Post on 20-Jan-2016
Axial Flow Compressors
Axial Flow Compressors• Elementary theory
Axial Flow Compressors
Axial Flow Compressors
Comparison of typical forms of turbine and compressor rotor blades
Axial Flow CompressorsAxial Flow CompressorsStage= S+RS: stator (stationary blade)R: rotor (rotating blade)First row of the stationary blades is called guide vanes
** Basic operation
*Axial flow compressors: 1) series of stages2) each stage has a row of rotor blades followed
by a row of stator blades.3) fluid is accelerated by rotor blades.
Axial Flow CompressorsIn stator, fluid is then decelerated causing change in the kinetic energy to static pressure.
Due to adverse pressure gradient, the pressure rise for each stage is small. Therefore, it is known that a single turbine stage can drive a large number of compressor stages.
Inlet guide vanes are used to guide the flow into the first stage.
Elementary Theory:Assume mid plane is constant r1=r2, u1=u2 assume Ca=const, in the direction of u.
12 www CCC
, in the direction of u.12 www CCC
Axial Flow CompressorsInside the rotor, all power is consumed. Stator only changes K.E.P static, To2=To3Increase in stagnation pressure is done in the rotor. Stagnation pressure drops due to friction loss in the stator:C1: velocity of air approaching the rotor.
1 : angle of approach of rotor.u: blade speed.V1: the velocity relative t the rotor at inlet at an angle 1 from the axial direction.V2: relative velocity at exit rotor at angle 2 determined from the rotor blade outlet angle.2: angle of exit of rotor.Ca: axial velocity.
Axial Flow CompressorsTwo dimensional analysis: Only axial ( Ca) and tangential (Cw). no radial component
13
13
22
222
a
2
11
C also
stagesimilar a togo air to prepare to
Cget u triangle & V
cosV ontained be V
C assuming
exit.at blade tangnt to
2
21
C
normally
then
Ccanthis
CC
V
VuC
a
aa
)(12
.
oop TTcmW
Axial Flow Compressors
Axial Flow Compressors
)tan-(tan)tan-(tan 2112
21 aaa CCC
(a) tantan/,tantanu/Ca 2211 aCu
from velocity triangles assuming
the power input to stage )(mW12
'ww CCu
rotors. theofexit andinlet at components l tangentiaare 21 ww andCC
where
or in terms of the axial velocity
From equation (a)
)tan-(tanmuCW 12a
Axial Flow CompressorsEnergy balance
pao
aoooopop
cuCT
uCTTcTTcTc
/)tan(tan
)tan(tan)()(
21
21
5
12135
pressure ratio at a stage
3 5
1 1
1
1
s
1
2
1 where, isentropic efficiency
Ex.
u 180 m/s, 43.9 , 0.85, 0.8,
150 / , 13.5, 288, 1.183 ,
higher due to centrifugal action
o s os
o o
os
a o s centrifugal
p Tp stage
p T
C m s T R R
Axial Flow Compressors
Degree of reaction
enthalpy rise in rotor
enthalpy rise in the stager
s
hstatic
static h
is the ratio of static enthalpy in rotor to static enthalpy rise in stage
For incompressible isentropic flow Tds=dh-vdp
dh=vdp=dp/ Tds=0h=p/ ( constant )Thus enthalpy rise could be replaced by static pressure rise ( in the definition of )
1obut generally choose =0.5 at mid-plane of the stage.
Axial Flow Compressors
=0: all pressure rise only in stator=1: all pressure rise in only in rotor=0.5: half of pressure rise only in rotor and half is in stator. ( recommend design)
3 1 aAssume C ,and C . ( for simplicity)C const
5
1 21 (tan tan ) / 2
o stagnation stage s
a
T T T T
C u
uCa /,tan1
1 2tan (tan tan ) / 2
Axial Flow Compressorsspecial condition
=0 ( impulse type rotor)from equation 3
1 2(tan tan ) / 2aC u 1=-2 , velocities skewed left, h1=h2, T1=T2=1.0 (impulse type stator from equation 1)=1-Ca(tan1+tan2)/2u, 2=1velocities skewed right, C1=C2, h2=h3T2=T3
1 2
1(tan tan )
2 2
2 1
2 12 1 1 2
3 1
; symmetric angles
V , ; P P
c V cP P
=0.5from 2
Axial Flow CompressorsThree dimensional flow2-D1. the effects due to radial movement of the fluid are ignored.2. It is justified for hub-trip ratio>0.83. This occurs at later stages of compressor.
3-D are valid due to 1. due to difference in hub-trip ratio from inlet stages to later-stages, the annulus will have a substantial taper. Thus radial velocity occurs.2. due to whirl component, pressure increase with radius.
Axial Flow Compressors
)1/(
11
3s
21
1213
12
12
11
2211
]1[R
stageper rise pressure
)tan(tan
)tan(tan
)tan(tan
)(
tantantantan
o
ss
o
o
p
a
ooooos
a
a
ww
a
T
T
p
p
c
UC
TTTTT
UCm
UCm
CCUmW
C
U
Axial Flow Compressors Design Process of an axial compressor
• (1) Choice of rotational speed at design point and annulus dimensions
• (2) Determination of number of stages, using an assumed efficiency at design point
• (3) Calculation of the air angles for each stage at the mean line
• (4) Determination of the variation of the air angles from root to tip
• (5) Selection of compressor blades using experimentally obtained cascade data
• (6) Check on efficiency previously assumed using the cascade data
• (7) Estimation on off-design performance• (8) Rig testing
Axial Flow Compressors Design process:• Requirements:• A suitable design point under sea-level static conditions
(with =1.01 bar and , 12000 N as take off thrust, may emerge as follows:
• Compressor pressure ratio 4.15• Air-mass flow 20 kg/s• Turbine inlet temperature 1100 K• With these data specified, it is now necessary to
investigate the aerodynamic design of the compressor, turbine and other components of the engine. It will be assumed that the compressor has no inlet guide vanes, to keep weight and noise down. The design of the turbine will be considered in Chapter 7.
Axial Flow CompressorsRequirements:• choice of rotational speed and annulus dimensions;
• determination of number of stages, using an assumed efficiency;
• calculation of the air angles for each stage at mean radius;
• determination of the variation of the air angles from root to tip;
• investigation of compressibility effects
Axial Flow Compressors Determination of rotational speed and annulus
dimensions:• Assumptions • Guidelines:• Tip speed ut=350 m/s• Axial velocity Ca=150-200 m/s• Hub-tip ratio at entry 0.4-0.6• Calculation of tip and hub radii at inlet• Assumptions Ca=150 m/s• Ut=350 m/s to be corrected to
250 rev/s•
Axial Flow Compressors
Equations• continuity
thus
•
rpstt NtU ***2
at
rta C
r
rrACm
2
211 1
1
)(
12
2
11
2 a
r
rC
mr
t
ra
t
trt
rrr
N /&rget tosolve,2
350t
at
rt C
r
rr
2
21 1
Axial Flow Compressors
procedure
311
1
111
21
1
1
1
/106.1
8.2762
150C
bar 01.1,288
11
21
1
1
mkgRT
P
PT
T
P
P
c
CTT
C
PPKTT
oo
p
o
a
aoao
Axial Flow Compressors
• From equation (a)
0.60.4 /r
2/350
1
03837.0
r
2
2
fromrassume
rtN
r
rr
t
t
r
t
tr rr /N
0.40.2137260.6
0.50.2262246.3
0.60.2449227.5
tr
Axial Flow Compressors
• Consider rps250• Thus rr/rt=0.5, rt=0.2262, ut=2rt*rps=355.3 m/s
7.385V 2
1
211t
at CuGet
1RTa
165.111
a
vM t
Is ok. Discussed later. Results r-t=0.2262, r-r=0.1131, r-m=0.1697 m
Axial Flow CompressorsAt exit of compressor
mmOutletmminlet
mrC
hrr
mh
rthushrhbut
CAmRT
PP
T
T
c
CTT
KT
P
P
T
Tbargiven
P
P
ma
mr
mm
a
op
ao
o
n
n
o
o
o
o
o
o
1491.0r ;1903.0r : ;1131.0r ;2262.0r :
1697.0 ;150 ;3.355u rps; 250N
results
m 1491.02
;19303.02
r ;0413.0)2(A
;044.0A , ;kg/m 03.3 bar; 84.3
P
P K; 3.441
2
;5.452 ;9.0 assume ,3174.1
4.01
n
1-n where
];19.4P [ 15.4
rtrt
t
t2
2223
2
222
12
o
22
2
2
1
o
22
2
1
2
1
2
2
1
2
Axial Flow Compressors
No. of stages To =overall = 452.5-288=164.5K• rise over a stage 10-30 K for subsonic• 4.5 for transonic• for rise over as stage=25• thus no. of stages =164.5/25 stages 7
- normally To5 is small at first stage
de haller criterion V2/V1 > 0.72
- work factor can be taken as 0.98, 0.93, 0.88 for 1st, 2nd, 3 rd stage and 0.83 for rest of the stages.
Axial Flow Compressors
Stage by stage design;
• Consider middle plane
• stage 1
• for no vane at inlet wo CuT cp
0 ,/ 9.76 1 smCw
smCC ww /9.76,0 21
m/s 266u thus,r2u mm
Axial Flow Compressors
• Angles
o
a
w
a
w
a
thus
C
C
C
Cu
C
u
98.8
bladesrotor in deflection the
14.27tan
67.51tan
64.60tan
21
22
2
22
2
11
check de Haller
0.72 than less is which 79.0cos
cos
cos
cos/
2
1
1/
2
1
2
a
a
C
C
v
v
Axial Flow Compressors
856.02
1
)tan(tan2u
C-1
308
249.11 assume
esefficienci cpoly tropi pressures
12
21a
513
3
1
1
5
1
3s
u
CC
equation
KTTT
pT
T
p
p
w
ooo
oo
o
o
o
Axial Flow Compressors• Second stage
05.41 ;06.11
tantan ;tantan
7.42&7.57(b) and (a)
)( 488.2tantan
7.0 take);tan(tan2
)2(
)( 6756.0tantan
)tan(tanc )1(
93.0,25
21
222
1
021
21
21
21
215p
5
aa
a
ao
o
C
u
C
u
solve
bu
C
a
uCT
KT
Axial Flow Compressors
0.28 ;24.51
)tan(tan2
);tan(tan
5.0 ,25T 0.88,
3
907.006.11cos
15.27cos
cos
cos
721.0cos
cos
V
V ; stage secondfor
bar 599.1308
251
33325308
21
21215
03
1
2
2
3
2
1
1
2
3
5.3
3
1
3
solvingu
CuCTc
K
stage
C
C
Hallerde
pP
P
T
aaop
os
o
o
o
Axial Flow Compressors
Kp
eperformanc
givingthus
take
o 35724333T bar; 992.1246.1599.1)(
246.1333
249.01
p
p
stage rd 3 of *
0.718 ofnumber Haller de ;65.28,92.50
685.0tantan24T
709.028cos
24.51cos
cos
cos
is no.Haller de
3o33
5.3
3o1
o
21
21o5
2
1
3
Axial Flow Compressors
:below summerized becan stages three theof eperformanc
).(71.27 38.51
7773.1150
6.26625.0)tan(tan
7267.01506.26683.0
10005.124tantan ;1
)tan(tan2
)tan(tanc
5,6 and 4 stages
/7.18492.50tan150 ;/9.8163.28tan150C
by given are s velocitie whirlthe.92.50
63.28 diagram velocity theofsymmetry From
120
1
21
3
21
21215p
21
012
21
1
5
1
3
the
andyielding
T
T
p
p
u
CuCT
smCsm
and
o
o
o
o
aao
Axial Flow Compressors
1oT
1
3
o
o
p
p
13 oo pp 3oT
3op
1oPStage456
1.9922.4472.968
357381405
1.2281.2131.199
2.4472.9683.560
381405429
0.4550.5210.592
Axial Flow Compressors• Stage 7• At entry to the final stage the pressure and
temperature are 3.56 bar and 429 K. the required compressor delivery pressure is 4.15*1.01=4.192 bar. The pressure ratio of the seventh stage is thus given by
KTgiving
th
p
p
os
o
o
8.22
177.1429
T0.901
from detrmined becan ratio pressure the
give torequired rise re temperatue
177.156.3
192.4
5.3
os
71
3
Axial Flow Compressors
• the corresponding air angles, assuming 50 per cent reaction, are then 1=50.98,
0.717. ofnumber Haller de
ry satisfacto a )(52.28 10
2 with
Design calculations using EES– "Determination of the rotational speed and annulus dimensions"
– "Known Information"
– To_1=288 [K]; Po_1=101 [kPa]; m_dot=20[kg/s]; U_t=350 [m/s]
– $ifnot ParametricTable
– Ca_1=150[m/s];r_r/r_t=0.5;cp=1005;R=0.287;Gamma=1.4
– $endif
– Gamr=Gamma/(Gamma-1)
– m_dot=Rho_1*Ca_1*A_1 "mass balance"
– A_1=pi*(r_t^2-r_r^2) "relation between Area and eye dimensions"
– U_t=2*pi*r_t*N_rps
– C_1=Ca_1
– T_1=To_1-C_1^2/(2*cp)
– P_1/Po_1=(T_1/To_1)^Gamr
– Rho_1=P_1/(R*T_1)
– $TabStops 0.5 2 in
Design calculations using EESDetermination of the rotational speed and annulus dimensions
Known Information
To1 = 288 [K] Po1 = 101 [kPa] m = 20 [kg/s] U t = 350 [m/s]
Ca1 = 150 [m/s] r r
r t = 0.5
cp = 1005 R = 0.287 = 1.4
Gamr =
– 1
m = 1 · Ca1 · A1 mass balance
A1 = · ( r t2
– r r2
) relation between Area and eye dimensions
U t = 2 · · r t · N rps
C1 = Ca1
T1 = To1 – C1
2
2 · cp
P1
Po1 =
T1
To1
Gamr
1 = P1
R · T1
Design calculations using EESCalculate radii at exit section
Choose (round) rotational speed as 250 rps
N rps = 250
Thus calc new value for tip speed
rt1 = 0.2262
U t = 2 · · rt1 · N rps
rm = 0.1697
Known Information
To1 = 288 [K]
P ratio = 4.15
Assumptions
Etta inf = 0.9
Ca2 = Ca1
Ca1 = 150 [m/s]
Gamr =
– 1
Design calculations using EESnratio =
1
Etta inf · Gamrnratio=(n-1)/n=(1/etta inf )/gamr
P ratio = Po2
Po1
To2
To1 =
Po2
Po1
nratio
m = 2 · Ca2 · A2
A2 = 2 · · h · rm
C2 = Ca2
T2 = To2 – C2
2
2 · cp
P2
Po2 =
T2
To2
Gamr
2 = P2
R · T2
r t = rm + h
2
r r = rm – h
2
Design calculations using EES
A2 = 0.04398 Ca1 = 150 [m/s] Ca2 = 150 [m/s] cp = 1005 [J/kgK] C2 = 150 [m/s] Ettainf = 0.9
= 1.4 Gamr = 3.5 h = 0.041 [m] m = 20 [kg/s] nratio = 0.3175 Nrps = 250 [rev per sec]
Po1 = 101 [kPa] Po2 = 419.2 P2 = 384 [kPa] Pratio = 4.15 R = 0.287 [kJ/kgK] 2 = 3.032
rt1 = 0.2262 [m] rm = 0.1697 [m] rr = 0.1491 [m] rt = 0.1903 [m] To1 = 288 [K] To2 = 452.5 [K]
T2 = 441.3 [C] Ut = 355.3 [m/s]
Design calculations using EESCalculate number of stages
Known Information
To1 = 288 [K] Po1 = 101 [kPa] m = 20 [kg/s]
P ratio = 4.15 Tooutlet = 452.5
Assumptions
delTstage = 25
Ca1 = 150 [m/s] cp = 1005 R = 0.287 = 1.4
Gamr =
– 1
delTov = Tooutlet – To1
Nstages = delTov
delTstage
Design calculations using EES
Ca1 = 150 [m/s] cp = 1005 [J/kgK] delTov = 164.5 delTstage = 25 = 1.4 Gamr = 3.5 m = 20 [kg/s] Nstages = 6.58
Po1 = 101 [kPa] Pratio = 4.15 R = 0.287 [kJ/kgK] To1 = 288 [K] Tooutlet = 452.5