# Centrifugal pumps

## Transcription

Centrifugal pumps
```Centrifugal pumps
Impellers
Multistage impellers
Cross section of high speed
water injection pump
Source: www.framo.no
Water injection unit
4 MW
Source: www.framo.no
Specific speed that is used to
classify pumps
nq  n 
Q
H
34
nq is the specific speed for a unit machine that
is geometric similar to a machine with the head
Hq = 1 m and flow rate Q = 1 m3/s
n s  51,55  n q
Affinity laws
Q1 n1

Q2 n 2
2
H1  u 1   n 1 
     
H2  u2   n2 
P1  n1 
  
P2  n 2 
2
3
Assumptions:
Geometrical similarity
Velocity triangles are the same
Exercise
• Find the flow rate, head and power
for a centrifugal pump that has
increased its speed
• Given data:
hh = 80 %
n1 = 1000 rpm
n2 = 1100 rpm
P1 = 123 kW
H1 = 100 m
Q1 = 1 m3/s
n2
1100
3
Q 2   Q1 
1  1,1 m s
n1
1000
2
 n2 
 1100 
H 2     H1  
 100  121 m
 1000 
 n1 
3
2
 n2 
 1100 
P2     P1  
 123  164 kW
 1000 
 n1 
3
Exercise
• Find the flow rate, head and power
for a centrifugal pump impeller that
has reduced its diameter
• Given data:
hh = 80 %
D1 = 0,5 m
D2 = 0,45 m
P1 = 123 kW
H1 = 100 m
Q1 = 1 m3/s
Q1   D1  B1  c m1 D1 n1



Q 2   D 2  B2  c m 2 D 2 n 2

D2
0,45
Q2 
 Q1 
1  0,9 m 3 s
D1
0,5
2
 D2 
 0,45 
  H1  
H 2  
 100  81 m
 0,5 
 D1 
3
2
 D2 
 0,45 
  P1  
P2  
 123  90 kW
 0,5 
 D1 
3
Velocity triangles
Reduced cu2
Slip angle
Friction loss
Slip angle
Impulse loss
Slip
Best efficiency point
Power
P  M
Where:
M = torque [Nm]
P    Q  r2  c 2  cos  2  r1  c1  cos 1   
   Q  u 2  c u 2  u1  c u1   
   Q  g  Ht
In order to get a better understanding of
the different velocities that represent the
head we rewrite the Euler’s pump
equation
u 2  c u 2  u 1  c u1
Ht 
g
w12  c12  u12  2  u1  c1  cos 1  c12  u12  2  u1  c u1
w 22  c 22  u 22  2  u 2  c 2  cos  2  c 22  u 22  2  u 2  c u 2
u 22  u12 c 22  c12 w 22  w12
Ht 


2g
2g
2g
Euler’s pump equation
u 2  c u 2  u 1  c u1
Ht 
g
u 22  u12 c 22  c12 w 22  w12
Ht 


2g
2g
2g
u 22  u12

2g
Pressure head due to change of
peripheral velocity
c 22  c12

2g
Pressure head due to change of
absolute velocity
w 22  w12 Pressure head due to change of
 relative velocity
2g
Rothalpy
Using the Bernoulli’s equation upstream and
downstream a pump one can express the
2
2
 p



c
p
c
H t  

 z   

 z 
 g 2g
2    g 2  g
1
The theoretical head can also be expressed as:
u 22  u12 c 22  c12 w 22  w12
Ht 


2g
2g
2g
Setting these two expression for the theoretical
head together we can rewrite the equation:
p2
w 22
u 22
p1
w12
u12





g 2g 2g g 2g 2g
Rothalpy
The rothalpy can be written as:
  r 
w
p
I


 cons tan t
g 2g
2g
2
2
This equation is called the
Bernoulli’s equation for
incompressible flow in a rotating
coordinate system, or the rothalpy
equation.
Stepanoff
We will show how a centrifugal pump is designed
using Stepanoff’s empirical coefficients.
Example:
H = 100 m
Q = 0,5 m3/s
n = 1000 rpm
b2 = 22,5 o
Specific speed:
nq  n 
Q
H
34
0,5
 1000 
 22,4
34
100
n s  51,55  n q  1153
K u  1,0
u2
Ku 
2g H
 u 2  K u  2  g  H  44,3 m s
2 n

60
We choose:
D2
u2  
2
Dhub  0,5  D1  0,17 m
u2  2
 D2 
 0,85 m

K m2  0,11
K m2
cm2
cm2

2g H
 c m 2  K m 2  2  g  H  4,87 m s
Q
Q
 
A   D2  d 2
w2
cm2

Q
d2 
 0,038 m
  D2  cm2
u2
c2
cu2
Until now, we have not considered the thickness
of the blade. The meridonial velocity will
change because of this thickness.
We choose:
s2 = 0,005 m
z = 5
s2
0,005
su 

 0,013 m
o
sin b 2 sin 22,5
cm2
Q
Q
 
A   D 2  z  s u   d 2

Q
d2 
 0,039 m
  D 2  z  s u   c m 2
K m1  0,145
c m1
K m1 
2g H
 c m1  K m1  2  g  H  6,4 m s
w1
c1= cm1
u1
Dhub
D1
 0,405
D2
D1
 0,405  D1  0,405  D 2  0,34 m
D2
We choose:
D1m
Dhub  0,5  D1  0,17 m
D12  D 2hub

 0,27 m
2
Q
Q
c m1 

A1   D1m  d1
Q
 d1 
 0,09 m
  D1m  c m1
Without thickness
the inlet
w1
b1
u1
D1
0,34
u1   
 104,7 
 17,8 m s
2
2
 c m1 
 6,4 
  a tan 
b1  a tan 
  19,8o
 17,8 
 u1 
s1
0,005
s u1 

 0,015 m
o
sin b1 sin 19,8
Q
d1 
 0,10 m
  D1m  z  s u1   c m1
Cm1=6,4 m/s
w2
c2
cm2=4,87m/s
b2=22,5o
u2=44,3 m/s
cm2
tan b 2 
u 2  cu 2
 cu 2
cu2
cm2
 u2 
 32,5 m s
tan b 2
u 2  c u 2  u1  c u1 44,3  32,5
H

 153 m
hh  g
0,96  9,81
w2
c2
cm2=4,87m/s
u2=44,3 m/s
H
u 2  cu 2
hh  g
 cu 2 
cu2
H  hh  g 100  0,96  9,81

 21,3 m s
u2
44,3
cm2
4,87
' b 2  a tan
 a tan
 11,9o
u 2  cu 2
44,3  21,3
bslip  bslip ' b2  22,5o 11,9o  10,6o
```