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Water Pumps andWater Pumps andPumping SystemsPumping Systems
ByBy
Dr. MohaDr. Mohammmedmed HaggagHaggag
Lecture 2Lecture 2
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Applications of the Darcy-Weisbach Equ.
Type I: To Determine Head LossGiven (d, v or Q, material)
Direct application
Type II: To Determine Velocity or Flow Rate
Given (d, hf, material)
Since V or (Q) is not known, Re and f can not bedetermined directly.
Iterative application
Type III: To Determine DiameterGiven (Q, limit of hf)
Since V or (Q) is not known, Re and f can not be
determined directly.Iterative application
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In addition to the continuous head loss along thepipe length due to friction, local head lossesoccur at changes in pipe section, bends, valves
and fittings.
Head loss due to
outlet, inlet, bends, elbows, valves, pipe sizechanges
Can be disregarded in long pipes
But significant for short lengths (< 50 m)
Most minor losses can not be obtainedanalytically, so they must be measured
Head Loss: Minor Losses
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Equivalent Length Technique
A fictions length of pipe is estimated that will cause thesame pressure drop as any fitting or change in pipe
cross section. This length is added to the actual pipelength
Proportional to Kinetic energy technique
The loss is considered proportional to kinetic energyhead given by the following formula
Head Loss: Minor Losses
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Head Loss: Minor Losses
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Flow expansions have high losses
Kinetic energy decreases across expansion
Kinetic energy ________ and _________
energy
Examples
________________________________
____________________________________
Losses can be minimized by gradualtransitions
Head Loss: Minor Losses
potential thermal
Hydraulic jump
Vena contracta
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Apply in direction of flow
Neglect surface shear
Divide by (A2)
Head Loss due to Sudden Expansion:
Conservation of Momentum
Pressure is applied over all ofsection 1.
Momentum is transferred overarea corresponding toupstream pipe diameter.V1 is velocity upstream.
sspp FFFWMM +++=+21
21
1 2
xx ppxx FF
2121 +=+
1
2
11 AVM x = 22
22 AVM x =
22212
2
21
2
1 ApApAVAV =+
g
AVV
pp 2
12
1
2
2
21
=
AA11AA22
xx
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Head Loss due to Sudden Expansion:
Conservation of Energy
1 2
l
tp
hg
Vzpg
Vzp +++=++22
2
22
2
2
2
11
1
1
lhg
VVpp
+
=
2
2
1
2
221
g
VVpph l
2
2
2
2
121 +
=
z1 = z2
What isp1
- p2
?
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Energy
Head Loss due to
Sudden Expansion
g
VVpph
l 2
2
2
2
121 +
=
g
AVV
pp 2
12
1
2
2
21
=
1
2
2
1
V
V
A
A
=
g
VV
g
V
VVV
h l 2
2
2
2
11
22
1
2
2
+
= g
VVVV
h l 2
2 21212
2 +
=
( )
g
VVh
l 2
2
21=
2
2
1
2
1 12
=
A
A
g
Vh l
2
2
11
=
A
AK
Momentum
Mass
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Contraction
V1 V2
EGL
HGL
vena contracta
g
VKh
cc2
22=
losses are reduced with a gradual contraction
Expansion!!!
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Entrance Losses
Losses can bereduced byaccelerating the
flow gradually andeliminating thevena contracta
Ke 0.5
Ke 1.0
Ke 0.04
he=Ke V
2
2g
reentrant
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Head Loss in Valves
Function of valve typeand valve position
The complex flow paththrough valves oftenresults in high head loss
What is the maximumvalue that K
vcan have?
_____
hv=KvV2
2g
How can K be greater than 1?
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Minor/Local Losses
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Questions
What is the headloss when a pipeenters a
reservoir?
Draw the EGL
and HGL
V
gV2
2
EGLHGL
2
2
1
1
= A
A
K
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Can the Darcy-Weisbach equation andMoody Diagram be used for fluids otherthan water? _____
Does a perfectly smooth pipe have head loss?
_____
Is it possible to decrease the head loss in a
pipe by installing a smooth liner? ______
Questions
Yes
Yes
Yes
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D=40 cmL=1000 m
D=20 cmL=500 m
valve100 m
Find the discharge, Q.What additional information do you need?Apply energy equationHow could you get a quick estimate? _________________Or spreadsheet solution: find head loss as function of Q.
lh
g
Vz
p
g
Vz
p+++=++
22
2
22
2
2
2
11
1
1
cs1
cs2
2
21 0 02
l
Vm h
g= +
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Quiz
In the rough pipe law region if the flow rate is
doubled (be as specific as possible)What happens to the major head loss?
What happens to the minor head loss?
Why do contractions have energy loss?
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Can you draw TEL and HGL?
hi
hfhe
V2/2g
Hs
Hs=hi+hf+he
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Hs=hi+hf+he
Thus
Static
head
Inlet head
loss
Friction head
loss
Exit head
loss
LossesHs =
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71
Hs=hi+hf+he
Can you get the required d to passa given flow in the system?
g
v
Keg
v
d
Lf
g
v
KiHs 22
.
2
222
++=
g
v
d
LfKHs
n
i 2)
.(
2
1
+= =
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Q
d
))4/.(2
).
(22
2
1
mindg
Q
d
lfkH
n
i
ors
+= =
g
v
d
LfKHs
n
i 2
).
(2
1
+= =
Q(eq. 2)fRnV=Q/AAd
Required d
Can you get the required d to
pass a given flow in the system?
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What if
d > available pipe
diameters
in the market
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In this case,
One could think of usinga pump
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Can you draw TEL and HGL?
Hp=Hs+hi+hfs+hfd+he
HpHs
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Can you get now Hp?
Hp=Hs+hi+hfs+hfd+he
g
v
d
Lf
d
LfKHH
d
dd
s
ssn
i
sp2
)..
(2
1
+++= =
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Can you draw TEL and HGL?
HpHs
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TEL Concept
For the shown Transmission Line Profile, Draw the TELand HGL
For the shown Transmission Line Profile, Draw the TELand HGL
What if we used amore powerful
pump?
Note also that:2- TEL Slope becomes steeper
as:
Q gets Bigger;D gets Smaller;
Pipe gets Rougher.
Note also that:2- TEL Slope becomes steeper
as:
Q gets Bigger;D gets Smaller;
Pipe gets Rougher.
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Answer the Following
Identify the flow direction for the given pipe:
PA =62 psi
PA =48 psi
ZA = 100 m
ZB = 112 m
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Q
Answer the Following
For the given TEL of the
shown transmission line:
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Determination of Pumping Head
Calculation of Hp,A numerical Example
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a) Calculate the required pumping head
b) Electric motor power (t =0.78)c) Monthly and annual cost of workingelectrical power if:
- Daily working hours = 12 hrs
- Cost of 1kw-hour=0.18 L.E.82
Calculation of Hreq & Electric Power
Example
(11.40)
(20.50)
60o, r/d=2
Q=100 l/s d= 300 mm L = 4.8 km
G.Steel pipe
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83
g
v
d
Lf
KHH
n
isp
2)
.
(
2
1++= =
mg
V
smAA
QV
mA
10211197.02
/415.11.0
07065.04
)3.0.(
2
22
=
===
==
Hs=20.5-11.4 =9.1 m
Pipe material Galvanized S. = 0.15 mm /d = 0.0005
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84
5
6102.4
10011.1
3.0415428.1.x
x
xdVRn ===
From Moody Diagram get f
f 0.0165
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85
Determination of Local Losses
KiType of local losses
Inlet
Exit
BendValves
Others
Ki
1.0
0.21x30.25+3
1.0
5.88
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g
v
d
Lf
KHH
n
isp 2)
.
(
2
1++= =
mH
H
p
p
6.37)6001.0448.27(1.9
1021119.0*)3.0
4800*0165.088.5(1.9
=++=
++=
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Electricenergy
Mechanicalenergy
Water
energyMotor Pump
Efficiencies
mpt
t
m
p
x
PHE
PHWPHE
PHMPHM
PHW
=
=
=
=
..
....
..
..
..
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Calculation of EHP Cost
WHP =75
.. pHQ =1000 kg/m3,
Q = 0.1 m3/s,
Hp=37.6m
WHP = 49.48
EHP= 49.48/0.78 = 63.44
EMP = EHPx0.746=48KWatt
Monthly cost of E.P.=E.M.PxTxCost
Monthly cost of E.P.=48x12(hrs)x30(day)x0.18
= 3100 L.E.
Annual Cost of E.P.= 3100x12(month)=37000 L.E.
(t =0.78)
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Any Questions?
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How to get f?
Moody Diagram (Chart)
dV.
d
0.025
0.02
0.0180.016
0.014
0.012
0.010
0.008
0.006
0.005
0.004
0.0030.0025
0.002
103
104
105
107
106
108
Friction Coefficient
Reynolds Number
Roughness Ratio
(k/d)
0.01
0.001
0.0001
0.00001
0.00004
0.0002
0.0006
0.002
0.004
0.006
0.02
0.04
0.06
4/f
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RoughnessHeight (mm)
Pipe Material
0.003PVC
0.03Glass fiber ReinforcedPlastic (GRP)
0.15Galvanized Steel
0.25Cast Iron
1.00Reinforced Concrete
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Inlet losses
Ki = 0.5 Ki = 1.0
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Exit losses
Ke = 1.0
Bend/Elbow losses
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Bend/Elbow losses
o, r/d
Bendr
o
Elbow
4321.51 r/d
0.080.080.090.10.1122.5
0.150.150.160.170.1945
0.190.20.210.220.2560
0.260.260.270.290.3390
0.350.350.350.360.41135
0.420.420.420.430.48180
907560453022.5
1.510.70.40.20.17Kelbow
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Valve LossesPartially OpenFully Open
Kgate = 0.25Gate Valve
Kcheck = 3
Check valve
(Non Return Valve)
Check valve Gate valve
Radial Flow Centrifugal Pump
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Radial Flow Centrifugal Pump
Settings
Suction level
Delivery level
Suction pipe, ls
Delivery pipe, ld
hs
Axial Flow Propeller Pump
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Axial Flow Propeller Pump
Settings
Delivery level
Delivery pipe, ld
Motor