Velocity of Liquid in Pipe Following Sudden Depressurisation

[Nosey]

We have a scenario where a 2 metre section of 3" pipework (call it 80mm ID) is pressurised up to 150 barg with water. If a quarter-turn ball valve is quickly opened (assume that it 0.25 seconds to fully open) then what is the maximum velocity that the water can attain in the downstream 3" pipework (assuming that there is no inflow to the original 2meter section) and the downstream pressure is atmospheric?

Taking the compressibility of water to be 0.00000000044 / Pa then the change in volume would be:
0.25*PI*(0.08^2)*0.00000000044 * 15,000,000 = 0.0000332 m3 change in volume.
If this is depressurised then the change in linear dimension will be = change in volume divided by cross sectional area = 0.0000332 / 0.25*PI*(0.08^2) = 0.0066m
as the valve is opened in 0.25 of a second the linear velocity of the water is 0.264 m/s.

Is this a valid methodology, or have I over-simplified it or fouled it up entirely?

Thanks in advance

 

[BigInch]

No. The volume expansion will be such a very very small part of the velocity downstream that I wouldn't even consider that. F/m = a will have to be considered.

Force is the (upstream pressure - downstream pressure) * X-sectional area,
Mass is the sum of the mass of the column of water being accelerated.

a is the accelaration of the water column.




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"Pumping accounts for 20% of the world’s energy used by electric motors and 25-50% of the total electrical energy usage in certain industrial facilities."-DOE statistic (Note: Make that 99% for pipeline companies)

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