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How do I work out what the pressure will be at a given point in a pipe?

I am working with a pipe of diameter 3.5cm at point 1 which then joins to a pipe of 1.4cm diameter at point 2. I know that the flow rate is constant at 270 litres/hour. I can assume steady incompressible flow and that the liquid is water. I am trying to find out if there will be a) a significant pressure increase due to the smaller diameter, and b) what the pressure is at either point 1 or two of the pipes.

So far, I have calculated the areas at both points, $A_1=9.62\,\mathrm{cm}^2$ and $A_2=1.54\,\mathrm{cm}^2$. I have used the flowrate to calculate the velocity at point 1, using $A_1$, to be 0.0779m/s. Then I used the continuity equation to determine the velocity at point 2, $v_2$ as 0.4866m/s. Then I rearranged the Bernoulli equation, neglecting differences in heights to work out $P_2 - P_1 = 115.316\ \mathrm{Pa}$. But how do I find what either $P_1$ or $P_2$ is?

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For an incompressible (or essentially incompressible) fluid, pressure is determined only to within an arbitrary constant. The only way to establish an absolute pressure is to match a boundary condition, which is not available in this problem specification.

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  • $\begingroup$ That makes sense as the force is just determined by the pressure gradient. $\endgroup$ – Virgo Nov 24 '17 at 0:36
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Momentum balance for a control volume from 1 to 2:

$$\int_S\rho v \ (v\cdot n)\ dS=-\int_S{P\cdot n}\ dS$$ where $v$ is velocity, $P$ pressure, $n$ the vector normal to the surface $S$. $\rho$ is the density of water under standard conditions. This results in the following equation.

$$\rho (-v_1^2A_1+v_2^2A_2)=P_1A_1-P_2A_2$$

Together with the Bernoulli equation you can solve the problem.

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  • $\begingroup$ This macroscopic momentum balance is not correct. You omitted the horizontal force exerted by the contraction in the pipe of the fluid flow. That horizontal force can be determined from the correct momentum balance. $\endgroup$ – Chet Miller Mar 20 '17 at 10:32

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