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 Compressors Axial Flow Compressors Stage= S+R S: 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 stages 2) each stage has a row of rotor blades followed by a row of stator blades. 3) fluid is accelerated by rotor blades.
Axial Flow Compressors In
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
C are C C used to guide the flow into the first guide vanes stage. Elementary Theory: Assume mid plane is constant r1=r2, u1=u2 assume Ca=const, in the direction of u. w
w2
C w C w C w 2
1
w1
, in the direction of u.
Axial Flow Compressors Inside the rotor, all power is consumed. Stator only changes K.E.P static, To2=To3 Increase 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 Compressors Two dimensional analysis: Only axial ( Ca) and tangential (Cw). no radial component
C 1 u V 1
V 2 tangnt toblade at exit.
assuming C a1 C a2 C a this V2 can be ontained V2 C a2 cos 2
then V2 & u triangle get C 2 normally 3 1 to prepare air to go to a similar stage
also C 3 C 1 .
W m c p (T o2 T o1 )
Axial Flow Compressors
Axial Flow Compressors from velocity triangles assuming
C a C a1 C a2
u/Ca tan 1 tan 1 , u / C a tan 2 tan 2 the power input to stage
'
(a)
W mu(C w2 C w1 )
where C w1 andC w2 are tangential components at inlet and exit of the rotors.
or in terms of the axial velocity
W muC a (tan 2 - tan 1 )
From equation (a) (tan 2 - tan 1 ) (tan 1 - tan 2 )
Axial Flow Compressors Energy balance c p T o5 c p (T o3 T o1 ) c (T o2 T o1 ) uC a (tan 1 tan 2 )
T o uC a (tan 1 tan 2 ) / c p 5
pressure ratio at a stage
s T o 1 ps po T o po3 1
5
1
1 where, s stage isentropic efficiency
Ex. u 180 m/s, 1 43.9o , s 0.85, 0.8, Ca 150m / s , 2 13.5, To1 288, Rs 1.183 Rcentrifugal ,
higher due to centrifugal action
Axial Flow Compressors Degree of reaction
is the ratio of static enthalpy in rotor to static enthalpy rise in stage hr static enthalpy rise in rotor static enthalpy rise in the stage hs For incompressible isentropic flow Tds=dh-vdp
dh=vdp=dp/ Tds=0 h=p/ ( constant ) Thus enthalpy rise could be replaced by static pressure rise ( in the definition of )
o 1
but 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)
Assume C3 C1 , and Ca const. ( for simplicity)
To Tstagnation Tstage Ts 5
1 Ca (tan 1 tan 2 ) / 2u 1 tan , C a / u tan (tan 1 tan 2 ) / 2
Axial Flow Compressors special condition =0 ( impulse type rotor)
from equation 3
Ca (tan 1 tan 2 ) / 2u
1=-2 , velocities skewed left, h1=h2, T1=T2 =1.0 (impulse type stator from equation 1) =1-Ca(tan1+tan2)/2u, 2=1 velocities skewed right, C1=C2, h2=h3 T2=T3 =0.5
from 2
2 1 ;
1 2
2
(tan 1 tan 2 )
symmetric angles
V2 c1 ,V1 c2 ;
P2 P1 P P
Axial Flow Compressors Three dimensional flow 2-D 1. the effects due to radial movement of the fluid are ignored. 2. It is justified for hub-trip ratio>0.8 3. 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 U C a
tan 1 tan 1 tan 2 tan 2
U (C w1 C w1 ) W m UC a (tan 2 tan 1 ) m UC a (tan 2 tan 1 ) m T os T o 3 T o1 T o 2 T o1
UC a c p
(tan 1 tan 2 )
pressurerise per stage Rs
po 3 po1
[1
s T s
T o1
] /( 1)
Axial Flow Compressors Design •
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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 Compressors Requirements: •
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
•
2 r r 2 1 r t 1 C a r t
U t 2 * * t t * N rps
2 r r 2 1C a A 1 r t 1 m r C a t 1
r t
m
2
r r 1C a1 1 2 r t
N
2
350 2 r t
(a)
, solve to get rt & r r / r t
Axial Flow Compressors procedure T o 1 T a 288K , Po Pa 1.01 bar 1
C1
C a 150 1
C 1
T 1 T o
2
2c p 2
1
276.8
T 1 1 P1 T o Po P1
1
1
1
P1 RT
1.106kg / m 3
Axial Flow Compressors •
From equation (a) r t 2
0.03837
r 2 1 r r t
N 350 / 2 rt assume rr / r t f rom 0.4 0.6 r r / r t
r t
N
0.4
0.2137
260.6
0.5
0.2262
246.3
0.6
0.2449
227.5
Axial Flow Compressors •
•
Consider rps250 Thus rr/rt=0.5, rt=0.2262, ut=2rt*rps=355.3 m/s Get V1t
u1t C a1 2
2
385 .7
a RT 1 M 1
v1t a
1.165
Is ok. Discussed later. Results r-t=0.2262, r-r=0.1131, r-m=0.1697 m
Axial Flow Compressors At Po2 Po1
exit of compressor
4.15 [ given Po 4.19bar ];
where
2
n -1 n
1
T 2 T o2
C a
0 .4 1 .4
P2
441.3 K;
P2 3.84 bar; 2
T o1
Po Po
2
1
Po 2
P2 RT 2
T 2 T o
2
1
2 A2 C a , A 2 0.044; 3.03 kg/m3 ; m
but A 2 h( 2 r m ) h 0.0413; thus rt r m r r r m
h
2
n
317, assume 0.9; T o 2 452.5 K ;
2
2c p
T o2
n 1
h
2
0.19303m;
0.1491 m
results N 250 rps; u t 355.3;
C a 150; r m 0.1697m
Axial Flow Compressors No. •
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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 7 stages - 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 1 st, 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
u 2 rm thus, u m 266 m/s
cpT o uC w C w 76 .9 m / s, 1 0 C w1 0, C w 2 76 .9m / s
Axial Flow Compressors •
Angles tan 1 tan 2 tan 2 thus
u
1 60.64
C a u
C w 2 C a
C w 2 C a
2 51.67
2 27.14
the deflection in rotor blades
1 2 8.98 o check de Haller C a / cos 2
v2
v
C
cos
cos 1
cos
0.79 which is less than 0.72
Axial Flow Compressors pressures poly tropic efficiencies
T o5 1 assume s 1 po3 1.249 po1 T o1 T o 3 T o1 T o 5 308K p o 3
equation 1 -
1
Ca 2u
C w 2 C 1
2u
(tan 1 tan 2 )
0.856
Axial Flow Compressors •
Second stage T o5 25K , 0.93 (1) c p T o5 uC a (tan 1 tan 2)
tan 1 tan 2 0.6756 (2)
C a
(a)
(tan 1 tan 2 ); take 0.7
2u tan 1 tan 2 2.488
(b)
solve (a) and (b) 1 57.7 & 2 42.7 u C a
tan 1 tan ; 2
u C a
0
tan 2 tan 2
11 06; 41 05
Axial Flow Compressors T o 3 308 25 333 Po3 Po1
s 25
1 308
3.5
p o3 1.599 bar
de Haller for secondstage ; C 3 C 2
cos 2 cos 1
cos 27.15 cos11.06
V2 V1
cos 1 cos 2
0.721
0.907
stage 3
0.88, T03 25K , 0.5 c p T o 5 uC a (tan 1 tan 2 );
C a
2u
(tan 1 tan 2 )
Axial Flow Compressors de Haller no. is cos 1 cos 2
cos51.24 cos 28
0.709
take To5 24 tan 1 tan 2 0.685 thus 1 50.92, 2 28.65; giving de Haller number of 0.718
* performance of 3 rd stage
p o3
3.5
0.9 24 1 1.246 p 333 o1 3 ( po 3 ) 1.599 1.246 1.992 bar; To3 3 333 24 357 K
Axial Flow Compressors From symmetryof the velocity diagram 1 2 28.63 and 2 1 50.92 .the whirl velocities are given by 0
C 1 150 tan 28.63 81.9m / s; C 2 150 tan 50.92 184.7m / s stages 4 and 5,6 c p T o5 uC a (tan 1 tan 2 ) po3 p o1
T o 1 T o
5
1
C a
2u
(tan 1 tan 2 )
3 24 1 . 005 10 ; tan 1 tan 2 0.7267 0.83 266.6 150
(tan 1 tan 2 ) 0.5 2
266.6 150
1.7773
yielding 1 51.38 and 2 27.71( 1 ). 0
Axial Flow Compressors Stage
Po1 T o1 p o 3
4
5
6
1.992
2.447
2.968
357
381
405
1.228
1.213
1.199
2.447
2.968
3.560
381
405
429
0.521
0.592
p o1
p o 3
T o 3
p o 3 p o1 0.455
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
p o 3 4.192 1.177 3.56 p o1 7 the temperature rise required to give
the pressure ratio can be detrmined from
1
0.90Tos 429
3.5
1.177
givingT os 22.8 K
Axial Flow Compressors •
the corresponding air angles, assuming 50 per cent reaction, are then 1=50.98,
2 28 .52 0 ( 1 ) with a satisfactory de Haller number of 0.717.
Design calculations using EES –
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"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 EES Determination of the rotational s peed and annulus dimens ions
Known Information To 1 = 288
[K]
Po 1 = 101
Ca 1 = 150
[m/s]
r r r t
Gam r = m
=
1
A1 = Ut =
P1 Po 1
1
=
· Ca 1 · A1
2 ·
2
2
– r r )
· r t · N rps
To 1 –
=
cp
=
= 20
1005
[kg/s] R =
U t = 350 0.287
mas s balance
Ca 1
T1 =
0.5
m
– 1
· ( r t
C1 =
=
[kPa]
T1 To 1 P1
C1
2
2 · cp Gamr
relation between Area and eye dim ens ions
=
[m/s] 1.4
Design calculations using EES Calculate radii at exit section
Choose (round) rotational speed as 250 rps N rps
=
250
Thus calc new value for tip s peed rt 1 =
0.2262
Ut =
2 ·
r m
0.1697
=
· rt 1 · N rps
Known Information To 1 = P ratio =
288
[K]
4.15
Assumptions Etta inf =
0.9
Ca 2 =
Ca 1
Ca 1 =
150
G
[m/s]
Design calculations using EES nratio
P ratio To 2
1
=
Etta inf · Gamr Po 2
=
Po 1 Po 2
=
To 1
Po 1
m
=
A2
=
2 ·
C2
=
Ca 2
T2
Po 2
2
2
=
P2
=
r t =
nratio
·
Ca 2 · A2 ·
C2
To 2 –
To 2 P2
R · r m +
2
2 · cp
T2
=
h · rm
T2 h 2 h
Gamr
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
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]
rt1 = 0.2262 [m]
r m = 0.1697 [m]
r r = 0.1491 [m]
r t = 0.1903 [m]
To1 = 288 [K]
T2 = 441.3 [C]
Ut = 355.3 [m/s]
= 1.4
2 = 3.032
To2 = 452.5 [K]
Design calculations using EES Calculate number of stages
Known Information To 1 = 288 P ratio =
[K]
4.15
Po 1 = 101 To outlet
[kPa]
=
452.5
cp =
1005
m
= 20
[kg/s]
Assum ptions delT stage
=
Ca 1 = 150 Gamr
=
[m/s]
– 1
delT ov = N stages
25
=
To outlet – To 1 delT ov delT stage
R =
0.287
=
1.4
