Suppose I know a static pressure at a valve when closed. Then suppose that the valve is opend slightly to give a flow rate and I manage to calculate the dynamic pressure at the valve.
Is the reduction in pressure at the valve calculated by taking the difference between the static and dynamic pressures? If not, can you point me in the right direction?
2007-05-03 02:49:41 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Engineering
Sorry I should have specified that I'm treating this as an ideal fluid
2007-05-03 04:44:23 · update #1
that is right
the reduction in pressure is the difference between the static and dynamic pressures
this reduction is transformed to another type of energy which is motion energy
2007-05-03 03:42:49 · answer #1 · answered by knight 3 · 1⤊ 0⤋
I suppose that you are measuring the pressure on the upstream side of the valve with low flow. The reduction in upstream static pressure will be negligible if the flow is silght.
If there is an appreciable flow, the static pressure at the upstream side of the valve will be reduced by the TWO combined dynamic pressure reductions of velocity head plus friction loss. These losses are all in the upstream piping. One can be partially recovered; the other (friction loss) cannot.
2007-05-03 16:32:09 · answer #2 · answered by Bomba 7 · 0⤊ 0⤋
The restriction to flow causes an increase in velocity and therefore a decrease in pressure. (Bernoulli).
The difference between the static and dynamic pressures can be used to calculate flow rate (along with other parameters like Temperature and Density).
2007-05-03 12:05:46 · answer #3 · answered by Norrie 7 · 1⤊ 0⤋
almost true, since the valve is not optimized there are turbulence in the air flow, those eventually will be transferred to heat and will reduce from the equation.
2007-05-03 11:31:21 · answer #4 · answered by eyal b 4 · 0⤊ 0⤋