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MET 212 Module 4

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MET 212 Module 4

1. 1. Module 4 Flow in pipes 1- Losses in pipesPressure 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 shownbelow.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
2. 2. 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 2Darcy formula for the loss of head in pipelinesAn alternative form of Darcy formula if Q is the discharge
3. 3. Example: page 104 information sheetIf Q = 2.73 m3/min , f = 0.01, L = 300 m , d = 150 mm 2- Shock losses P1 P2 V1 V2 A1 A2Head lost at enlargement hLFor continuity of flow A1V1 = A2V2 V2 = hLExample 1 page 96A pipe increases suddenly in diameter from 0.5 m to 1 m . a mercury U-tube has one legconnected just upstream of the change and the other leg connected to the larger section ashort distance downstream. If there is difference of 35 mm in the mercury levels the rest ofthe gauge being filled with water find the discharge.SolutionSudden contraction P1 P2 V1 V2 A1 A2
4. 4. Example 2 page 101A pipe carrying 0.06 m3 /s suddenly contracts from 200 mm to 150 mm diam. Assuming thata vena contract is formed in the smaller pipe calculate the coefficient of contraction if thepressure head at a point upstream of contraction is 0.655 m greater than at a point just downstream of the vena contract.SolutionApplications on pipelines problemsExamples 5, 6, 7, 8 and 9 page 106 - 113