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CHAPTER 5:
PIPING SYSTEM
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EL AND HGL
Energy line (EL) shows totalhead for a certain crosssection in system.
p V2
EL 1 1+ +z 2 g 1
Hydraulic Grade Line (HGL)shows piezometric headfora certain cross section in
system.p1
HGL +z
1
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EL AND HGL# Example 1
EL & HG L
V22/2g
P2/g
z1 z2
Water flows from tank to atmosphere4
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EL AND HGL# Example 2EL & HGL
EL
HGL
z
V22/2g
hturbine
turbine
drop immediately in total energy head
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EL AND HGL # Example 3
EL
HGLV22/2g
Velocity
increased when
there is a
contraction in
diameter of pipe
Fluid flows through a nozzle
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EL AND HGL# Example 4
Water flows through 2 types of diameter pipes7
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EL AND HGL# Example 5EL
Energy equationHGL
Apam
zA
B
turbin
zB
from A to B:
DatumEnergyhead
2 2 lossesp V p VA A+ +z +( h Bh )= B+ +z +h
2 gA pam turbin B L
2 g
h hf hmmachine
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EL AND HGL Conclusion
EL always higher or same level with HGL. The difference between 2 lines is the value ofkinetic energy head, V2/2g.
Major losses - cause EL and HGL to decreasegradually.
Minor losses - cause EL and HGL to decreaserapidly.
Use of pump - Lines increased immediately Use of turbine - Lines decrease immediately
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FLOW TO ATMOSPHEREEL and HGL
Or z
A
V22/2gEL
HGL Hs= zA- zB
to atm
B
Energy equation from A to B:
p
AV
+2
2
A + zg A
2p VB B= + +
2gz +h +
B fh
m
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FLOW TO ATMOSPHERE
p V2 2p V
A
+ A +z
2gA
B B= + +z +hB L
2g
zA
z =B
2
V
2
2
B + h + hf mg
2 2
hS
V=
2B +g
fLV 0.5V+
2gD 2g
hS=2VB
2g
4fL1+ + 0.5
D
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FLOW TO ATMOSPHERE
V 2 fL 2 ghBh = 1+ s+ 0.5 V =S
2g DB
Let assume
as K
1.5+fLD
2ghQ =AV =A
k
s V =B2ghs
k
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EXAMPLE 5.1
Determine the discharge of water flows in the system.
A
Hs10 m
f = 0.005ke atm
L=100 m, d = 0.5 cm B
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EXAMPLE 5.1: SOLUTION
Taken from previous slide2V =
gh
ks
In test, the formula must be derivedfrom Bernoulli Equation
hS = zA+ zB= 10 cm , f = 0.005 , L = 100 m , D = 0.5 cmk = 1.5 + fL/D = 1.5 +(0.005x100)/0.005 = 101.5
V=2(9.81)(10)
101.5
=1.39m/s
5Q =AV =4
(0.005)2x1.39= 2.73x10 m3 /s
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FLOW WHICH CONNECTING 2
RESERVOIRSA
Hs
Energy equation
p 0 V
0 02
p V
B
02
from A to B:
Head loss atA A+ +z B B= + + z +h +h entrance and
2gA
2gB m f
submergeddischarge
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FLOW WHICH CONNECTING 2
RESERVOIRS
zA
z =B
0.5V2 g
2
2 2
fLV+ +2 gD
2
V2g
2ghV B fL sV =hS
= 0.5 +2g
+1D
Let assumeas K
B
1.5+fLD
Q =AV =A 2gh s Vk
2ghB =k
s
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EXAMPLE 5.2
Determine the difference level between 2 reservoirswhich is connected by single pipe of cast iron for 1km. Discharge and diameter is given as 0.01 m3/sand 5 cm respectively. ( =1.14 x 10-6m2/s)
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EXAMPLE 5.2: SOLUTION
V= QA
=
4
0.01
(0.05)
VD
=5.09m/s2
5.09x0.05 5
e
NR= =
v0.25
61.14x10
=2.23x10
=D
= = 0.00550
From Moody Chart , f =0.03
k=1.5+fL
D
0.03x1000= 1.5+
0.05=601 .5
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EXAMPLE 5.2 :SOLUTION
2V =
gh
ks
hs =kV
2g
2
601.5(5.09)2
hs
= = 794.28m2x9.81
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EXAMPLE 5.3 (Use Pump)
The tanks, pump and pipelines in the figure havethe characteristics noted. The suction lineentrance from the pressure tank is flush and thedischarge into the open tank is submerged. If the
pump P puts 2.0 hp into the liquid. (a) Determinethe flow rate (b) find the pressure in the pipe onthe suction side of the pump.
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EXAMPLE 5.3 (Use Pump)
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EXAMPLE 5.3a: SOLUTION
Qh 52QhP= p =2.0=
550
p Thus h550
p
21.2=
Q
Energy Equation, 1 to 3
p V2 2
p V6 6+ +z 8+h = + 8 +z + h + h 2g
5(144) 21.2
6 p
2V6
8
2g
50
m f
2
V62 2200 V V
8 8
52+ =10+0.5
Q 2 g
Q
+0.0256/12 2g
+0.03 8/12 2g 2g
QV=6 and V=80.1963 0.349
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EXAMPLE 5.3a: SOLUTION
Substituting for hp, V6, V8
23.8+21.2
Q
22.48Q = 2.48Q0
3 23.8Q21.2=0
Solving this by trial and error or equation polynomialsolver
Q=3.47cfs
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EXAMPLE 5.3b: SOLUTION
To obtain the pressure P2at the suction side of the pump
V=63.47
0.1963=17.68
Energy equation, 1 to 2
2 2 25(144) P=
17.682
+15+17.68
+0.550 17.68
+0.02552
P
2(32.2) 2(32.2)
52
6/12 2(32.2)
2
=20.6ft P2 =20.6 =7.43 psi
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NONRIGOROUS HEADLOSS EQNS
The empirical equation which are Hazen-William,Manning and Darcy Weisbach were rearranged intothe form of head loss equation.
Darcy - Weisbach2
Hazen - William Manning
hf 8fLQ= 2 5 10.67LQh=
1.85
2 210 .29LnQ gD
8fL
f 4.87C DH
hf
=16 .3D
210 .29Lnk=
, n = 22 5gD
10.67Lk=
4.87, n = 1.85 k=
D 16.3CHD, n = 2
As we can see that hf= kQn
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BRANCHING PIPE
Let us consider 3 pipes connected to 3 reservoirsand branching together at common junction point J.
We shall assume that all the pipesare sufficient long that we can
neglect minor losses and velocityhead so hL= hfwhich we shalldesignate as h
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BRANCHING PIPE
As there are no pumps, the elevation of P must liebetween the surfaces of reservoirs A and C.
If P is below the surface of reservoir B then h2andQ2are both zero.
If P is above the surface of reservoir B then watermust flow into B and Q1= Q2+Q3.
If P is below the surface of reservoir B then watermust be out of B and Q1+Q2 = Q3.
There are several different method of solutions
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EXAMPLE 5.4
The elevations of water surfaces in reservoirs A andC are 250 ft and 160 ft, respectively and thedischarge Q2into reservoir B is 3.3 cfs. Find thesurface elevation of reservoir B.
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EXAMPLE 5.4
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EXAMPLE 5.4: SOLUTION
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EXAMPLE 5.4: SOLUTION
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EXAMPLE 5 5
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EXAMPLE 5.5
With the sizes, length and material of pipes given asexample 5.4, suppose that the surface elevation ofreservoirs A, B and C are 525 ft, 500 ft and 430 ftrespectively.
(a) Does water enter or leaves reservoir B?(b) Find the flow rates of 60oF water in each pipe.
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EXAMPLE 5.5: SOLUTION
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EXAMPLE 5.5: SOLUTION
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EXAMPLE 5.5: SOLUTION
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EXAMPLE 5.6
The surface elevation of reservoirs A, B and C are160 m, 150 m and 120 m respectively.
(a) Does water enter or leaves reservoir B?(b) Find the flow rates in each pipe.
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EXAMPLE 5 6
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EXAMPLE 5.6
ZMA/S20607 37
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EXAMPLE 5.6: SOLUTION
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EXAMPLE 5.6: SOLUTION
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PIPES IN SERIESIf a pipeline is made up of lengths of differentdiameters, conditions must satisfy the continuity andenergy equations namely,
If given the rate ofdischarge Q, we findthat hL= Q
Using the nonrigorous eqn,h =h =kQ
n n n+kQ +kQ +L f 1 1 2 2 3 3
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PIPES IN SERIESBut since all the Qs are equal, this becomes
h =( k + k+k + ....) Qn n=(k)QL 1 2 3 3If substituting from Darcy eqn into hL(for series) and
including minor losses if we wish (usually L/D
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EXAMPLE 5.7
Referring to figure (previous slide) and data givenfor new cast iron pipes, find the rate of flow from Ato B in conveying 15oC water. (z or hs= 10 m andneglect minor losses)
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EXAMPLE 5 7 :SOLU ION
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EXAMPLE 5.7 :SOLU ION
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EXAMPLE 5 7 :SOLU ION
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EXAMPLE 5.7 :SOLU ION(Sample Problem 8.17)
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EXAMPLE 5.8
2 reservoirs were connected by a pipeline 1 (150mm diameter,6 m length) and pipeline 2 (255 mmdiameter and 15 m length). Inlet and outlet ofpipes are categories as a sharp and z is 6 m. Tableall losses, calculate the rate of flow and sketch EL
and HGL for the system. Take f = 0.01 for bothpipes.
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EXAMPLE 5.8: SOLUTION
Continuity Equation:Q = A1V1= A2V2 V1= (A2/A1)2V2 = 2.89V 2
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EXAMPLE 5.8: SOLUTION
Pers. Bernoulli diantara 1 dan 2,
p
1V
+2
2
1 +zg
1
2p V2 2= + +
2g
12 .94V
z +h2 L
22 V =3.02m/s6=
2g
(0.255 )
2
2
3Q =AV =2 2
x3.02 =0.154m /s4
V1= 2.89V 2 = 2.89 (3.02) = 8.73 m/s
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EXAMPLE 5.8: SOLUTION
EL and HGL
A
EL
HGL
d1= 0.15 m
L1= 6 m
d2 = 0.255 m
L2= 15 m
The sketches of EL and HGL
6 m
B
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PIPES IN PARALLEL
If a pipeline is made up of lengths of differentdiameters, conditions must satisfy the continuity andenergy equations namely,
If we are given the total flow Q and want to find hL
1/n 1/nhh=kQ n Q
hf1 f2= + +
fk k
1 2
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PIPES IN PARALLEL
But since all the hfs (= hLs) are equal, this becomes1/n 1/ n 1/n
1/n 1 1 1/ n 1Q=(h) + + =(h)
fk k
f
k 1 2
To find hLand the individual Qs, and including minor losses,
can write the head loss eqn as
hL
=L
f +kD
2V
2 g
Where k is the sum of the minor
loss coefficient
We usually neglect if the pipe is longer than 1000
diameters. Solving for V and then Q for pipe 1.
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PIPES IN PARALLEL
2ghLQ =AV =A =C h Where C1=constant1 1 1 1f(L/D) +
1 Lk
1 1 1 1 for given pipe
We can similarly express the flows in the other pipesusing assumed values of f. Finally becomes
Q =C h +C h +C h =(C +C +C) h1 L 2 L 3 L 1 2 3 L
This enables us to find a first estimate of hLand distribution offlows and velocities in the pipes. Using these, improvements to
the f values can be made. If necessary repeat them until wefinally obtain a correct determination of hLand distribution of
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EXAMPLE 5.9
There pipes A, B and C are interconnected as in figurebelow. The pipe characteristics are as follows:
Find the rate at whichwater will flow in each
pipe. Find also the
pressure at point P. Allpipe lengths are much
greater than 1000diameters, so neglect
minor losses
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EXAMPLE 5 9 S OLUTION
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EXAMPLE 5.9: S OLUTION
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EXAMPLE 5 9: SOLUTION
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EXAMPLE 5.9: SOLUTION(Sample Problem 8.18)
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EXAMPLE 5.9: SOLUTION
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EXAMPLE 5.9: SOLUTION(Sample Problem 8.18)
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EXAMPLE 5.10
Pipe 2 has a length 4 times longer thanpipe 1. Both pipe has a same diameterwhich is 305 mm. If discharge in pipe 2 is1 m3/s, find discharge for pipe 1. Calculate
flow rate of system if f equal to 0.01 andneglect minor losses.
2
1
A B
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EXAMPLE 5.10
V Q 1= =13.69m/s2=A
24 (0.305)
2
In parallel pipe:
hL1= hL2 and h f1 = h f2
2 22 2 f LV f LVV1
V1
=1x4x1x13.69=27.3m/s
1 1
2gD1 =1
2 2 2
2gD2
where L2/L1= 4
Q1=A1V1=
(0.305 )
4
2
x27.30=3
1.99m / s
Qsystem = Q1+ Q2= 1.99 + 1 = 2.99 m3/s57