Module 4

Flow in pipes

1- Losses in pipes

Pressure loss due to friction in a pipeline.

In a pipe with a real fluid flowing, at the wall there is a shearing stress retarding the flow, as shown below.

The pressure at 1 (upstream) is higher than the pressure at 2.

We can do some analysis to express this loss in pressure in terms of the forces acting on the fluid. Consider a cylindrical element of incompressible fluid flowing in the pipe, as shown

The pressure at the upstream end is p, and at the downstream end the pressure has fallen by p to (p-p).The driving force due to pressure (F = Pressure x Area) can then be written

driving force = Pressure force at 1 - pressure force at 2

Darcy formula for the loss of head in pipelines

h f=4 fLdv 2❑

2gAn alternative form of Darcy formula if Q is the discharge

h f=fL

3d5

Q2❑

❑

Example: page 104 information sheet

If Q = 2.73 m3/min , f = 0.01, L = 300 m , d = 150 mmh f=0.01x 300x ¿¿

2- Shock losses

P1 P2 V1 V2

A1 A2

Head lost at enlargement hL

hL=(V 1−V 2)

2

2gFor continuity of flow A1V1 = A2V2

V2 = A1

A2

V 1

hL¿kV 2

1

2 g

k=(1−A1

A2

)2

Example 1 page 96A pipe increases suddenly in diameter from 0.5 m to 1 m . a mercury U-tube has one leg connected just upstream of the change and the other leg connected to the larger section a short distance downstream. If there is difference of 35 mm in the mercury levels the rest of the gauge being filled with water find the discharge.

Solution

Sudden contraction

P1 P2 V1 V2

A1 A2

hL=(V 1−V 2)

2

2g

Example 2 page 101A pipe carrying 0.06 m3 /s suddenly contracts from 200 mm to 150 mm diam. Assuming that a vena contract is formed in the smaller pipe calculate the coefficient of contraction if the pressure head at a point upstream of contraction is 0.655 m greater than at a point just down stream of the vena contract.

Solution

Applications on pipelines problems

p1

γ+v1

2

γ+z1=

p2

γ+v2

2

γ+ z2+shock loss+ friction loss

Examples 5, 6, 7, 8 and 9 page 106 - 113