Transcript Slide 1

Lecture 6.3:
1.
2.
3.
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
Performance Parameters
Hydraulic / Fluid Stream
VS
Impeller
VS
Mechanical/Shaft
Angular Momentum
1.
2.
4.
C-Angular Momentum and Turbomachines
Angular Momentum About A (Fixed) Point C
of A Particle
VS
of A Continuum Body
Angular Momentum and RTT
Moment of Forces:
Point Force VS
Distributive Force
abj
1
1.
Moment of Surface Forces: Pressure and Shear Forces
5.
6.
7.
abj
C-Angular Momentum
1.
C-Angular Momentum (MV)
2.
C-Angular Momentum (CV)
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
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
5.
The Associated Power Equation
VS Impeller Torque
VS
Mechanical Torque at Shaft
2
8.
9.
abj
Analysis for The Performance of Idealized Turbomachines
1.
Problem
2.
Blade b (and Flow a) 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-q-z: Axial-Flow Machine
9.
Basic of Velocity Diagram
Appendix and Review
1.
Recall:
The Terminologies:
Hydraulic / Fluid Stream
2.
Recall:
Free-Body-Diagram (FBD) Concept
VS
Mechanical / Shaft
3
Very Brief Summary of Important Points and Equations [1]
1. Euler Turbomachine Equation:


 




Ts  Th  :  (r  V )dm   (r  V )dm 
 A

A
2
1






 m 2 (r2  V2 )  m 1 (r1  V1 )
2. The Associated Power Equation:


 
 
 

 2 (U 2 V2 )  m
 1 (U1 V1 )
W : Th    m
3. Relative Velocity Relation and Velocity Diagram:
  
V  U  Vrb
abj
4
Introduction to
Fluid Machinery and Turbomachine
abj
5
Classification of Fluid Machinery
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 stream
Extract energy from fluid stream
Pump, fan, compressor
Hydraulic/Wind/Gas/Steam turbine
Classification by Direction/Path of Flow
(as it passes through the blade passage.)
abj
Radial-Flow
Mixed-Flow
Axial-Flow
6
Turbomachines and Their Energy Transfer Aspect
Energy extracted from a high-energy fluid stream to
be converted to useful shaft work
1
A fluid stream
Energy added to a low-energy fluid stream to raise
the energy level of the stream
PE of a fluid stream is extracted and
converted to shaft work OUT
W s
Energy is added to a fluid stream
as shaft work IN
W s
2
Hydraulic turbine installation:
Low E fluid
stream
http://lmhwww.epfl.ch/Publications/Theses/Mauri/t
hesis_html/node8.html
A fluid stream
Streamtube
High E fluid
stream
1
2
1
Centrifugal fan (Radial-flow)
http://www.nyb.com/frames/products_fr.htm
A fluid stream
High E fluid
stream
2
Low E fluid
stream
W s
KE of a fluid stream is extracted and
converted to shaft work OUT
http://www.globalenergyequipment.com/
abj
7
Axial-Flow Machine
eˆ z
The dominant velocity components are
• z (axial)
•
– for carrying the flow through the blade passage/machine, and
q (tangential) – for change in angular momentum 

V  Vz eˆz  Vq eˆq
eˆ r
z
torque
eˆz  eˆr  eˆq
Reference
axis
zeˆ z
reˆr
eˆq

r  reˆr  zeˆ z
q
c
abj
Gas Turbine (Axial Flow)
PW-4000 Commercial High Bypass Turbofan Engines
(From http://www.aircraftenginedesign.com/pictures/PW4000.gif)
Axial Flow: Blade Rows
(From Fluid Dynamics and Heat Transfer of Turbomachinery,
Lakshminarayana, B., John Wiley & Son., 1996, p. 9.
8
Photograph courtesy of FIAT.)
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.)
abj
Turbine Cascade (From
http://www.sm.go.dlr.de/SMinfo/IADi
nfo/IAD.html#Projects)
9
Duct fan (Axial-flow)
(From http://www.nyb.com/frames/products_fr.htm)
abj
Wind turbine
(From http://www.windpower.org/en/pictures/multimeg.htm)
10
Radial-Flow Machine
The dominant velocity components are
• r (radial) – required for carrying the flow through the blade passage/machine, and
•
q (tangential) – required for change in angular momentum 

V  Vr eˆr  Vq eˆq
torque
eˆz  eˆr  eˆq
eˆ r
eˆq
eˆ z

r  reˆr  zeˆ z
q
reˆr
zeˆ z
c
Reference
axis
z
abj
Centrifugal fan (Radial-flow)
(From http://www.nyb.com/frames/products_fr.htm)
11
Centrifugal pump
(From http://www.peerlessxnet.com/documents/8175_Prod.bmp)
abj
12
Mixed-Flow Machine

V  Vz eˆz  Vr eˆr  Vq eˆq

V  Vz eˆz  Vq eˆq

V  Vr eˆr  Vq eˆq
abj
Radial/Mixed compressor impeller
(From http://turbo.mech.iwateu.ac.jp/Fel/turbomachines/stanford/images/turbo/radial1.html)
13
Turbocharger
abj
(From http://turbo.mech.iwate-u.ac.jp/Fel/turbomachines/stanford/images/turbo/radial6.html)
14
Dominant Velocity Components in Axial- and Radial-Flow Machines

Axial-Flow Machines V

Radial-Flow Machines V
 Vz eˆz
 Vq eˆq
 Vr eˆr
 Vq eˆq

V
 Vn eˆn
 Vq eˆq

Vq

 
 Vz
Vn   

Vr
n
(normal)
required for carrying the flow
through the blade passage/
machine.
abj
(axial)
(radial)
q
Axial Flow: Blade Rows
(From Fluid Dynamics and Heat Transfer of
Turbomachinery, Lakshminarayana, B., John Wiley &
Son., 1996, p. 9. Photograph courtesy of FIAT.)
(tangential)
required for change in angular
momentum  torque
Radial Flow
(Centrifugal fan)
15
(From http://www.nyb.com/frames/products_fr.htm)
Performance Parameters
Hydraulic / Fluid Stream
abj
VS
Impeller
VS
Mechanical/Shaft
16
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 ]
Impeller Power
Hydraulic Power
W f
W s
Shaft Power
W s
1
1

 (me)dm   (me)dm
A2
1
me : pv  V 2  gz
2
1
2
W f
 

Ws  T  
A1
Properties of fluid stream, not
those of solid shaft, e.g., pressure
p of fluid, velocity V of fluid, etc.
Hydraulic Power
The actual amount of mechanical energy me that the fluid
stream receives/gives up from inlet 1 to exit 2.
abj
2
ME2  ME1
2
HydraulicP ower: ME2  ME1
W f
Impeller work
Shaft Work
[Mechanical] Energy transfer
as work (between the moving
solid impeller and the fluid
stream) at the moving solid
impeller surface. (Solid-Fluid
Interaction]
[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.
17
Performance Parameters
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.)

2.
Power input/output.

3.
abj
Hydraulic Torque VS Impeller Torque VS Mechanical / Shaft Torque
Hydraulic Power VS Impeller Power VS Mechanical/Shaft Power
Other parameters of interest are, e.g.,
Recall Terminologies:

flowrate,

hydraulic head,

total and static pressure rise,

etc.
1.
2.
Hydraulic/Fluid Stream Quantities
•
Properties of a fluid stream,
•
Evaluated from fluid stream properties
Mechanical / Shaft Quantities
•
Properties of a solid shaft
•
Evaluated from shaft
See also Appendix and Review:
Recall: The Terminologies:
Hydraulic / Fluid Stream VS
Mechanical / Shaft
18
Summary of Important Quantities:
Hydraulic / Fluid Stream VS
Mechanical/Shaft
Mechanical / Shaft Side
(subscripted s)
Hydraulic / Fluid Stream Side
(subscripted h)
Torque

Th :


A1
Hydraulic Power
W h :

(me)dm 
A2
Head


 

(
r

d
F
Ts :
 )



A (soild shaft
cross section)


Shaft Power

 

Ws : Ts  
(me) dm
A1




ME 2
ME1
Hydraulic Head
H :
W h
m g

Euler Turbomachine Equation: Ts

(for idealized machines)
Associated Power Equation:
abj

Ts
 
 
(r  V )dm  (r  V )dm
A2
Power
Shaft Torque

Hydraulic Torque
shaft torque


Th

hydraulic torque

 :



 
 
(r  V )dm  (r  V )dm 

A2
A1



 
= Hydraulic torque x shaft angular velocity
W  Th  
 
 
 m 2U 2  V2  m 1U1  V1 (mixed quantity)
19
Angular Momentum
abj
20
Angular Momentum About A (Fixed) Point C
of A Particle
VS
of A Continuum Body
Particle
m

r
Continuum body
dm  dV


P  mV
C
y
 
dP  Vdm



dH c : r  dP
 
 r  Vdm

r
C
Observer A
x


  
H c  r  P  r  mV
y
Observer A
x

  

dH c : r  dP  r  Vdm


H c :
 
 r  Vdm
MV (t )
[ Length Momentum]
[ Length Momentum]
 Angular momentum of a body about a point c is defined as the
moment of linear momentum of the body about the point c.
abj
21
Angular Momentum and [Recall] The RTT
MV(t) and CV(t)
dm  dV

r
Coincident MV and CV
C
y
Observer A
x
Reynolds Transport Theorem (RTT):

dH MV ,c (t )
dt


Time rate
 of change
of H c of MV (t )

dH CV ,c (t )
dt



  
N  H c  r  P  r  mV ,

Time rate
 of change
of H c of CV (t )
:

H V ,c 
 
(
r
 V )(dV ),


 
(
r

V
)(

V

d
A
),
f /s

CS (t )

 
dP  Vdm



dH c : r  dP
 
 r  Vdm


  r V
 AngularMomentum


Time



Net convection efflux of H c
through CS (t )
V (t )  MV (t ) or CV (t )
V (t )
abj
22
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.
abj
23


V2  V1
y
x
c
c
m
c
r2
+
r1
r

V1
m

V2
r
+
r
?
dt
M c, z  ?

V2
r
+
dHMV ,c, z

V2
m

V1

V1

V2
tangential exit

V2
radial exit
r
r
axial inlet
axial inlet
m
abj

V1
+
c
z
m

V1
+
c
z
24
Moment of Forces
Point Force VS
abj
Distributive Force
25
Moment of Forces about A Point C:
Point Force VS Distributive Force
Point Force

r
Distributive Force

r

F
C

r
C

 
dM c  r  dF

 
M c : r  F
[ Length Force]
Surface Force:

dA


dFS  TdA



dFB  gdm  ( g )dV

 

M c  r  dF

[ Length Force]
 
dF  TdA

Mc 

results in
 
r  TdA Surface Integral
A ( open or closed )
Volume Force:
abj





dF  ( g )dV M c  r  ( g )dV

V
results in
Volume Integral
26
Because the evaluation of the moment of distributive forces results in



Surface Force: M c   r  dF
A
Volume Force:



M c   r  dF
 
  r  TdA
Surface Integral
A


  r  ( g )dV
Volume Integral
V
It is to be understood that when we simply write

 
M c  r  dF

we refer to the corresponding surface or volume integral.
abj
27
Moment of Surface Forces: Pressure and Shear Forces


dFp   pdA
dA


dFp   pdA

dA

dF

r
c
A
A
C
Open surface
Closed surface
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.
Moment about c due to pressure force:





dM c , p  r  dF p  r  ( p )dA

dF

r



M c, p   r  dF p

A


   r  pdA
A
Moment about c due to shear/frictional force:



dM c ,  r  dF




M c ,   r  dF
A
Note: The shear force – and in fact all the surface forces - can be written in terms of the area vector as
abj


dF  dA 
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.
28
C-Angular Momentum [ MV ]
Here, we shall limit ourselves to an observer in IFR only.
abj
29
C-Angular Momentum [MV]
FBD
Distributive force
Pressure p
Shear 

Concentrated force Fi

V
(again)

r
c
Solid part
y
Observer A, IFR
x


gdm  g ( dV )
 Physical Law (for an observer in an IFR and a fixed point C)

Mc

:

Mc =
abj

dH MV ,c (t )
 AngularMomentum  Length Momentum

, 

Time
Time

 

,
dt

H MV ,c 

 
(r  V )( dV )
MV (t )
Net/Resultant moment of all external forces/moments (hence, FBD again)
on the MV(t) about a fixed point c.
30

Mc
Net/Resultant moment of all external forces/moments
(hence, FBD again) on the MV(t) about a fixed point c

It is emphasized that M c in the physical law


dH MV ,c (t )
Mc 
dt
is
Recall the FBD Concept
the net/resultant moment of all external forces/moments (hence, FBD again)
on the MV(t) about a fixed point c.
In C-Angular Momentum, we need to sum all the external moments of all the external
forces/moments about C on the MV,

Mc
abj
FBD
Concept
31
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

F , and

2. resultant moment
Mc
1. resultant force


Recall in case of fluid flow: Forces in fluids (or solid – i.e., continuum, - for that matters)
 Forces in Fluid:
 Surface forces
abj
Line Force
=
+
Pressure/Normal
Surface Force
+
+
Body Force
Friction/Tangential
 Need to sum the contributions due to surface forces on all surfaces of the MS.
32
Examples



M c , p  r  dF p

Distributive force
in fluid
A
Concentrated force
Pressure p

Fi
Shear 
c
surface force  surface integral



M c,  r  dF

A
Pressure p

r


gdm  g ( dV )
Shear 
Solid part
Normal p
Cross section
of a solid shaft

Shear  T
s
c


gdm  g ( dV )
Solid part

Ts
Distributive force
in solid

r
Normal p
Shear 



  
M c  M c, p  M c ,   r  Fi  Ts 
i
 
r
  gdm   ...



M c  M c , p  M c , 
 
r
  gdm   ...
MV ( t )
MV (t )
• 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
abj

M c, p

T s is in fact the resultant moment/couple of shear stress (surface force) distribution
33
over a cross section of a solid shaft.
Note on Notations
• Thus, depending on the forces and moments that act on the current system of interest, the specific

form for M c depends on that specific force/moment system.
• According to the physical law of C-Angular Momentum (hence, FBD)

Mc
= sum of all external moments due to all external forces/moments on MV
• Generically, though, we may simply write



M c  M c , p  M c , 
 
r
  gdm
  ...
or even



M c  M c, p  M c , 
 
r
  gdm
MV (t )
MV ( t )
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 M c , e.g.,



 
M c  M c , p  M c ,   r  Fi 
i
abj
 
r
  gdm  ...
MV (t )
34
C-Angular Momentum [ CV ]
abj
35
C-Angular Momentum [CV]
Distributive force
Pressure p
Concentrated force
Shear 
Coincident MV(t) and CV(t)
c
Solid part
 Physical Law (for an observer in IFR
 and a fixed point C)
 RTT

dH MV ,c (t )

dH MV ,c (t )

dt
abj

:

dH MV ,c (t )

dt

H V ,c 
CV(t)

dH CV ,c (t )
dt



 
(
r

V
)(

V

d
A
)
f /s

CS (t )



 
M c  M c , p  M c ,   r  Fi 
Physical Laws

Mc


gdm  g ( dV )
MV(t)
dt





 
N  H c  r  P  r  mV ,   r V
 C-Angular Momentum

V

V

r

Mc

Fi

dH CV ,c (t )
dt
i

 
r
  gdm
MV (t )


 
 (r V )(V f / s  dA)
CS (t )
RTT
 
 (r V )(dV ), V (t ) is MV (t ) or CV (t )
V (t )
36
Force, Torque, and Energy Transfer as Work in
Turbomachines
abj
37
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
  
V  r

3. Ts

1. dF p
c
z

4. T f

dA

r

2. dF
Impeller and shaft as a system
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.

dF p ,



dM c , p  r  dF p , Force results in Torque
 
Force results in Energy Transfer as work
W p  V  dF p



 



 (  r )  dF p  (r  dF p )    dM c , p  

dF ,



dM c,  r  dF ,
 

W  V  dF

 

 


 (  r )  dF  (r  dF )    dM c,  
abj Centrifugal fan (Radial-flow, from http://www.nyb.com/frames/products_fr.htm)
38
Impeller Torque VS

Ts
Shaft Torque
Impeller Torque

M c,impeller
defined as the net moment due to surface force on the impeller



M c ,impeller  M c , p  M c ,


  r  dF p 

Tf
c
impeller

r
Note that

dF
z

dF p
1.

shaft torque, T s
2.
impeller torque,


 r  dF
impeller
and

M c,impeller
are related through the FBD of the section of the solid impeller as
shown.
However, they are generally not equal due to, e.g.,


Ts  M c,impeller
frictional torque, at the bearings
abj

Tf
.
39
Euler Turbomachine Equation,
Hydraulic Torque,
and The Associated Power Equation
abj
40

MV / CV and FBD M c
Various MV/CV’s and
The Corresponding Net/Resultant Moment for Turbomachine Analysis
CV2/MV2: CV includes fluid stream only, no solid part.
CV includes fluid stream, impeller, and
part of solid shaft
Recall the concept of FBD
CV1/MV1:

dF p

r

Ts

dF p

dF
= Shaft torque
z


dF dA

M c, p 

M c, 



 r  dFp 


 r  dF 
MS CS

 dFp 
 dA
 z
gdm

dF p
2

Mc 
abj

Ts

dF
c

 
r

g
dm

T
s

MV (t )

 
 r  gdm  Ts
MV CV
(Recall the outward
normal to the system of
interest)
No shaft torque for CV2/MV2

Mc 


M c, p



 r  dFp 
MS CS
 
r dF
Moving solid
impeller surface
2
Shaft torque for CV1/MV1
MS CS
1

dA

r

gdm
c

Ts
1

dA

M c,



 r  dF 
MS CS
 
r
  gdm
MV (t )
 
r
  gdm
MV CV
41

Example 1: Radial-Flow Machines [1] MV / CV and FBD M c
Centrifugal fan (Radial-flow)
(From http://www.nyb.com/frames/products_fr.htm)

dF p

dA
z

dF p

V2

dF

dA
1
1

dF

dA

dF p
c
Exit 2 = Outer cylindrical surface

T s at solid shaft cross section
CV/MV: CV includes fluid stream, impeller (and its
back plate), and cuts across the solid shaft at the
back plate.
abj

gdm
Inlet 1 = Inner cylindrical surface
r

gdm
2

dF

V1 
2

V2
View this way

Mc 

M c, p 

M c, 



r

d
F
p 



r

d
F
 

MS CS
MS CS
r1
Ts

dF

dF p

V1

 
r

g
dm

T
s

MV (t )

 
r

g
dm

T
s

MV CV
42
Note on Notations
To avoid messy diagram, it is to be understood that

to evaluate M c , 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
dF p

dA
the area vector
and only
abj

dA
or simply
on some surfaces

in place of the more complete one dA
instead of

dF
on all surfaces.
43

Example 2: Radial-Flow Machines [2] MV / CV and FBD M c
We can also include the fluid stream only.

• This time we will not see the shaft torque T s
• 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 M c, p and M c, .
2
2
For clarity, forces on other surfaces
are omitted from the diagram.
1
1
r1
Ts
r1

dF
Example 1 CV/MV

Mc 
abj

M c, p 

M c , 
CV/MV:

 
r

g
dm

T
s

MV (t )

Mc 

dA

dF p
CV includes fluid stream only, no solid part.

M c, p 

M c , 
 
r
  gdm
MV (t )
44

MV / CV and FBD M c
Example 3: Radial-Flow Machines [3]
[Compare to Example 1]

V2

dA
z
2

V1
View this way
z

Ts

gdm

dF p
 r
r
1

V1

V2

dA

dF
c

dA

gdm
View this way
c
Inlet 1 = Inlet duct circular cross section
abj
Exit 2 = Outer cylindrical surface

Mc 

M c, p 

M c, 



r

d
F
p 



r

d
F
 

MS CS
r1

dF

dF p

dA
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.

Ts
MS CS

 
 r  gdm  Ts
MV (t )

 
r

g
dm

T
s
45

MV CV

MV / CV and FBD M c
Example 4: Axial-Flow Machines

dA

V1
A single blade

V2

Ts
c

dA
Annular inlet plane
z
Annular exit plane
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.

Mc



M c , p  M c, 

 
 r  gdm  Ts
MV (t )
Turbine Cascade (From
abj http://www.sm.go.dlr.de/SMinfo/IADinfo/IAD.html#Projects)
46
Euler Turbomachine Equation

dF p
Coincident MV(t) and CV(t)

dF p
 
Ts , 

Tf

r
c
1

dA

dA
2

z dF

dF

gdm

V2
1
z
2

gdm
r
T s1

dF

dF p

V1
CV/MV
CV/MV [CV in Example 1]
includes fluid stream, impeller, part of solid shaft,
and inlet and exit ducts. It cuts across a cross
section of the solid shaft.
includes fluid stream, impeller (and its back
plate), and cuts across a cross section of the
solid shaft (and no inlet and exit ducts).
This CV illustrates the global nature of the Euler
Turbomachine equation - without having to deal with the
abj
blade geometry, i.e., torque = net angular momentum
efflux.
Later, this CV is used in developing the associated power
equation.
47
Assumptions

dF p
 
Ts , 

Tf
1

dA

dF

r

gdm
c
1.
All flow properties are steady in mean.
2.
Evaluate mean properties.
3.
Incompressible flow
4.
Neglect all other torques, except shaft torque.
z
[Neglect torques due to surface force
(pressure + shear), body force (mg),
frictional torque at bearings, etc.].
2
 C-Angular Momentum

Mc

:
abj

dH MV ,c (t )

dt

H V ,c 

dH CV ,c (t )
dt
 
(
r
 V )(dV ),
V (t )



 
 (r V )(V f / s  dA)
CS (t )
V (t ) is MV (t ) or CV (t )
48

dF p
 
Ts , 

Tf

r
c

dF p
1

dA

V2

dA

dF
2

dF
1
z

gdm

gdm
2
r1
Ts

dF

dF p

V1

d   
[steady in  mean properties]
  (r V )(dV )  0
dt CV



 
 
 
Net Convection Efflux Term:
 (r  V )(V f / s  dA)   (r  V )dm   (r  V )dm
Unsteady Term:
CS ( t )
Net Moment about c:


M c  Ts
A2
A1
[ Neglect all other torques,except shaft torque.]
C-Angular Momentum becomes
Euler Turbomachine Equation:
abj

Ts 
 
 
 (r  V )dm   (r  V )dm
A2
A1
49
Hydraulic Torque
Euler’s Turbomachine Equation:

Ts 
 
 

 (r  V )dm   (r  V )dm
A2
A1
In a similar manner as hydraulic power, we define the RHS (and the RHS only)
as the hydraulic torque.


Ts 

 
 
 (r  V )dm   (r  V )dm
A2
Hydraulic Torque
:

HydraulicTorque(Th )
A1
[CV Viewpoint]

HydraulicTorque (Th )
:
 
 
 (r  V )dm   (r  V )dm
A2
A1
= Net convection efflux of angular momentum through CS.
Hydraulic Torque
Hydraulic Torque
[MV Viewpoint]
:=
Like hydraulic power, through the RTT, we can
see that the two views are equivalent.
the time rate of change of angular momentum of
the fluid stream as it flows through CV [from CV
abj
inlet 1 to CV exit 2].
50

HydraulicTorque (Th )
:
 
 
 (r  V )dm   (r  V )dm
A2
A1
Properties of fluid stream
Hydraulic Torque is
• the property of fluid stream, not shaft
• evaluated from the properties of fluid stream, not shaft.
abj
51
Euler Turbomachine Equation


Ts  Th
Ideal Case
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’s Turbomachine Equation:
Shaft Torque = Hydraulic Torque

Ts

Shaft Torque
abj


Th

Hydraulic Torque


 


: (r  V )dm  (r  V )dm 
 A

A
1
 2



52

dF p

Tf



Ts

r
c
Shaft Torque
1

dA

dF

gdm
Hydraulic Torque

Th 
z
 
 
 (r  V )dm   (r  V )dm
A2
A1
2


Ts  Th
In a real case, however, since there are other torques Real Case
•
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.
abj
53
 
Further Assume Uniform (r V )
Assumption 5:

dF p
 
Ts , 

Tf

r

dF

Ts 

gdm
c
Assumption 5:
1

dA


A2
2
Uniform (r  V ) at each cross section. Or, evaluate
some reference radius.
Euler Turbomachine Equation:
C-Mass also gives
abj
z
 
 
 (r  V )dm   (r  V )dm
1  m
 2 : m

m
A1
 
(r V ) at

 
 
 2 (r2 V2 )  m
 1 (r1 V1 )
Ts  m

V = Velocity of fluid at inlet/exit
54
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 ]
Impeller Power
Hydraulic Power
W f
W s
Shaft Power
W s
1
1

 (me)dm   (me)dm
A2
1
me : pv  V 2  gz
2
1
2
W f
 

Ws  T  
A1
Properties of fluid stream, not
those of solid shaft, e.g., pressure
p of fluid, velocity V of fluid, etc.
Hydraulic Power
The actual amount of mechanical energy me that the fluid
stream receives/gives up from inlet 1 to exit 2.
abj
2
ME2  ME1
2
HydraulicP ower: ME2  ME1
W f
Impeller work
Shaft Work
[Mechanical] Energy transfer
as work (between the moving
solid impeller and the fluid
stream) at the moving solid
impeller surface. (Solid-Fluid
Interaction]
[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.
55
In the Same Manner to Power
Hydraulic Torque VS Impeller Torque

Hydraulic Torque T h
VS
Mechanical Torque at Shaft
Impeller Torque

M c,impeller

Ts
Shaft Torque

Ts
• With the concept of resultant moment,

we can see that T can be written in terms

s
of stress in the same manner as M
c,impeller
c
1
2
• The area integral in this case is the solid
shaft cross section.

r

Th

dF

dF p
z

Th :
 
 
 (r  V )dm   (r  V )dm
A2
A1
• For this generic definitions, we leave the
reference point and axis unspecified first.



M c ,impeller  M c , p  M c ,


  r  dF p 
impeller


r

d
F



Ts
impeller
Hydraulic Torque
Impeller Torque
Shaft 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].
defined as the net moment
due to surface force on the
impeller
Net moment due to
surface stress (shear
stress) distribution over a
cross section of a solid
56
shaft
abj
The Associated Power Equation

V2

U2
If we focus on the impeller and look at the CV below:
2

U1
Euler’s Turbomachine Equation:


Ts  Th
1

V1
r1
Ts


 
 
 2 (r2 V2 )  m
 1 (r1 V1 )
 m
With the Euler’s Turbomahine Equation, which gives an ideal shaft
torque, we can find the mechanical power of the shaft as

 
 

W  Th    Ts  

 
 
 m 2 (r2  V2 )  m 1 (r1  V1 )  
    
  
:
(r  V )    (  r )  V  U  V ,
 
 



W  m2 (U 2  V2 )  m1 (U1  V1 )





 
U :   r
is the tangential velocity (of the blade).

 
 
 

The Associated Power Equation:W : T    m
 2 (U 2 V2 )  m
 1 (U1 V1 )
h



Essentially, we define the associated power W from the hydraulic torque T h not shaft torque T s
abj
57
Assumptions
In Summary
 
Ts , 

Tf

r
c

V

 
U i    ri

dF p
1

dA

dF

gdm
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.].
z

V2

U2
2

U1
2
=
Velocity of fluid at inlet/exit
=
The tangential velocity (of the blade).
1
r1
Ts

V1


 




   (r  V )dm 
Euler Turbomachine Equation: Ts  Th  :  (r  V )dm
 A

 
A1
2


Assumption 5: Uniform (r V ) at each cross





section. Or, evaluate (r  V ) at some


 m2 (r2  V2 )  m1 (r1  V1 )
reference radius.
abj


 
 
 

 2 (U 2 V2 )  m
 1 (U1 V1 )58
The Associated Power Equation: W : Th    m
Analysis for The Performance of
Idealized Turbomachines
abj
59
Problem: Idealized Turbomachine Performance Analysis

V2

U2
Euler Turbomachine Equation:
2

U1

 
 
 2 (r2 V2 )  m
 1 (r1 V1 )
Ts  m
1

V1
r1
Ts

The Associated Power Equation:
 
 

 2 (U 2 V2 )  m
 1 (U1 V1 )
W  m

Given
1.
geometry and geometric parameters
•
blade angles
•
radii
( r1 , r2 )
2.
kinematical parameter:
3.
mass flowrate
( b1 , b 2 )

angular velocity 
m
Questions
1. Sketch the blade shape, and
Find the idealized
Need to find the two
kinematical unknowns
 
(V1 , V2 )
2. shaft torque (= hydraulic torque)
abj
3. shaft power (= hydraulic power,  = 1)
4. hydraulic head.
60
How to tell the geometry of the blade:
Blade b (and Flow a Angles Convention

Vrb
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, Vrb .
tangent to blade
at inlet/exit
b
CS

U
a

V
 

Vn (Vz or Vr )

U

V

Vrb

Vn
b
= 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 V
= blade angle
= the angle that the blade tangent at inlet/exit


makes with U (measured away from Vn )
a
= flow angle

= the angle that the absolute flow velocity V
abj


makes with Vn (measured towards U
)
61
Example of Blade Angles in Axial- and Radial-Flow Machines

U
Axial-flow machine

U

U

Vn


Turbine Cascade (From
http://www.sm.go.dlr.de/SMinfo/IADi
nfo/IAD.html#Projects)
b1
1
2
b 2 U
2
1

Vn
Flow

U
2
Radial-flow machine

Vn
b2


abj
r1
1
2

Vn
1
Flow
Centrifugal fan
(From http://www.nyb.com/frames/products_fr.htm)

U
b1

U
62
Example: Sketch The Blade Shape: Axial-Flow Machine
Example: Sketch an axial-flow machine blade with b1 = 30o , b2 = 60o
flow
flow
b1

U
b1

U


Vn
Slope/angle increases from 30o to 60o

Vn
Curvature
b2

U
z
b2

U

Vn

Vn
b 2  b1 Blade concave towards

Example: Sketch an axial-flow machine blade with b1 = 60o , b2 = 30o
the direction of U
flow

U

U
b1

Vn

b2

U
abj
z
b1
flow
Slope/angle decreases from 60o to 30o

Vn
Curvature
b2

Vn

U

Vn
b 2  b1 Blade convex towards

63
the direction of U
Example: Sketch The Blade Shape: Radial-Flow Machine
Example: Sketch a radial-flow machine blade with b1 = 90o, b2 = 45o
b1

U
q
Backwardly-curved blade


Vn
b2  q
(wrt the direction of angular rotation)

Vn

U
Example: Sketch a radial-flow machine blade with b1 = 90o, b2 = 135o
q
b1

U

U

Vn

Forwardly-curved blade
(wrt the direction of angular rotation)

Vn
abj
64
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
1

Vn
b2
2
abj

Vn

U
b1

U

U
65
Shockless-Entry/Exit Condition
Further Assumption for Idealized Turbomachines
Shockless-Entry/Exit Condition:

Vrb is tangent to the blade at inlet and exit.
1

Vn

U
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.

Vrb1
b1

U
Assumptions (Idealized Turbomachines)
b2
2
abj

Vn

Vrb 2

U
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.].
5.
Shockless entry/exit condition
66
Relative Velocity Relation and Velocity Diagram
Recall our Problem

U2

V2
Euler Turbomachine Equation:
2

U1

 
 
 2 (r2 V2 )  m
 1 (r1 V1 )
Ts  m
1

V1
r1
Ts

The Associated Power Equation:
 
 

 2 (U 2 V2 )  m
 1 (U1 V1 )
W  m

Given
1.
geometry and geometric parameters
•
blade angles
•
radii
( r1 , r2 )
2.
kinematical parameter:
3.
mass flowrate
( b1 , b 2 )

angular velocity 
m
Questions
1. Sketch the blade shape, and
Find the idealized
Need to find the two
kinematical unknowns
 
(V1 , V2 )
2. shaft torque (= hydraulic torque)
abj
3. shaft power (= hydraulic power,  = 1)
4. hydraulic head.
67
1

Vn
b2
2

Vn

Vrb 2

U

Vrb1
b1

U
  
V  U  Vrb

U
1. Euler Turbomachine Equation:

 
 
 2 (r2 V2 )  m
 1 (r1 V1 )
Ts  m
2. The Associated Power Equation:
 
 



W  m2 (U 2 V2 )  m1 (U1 V1 )
3. Relative Velocity Relation:
  
V  U  Vrb
abj
68
  
V  U  Vrb
Velocity diagram can be used as a graphical/geometrical
aid in solving the relative velocity vector relation.
Inlet
1

Vn

U

Vrb1
b1

U
2

Vrb 2
abj

V1
b1

U1
Exit
b2

U

Vn

Vrb1

Vn1

Vn 2

V2

Vrb 2
b2

U2
69
Velocity Diagram for Axial-Flow Machines

V1

V2
r1  r2



Vn1  Vn 2
A1  A2 
1

Vn
b2

 U
Vn
It is then recommended that the velocity diagrams at inlet and exit

be superimposed on the same base U and the velocity diagrams
look like below.
2
1

U

Vrb1
b1
Inlet

Vrb 2

U

V2
  
V  V2  V1

Vrb1

V1


Vn1  Vn 2
b 2 b1

Vrb 2
abj



U1  U 2 : U
z 2. From C-Mass, we also find
c
2
1. For simplicity we evaluate the properties at the mean radius of
the CV shown on the left. Thus, since
Exit

U
70
Is it a pump or a turbine?

T
Recall that if
abj
System / MV

T
  0,
  
W  T   
  0,

is an external torque acting on the system.
Energy is input into the system

Energy is extracted from the system 
Pump
Turbine
71

V1

V2

Ts
c

U1
1

U2

V2

U2
z
2

U1
1
r1
Ts

V1
2
From the associated power equation, then for an idealized machine


 
 
 



W : Th    m2 (U 2  V2 )  m1 (U 1  V1 )
abj

  0 Energy is input into the stream


  0 Energy is extracted from the stream 

Pump
Turbine
72
Example: Axial-Flow Machine: Is it A Pump or A Turbine?


b 2  b1 (concave in the directionof U )
b 2  b1 (convex in the directionof U )
b1
1
b1
1

U

U
b2
b2
2


Vn1  Vn 2

Vrb 2
b 2 b1


  
V  V2  V1


Vrb1
V2
  
V  V2  V1


V1
Vrb 2
2

V1


Vn1  Vn 2
b1 b 2

U
 
 
 

W : Th    m 2 (U 2  V2 )  m 1 (U 1  V1 )
  
 m U  (V2  V1 )


abj
 m U  V  0 
pump

Vrb1



V2

U
 
 


W : Th    mU  V  0

turbine
73
[Ideal] Hydraulic Power and Hydraulic Head
1. Euler Turbomachine Equation:

 
 
 2 (r2 V2 )  m
 1 (r1 V1 )
Ts  m
2. The Associated Power Equation:
 
 

 2 (U 2 V2 )  m
 1 (U1 V1 )
W  m
If there is no me loss in the machine
3. Relative Velocity Relation:
  
V  U  Vrb
( = 1), we have
HydraulicP ower  Shaft P ower
 
 
 m 2 (U 2  V2 )  m 1 (U 1  V1 )
 
 



mgH  m (U 2  V2 )  m (U 1  V1 )

abj
 
1  
HydraulicHead H 
(U 2  V2 )  (U 1  V1 )
g

74
In Circular Cylindrical Coordinates r-q-z: Axial-Flow Machine
Right-handed coordinates:
Recall in circular cylindrical coordinates:
Position vector:
eˆ r
Reference axis
abj
1


U

r
q c
z

Ts

V1
eˆr  eˆr (q )
eˆq
eˆ z

r
eˆz  eˆr  eˆq

r  reˆr  zeˆz ,
2

V2
1
2
Axial- Flow Machine

Ts



U

V
 Ts eˆ z
  eˆ z
 U eˆq
 V z eˆ z  Vq eˆq
75
In the Circular Cylindrical Coordinates r-q-z: Radial-Flow Machine
Recall in circular cylindrical coordinates:
Position vector:
eˆ r
eˆq
eˆ z


z
Right-handed coordinates:

V2

U2

r
r
q
Reference axis
Radial- Flow Machine
1
r1
Ts
c
eˆr  eˆr (q )
2

U1
2
 
T s V1 1
abj
eˆz  eˆr  eˆq

r  reˆr  zeˆz ,

V2

V1


Ts



U

V
 Ts eˆ z
  eˆ z
 U eˆq
 Vr eˆr  Vq eˆq
76
In the Circular Cylindrical Coordinates r-q-z
1. Euler Turbomachine Equation:

 
 
Ts  Ts eˆ z   m 2 (r2  V2 )  m 1 (r1  V1 )
 
:
r V
 (reˆr  zeˆ z )  (Vr eˆr  Vq eˆq  Vz eˆ z )
 
:
(r  V ) z  (rVq )eˆ z
Ts  m 2 r2V2q  m 1r1V1q
 
 



W  m2 (U 2  V2 )  m1 (U1  V1 ),

:
U  Ueˆq
W  m U V  m U V
2. The Associate Power Equation:
2
3. Relative Velocity Relation:
  
V  U  Vrb
:
:
:
2 2q
1 1 1q

V  Vn eˆn  Vq eˆq

U  Ueˆq

Vrb  Vrb , n eˆn  Vrb ,q eˆq
Vn eˆn  Vq eˆq  Vrb , n eˆn  U  Vrb ,q eˆq
abj
77
Basic of Velocity Diagram
eˆb
tangent to blade at i

Vrb
eˆ n
2.

V

Vn

Vq
b

U
eˆq
Decomposition into two systems of coordinates.
a.
eˆn  eˆq
b.
eˆb  eˆq
  
V  U  Vrb
Vn eˆn  Vq eˆq  U eˆq  Vrb eˆb
: eˆb  sin b eˆn  cos b eˆq
The relations between velocities can be illustrated as
Vn eˆn  Vq eˆq  U eˆq  Vrb sin b eˆn  cos b eˆq 
 Vrb sin b eˆn  U  Vrb cos b eˆq
follows.
1. Relative velocity relation:
Vn  Vrb sin b
Regardless of whether it is
•
a radial-flow or an axial-flow machine
•
at inlet or exit,
Vq  U  Vrb cos b
•
U , b , Vn , Vq , Vrb
the relative velocity relation, which is a kinematic relation,
holds. Thus, we have
  
V  U  Vrb
abj
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, U , b
are known.
•
Vn may be found from C-Mass (e.g.,
flowrate is given).
78
From Fox, R. W., McDonald, A. T., and Pritchard, P. J., 2004, Introduction to Fluid Mechanics, Sixth Edition, Wiley, New York.
abj
79
abj
From Fox, R. W., McDonald, A. T., and Pritchard, P. J., 2004, Introduction to Fluid Mechanics, Sixth Edition, Wiley, New York.
80
Appendix and Review

Recall:
The Terminologies:
Hydraulic / Fluid Stream

abj
Recall:
VS
Mechanical / Shaft
Free-Body-Diagram (FBD) Concept
81
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,
.
HydraulicP ower (W h ) : ME2  ME1 

HydraulicTorque (Th ) :
 (me)dm   (me)dm
A2

A2
 
 
(r  V )dm  (r  V )dm

A1
A1
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
 
.
W s  Ts   Subscript
s
abj Recall that that the hydraulic quantities and the mechanical quantities may be related but, e.g., due to friction, they are
generally not equal.
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Recall: Free-Body-Diagram (FBD) Concept
1. The Free-Body-Diagram (FBD) concept is related to the Newton’s



dPMV (t )
F (or simply F for simplicity) 
dt



dH MV ,c (t )
M c (or simply M c for simplicity) 
dt
Second Law:


where




F (or simply F )
=
resultant/net external force on MV
=
sum of all external forces on MV


M c (or simplyM c ) =
=
abj
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
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2.
Hence, in the application of these laws, we need to be able to find

3.


F (or simply F )

and


M c (or simplyM c )
Hence, “FBD” here does not refer to the diagram per se, but refers to the fact that we need to be
able to
FBD Concept
1.
Define the system of interest clearly.
2.
Know and Recognize various types of forces (/moments, and their natures) that can act on
the system.
1.
[Recall that moment is an effort of force in causing angular motion.]
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


F
and their resultant moment


Mc
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.
84 into
• abj
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
account and sum all the external forces/moments on MV] should be kept in mind.
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

F , and

2. resultant moment
Mc
1. resultant force
abj


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