Part 1: Fundamental equations Part 2: Turbopumpsatm.ulb.ac.be/CoursATM/Turbo/Turbomachinery_Part1and2_Text_vfin2.… · 3 CHAPTER 1 : BASICS and PRINCIPLES 1. INTRODUCTION An important

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TURBOMACHINERY

Part 1: Fundamental equations Part 2: Turbopumps

Theory

Professor Patrick HENDRICK

MECA-H-402

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PART 1

FUNDAMENTAL EQUATIONS OF TURBOMACHINERY

3

CHAPTER 1 : BASICS and PRINCIPLES

1. INTRODUCTION

An important category of machines is the one where the energy transfer is taking place between a fluid and a rotor in a continuous way (not in an intermittent or alternative way). This rotor is built with one or more wheels/discs mounted on a shaft, with blades inserted or attached on these discs, the whole assembly being in rotation around the shaft of the machine. Between the different discs or before and/or after a single disc, there are also static components equipped with non-rotating blades called vanes. The fluid flows through the channels / passages that are between the blades (rotating channels) and between the vanes of the stators. An energy transfer happens between the fluid and the rotor. These machines / systems are called turbomachines or turbomachinery. 2. CLASSIFICATION OF TURBOMACHINES

The fluid (water, steam, air, gas, ...) flowing from upstream to downstream goes through the machine itself often entering via a distributor and leaving the machine via a diffuser. If the role of the machine consists in transferring energy from the fluid to the rotor (and in fact delivering torque, so mechanical energy, on its shaft), one speaks about a turboproducers or turbomotors or more generally of turbines. The energy level of the fluid downstream of the machine is lower than upstream. A few examples of turboproducers are:

- Hydraulic turbines - Steam turbines (electric power generation) - Gas turbines (aircraft propulsion systems and electric power generation) - Wind turbines (electricity production).

If the turbomachinery is used to transform mechanical energy, applied to a rotor shaft, into energy transferred to a fluid in such a way that the energy of the fluid downstream is higher than upstream, the machine is called a turboabsorber or turbogenerator. This is for example the case in:

- Turbopumps or centrifugal pumps (water distribution or pipelines) - Ventilators or fans (air conditioning systems) - Turbocompressors (pressurized air; turboreactors) - Propellers (in ships or in aircraft).

3. GENERAL ORGANISATION OF TURBOMACHINES

Figure 1.1 gives a schematic of an axial turbomachinery cutaway. On Figure 1.2, it is a cutaway in a centrifugal turbopump that is presented. The main components of a turbomachinery are:

- The distributor (not always present), sometimes equipped with non-rotating blades (vanes or nozzles). It receives the fluid from upstream and orientate it towards the rotating blades with an adequate direction and the right velocity/Mach number;

- The wheel (with the disc and the blades), fixed on the shaft or at least moving with the shaft.

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Through it, the energy transformation from the fluid into mechanical power or vice-versa is taking place;

- The diffuser (not always present), installed between the wheel and the exhaust plenum. It is the place

where the transformation of kinetic energy into pressure energy (pressure head) or into mechanical power is done and it orientates the fluid correctly towards the exhaust plenum / user;

- The carter, making the link between the distributor and the diffuser, used as the external envelope of the rotor blades but also as the internal support of the machine. The carter (also sometimes called stator) is equipped with journal bearing or roller bearings, used to support and maintain the shaft and the rotor, but also with air/air and air/oil seals (labyrinth or brush), used to avoid air or oil leaks from or into the system. The bearings are outside of the carter itself.

The rotor can be made of several wheels / discs mounted on a same shaft or even two or three shafts. Then fixed components (that are in fact the distributors of the downstream wheel or the diffusers of the upstream wheel), mounted in and supported by the carter, change the direction of the flow in between two consecutive wheels (Figures 1.3 and 1.4). At the root of these stator blades (vanes), there are also seals present, in front of the shaft.

4. NOTATIONS

The mechanical and thermodynamic variables that characterize the condition of the fluid are in this course receiving the following indexes (Fig. 1.1.):

i or e : upstream plenum; 0 : inlet section of the distributor; 1 : exhaust section of the distributor so also inlet section of the rotor; 2 : exhaust section of the rotor and so inlet section of the diffuser; 3 : exhaust section of the diffuser; o or s : exhaust section of the machine or downstream plenum.

The velocities of a fluid element / particle are written as follows:

v = absolute velocity u = frame tangential (entrainment) velocity w = relative velocity

These velocities are to be given in m/s. The Mach number related to each of these velocities is normally the important parameter not so much the velocity itself. 5. VELOCITY TRIANGLES

Following the definitions used in rational mechanics:

wuv +=

5

The angles between the velocity vectors, measured from the frame tangential speed vector, are defined as follows (Fig. 1.5):

- ! : angle measured, in the trigonometric direction, between the vectors u and v

- ! : angle measured, in the trigonometric direction, between the vectors u and w .

The following relations exist between these velocities and angles:

!" coscos wuv +=

!cos2222 uvvuw ++=

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CHAPTER 2 : FUNDAMENTAL EQUATIONS OF THE FLOWS

6. ASSUMPTIONS

One will only examine steady state working conditions (ratings). That means that: - The rotation speed of all rotating components is constant (ω = Cte) around a shaft that is in a

fixed position;

- The fluid flow is permanent, i.e. in a given spatial point of machine all the characteristics of the flow remain constant, they are only function of the relative position of the point and independent of time.

Moreover we admit that the flow takes places in parallel slices (neglecting the presence of the boundary layer).

7. EQUATIONS OF THE FLOW IN A FIXED FRAME

One applies the fundamental equations of thermodynamics to the fluid that is, at a given time, between the sections Ao and Al of a fixed frame, as represented on Figure 1.6. One obtains, per kg of fluid, the following relations: 7. 1. Energy equation

( )01

20

21

01 2zzgvvhhq !+

!+!=

q : the quantity of heat exchanged, per unit of masse of fluid, with the outside world [J/kg] h : the enthalpy [J/kg] g : the acceleration of gravity [ 2/ sm ] z : the altitude compared to a reference level [m]

7.2. Equation of kinetic energy

( ) ! ""="+" 1

0

'01

20

21

2

p

pfwvdpzzgvv

p = pressure [ ]2/mN ! = specific volume (=1/ρ) [ ]kgm /3 w’f work resulting from all friction effects present between sections 0A and 1A [ ]kgJ / .

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7.3. Global thermodynamic equation

!""=+1

0

01'

p

pf vdphhwq

7.4. Mass flow rate equation

1

11111000 v

vAvAvAvAm ==== !!!&

m& = mass flow rate (kg/s) A = surface of a section normal to the velocity vector (m2) ! = volumetric mass (kg/m3)

v = specific volume (m3/kg). 8. EQUATIONS OF THE FLOW IN A MOVING (rotating) FRAME

This type of flow concerns the fluid flowing in the channels between the blades of a wheel.

If one directly applies the energy equation, or the kinetic energy equation to a flow in a moving frame, one faces a large hurdle: the calculation of the work .seW ! . To solve this difficulty, one applies the equations in a relative frame that is fixed compared to the wheel that is supposed to be in a uniform rotation (! = cte) around a fixed shaft.

8.1. Equation of kinetic energy (in a relative space) From the definition in rational mechanics:

( )cpsirr LFdWmvd !+!++="#

$%&

' (( 2

21

m : mass of a material point of the system

rv : relative velocity of this mass m

F : external force acting on a point (contact force, mass force or external linking force)

iL : internal linking force

e! : apparent driving force

cp! : apparent force of Coriolis

Wr : work developed by the forces in the relative frame

! : sum for all points of the system

The work due to the apparent forces of Coriolis in the relative frame is zero (prove it).

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The system to be considered is the fluid at a given time t between the sections a and b of a channel. A t + dt, this fluid will be between the sections a' and b' (Figure 1.7).

We need to calculate the term: ( )! " srdW .

The apparent driving force is reduced to the centrifugal force ee am!=" .

ea is the centripetal acceleration of the point, rae2!"= (r is the distance from the rotation

axis). The elementary work in the relative frame is the scalar product of e! with the elementary relative displacement rsd :

( ) ( ) drsddW ereer !=!=!

Then:

( ) ( ) !"

#$%

&===' 2222

21

21 mudrmdrdrmdW er ((

because ru !=

( )abba

er mumudW !"

#$%

&'!"

#$%

&=( ))) 2

''

2

21

21

As the flow is permanent in the relative frame:

abba

mumu !"

#$%

&=!"

#$%

& '' 2

""

2

21

21

So that in

""

2

''

2

21

21

baba

mmu !"

#$%

&'!"

#$%

& (( )

the kinetic energy of the common part a'b" disappears and thus:

( )! !! "#

$%&

'("#

$%&

'=) '"21

21 2

'"

2 aamumudWbb

s

As the flow is also permanent in the relative frame, one finds also:

'"

2

'"

22

21

21

21

aabbr mwmwvmd !

"

#$%

&'!"

#$%

&=!"

#$%

& (( (

9

The work due to the external forces F is reduced to: - work due to the pressure forces that are present on the inlet and outlet sections of the channel; - work due to the mass forces ( zmg!" , the rotation shaft being fixed).

The work due to the internal liaison forces L comprises:

- on one side, the work due to the elastic forces ! pdv

- on the other side, the work due to the internal friction forces that we note: "fw!

The equation of the kinetic energy, per kg of fluid, can be written as follows:

( )22

21

22"

122211

21

22

2

1

uuwpdvzzgvpvpwwf

v

v

!+!+!!!=

!"

or

!

w22 " w1

2

2+ g z2 " z1( ) = " vdp

p1

p2

# " w f" +

u22 " u1

2

2

8.2. Global thermodynamic equation This equation does not contain any term relative to the work done by the external forces or related to the kinetic energy. The fact that the channel is fixed or mobile is without any influence.

!""=+2

1

12"

p

pf vdphhwq

8.3. Energy equation When combining the two previous equations, one gets:

!

q +u22 " u1

2

2= h2 " h1 +

w22 " w1

2

2+ g z2 " z1( )

8.4. Mass flow rate equation A being a surface of a section normal to the relative velocityw , the mass flow rate is written:

222111 !!! wAwAAwm ===&

9. REMARK

Il is interesting to compare the equations of the flow in a fixed frame and in a rotating frame.

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CHAPTER 3: FUNDAMENTAL EQUATIONS OF TURBOMACHINERY

10. INTRODUCTION

The objective is to establish the general equations that will be after used in the study of each particular type of machine as the turbopumps, the turbines, the turbocompressors. We shall look successively at the cases of the turbo-absorbers and of the turboproducers.

A. COMPRESSORS - TURBO-ABSORBERS - TURBOGENERATORS

11. DESCRIPTION

The turbo-absorbers are realized in very different ways, i.e. the flow of fluid in these machines can be very different. This is also true for all losses, internal as well as external. The formulas that we shall establish will be valid for a very general architecture of the machine, something that can be applied to all types of turbo-absorbers. A turbo-absorber is always organized as shown on Figure 1.8. It is important de have a clear understanding of this figure in order to have a good understanding of the working principle of such a machine, from where we shall deduce the related equations. The machine sucks the flow in from the upstream plenum where the energy level is low. This fluid, after its passage through the machine, is expelled, with a higher energy level, into the downstream plenum. The driving power P, applied on the shaft, is used to raise the energy level of the fluid, considering also that the evolution of the fluid through the machine is taking place with some friction losses and other mechanical losses.

12. INTERNAL LOSSES The mass of fluid flows through the inlet section i, via the distributor, towards the wheel. There exists there a first energy loss called '

Fw due to the friction forces inside the static distributor. The fluid flows then through the interblade channels of the rotor. Its energy increases but it is together with a loss called "

fw due to the friction forces inside these rotating channels.

Another loss is also present in the rotor. It is due to the clearance existing between the tip of the rotor and the carter and to the fact that the energy at the exit of the rotor blades is higher than the energy at the inlet of these blades. This creates an internal leak flow called im& , between the rotor and the carter, from the exit of the wheel towards the entrance of the wheel. When designating the mass flow rate through the wheel with Rm& , one can write:

iR mmm &&& +=

The largest part of the flow goes, via the diffuser into the exhaust plenum through the exhaust

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section 0 but a small leak flow called em& is also appearing from the system towards the outside world, through the seals. This em& constitutes an external loss (cf. paragraph n°13).

The volume due to the distance existing between the rotor and the internal carter is filled in by a part of the fluid. This fluid is then not participating anymore to the cycle. It is a non-active fluid that is stagnating and it is not participating to the energy exchanges in the machines. This fluid is nevertheless introducing a extra loss as a part of the power applied on the machine shaft will be dissipated by friction of the back of the wheel on this non-active fluid. This lost power is called frP and will be estimated later on.

13. EXTERNAL LOSSES The pressure difference between the upstream plenum and section 1 is generally small, what allows to consider that the leak flow through the forward seal is very limited. This of course not the case for the backward seal where a leak flow em& must be considered as the pressure in 2 is largely higher than the atmospheric pressure. This leak flow can be largely reduced when using highly efficient seals like brush seals on new turbomachinery. Relatively this leak is quite limited compared to the main mass flow rate. In our study, this loss with be neglected. The rotor is supported by bearings where friction losses are also present. A small part of the power applied on the machine shaft is dissipated in these bearings and the related pumps and other components of the lubrication circuit. There are also other accessories of the machine that are absorbing mechanical power from the rotor shaft (as the fuel pump for example). We note faP the power absorbed by all these losses. 14. ENERGY EQUATION One applies this equation to the machine rotating at a given rating ( Cte=! ), with the rotor limited to the portion of the machine between the forward and aft seal, and with the fluid contained at a given moment t between the inlet section iA and the outlet section 0A . Figure 1.9 shows the system. Per kg of fluid in the system, the energy equation can be written:

!

q +We" s = u0 # ui +v02 # vi

2

2+ g(z0 # zi)

Q = Quantity of heat exchanged with the outside world

seW ! = Work of the external forces acting on the system and including: - the work due to the fluid pressure 00vpvp ii !

- the work applied on the rotor shaft mPP fa

&

!=mPi&

(Pi is called the internal power)

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iuu !0 = Variation of the intern energy

2

220 ivv ! = Variation of the kinetic energy

( )izzg !0 = Variation of the position / height (all variables are expressed in J/kg).

When introducing the enthalpy h :

!

q +Pi

˙ m = h0 " hi +

v02 " vi

2

2+ g z0 " zi( )

Quite often the heat exchanges between the external world and the system are very limited (adiabiatic evolution) and the term ( )izzg !0 is extremely small compared with the other terms, thus:

( )titi

iifa hhmvvhhmPPP !=""#

$%%&

' !+!==! 0

220

0 2&&

( th = total enthalpy) The value of the total enthalpy depends on the thermal energy and the mechanical (kinetic) energy

of the fluid. The above formula shows thus that the work mPi&

applied on the shaft, between the

seal positions, is used to increase the total energy of the fluid (energy conservation). 15. EQUATION OF KINETIC ENERGY Applying this equation to the system defined in paragraph 14 gives, per kg of fluid in the machine:

( ) f

v

vi

iifa PzzgvdpvvmPPP

i

+!!"

#

$$%

&'++

'==' (

0

0

220

2&

fP is the total power lost due to the internal friction losses with the active as well as the non-

active fluid. 16. ENERGY TRANSFERED TO THE FLUID We define:

( )! "++"

=0

0

220

2

p

pi

i

i

zzgvdpvve

13

that is the energy transferred to the fluid through its passage in the turbo-absorber.

fi PemP += & Explain the physical meaning of this formula ? 17. A FIRST DISTRIBUTION OF THE POWER – EFFICIENCIES The previous equations have been established considering the turbomachinery as a whole, with inside, an energy transformation that is happening together with a degradation of the energy due to all friction effects. Figure 1.10 shows the distribution of the power as it appears in the three previous equations (the external losses are neglected). This distribution allows the definition of the global efficiency g! :

Pem

g&

=!

This expression can also be written as:

PP

Pem i

ig

&=!

iiPem

!=&

= internal efficiency

mi

PP

!= = external mechanical efficiency.

18. ENERGY TRANSFER TO THE ROTOR (Equation of Euler-Rateau) The fluid absorbs energy only during its passage through the interblade channels of the wheel, i.e. between sections 1 and 2. The study of this energy transfer is important. One can also see what is the shape to give to the blades of the wheel to transfer the maximum of energy with a minimum of losses. The wheel being in rotation around the shaft, one uses the equation of the kinetic moment, or moment of the quantity of movement, related to this shaft:

( ) eaxeaxe FMvmMdtd

!! =

14

and apply it to the system made by the wheel and its shaft between the front and aft seal (Fig. 1.11), and the fluid that it contains, between the inlet section 1 and the outlet section 2, at a given moment. From time t to time t+dt, the wheel turns and the fluid moves as well. The variation of the kinetic moment of the wheel is zero as this one rotates at a constant speed. For a fluid element with a mass m, at a distance r from the shaft, the quantity of movement is equal to m v . The velocity v is composed by three components:

- one parallel to the shaft, - one in the radial direction, - one tangential (directed along the tangential direction with the circumferential trajectory of

m). Only the tangential component delivers a moment:

( ) !cosvrmvmMaxe =

As the flow is permanent, the variation of the quantity of movement between time t and time t + dt (Fig. 1. 12), is reduced to:

( ) ( )!! ""'"'

coscosaabb

vrmvrm ##

If the mass flow rate is constant and if dt is very small, this variation is written as:

( ) ( )uu vrvrmvrvrm 1122'

111222' coscos !=! ""

When accepting that uu vrvr 1122 .. ! is the same for all channels of the wheel, one obtains by summing all the channels:

( ) ( )uuRuu vrvrdtmmvrvr 11221122 ' !=! " & The first term of the equation is equal to:

( )dtvrvrm uuR .1122 !&

To calculate the second term, the external forces acting on the system must be detailed:

- the weight of the wheel: for symmetrical reasons, the moment is zero - the weight of the fluid: idem - the pressure on the inlet section, the exit section and the side walls: idem

- the driving torque

!

Mi =Pi"

applied on the rotor shaft

- the resistive torque frM! due to the friction of the wheel on the non-active fluid. The equation is then written:

!

˙ m R r2v2u " r1v1u( ) = Mi " M fr

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If one multiplies the two terms of the equation by the rotation speed ! : ( )uuRRfrfafri vuvumPPPPPP 1122 !==!!=! & , it becomes the equation of Euler-Rateau written

under a format that we shall often use. This equation expresses that the power PR available on the rotor shaft, in order to me transferred to the mass flow Rm& , must show an increase of the quantity !cos... vrvr u = . But uvr. is in direct link with the velocity triangles, so that this formula tells us how the velocity triangles at the inlet and at the outlet of the wheel should be. 19. OTHER FORMULATION OF Euler-Rateau

As uuvvuw 2222 !+= , one can also write:

!

PR = ˙ m R (v22 " v1

2

2+

u22 " u1

2

2"

w22 " w1

2

2)

20. ENERGY GENERATED BY THE ROTOR The equation of the kinetic energy in a rotating frame applied to the wheel / rotor gives:

( )22

21

22"

12

21

22

2

1

uuwvdpzzgww p

pf

!+!!=!+

!"

Introduced in the previous equation, this one becomes:

"12

21

22

2

1

)(2 fR

p

pRR wmzzgvdpvvmP && +

!!"

#

$$%

&'++

'= (

The quantity between brackets represents the energy transferred to the fluid through the channels on the rotor, between the inlet section 1 and the exit section 2:

! "++"

=2

1

)(2 12

21

22

p

pR zzgvdpvve

"fRwm& is the power "

fP absorbed by the friction effects in all the channels.

"fRRR PemP += &

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21. OTHER WAY TO OBTAIN THE EQUATION OF Euler-Rateau One can also establish the formula of Euler-Rateau when applying the fundamental equations of thermodynamics. For that, one applies the equation of the kinetic energy to the system made with the rotor and the fluid that it contains at a given time t, between the inlet section 1 and the outlet section 2. The equation is for a flow in a rotating frame. One gets finally the equation given in paragraph n° 18. Ii is also normal that the equation of the kinetic moment and the one of the kinetic energy, applied on the same system, deliver the same result, as both of them are the translation of the same expression principle, i.e. the one of Newton. 22. DETAILED DISTRIBUTION OF THE POWER In paragraph n° 17, we have defined a first balance of the power where all the internal losses were included under a common name Pf . Thanks to the formula of Euler-Rateau, we are able to detail these losses in order to understand them better. We have established that:

fai PPP != (Fig. 1.10) [ ]W

friR PPP != (N° 18) [ ]W "fRRR PPem !=& (N°20) [ ]W

mmm Ri &&& != (N°12) !"

#$%

&skg

With the internal leak flow im& corresponds a power loss equal to Ri em& . The mass flow rate m& absorbs through the wheel an energy Re . But from the power de la Rem& , a part is lost due to the friction losses inside the distributor and the diffuser. This lost power in a fixed frame is noted:

''ff wmP &=

The detailed distribution of the power in the whole turbo-absorber is therefore given on Figure 1.13. The detailed understanding of the losses (origin, importance, location) helps the engineer and the manufacturer to adopt the requested design measures to reduce these losses to a minimum with the objective to increase the global efficiency.

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B. TURBINES – TURBOMOTORS - TURBOPRODUCERS

23. INTRODUCTION The study of the turboproducers is similar to the one of the turbo-absorbers. The way to analyze the machine is identical. Only the working principle of the machine, the leak flows and the losses are organized in another way. When these one are understood from a physical point of view, the write-up of the equations does not show any special extra difficulty compared with what has been seen in the turbo-absorbers. Therefore, after the explanation of the organization and of the working principle of this machine type, we shall only give the related fundamental formulas and the power balance. 24. DESCRIPTION

The fluid that goes through a turboproducer flows from upstream where the energy level is high to the downstream plenum at a low energy level (Fig.1.14). The fluid gives a part of its energy to the rotor and leaves the machine with a energy level lower than when it entered into the system. As the energy level, often due to the pressure head, is higher in section 1 than in section 2, a part of the flow m& , leaving the distributor, goes through the clearance between the rotor blade tips and the carter, this is the internal leak flow im& .

External leak flows em& can also be present through the seals. They are higher through the front seal than on the aft seal. But both of them are low and can be neglected. The flow through the rotor blades is iR mmm &&& += . The mass flows Rm& and im& join themselves downstream of the rotor and are again there at the same energy level. As they had the same energy level in 1, the energy drop of the two flows has been identical. The difference is that the leak flow could not transfer any power to the rotor. The rotor is supported by bearings and can also drive some extra accessories. One notes Pfa the power lost by this way.

The available mechanical power for the load (on the shaft) is written Pu (useful power). This one can be used to drive a compressor, an electric generator or any type of pump. The flow is characterized by pressure losses in fixed pipes (distributor, diffuser) that we note '

fw and 'fP

for respectively the lost energy and the lost power and in rotating channels that we note "fw and "

fP for

respectively the lost energy and the lost power. Finally, the rotor turns in some non-active fluid and the power lost due to this is Pfr.

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25. ENERGY EQUATION

This equation is applied on the system made of the machine and the fluid that it contains at a given moment t, between the inlet sections Ai and the outlet section Ao. This equation becomes:

( ) ( )iii zzgvvhhmPq !+

!+!=+ 0

21

20

0 2&

If the evolution is adiabatic and if izz !0 is negligible:

( ) fauttii PPhhmP +=!= 0&

Explain the physical meaning of this equation ? 26. EQUATION OF KINETIC ENERGY This equation leads to:

( ) ifi

p

p

i PPzzgvdpvvm =!""#

$

%%&

'!++

!( 0

20

2 1

02

&

fP is the power lost internally due to friction effects.

27. ENERGY TRANSFERED FROM THE FLUID

It is defined as: ( )020

2 1

02

zzgvdpvve i

p

p

i !++!

= "

From that it follows that: fiï Pem +=&

28. FIRST DISTRIBUTION OF THE POWER - EFFICIENCIES

A first distribution of the power is given on Figure 1.16.

The global efficiency is: emPu

g &=!

Here em& is in the denominator.

imi

i

ug x

emP

PP

!µ! ==&.

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29. EQUATION OF Euler-Rateau

One can establish the following:

( ) RfriR PPPvuvum =+=! 222111 coscos ""&

PR is the power transferred from the fluid Rm& to the rotor. 30. ENERGY ABSORBED BY THE ROTOR

This energy is due to the flow of the fluid in the rotating channels of the rotor. The formula of Euler-Rateau combined with the one of the flow in a rotating frame gives:

( ) RfR

p

pR Pwmzzgvdpvvm =!

""#

$

%%&

'!++

+( "

21

22

21

1

22

&&

When designating with Re the energy transferred from the fluid to the rotor, on obtains:

"fRR PemP != &

with ""ffR Pwm =& which is the power lost in the channel of the rotor.

31. DETAILED DISTRIBUTION OF THE POWER

This power balance is given on Figure 1.17 where the distribution of the losses Pf due to friction and internal losses are also shown. This balance is based on the description of the losses (paragraph n°24) and on Figure 1.15.

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CHAPTER 4: AXIAL THRUST and DISC FRICTION 32. DEFINITION The wheel of a turbomachinery is clamped on the shaft. The pressure acting on both sides of the wheel is different. Also the entrance and exit speeds of the fluid in the channels between the blades are different. Therefore the wheel subjected to an axial force tends to make the shaft moving in the same direction as this force. A thrust bearing needs to be selected and installed at the end of the shaft in order to support the axial load. It prevents the shaft to slip away, what could lead to contact and friction of the rotating blades against the stator vanes. 33. EQUATION OF THE AXIAL THRUST We apply the equation of the quantity of movement to the wheel, without the shaft, and to the fluid that it contains at a given time t, between the inlet section 1 and the outlet section 2:

( ) ( )!! = dtFvmd e

As we want to obtain the axial thrust, we make a projection on the axis of the shaft (Fig. 1.18) orientated along Ox. The system as defined before and previously isolated is subjected to the external forces:

Due to the pressure:

11 RAp et 22 RAp

-AR1 and AR2 are the complete lateral surfaces (disc + blade crown) of the rotor, - p1 and p2 are the mean values of the pressure applied on these surfaces. Due to the mass of the rotor and of the fluid:

If this weight is G , its projection on Ox is !cosG with ! the angle between the machine axis and vertical direction. Due to the reaction of the shaft on the rotor:

If aF is this reaction, with a projection Fax on Ox (- axF is the axial thrust that we want to find).

The driving torque and the resistant torque: Their projections on Ox are zero. The projection on Ox of the change of the quantity of movement becomes (to be demonstrated):

21

( )axaxR vvdtm 12 !&

(vax indicates the component, along Ox, of the speed v). The equation of the quantity of movement, when projecting on Ox, becomes:

( ) ( )[ ]dtGFApApvvdtm axRRaxaxR !cos221112 ++"="&

That means that the force applied by the rotor on the shaft and that is equal to –Fax (action and reaction) is:

( ) ( ) !cos112212 GApApvvmF RRaxaxRax ""+"= &

This formula of the axial thrust is to be used to select the thrust bearing that the engineer needs to mount at the end of the rotor shaft. 34. FRICTION OF THE DISC

When establishing the theorem of Euler-Rateau, we have mentioned the friction losses on the disc due to the friction of the lateral surfaces of the disc in the non-active fluid. The case of a disc (Fig. 1.19) rotating in a closed atmosphere has been extensively studied experimentally. It is obvious that the friction torque due to the fluid will be higher if the disc is larger (r high), if the rotational speed N is higher (u large) and the volumetric masse of the fluid is important. Considering an annular portion of the disc limited by two circumferences with respectively the radii r et r+dr rotating at the speed N in a fluid with a specific weight ! .

The elementary surface is equal to drr..2! and the elementary friction force created by the fluid on the disc is:

rdruKdFfr !" 22=

The elementary friction torque is equal to:

2222 udrrkdC fr !"=

(the second member is to be multiplied by 2 because the two faces of the disc contribute to friction). The total friction torque is obtained by integrating the elementary torque from the internal radius of the disc r up to the external radius r1:

51

2 rKC fr !"=

(r5 being neglected compared with 51r ).

22

The power absorbed by friction is written: 51

3 rKCP frfr !"! ==

On some machines, this power can become rather high. If the sidewalls of the disc are very smooth, the value of the constant K will depend on the Reynolds number of the flow:

µ!" 2

1rRe =

with µ the dynamic viscosity of the fluid. If these surfaces are rough, the relative roughness will be the parameter fixing the value to be given to the parameter K as soon as the Reynolds number Re will have increased over its critical value, as it is also the case for the friction losses in non-smooth pipes.

23

PART 2

TURBOPUMPS – CENTRIFUGAL PUMPS

24

Chapter 1

GENERALITIES 1.1 INTRODUCTION

The turbopumps (TP) are turbo-absorbers that have to transfer energy to an incompressible fluid, as water, oil, fuel, ...

The equations established in Part 1 are applicable and will be applied to this type type of machine. As the fluid is incompressible, it is to be expected that the theories developed for the study of the turbopumps will be easier than the ones for the turbocompressors.

For this reason, we study first the turbopumps that are also currently used in the industry and in houses. A special application of the TP is certainly in the cryogenic rocket engines with the liquid hydrogen and liquid oxygen TPs.

The objective is not so much to design and build a turbopump but mainly to examine this machine from the point of view of the users.

1.2 DESCRIPTION - TYPE OF TURBOPUMP

Figures 2.1 and 2.2 show two types of pump: centrifugal and axial. Figure 2.3 shows that in between these two extreme types there exist a lot of intermediate types: helico-centrifugal or helico-radial. The necessity to go for one or the other type depends on the requirements from the mass flow rate of the fluid or from the energy to be transferred.

Each turbopump is equipped with a rotor made of one or several wheels, clamped on the same shaft. These wheels can be assembled in parallel or in series. The parallel mounting has the objective to increase the mass flow rate of the fluid, while the configuration in series is intended to transfer a lot of energy to the fluid, with a mass flow that can be limited (Figure 1.3). A solution that is often used consists in equipping the machine with two symmetric wheels that are put against each other with each of them using their own inlet section (Figure 2.4) and working in parallel. This solution presents the advantage of canceling, in principle, any axial thrust and to reduce the friction of the disc in the non-active fluid stagnating in the machine.

1.3 INSTALLATION OF A TURBOPUMP

Figure 2.5 shows the installation schematics of a pump intended to be used to transfer water from a lower reservoir (level z') up to a higher reservoir (level z"), for example from a river up to water tower.

The pump, driven by an electric motor or any other driving system, sucks the water through an aspiration pipe, equipped with a filter at its entrance and a valve. This valve gets automatically open as soon as the low pressure region at the entrance i of the pump succeeds

25

in aspiring the water. This valve is instantaneously closed as soon as the pump stops running. Its functionality is to maintain the pumping circuit full of liquid, ready to be transferred to the upper reservoir and to avoid that, when stopped, the upper reservoir does not empty itself into the lower one just by syphoning.

An exhaust pipe is connected on the outlet section o of the pump. It is equipped with a control valve or a control tap and it exits into the upper tank.

1.4 ENERGY DEVELOPED BY THE TURBOPUMP – FLOW RATE See Figure 2.6 As seen in the fundamental equations, the energy transferred to the fluid during its passage in the turbo-absorbers is given by:

( )

( )iii

po

pii

i

zzgvvpp

e

zzgvdpvv

e

!+!

+!

=

!++!

= "

0

2200

0

220

2

2

#

The term with the pressure energy !

io pp " is usually the largest one.

The velocities must stay limited in order to minimize the head (pressure) losses. The energy e can be determined experimentally by installing a manometer at the entrance and at the exit of the pump, by measuring the volumetric flow rate and the level difference izz !0 between the inlet and the outlet of the pump.

This e can be written:

!

e =pa " pi#

"pa " po#

+(Ai

2 " A02)

2A02Ai

2 Q2 + g z0 " zi( )

The mass flow rate is given by:

!! iioo AvAvm ==&

In the international measuring system, the mass flow rate m& is expressed in kg/s and the volumetric flow rate Q in m3/s. But technically, centrifugal pumps are often detailed using practical units as m3/h, m3/min, 1/min or 1/s. This needs to be carefully checked in the numerical applications.

26

1.5 USEFUL POWER OR HYDRAULIC POWER The hydraulic power is the power transferred to the fluid from the entrance of the pump and its exit. It is equal to [ ]WeQem ... !=& . 1.6 GLOBAL EFFICIENCY OF A TURBOPUMP mP is the mechanical power needed to drive the pump, em.& is the useful power, the global

efficiency is given by:

mm PeQ

Pem ... !

" ==&

1.7 WORKING POINT OF A TURBOPUMP We apply the equation of the kinetic energy (Bernoulli) to the aspiration piping parts and the exhaust piping part of the pump (Figure 2.5). Between the free level z' and the entrance section i:

!

vi2 " v ' 2

2+ g zi " z

'( ) = "pi " p

'

#" w fa

'

'faw represents the energy loss (head losses) due to the friction in the aspiration pipe.

Between the outlet section o and the free level z":

( ) 0"2

'"2

02"

=+!+!

+!

froo wzzgppvv

"

w’fr is the head loss in the exhaust pipe. If one considers that the aspiration tank and the exhaust tank have quite large dimensions ( )0"' != vv and that their free surface levels are quite similar ( )appp == "' , the two above relations can be recombined and written like this:

e = en

The first term represents the energy e transferred to the fluid by the pump, the second, what is requested by the circuit on which the pump is mounted. It is written en :

( ) ''"

fn wzzge +!= 'fw is the total head losses ( )''

fafr ww += .

27

In steady state regime (rating), we always have that nee = . 1.8 CHARACTERISTIC OF THE HYDRAULIC CIRCUIT The energy ne is function of the level difference '" zz ! and of the mass flow rate because the head losses '''

ffafr www =+ are proportional to the square of the flow velocity and thus to the square of the volumetric mass flow rate.

The value of the pressure loss coefficient k depends on the length and of the internal condition of the aspiration and exhaust circuits as well as on the presence of turns, a valve, the filter and its position of the system, i.e. some accessories.

If one presents in a diagram, the parameter e in function of Q, for a given position of the valve, one gets a branch of a parabolic curve (Figure 2.7). The slope of this curve depends on the value of k, thus on the closing degree of the valve: the closer the valve the higher the slope.

1.9 PERFORMANCE CURVE OF A PUMP

Let’s consider a turbopump with a rotation at constant speed. We have installed a control valve on the exhaust pipe. We can establish for each of the valve position a working regime and fix at this engine rating:

- the energy e developed by the pump; - the mass flow rate Q. When we put in a diagram the values of e in function of the flow rate Q , we obtain (Figure 2.7) the working curve of the pump at a speed N. For each value of N, there is another working curve.

1.10 WORKING REGIMES

The experimental study of a turbopump (centrifugal pump) shows that such a pump is able to work at different ratings but when two of the three parameters e, Q, n are fixed, the working point is also fixed. That can also be written in this way:

( ) 0,, =nQef

1.11 PRACTICAL UNITS

In the SI system, the energy is expressed in J and here in J/kg as it is the energy per kg of fluid that we are considering. This energy is also equal to the constant g multiplied by a delta z so a height in m. This is the reason why, practically, it is not e that is used in turbopumps but e divided by g, that is called H and is expressed in meters. The energy transferred to the fluid by the pump is therefore very often called the “height”

28

transferred by the pump to the liquid or the “pressure head” of the fluid / pump. H is with the following equation:

ioioi zz

gpp

gvvH !+

!+

!=

"2

220

It is exactly the same for the head nH requested by the circuit.

The characteristic curves of a pump and of a circuit are delivered under the shape of H or

nH (in m) in function of the volumetric flow Q, the « Head capacity characteristic ».

The switch between H (in m) and e (in J/kg) is easy: gHe = . Practically, other units, as British units are also sometimes used. An example is shown on Figure 2.8. It must be noted that this pump is rotating at 1750 tr/min (RPM). The efficiency shown is between 0 and 70%. It is obvious that a pump must be used in working conditions corresponding to a maximum efficiency or not too far from it and that the choice made for a pump will depend on the type of the circuit where the pump needs to be installed and used and the demand on this circuit (flow rate and height).

Figure 2.9 shows working curves of a family of centrifugal turbopumps rotating at 2900 RPM. These are curves as they are provided by a manufacturer. The two final digits in the model indicate the external radius (in cm) of the wheel.

29

Chapter 2

THE CENTRIFUGAL PUMPS

2.1 INTRODUCTION

We are now examining the way the turbopumps are designed and build in order to fix the relations that must exist between the performance required from a pump (energy and flow rate) and the geometry and the dimensions to be given to the machine. The application of all existing theories to all types of turbopumps would be much too long. We shall limit ourselves to a 2D theory applied on the most common type: the centrifugal turbopump with a single rotor (and, by extension, the helico-centrifugal). 2.2 ORGANISATION OF A CENTRIFUGAL TURBOPUMP The pump is composed with the following components (Figures 2.1 and 2.10): - an inlet distributor D (or inlet) that leads the fluid from the entrance section i of the

machine towards the entrance section 1 of the rotor. This fixed distributor is usually not equipped with any stator vanes / guide vanes.

- a rotor R (or impeller) made of a wheel clamped on a shaft. The rotor is made by one or

two discs, one which or between which the blades are arranged starting at the section 1 (at an external radius r1) until an exit section 2 (radius r2). In the first case, the wheel is of the type « open rotor », in the second, it is a « closed rotor ».

- a fixed diffuser d made of two discs (flanges) parallel or slightly divergent and between

which stator vanes are sometimes present. The diffuser surrounds the exhaust from the rotor and the section available for the flow between 2 and 3 is increasing. Sometimes the diffuser is not existing.

- a volute or fixed collector c, with the shape of a snail surrounding the diffuser (or directly

the rotor). It collects the fluid that flows out of it and it directs it towards exit section o of the machine.

2.3 THE DISTRIBUTOR

Without guide vanes in the distributor, the fluid penetrates axially into the rotor where it takes then a radial direction symmetrically as we assume that no fluid particle is participating to the rotating movement of the wheel before it is inside the part of the rotor that is equipped with blades.

So, in the distributor and in the entrance of the rotor, each fluid particle is supposed to move in a radial plan (plan going through axis of the rotor) with an absolute velocity that is first axial and then radial.

If the distributor is equipped with guide vanes, the direction of the velocity, at the entrance of

30

the wheel, is fixed by the geometry of the vanes. We suppose here that there are no guide vanes. The absolute velocity 1v is radial for a fluid element entering into section 1 (radius 1r ). The velocity triangles in this section, in a plan perpendicular to the shaft, are as represented on Figure 2.11.

The application of the equation of the kinetic energy to the fluid between the sections i and 1 gives:

( ) '1

121

2

2 fDiii wzzgppvv

=!+!

+!

"

or, as 1zzi ! is quite small:

'21

21

2 fDii wvvpp

+!

!=!"

'fDw represents the pressure losses in the distributor. These losses are proportional to the

square of the flow rate Q and thus also to 21v :

2

21' vKw DfD = with 310.5 !"DK

The value of KD depends on the shape and the dimensions of the distributor and among others of its exit section that is the minimum section seen by the flow of fluid. This is in this section, called the « throat », that the flow speed is the highest, so the pressure loss maximum and the static pressure minimum.

The experience showed that flow velocities in the throat are not allowed to go over 2 to 3m/s, exceptionally 5m/s.

2.4 THE ROTOR

The wheel (impeller) itself starts at section 1 (radius 1r ) and stops at section 2 (radius 2r ). The width / height of the rotor blades is respectively 1b and 2b . The sections useful for the flow are:

1111 2 ebrA != (2.1)

2222 2 ebrA != (2.2) 1e and 2e are coefficients lower than one (0,9 - 0,95), used to consider the reduction of the

section due to the thickness of the blades (blockage coefficient). The values of 111 ,, ebr are known, as well as the flow rate QR and the rotational speed ! (or N) of the rotor, the elements of the velocity triangles at the entrance of the rotor can be draught or calculated.

31

The absolute velocity 1v is given by:

1111 2 ebr

Qv R

!= (2.3)

and is radial. The frame velocity 1u equals:

111 602 rnru !

" == (2.4)

with =1! 90°, thus 1w and the fluid angle 1! can be draught as on Figure 2.11 and thus can be calculated. At the exit, 2r 2b , 2e , RQ , ! or N are known as well as the angle 2! made by the velocity 2w tangent to the blades at the exit with the tangential velocity 2u , the elements of the velocity triangles can be represented and calculated as follows:

222 602 rnru !

" == (2.5)

22222 sin2 !" wbreQR = (2.6)

2.5 ANGLES OF THE BLADES In order to reduce the total pressure losses, due to the shocks and flow separation, at the entrance on the rotor blades, the relative velocity 1w must be tangent to the blades. In the case where the flow rate RQ is fixed, this condition is satisfied by giving to the blades the entrance solid angle 1! determined with the velocity triangle at the entrance (the solid angle 11 !! = the fluid angle). The pump is thus build to realize a flow rate RQ with a rotational speed N and if during the operations, RQ or N take another value than the one of the design conditions, the velocity triangle at the entrance is changed and shocks take place against the leading edge of the blades or flow separation (Figure 2.12) with 11 !! " .

Losses are induced due to the shocks and to the flow separation on the blades (there is a flow expansion followed by a recompression as the cross section for the flow is first decreasing, leading to an expansion, and after that is becoming larger, leading to a compression).

The manufacturer chooses the angle 2! of the blades at the exit. With the same radial velocity,

222 ,sin !!w influences strongly the absolute velocity 2v and thus the diffuser (Figure 2.13). In reality, the angle 2! is always selected much higher than 90° because with a lower value, the absolute velocity 2v at the exit of the rotor is high and there is then a need to build a larger

32

diffuser to transform the kinetic energy of the fluid available at the exit of the rotor in pressure energy, with a passage of the fluid through the diffuser that goes with large losses. Into practice, 2! is always chosen between 145° and 165°.

2.6 NUMBER AND SHAPE OF THE BLADES

To be sure that the velocity w2 is perfectly tangent to the rotor blades at the exit (radius r2), the rotor should be equipped with an infinite number of blades. Of course this is impossible and the manufacturer needs to make a trade-off between a large number of blades providing a very good guiding to the flow but large pressure losses and a smaller number of blades leading to a less good guidance but with much lower pressure losses. The number of blades is usually around 6 and 12 and is selected so that the entrance channels into the wheel are with a square section perpendicular to the relative speed 1w . The profile of the blades (airfoil section) must be so that the angles 1! and 2! are respected.

Very often designers use blades / airfoils with a shape in « arc of circle », i.e. the active face of the blades has a circular shape and makes the angles 1! and 2! with the tangential velocities

1u et 2u .

If the leading edge of the blade is made out of two surfaces making a large angle between them (Figure 2.14), the velocity 1w cannot be simultaneously oriented towards the tangent to both faces. Important shocks can result from this, mainly when the flow angle 1! will deviate from the solid 1! of the rotor blades.

There is a trend to equip the wheels of modern centrifugal turbopumps with blades with the leading edge similar to the leading edge of a subsonic wing: thick and rounded. These blades permit a large excursion of the speed regime compared with the design value, without the presence of too heavy shocks at the entrance.

2.7 HEAD OF EULER OF THE ROTOR

Using the equation of Euler-Rateau and the distribution of the power:

!

PR = ˙ m R u2v2u "u1v1u( ) = ˙ m R eR + ˙ m R w fR

" (2.7) Or also:

!

eR = u2v2u " u1v1u " w fR" (2.8)

As 0,90 11 =° uv! and

!

eR = u2v2u " w fR" (2.9)

the term uvu 22 is often called the energy of Euler and represents the energy that the rotor would deliver to the fluid if there would not be any losses

!

w fR" . This is thus the energy that

33

theoretically the rotor can transfer to the fluid.

The head of Euler corresponds to: gvu u22 (2.10)

This energy or head is function of the volumetric flow in the rotor RQ , of the rotation speed N and of the geometric (solid) angles 21 ,!! of the blades. If the guidance of the flow is perfect:

22 !! = (2.11)

22222 sin2 !" wbreQR = (2.12)

As: 22222 coscos !" wuv += (2.13) 222

22222 coscos !" wuuvu += (2.14)

!

eR = u22 +

u2QR

2" e2r2b2 tan#2 (2.15)

At constant rotation speed N, uvu 22 is a linear function of RQ that is: - increasing if °< 902! - decreasing if .902 °>! Its y-axis value is equal to 2

2u (Figure 2.15). Most of the time 2! > 90° and uvu 22 will therefore be, generally, a linearly decreasing function of RQ with the slope depending on the angle 2! . The shape of the characteristic curve e-Q (or H-Q) of the pump is to be, we shall see later, as sera, deduced from the one of ( Ru Qvu !22 ) by substracting the losses.

2.8 INFLUENCE OF THE LIMITED NUMBER OF BLADES

The velocity triangles have been draught until now with the assumption that the flow in the rotor was axisymmetric, what would mean that the rotor would be equipped with an unlimited number of blades. We now have to look at the influence of this limited number of blades on the flow inside the rotor and on the velocity triangles. The energy is transferred from the rotor to the fluid. That means that the pressure on the active side of the blades (intrados or pressure side) is higher than the pressure on the non-active side (extrados or suction side). These pressure differences are in fact the origin of the

34

resistive torque applied by the fluid on the rotor. This pressure difference induces that the relative velocity 2w near the suction side is higher than near the pressure side. To prove this, one can use the equation of the kinetic energy to a flow in a rotating frame (Figure 2.16). A and B are located at an equal distance from the machine axis and the physical parameters in these points are the same. A' and B' are also at the same distance from the axis but ''

BA pp > . That induces that 2w in A' is lower than 2w in B' and that the flow leaving the rotor is not uniform. On Figure 2.17 it can be observed that where w2 increases the energy 222 cos!vu decreases and vice-versa. Moreover, due to the change in static pressure, the mass flow rate near the extrados is lower than near the intrados and so the power calculated with the integral:

!= mdvuPR &222 cos" (2.16) extended to the whole exhaust section of the rotor is lower than the theoretical one:

222 cos!vumP RtheorR &= (2.17) Calculated with a mean 2w that is uniform and satisfies to the following equation:

22222 sin2 !"#

webrmQ RR ==

& (2.18)

as Rmmd && =!

theorRR

R em

mdvue <= !

&

&222 cos" (2.19)

The limited number of blades causes also a recirculation relative to the fluid between two successive blades (Figure 2.18). This recirculation is due to the inertia effects of the fluid particles not sticked to friction. Different theories have been elaborated in order to calculate this recirculation. They are vased on the concept of a non-turbulent flow of a non-viscous fluid, i.e. based on the conservation of the total circulation. This relative recirculation that is created in each channel between two blades induces a deviation of the relative velocity 2w in the opposite direction of 2u . The angle of the flow

2! will thus be higher than the solid angle 2! of the blades at their exit (Figure 2.19). The parameters 22 ,vw et 2! are the obtained characteristics with the relative velocity that is

35

supposed to be tangent to the blades at the exit. The parameters 22 ,vw et 2! are the real characteristics. The conservation of the flow rate imposes that the radial component of the velocities remains unchanged:

2222 sinsin !! ww = (2.20)

A formula, established experimentally by Stodola, allows the calculation of the vu change:

uuu vvv 22 !=" (2.21)

This formula is:

[ ]smN

ukvu /sin 22 !"=# (2.22)

in which: - k is a coefficient with a value between 0.8 and 1.2 - N is the number of blades on the rotor.

Pfeiderer proposes another formula:

( )2122

222

2

21

1

rrNrv

v

u

u

!+

=

"

(2.23)

The value of the coefficient ! is between 0.8 and 1. The velocity triangles show the influence due to this deviation upon the Euler head that decreases again as:

( ) ( )reeltheor vuvu 222222 coscos !! " (2.24) The equal sign corresponds to the case where the flow is zero (Figure 2.15). It is important to note that these two phenomena: - difference in the values of 2w on the extrados and the intrados - inclination of the flow backwards are not losses because if the energy absorbed by the fluid decreases, the one to be transferred to the rotor, and therefore also to the pump, decreases too.

( ) RR mvuP &222 cos!= (2.25)

36

2.9 LOSSES IN THE ROTOR The flow rate, Rm& or RQ , flows through the channels between the rotor blades. We have formulated in the general equations of turbomachinery that:

"fRRR PemP += & (2.26)

because

""fRf wmP &= (2.27)

"fw is the energy lost by friction in the rotating channels. The term "

fw can be considered as the sum of two energy losses:

- a first loss, due to the development of the boundary layer on the sidewalls of the channels, is, as for all flows in a pipe, proportional to

!

QR2 and can be written with the following

equation:

2

212

1wKQk RR = with 025,0!RK (2.28)

- a second one, due to possible shocks against the edges of the blades or due to a boundary

layer separation, each time that the relative velocity at the entrance of the rotor 1w is not tangent to these blades as shown on Figure 2.20 where:

- 1! is the angle between 1u and the tangential direction compared to the blade at a radius

1r

- 1! is the angle between 1u and 1w . If w! is the component of 1w following the direction perpendicular to the tangent to the blade, one can accept that the losses dues to shocks are proportional to ( )2w! . This is quite well checked experimentally. For a constant value of u , w! is proportional to

1v! , that on its turn is proportional to RDR QQ ! (where RDQ is the flow rate when we are working in the design conditions or also called in the design point, i.e. for which 11 !! = ).

DRD vebrQ 11112!= (2.29)

11112 vebrQR != (2.30)

37

This second loss satisfies then to the following expression:

( )22 RDR QQk ! (2.31)

We thus see that "fw is the sum of these two losses that are on Figure 2.21 represented by

branches of parabols.

2.10 PERTE DUE AU DEBIT DE FUITE INTERNE The internal leak flow im& depends on the size of the pump and on its architecture (type of rotor: open or closed, clearances between rotor and carter). im& is between 1 and 10% of Rm& , the smallest value corresponds to large pumping units and

the largest one to small machines. This is due to the fact that the clearance between rotor and carter cannot be reduced under a certain absolute size and that the sealing (with labyrinth seals or even brush seals with different advanced materials) in large machines cannot be more efficient and better organized than in small ones. The loss in power linked with the internal flow leak is:

!

˙ m ieR = 0,01 à 0,10( ) ˙ m ReR W[ ] (2.32) 2.11 FRICTION OF THE DISC ON THE NON-ACTIVE FLUID The power lost by friction of the rotor disc on the non-active fluid is one of the main losses. Its general expression has been given in Chapter 1. Following Gibson and Ryan, one can see that, per side of the disc, it can be estimated using the following formula:

[ ]chDuPfr22

32

310.21,1 != (2.33) with 2u in m/s and 2D in meter.

2.12 THE DIFFUSER

2.12.1. Energy transformation in the diffuser The particles of fluid leave the rotor with a velocity 2v that is generally too high for any

customer / application behind the pump. In the diffuser, a part of the kinetic energy 2

22v is

38

transformed in pressure energy. The velocity at the exit of the diffuser is 3v , lower than 2v .

It must just be high enough to allow the flow of fluid to leave the pump. The energy transfer in the (fixed) diffuser is based on the theorem of the kinetic energy:

( ) '2323

22

23

2 fdwppzzgvv

!!

!=!+!

" (2.34)

As 023 =! zz , one has:

!

p3 " p2#

=v22 " v3

2

2" w fd

' (2.35)

where '

fdw are the pressure losses in the diffuser. An estimate is given with:

02,0 - [ ]kgJv /2

03,022 (2.36)

This formula clearly shows that it exists a relation, between the velocity reduction and the increase in pressure between the entrance and the exit of the diffuser. The question is now to determine what is the optimal shape to give to the diffuser and its dimensions to reduce 2v to a value 23 vv < (v2 is fixed) with a minimum of pressure losses ?

2.12.2. THE DIFFUSER WITH FLANGES (without vanes) In its most simple architecture, a diffuser is made of two circular flanges put around and in the continuity of the exhaust of the rotor. The fluid freely flows between these two discs. To study the trajectory in the diffuser of a fluid element with a mass m, one applies to this element, at a distance r from the machine axis, the equation of the kinetic moment:

eaxeaxe FMvmMdtd

= (2.37)

One may suppose that each fluid particle is moving in a plan perpendicular to the axis and that the flow in the diffuser is axi-symmetrical. If one neglects the weight mg , all the fluid

particles at the same distance r from the axis, are facing, for symmetrical reasons, the same radial pressure gradient (Figure 2.22) and describe identical trajectories. Therefore, one can write:

39

( ) 0cos... =!rvmd (2.38)

or Ctevrvr u == .cos.. ! (2.39)

where uv is the tangential component of v. The condition Ctevr u =. is the one of a non-

rotational flow (« free vortex flow »). As the flow equation imposes that:

Cterbv =!" sin2 (2.40)

one obtains, when dividing member by member, that:

Cteb =!tan (2.41) If the flanges of the diffuser are parallel, b is constant and ! is also constant. The angle between the radial direction and the direction of the tangent to the trajectory is constant, that means that the trajectory of a fluid element in the diffuser is a logarithmic spiral. The flow equation allows the calculation of the radius 3r to be given to the diffuser in order to reduce the velocity from 2v at the entrance to 23 vv < at the exit:

33332222 sin2sin2 !"!" vbrvbr = (2.42)

That gives then:

3

223 rrvv = (2.43)

To increase the diffusing effect without increasing any further 3r , the flanges can be mounted

slightly divergent. In this case, b increases between the entrance and the exit of the diffuser and ! decreases as we have

Cteb =!tan (2.44) The flow equation is written as:

33332222 sin2sin2 !"!" vbrvbr = (2.45)

40

That gives then:

33

222

33

22

3

223 cos

cossinsin

!!

!!

rrv

bb

rrvv == (2.46)

As we have 3232 coscos, !!!! <> and the diffusing effect is enhanced compared with the

diffuser with parallel flanges. The divergence angle of the discs cannot go over 6-7° in order to avoid the boundary layer separation on the sidewalls of the diffuser. 2.12.3. THE DIFFUSER WITH BLADES (Figure 2.1) The previous formulae show that to obtain a velocity given 3v , one needs, with a fixed 2r to have a radius 3r that is higher when 2v is higher, even when we use a diffuser with divergent

discs:

3

2

2

3

vv

rr= (2.47)

To increase the diffusing effect without increasing too much the used volume, another possible solution is offered by diffusers equipped with stator vanes put between the flanges. These vanes have a shape selected so that the fluid particles going through the diffuser are forced to follow a trajectory with an angle ! that is increasing. Then we have that 23 !! > .

The flow equation gives then:

333332222 sin2sin2 !"!" vbrevbrQ == (2.48)

In the case of parallel discs ( )32 bb = :

23

22

32

3

sinsin1

!!

vv

err= (2.49)

To obtain a given velocity reduction 2

3

vv

, 3r will become smaller and smaller when 3! is

chosen larger and larger compared with 2! (but < 90°).

To avoid shocks at the entrance of this diffuser equipped with vanes:

41

- The distance between the two discs at the entrance must be 1 to 2 mm larger than the width 2b at the exit of the rotor,

- The diffuser vanes must be tangent to the real velocity 2v at the exit of the rotor.

One gives to the angle 2! of the vanes at the entrance of the diffuser the value of the angle

2! calculated in the design conditions with the flow rate DQ and the rotation speed Dn . The shape of the vanes is then chosen so that ! increases from 2! up to 3! , without any

boundary layer separation. After the manufacturing of the machine, when the working conditions are different from the design conditions (« off-design » conditions), 2! is not anymore equal to the solid angle 2! . Then shocks and separations are taking place.

The leading edges of the diffuser vanes are always a bit separated (a few mm) from the trailing edge of the blades at the exit of the rotor; the objective is to reduce the noise and the vibrations introduced by the interference between the blades when they are passing in front of each other. Moreover, to avoid a possible phase-tuning of these periodic perturbations, the number of blades in the rotor and of vanes in the diffuser are selected as prime between them. For what concerns the energy losses in the diffuser, they are as in the rotor, i.e. the sum of a loss proportional to 2Q (a loss that is the only one present when the fluid angle 2! is equal to the solid angle 2! ) and another one proportional to (Q-QD)2 that takes place only when there are shocks at the entrance or a separation, when 2! is different from 2! .

2.12.4. CONCLUSIONS

For a given ratio 2

3

vv

, the dimensions of the diffuser equipped with vanes are smaller than

with simple diffuser. Nevertheless the off-design performance of a bladed diffuser can give problems, what never the case with a simple diffuser that shows losses purely proportional to

2Q .

2.13 THE VOLUTE OR COLLECTOR

The fluid needs to be progressively collected at the exit of the diffuser and directed towards the exhaust piping. This is the role of the volute or collector. This component is directly put around the rotor when the pump is not equipped with a diffuser (Figure 2.23). Normally the volute is calculated so that the mean velocity is constant in all radial sections starting from the beginning of the collector until its exit. For that, it is required that the flow

42

section BC increases, from a point A, proportionally to the length of the bow AB. The flow in a radial section is not uniform but it follows a non-rotational flow law:

Ctervrv uu == 33 (2.50)

(or 22rvu if there is no diffuser; in this case, a certain diffusing effect can be obtained by

putting a short divergent just before the coupling connecting the pump with the rejection pipe). The losses in the diffuser and the volute are estimated with:

(0,02 and 0,03) [ ]kgJv /2

22 (2.51)

2.14 CHARACTERISTIC CURVES OF A CENTRIFUGAL PUMP

In paragraphs 2.7 and 2.8, we have calculated the theoretical energy or head of Euler

222 cos!vu in function of the flow RQ in the rotor. For a given rotation speed, it is a linear

function that is decreasing with QR as already represented on Figure 2.15. The energy Re (or the head RH ) transferred by the rotor to the fluid is the head of Euler reduced by the losses "

fw in the rotor, that show, in function of RQ , a parabolic shape

presenting a minimum for RQ not too far from RDQ .

The resulting curve RR QH ! will be, depending on the slope of 222 cos!vu (that changes with the solid angle 2! ) and the shape of the curve of "

fw , a function that is first decreasing

and after that increasing with RQ (thus with a maximum) or a function strictly decreasing.

To deduce the curve of the pumping head H generated by the pump in function of the flow Q, one still needs:

- To consider the internal leak flow im& by changing the scale of the x-axis with:

!i

RmQQ&

+= (Fig. 2.24) (2.52)

- To substract the losses 'fDw and '

fdw that are function of 2Q and of ( )2DQQ ! when the

diffuser is equipped with vanes).

When the rotation speed increases, the curve moves up (and normally to the right). This rational gives us the way to predetermine qualitatively the shape of the curve QH !

and the influence of some manufacturing options like the inclination of the blades at the exit

43

of the rotor (the angle 2! ).

The precise calculation of H in function of Q for a machine in design is quite difficult as we have problems in estimating precisely the losses "' , ff ww , the leak flow im& and the slip

coefficient. When the pump is built (for a prototype or for a series production), one measures experimentally the characteristic curve H-Q when installing it by measuring the parameters indicated in section 1.4. For a prototype, the experimental curve H-Q that is obtained will be closer to the one predicted if the data used for the project are themselves well estimated.

44

CHAPTER 3

PERFORMANCE CURVES OF CENTRIFUGAL PUMPS NON-DIMENSIONAL ANALYSIS

3.1 THEORY OF VACHY-BUCKINGHAM Let’s consider a series of machines that are geometrically identical. A machine of this series is characterized by one of its geometric dimensions, for example, in the case of the centrifugal pumps, the external radius of the rotor 2r . Its working condition is characterized by a certain number of physical parameters as mn &, , P, e, η, …

To know all these variables it is not necessary to fix without any confusion the working point: only p from them may be enough, the others being deduced from them. In this case, p is the number of independent physical variables that characterize the machine when it is working. When these p independent variables contain q fundamental parameters (mass, length, time, …), it is possible to build between them (p-q) independent non-dimensional variables (reduced variables). Any other combination without dimension formed with these variables, independent or dependent, characterizing the system will be a function of the (p-q) independent reduced variables. 3.2 APPLICATION TO TURBOMPUMPS The physical variables that characterize the working point of a pump from a series of pumps that are geometrically identical are:

2rr = : outside radius of the rotor, dimension [ ]L n : rotation speed,

!

T "1[ ] m& : mass flow rate,

!

MT "1[ ] ! : specific weight of the fluid,

!

ML"3[ ] µ : dynamic viscosity of the fluid,

!

MT "1L"1[ ]

The physical variables r and n can be replaced by r and .602 rnu !

=

The dynamic viscosity µ is included in the Reynolds number, an already non-dimensional

45

parameter, that in the case of pumps will be written like this:

DvQ

Dm

rmRe ===

µµ

&&

2 (3.1)

If we limit the study/working to a well defined fluid, for example, water at ambiant temperature or around this one, the dynamic viscosity µ cannot change too much and its influence upon eR is so limited that the variations of eR are without any influence on the

working point. This does not mean that the viscosity effects are to be neglected. They exist (via the losses terms fw ) but are just slightly influenced by the small variations of eR . It is the variation of the mass flow rate and of the rotation speed that are acting on the losses fw .

In order to simplify the next rational, one supposes that the used fluid is water without forgetting that the performance curves, that we shall define, could change a lot if the used fluid would change of viscosity so that the Reynolds number would largely affected. The p physical variables are then reduced to 4:

r [ ]L u

!

LT "1[ ] m& [ ]1!MT !

!

ML"3[ ]

The number q of fundamental parameters being equal to 3 [ ]TML ,, , there exists just one single independent reduced variable. This non-dimensional variable is obtained by making the product:

wzyx mur !" &= (3.2)

and choosing the exponents so that when replacing the variables with their dimensions, this product will be of the type:

000 TLM (3.3) It is therefore requested that:

( ) ( ) ( ) ( ) 000311 ... TLMMLMTLTLwzyx =!!! (3.4)

from where the 3 equations:

46

0003

=+

=!!

=!+

wzzywyx

This linear system of 3 equations contains 4 variables. One have then to decide about the value to be given to one of these variables de. Here we choose z = 1 as we want to obtain a factor where the flow rate m& is present at the first power. The solution of the system is thus:

!

y = "1;w = "1;x = "2 (3.5)

and the independent reduced variable is:

22 ruQ

rum

=!

& (3.6)

Other reduced variables but dependent are obtained when combining e (or H), P, g! … and the 4 independent physical variables !,,, mur in the non-dimensional products in such a way that:

fdcba mure !" &=1 (3.7)

!

"2 = pa 'rb'uc ' ˙ m d '# f ' (3.8) """""

3fdcba

g mur !"# &= (3.9)

The energy e is in J/kg and its dimension is

!

L2T "2 , the product 1! will thus be without dimension if:

A = 1 ; e = -2 ; b = d = f = 0 (3.10) It leads therefore to the following reduced variable:

22 ugHou

ue (3.11)

that is only function of:

2ruQ (3.12)

Also one can define:

getruP

!" 23 (3.13)

that are also only function of:

47

2ruQ (3.14)

3.3. PERFORMANCE CURVES OF A TYPE OF PUMPS

The performance curves of a series of pumps geometrically identical are showing:

gruP

ugH

!"

,, 232 (3.15)

in function of: 2ruQ (3.16)

They are as shown on Figure 2.25.

These 3 curves are not independent of each other as when one knows two of them the third one can be derived from them.

For example, if 2ugH and g! are known in function of 232 , ru

PruQ

! can be build too because:

2223 ..ue

urm

emP

ruP

!!

&

&= (3.17)

22 ..1ugH

ruQ

g!= (3.18)

3.4 CHARACTERISTIC CURVES OF A PUMP

A pump of a series is determined by its outside rotor radius 2r . When it is fixed, one can deduce the characteristic curves of the so selected pump from the curves of the pump type. One just needs to replace in the parameters the factor r with its value.

Therefore the curves of:

!"

#$%

&nQfen

nP

nHou

ne

g'(;; 322 (3.19)

show the performance of the pump with a rotor radius equal to 2r .

For a determined fluid with a fixed average volumetric de mass! (one needs to be careful if one changes the fluid), when N is fixed, one can deduce from the previous ones the curves of:

48

e or H, P and η en f (Q)

that are the performance curves of the pump at a rotation speed N (Figures 2.9 et 2.26). The change of the rotation speed plays an important influence upon the shape of the performance curves. Figures 2.27 and 2.28 give the performance curves of centrifugal pumps rotating at respectively 2.900 and 1.450 RPM. Please examine the relative position of the curves and calculate the global efficiency in a few points for a given pump. 3.5 INFLUENCE OF THE FLUID VISCOSITY

If it is required to consider the viscosityµ , there now are 5 independent physical variables that gives, with the same 3 fundamental parameters, 2 independent reduced variables. The products ! contains 6 factors, for example:

µ!" fdcba mure &=1 (3.20) and are depending on 2 independent reduced variables. A solution is for example:

!!"

#$$%

&= 22 ; u

gHruQf

Dmµ

& (3.21)

The performance curves of a pump type are not anymore given with a simple presentation (2

independent reduced variables). But, as already mentioned, the variation of µrm& is almost

always negligible and, in any case very limited when one consider a given and same fluid.

3.6 STABILITY OF A TURBOPUMP When one speaks about the stability of a centrifugal pump, two aspects must be clearly differentiated: the first one concerns the stability of the motor + pump assembly, the second is the one of the pump on the network. 3.6.1 Stability of the motor + pump assembly This has already been studied in the course « Kinematics and Dynamics of Machines ». This is based on the shape of the characteristic curves: the one from the driving torque of the motor and the one of the resistive torque from the pump, in function of the rotation speed. Let’s consider that the pump is installed on fixed network, i.e. so that the level difference

'" zz ! is constant and that the position of the regulation valve is fixed. It follows too that the characteristic curve of the network H(Q ) is fixed too. The resistive torque pM of the pump is function of the rotation speed and of the flow rate:

( ) 0,, =QnMf p (3.22)

For 1nn = , the shape of the curve ( )QM p can approximately be determined from the formula of Euler:

uRR vumP 22&= (3.23) or

49

!!"

#$$%

&+=

222

222 tan2

1'()*

+* b

QuuQP nR

R (3.24)

Explain, using this formula, the shape and the position of the curves of Qrp ! for n with increasing values ...,, 321 nnn (Figure 2.29). Using the characteristics ( )nQM p , and ( )nQH , of the pump and the one of the network ( )QHn , one can determine for nHH = the resistive torque pM of the pump installed on the

network. This torque is designed with netpM . . In function of the rotation speed n, this torque is represented by a curve with a positive slope. The characteristic curves of the driving motor M(n) that depends on the opening of the control organ (opening of the throttle of a gasoline engine; position of the rheostat of an electric motor; position of the admission valve for the steam in a steam turbine) are always curves with a negative slope (very rarely horizontal) This is by superposition of the curve of netpM . and the one of the motor M in function of n, that one sees if the motor-pump assembly is stable.

3.6.2 Stability of the pump on the circuit

The explanation of the stability and instability phenomena will be given using a qualitative rational that consists in examining the working of a pump that is replenishing a water tower; the circuit is ideal (zero pressure losses). Figure 2.30.a shows a schematic of the installation. The pressure losses in the pipes are zero, the characteristic of the circuit is horizontal (ordinate at the origin: '" zz ! ). Two cases can take place depending on the shape of the performance curve of the pump that can be purely decreasing (Figure 2.30.b) or presenting a maximum (Figure 2.30.c). The working point is in A. Let’s examine what is taking place when the requested flow rate decreases. The level in the water tower increases because when the consumption decreases the pump delivers a flow rate higher than the requirement. The characteristic curve of the circuit is going up. The flow rate delivered by the pump decreases and the working point is repositioned in A’ with the requested flow rate. In the case of working curve that is purely decreasing, all the flow rates are available from AQ down to zero 0. But if the working curve presents a maximum (Figure 2.31.c), difficulties will be appearing as soon as A’ reaches B. Indeed, if when having BQ , the consumed flow rate decreases further, the water level in the tower cannot become higher, as the pump is not able to provide a head > BH . Moreover, the characteristic of the circuit is going up, there is no intersection point anymore between H et

nH . The hydraulic equilibrium cannot be obtained anymore. The pump stops delivering any flow rate (but continues rotating). The flow has even the trend to reverse but the anti-return valve is entering into action. The new working point goes to en C (zero flow rate) and stays there until the water level in the tower goes lower and becomes again equal to CH . At this moment, the pump suddenly starts again delivering a flow and the working point moves to D. If the consumption becomes then lower then DQ , the level in the tower increases and the same phenomena can happen again if the consumption is < than BQ .

50

This instability, called also a « surge » or a “stall”, is a phenomenon of large amplitude and low frequency, depending on the size of the upper reservoir, on the difference between the requested flow rate and DQ and also on the shape of the performance curve. The flow rate of the pump cannot exist for values lower than BQ . In reality, the characteristic of the circuit is not horizontal but is a branch of a parabol. The rational here above remains nevertheless valid. We need to say that surge can only take place if the level difference '" zz ! can become higher than CH (for example this requires that the height of the reservoir is large enough) and if the requested flow rate is lower than BQ . This periodic phenomenon never shows up when the performance curve of the pump is decreasing. The characteristic frequency is in the region of 1 Hz.

3.7 PUMP CAVITATION If the pump is installed above the free level of the aspiration level, the pressure at the entrance of the pump is lower than the atmospheric pressure. This is in the minimum section (called the throat or the “ear” of the pump) and that is normally at the exit of the distributor that the underpressure is maximum. But there can also be zones of underpressure at certain places on the des blades where there is a local increase of the speed of the flow. If in these locations the static pressure goes lower than the saturation pressure corresponding to the temperature of the liquid, there is evaporation with formation of gas bubbles or of vapor pockets. This evaporation is enhanced by the presence of gas dissolved in the liquid. This formation at local level of bubbles or pockets constitutes a phenomenon of cavitation. A first consequence is a reduction of the flow rate due to a narrowing of the fluid section. A second consequence is the noise very characteristic of the phenomenon and changing with the intensity of the cavitation: a crackling first, increasing in intensity together the vibration level of the machine, becoming after a louder and continuous noise with a decrease of the flow rate. A third consequence is the degradation caused to the blades by the mechanical action of the bubbles that are condensating on some solid walls what is leading on these places to high stresses that can lead to erosion or pitting and even sometimes to corrosion of the metal of the blades. In order to compute the static pressure mp existing in the minimum section m (at the entrance of the pump), one applies the equation of the kinetic energy to the fluid that is between the free level of the liquid and this minimum section (Figure 2.31):

( ) 2'22

20 kQppzzgv am

mm !

!!=!+

!

" (3.25)

As mmm AvAQ (= = minimum section):

!

pm = pa " g#(zm " z' ) " # k +

12Am

2

$

% &

'

( ) Q2 (3.26)

Cavitation will take place if sm pp ! with ps the saturation pressure corresponding to the liquid temperature. The pressure p decreases rapidly when the flow rate Q increases. There will thus be more easily cavitation when the flow rate is high than at low flow rates. An important factor inducing more easily cavitation is also the Gas-To-Liquid ratio (GTL) that largely changes from one fluid to another (the GTL is very large for water).

51

The formula of pm shows also that at a given flow rate the aspiration height 'zz ! should also be limited.

!

(zm " z' )max =

pa " psg#

"1gk +

12Am

2

$

% &

'

( ) Q2 (3.27)

Experience has shown that the value of mp cannot be too much just above sp and that a safety operational margin is to be taken. The table here after, valid for water, gives the maximum values and the normal ones (i.e. commonly admited) that the difference between patm and ps can take:

!

pa " pm = g#(zm " z' ) + # k +

12Am

2

$

% &

'

( ) Q2 (3.28)

Compare these values to the maximum ones theoretically admitted ssata pp !, given in the last column.

Water temperature pa - pm pa,sat - ps

(°C) (Pa) (Pa)

NORMAL MAXIMUM

25 0,68.105 0,78.105 0,98.105

50 0,60.105 0,69.105 0,89.105

70 0,41.105 0,51.105 0,70.105

80 0,24.105 0,34.105 0,54.105

90 0,02.105 0,118.105 0,31.105

92 -0,039.105 0,059.105 0,25.105

94 -0,098.105 0 0,196.105

96 -0,15.105 -0,059.105 0,137.105

98 -0,22.105 -0,125.105 0,0685.105

100 -0,29.105 -0,195.105 0

We have to note that at high temperature of the fluid, ma pp ! must be negative, what requests that am pp > or that the pump must be installed under the free surface z’ (immerged pump). In pratice, it is required to:

- Install the pump as closest as possible to the aspiration level and even lower if possible; - Use the shortest possible aspiration pipe and with the lowest possible pressure losses; - Never install any valve (except the anti-return valve) on the aspiration pipe.

52

3.8 USEFUL PORTION OF THE PERFORMANCE CURVE In principle, the pump must be used at working ratings that allow that the global efficiency is maximum or near the maximum. It follows that important modifications brought to a circuit where a pump is installed can cause a movement of the working point and can bring it from an optimal initial position either in the zone of the low flow rates and low efficiency, with a danger of surge, or in the zone of the high flow rates with a low efficiency too with the risks of cavitation. It is therefore always necessary to check the exact consequences of the modifications of the circuit that can degrade the performance compared to what had been calculated / planned. 3.9 NPSH In order to raise a water column at a height of 1m, there is a need for a suction force at the top of the column at least equal to the pressure (generated by the weight of the water) at the root of this water column. If the required suction to raise this column makes the liquid pressure fall lower than the treshhold of the saturation pressure sp , cavitation will take place. The ‘water head at the aspiration’ is thus limiting the underpressure that can be created without cavitation. If the water head to be raised or to be sucted in is too important to be elevated without cavitation, it is required to place one or several intermediate pumps. This maximum possible underpressure before cavitation is called the available NPSH1. This NPSH is related to the water head at aspiration. On top of raising this water column, it is needed to give a flow speed to this water column. This pump must also generate an extra suction force so that the liquid moves with a certain flow rate. This underpressure will thus need to be higher when the flow rate is higher. This extra underpressure related to the pump flow rate is called the requested NPSH. It is expressed in m of column of fluid and decreases with Q. A typical accepted value of this “design margin” is equal to 0.50 m. One can indeed write:

eP must be > SP or gP

gP Se

!!>

!

NPSHavail = He "PS# g(> 0)

The condition of non-cavitation is:

!

NPSHavail > NPSHrequested (3.29)

___________________________________________

1NPSH = Net Positive Suction Head

53

Chapter 4

THE NETWORK AND ITS CHARACTERISTIC CURVE 4.1 SIMPLE CIRCUIT The circuit is called simple when the pump is installed on the only pipe between the downward domain ( )''' ,, vzp of the aspiration and the upward domain at the exhaust ( )""" ,, vzp (Figure 2.5). The application of the equation of kinetic energy between the downward free level and the entrance i of the pump gives:

!

v ' 2 " vi2

2+ g zi " z

'( ) = "pi " p

'

#" w f ,asp$ (4.1)

! aspfw , contains all the pressure losses due to friction through the aspiration pipe and in all the accessories present on it. We do the same between the exit section 0 of the pump and the free level of the exhaust domain:

!

v"2 " v02

2+ g z" " z0( ) = "

p" " p0#

" w f .ref$ (4.2)

When combining these two equations:

!

v"2 " v ' 2

2g+p" " p'

# g+ z" " z'( ) +

w f$g

=v02 " vi

2

2g+p0 " pi# g

+ z0 " zi( ) (4.3)

[ ]mHHn = (4.4) and een = [ ]kgJ / (4.5) This is the working condition of the pump on the circuit. How is this characteristic curve of the circuit ( )QHn in common cases ? In the case of a water pumping installation as represented on Figure 2.5:

0'" != vv and appp == '" (4.6)

so that:

( ) !+"= fn wzzge '" or ( )gw

zzH fn

!+"= '" (4.7)

The pressure losses in the elements of the installation are proportional to the square of the flow rate.

In other applications, one cannot neglect neither gvv

2

2'2" ! nor gpp

!

'" " and nH contains

two terms independent of Q . '" zz !

gpp

!

'" "

and two terms function of 2Q :

54

gvv

2

2'2" !

gwf!

The current shape of the characteristic curve ( )QHn of a simple circuit is an increasing branch of

a parabol with an ordinate at the beginning equal to gppzz

!

'"'" "+" . The requested flow rate is

DQ and the working point can be defined on this curve. 4.2. CALCULATION OF THE CHARACTERISTIC CURVE OF A CIRCUIT

4.2.1 Losses in the circuit

We limit our study to the most common case where the circuit is composed with elements of non-incrusted pipes with a circular section and the flow regime is turbulent. This regime will always be turbulent when the Reynolds number:

vvDRe = (4.8)

in which:

- v is the average speed in m/s

- D is the internal diameter in m

- v is the kinematic viscosity in

!

m2 /s 1

is higher than 2500. There will be pressure losses aura in:

- the straight pipes,

- the accessories (filters, turns, valve, …)

- shape modifications (abrupt restrictions or openings).

1 The dynamic viscosity is noted µ (in saP . or Poiseuille ( oP )). The kinematic viscosity is noted

v (in m2/s or Stoke (St)) and is the ratio !µ .

1m2/s = 106 cSt 1Pl = 1kg/ms 1Pl = 10 Po

55

4.2.2. Regular pressure losses (in straight pipes)

They are proportional to the length L of the pipe, to the flow rate Q (and thus to v ), to the viscosity v of the fluid and inversely proportional to the section (thus to )2D following formulae for which the graphical translation, in the case of water at 15°C or of a liquid with the same viscosity, is given on Figure 2.32. 4.2.2 Singular pressure losses (in all accessories)

The accessories such as anti-return systems, turns, valves introduce losses that are depending on their shape, internal diameter, roughness, … that must be added to the losses in straight pipes. The preference will be given, if possible, to accessories with low losses as turns with a large radius of curvature, gliding valves, … To facilitate the calculations, each accessory can, using a table as the one here after, be replaced by an equivalent length of straight pipe (equivalent from the point of view of the pressure losses). These equivalent lengths are given in meters, by multiplying the diameter in cm by the indicated coefficient.

Accessories Coefficient Turn 45° Turn 90° Turn 90° with large radius Turn 180° Gliding valve open Gliding valve ½ closed Poppet valve open Anti-return valve + filter

0,15 0,7 0 ,3 0,75 0,2 4,0 3,5 0,7

4.2.3 Losses due to abrupt shape modifications

They are calculated in m using the formula gv2

2

! (4.9)

v is the downward speed in the case of a opening of the section and the upward speed in the case of a restriction. In the case of a section increase:

! =2

2

2

1!!"

#$$%

&'

amont

aval

DD

(4.10)

In the case of a restriction, the value of ! is to be determined using a graph as on Figure 2.33. 4.2.4 Example of a calculation

The installation schematized on Figure 2.34 is intended to deliver water to a water tower using pipes of non-incrusted steel with an internal diameter of 100 mm.

56

It contains: 127,5 m of straight pipes 2 turns of 90° with a large radius of curvature 1 straight exhaust (coefficient = 0,75) 1 filter + anti-return valve 1 gliding valve The equivalent length due to the accessories is: 2 turns large radius = 2x10x0,3 = 6,0 m 1 exhaust = 1x10x0,75 = 7,5 m 1 filter + anti-return = 1x10x0,2 = 2,0 m 1 gliding valve = 1x10x0,2 = 2,0 m Total = 22,5 m The total equivalent length of the circuit = 22,5 + 127,5 = 150,0 m. Using graph 2.32, we calculate the values of nH for flow rates from 300 up to 1500 l/min:

Q Pressure losses over 100 m (Ø 100m)

Losses for 150 m

of .equivL

'" zz ! nH

l/min m OH 2 m OH 2 m M 300 0,58 0,87 40 40,87 400 1,00 1,50 40 41,50 500 1,50 2,25 40 42,25 600 2,10 3,15 40 43,15 700 2,80 4,20 40 44,20 800 3,80 5,70 40 45,70

1000 5,50 8,25 40 48,25 1500 12,00 18,00 40 58,00

This allows to trace the curve of ( )QHC with the ordinate at the origin equal to mzz 40'" =! (Figure 2.35). For all used flow rates, eR is by far larger than 2500 because for the lowest flow arte of 300 l/min (the flow speed is only 0,65 m/s), it is already reaching:

6000010.1,11,065,06 !="

xRe (4.11)

The kinematic viscosity of water at 15° C is equal to 1,1cSt or 1,1.10-6 m2/s. 4.2.6 Remark

For rough pipes or of an extreme length and for very viscous liquids, the pressure losses must be calculated in a more rigorous way by considering the roughness of the walls and the right viscosity.

57

4.3. COMPLEX CIRCUITS

Let’s consider the case of a pump aspirating a fluid from a tank E and delivering it to two tanks E’ and E’’ using 2 pipes put in parallel (Figure 2.36). The determination of the global characteristic of the circuit is easy in the case where the pump is installed just near the tank E and of the separation towards E’ and E’’ so that the pressure losses between E and the pump can be neglected, so that the pipes towards E’ and E’’ are identical and also so that p’ = p’’, v’ = v and z’ = z’’. In this case, the characteristic curves of the two exhaust circuits are identical. As soon as the head delivered by the pump will be sufficient to deliver a flow rate in one of the two circuits, it will also deliver flow to the other and, from this moment, the flow rate through the pump will need to be large enough to insure the same flow rate in both derived pipes. The global characteristic of the circuit is obtained by doubling horizontally the abscissae (Figure 2.36). If the tanks E’ and E’’ have different levels and if the pipes are not identical, we suppose "' zz < ,

'nEH the characteristic curve of circuit 1, "nE

H the curve of circuit 2 (Figure 2.37). When the head of the pump is > z’ but < "z , only the tank E’ is filled in. The global characteristic curve ie 'nE

H . When it becomes > "z , the pump delivers water to both circuits and the global characteristic is

nH obtained by adding the abscissae of 'nEH and "nE

H . If it is needed to consider the pressure losses at aspiration and at the exhaust between the exit of the pump and the flow separation point, one needs to add to the ordinates of the curve nH the one of the curve of ( ) 2.Qkk ra + and one makes the sum of the pressure losses along these two pipings.

58

Chapter 5

SELECTION OF A PUMP OR OF A PUMP TYPE 5.1 USERS OF PUMPS We limit our study to a preliminary design of a pump to be selected and installed, that does not means the manufacturing of a new pump of a new model but only the best choice to be made, among the pumps from different existing types with existing performance curves, for a well defined application. This is the problem of a plant or process engineer not of a designer. From the analysis of the circuit on which the pump needs to be installed, we extract the essential inputs for our choice: the head H to be realised by the pump at the requested flow rate Q. The type of driving motor influences in most cases the choice of the rotation speed. An important feature is that motors and pumps running at a higher speed are more compact and could be less expensive. It must be emphasized that driving the pump at a variable rotation speed can be extremely useful in order to optimize the pump efficiency over a wider range of volumetric flow rates. We accept as fundamental elements leading to the choice of a pump:

- The definition of the circuit on which the pump must be installed; - The nature of the fluid ( ! ) and the design flow rate DQ normally desired as well as its

extreme values; - The rotation speed N or, at least, its order of magnitude.

5.2 GENERAL CASE We have the performance curves available:

!

Wm

" u3r2, eu2

or 2ugH and g! in !!

"

#$$%

&2ruQf

For different types of centrifugal pumps with an about equal global efficiency for all of them. The question that we have is: first, what type of pump to select and second, in this series of pumps of this type, which is the pump to be selected to deliver this DQ at a head DH with the imposed N (or a N not too far from that). 5.2.1 CHOICE of the PUMP TYPE

We can, for each type of pumps, thanks to an ordinate change, build the curve of 23

2 !"

#$%

&ue in

function of 2ruQ , and put these curves on the same figure (Figure 2.38) and identify on each of

them the zone where g! is maximum or does not goes too far away of this maximum, let’s say by not more than 1 or 2 %. If we join a point P on one of these curves with the origin, one define the angle ψ:

59

( ) 2

22/3

2

2/32 )(/)/(tan

ur

QgH

ruQue

==! (5.1)

and with rnru602!

" == (5.2)

2

2/32 )(260tan

nQgH

!"

#$%

&='

( (5.3)

If, in this formula, one replaces H, Q, n with their requested value, ! takes a value D! . The half straight line Dd making with the axis x the angle D! and intersecting the curve in the zones with the requested efficiency indicates us the type of pump to be selected (here the type 1). 5.2.2 CHOICE OF THE PUMP The type of pump has now been fixed, we have to, with this type, determine the pump to select by calculating the external radius r2 of the rotor. The value of:

32 260

rnQ

ruQ

!= (5.4)

is fixed by the intersection of Dd with the curve 1, and so we can calculate Dr by introducing DQ and Dn . Very often, there is no pump of this type having exactly the same rotor radius Dr . One will then normally choose the existing pump with the radius just above the calculated one except if we can work with a flow rate a bit lower than DQ that would be obtained with a pump a bit smaller than the calculated one. DH and DQ are indeed often desired and not absolutely requested. A general rule does not exist, it is each time a different case. If the selected pump is too large, one will adapt the flow rate using a regulation valve. With the value of r, one calculates:

rnu DD 60

2!= (5.5)

and 2ruQ

D

D (5.6)

One reads also on the curves the value of g! at the working point. One can now calculate the requested (mechanical) power that is required to drive this pump:

g

DDm

HQgP

µ!

= (5.7)

5.3 PARTICULAR CASE One has the performance curves of a type of pumps and one has to select the pump in this series to deliver a flow rate DQ at a head DH , without any constraint on N.

60

In principle, the selected pump must work at its maximum efficiency. For g! max, one reads the

corresponding values of 2/ ruQ and 2/ugH . The second value with DHH = gives u that, when introduced in the first value with DQ , gives the value that the rotor radius r should have. It there does not exist any rotor with exactly the calculated r, one takes the ready available value generally a bit higher. One calculates as follows the working conditions. One builds first, using the available data, a diagram of )/(/ 22 ruQfugH = . For 1 point on this curve:

4242

2

2 /tan rQgH

ruQ

ugH

==! (5.8)

For DD QQHH == , and the selected r, ! takes a value for which one can read:

- the value of ( )22/ ruQ from where we have 2/ ruQ and then g!

- the value of 222 // rgHugH != from where we have ! and then n .

One can also calculate the required mechanical:

g

DDm

HgQP

!"

= (5.9)

5.4 GROUP OF PUMPS 5.4.1 Problem For low flow rates and large heads, tan D! (Figure 2.38) is high and the half line Dd is almost vertical and will maybe not cut any of the curves in the zone of the admitted efficiencies. One will then adopt either several pumps put in a serial arrangement or a multi-cellular pump with several identical stages disposed in a serial way (Figure 1.3). For high flow rates and low heads H, Dd is almost horizontal and a similar problem to the previous one appears. One solves it thanks to several pumps working in parallel or using pumps with a double inlet (Figure 2.4). 5.4.2 Serial grouping of pumps Let’s consider (Figure 2.39) the pumps 1 and 2 put in series on the same circuit. Everything that goes through one of them has to go to the other(s): their flow rates are identical. The heads will cumulate and the working point will be at the intersection of 21 HH + and CH . Indeed, we can apply the equation of the kinetic energy to the 3 parts of pipes. Between E and i1 :

( ) fai

ii wppzzgvv

!!

!=!+!

"1'

1

221

2 (5.10)

Between 1o and 2i :

( ) fboi

oioi w

ppzzg

vv!

!!=!+

!

"12

12

21

22

2 (5.11)

Between 2o and 'E :

61

( ) feo

oo w

ppzzg

vv!

!!=!+

!

"2

'

2'

22

2'

2 (5.12)

By addition and then grouping, one sees that H1 + H2 = Hn . (5.13) DISCUSSION (Figure 2.40)

- If the circuit characteristic is 1nH : none of the 2 pumps considered separately is able to deliver anything on the circuit : the grouping in series is mandatory.

- If the characteristics is 2nH : pump 1 can deliver something on the circuit but it would give a flow rate and a head that are lower than the pumps grouped in series.

- If the characteristics is 3nH : each of the pumps can work alone but delivering a flow rate and a head lower than the group.

- If the characteristics is 4nH : the grouping is not suited because pump 2 alone can deliver a flow rate with a higher head than the group. There is here no existing working point for the series configuration above the Qmax of pump 1.

CONCLUSION The grouping in series must be reserved for large heads (and low flow rates). REALIZATION OF MULTI-CELLULAR PUMPS In principle, it is not important that the pumps put in series be installed near each other and run at the same rotation speed. But it is highly probable that pumps, mounted on the same support/basement, coupled with each other and driven by one single engine, will be less expensive at the installation. A even more interesting solution is to use a multi-cellular pump where, in one single pump body, we group in series a certain number of pumps with each of them being a stage of the whole pumping system (Figure 1.3). The first stage of the pump comprises:

- 1 distributor - 1 rotor - 1 diffuser

The last stage contains: - 1 return pipe - 1 rotor - 1 collector

Each intermediate stage contains: - 1 return pipe - 1 rotor - 1 diffuser.

The return pipes are fixed channels (very often with vanes) and that have a double function: - maintain the speed constant between the exit of the diffuser and the entrance of the next

rotor; - guide the liquid when entering into the wheel with a given direction.

62

STUDY OF THE MULTI-CELLULAR TPs The theoretical study of multi-cellular TPs is done in the same way as the grouping in series of several identical mono-cellular TPs. One determines the head generated by one stage and one multiply it by the requested number of cells to obtain the head to be provided by the pump. 5.4.3 Grouping of pumps in parallel Let’s consider (Figures 2.41 and 2.42) the pumps 1 and 2 installed in parallel. Curves 1H and 2H are the performance curves of these two pumps (NRV = Non-Return Valve). In steady state, the heads generated by these pumps are equal whereas their flow rate is equal to the sum of the flow rates of the two machines.

nHHH == 21

!

Q =Q1 +Q2 (5.14) The performance curve of the assembly is )(QH on Figure 2.42. It is obtained by making the sum of the flow rates Q for each value of H . Discussion (Figure 2.42)

- If the characteristic of the circuit is 1CH : pump 1 alone can work but pump 2 cannot. To avoid that the flow is reversed in pump 2, one needs to foresee an anti-return valve and then pump 1 works alone.

- If the characteristic is 2CH : pumps 1 and 2 taken separately can deliver on the circuit and the sum of the flow rates will be higher than the flow rate of the pumps grouped in parallel.

- If the characteristic is 3CH : each of the pumps can deliver something individually but with a flow rate lower than the one obtained with the grouping. The grouping in parallel is here highly recommended.

Conclusion The grouping in parallel must be reserved to large flow rates (and low heads). Realization Quite often the pumps are identical and driven, at the same rotation speed, with a single engine. A nice solution is the one of pumps with their rotors made of two symmetrical parts with regard to a plane perpendicular to the shaft and that present two axial inlets (ears) also symmetrical. In addition to their high flow rate, these pumps offer the advantage that they are equilibrated in axial thrust and that the losses frP due to the friction of the rotor with the non-active fluid are about zero.

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Part 1: Fundamental equations Part 2: Turbopumpsatm.ulb.ac.be/CoursATM/Turbo/Turbomachinery_Part1and2_Text_vfin2.… · 3 CHAPTER 1 : BASICS and PRINCIPLES 1. INTRODUCTION An important