A pipe of 150mm bore is delivering water at a rate of 7500 liters per min at a pressure of 820kpa it connects via a horizontal gradually expanding pipe to a main of 300mm in diameter. Calculate the fluid velocity both in the main body of the pipe and at the outlet.

Equation p1 * P2 / T1

Any help would be great.

Thanks in advance

Flow is the cross sectional area times the logitudinal flow rate (length/time).

Cross sectional area A = \pi r^2 where r is the radius of the pipe.

You know the flow in liters/min. Convert liters to m^3. (1 liter = 0.001 m^3). Convert the radius from millimeters to meters so that you're in the same dimensions. Divide the volumetric flow rate by the cross sectional area. The velocity will be in meters/min.

\frac {m^3}{min} \times \frac {1}{m^2} = \frac {m}{min}

The pressure is irrelevant insofar as this is concerned since you know the volumetric flow rate.

I converted the flow rate from 7500 litters per min to 7.5m^3

I then worked out the cross sectional area being for 150mm diameter is 0.017 &
for the 300mm diameter pipe 0.070

so for 150 mm diameter I get 441.17 m/mm

and for 300mm diameter I get 10623 m/mm

Is this correct.

A pipe of 150mm bore is delivering water at a rate of 7500 liters per min at a pressure of 820kpa it connects via a horizontal gradually expanding pipe to a main of 300mm in diameter. Calculate the fluid velocity both in the main body of the pipe and at the outlet.


150 mm = 0.15 m; r = 0.075 m

A=\pi (0.075\,m)^2=0.001767\,m^2

7.5\,\frac {m^3}{min} \times \frac {1}{0.001767\,m^2}=424.4\,\frac {m}{min}

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300 mm = 0.3 m; r = 0.15 m

A=\pi (0.15\,m)^2=0.007686\,m^2

The linear velocity in the larger pipe must be lower than the linear velocity in the smaller pipe.

7.5\,\frac {m^3}{min} \times \frac {1}{0.007686\,m^2}=106.1\,\frac {m}{min}

Thanks for the help very pleased thank you.