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?

  • $\begingroup$ Your question seems to assume that there is a pressure loss due to the loss of fluid. That may not be the case depending on the assumptions. Furthermore, you're going to need to establish which pressure you're talking about. At the end of the pipe? And we haven't even established if this is a time dependent problem or a flow problem. $\endgroup$ Feb 22 '12 at 3:06
  • $\begingroup$ IMHO this really depends upon the orientation of the holes. And you'll have to take turbulence into account. Bernoulli's equation refuses to be applied here (Since the velocity of fluid decreases after crossing a hole, while writing Bernoulli's for the water at the hole, do we use $\rho v_1^2$ or $\rho v_2^2$). $\endgroup$ Feb 22 '12 at 3:29
  • $\begingroup$ Can assume it is a steady state problem, and the pressure loss I'm concerned is that of the main pipe flow, due to water squirting out from the side wall holes. $\endgroup$
    – Raymond
    Feb 22 '12 at 4:45

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.


This answer may be incorrect. I tried to neglect turbulence, but I personally don't like what crops up. At most, it can give a good approximation.


  • No turbulence
  • No compressibility
  • Horizontal pipe
  • Small pipe (effect of gravity neglected)
  • No viscosity


Let the pipe have area $A$, and the hole $a$. Let the input pressure be $p_i$, and input velocity be $v_i$. Use same notation for output velocity with subscript f. Let velocity of fluid leaving the hole be $v_h$, and atmospheric pressure be $p_0$. Density is $\rho$.


I'm shying away from Bernoulli's equation here, it has some issues when applied to systems like this. I'll go from a more fundamental POV.

Let's take a small time $dt$. In this time, let a fluid column of length $dx_i$ enter the pipe, $dx_f$ leave the pipe, and $dx_h$ leave the hole. It is obvious that $dx_i=v_idt$ &c. Now, work done by pressure is $p_iAdx_i$ at the inlet, $-p_fAdx_f$ at the outlet, and $-p_hadx_h$ at the hole. So, work done $$w=p_iAdx_i-p_fAdx_f-p_0adx_h$$ Now, water of mass $\rho A dx_i$ enters with velocity $v_i$ &c. Thus, change in kinetic energy $$\Delta KE=\frac{1}{2}\rho(Adx_fv_f^2+adx_hv_h^2-Adx_iv_i^2)$$. Applying work-energy theorem, $$p_iAdx_i-p_fAdx_f-p_0adx_h=\frac{1}{2}\rho(Adx_fv_f^2+adx_hv_h^2-Adx_iv_i^2)\qquad \cdots (1)$$

Now, conserving mass, the amount of water entering is $\rho Adx_i=\rho Av_idt$. Equating this with the amount of water leaving, and cancelling $dt$, $$Av_i=av_h+Av_f\qquad \cdots (2)$$ (IIRC this is called the continuity equation)

Now, replacing $v_idt=x_i$ &c in (1), cancelling $dt$ we get: $$p_iAv_i-p_fAv_f-p_0av_h=\frac{1}{2}\rho(Av_f^3+av_h^3-Av_i^3)\qquad \cdots (3)$$

I am now applying horizontal momentum conservation. This is wrong, as the edges of the hole can exert a horizontal force on the system, but I feel that turbulence will then come into the picture. Since we're neglecting turbulence, it's OK-ish to apply momentum: $$\rho Adx_iv_i=\rho Adx_fv_f$$, or $$v_i^2=v_f^2 \implies v_i=v_f \qquad(4)$$ This contradicts (2), but we can proceed by assuming that $v_i\approx v_f$. This is justifiable only if $A>>a$. Then, we can replace all instances of $v_i^3-v_f^3$ with $3v_i^2\Delta v$, where $\Delta v=v_i-v_f=\frac{av_h}{A}$. All other instances of $v_f$ can be replaced by $v_i$.

If we had one more equation, we could probably solve it without any assumptions, but I can't manage to think of another equation. I strongly suspect that the situation is impossible without considering turbulence (and possibly the thickness of the hole). Then I guess we may need to apply Navier-Stokes (I'm no good at that).

With multiple holes, the method will be similar. Because $v_i\approx v_f$, each hole only changes the pressure, so you can recursively apply whatever formula you get for $p_f=f(p_i)$.

  • $\begingroup$ Your equation (4) can be an assumption that the mass flow out of the hole is negligible compared to the total flow-rate. However, you have terms with $v_i-v_f$, so this becomes a bit dangerous $\endgroup$
    – Bernhard
    Feb 22 '12 at 8:51
  • $\begingroup$ Its more of a half-approximation: only substitute it for $v_i+v_f$. Wherever we have difference terms, all t he other terms are also differences or $v_h$ terms. Cant be neglected there. $\endgroup$ Feb 22 '12 at 9:20
  • $\begingroup$ But yes, that's why I don't like this. If you can cook up an alternative third equation, it might work. $\endgroup$ Feb 22 '12 at 9:21
  • $\begingroup$ Using eqn(4) result into eqn(2), v_h will have to be 0. Then, it seems that Eqn (3) will be reduced to simple Bernoulli flow in a normal pipe with no loss. $\endgroup$
    – Raymond
    Feb 22 '12 at 9:23

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