Calculate fluid pressure loss due to hole leakage 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?
 A: 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.
A: 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.
Assumptions


*

*No turbulence

*No compressibility

*Horizontal pipe

*Small pipe (effect of gravity neglected)

*No viscosity


Variables
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 $\rho$.
Derivation
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 $p_f=f(p_i)$. 
