va2
Inlet and exit velocity triangles are
( )UcosαV4U 11 −=
)2
1
1
V
Ucosα −
132.8m/s
)
026450
70535.817635.8
cosβ
70535.8230132.8
0.5
2
22
2
r1
=
=
=+
=
For a 50% degree of reaction axial flow turbomachine, the inlet fluid ˚; inlet rotor angle
˚ and the outlet rotor angle is 25˚. Find the utilization factor, axial
ADARSHA H G
0.822126.360.5230
126.36230
RVV
VVε
126.36m/sV
70535.8VV
22
22
2
2
2
1
2
2
2
1
2
2
2
2
r2
=×−
−=
−
−=
=
=+
( )
( )
s
kg
N16.28
m
F
sinβVSinαVVVm
F
VVmThrustAxial
ax
2r211a2a1ax
a2a1
=
−=−=
−=
&
&
&
( )
( ) 37KJ/kg/s78.91199.98132.8m
P
78.91m/sUcosβVV
199.98m/scosαVV
VVUmP
2r2u2
11u1
u2u1
=+=
=−=
==
−=
&
&
ADARSHA H G
UNIT 4
Thermodynamics of fluid
Sonic velocity and Mach number
1.Sonic velocity is the speed of propagation of pressure wave in a medium.
2.The speed of sound in a fluid at a local temperature for isentropic flow is given
by where γ is the ratio of specific heats = 1.4, R is characteristics gas constant =
287 J/kg K and T is the local temperature in kelvins. At 15˚ the speed of sound is
340 m/s.
3.As altitude increases temperature decreases and speed of sound decreases.
4.Mach Number is defined as the ratio of local velocity of fluid to the sonic
velocity of sound in that fluid
5. Many turbines and compressors experience high Mach numbers
6.High Mach numbers give rise to some special problems such as shock waves
which leads to irreversibility and cause loss in stagnation pressure and increase in
entropy
7.Using continuity equation, Euler’s equation and isentropic equation following
two equations are derived
8.The above equations decide the variations in velocity, pressure and area for
different Mach numbers
9.The three basic classifications are a) Subsonic flow M<1
b) Sonic flow M=1 c) Supersonic flow M>1
γRT
V
c
VM ==
γRTc =
2γM
M1
p
dp
A
dA
1Mγp
dp
V
dV
2
2
2
−−−
−=
−−−−=
ADARSHA H G
Classification fluid flow based on Mach number
a) Subsonic flow (M<1) :
Nozzle
Diffuser
anddecreasesarea
negativeisp
dpwhen
andincreasesarea
increasesisp
dpwhen
Classification fluid flow based on Mach number
Subsonic flow (M<1) :
increasesvelocityand
negativeisA
dAnegative
γ
1
p
dp
A
dA
γp
dp
V
dV
=
−=
decreasesvelocityand
positiveisA
dAincreases
Classification fluid flow based on Mach number
2γM
M1
1M
dp
2
2
2
−−−
−
−−−
ADARSHA H G
b) Supersonic flow (M>1) :
Divergent nozzle
Convergent Diffuser
velocityandincreasesarea
negativeisp
dpwhen
velocityanddecreasearea
positiveisp
dpwhen
Supersonic flow (M>1) :
γM
1
p
dp
A
dA
Mγp
dp
V
dV
−=
−=increasesvelocity
positiveisA
dA
decreasesvelocity
negativeisA
dApositive
2γM
M
1M
dp
2
2
2
−−−
−
−−−
ADARSHA H G
C) Sonic flow (M=1) :
At the throat portion of the convergent divergent nozzle the velocity is sonic.
Static and Stagnation states
Turbomachines involve the use of compressible and incompressible fluids. In compressible TM’s fluid move with velocities more than Mach one at many locations. In incompressible TM’s the fluid velocities are generally low, however, K.E. and P.E. of the moving fluid are very large and cannot be neglected To formulate equations based on actual state of the fluid based on laws of thermodynamics two states are used.
The states are static state and stagnation states.
Static State If the measuring instrument is static with respect to the fluid, the measured quantity is known as static property. The measured static property could be pressure, velocity temperature, enthalpy etc.,. The state of the particle fixed by a set of static properties is called static state.
Stagnation State
It is defined as the terminal state of a fictitious, isentropic work-free and steady flow process during which the final macroscopic P.E. and K.E. of the fluid particle are reduced to zero. Real process does not lead to stagnation state because no real process is isentropic. Stagnation property changes provide ideal value against which the real machine performance can be compared. It is possible to obtain stagnation properties in terms of static properties by using the definition of stagnation state. Consider the steady flow process given by the first law of thermodynamics.
sonicisvelocityandconstantisarea
zeroisA
dA
2γM
M1
p
dp
A
dA
1Mγp
dp
V
dV
2
2
2
−−−
−=
−−−−=
ADARSHA H G
Static state is the initial state in a fictitious isentropic work free , steady flow process and the stagnation state is the terminal state in which the ke and pe are reduced to zero, one can define a stagnation state at the initial static state.
In the above eq. subscript o represents stagnation state and subscript i represents initial static state If subscript i is removed from the initial static state Thus stagnation state has been expressed as the sum of three static properties. Since the process is isentropic, the final entropy is same as initial enthalpy. Final entropy is stagnation and initial entropy is static. Any two independent properties at a specific state is sufficient to fix the state of simple compressible substance by using thermodynamic relations.
a) Incompressible Fluid: (Density is constant)
Final state is stagnation with subscript o and initial state is Static without subscript
∆pe∆ke∆hwq
gZ2
VhwgZ
2
Vhq o
2
ooi
2
ii
++=−
+++=
+++
∆pe∆ke∆hwq ++=−
iooo hh∆h0;pe0;ke0;w0;q −=====
( ) 1−−−++= iiio pekehh
( ) 2pekehho −−−++=
3---sso =
∫∫ =
=
==
−=⋅
oo
o
dpρ
1dh
vdpdh
ssbecause0dsbut
vdpdhdsT ADARSHA H G
Thus stagnation pressure of an incompressible fluid is expressed in terms of static
pressure, velocity and height above a datum line.
Thus using thermodynamic relations stagnation pressure, Stagnation pressure,
stagnation temperature and stagnation internal energy are found.
Problem 1
Liquid water at standard density flows at a temperature of 20˚, a static pressure of 10 bar and a velocity of 20m/s. Find the total pressure and total temperature of the water. Given: T=20˚, p=10 bar and V=20m/s, To=? and Po=? , Water is incompressible with ρ = 1000 kg/m3
( )
( )( )
pgZ)2
Vρ(p
phpekehρp
phhρp
ppρ
1hh
2
o
o
oo
oo
++=
+−++=
+−=
−=−
TT0;dT
dTCduAlso,
uu0;du
constantρ0dv
pdvdu
ssas0ds
pdvdudsT
o
v
o
o
==∴
=
==∴
==
−=
==
+=⋅
o20TT
12barp
10102
201000p
pgZ)2
Vρ(p
o
o
52
o
2
o
==
=
×+
=
++=ADARSHA H G
Problem 2
A turbomachine handling liquid water is located 8m above the sump level and
delivers the liquid to a tank located 15m above the pump. The water velocities in
the inlet and outlet pipes are 2m/s and 4m/s respectively. Find the power required
to drive the pump if it delivers 100 kg/min of water.
Given: z1=8m; z2=15m; V1=2m/s; V2=4m/s; mass flow rate=100 kg/min
Find the Power P=?
b) Perfect gas
( )
( )
W386wmPPower
J/kg231.6pumponwork
J/kg231.68159.812
240w
zzg2
VV
ρ
ppw
ρ
∆p∆hqw
22
12
2
1
2
212
oo
===
=
−=
++
−+−=
−+
−+
−−=
−=−=
&
1γ
Rγc
cgeliminatin
Rccandγc
c
p
v
vp
v
p
−=
=−=
p
2
o2c
VTT +=
ADARSHA H G
Substituting the value of cp in the above equation and simplifying
Efficiencies of Turbomachines
Efficiency of a turbomachine is given by
( )
−+=
2
M1γ1TT
2
o
( )
( ) 1γ
γ2
o
γ
1
o
o
2
oo
ooo
o
oo
2
M1γ1
p
p
p
p
v
v
2
M1γ1pvvp
pv
vp
T
T
T
pv
T
vp
−
−+=
=
−+=
=
=
a
m
ma
ma
ηη
100%almostareη
efficiencymechanicalη;efficiencyadiabaticηwhere,
ηηη
≈∴
==
= ADARSHA H G
The adiabatic efficiency of a TM can be calculated from the h
the expansion and compression process. The ideal
using either static or stagnation states.
a) Power Generating Turbomachines (PGTM)
Actual work output for PGTM =
The proper equation is determined by the conditions of Turbomachine in question.
For example in a turbine if the inlet
production of mechanical energy somewhere else, then static to total definition is
used. If the exit ke is wasted then static to static definition is used.
48;h50;hLet 101 ==
35
30
hh
hhη
02'01
0201tt ==
−
−=−
33
30
hh
hhη
02'1
0201ts =
−
−=−
The adiabatic efficiency of a TM can be calculated from the h-s diagram for both
the expansion and compression process. The ideal work input or output can be
r static or stagnation states.
Power Generating Turbomachines (PGTM)
Actual work output for PGTM =
The proper equation is determined by the conditions of Turbomachine in question.
For example in a turbine if the inlet ke is negligible and exit ke is used for
production of mechanical energy somewhere else, then static to total definition is
is wasted then static to static definition is used.
0201 hh −
1
01s-s2'1ss
1
01ts02'1ts
01
01st2'01st
01
01tt02'01tt
h
hηhhw
h
hηhhw
h
hηhhw
h
hηhhw
=−=
=−=
=−=
=−=
−
−−
−−
−−
h10;h15;h20;h48; 202'02 ===
85.71% 66.67%45
30
hh
hhη
2'01
0201st ==
−
−=−
90.9%= 69.76%43
30
hh
hhη
2'1
0201s-s ==
−
−=
diagram for both
work input or output can be
The proper equation is determined by the conditions of Turbomachine in question.
is used for
production of mechanical energy somewhere else, then static to total definition is
2'1
0201
02'1
0201
2'01
0201
02'01
0201
h
h
h
h
h
h
h
h
−
−
−
−
−
−
−
−
5h2' =
66.67%
69.76%
ADARSHA H G
b) Power Absorbing Turbomachines (PATM)
Actual work input for PATM =
Problem 1
Air as a perfect gas flows in a duct at a velocity of 60 m/s, a static pressure of 2 atm., and a static temperature of temperature of air at this point inRepeat the problem with a flow velocity of 500 m/s.Given: (a) V=60 m/s, p=2atm., T=300K (b) V=500 m/s, p=2atm., T=300K Find (a) To = ? and po=? (b) To = ? and po=? Solution: Equations used are
p
2
o2c
VTT += o
p
p
bar2.07p
2871.4
60
γRT
VM
301.79K10052
60300T
0
2
0
=
×==
=×
+=
Power Absorbing Turbomachines (PATM)
Actual work input for PATM =
Air as a perfect gas flows in a duct at a velocity of 60 m/s, a static pressure of 2 atm., and a static temperature of 300 K. (a) Find total pressure and total temperature of air at this point in the duct. Assume ratio of specific heats as 1.4. (b) Repeat the problem with a flow velocity of 500 m/s. Given: (a) V=60 m/s, p=2atm., T=300K (b) V=500 m/s, p=2atm., T=300K Find (a) To = ? and po=? (b) To = ? and po=? Solution: Equations used are
0102 hh −
s-s12'ss
ts102'ts
st012'st
tt0102'tt
h
hηhhw
h
hηhhw
h
hηhhw
h
hηhhw
=−=
=−=
=−=
=−=
−
−−
−−
−−
( ) 1γ
γ2
2
M1γ1
−
−+=
0.1728300
301.79K
=×
Air as a perfect gas flows in a duct at a velocity of 60 m/s, a static pressure of 2 300 K. (a) Find total pressure and total the duct. Assume ratio of specific heats as 1.4. (b)
Given: (a) V=60 m/s, p=2atm., T=300K (b) V=500 m/s, p=2atm., T=300K
0102
12'
0102
102'
0102
012'
0102
0102'
hh
hh
hh
hh
hh
hh
hh
hh
−
−
−
−
−
−
−
−
ADARSHA H G
Problem 2 Air enters a compressor at a static pressure of 15 bar, static temperature of 15and flow velocity of 50 m/s. At exit, the static pressure is 30 bar, static temperature of 100˚C and flow velocity of 10isentropic change in total enthalpy and b) Actual change in total enthalpy.Given: p1=15 bar; T1=288K; V=50 m/s; p2=30 bar; T2=100z1=1m. Find a) Isentropic change in total enthalpySolution: Plot the h-s or T-s diagram as shown in fig.p1=15 bar; T1=288K; V=50 m/s; p2=30 bar; T2=100
bar6.9p
2871.4
500
γRT
VM
424.38K10052
500300T
0
2
0
=
×==
=×
+=
(
++=
−=−
p
2
p
2
22'p
0102'p0102'
c
gz
2c
VTc
TTchh
inchangeIsentropic
T;p
p
T
T2'
γ
1γ
1
2
1
2'
=
−
(Tchh 02'p0102' −=−
Air enters a compressor at a static pressure of 15 bar, static temperature of 15and flow velocity of 50 m/s. At exit, the static pressure is 30 bar, static temperature
˚C and flow velocity of 100 m/s. The outlet is 1 m above the inlet. Find a) isentropic change in total enthalpy and b) Actual change in total enthalpy.Given: p1=15 bar; T1=288K; V=50 m/s; p2=30 bar; T2=100˚C; V2=100m/s; z2
Find a) Isentropic change in total enthalpy, b) Actual change in total enthalpys diagram as shown in fig.
p1=15 bar; T1=288K; V=50 m/s; p2=30 bar; T2=100˚C; V2=100m/s; z2
1.44300287
500
424.38K
=×
)
++−
−=
p
1
p
2
11
01
0102'
c
gz
2c
VT
hhenthalpytotalin
351.07K=
) KJ/kg67084T01 =−
Air enters a compressor at a static pressure of 15 bar, static temperature of 15˚C and flow velocity of 50 m/s. At exit, the static pressure is 30 bar, static temperature
0 m/s. The outlet is 1 m above the inlet. Find a) isentropic change in total enthalpy and b) Actual change in total enthalpy.
˚C; V2=100m/s; z2-
tual change in total enthalpy
˚C; V2=100m/s; z2-z1=1m
ADARSHA H G
Finite stage efficiency
1. A stage with a finite pressure drop is a finite stage 2. In a multi-stage turbine along with the overall isentropic efficiency the
efficiencies of individual stages are important 3. On account of large pressure drop and associated thermodynamic effect the
overall isentropic efficiency is not a true index of aerodynamic or hydraulic performance of machine
4. Different stages with the same pressure drop located in different regions of h-s plane will give different values of work output
Effect of Reheat (Turbines)
Consider four number of stages between two states as shown in fig. It is assumed that the pressure ratio and stage efficiency are same for all the four stages.
The actual work during expansion from state 1 to state 2 is
( )
++−
++=
−=−
−=
p
1
p
2
11
p
2
p
2
22p
0102p0102
0102
c
gz
2c
VT
c
gz
2c
VTc
TTchh
hhenthalpytotalinchangeActual
( ) KJ/kg89099.8TTchh 0102p0102 =−=−
s
aT
2
z
z
y
y
x
x
1
W
WηefficiencyOverall
1---Constantp
p
p
p
p
p
p
p
==
====
2WηW sTa −−−=
ADARSHA H G
The values of ideal or isentropic work in the stages are
The total value of actual work done in these stages are
Equating eq. 2 and eq. 3
The slope of constant pressure lines on h
s2s1 ∆,∆W,∆W
η∆WW staa ==∑2WηW sTa −−−=
W
∆W
ηη
ηWηW
s
4
1istT
stsTa
=
==
∑=
Ts
h
p
−=
∂
∂
The values of ideal or isentropic work in the stages are
The total value of actual work done in these stages are
Equating eq. 2 and eq. 3
The slope of constant pressure lines on h-s plane is given by
s4s3 ∆W,∆W
( )∆W∆W∆W∆W s4s3s2s1st −−−+++
2
4
W
∆W
si
4
1i
sist
−−−
∑=
5−−−
3−
ADARSHA H G
The above equation shows that constant pressure lines must diverge towards the
right
This makes the overall efficiency of the turbine greater than the individual stage
efficiencies
The quantity is known as reheat factor and is always greater than
one.
Reheat is due to the reappearance of stage losses as increased enthalpy during
constant pressure heating process.
Infinitesimal stage efficiency or Polytropic efficiency
(Turbines)
• To obtain the true aerodynamic performance of a stage the concept of small or infinitesimal stage is used
• This is an imaginary stage with infinitesimal pressure drop and is therefore independent of reheat effect
• Fig. shows a small stage between pressures p and p-dp The efficiency of this stage is
6---1W
∆W
s
4
1i
si
>∴∑
=
stT ηη >
s
4
1i
si
W
∆W∑=
1>=stη
ηRF
s
pdT
dT
dropetemperaturisentropic
dropetemperaturactualη ==
ADARSHA H G
For infinitesimal isentropic expansion
Expanding the terms on r.h.s. using bionomial expression and neglecting terms
beyond second
γ
1γ
s
γ
γ
s
p
dp1
T
dT1
p
dpp
T
dTT
−
−
−=−
−=
−
−−=−
γ
1γ1
T
dT1 s
ηp
dp
γ
1γ
T
dT
dp1γ
γ
T
dTη
1dT
dTηBut
p
dp
γ
1γ
T
dT
p
s
p
s
−=
−=∴
−−−=
−=
For infinitesimal isentropic expansion
the terms on r.h.s. using bionomial expression and neglecting terms
1
γ
1−
p
dp1
2η
dp
p
1
p −−−
the terms on r.h.s. using bionomial expression and neglecting terms
ADARSHA H G
Integrating the eq.2 between limits 1 and 2
The irreversible adiabatic (actual) expansion process can be considered as equivalent to a polytropic process with index n.
When ηp=1, n=γ the expansion line coincides with the isentropic expansion. The efficiency of a finite stage can now be expressed in terms of small stage efficiency. Taking static values of Tand p and assuming perfect gas
4p
p
T
T
3
p
plog
T
Tlog
1γ
γ
η
p
plogη
γ
1γ
T
Tlog
p
dp
γ
1γη
T
dT
p
2
1
2
1
ηγ
1γ
1
2
1
2
1
2e
1
2e
p
1
2ep
1
2e
p
p
p
T
T
−−−
=
−−−
−=
−=
−=
−
∫∫
( )6
η1γγ
γn
isprocessactualinexpansionofindexThe
51γ
γ
n
1nη
n
1nη
γ
1γ
indicesEquating
p
p
p
p
T
T
p
p
p
n
1n
1
2
ηγ
1γ
1
2
1
2
p
−−−−−
=
−−−
−
−=
−=
−
=
=
−
−
ADARSHA H G
Equations 7 and 8 can give the efficiencies of various finite expansion processes
with different values of pressure ratio and small stage efficiency
Problem 1
The overall pressure ratio through a 3 stage gas turbThe temperature at inlet is 1400 K. If the temperature rise in each stage is same, determine for each stage (a) pressure ratio (b) stage efficiency.Given: T1=1400 K; p1/p2=10; To find: pressure ratio for each stage and Draw the T-s diagram and assume the gas to be perfect gas
−=−
−
−=
121
2s1
21st
1TTT
TT
TTη
γ
γ
γ
1γ
st
12s1
pr1
pr1η
1TTT
−
−=
−=−
−
−−
ratiopressureif
Equations 7 and 8 can give the efficiencies of various finite expansion processes
with different values of pressure ratio and small stage efficiency
The overall pressure ratio through a 3 stage gas turbine is 10 and efficiency is 86%. The temperature at inlet is 1400 K. If the temperature rise in each stage is same, determine for each stage (a) pressure ratio (b) stage efficiency. Given: T1=1400 K; p1/p2=10; η=86%; ∆Tst= constant To find: pressure ratio for each stage and ηst
s diagram and assume the gas to be perfect gas
( )
−=
−
−pη
1γ
γ1
1
2
Pr
11T
T
T
( )
2
1
γ
1
η1
γ
1γ1
1
2s
p
pprwhere-7-;-
pr
11T
T
T
p
=
−=
−
−
−
8
pr1
pr1η;
p
pprratio
γ
1γ
ηγ
1γ
st
1
2
p
−−
−
−==
−
−
Equations 7 and 8 can give the efficiencies of various finite expansion processes
ine is 10 and efficiency is 86%. The temperature at inlet is 1400 K. If the temperature rise in each stage is same,
8
ADARSHA H G
First stage efficiency is given by
1013.1KTand1206.55KT
193.45K3
580.37∆T
sameisstageeachinriseeTemperatur
580.37KTT
819.6KT
725.1KT;10
1
p
p
T
T
0.86TT
TTη
yx
st
21
2
2'
1.4
0.4γ
1γ
1
2
1
2'
2'1
21
==
==
=−
=∴
=
=
=
=−
−=
−
( )
1.895p
p
0.5271400
1206.55
T
T
p
p
p
p
T
T
0.814
10
1log
1400
819.6log
0.4
1.4
p
plog
T
Tlog
1γ
γ
η
x
1
.40.814
1.4η1γ
γ
1
x
1
x
ηγ
1γ
1
x
1
x
e
e
1
2e
1
2e
p
p
p
=
=
=
=
=
=
×=
−=
×−
−
0.8η
p
pprwhere;
pr1
pr1η
st
x
1
γ
1γ
ηγ
1γ
st
p
=
=
−
−=
−−
−−ADARSHA H
G
Pressure ratio and stage efficiency of 2 stage is given by
Pressure ratio and stage efficiency of 3 stage is given by
Problem 2 In a 3 stage turbine the pressure ratio of each stage is 2 and stage efficiency is 75%. Calculate the overall efficiency and power developed if the air is initially at a temperature of 600˚C and flows through it at the rate of 25kg/s. Also Find the reheat factor. Given: ηst=75%; 3 stage turbine; pressure ratio across each stage=2; T1=600˚C; mass flow rate = 25 kg/s To find: η=?; P=?; RF=? Draw the T-s diagram
1013.1KT1206.55K;T yx ==
( )
0.4711206.55
1013.1
T
T
p
p .40.814
1.4η1γ
γ
x
y
x
y p
=
=
=
×−
2.12p
p0.471;
p
p
y
x
x
y==
y
x
γ
1γ
ηγ
1γ
stp
pprwhere;
pr1
pr1η
p
=
−
−=
−−
−−
0.797ηst =
0.7938η
2.487p
p0.402
p
p
819.6KT1013.1K;T
st
2
y
y
2
2y
=
==
==
ADARSHA H G
Finite stage efficiency (Compressor)
• A compressor with a finite pressure rise is known as a finite stage • Stage work is a function of initial temperature and pressure ratio • For the same pressure ratio, a stage requires a higher value of work with
higher temperature • Thus compressor stages in the higher temperature region suffer on account
of this The above factors have a cumulative effect on the efficiency of multistage compressor
( )( )73.07%η
0.750.51
0.5-1
0.75
pr1
pr1η
p
0.2857
η.2857
γ
1γ
ηγ
1γ
st
p
p
=
=−
=
−
−=
×
−
−
( )
1.0480.75
0.7862
η
ηRF
7724.25KW8
118731.00525TTcmP
78.62%η
p
p1
p
p1
p
p
p
p
p
p1
p
p
p
p
p
p1
TT
TTη
st
.2857.7307
21p
3γ
1γ
1
x
3ηγ
1γ
1
x
γ
1γ
y
2
x
y
1
x
ηγ
1γ
y
2
x
y
1
x
2'1
21
pp
===
=
−×××=−=
=
−
−
=
××−
××−
=−
−=
×
×−
×−
−
−
&
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Effect of preheat (Compressor)Consider a compressor with four stages as shown. It is assumed that all the stages have the same efficiencies and pressure ratios.
The total isentropic work from state 1 to 2s is WsThe isentropic work in the individual stages are The overall efficiency of the compressor is
This makes the overall efficiency of the compressor smaller than stage efficiency
This is due to the thermodynamic effect called ‘Preintentionally heated( preheated) at the end of eachin small constant pressure processes is only an internal phenomena and the compression process still remains an adiabatic process.
1
∆W
Wbut
∆Wη
1W
W
∆WηHowever
4
1i
si
s
4
1i
si
st
a
1x
s1st
<
=
==
∑
∑
=
=
η
Effect of preheat (Compressor) Consider a compressor with four stages as shown. It is assumed that all the stages have the same efficiencies and pressure ratios.
otal isentropic work from state 1 to 2s is Ws. The isentropic work in the individual stages are ∆Ws1, ∆Ws2, ∆Ws3 and The overall efficiency of the compressor is η=Ws/Wa
This makes the overall efficiency of the compressor smaller than stage efficiency
This is due to the thermodynamic effect called ‘Pre-reheating ; the gas is not intentionally heated( preheated) at the end of each compression stage. The preheat in small constant pressure processes is only an internal phenomena and the compression process still remains an adiabatic process.
W
∆W
W
∆W
W
∆W
z2
s4
yz
s3
xy
s2 == ;
( s2s1
st
a ∆W∆W∆Wη
1W ++==
stηη <
Consider a compressor with four stages as shown. It is assumed that all the stages
Ws3 and ∆Ws4
This makes the overall efficiency of the compressor smaller than stage efficiency
reheating ; the gas is not compression stage. The preheat
in small constant pressure processes is only an internal phenomena and the
)s4s3 ∆W∆W +
ADARSHA H G
Infinitesimal or Polytropic Efficiency (Compressor)• A finite compressor stage can be made up of infinite number of small stages • Each of these infinitesimal stages have an efficiency
efficiency or infinitesimal stage efficiency• It is independent of thermodynamic effect and is therefor
the aerodynamic performance of the compressor• Consider a stage in which air is compressed from state 1 to state 2. It also
shows an infinite stage operating between pressures
T
dT
T
dT1
using
T
dT1
T
dTT
dT
dTη
s
s
s
s
p
=
+
+
+
=
Infinitesimal or Polytropic Efficiency (Compressor)A finite compressor stage can be made up of infinite number of small stages Each of these infinitesimal stages have an efficiency ηp called polytropic efficiency or infinitesimal stage efficiency It is independent of thermodynamic effect and is therefore a true measure of the aerodynamic performance of the compressor Consider a stage in which air is compressed from state 1 to state 2. It also shows an infinite stage operating between pressures p and p+dp
p
dp
γ
1γ
p
dp
γ
1γ1
expansionbionomial
p
dp1
p
dpp
1dT
dT
γ
1γ
γ
1γ
s
−
−+=
+=
+=
−−
−
−
Infinitesimal or Polytropic Efficiency (Compressor) A finite compressor stage can be made up of infinite number of small stages
called polytropic
e a true measure of
Consider a stage in which air is compressed from state 1 to state 2. It also p+dp
ADARSHA H G
Substituting the value of dTs into equation 1
Integrating eq.2 between state 1 to 2
Assuming the irreversible adiabatic compression as equivalent to a polytropic
process with index n, equation 3 can be written as
The efficiency of finite compressor stage can be related to small stage efficiency
The actual temperature rise is given by
2η
1
p
dp
γ
1γ
T
dT
p
dp
γ
1γ
T
dTη
p
p
−−×
−=
−=
3
T
Tlog
p
plog
γ
1γ
ηorp
plog
η
1
γ
1γ
T
Tlog
1
2e
1
2e
p
1
2e
p1
2e −−
−
=
−=
( )4
η1γ1
γη
1n
n
γ
1γη
n
1n
γη
1γ
p
p
p
p
p
p
p
p
n
1n
2
1γη
1γ
2
1p
−−−−−
=−
×−
=
−=
−
=
−−
5
1pr
1pr
1T
T
1T
T
TT
TTη
p
pprwhere1prT1
T
TTTT
p
p
γη
1γ
γ
1γ
1
2
1
2s
12
12sst
1
2γη
1γ
1
1
2112
−−−
−
−=
−
−
=−
−=
=
−=
−=−
−
−
−ADARSHA H G
Problem1
A 16 stage axial flow compressor is to have a pressure ratio of 6.3, with a stage efficiency of 89.5%. Intake conditions are 288K and 1 bar. Find (a) Overall efficiency (b) Polytropic efficiency ( c) Preheat factor. Assume pressure ratio per stage is same. Given: 16 stage axial flow compressor, pressure ratio = 6.3, ηst=89.5%, T1=288K and p1=1bar Solution:
Problem 2 An air compressor has 8 stages of equal pressure ratios of 1.3. The flow rate through the compressor and its overall efficiency are 45 kg/s and 80% respectively. If the conditions of air at entry are 1 bar and 35˚C determine (a) State of compressed air at exit (b) polytropic efficiency (c) Stage efficiency Given: 8 stages of equal pressure ratio of 1.3, mass flow rate of 45 kg/s, η = 80%, p1= 1bar, T1= 35˚C To find: (a) p2=?, T2=? (b) ηp=? (c) ηst=?
1.1219stageperratiopressurex
x6.3p
p
xconstantp
p
p
p...
p
p
p
p
16
1
17
1
2
2
3
15
16
16
17
==
==
======
( )
( )
( )( )
( )
( )
( )( )
0.970.9095
0.8775
η
ηfactorPreheat
87.75%16.3
16.3
1pr
1prη
0.9045η
11.1219
11.12190.895
1pr
1prη
st
0.2857/η
0.2857
γη
1γ
o
γ
1γ
o
p
0.2857/η
0.2857
γη
1γ
γ
1γ
st
p
p
p
p
===
=−
−=
−
−=
=
−
−==
−
−=
−
−
−
−
ADARSHA H G
Stage pressure ratio = 1.3; Overall pressure ratio = (1.3) = 8.157
8.157barp624.26KT
308T
3085610.8
TT
TTη
561KT
1
8.157
p
p
T
T
22
212
12'
2'
0.2857γ
1γ
1
2
1
2'
==
−
−==
−
−=
=
=
=
−
84.88%
308
624.26log
1
8.157log
0.2857
T
Tlog
p
plog
γ
1γ
η
e
e
1
2e
1
2e
p =
=
−
=
( )
( )84.3%
11.3
11.3
1pr
1prη
0.8488
0.2857
0.2857
γη
1γ
γ
1γ
st
p
=
−
−=
−
−=
−
−
ADARSHA H G