ASE 376K Propulsion

© Philip L. Varghese, UT Austin, 2001 1

Designing rotor blades for radial equilibrium(cr ≈ 0)

The radial equilibrium condition is:

drdsT

drdh

drrcd

rc

drdc

crc

drdc

cdrdc

c ozz

zz −=+=++

)(2θθθθ

θ (1)

We assume in this analysis that 0≈drds

and choose to design for 0=drdho , i.e. constant work

input (compressor) or output (turbine) per unit length of rotor blade. Thus we have

0)(

=+drrcd

rc

drdc

c zz

θθ , (2)

which implies that the radial variations of cz and cr are coupled, if we want cr≈0.

Consider a compressor rotor and let us design our blade twist variations in the rotor (β1(r), β2(r)) such that:

+

=

−

=

rrb

rrarc

rr

brr

arc

m

n

m

m

n

m

)(

)(

2

1

θ

θ

rm is the mean radius; a, b > 0. (3)

The exponent n is a design choice. A similar set of equations can be used for a turbine rotor, by replacing subscript 1 with 3. Note that c2θ > c1θ for a compressor and c2θ > c3θ for a turbine.

From Eqs. (3) we see that

=−

rr

brcrc m2)()( 12 θθ , so that

( ) )(222)()()( 12 rfbUrbrr

brrcrcrUh mmm

o ≠=Ω=Ω=−=∆ λλλλ θθ . (4)

Here Ω is the angular velocity of the rotor. We see that the choice of tangential velocity

distribution (3) has guaranteed that ∆ho ≠f(r), i.e. 0=drdho , for any a, b, n (provided the fluid

follows the blades). Equation (4) determines b, provided ∆To,stage, Um and λ are known for the stage

λλ m

stageop

m

o

U

Tc

Uh

b22

,∆=

∆= . (5)

For example, if ∆To,stage ≈50K, cp ≈ 1000 J/kgK, Um ≈ 200 m/s, λ≈1, then b≈125m/s.

Using 222222zzr cccccc +=++= θθ for radial equilibrium, the degree of reaction for a

compressor

o

zz

ooo hcc

hcc

hrcrc

h

hhrR

∆−

−∆−

−=∆−

− =∆

=′ −

221

2)()(

1 )(21

22

21

22

21

2212 θθ

ASE 376K Propulsion

© Philip L. Varghese, UT Austin, 2001 2

Because ))(( 12122

122 θθθθθθ cccccc +−=− and )( 12 θθλ ccUho −=∆ , we get

o

zz

hcc

Ucc

rR∆−

−+

−=′22

1)(21

2212

λθθ (6)

It is convenient to normalize by the mean radius rm and introduce a dimensionless variable

mrr

x ≡ in equation (3):

xb

axcxb

axc nn +=−= θθ 21 ; (3a)

so

112

22

2−==

+ n

mm

n

xUa

xUax

Ucc

λλλθθ . (7)

Now we need to determine cz(r), or equivalently cz(x), from the radial equilibrium condition (2).

( )

( )

( )∫

∫

∫∫

≡−=−

−=−

−=

r

rzmz

r

r

zmz

r

r

r

r

z

m

m

mm

drrcdrd

rc

IIcc

drrcdrd

rccc

drrcdrd

rc

drc

drd

θθ

θθ

θθ

;2

22

2

22

22

2

(8)

Using the definitions of c1θ, c2θ in Eq. (3a), it is easy to see that the integral I in Eq. (8) is

( )

( ) ( )

( ) 1;)ln(12

2

0;11ln

1,0;11

12

)1(

)1(

2

12

1

212

=

−=

=

−±=

≠

−

−−+=

+=

−

−−∫

nxbxa

a

nx

bxa

nxn

bx

na

na

dxbxaxnaI

nn

x nn

m

m

m

(9)

The distributions with n=0,±1 are often used. They are called exponential (n=0), free-vortex (n=−1), and first-power (n=1). Hill and Peterson (p. 335) refer to the case n=1 as “constant reaction” but we shall see that the degree of reaction is not constant in this case. (Note that Hill and Peterson define the distribution for rcθ, rather than cθ.) Since blades are made by numerical machining there is no reason to stay with integral n. Although it gives mathematicians fits, the cases n=0 and n=1 can be considered under the general case using n=ε, and n=1±ε respectively, where ε is a small number like 10-5

. The numerical result for the blade angles, degree of reaction,

ASE 376K Propulsion

© Philip L. Varghese, UT Austin, 2001 3

etc. computed using this approximation are the same (to better than 1%) as the exact mathematical results for these special cases, and the fluid does not follow the specified angles exactly anyway.

Then, confining ourselves to the general case, we get:

( ) ( )

( ) ( )

−

−−−+−=−

−

−+−+−=−

−

−

11

12

)1(2

11

12

)1(2

1222

21

1221

22

nnzmz

nnzmz

xn

bxnanacc

xn

bx

na

nacc (10)

Assuming that c1zm= c2zm= czm (common design practice), then

( )1)1()1(

4 121

22 −

−+

−=− −nzz x

nn

abcc

Using ∆ho = 2bUmλ, we obtain ( )1)1()1(

21

21

22 −

−+−=

∆− −n

mo

zz xnn

Ua

hcc

λ, so finally from Eq. (6)

)1(

)12(1)(1

−−−+=′

−

nnx

UaxR

n

mλ. (11)

At the mean radius (x=1, r=rm) mm

m Ua

nn

Ua

RRλλ

−=−

−−+=′≡′ 1

)1()12(

1)1( , so the second

constant in the velocity expression is determined from

)1( mm RUa ′−= λ (12)

for all n. For mR′ ≈0.5, Um≈200 m/s, λ≈1, we have a≈100 m/s. In general R' increases with radius, so one has to choose mR′ sufficiently high that degree of reaction is not too low near the hub (problems of high adverse pressure gradient and flow separation near hub in stator). Conversely mR′ must not be so high that there is a high adverse pressure gradient and flow separation in the rotor near the tip. Examination of the figure shows why rotors commonly have relatively high hub-to-tip ratios, ζ, because this limits the variation in R' (and all other flow properties) from their mean radius values.

Special cases:

)1ln2)(1(11

21)1(10

1)1(11 2

−′−+=′=

−′−+=′=

′−−=′−=

xRRnx

RRn

xRRn

m

m

m

0.0

0.2

0.4

0.6

0.8

1.0

0.6 0.8 1.0 1.2 1.4

n=-1 (free vortex)n=0 (exponential)n=1 (first power)

Deg

ree

of r

eact

ion,

R'

x

R'm=0.6

ASE 376K Propulsion

© Philip L. Varghese, UT Austin, 2001 4

Examining Eq. (9) we see that for free-vortex blading (n=−1), the integral I is identically 0, so from Eq. (10) cz=czm for all x (or r). However in this case the degree of reaction falls quite rapidly for small x, and tends to go negative near the hub which limits the lower value of hub-to-tip ratio for the blade.

For a mean radius (rm) and hub-to-tip radius ratio ζ≡rh/rt, and designer’s choice for n, the values of mR′ , rotational speed (and thus Um), and stage stagnation temperature rise ∆To,stage are constrained by the need to maintain reasonable reaction at all radii, and limitations on the amount of turning that can be achieved in the rotor. These then determine the constants a and b. It must be remembered that while one can specify a flow turning angle in the rotor the flow will not necessarily follow the blade if the turning is too specified too high.