Abj1 Lecture 6.3: C-Angular Momentum and Turbomachines 1.Introduction to Fluid Machinery and Turbomachine 1.Classification of Fluid Machinery 2.Turbomachines

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Lecture 6.3: C-Angular Momentum and Turbomachines

1. Introduction to Fluid Machinery and Turbomachine

1. Classification of Fluid Machinery

2. Turbomachines and Their Energy Transfer Aspects

3. Examples of

Axial-Flow Machine

Radial-Flow Machine

Mixed-Flow Machine

Dominant Velocity Components in Axial- and Radial-Flow Machines

2. Performance Parameters

Hydraulic / Fluid Stream VS Impeller VS Mechanical/Shaft

3. Angular Momentum

1. Angular Momentum About A (Fixed) Point C

of A Particle VS of A Continuum Body

2. Angular Momentum and RTT

4. Moment of Forces: Point Force VS Distributive Force

1. Moment of Surface Forces: Pressure and Shear Forces

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5. C-Angular Momentum

1. C-Angular Momentum (MV)

2. C-Angular Momentum (CV)

6. Force, Torque, and Energy Transfer as Work in Turbomachines

1. Relation between

1) Surface Force (Pressure + Shear) on the Moving/Rotating Impeller Surface

2) Energy Transfer as Work at the Impeller Surface, and

3) Impeller Torque and Shaft Torque

2. Impeller Torque VS Shaft Torque

7. Euler Turbomachine Equation, Hydraulic Torque, and The Associated Power Equation

1. Various CV’s and The Corresponding Net/Resultant Moment for Turbomachine Analysis

2. Euler Turbomachine Equation

3. Hydraulic Torque

4. Hydraulic Torque VS Impeller Torque VS Mechanical Torque at Shaft

5. The Associated Power Equation

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8. Analysis for The Performance of Idealized Turbomachines

1. Problem

2. Blade (and Flow ) Angles Convention

3. Sketch the Blade Shape

4. Shockless-Entry/Exit Condition

5. Relative Velocity Relation and Velocity Diagram

6. Is it a pump or a turbine?

7. Hydraulic Power and Hydraulic Head

8. In Circular Cylindrical Coordinates r--z: Axial-Flow Machine

9. Basic of Velocity Diagram

9. Appendix and Review

1. Recall: The Terminologies: Hydraulic / Fluid Stream VS Mechanical / Shaft

2. Recall: Free-Body-Diagram (FBD) Concept

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Very Brief Summary of Important Points and Equations [1]

2. The Associated Power Equation:

)()(: 111222 VUmVUmTW h

)()(

)()(:

111222

12

VrmVrm

mdVrmdVrTTAA

hs

1. Euler Turbomachine Equation:

3. Relative Velocity Relation and Velocity Diagram:

rbVUV

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Introduction to

Fluid Machinery and Turbomachine

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Fluid Machinery

Positive Displacement(Confined volume)

Dynamic/Turbomachines(Dynamic effect between fluid stream and solid

component/rotor)

Classification by Direction of Energy

Input energy into fluid streamPump, fan, compressor

Extract energy from fluid streamHydraulic/Wind/Gas/Steam turbine

Radial-Flow Axial-FlowMixed-Flow

Classification by Direction/Path of Flow (as it passes through the blade passage.)

Classification of Fluid Machinery

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Centrifugal fan (Radial-flow)http://www.nyb.com/frames/products_fr.htm

Turbomachines and Their Energy Transfer Aspect

Hydraulic turbine installation:

http://lmhwww.epfl.ch/Publications/Theses/Mauri/thesis_html/node8.html

PE of a fluid stream is extracted and converted to shaft work OUT

sW

http://www.globalenergyequipment.com/

Streamtube

sW

KE of a fluid stream is extracted and converted to shaft work OUT

12

A fluid stream

Energy extracted from a high-energy fluid stream to be converted to useful shaft work

1

2

A fluid stream

A fluid stream

1

2

High E fluid stream

Low E fluid stream

Low E fluid stream

High E fluid stream

Energy is added to a fluid stream as shaft work INsW

Energy added to a low-energy fluid stream to raise the energy level of the stream

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Axial-Flow Machine

Gas Turbine (Axial Flow)PW-4000 Commercial High Bypass Turbofan Engines

(From http://www.aircraftenginedesign.com/pictures/PW4000.gif)

The dominant velocity components are

• z (axial) – for carrying the flow through the blade passage/machine, and

• (tangential) – for change in angular momentum torque

Axial Flow: Blade Rows(From Fluid Dynamics and Heat Transfer of Turbomachinery,

Lakshminarayana, B., John Wiley & Son., 1996, p. 9.

Photograph courtesy of FIAT.)

zr ezerr ˆˆ eee rz ˆˆˆ

re

e

c

Reference

axisrer ˆ

zez ˆ

z

ze

eVeVV zz ˆˆ

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A Closer View of Turbine Cascade in Axial-Flow Machine

Flow through a turbine cascade, outlet Mach number = 0.68 (~ 200 m/s).

(From Visualized Flow, The Japan Society of Mechanical Engineers,

Pergamon Press, 1988, p.101.)

Turbine Cascade (From http://www.sm.go.dlr.de/SMinfo/IADinfo/IAD.html#Projects

)

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Wind turbine

(From http://www.windpower.org/en/pictures/multimeg.htm)

Duct fan (Axial-flow)

(From http://www.nyb.com/frames/products_fr.htm)

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Radial-Flow Machine

Centrifugal fan (Radial-flow)(From http://www.nyb.com/frames/products_fr.htm)

The dominant velocity components are

• r (radial) – required for carrying the flow through the blade passage/machine,

and

• (tangential) – required for change in angular momentum torque

eVeVV rr ˆˆ

zr ezerr ˆˆ ze

ree

c Reference

axis

rer ˆ

zez ˆ

z

eee rz ˆˆˆ

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Centrifugal pump

(From http://www.peerlessxnet.com/documents/8175_Prod.bmp)

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Mixed-Flow Machine

Radial/Mixed compressor impeller

(From

http://turbo.mech.iwate-u.ac.jp/Fel/turbomachines/stanford/images/

turbo/radial1.html)

eVeVV zz ˆˆ

eVeVV rr ˆˆ

eVeVeVV rrzz ˆˆˆ

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Turbocharger(From http://turbo.mech.iwate-u.ac.jp/Fel/turbomachines/stanford/images/turbo/radial6.html)

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Dominant Velocity Components in Axial- and Radial-Flow Machines

Axial Flow: Blade Rows

(From Fluid Dynamics and Heat Transfer of

Turbomachinery, Lakshminarayana, B., John Wiley &

Son., 1996, p. 9. Photograph courtesy of FIAT.)

Radial Flow (Centrifugal fan) (From http://www.nyb.com/frames/products_fr.htm)

(tangential)

required for change in angular

momentum torque

eVeVV zz ˆˆ

Axial-Flow Machines

eVeVV rr ˆˆ

Radial-Flow Machines

eVeVV nn ˆˆ

n (normal)

required for carrying the flow

through the blade passage/

machine.

)(

)(

radialV

axialVV

r

zn

V

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Performance ParametersHydraulic / Fluid Stream VS Impeller VS Mechanical/Shaft

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Recall: Hydraulic Power VS Impeller Power VS Mechanical Power at Shaft [See also Appendix and Review: Recall The Terminologies: Hydraulic / Fluid Stream VS Mechanical / Shaft ]

Shaft Work

[Mechanical] Energy transfer

as work (between one part of

the solid shaft to another part

of the solid shaft) at the solid

cross section of a shaft.

TWs

1 2

sWsWShaft Power

Impeller work


as work (between the moving

solid impeller and the fluid

stream) at the moving solid

impeller surface. (Solid-Fluid

Interaction]

fW

1 2fW

fWImpeller Power

Hydraulic Power

The actual amount of mechanical energy me that the fluid

stream receives/gives up from inlet 1 to exit 2.

Hydraulic Power

12

)()(:PowerHydraulic 12

AA

mdmemdmeMEME

gzVpme 2

2

1v: Properties of fluid stream, not

those of solid shaft, e.g., pressure p of fluid, velocity V of fluid, etc.

1 2 12 MEME

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With the intended uses of turbomachines, the following performance

parameters are of interest. [See summary next slide.]

1. Torque input (pump, fan, etc.) or torque output (turbine, etc.)

Hydraulic Torque VS Impeller Torque VS Mechanical / Shaft Torque

2. Power input/output.

Hydraulic Power VS Impeller Power VS Mechanical/Shaft Power

3. Other parameters of interest are, e.g.,

flowrate,

hydraulic head,

total and static pressure rise,

etc.

Performance Parameters

Recall Terminologies:

1. Hydraulic/Fluid Stream Quantities

• Properties of a fluid stream,

• Evaluated from fluid stream properties

2. Mechanical / Shaft Quantities

• Properties of a solid shaft

• Evaluated from shaftSee also Appendix and Review:

Recall: The Terminologies:

Hydraulic / Fluid Stream VS Mechanical / Shaft

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Summary of Important Quantities: Hydraulic / Fluid Stream VS Mechanical/Shaft

Hydraulic / Fluid Stream Side(subscripted h)

Mechanical / Shaft Side(subscripted s)

section)crossshaft (soild

)(:A

ss FdrTT

12

)()(:AA

h mdVrmdVrT

Hydraulic Torque Shaft TorqueTorque

1

1

2

2

)()(:

ME

A

ME

A

h mdmemdmeW Hydraulic Power

ss TW :

Shaft PowerPower

Hydraulic HeadHead

gm

WH h

:

Associated Power Equation:

111222 VUmVUm

TW h

12

)()(:

torquehydraulicueshaft torq AA

hs mdVrmdVrTT

Euler Turbomachine Equation:

(for idealized machines)

= Hydraulic torque x shaft angular velocity

(mixed quantity)

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Angular Momentum

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Angular Momentum About A (Fixed) Point C

of A Particle VS of A Continuum Body

Angular momentum of a body about a point c is defined as the

moment of linear momentum of the body about the point c.

Particle Continuum body

r

dmVr

PdrHd

dmVPd

c

:

dVdm

x

yObserver A

C

)(

::tMV

cc dmVrHdmVrPdrHd

][ MomentumLength

r VmP

m

x

yObserver A

C

VmrPrH c

][ MomentumLength

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Angular Momentum and [Recall] The RTT

)(or)()(,))((:

,))(()()(

)(

,

)(through ofeffluxconvectionNet

)(

/

)( of of change of rate Time

,

)( of of change of rate Time

,

tCVtMVtVdVVrH

Time

MomentumAngularAdVVr

dt

tHd

dt

tHd

tV

cV

tCSH

tCS

sf

tCVH

cCV

tMVH

cMV

ccc

Reynolds Transport Theorem (RTT): VrVmrPrHN c

,

Coincident MV and CV r

dmVr

PdrHd

dmVPd

c

:

dVdm

x

yObserver A

C

MV(t) and CV(t)

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Example 1: Finding The Time Rate of Change of The Angular Momentum of an MV By The Use of A Coincident CV and The RTT

Problem: Given that the velocity field is steady-in-mean and the flow is

incompressible, and we evaluate the mean properties.

1. State whether or not the time rate of change of the angular momentum about the z-

axis passing through point c (Hc,z) of the material volume MV(t) that instantaneously

coincides with the control volume CV shown below vanishes.

2. if not, state also

- whether they are positive or negative, and

- whether there should be the corresponding net moment (Mz) acting on the

MV/CV, and

- whether the corresponding net moment is positive or negative.

3. Find it.

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x

y

c+

1V

2V

1r

2r

m

12 VV

1Vm

2V

c+

r

r

1Vm

2V

c+

r

r

tangential exit

axial inlet

+ c

1Vm z

r

2V

radial exit

axial inlet

+ c

1Vm z

r

2V

?

?

,

,,

zc

zcMV

Mdt

dH

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Moment of Forces

Point Force VS Distributive Force

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Moment of Forces about A Point C: Point Force VS Distributive Force

Point Force

FrM c

:

][ ForceLength

r

F

C

dATFd

Surface Force: )( closedoropenA

c dATrM results in

Surface Integral

dVgFd )(

Volume Force: V

c dVgrM )(

results in

Volume Integral

Distributive Force

FdrMFdrMd cc

][ ForceLength

rC

r dATFd S

dVgdmgFd B )(

Ad

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FdrM c

Surface Force:

Volume Force:

AA

c dATrFdrM

Surface Integral

V

c dVgrFdrM )(

Volume Integral

Because the evaluation of the moment of distributive forces results in

It is to be understood that when we simply write

we refer to the corresponding surface or volume integral.

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Moment of Surface Forces: Pressure and Shear Forces

Moment about c due to pressure force:

AA

ppcppc ApdrFdrMAdprFdrMd

,, )(

Moment about c due to shear/frictional force:

A

cc FdrMFdrMd

,,

Note: The shear force – and in fact all the surface forces - can be written in terms of the area vector as

where is the stress tensor corresponding to that force.

Because it is a little more complicated at this point, we shall leave it at that.

AdFd

c

ApdFd p

Ad

Fd

r

AOpen surface

Closed surfaceC

r

ApdFd p

Ad

Fd

A

To find the net moment of a surface force (pressure or

shear) on an MV, we just simply sum (integrate) it over the

whole MS.

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C-Angular Momentum [ MV ]

Here, we shall limit ourselves to an observer in IFR only.

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C-Angular Momentum [MV]

Physical Law (for an observer in an IFR and a fixed point C)

)(

,

,

))((:

,,)(

tMV

cMV

cMVc

dVVrH

Time

MomentumLength

Time

MomentumAngular

dt

tHdM

cM

= Net/Resultant moment of all external forces/moments (hence, FBD again)

on the MV(t) about a fixed point c.

x

yObserver A, IFR

FBD(again)

r

Pressure pShear

iF

)( dVgdmg

c

Concentrated force

Solid part

Distributive force

V

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Net/Resultant moment of all external forces/moments

(hence, FBD again) on the MV(t) about a fixed point c

cM

It is emphasized that in the physical law

is

the net/resultant moment of all external forces/moments (hence, FBD again)

on the MV(t) about a fixed point c.

dt

tHdM cMV

c

)(,

cM

Recall the FBD Concept

In C-Angular Momentum, we need to sum all the external moments of all the external

forces/moments about C on the MV,

cM FBD

Concept

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Recall in case of fluid flow: Forces in fluids (or solid – i.e., continuum, - for that matters)

Forces in Fluid: Line Force + Surface Force + Body Force

Surface forces = Pressure/Normal + Friction/Tangential

Need to sum the contributions due to surface forces on all surfaces of the MS.

FBD Concept

1. Define the system of interest clearly.

2. Know and Recognize various types of forces (/moments).

3. With 2, recognize all forces (/moments) that act on the current system

of interest.

4. Know how to find their

1. resultant force , and

2. resultant moment

F

cM

FBD Concept

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Examples

sT

...)(

,, tMV

si

icpcc dmgrTFrMMM

r

Pressure p

Shear

)( dVgdmg cSolid part

Normal pShear

r

Pressure p

Shear iF

)( dVgdmg c

Concentrated force

Solid part

Distributive forcein fluid

Cross section of a solid shaft

sT

Normal p

Distributive forcein solid

Shear

• For surface force, we need to sum the contributions of surface forces over all surfaces of MV, and

both solid and fluid parts.

• For solid part, p refer to normal stress and it must also be included in

• Recognize that is in fact the resultant moment/couple of shear stress (surface force)

distribution over a cross section of a solid shaft.

sT

pcM ,

...)(

,, tMV

cpcc dmgrMMM

A

c

A

ppc

FdrM

FdrM

,

,

surface force surface integral

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Note on Notations

• Thus, depending on the forces and moments that act on the current system of interest, the specific

form for depends on that specific force/moment system.

• According to the physical law of C-Angular Momentum (hence, FBD)

= sum of all external moments due to all external forces/moments on MV

• Generically, though, we may simply write

or even

to emphasize the distributive nature of forces in fluid.

• However, for example, if there are concentrated forces on the MV, it is to be understood that all

these forces must be taken into account in , e.g.,

cM

...)(

,, tMV

cpcc dmgrMMM

cM

...)(

,, tMVi

icpcc dmgrFrMMM

cM

)(

,,

tMV

cpcc dmgrMMM

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C-Angular Momentum [ CV ]

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C-Angular Momentum [CV]

dt

tHdM cMV

c

)(,

Physical Law (for an observer in IFR and a fixed point C)

)(or)( is)(,))((:

))(()()(

)(

,

)(

/,,

tCVtMVtVdVVrH

AdVVrdt

tHd

dt

tHdM

tV

cV

tCS

sfcCVcMV

c

Physical Laws

RTT

C-Angular Momentum

)(

/,, ))((

)()(

tCS

sfcCVcMV AdVVr

dt

tHd

dt

tHd

VrVmrPrHN c

, RTT

Coincident MV(t) and CV(t)

)(

,,

tMViicpcc dmgrFrMMM

r

Pressure pShear iF

)( dVgdmg c

Concentrated force

Solid part

Distributive force

V

MV(t)

CV(t)

V

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Force, Torque, and Energy Transfer as Work in

Turbomachines

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Relation between 1) Surface Force (Pressure + Shear) on the Moving/Rotating Impeller Surface2) Energy Transfer as Work at the Impeller Surface, and 3) Impeller Torque and Shaft Torque

Centrifugal fan (Radial-flow, from http://www.nyb.com/frames/products_fr.htm)

,

,

)()(

,

,

c

c

MdFdrFdr

FdVW

FdrMd

Fd

According to the Energy-Work Principle:

A moving solid body in a fluid stream (in this case, a

moving/rotating solid impeller) is required in order to

have energy transfer as work between the solid body

and the fluid stream.

pcpp

pp

ppc

p

MdFdrFdr

FdVW

FdrMd

Fd

,

,

)()(

,

,

Force results in Energy Transfer as work

Force results in Torque

rV

c

r

Ad

sT

.3

z

Impeller and shaft as a system

Fd

.2

pFd

.1

fT

.4

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Note that

1. shaft torque, and

2. impeller torque,

are related through the FBD of the section of the solid impeller as

shown.

However, they are generally not equal due to, e.g.,

frictional torque, at the bearings .

sT

impellercM ,

fT

Impeller Torque VS Shaft Torque

impellercs MT ,

Impeller Torque

defined as the net moment due to surface force on the impeller

impellercM ,

impellerimpeller

p

cpcimpellerc

FdrFdr

MMM

,,,

fT

sT

pFd

Fd

z

c

r

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Euler Turbomachine Equation,

Hydraulic Torque,

and The Associated Power Equation

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Various MV/CV’s and The Corresponding Net/Resultant Moment for Turbomachine Analysis

CV1/MV1: CV includes fluid stream, impeller, and

part of solid shaft

CV2/MV2: CV includes fluid stream only, no solid part.

sT

= Shaft torque

s

CVMVCSMSCSMS

p

s

tMV

cpcc

TdmgrFdrFdr

TdmgrMMM

)(

,,

CVMVCSMSCSMS

p

tMV

cpcc

dmgrFdrFdr

dmgrMMM

)(

,,

c zdmg

sT

Ad

pFd

Fd

r

1

2

r

c z

AdpFd

Fd

dmg

Ad

pFdFd

1

2

r

Ad

pFd

Fd

sT

Moving solid

impeller surface

(Recall the outward

normal to the system of

interest)

Recall the concept of FBD

No shaft torque for CV2/MV2Shaft torque for CV1/MV1

cMFBDCVMV

and/

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Example 1: Radial-Flow Machines [1]

Centrifugal fan (Radial-flow)(From http://www.nyb.com/frames/products_fr.htm)

Inlet 1 = Inner cylindrical surface

Exit 2 = Outer cylindrical surface

r1 1V

2V

sT

pFd

Fd

Ad

pFdFd

dmg

1

2

z

c

r

Ad

pFd

Fd

dmg

Ad

pFdFd

View this waysT

at solid shaft cross section

1V

2V

1

2

CV/MV: CV includes fluid stream, impeller (and its

back plate), and cuts across the solid shaft at the

back plate.

s

CVMVCSMSCSMS

p

s

tMV

cpcc

TdmgrFdrFdr

TdmgrMMM

)(

,,

cMFBDCVMV

and/

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Note on NotationsTo avoid messy diagram, it is to be understood that

to evaluate , all moments due to all forces must be taken into account.

• For surface forces, need to sum the surface forces on all surfaces of the MS/CS

• Surface forces = Pressure/Normal + Friction/Tangential

Hence, to remind us once in a while or to focus on some surfaces, we shall simply use the

figure of

the area vector or simply in place of the more complete one

and only on some surfaces instead of on all surfaces.

cM

pFd

Ad

FdAd

Ad

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Example 2: Radial-Flow Machines [2]

r1

sT

1

2

Example 1 CV/MV

s

tMV

cpcc TdmgrMMM

)(

,,

We can also include the fluid stream only.

• This time we will not see the shaft torque

• But we see instead additional forces (pressure and shear) on the impeller surface (fluid as a

system) whose moment must be taken into account in and .

sT

pcM ,

,cM

For clarity, forces on other surfaces

are omitted from the diagram.

)(

,,

tMV

cpcc dmgrMMM

CV/MV: CV includes fluid stream only, no solid part.

r1

1

2

pFd

Fd

Ad

cMFBDCVMV

and/

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Example 3: Radial-Flow Machines [3] [Compare to Example 1]

z

1V

View this way

r1

2V

sT

pFd

Fd

Ad

pFdFd

dmg

Inlet 1 = Inlet duct circular cross section

Exit 2 = Outer cylindrical surface

CV/MV: CV includes fluid stream, impeller (and its

back plate), and cuts across the solid shaft at the

back plate. It also includes part of the axial inlet duct.

s

CVMVCSMSCSMS

p

s

tMV

cpcc

TdmgrFdrFdr

TdmgrMMM

)(

,,

z

c

r

Ad

dmg

Ad

View this way

sT

2V

2

1

1V

cAd

r

cMFBDCVMV

and/

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Example 4: Axial-Flow Machines

Turbine Cascade (From http://www.sm.go.dlr.de/SMinfo/IADinfo/IAD.html#Projects)

CV covers the whole rotor and cuts through the cross section

of a solid shaft.

Note, however that the inlet and exit planes are annular.

Annular exit planeAnnular inlet plane

z2V

1V

csT

Ad

Ad

A single blade

s

tMV

cpcc TdmgrMMM

)(

,,

cMFBDCVMV

and/

abj 47

Euler Turbomachine EquationCoincident MV(t) and CV(t)

z

r1 1V

2V

sT

pFd

Fd

Ad

pFdFd

dmg

1

2

fT

c zdmg

,sT

Ad

pFd

Fd

r

1

2CV/MV

includes fluid stream, impeller, part of solid

shaft, and inlet and exit ducts. It cuts across a

cross section of the solid shaft.

This CV illustrates the global nature of the Euler

Turbomachine equation - without having to deal with the

blade geometry, i.e., torque = net angular momentum

efflux.

CV/MV [CV in Example 1]

includes fluid stream, impeller (and its back

plate), and cuts across a cross section of the

solid shaft (and no inlet and exit ducts).

Later, this CV is used in developing the associated power

equation.

abj 48

)(or)( is)(,))((:

))(()()(

)(

,

)(

/,,

tCVtMVtVdVVrH

AdVVrdt

tHd

dt

tHdM

tV

cV

tCS

sfcCVcMV

c

C-Angular Momentum

Assumptions

1. All flow properties are steady in mean.

2. Evaluate mean properties.

3. Incompressible flow

4. Neglect all other torques, except shaft torque.

[Neglect torques due to surface force

(pressure + shear), body force (mg),

frictional torque at bearings, etc.].

fT

c zdmg

,sT

Ad

pFd

Fd

r

1

2

abj 49

12

)()())(()(

/

AAtCS

sf mdVrmdVrAdVVr

]propertiesmeaninsteady[0))((

CV

dVVrdt

d

Unsteady Term:

Net Convection Efflux Term:

Net Moment about c: que.]shaft torexcept ques,other tor allNeglect [sc TM

C-Angular Momentum becomes

12

)()(AA

s mdVrmdVrT


fT

c zdmg

,sT

Ad

pFd

Fd

r

1

2

r1 1V

2V

sT

pFd

Fd

Ad

pFdFd

dmg

1

2

abj 50

Hydraulic Torque

12

)()(AA

s mdVrmdVrT

Euler’s Turbomachine Equation:

In a similar manner as hydraulic power, we define the RHS (and the RHS only)

as the hydraulic torque.

Hydraulic Torque : = the time rate of change of angular momentum of

the fluid stream as it flows through CV [from CV

inlet 1 to CV exit 2].

Hydraulic Torque [MV Viewpoint] Like hydraulic power, through the RTT, we can see that the two views are equivalent.

Hydraulic Torque [CV Viewpoint]

12

)()(:)(TorqueHydraulicAA

h mdVrmdVrT

= Net convection efflux of angular momentum through CS.

)( TorqueHydraulic:)()(

12

h

AA

s TmdVrmdVrT

abj 51

Hydraulic Torque is

• the property of fluid stream, not shaft

• evaluated from the properties of fluid stream, not shaft.

12

)()(:)(TorqueHydraulicAA

h mdVrmdVrT

Properties of fluid stream

abj 52

In an ideal case (idealized machine)

where the flow has

• steady-in-mean flow properties

• no other torque except shaft torque,

the Euler Turbomachine Equation states that

Euler Turbomachine Equation

12

)()(:

TorqueHydraulicTorqueShaft AA

hs mdVrmdVrTT


Shaft Torque = Hydraulic Torque

hs TT

Ideal Case

abj 53

In a real case, however, since there are other torques

• moments due to pressure and friction on MS/CS,

• frictional torque at bearings, etc.,

acting on the MV/CV, the two torques are not equal.

fT

c zdmg

Ad

pFd

Fd

r

1

2

12

)()(AA

h mdVrmdVrT

sT

Shaft Torque

Hydraulic Torque

hs TT

Real Case

abj 54

Assumption 5: Further Assume Uniform

Assumption 5: Uniform at each cross section. Or, evaluate at

some reference radius.

)( Vr )( Vr

12

)()(AA

s mdVrmdVrT

)()( 111222 VrmVrmTs


V

= Velocity of fluid at inlet/exit

)( Vr

fT

c zdmg

,sT

Ad

pFd

Fd

r

1

2

mmm :21 C-Mass also gives

abj 55

Recall: Hydraulic Power VS Impeller Power VS Mechanical Power at Shaft [See also Appendix and Review: Recall The Terminologies: Hydraulic / Fluid Stream VS Mechanical / Shaft ]

Shaft Work


as work (between one part of

the solid shaft to another part

of the solid shaft) at the solid

cross section of a shaft.

TWs

1 2

sWsWShaft Power

Impeller work


as work (between the moving

solid impeller and the fluid

stream) at the moving solid

impeller surface. (Solid-Fluid

Interaction]

fW

1 2fW

fWImpeller Power

Hydraulic Power

The actual amount of mechanical energy me that the fluid

stream receives/gives up from inlet 1 to exit 2.

Hydraulic Power

12

)()(:PowerHydraulic 12

AA

mdmemdmeMEME

gzVpme 2

2

1v: Properties of fluid stream, not

those of solid shaft, e.g., pressure p of fluid, velocity V of fluid, etc.

1 2 12 MEME

abj 56

In the Same Manner to PowerHydraulic Torque VS Impeller Torque VS Mechanical Torque at Shaft

Shaft Torque

Net moment due to

surface stress (shear

stress) distribution over a

cross section of a solid

shaft

sT

sT

Shaft Torque

Impeller Torque

defined as the net moment

due to surface force on the

impeller

impellercM ,

Impeller Torque

impellerimpeller

p

cpcimpellerc

FdrFdr

MMM

,,,

sT

pFd Fd

z

cr

Hydraulic Torque

the time rate of change of angular momentum of the

fluid stream as it flows through CV [from CV inlet 1 to

CV exit 2].

12

)()(:AA

h mdVrmdVrT

1 2 hT

Hydraulic Torque hT

• With the concept of resultant moment,

we can see that can be written in terms

of stress in the same manner as

• The area integral in this case is the solid

shaft cross section.

• For this generic definitions, we leave the

reference point and axis unspecified first.

sT

impellercM ,

abj 57

The Associated Power Equation

)()( 111222 VrmVrmTT hs


With the Euler’s Turbomahine Equation, which gives an ideal shaft

torque, we can find the mechanical power of the shaft as

The Associated Power Equation: )()(: 111222 VUmVUmTW h

)()(

:,)()(:

)()(

111222

111222

VUmVUmW

rUVUVrVr

VrmVrm

TTW sh

is the tangential velocity (of the blade).

If we focus on the impeller and look at the CV below:

r1 1V

2V

sT

1

22U

1U

Essentially, we define the associated power from the hydraulic torque not shaft torque sT

hT

W

abj 58

In SummaryAssumptions





[Neglect torques due to surface force (pressure + shear),

body force (mg), frictional torque at bearings, etc.].

The Associated Power Equation: )()(: 111222 VUmVUmTW h

)()(

)()(:

111222

12

VrmVrm

mdVrmdVrTTAA

hs


Assumption 5: Uniform at each cross

section. Or, evaluate at some

reference radius.

)( Vr

)( Vr

V

= Velocity of fluid at inlet/exit

ii rU

= The tangential velocity (of the blade).

fT

c zdmg

,sT

Ad

pFd

Fd

r

1

2r1 1V

2V

sT

1

22U

1U

abj 59

Analysis for The Performance of

Idealized Turbomachines

abj 60

Problem: Idealized Turbomachine Performance Analysis

The Associated Power Equation:

)()( 111222 VUmVUmW



Need to find the two

kinematical unknowns

),( 21 VV

Given

1. geometry and geometric parameters

• blade angles

• radii

2. kinematical parameter: angular velocity

3. mass flowrate

Questions

1. Sketch the blade shape, and

Find the idealized

2. shaft torque (= hydraulic torque)

3. shaft power (= hydraulic power, = 1)

4. hydraulic head.

),( 21

m

),( 21 rr

r1 1V

2V

sT

1

22U

1U

abj 61

How to tell the geometry of the blade:

Blade (and Flow Angles Convention

For an idealized machine where the shockless entry/exit

condition is applied, this is the direction of the relative velocity

of fluid wrt an observer moving with the blade, .rbV

CS

V

)or( rzn VVV

U

tangent to blade

at inlet/exit

rbV

V

U

= solid blade tangential speed

= absolute fluid velocity (relative to IFR)

= relative velocity of fluid wrt moving blade

= the normal (to CS) component of absolute fluid velocity

= blade angle

= the angle that the blade tangent at inlet/exit

makes with (measured away from )

= flow angle

= the angle that the absolute flow velocity

makes with (measured towards )

V

U

nV

nV

nV

U

VrbV

abj 62

Example of Blade Angles in Axial- and Radial-Flow Machines

r1

1

2

nV

U

2

nV

U

1

Radial-flow machine

Centrifugal fan

(From http://www.nyb.com/frames/products_fr.htm) 1

2

Flow

U

U

nV

U

nV

U

1

2

1

2

Axial-flow machine

Turbine Cascade (From http://www.sm.go.dlr.de/SMinfo/IADinfo/IAD.html#Projects

)

1

2

Flow

U

abj 63

Example: Sketch The Blade Shape: Axial-Flow Machine

flow

z

Slope/angle increases from 30o to 60o

CurvatureU

nV

U

nV

U

nV

U

nV

1

1

2 2

Example: Sketch an axial-flow machine blade with 1 = 30o , 2 = 60o

flow

flow

z

Slope/angle decreases from 60o to 30o

CurvatureU

nV

U

nV

U

nV

U

nV

1

1

2 2

Example: Sketch an axial-flow machine blade with 1 = 60o , 2 = 30o

flow

12 Blade concave towards

the direction of U

12 Blade convex towards

the direction of U

abj 64

U

nV

1

U

2

Example: Sketch a radial-flow machine blade with 1 = 90o, 2 = 45o

Example: Sketch a radial-flow machine blade with 1 = 90o, 2 = 135o

2

nV

U

nV

1

U

2

nV

Backwardly-curved blade

(wrt the direction of angular rotation)

Forwardly-curved blade

(wrt the direction of angular rotation)

Example: Sketch The Blade Shape: Radial-Flow Machine

abj 65

Whether it is an axial- or radial-flow machine, we can represent the blade and the

kinematics of the flow through the blade by the cascade diagram

U

nV U

1

2

1

2

nV U

abj 66

Shockless-Entry/Exit ConditionFurther Assumption for Idealized Turbomachines

Assumptions (Idealized Turbomachines)





[Neglect torques due to surface force (pressure + shear),

body force (mg), frictional torque at bearings, etc.].

5. Shockless entry/exit condition

Shockless-Entry/Exit Condition:

is tangent to the blade at inlet and exit.rbV

The direction of the relative velocity of the fluid

with respect to the moving blade (wrt an

observer moving with the blade) is tangent to

the blade at inlet/exit.

U

nV U

1

2

1

2nV U

1rbV

2rbV

abj 67

Relative Velocity Relation and Velocity DiagramRecall our Problem

Need to find the two

kinematical unknowns

),( 21 VV

Given

1. geometry and geometric parameters

• blade angles

• radii

2. kinematical parameter: angular velocity

3. mass flowrate

Questions

1. Sketch the blade shape, and

Find the idealized

2. shaft torque (= hydraulic torque)

3. shaft power (= hydraulic power, = 1)

4. hydraulic head.

),( 21

m

),( 21 rr

r1 1V

2V

sT

1

22U

1U

The Associated Power Equation:

)()( 111222 VUmVUmW



abj 68

U

nV U

1

2

1

2nV U

1rbV

2rbV

rbVUV


)()( 111222 VUmVUmW



3. Relative Velocity Relation:

rbVUV

abj 69

rbVUV

Velocity diagram can be used as a graphical/geometrical

aid in solving the relative velocity vector relation.

2U

U

nV U

1

2

1

2nV U

1rbV

2rbV

1

1U

1rbV

1V

Inlet

1nV

2

2rbV

2V

Exit

2nV

abj 70

Velocity Diagram for Axial-Flow Machines

U

nV U

1

2

1

2 nV U

1rbV

2rbV

z

2V

1V

c

1 2

Exit

2

2rbV

2V

1

U

1rbV

1V

Inlet

21 nn VV

12 VVV

1. For simplicity we evaluate the properties at the mean radius of

the CV shown on the left. Thus, since

2. From C-Mass, we also find

It is then recommended that the velocity diagrams at inlet and exit

be superimposed on the same base and the velocity diagrams

look like below.

UUUrr

:2121

2121 nn VVAA

U

abj 71

Is it a pump or a turbine?

Recall that if

,0

,0

TW

Energy is input into the system Pump

Energy is extracted from the system Turbine

System / MV

T T

is an external torque acting on the system.

abj 72

r1 1V

2V

sT

1

22U

1Uz

2V

1V

csT

1 2

2U

1U

0

0

)()(: 111222 VUmVUmTW h

Energy is input into the stream Pump

Energy is extracted from the stream Turbine

From the associated power equation, then for an idealized machine

abj 73

Example: Axial-Flow Machine: Is it A Pump or A Turbine?

U

1

2

1

2

2

2rbV

2V

1U

1rbV

1V

21 nn VV

12 VVV

)ofdirectiontheinconcave(12 U

pump0

)(

)()(:

12

111222

VUm

VVUm

VUmVUmTW h

U

1

2

1

2

1

1rbV

1V

2U

2rbV

2V

21 nn VV

12 VVV

)ofdirectiontheinconvex(12 U

turbine0: VUmTW h

abj 74

[Ideal] Hydraulic Power and Hydraulic Head


)()( 111222 VUmVUmW




rbVUV

If there is no me loss in the machine

( = 1), we have

)()(1

Head Hydraulic

)()(

)()(

PowerShaftPowerHydraulic

1122

1122

111222

VUVUg

H

VUmVUmgHm

VUmVUm

abj 75

In Circular Cylindrical Coordinates r--z: Axial-Flow Machine

Axial- Flow Machine

eVeVV

eUU

e

eTT

zz

z

zss

ˆˆ

ˆ

ˆ

ˆ

Recall in circular cylindrical coordinates:

)(ˆˆ,ˆˆ rrzr eeezerr eee rz ˆˆˆ Right-handed coordinates:

Position vector:

ree

Reference axis

r

zc

sT

1 2

zer

2V

1V

U

1

2

abj 76

In the Circular Cylindrical Coordinates r--z: Radial-Flow Machine

Radial- Flow Machine

eVeVV

eUU

e

eTT

rr

z

zss

ˆˆ

ˆ

ˆ

ˆ

r1 1V

2V

sT

1

2

1U

2U

z

c

2Vre

eze

Reference axis

r

sT

1V

1

2

r

Recall in circular cylindrical coordinates:

)(ˆˆ,ˆˆ rrzr eeezerr eee rz ˆˆˆ Right-handed coordinates:

Position vector:

abj 77

In the Circular Cylindrical Coordinates r--z

2. The Associate Power Equation:

111222

111222

ˆ:

),()(

VUmVUmW

eUU

VUmVUmW

111222

111222

ˆ)()(:

)ˆˆˆ()ˆˆ(:

)()(ˆ

VrmVrmT

erVVr

eVeVeVezerVr

VrmVrmeTT

s

zz

zzrrzr

zss



eVUeVeVeV

eVeVV

eUU

eVeVV

VUV

rbnnrbnn

rbnnrbrb

nn

rb

ˆˆˆˆ

ˆˆ:

ˆ:

ˆˆ:

,,

,,

abj 78

2. Decomposition into two systems of coordinates.

a.

b.

• This is a system of 2 equations in 5 unknowns:

• Require the knowledge of 3 to solve for 2. • For example:

• If geometry and speed are given,

are known. • may be found from C-Mass (e.g.,

flowrate is given).

The relations between velocities can be illustrated as

follows.

1. Relative velocity relation:

Regardless of whether it is

• a radial-flow or an axial-flow machine

• at inlet or exit,

the relative velocity relation, which is a kinematic relation,

holds. Thus, we have

Basic of Velocity Diagram

rbVUV

ee ˆˆ een ˆˆ

rbVUV

rbn VVVU ,,,,

,U

U

nV

rbV

V

V

e

tangent to blade at ie

ne

nV

eee n ˆcosˆsinˆ:

cos

sin

rb

rbn

VUV

VV

eVUeV

eeVeUeVeV

rbnrb

nrbnn

ˆcosˆsin

ˆcosˆsinˆˆˆ

eVeUeVeV rbnn ˆˆˆˆ

abj 79

From Fox, R. W., McDonald, A. T., and Pritchard, P. J., 2004, Introduction to Fluid Mechanics, Sixth Edition, Wiley, New York.

abj 80From Fox, R. W., McDonald, A. T., and Pritchard, P. J., 2004, Introduction to Fluid Mechanics, Sixth Edition, Wiley, New York.

abj 81

Appendix and Review Recall: The Terminologies:

Hydraulic / Fluid Stream VS Mechanical / Shaft

Recall: Free-Body-Diagram (FBD) Concept

abj 82

Recall: The Terminologies: Hydraulic/Fluid Stream VS Mechanical/Shaft

Hydraulic / Fluid Stream: The term is used to refer to the quantities that are

• associated with, and evaluated from, the properties of the fluid stream (fluid stream side).

They are properties of the fluid stream.

Examples:

• Hydraulic torque is the torque that is evaluated from – or equivalent to - the change in the angular momentum flux of a fluid

stream.

• Hydraulic power is

• the mechanical power gained by a fluid stream (pump), or

• the mechanical power removed from a fluid stream (turbine).

In the case of steady incompressible stream, .

12

12

)()(:)(TorqueHydraulic

)()(:)(PowerHydraulic 12

AA

h

AA

h

mdVrmdVrT

mdmemdmeMEMEW

Subscript h

Mechanical / Shaft: The term is used to refer to the quantities that are associated with the mechanical / shaft side.

Examples:

• Mechanical / shaft torque is the torque that is evaluated/measured at the shaft (resultant moment of shear stress

distribution over a cross section of a solid shaft) .

• Mechanical power at shaft is .

Recall that that the hydraulic quantities and the mechanical quantities may be related but, e.g., due to friction, they are

generally not equal.

ss TW Subscript s

abj 83

1. The Free-Body-Diagram (FBD) concept is related to the Newton’s Second Law:

Recall: Free-Body-Diagram (FBD) Concept

dt

tHdMM

dt

tPdFF

cMVcc

MV

)()simplicityfor simply or (

)()simplicityfor simply or (

,

where = resultant/net external force on MV

= sum of all external forces on MV

= resultant/net external moment about a fixed

point c of all external forces/moments on MV

= sum of all external moments about

a fixed point c of all external forces/

moments on MV

)simply or ( FF

)simply or ( cc MM

abj 84

2. Hence, in the application of these laws, we need to be able to find

and

3. Hence, “FBD” here does not refer to the diagram per se, but refers to the fact that we need to be

able to

)simply or ( FF

)simply or ( cc MM

F

cM

FBD Concept


2. Know and Recognize various types of forces (/moments, and their natures) that can act on

the system.

1. We then classify various types of forces (/moments) in order to be able to take them

into account in the application of FBD and the Newton’s second law systematically

and effectively.

[e.g., line, surface (pressure + friction), and body forces, etc.]

3. With 2, recognize all forces (/moments) that act on the current system of interest.

4. Know how to find their resultant force and their resultant moment

Note:

• Due to the complexity of the surface geometry of turbomachines, the surface forces may not be drawn on all surfaces in the diagram for FBD.

• However, the concept of the FBD in relation to the application of the Newton’s second law as mentioned above [i.e., we need to take into

account and sum all the external forces/moments on MV] should be kept in mind.

[Recall that moment is an effort of force in causing angular motion.]

abj 85

FBD Concept


2. Know and Recognize various types of forces (/moments).

3. With 2, recognize all forces (/moments) that act on the current system

of interest.

4. Know how to find their

1. resultant force , and

2. resultant moment

F

cM

Documents

Abj1 Lecture 6.3: C-Angular Momentum and Turbomachines 1.Introduction to Fluid Machinery and Turbomachine 1.Classification of Fluid Machinery 2.Turbomachines