Q & A – Fluid Mechanics – Solving Pipe Networks using Quantity Balance Method – Q.6

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Q & A – Fluid Mechanics – Solving Pipe Networks using Quantity Balance Method – Q.6

Q & A – Fluid Mechanics – Solving Pipe Networks using Quantity Balance Method – Q.6

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Question No.6

Topic : Solving Pipe Networks using Quantity Balance Method

In the system illustrated in the following figure, a pump is installed in pipe BC to provide a flow of 40 lit/sec to reservoir C. Neglecting minor losses, calculate the total head to be generated by the pump and the power consumption assuming an overall efficiency of 60 per cent. Determine also the flow rates in the other pipes.

Data :

Pipe

Length (m)

Diameter (mm)

AB

10,000

400

BC

4,000

250

BD

5,000

250

Roughness size of all pipes = 0.06 mm

Given :

  • Roughness  heightk=0.06  mm
  • ZA=150  mZC=145  mZD=90  m
  • QBC=40  lit/sec
  • ηtotal=0.60

Steps of solution :

Step (1) - Estimate ZB (pressure head elevation at B) = 130 m (Note that the elevation of the pipe junction itself does not affect the solution.)

ZB=130  m

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

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 :

ZIZB

Step (3) – Calculate the correction value for the estimated 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 ZB value :

ΔZB=2(ΣQIBFB)ΣQIBhL,IB

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

FB=40  lit/sec=0.04  m3/sec

Therefore :

ΔZB=2(ΣQIB0.04)ΣQIBhL,IB

For the first iteration :

Correction to ZB :

ΔZB=2(0.0510.04)0.0083=2.582  mZB=130+2.582=132.582  m

For the second iteration :

Correction to ZB :

ΔZB=2(0.0390.04)0.0086=0.199  mZB=132.5820.199=132.383  m

For the third iteration :

Correction to ZB :

ΔZB=2(0.0400.04)0.0086=0.011  mZB=132.383+0.011=132.394  m

For the fourth iteration :

Correction to ZB :

ΔZB=2(0.0400.04)0.0086=0.001  mZB=132.3940.001=132.394  m

Final discharges :

Final discharges (after 4 iterations) :

Pipe

Q (l/sec)

Flow Direction

AB

118.896

A to B

BC

40.000

B to C

BD

78.896

B to D

To calculate the total head generated by the pump :

ZBZC+Hpump=KBCQBC2

vBC=QBCABC=QBCπ4DBC2=0.04π4(0.25)2=0.815  m/sec

To calculate Reynolds number :

Re=vDν=0.8150.251.011106=201,502

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

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

1f=log0.06/2503.7+5.1286201,5020.89

f=0.0174

To calculate the head loss coefficient K for pipe BC :

KBC=8fBCLBCπ2gDBC5=80.01744,000π29.81(0.25)5=5,892.25

ZBZC+Hpump=KBCQBC2

132.394145+Hpump=5,892.250.042

Therefore, the total head provided by the pump is :

Hpump=22.034  m

To calculate the power consumption :

Output  Power=γwQpumpHpump=9,8100.0422.034=8,646.14  Watt

ηtotal=Output  PowerInput  Power 0.60=8,646.14InputPower

Input  Power=14,410.236  Watt=14.41  kW

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