Q & A – Fluid Mechanics – Flow of Incompressible Fluids in Pipelines – Q.2

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Q & A – Fluid Mechanics – Flow of Incompressible Fluids in Pipelines – Q.2

Q & A – Fluid Mechanics – Flow of Incompressible Fluids in Pipelines – Q.2

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Question No.2

Topic : Flow of Incompressible Fluids in Pipelines

A uniform pipeline, 5000 m long, 200 mm in diameter and roughness size 0.03 mm, conveys water at 15oC between two reservoirs, the difference in water level between which is maintained constant at 50 m. In addition to the entry loss of 0.5 v2 / 2g a valve produces a head loss of 10 v2 / 2g. α = 1.0

Determine the steady discharge between the reservoirs using :

  • the Colebrook-White equation
  • the Moody Diagram
  • the Hydraulics Research Station Charts
  • the Barr's equation

Given :

  • L=5,000  m
  • D=200  mm=0.2  m
  • k=0.03  mm
  • Temperature : 15oC
  • zAzB=50  m
  • hentry=0.5v22g
  • hvalve=10v22g
  • α=1

Solution :

Consider two points, Point A at the water surface of tank A and Point B at the water surface of tank B. Apply Bernoulli's equation (Conservation of Energy Principle) between the two points, therefore :

T.EA=T.EB+hlAB

zA+PAγ+vA22g=zB+PBγ+vB22g+hlAB

hlAB=hentry+hf+hvalve+hexit

hlAB=0.5v22g+fLDv22g+10v22g+v22g

hlAB=0.5+fLD+10+1v22g hlAB=fLD+11.5v22g

zA+PAγ+vA22g=zB+PBγ+vB22g+hlAB

50+0+0=0+0+0+hlAB

hlAB=50  m

fLD+11.5v22g=50 1

Method 1 - Using the Colebrook-White equation :

1f=2logk/D3.7+2.51Ref

The solution to this problem is obtained by solving equation 1 and Colebrook-White equation simultaneously. However, direct substitution of the friction coefficient f from equation 1 into the Colebrook-White equation yields a complex implicit function in flow velocity v which can only be evaluated by trial or graphical interpolation.

A simpler computational procedure is obtained if terms other than the friction head loss in equation 1 are initially ignored; in other words the gross head is assumed to be totally absorbed in overcoming friction. Then, the following equation can be used to obtain an approximate value of flow velocity v :

v=22gDhfLlogk/D3.7+2.51νD2gDhfL

hf=H=zAzB=50m hfL=505,000=0.01

v=229.810.20.01log0.03/2003.7+2.511.131060.229.810.20.01=1.564  m/sec

The terms other than the friction head losses in equation 1 can now be evaluated :

hminor=hentry+hvalve+hexit

hminor=0.5v22g+10v22g+v22g

hminor=0.5+10+1v22g hminor=11.5v22g

hminor=11.51.564229.81=1.435  m

A better estimate of the friction head losses can now be obtained :

hf=Hhminor=501.435=48.565  m hfL=48.5655,000=0.009713

v=22gDhfLlogk/D3.7+2.51νD2gDhfL

v=229.810.20.009713log0.03/2003.7+2.511.131060.229.810.20.009713=1.544  m/sec

Repeating until successive values of v are sufficiently close yields :

v=1.541  m/secQ=48.41  l/sechf=48.61  mhm=1.39  m

Method 2 - Using the Moody Diagram :

The use of the Moody Diagram involves the determination of the friction factor f. In this case, the minor losses need not be neglected initially. However, the solution is still iterative and an estimate of the mean velocity is needed.

Assume the flow velocity in the pipe :

v=m/sec

Then, calculate the Reynolds number :

Re=vDν=20.21.13106=3.54105

Calculate the relative roughness for the pipe :

k/D=0.03/200=0.00015

Using the Moody Diagram :

f=0.015

By substituting in equation 1, the flow velocity is obtained :

fLD+11.5v22g=50 0.0155,0000.2+11.5v22g=50

v=1.593  m/sec

Then, calculate the revised Reynolds number :

Re=vDν=1.5930.21.13106=2.82105

Using the Moody Diagram :

f=0.016

By substituting in equation 1, the flow velocity is obtained :

fLD+11.5v22g=50 0.0165,0000.2+11.5v22g=50

v=1.54  m/sec

Further change in friction factor f due to small change in flow velocity v will be undetectable in the Moody Diagram.

Therefore, the final values are :

v=1.54  m/sec,Q=48.41  l/sec

Method 3 - Using the Hydraulics Research Station Charts :

The solution by the use of the Hydraulics Research Station Charts is basically the same as the first method (Colebrook-White equation) except that values of flow velocity v are obtained directly from the chart instead of the following equation :

v=22gDhfLlogk/D3.7+2.51νD2gDhfL

Starting with the initial assumption that the friction head losses is equal to the elevation difference between the two reservoirs A & B :

hf=H=50  m

The hydraulic gradient is :

100hL=100505,000=1

Using the Hydraulics Research Station Chart, the flow velocity is obtained :

v=1.55  m/sec

hminor=11.5v22g hminor=11.51.55229.81=1.41  m

A better estimate of the friction head losses can now be obtained :

hf=Hhminor=501.41=48.59  m 100hL=10048.595,000=0.972

Using the Hydraulics Research Station Chart, the flow velocity is obtained :

v=1.5  m/sec,Q=47.1  lit/sec

Note that the loss of fine accuracy is due to the graphical interpolation.

Using Barr's equation :

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

The solution is still iterative and an estimate of the mean velocity is needed.

Assume the flow velocity in the pipe :

v=m/sec

Then, calculate the Reynolds number :

Re=vDν=20.21.13106=3.54105

Calculate the relative roughness for the pipe :

k/D=0.03/200=0.00015

Method 4 - Using the Barr's equation :

f=0.0156

By substituting in equation 1, the flow velocity is obtained :

fLD+11.5v22g=50 0.01565,0000.2+11.5v22g=50

v=1.563  m/sec

Then, calculate the revised Reynolds number :

Re=vDν=1.5630.21.13106

Using the Barr's equation :

f=0.0161

By substituting in equation 1, the flow velocity is obtained :

fLD+11.5v22g=50 0.01615,0000.2+11.5v22g=50

v=1.54  m/sec

Therefore, the final values are :

v=1.54  m/secQ=48.41  l/sec

 

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