1

Design of the Rotor of a Turbomachine

2

• The following data of a new design are normally known:

• The type of rotor to be used follows the shape number:

3m kgvolume V or mass flowMs s

3/4shapen VNY

If n is given the type of rotor to be used follows from this equation.

2

2

mspecific work Y or Head ms

1in somecases alsothe speed ns

If n is not known the type of rotor has to be assumed and the shape number determines the speed n.

3

• The shape number also indicates if the machine can be built with a single stage or if a multi-stage design is needed.

• Example: Data given for pump with slow-running impeller, Nshape ≈ 0.033 to 0.12.

• An initial calculation of the diameter of the shaft follows from:

2 p tNTorqu NT

ne Z

3

2 sec mod

sec16

p

t

where Z tional ulus

d for circular cross tion

of the shafttorsion stress

4

• thus,

• The permissible torsion stress may be assumed as:

• For multi-stage machines a large diameter (lower value of the permissible torsion stress) is required as the critical speed and the permissible bending of the shaft has also be considered.

Drive Power

Where the overall efficiency η has to be estimated.

3 3 316 1.72 0.93

t t t

T T Ndn

52(80 200)10 .42.11 .50.11t perm

Nto for steel St to Stm

1N VY

5

Diameter of hub dn ≈ 1.1 to 1.3(to 1.5) d

Suction Diameter DS: given by

The velocity Com is estimated using the inlet number ε,

With assumption of ε the angle βoa is pre-determined

2 2' :4 S n SV D d C where

' 1.02 1.1V V V to V

1

1 ,S oma oma omC C normallyC Cf

2omC Y

Considering the leakage loss

6

Diameter D1:

Where the factor f2 = 1.1 to 1.0 according to the way how far the suction edge of the vane is drawn into the eye of the impeller. :

Width b1: is given by the equation:

Velocity triangle at point 1: The velocity triangle is determined if three values of its three sides and three angles are known.

1 2 SD f D

1 1

' om om

om

V C area toC

b DC

If the vane edge at 1 is parallel to axis

7

• Triangle with α0 = 900,(no pre-rotation): This may be the common design of a single stage machine.

• Where Com was already determined by

1 1U D n

11 1 1 1

1 1

sin sinmSC C Com

S

Peripheral velocity

Angle α1=900

Velocity

2omC Y

1

1 1

1.1 1.2S toS

This contraction factor has to be checked after the design is finalized and if necessary Clm and β1have to corrected.

8

Vane angle β1: this angle can be determined either from the velocity triangle or from the equation:

The angle β1 should be in the optimum range, in the case of αo=900:

If the obtained angle β1 differs from the range of optimum values, the calculation has to be repeated assuming an altered value of ε.

111

11 cos

tan

CUC m

frictionlowvelocitysonicto

Cavitationto

,4025

25150

1

01

9

Velocity triangle at point 2: is given by the fundamental equation which reads for U2.

Where in the case of the pump:

The hydraulic efficiency ηh is to be assumed in accordance with the estimated overall efficiency

The slip-power factor p, is to be calculated under pre-assumption β2 of and the number of vanes z.

oublademm CUYCCU 1

2

2

2

2

22 tan2tan2

PYPYYh

bladeblade 11

machineeldesignedWell

smallpoorly

h toarg

8.06.0

10

• The vane angle β2 is either chosen in accordance with the statements given in previous chapters, i.e. in the case of the pump it is β2=15 to 400; or

• The diameter ratio D2/D1 is to be chosen in accordance with the shape of the rotor as determined by the shape number, where D2/D1 ≈2 is commonly used for slow-running rotors.

• The velocity C2m may be chosen as C2m ≈(0.8 to 1.0) C0m which effects a light or no deceleration of the meridian velocity while passing the impeller.

With the knowledge of the values of U2, β2 and C2m the velocity triangle at point 2 can be drawn:

11

• The so obtained vane angle β2 has to be in the range as mentioned in previous chapters. If the calculation results in another value the calculation has to be repeated with new assumptions.

Width b2 is given by

Where contraction factor,

• Finally, the course of the vane has to be determined in such a way that β1 and β2 are obtained. A circular vane course is often used:

2

22222'

ssCDbV m

122

2 ss

12

Construction of a circular vane course in the case of a radial flow impeller:

a. Draw

b. Draw at angle

c. Extend to B

d. Bisect obtaining B

e. Erect the perpendicular at M

f. Draw line at angle β2 at A, obtain C which is the center of the circular vane with radius ρ

Analytical determination of ρ

OA

OD 21

AD

AB

1122

21

22

coscos21

rrrrCBCA

13

Construction of a circular vane course in the case of an axial flow impeller:

L or e has to be chosen: U1=U2

21

2112

21

coscos2

sin2

m

and

eL

14

The Guide Mechanism of Pumps and Blowers

15

• Pumps and blowers need a guide mechanism at the pressure end of the impeller to collect the flow medium discharged from the impeller and to convert part of the velocity energy C2

3/2 into pressure energy.

where

• There are three main designs of guide mechanisms:

22

23

23

23 mu CCC

323

00

23 ,90

bDVCif

UYC mblade

u

16

• Neglecting the friction, the flow in the guide ring is given by:– The vortex rcu = const. and,– The equation of continuity V =bπDcm or bDcm =

const.

Thus

• Because D4>D3 therefore C4u<C3u and C4m<C3m; i.e. both components Cu and Cm are reduced while the medium passes from point 3 to 4.

• The kinetic energy converted into pressure energy in the guide ring is

mmuu CDbCDbandCDCD4443234

43

2

22

22

24

23 CC

17

• The course of the absolute flow line passing through the guide ring is given by the angle α where α can be calculated for any diameter by the equation:

• The flow lines are logarithmic spirals in a guide ring with parallels side walls (b = const.); equation of the log. Spiral.

33

3

33 tantan bb

CC

bb

CC

u

m

u

m If frictionless flow is assumed

tanln180 20 rroraer m Where tan α = const. for all r

233

4tantan rr

bbb

CC

u

m If flow with friction is assumed

Friction coefficient λ≈0.04

18

+ applies to a flow outward from the centre (as is normally case for pumps and blowers)

- applies to a flow towards the centre of the guide ring (as normally the case for turbines but also for return channels of pumps an blowers)

• The flow lines in the guider ring are spiral lines. The guider ring without vanes may be used if α3≥150 as for smaller α3 the spiral lines are too long from 3 to 4 and, consequently the frictional loss of the flow in the guide ring would be too high.

• Advantage of using a guide ring without vanes: easy and cheap manufacturing, no easy blockade due to impurities in the flow.

19

Spiral Casing• The spiral casing does not only convert kinetic energy into

pressure energy but collects also the flow which is discharge at the circumference over 360 degrees and guides it into the pressure pipe.

• Spiral casing are used for single-stage machines and for the last stage of multi-stage of multi-stage machines.

20

• Alternative diffuser systems used in pumps and compressors: (a) simple volute casing with (i) tangential discharge (ii) radial discharge: (b) diffuser system with (i) vaned diffuser (ii) vaneless diffuser.

21

22Sauchelle-Huebra, SPAIN

23

• The flow condition in the spiral casing is given by the vortex flow with which the flow enters the spiral casing.

• The outer wall of the spiral casing can be considered as one of the vortex streamlines where this streamline starts at radius r’ at the point X which is called the theoretical tongue.

• Actual tongue of the casing is often well rounded in order to reduce shocks due to flow conditions below or above normal discharges.

• The vortex flow condition for a frictionless flow is given by

KconstrCU . Where K may be determined from the known vortexAt radius r2:

24

• The amount of flow passing through a cross-section of the spiral casing at angle φ from the theoretical tongue is:

• Thus, the function ϕo=f(R) which refers to the streamline coinciding with the outer spiral casing wall is given by:

• The function b = f(r) is chosen and the corresponding angle ϕo

for a radius R under consideration is determined.

vanesguideno

ifcrYK oou

blade ,9032

rfbwithVbdrrKdfcV o

oR

r

R

r u 360''

R

r

oo dr

rb

VK

'

360

25

• In the case of pumps with cast casings, spiral casings with circular cross-section are in use.

• The calculation of such spiral casings may follow the method as mentioned above in spite of the fact that the circular cross-section of the spiral disturbs the assumed vortex condition.

,,2

222

thusarb

2222

'22

aa

rdrardr

rb a

a

R

r

If a = ri + ρ iii

oo rrr

VK 2720 2

VKCwithCr

C

oo

i

o 7202

For fo = 360o KVr

KV

i

2max

26

Guide Vanes at the Suction End of Pumps and Blowers

• Pre-rotation in or opposite to the direction of impeller rotation can be obtained by means of guide vanes at the suction end of the impeller.

• As cou and u are in the same direction, the product Couu1 is positive :

• Yblade = u2c3u-u1cou is smaller than in the case αo=90o with u1cou = 0

αo<90o: Pre-rotation in direction of the impeller rotation.

11 1 UCour

27

• A slight pre-rotation in direction of U may have some advantage as far as avoiding cavitation or sonic velocity is concerned.

• As Cou and u are in opposite direction, the product U1Cou is negative thus

• Yblade = u2c3u-u1cou is larger than in the case αo=90o.

• The application of this pre-rotation is limited by cavitation or sonic velocity.

αo>90o: Pre-rotation in direction opposite to the impeller rotation.

11 1 UCour

28

Axial Thrust by Radial-flow Turbomachines

29

• Axial thrust is the resulting force which acts on the rotor in axial direction.

• If there is such a resulting force this force has to be transferred to the casing by means off an axial thrust bearing.

• The axial thrust results from three sources, namely

1. Pressure acting on the outside surfaces of the shrouds (force A1)

2. Impulse force due to the deflection of the flow from axial to radial direction or vise versa (force A2) and

3. Weight of the rotor if shaft is inclined or vertical (force A3).

Fig. Page 105

30

31

1. Pressure acting on the outside surface of the shrouds (force A1):

• The component of the static pressure acting in axial direction on the inside surfaces of the shrouds from rs2 to r2 may be considered to be zero.

• A resulting force A1 remains from the acting of the static pressure on the outside surfaces of the shrouds.

• The static pressure acting on the outside surface of the shrouds is Δp= p–p0 = f(r) ; where at radius r2 the static pressure P3 exists thus at radius r2: Δp= p3 –p0.

32

• In order to determine the function Δp =f(r) one may assume that the medium in the space between the shrouds and the casing rotates on an average with half of the rotation ω of the rotor ( particles on the shrouds rotate with ω of the rotor, particles on the casing do not rotate).

• The rotation of the medium in the space between shrouds and casing generates a pressure gradient resulting in higher pressure at larger radii.

• Te pressure at r2 has to be in equilibrium with p3 by the action of the rotor.

Pressure Δpat any radius r:

22

2

222rPVPP axisaxis

ruV

ismediumofvelocityaverageif

21

21

33

= f(r) which is a parabolic function of r• Δp may be expressed also by using the specific static rotor

work

Pressure Δpat r2: 03

22

2

222PPrPP axisr

222

2

032

22

2

2

03 222222rrPPrrPPP

thenPPYp ,103

22

2

2

22 rrYP p

Pressure Δpat any radius r:

34

• The resulting force A1 follows from the integration of the pressure difference Δp:

Area given by Force

0 < r < rs1 0 because (Δp)front=(Δp)back

rs1 < r < rs2

rs2 < r < r2= 0 because (Δp)front-shroud=(Δp)back-shroud

Sum0 < r < r2

28

222

212

2

221

22

2

1ss

pss

rs

rs

rrrYrrPrdr

28

22

212

2

221

221

sspss

rrrYrrA

35

2.Impulse force due to deflection of flow from axial to radial direction (force A2):

• The force A2 may be determined with the help of the impulse theorem applied to the eye of the rotor in which the flow is deflected ( Control Area C):

RPKdVC

isdirectionxinforceimpulseThe

C x

:

Where K = gravity forceP = static pressure force, here P = 0 as pressure prevailing in control area may be assumed everywhere equal to Po

R=reaction force acting from wall to liquidR= -A2.

2ARVCdVC sC x or VcA s2

36

3. Weight of the rotor (force A3):• As long as the shaft is horizontal the total resulting

force is A = A1 + A2. In the case of machines with inclined or vertical shafts a trust A3 has to be added which results from the weight (minus lifting force due to the surrounding medium) of the rotor.

• The medium may rotate in a different way (from ω/2) according to the distance and surface of shroud and casing and to the roughness of the wet surfaces. Therefore, the axial thrust is often calculated using empirical equations, for instance:

shapeNYVnkwithYDkA 2.12.1 4/3

22

37

Balancing of the Axial Thrust• Balancing of axial thrust means reducing o eliminating the resulting

force acting on the rotor in axial direction.

• Balancing is often necessary, especially in multi-stage machines,

a. Different clearance between shrouds and casing:

The axial thrust Ai = A1’ –A1’’ can be reduced if the clearance between back shroud and casing is made smaller than those between front shroud and casing.

The medium rotates less in a space withlarger clearance, the radial pressure gradient is, thus also less. Hence P’’axis > p’axis

38

b. Radial ribs on the back shroud:

The medium in the space between backshroud and casing rotates approximately with ω instead with ω/2, thus A1’ is reduced.

c. Pressure equalization at the back shroud

Equalization of the pressure acting on both sides of the eye portion of the back shroud by means of an outer seal and relieving holes.

The holes lead to stress concentration, hence the strength of the back shroud is reduced

39

Equalization of the through a conduit around the casing

This method is more effective than the relieving holes but also more expensive.

However, this design has to be applied if the strain in the back shroud is already high.

d. Arrangement with opposed stages:

Serial arrangement with multi-stage machines Parallel arrangement with double-

suction machines.

40

e. Balancing drum:

• The pressure in space A is given b the pressure P3 of the last stage and may be assumed as approximately equal to p3.

• Space B is connected to the suction end of the machine. Hence, the pressure in space B is approximately equal to PS.

• The resulting pressure force acting on the drum is directed towards the right.

• This force balances part of the resulting pressured force acting on the different back shrouds and directed towards the left.

41

f. Automatic balancing disc:

• An axial thrust due to P3 shifts initially the rotor to the left, Hence, the clearance x is reduced. The leakage through the unaffected clearance z increases the pressure in the space A. Thus the pressure difference at both sides of the disc (PatA- PS) is increased.

• The resulting force sides acting on the disc due to this pressure difference shifts the rotor to the right.

• Finally, the clearance x adjust so that a complete balancing is obtained.