I have a cylindrical pipe of internal diameter of around 5mm, with pressurised fluid flowing though it. If I have holes at the wall of the pipe, how do I calculate the pressure loss due to water leaking out through those holes?
[Physics] Calculate fluid pressure loss due to hole leakage
fluid dynamics
Related Solutions
Assumptions:
- The hole is in the region below the air pocket (so water, not air, is leaking)
- Air pocket volume is $V_p$ when the pressure is $P$
- Isothermal process (slow expansion: temperature constant)
- Volume of container doesn't change with pressure (probably not true… - this will underestimate the leakage rate)
you can write the rate of change of the volume of the air pocket as a function of pressure:
$$\frac{PV}{T} = const\\ P_1V_1 = P_2V_2$$ Differentiating $PV = constant$: $$P dV + V dP = 0$$ Dividing by $dT$ and rearranging: $$\frac{dV}{dt}=-\frac{V}{P}\frac{dP}{dt}$$
From this you compute the flow rate from the pressure change. As you can see, the smaller the volume $V$, the smaller the volume change $dV$ that you can calculate for a given change in pressure.
This leaves you with the problem of calibrating the air pocket. This is best done by having an air filled capillary somewhere near the top of your system: you will be able to see the liquid rise in this capillary as the system is pressurized, and from the rate at which it drops you can determine the leakage rate immediately - if you know the diameter of the capillary, no other math is required…
Do note that a heat exchanger is likely to expand when pressurized - you should be able to determine how much this affects the result by having a small calibrated plunger (ideally the same size as the capillary) with which you can inject a small known amount of additional liquid into the system. If the liquid plus heat exchanger were truly incompressible (constant volume, constant density) then the air in the capillary should rise as you push the plunger down. When this does not happen (say the capillary rises half as much) then you know what volume of liquid corresponds to what volume change in the capillary, and this gives you the calibration from capillary volume to liquid volume.
In an ideal fluid, assuming the diameter of the pipe after the contraction is the same as the diameter of the pipe before the contraction, $P_2^\prime = P_2$. There is no effect of the contraction downstream from the contraction itself.
If we consider an inviscid, isentropic, incompressible flow, the total pressure in the flow is constant along streamlines (and since all streamlines originate from the same source, it is also constant throughout the flow). This means $P_0 = P + 1/2 \rho U^2 = \text{const}$. We also know that $\rho_1 U_1 A_1 = \rho_2 U_2 A_2$ for any positions 1 and 2 where $\rho$ is the density, $U$ is the velocity and $A$ is the area of the pipe.
So, we know density is the same because it is incompressible. We know $A_{inlet} = A_{outlet}$ because the pipe is the same diameter away from the contraction. Therefore, the velocities are the same $U_1 = U_2$. And since the total pressure is constant, the values of $P_2$ and $P_2^\prime$ must be the same also.
This is probably counter-intuitive. You would expect something to be different. And in a real fluid, it very well likely would be. Turbulence, separation, friction, heating, all kinds of losses will add up to have an influence. But in an ideal world where none of those things happen, there is no effect of the contraction away from the contraction.
Best Answer
A given section of pipe will have a flow equation like: $\Delta P = \gamma w^2$ where $\Delta P$ is the loss of head pressure over the length of pipe, $\gamma$ is a coefficient related to length, diameter, Reynold's number etc., and $w$ is the flow rate. You can find expressions for these in engineering books for example.
Now imagine we have a pipe and we want to put one small hole in it. We can analyze it by analogy with an electric circuit, where the hole is represented by a large shunt resistance added at that point going to ground. Let $P_0$ represent the fixed pressure driving the flow (analogous to a voltage source), $P_1$ be the pressure at the point of interest near the hole, and let $\gamma_{1,2,3}$ be the flow coefficients for the section of pipe upstream of the hole, downstream of the hole, and the hole itself. Let $w$ be the flow rate upstream of the hole, $w_2$ downstream, and $w_3$ be the flow through the hole itself. Then you have a system of equations you can solve for $P_1$:
$P_0 - P_1 = \gamma_1 w^2$
$P_1 - 0 = \gamma_2 w_2^2$
$P_1 - 0 = \gamma_3 w_3^2$
$w_2+w_3 = w$ (Assuming incompressible flow)
The case of no hole is given by $\gamma_3 = \infty$. Since the hole is small, you can assume $\gamma_3 >> \gamma_2, \gamma_1$ and derive a perturbative expression for $\Delta P_1$ in terms of $\gamma_3^{-1}$. I will leave this as an exercise for the reader. Anyhow, this will tell you the incremental pressure loss for a single hole.
Of course, I have assumed the flow is driven by a fixed idealized source pressure. In reality the source pressure may be a function of flow rate (analogous to the impedance of a voltage source) which would also effect the result, but which could be added into the above equations in a straightforward manner.