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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
=
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
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 =⋅
=⋅
=
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
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=
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 =
=
=β
( ) 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