Transcript Kaskadeteori

Cascade theory
The theory in this lecture comes from:
Fluid Mechanics of Turbomachinery
by George F. Wislicenus
Dover Publications, INC. 1965
dc
0
dt
Y
c = c∞+Dc
p0  p 
c
2
 konst .
2
FY
c∞
FX
X
Contour
Dc is the change of
velocity due to the vane
The contour is large
compared to the
dimensions of the vane
Decompose the velocity in the
normal and the tangential direction
of the contour
c  c   cos   D c n   c   sin   D c s 
2
2
2


c  c   cos   sin
2
2
2
2

  2  v    D c n  cos   D c s  sin    D c n  D c s
2

c  c   2  c    D c n  cos   D c s  sin    D c
2
2
2
2
Bernoulli’s equation
p0  p 
 c
2
2

p0  p 

2

 c   2  c    D c n  cos   D c s  sin    D c
2
2

Forces in the x-direction
The forces in the x-direction acting on the element ds can be calculated as a
force coming from pressure and impulse.
dF x   p  ds  cos 

   c
 
  ds  c

sin
   c   cos   D c n  ds  c   cos   D c n cos 

 cos   D c n
Flow Rate, Q

 sin   D c s

Velocity in x-direction, cx
Forces in the x-direction
We insert the equation for the pressure, p from
Bernoulli’s equation.
 p  p0 

2

 c   2  c    D c n  cos   D c s  sin    D c
2
dF x   p  ds  cos 

   c
 
  ds  c

sin
2
   c   cos   D c n  ds  c   cos   D c n cos 

 cos   D c n

 sin   D c s


Forces in the x-direction
We insert the equation for the pressure, p from
Bernoulli’s equation.
 p  p0 


 c   2  c    D c n  cos   D c s  sin    D c
2
2
2

dF x   p 0  ds  cos 


2

 c   2  c    D c n  cos   D c s  sin    D c

   c
2
 
  ds  c

sin
2
  ds  cos 
   c   cos   D c n  ds  c   cos   D c n cos 

 cos   D c n

 sin   D c s

Forces in the x-direction
dF x   p 0  ds  cos 
2
 c 2

Dc
2
   ds  
 cos   c   D c n  cos   c   D c s  cos   sin  
 cos  
2
 2


   ds  c
   ds  c   cos   D c n  cos   2  c   D c n  cos 
2
3
2

 cos   sin
2
2
2

  D c n  D c s  sin   c   D c s  cos   sin   c   D c n  sin
2


Forces in the x-direction
The change of velocity, Dc is very small because the large distance from the
airfoil to the contour. We neglect the terms that has the second order of Dc.
dF x   p 0  ds  cos 
2
 c 2

Dc
2
   ds  
 cos   c   D c n  cos   c   D c s  cos   sin  
 cos  
2
 2


   ds  c
   ds  c   cos   D c n  cos   2  c   D c n  cos 
2
3
2

 cos   sin
2
2
2

  D c n  D c s  sin   c   D c s  cos   sin   c   D c n  sin
2


Forces in the x-direction
 2
1
2
dF x   p 0  ds  cos     ds   c   cos     cos   sin
2

2

2
     ds  c   D c n  cos   sin

2

2
dF x   p 0  ds  cos    
c
2
 cos   ds    c   D c n  ds
This is the force acting in the x-direction on a small element, ds of the contour.

 

Forces in the x-direction
2
dF x   p 0  ds  cos    
c
2
 cos   ds    c   D c n  ds
By integrating around the contour, we will find the total force acting in the
x-direction.
2
Fx   p 0   cos   ds   

Fx     c    D c n  ds
c
2
  cos   ds    c    D c n  ds
d’Alembert paradox
Fx     c    D c n  ds  0
The term Dcn·ds is the flow rate through the contour. If the flow is
incompressible, the integral of the term Dcn·ds around the contour will be
zero.
A body in a two-dimensional and nonviscous flow with constant energy will not
exert a force in the direction parallel
undisturbed flow, c∞
Forces in the y-direction
The forces in the y-direction acting on the element ds can be calculated as a
force coming from pressure and impulse.
dF y   p  ds  sin 

   c
 
  ds  c

  cos 
   c   cos   D c n  ds  c   cos   D c n  sin 

 cos   D c n

 cos   D c s
Forces in the y-direction
2
dF y   p 0  ds  sin    
c
2
 sin   ds    c   D c s  ds
This is the force acting in the y-direction on a small element, ds of the contour.
Forces in the y-direction
2
dF y   p 0  ds  sin    
c
2
 sin   ds    c   D c s  ds
By integrating around the contour, we will find the total force acting in the
y-direction.
2
Fy   p 0   sin   ds   

Fy     c    D c s  ds

c
2
  sin   ds    c    D c s  ds
Lift
Fy     c    D c s  ds

Circulation
 
D
c

ds
 s

Lift
Fy     c   
The law of the circulatory flow about a
deflecting body
In the absence of any deflecting
body inside the hatched area of
the contour the force in ydirection must necessarily be
zero. This leads to the theorem
that:
For a flow of constant energy,
the circulation around any
closed contour not enclosing
any force-transmitting body
must be zero.
The law of the circulatory flow about a
deflecting body
Let the circulation around the
outer contour in the figure be:
1 
c
 ds
s

cs
Let the circulation around the
inner contour in the figure be:
2 
c

s
 ds
The law of the circulatory flow about a
deflecting body
Let the circulation around the
inner and outer contour be
connected along the line A-B.
cs
The circulation around the
hatched area can now be
written as:
B
1  2  1 
c
A
D
s
 ds   2 
c
C
s
 ds
The law of the circulatory flow about a
deflecting body
From the figure we can see that:
B
c
A
cs
D
s
 ds    c s  ds
C
The circulation around the
hatched area can now be
written as:
1  2  1   2
The law of the circulatory flow about a
deflecting body
Since we do not have any body
inside the hatched area:
1  2  1   2  0
cs
Which gives:
1   2
The law of the circulatory flow about a
deflecting body
1   2
This leads to the theorem:
cs
For a given flow condition
(with constant energy), the
circulation around the
deflecting body is
independent of the size and
shape of the contour along
which the circulation is
measured.
The law of the circulatory flow about a
deflecting body
The mean velocity for the
circulation around a contour
having the length s is:
  s  c sm 
c
s
 ds

cs
For a constant value of the
circulation, the mean velocity,
csm has to decrease if the
length s increases.
The circulation is in inverse
ratio to the distance of the
contour
Circulation about several deflecting bodies
We have 3 wing profiles in a twodimensional cascade and makes a
contour around the whole cascade.
This contour is marked ABGDEF.
E
c
1 
 ds 
s
AEF
A
c
s
 ds   c s  ds
A
E
B
c
2 
s
 ds 
ABDE
D
c
s
 ds 
A
c
c
BGD
s
 ds 
 ds 
s
B
D
3 
E
c
B
D
B
s
 ds 
c
D
c
s
 ds
A
s
 ds 
c
E
s
 ds
Circulation about several deflecting bodies
From the figure we can see that:
E
A
c
s
 ds    c s  ds
A
E
D
c
B
B
s
 ds    c s  ds
D
Circulation about several deflecting bodies
Circulation around 3 wing profiles in a
cascade becomes:
A
1   2  3 
c
E
B
s
 ds 
c
A
D
s
 ds   c s  ds 
B
1  2  3  0
E
c
D
s
 ds
Cascade in an axial flow turbine
Let us look at the cylindrical section AB through the
axial flow turbine.
Cascade in an axial flow turbine
By unfolding the cylindrical section AB from the last slide,
we can look at the blades in a cascade
Cascade in an axial flow turbine
Circulation around the blades is: (where Z is the number of blades)
b
  Z  i 
c
b
a
s
 ds 
c
b
a
s
 ds 
c
a
b
s
 ds 
c
a
s
 ds
Cascade in an axial flow turbine
b
From the figure we can see that:
c
s
 ds  2    r  c u 2
s
 ds   2    r  c u 1
b
a
c
a
Cascade in an axial flow turbine
a
  Z  i  2    r  c u 2 
c
b
b
s
 ds  2    r  c u 1 
c
a
s
 ds
Cascade in an axial flow turbine
a
  Z  i  2    r  c u 2 
c
b
s
 ds  2    r  c u 1 
b
a
b
From the figure we can see that:
c
c
a
a
s
 ds    c s  ds
b
s
 ds
Cascade in an axial flow turbine
The circulation becomes:
  2    r  c u 2  2    r  c u1
Cascade in an axial flow turbine
The change of angular momentum is related to the vane
circulation by the equation:
Z 
2
 r  c u 2  r  c u1
Cascade in an axial flow turbine
By multiplying the change of angular momentum from the
upstream to the downstream side of a turbine runner is the
torque acting on the turbine shaft with the angular velocity of
the runner we will recognize Euler’s turbine equation.
E    r  cu 2    r  cu1   

E  u 2  c u 2  u1  c u1
Z 
2