Pipeflow
Pipe#6
Civil Engineering Fluid Mechanics
Covered in this lecture:
1. Hazen Williams Equations 2. Hardy Cross Methods Hazen William Formula
Historically, and number of engineering formula have been used to compute head loss. One widely used approach is the Hazen − Williams formulation 0.54 S ( FSS units) : V = 1.318 C hw R 0.63 h 0.54 ( SI units) : V = 0.849 C hw R 0.63 S h
Wher Wheree Rh is the hy hyddraul raulic ic radi radiuus (A/w (A/wpp), S is the the head ead loss loss per un unit it length, and Chw is the roughness coefficient. Hazen−Williams Coefficient Chw Pipes extremely straight & smooth Pipes very smooth Smooth wood, smooth masonry New riveted steel, vitrified clay Old cast iron, ordinary brick Old riveted steel Old iron in poor condition
140 130 120 110 100 95 60−80
ThereisarelationshipbetweentheDarcy f and and the the fo form rmul ula, a, assu assumming water at 70 deg F f =
1090 C 1.85 d 0.015 R 0.15 e hw
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Pipeflow
Pipe#6
Civil Engineering Fluid Mechanics
Advantages to Hazen Williams approach
1. Coefficient Chw is rough measure of relative roughness 2. Effect of Reynolds number is included in formula 3. Effect of roughness on velocity are given directly Disadvantages to Hazen Williams approach
1. Empirical 2. Can not be applied to all fluids in all conditions Example: If 90 gal/min of water flows through smooth 3 in pipe,
calculate the head loss is 6,000 ft of pipe. Given: Q = VA R h = d ∕4 0.54 V = 1.318 C hw R 0.63 S h
Q = 90 gal/min
d= 3 in = 0.25 ft
l = 6000 ft
Solution: V =
90 × π 13 2 = 4.08 ft ∕s 60 × 7.48 ( )( ) 4 12
R h = ( 3 )∕4
12
= 0.0625 ft
C hw = 140
4.08 = 1.138 × 140 × (0.0625) 0.63 S 0.54 S
= 0.0218 = h l ∕6000
2
h l = 131 ft of water
Pipeflow
Civil Engineering Fluid Mechanics
Pipe#6
Hardy Cross Method
Multiple pipes reach greatest complexity in distribution problems, e.g. city water supply
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Basic principles are presented here
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Pipe network is the aggregation of connected pipes used to distribute water
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Network consists of various size pipes, geometric orientations, hydraulic characteristics, plus pumps, valves, fittings, etc
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A simple pipe network is presented below
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Pipe Junctions denoted by capital letters A −H
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Individual pipes numbered 1 −10
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Closed circuits given Roman Numerals I −III 3
Pipeflow
Civil Engineering Fluid Mechanics
S
Flows are assumed to be clockwise around each loop Pipes 1, 3, 4 and 2 comprise Loop 1
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Pipes 2, 8, 10, 7 comprise Loop 2
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Pipe 4 is common to both loops
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S
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Solution to any network problem must satisfy continuity and energy principles throughout the network Bernouli principle requires that at any junction only one EGL is possible
Any head loss around any single loop must be zero.
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Applying this to each loop and junction results in a series of simultaneous equations
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Pipe#6
Pipeflow
Pipe#6
Civil Engineering Fluid Mechanics
S
Equations for loop I are
Q = −Q
A
+ Q2 − Q1 = 0
A
Q = Q Q = Q
1 + Q F − Q 3
=0
3 − Q4 − Q8
=0
F
E
Q = −Q
2 + Q4 + Q7 + Q5
=0
B
h
n n n n = + + + K Q K Q K Q K Q L 1 1 3 3 4 4 2 2=0
I
There are like equations for the other loops in the network. S
Flow directions are assumed, may actually be different
Assumed that pipe sizes, length, hydraulic characteristics, inflows and outflows, elevations, are known.
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Unknown flowrates Qi, i= 1 to 10 are to be determined
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The solution is by trial and error
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The simplest method is known as Hardy−Cross Starts with reasonable set of guesses of Qi that satisfy continuity and then iterate until head loss is satisfied
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Pipeflow
Civil Engineering Fluid Mechanics
Pipe#6
If first guesses are relatively accurate, then the actual flowrates Qi should be only a small ∆L different from t he guesses, Q0i in each loop
S
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For a pipe that is only in one loop, then Q i = Q 01 Δ L
where (+ −) depends on the directions for each assumed Qi and ∆L To maintain continuity, the correction ∆L is applied to every pipe in a loop
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For example
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Q 2 = Q 03 + Δ I Q 8 = Q 08 + Δ II Q 4 = Q 04 + Δ I − Δ II
because ∆L is always assumed to be clockwise around each loop. In general the head loss takes is expressed as
h
L
L
=
() K Q i
n i
=0
L
where (+−) depends on the direction of the flow and the Qi are taken as the magnitudes of the flow. For the assumes flow directions
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Pipeflow
Pipe#6
Civil Engineering Fluid Mechanics
and loops given in the figure, the head loss equations are n + K 1Qn1 + K 3Q n3 + K 4Q N K Q 2 2=0 4 n − K 4Qn4 + K 8Q n8 + K 10Q N + K Q 7 7=0 10 n + K 5Q n5 + K 7Q n7 + K 9Q N K Q 6 6=0 9
and related to our flow rate guess adjustment equation
h
L
=
L
() K (Q i i
01
Δ L) n = 0
L
Expanding by the binomial theorem and neglect all terms with small products of ∆L, e.g. ∆L 2, ∆L 3 etc because ∆L is small, then
h
L =
L
() K (Q i i
n
01
1Δ ) = 0 nQ n0− L i
L
Solving for ∆L
() K Q L = − nK Q − | i i
Δ
N 0i
L
i
n 1 0i
L
This equation is used to correct to guessed flows in each loop. Process is continued until continuity is satisfied and ∆L is very small
S
Process can be complicated by boundary conditions, pumps, etc. You will use WaterCad to solve these equations.
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