Q & A – Fluid Mechanics – Solving Pipe Networks using Hardy-Cross Method – Q.5

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Q & A – Fluid Mechanics – Solving Pipe Networks using Hardy-Cross Method – Q.5

Q & A – Fluid Mechanics – Solving Pipe Networks using Hardy-Cross Method – Q.5

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Question No.5

Topic : Pipe Network Analysis (Hardy-Cross Method)

In the system shown in the following figure, a pump is installed in BC to boost the flow to C. Neglecting minor losses, determine the flow distribution and head elevations at the junctions of the pump delivers a head of 15 m. All pipes have the same roughness size of 0.03 mm. The outflows at the junctions are shown in l/sec.

Data :

Pipe

AB

BC

CD

DE

EA

BE

Length (m)

500

600

200

600

600

200

Diameter (mm)

200

150

100

150

200

100

Pressure head elevation at A = 60 m

Given :

  • Roughness  heightk=0.03  mm
  • ZA+PA/γ=60  m
  • Hpump=15  m

Steps of solution :

Step (1) - Identify loops. When using hand calculations, the simplest way is to employ adjacent loops.

Loop 1 : ABEA

Loop 2 : BCDEB

The inlet flow at junction A is equal to the sum of the outlet flow at the other junctions, therefore the inlet flow at junction A is equal to :

= 60 + 50 + 50 + 40 = 200 l/sec

Step (2) - Allocate estimated flows in the pipes. Only one estimated flow in each loop is required; the remaining flows follow automatically from the continuity condition at the nodes.

Assume that water is flowing in pipe AB with a flowrate of 120 lit/sec (from junction A to junction B) and flowrate in pipe BC is 40 lit/sec (from junction B to junction C).

Therefore, the initial estimates of flowrates through the network are :

Pipe

Q (l/sec)

AB

120

BC

40

CD

-10

DE

-50

EA

-80

BE

+20/-20

Step (3) – Evaluate the head loss coefficient K for each pipe. The friction coefficient f is obtained through Moody chart or using Barr's equation. The calculations proceed in tabular form. Note that Q is written in liter/sec simply for convenience; all computations are based on Q in m3/sec.

To calculate the velocity at each pipe :

v=QA=Qπ4D2

To calculate Reynolds number :

Re=vDν

To calculate the friction coefficient, Barr's equation is used :

1f=2logk/D3.7+5.1286Re0.89

Barr's equation is based partly on an approximation to the logarithmic smooth turbulent element in the Colebrook-White equation.

To calculate K for each pipe (using Darcy-Weisbach equation) :

K=8fLπ2gD5

To calculate the correction required in the discharge value :

ΔQ=ΣKQQ2ΣKQ

For the 1st iteration :

ΔQ1=11.948  l/sec,ΔQ2=13.298  l/sec

For the 2nd iteration :

ΔQ1=1.400  l/sec,ΔQ2=0.591  l/sec

For the 3rd iteration :

ΔQ1=0.219  l/sec,ΔQ2=0.127  l/sec

For the 4th iteration :

ΔQ1=0.050  l/sec,ΔQ2=0.015  l/sec

For the 5th iteration :

ΔQ1=0.007  l/sec,ΔQ2=0.007  l/sec

For the 6th iteration :

ΔQ1=0.002  l/sec,ΔQ2=l/sec

Final discharges (after 6 iterations) :

Pipe

Q (l/sec)

Head loss (m)

AB

106.373

21.405

BC

52.558

13.337

CD

2.558

0.258

DE

37.442

14.876

EA

93.627

20.123

BE

6.185

1.282

To calculate the pressure head elevations at different junctions, use the energy equation between point A and different points ,total energy at any point can be calculated through :

T.E=Z+Pγ+v22g

Since the minor losses can be neglected, the velocity head term is negligible (equal to zero), therefore the total energy at point A is equal to :

T.EA=ZA+PAγ+vA22g T.EA=60+0=60m

To calculate the pressure head elevation at point B, apply Bernoulli's equation between points A & B :

T.EA=T.EB+hlAB

T.EB=ZB+PBγ+vB22g T.EB=ZB+PBγ+0

60=ZB+PBγ+21.405 ZB+PBγ=38.595m

Similarly, the pressure head elevations at different points is summarized in the following table :

Junction

T.E (m)

A

60

B

38.595

C

25.258

D

25.001

E

39.877

 

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