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?