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While doing pressure drop calculation for pipelines, by choosing any 2 arbitrary points for instance in the pump discharge, let us see the following example.
1. choose a point 1 in pump discharge (assume pipe discharge dia to be 6" ) and another point 2 in the pipeline discharge with same dia just 100 m away from the first point without any change in elevation (static head difference between point 1 & 2 is nil). In that case, assuming a pressure drop of 0.1 kgf/cm2 in the span of the above said 100m, the change in pressure head between point 1 & 2 has to be accomodated else where according to Bernoulli's equation.
Pressure drop obviously occured due to frictional losses. In that case, since it is a steady state flow(pump discharge is normally steady state flow) , the velocity head factor should remain constant since there is no change in diameter of the pipe.
In this case, can anyone explain where the change in pressure will be exactly accomodated?( as there are only 3 heads viz,, the pressure head, velocity head and the static head , the change or drop in any one of the heads leads to an increase in any of the other heads as the sum total of the 3 heads any point in agiven pipe line for a steady state flow needs to be the same as per Bernoulli's equation.)
Kindly throw some light on it.
Regards,
N.S.Balaji
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Pressure Drop Calculationin Pipes
Started by , Jul 10 2012 11:47 PM
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katmar
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Posted 11 July 2012 - 01:02 AM
Bernoulli's theorem assumed frictionless pipe. In real pipe, pressure head is lost to frictional resistance. The "lost" energy is converted to heat, but it is usually so small that we are not aware of it.
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