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14. Pipe flow II (6.4, 7.1-7.4)

• Pipes in parallel• Three reservoir problem• Quasi stationary pipe flowExercises: D21, D26, and (D27)

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Pipe systems – pipes in parallel

Solution• Energy equation ⇒ hf1 + Σhlocal,1 = hf2 + Σhlocal,2• Continuity equation ⇒ Q = Q1 + Q2

(elevation z in reservoirs same for both pipes, velocity V in reservoirs equal to zero or otherwise same for both pipes)

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D21 A 0.6 m pipeline branches into a 0.3 m and a 0.45 m pipe, each of which is 1.6 km long, and they rejoin to form a 0.45 m pipe. If 0.85 m3/s flow in the main pipe, how will the flow divide? Assume that f = 0.018 for both branches.

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Example – parallel pipes. A 300 mm pipeline 1500 m long (f = 0.020) is laid between tworeservoirs having a difference of surface elevation of 24 m. What is the maximum obtainable flowratethrough this line (with all the valves wide open)? When this pipe is looped with a 400 mm pipe 600 m long (f = 0.025) laid parallel and connected to it, what increase of maximum flowrate may be expected? Assume that all local losses may be neglected.

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Pipe systems – branched pipe systems

Solution3 Possible flow situations:1) From reservoir 1 and 2 to reservoir 32) From reservoir 1 to reservoir 2 and 33) From reservoir 1 to 3 (Q2 = 0)For the situation as shown:Energy equation ⇒

HJ = PJ/w + zJ + V2J/2g

hf1 + Σhlocal,1 = z1 – HJhf2 + Σhlocal,2 = z2 – HJhf3 + Σhlocal,3 = HJ – z3

Continuity equation ⇒ Q3 = Q1 + Q2

As HJ (HJ is total head at J) is initiallyunknown, a method of solution is as follows:

1) Guess HJ2) Calculate Q1, Q2, and Q33) If Q1 + Q2 = Q3, then the solution is

correct4) If Q1 + Q2 ≠ Q3, then return to 1).

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D26 A 900 mm pipe divides into three 450 mm pipes at elevation 120. The 450 mm pipes (length, see table) runs to reservoirs with elevations according to table. When 1.4 m3/s flows in the big pipe, how will the flow divide? Assume f = 0.017 in all pipes.

Reservoir Elevation (m) Pipe length (m)A 90 3200B 60 4800C 30 6800

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NON-STATIONARY PIPE FLOW -OUTFLOW FROM RESERVOIR UNDER VARYING PRESSURE LEVEL

• When water flows under varying pressure levels the outflow from the reservoir willvary accordingly.

• In the figure V represents the volume in the reservoir at a certain time. There is also an inflow, Qi, to and an outflow, Qo, from the reservoir.

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NON-STATIONARY PIPE FLOW, cont.• Volume change in the reservoir during a small time interval dt, can

be expressed as:dV = (Qi – Qo) · dt and dV = As · dz →As · dz = (Qi – Qo) · dt where both inflow and outflow can vary in time.

The outflow can normally be determined by the energy equation that gives outflow as a function of z. For example outflow through a hole: Qo = Ahole · CD · (2gz)1/2 (Eqn. 5.12)

If time is to be estimated that changes the water level from z1 to z2integration of dt = (As /(Qi – Qo)) dz gives:

t = z1∫z2 (As /(Qi – Qo)) dz this expression can be derivated if Qi = 0 or if Qi

= constant and Qo can be re-written as a function of z. Qo can be determined by the energy equation during short time periods assuming stationarity. If water level changes quickly an acceleration term has to be included though.

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Example – unsteady pipe flow

The open wedge-shaped tank in the figurebelow has a length of 5 m perpendicularto the sketch. It is drained through a 75mm diameter pipe, 3.5 m long whosedischarge end is at elevation zero. Thecoefficient of loss at the pipe entrance is0.5, the total of the bend loss coefficientsis 0.2, and the friction factor is f = 0.018.Find the time required to lower the watersurface in the tank from elevation 3 m to1.5 m. Assume that the accelerationeffects in the pipe are negligible.

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D27 Three reservoirs are connected with pipes via a connection point O at elevation 120. With the data according to the table calculate the flowrates in the lines. Assume f = 0.020.

Reservoir Elevation (m) Pipe length(m)

Diameter (mm)

A 150 1600 300B 120 1600 200C 90 2400 150