Working out pressure drop from leak rate

Question

I am trying to work out the acceptable pressure drop/increase on a leak test rig for a given leak rate. A simplified version of the test rig looks like this:

The gas (air) supply is at 1 bar above atmospheric pressure, and the inlet & outlet control valves + pressure sensor are part of the test rig and can be assumed to be leak-free. I am also working on the assumption that the process is isothermal.

Test 1

The first test is to open the inlet control valve, with the outlet control valve closed and the valve of the thing under test also closed, and look for a pressure increase on the pressure sensor. At the start of the test, the pressure sensor is at atmospheric pressure.

Here's my attempt, using $PV=nRT$:

Moles of gas gained downstream of the valve under test due to leakage past the valve:

$n_2 = n_1 + \frac{LR * \Delta t * P_1}{RT}$ (1)

where $P_1$, $V_1$ and $n_1$ are the initial conditions ($P_1 = P_{atm}$), $LR$ is the leak rate past the valve under test and $\Delta t$ is the duration of the test.

The final pressure can then be worked out as:

$P_2 V_2 = n_2 RT = (n_1 + \frac{LR * \Delta t * P_1}{RT}) RT$

with $V_2 = V_1$ (the volume hasn't changed), we get:

$P_2 = \frac{P_1 V_1 + LR * \Delta t * P_1}{V_1}$

and then obviously:

$\Delta P = P_2 - P_1 = \frac{LR * \Delta t * P_1}{V_1}$

Question

Is that the right approach? Should I use $P_1 + 1 \text{bar}$ instead of $P_1$ in equation (1)?

Test 2

Now the whole thing is pressurised to 1 bar by opening the inlet control valve, the valve under test and closing the outlet control valve, before closing the inlet control valve, which is when the test starts. This is to look at leakage out of the "thing" through pneumatic fittings, etc... We are now looking at a pressure drop from 1 bar, instead of a pressure increase as in test 1.

I am using essentially the same approach, except that we're now looking at a number of moles of air lost in equation (1), which becomes:

$n_2 = n_1 - \frac{LR * \Delta t * P_1}{RT}$ (1a)

The rest follows as per test 1, so that:

$\Delta P = P_2 - P_1 = -\frac{LR * \Delta t * P_1}{V_1}$

of course, this time $P_1 = P_{atm} + 1 \text{bar}$ and $LR$ is the leak rate out of the "thing" rather than past the valve.

Question

Again, is that the right approach?

Answer 1

The rate of gas leakage depends on the pressure difference, and in your setup it looks like the pressure difference is constantly decreasing as the test is running. This means that the leak rate parameter ($LR$) in your derivation is the average leak rate over the duration of the test. In my experience, leak rate is usually specified as an instantaneous value at steady-state conditions (constant pressure difference). I'm not sure how your leak rate spec is meant to be interpreted, but measuring it with an unsteady-state test rig is probably more difficult than with a steady-state setup.

Anyway, here's an equation you can use to correlate leak rate to pressure difference and hole area:

$$ \dot m = \mu \cdot A \cdot \sqrt{\frac{\gamma}{\gamma -1} \left[ \left( \frac{p_1}{p_2} \right)^\frac{2}{\gamma} - \left( \frac{p_1}{p_2} \right)^\frac{\gamma + 1}{\gamma} \right]} \cdot \sqrt{2\cdot p_2 \cdot \rho_2} $$ Source: MACEAS

• $\dot m$ is mass flow rate
• $A$ is area of the leak
• $\gamma$ is the specific heat ratio of the gas (air = 1.4)
• $p_1$, $p_2$ are downstream and upstream pressures, respectively
• $\mu$ is a dimensionless friction coefficient that needs to be determined experimentally

If you want to use this in your current test setup, you'll need to integrate it over time and replace downstream pressure with an equation to account for the changing pressure. Also take a look at this other, related question regarding the calculation of unsteady flow.