Q & A – Fluid Mechanics – Solving Pipe Networks using Quantity Balance Method – Q.4

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Q & A – Fluid Mechanics – Solving Pipe Networks using Quantity Balance Method – Q.4

Q & A – Fluid Mechanics – Solving Pipe Networks using Quantity Balance Method – Q.4

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Question No.4

Topic : Solving Pipe Networks using Quantity Balance Method

If in the network shown in the following figure the flow to C is regulated by a valve to 100 lit/sec, calculate the effect on the flows to the other reservoirs; determine the head loss to be provided by the valve.

Data :

Pipe

Length (m)

Diameter (mm)

AJ

10,000

450

BJ

2,000

350

CJ

3,000

300

DJ

3,000

250

Roughness size of all pipes = 0.06 mm

Given :

  • Roughness  heightk=0.06  mm
  • ZA=200  mZB=120  mZC=100  mZD=75  m
  • QCJ=100  lit/sec

Steps of solution :

Step (1) - Estimate ZJ (pressure head elevation at J) = 150 m (Note that the elevation of the pipe junction itself does not affect the solution.)

ZJ=150  m

Step (2) - Since the flow in CJ is prescribed, it is simply treated as an external outflow  at J.

For automatic computer analysis, Darcy-Colebrook-White combination can be used :

Q=2A2gDhfLlogk/D3.7+2.51νD2gDhfL

For each pipe (friction head loss) is initialized to :

ZIZJ

Step (3) – Calculate the correction value for the assumed pressure head elevation. 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 correction required in ZJ value :

ΔZJ=2(ΣQIJFJ)ΣQIJhL,IJ

For this problem, the flow in CJ is considered as an external outflow at junction J, therefore :

FJ=100  lit/sec=0.1  m3/sec

Therefore :

ΔZJ=2(ΣQIJ0.1)ΣQIJhL,IJ

For the first iteration :

Correction to ZJ :

ΔZJ=2(0.1150.1)0.0160=26.786  mZJ=15026.786=123.214  m

For the second iteration :

Correction to ZJ :

ΔZJ=2(0.1610.1)0.0316=3.879  mZJ=123.214+3.879=127.093  m

For the third iteration :

Correction to ZJ :

ΔZJ=2(0.1060.1)0.0240=0.501  mZJ=127.093+0.501=127.594  m

For the fourth iteration :

Correction to ZJ :

ZJ=127.5940.019=127.575  m

For the fifth iteration :

Correction to ZJ :

ZJ=127.576  m

Final Discharges :

Final discharges (after 5 iterations) :

Pipe

Q (l/sec)

Flow Direction

AJ

340.565

A to J

BJ

125.365

J to B

CJ

100.000

J to C

DJ

115.201

J to D

To calculate the head loss provided by the valve :

ZJZC=KCJQCJ2+hvalve

vCJ=QCJACJ=QCJπ4DCJ2=0.1π4(0.3)2=1.415  m/sec

To calculate Reynolds number :

Re=vDν=1.4150.31.011106=419,795

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

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

1f=2log0.06/3003.7+5.1286419,7950.89

f=0.0158

To calculate the head loss coefficient K for pipe CJ :

KCJ=8fCJLCJπ2gDCJ5=80.01583,000π29.81(0.3)5=1,610.271

ZJZC=KCJQCJ2+hvalve 127.576100=1,610.2710.12+hvalve

Finally, the head loss provided by the valve is :

hvalve=11.473  m

 

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