29
Centrifugal pumps

Centrifugal Pumps - NTNU · Cross section of high speed water injection pump Source:

Embed Size (px)

Citation preview

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 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

43q HQnn ⋅=

qs n55,51n ⋅=

Affinity laws

2

1

2

1

nn

QQ

=

2

2

1

2

2

1

2

1

nn

uu

HH

=

=

3

2

1

2

1

nn

PP

=

Assumptions:Geometrical similarityVelocity triangles are the same

Exercise

sm1,1110001100Q

nnQ 3

11

22 =⋅=⋅=

m12110010001100H

nnH

2

1

2

1

22 =⋅

=⋅

=

kW16412310001100P

nnP

3

1

3

1

22 =⋅

=⋅

=

• Find the flow rate, head and power for a centrifugal pump that has increased its speed

• Given data:ηh = 80 % P1 = 123 kW n1 = 1000 rpm H1 = 100 mn2 = 1100 rpm Q1 = 1 m3/s

Exercise• Find the flow rate, head and power

for a centrifugal pump impeller that has reduced its diameter

• Given data:ηh = 80 % P1 = 123 kW D1 = 0,5 m H1 = 100 mD2 = 0,45 m Q1 = 1 m3/s

sm9,015,045,0Q

DDQ

nn

DD

cBDcBD

QQ

31

1

22

2

1

2

1

2m22

1m11

2

1

=⋅=⋅=

==⋅⋅⋅Π⋅⋅⋅Π

=

m811005,045,0H

DDH

2

1

2

1

22 =⋅

=⋅

=

kW901235,045,0P

DDP

3

1

3

1

22 =⋅

=⋅

=

Velocity triangles

Slip angle

Reduced cu2

Slip angle

Slip

Best efficiency point

Friction loss

Impulse loss

Power

ω⋅= MPWhere:

M = torque [Nm]ω = angular velocity [rad/s]

( )( )

t

1u12u2

111222

HgQcucuQ

coscrcoscrQP

⋅⋅⋅ρ=ω⋅⋅−⋅⋅⋅ρ=

ω⋅α⋅⋅−α⋅⋅⋅⋅ρ=

gcucuH 1u12u2

t⋅−⋅

=

In order to get a better understanding of the different velocities that represent the head we rewrite the Euler’s pump equation

1u121

21111

21

21

21 cu2uccoscu2ucw ⋅⋅−+=α⋅⋅⋅−+=

2u222

22222

22

22

22 cu2uccoscu2ucw ⋅⋅−+=α⋅⋅⋅−+=

g2ww

g2cc

g2uuH

21

22

21

22

21

22

t ⋅−

−⋅−

+⋅−

=

Euler’s pump equation

gcucuH 1u12u2

t⋅−⋅

=

g2ww

g2cc

g2uuH

21

22

21

22

21

22

t ⋅−

−⋅−

+⋅−

=

=⋅−g2uu 2

122 Pressure head due to change of

peripheral velocity

=⋅−g2cc 2

122

=⋅−

g2ww 2

122

Pressure head due to change of absolute velocity

Pressure head due to change of relative velocity

RothalpyUsing the Bernoulli’s equation upstream and downstream a pump one can express the theoretical head:

1

2

2

2

t zg2

cg

pzg2

cg

pH

+

⋅+

⋅ρ−

+

⋅+

⋅ρ=

g2ww

g2cc

g2uuH

21

22

21

22

21

22

t ⋅−

−⋅−

+⋅−

=

The theoretical head can also be expressed as:

Setting these two expression for the theoretical head together we can rewrite the equation:

g2u

g2w

gp

g2u

g2w

gp 2

1211

22

222

⋅−

⋅+

⋅ρ=

⋅−

⋅+

⋅ρ

Rothalpy

The rothalpy can be written as:

( )ttancons

g2r

g2w

gpI

22

=⋅

⋅ω−

⋅+

⋅ρ=

This equation is called the Bernoulli’s equation for incompressible flow in a rotating coordinate system, or the rothalpyequation.

StepanoffWe will show how a centrifugal pump is designed using Stepanoff’s empirical coefficients.

Example: H = 100 mQ = 0,5 m3/sn = 1000 rpmβ2 = 22,5 o

4,22100

5,01000H

Qnn 4343q =⋅=⋅=

1153n55,51n qs =⋅=

Specific speed:

This is a radial pump

0,1Ku =

sm3,44Hg2KuHg2

uK u22

u =⋅⋅⋅=⇒⋅⋅

=

srad7,10460

n2=

⋅Π⋅=ω

m85,02uD2

Du 22

22 =

ω⋅

=⇒⋅ω=

We choose: m17,0D5,0D 1hub =⋅=

11,0K 2m =

sm87,4Hg2KcHg2

cK 2m2m2m

2m =⋅⋅⋅=⇒⋅⋅

=

m038,0cD

Qd

dDQ

AQc

2m22

222m

=⋅⋅Π

=

⋅⋅Π==

u2

c2w2

cu2

cm2

Thickness of the blade

Until now, we have not considered the thickness of the blade. The meridonial velocity will change because of this thickness.

( )

( ) m039,0cszD

Qd

dszDQ

AQc

2mu22

2u22m

=⋅⋅−⋅Π

=

⋅⋅−⋅Π==

We choose: s2 = 0,005 mz = 5

m013,05,22sin

005,0sin

ss o2

2u ==

β=

145,0K 1m =

sm4,6Hg2KcHg2

cK 1m1m1m

1m =⋅⋅⋅=⇒⋅⋅

=

u1

w1c1= cm1

405,0DD

2

1 =

m34,0D405,0D405,0DD

212

1 =⋅=⇒=

m09,0cD

QddD

QAQc

1mm11

1m111m =

⋅⋅Π=⇒

⋅⋅Π==

We choose:

Dhub

m17,0D5,0D 1hub =⋅=

m27,02DDD

2hub

21

m1 =+

=

Without thickness

Thickness of the blade at the inlet

m015,08,19sin

005,0sin

ss o1

11u ==

β=

u1

w1Cm1=6,4 m/s

sm8,17234,07,104

2Du 1

1 =⋅=⋅ω=

β1

o

1

1m1 8,19

8,174,6tana

uctana =

=

( ) m10,0cszD

Qd1m1um1

1 =⋅⋅−⋅Π

=

m15381,996,05,323,44

gcucuH

h

1u12u2 =⋅⋅

=⋅η

⋅−⋅=

u2=44,3 m/s

c2w2

cm2=4,87m/s

β2=22,5o

cu2

sm5,32tancuc

cuctan

2

2m22u

2u2

2m2 =

β−=⇒

−=β

u2=44,3 m/s

c2w2

cm2=4,87m/s

cu2

sm3,213,44

81,996,0100u

gHcg

cuH2

h2u

h

2u2 =⋅⋅

=⋅η⋅

=⇒⋅η⋅

=

o

2u2

2m2 9,11

3,213,4487,4tana

cuctana' =

−=

−=β

ooo2slipslip 6,109,115,22' =−=β−β=β