Using pressure drop to correlate mass flow

Question

I am looking for a way to double check readings I am getting through a mass flow meter that appear to be reading lower than expected.

Once flow starts in the system, the pressure regulator at the supply tank drops in pressure readout. This drop should be due to the dynamic pressure of the flow of gas through the regulator so I was thinking it may be possible to correlate the mass flow meter's readings with the calculated mass flow resulting in the drop in pressure at the regulator. I started by using the ideal gas law (low enough pressures) to calculate the density of the gas at the static pressure read by the regulator.

$$\rho=\frac{PM}{RT}$$

Then I took the pressure difference read from the tank regulator between the static no-flow to flowing as the dynamic pressure.

$$P_d=\frac{1}{2}\rho v^2$$

Using the above equation I calculated the velocity of the fluid flow, and then just used the cross sectional area to calculate the mass flow.

$$\dot{m}=\rho A_c v$$

When calculating the density I am getting an unreasonable value for fluid velocity so I was wondering if this procedure makes sense?

Answer 1

It is not entirely clear from the question, where the instrument readings take place. The following equations might need to be adapted in order to reflect the sensor positions.

The basic equation for the mass-flow $\dot{m}$ is:

$\dot{m} = A \cdot \rho \cdot v$

$A$, $\rho$, and $v$ need to be evaluated or measured at the same location.

The ideal gas law can be used here but be aware the static values for temperature $T$ and pressure $p$ have to be used here:

$\rho = \frac{p_\mathrm{s}}{RT_\mathrm{s}}$

The specific gas constant for air is something around $287.058\frac{J}{kgK}$. It changes with temperature but this does not seem to be the problem right now.
Usually the problem is now to compute $T_\mathrm{s}$.

The following equations use an isentropic process to compute first the Mach-Number and then the static Temperature. Step 1 is to solve the equation for $M$:

$\frac{p_\mathrm{s}}{p_\mathrm{t}} = \left( 1 + \frac{\gamma - 1}{2} M^2\right)^{\frac{\gamma}{\gamma-1}}$

Here $\gamma$ is the isentropic coefficient which is around $1.4$ for air depending on the humidity. $p_\mathrm{t}$ and $p_\mathrm{s}$ are the static and total pressure. Their relation is:

$p_\mathrm{t} = p_\mathrm{s} + p_\mathrm{d}$

$p_\mathrm{d}$ is often calculated with $\frac{1}{2}\rho u^2$ this simplification creates an error in the density $\rho$ of 5% for a Mach-Number $M=0.3$. So it is possible with the result of step 1 to check if this was the source of the problem.

With the calculation of the Mach-Number the isentropic relation of total and static temperature can be used to calculate the static temperature in step 2:

$\frac{T_\mathrm{s}}{T_\mathrm{t}} = \left( 1 + \frac{\gamma - 1}{2} M^2\right)^{-1}$

Here the total temperature $T_\mathrm{t}$ is the temperature of the gas in the pressure supply.

The final step 3 is the calculation of the velocity $v$. This will be done using the equation for the Mach-Number:

$M = \frac{u}{a}$

Here the only unknown is $a$ (the speed of sound) which can be computed using the following relation:

$a = \sqrt{\gamma R T_\mathrm{s}}$

Doing these calculations by hand is laborious but it's easy to set up a script or sheet to do these calculations. There are three constants:

• Through flow area, $A$
• Specific gas constant, $R$
• Isentropic coefficient, $\gamma$ (sometimes its $\kappa$)

And there are three measurement values:

• Total pressure, $p_\mathrm{t}$
• Static pressure, $p_\mathrm{s}$
• Total temperature, $T_\mathrm{t}$

Answer 2

If I am not mistaken, you are looking at the drop in pressure right at a pressure reducing regulator when flow initiates, and wanting to infer mass flow from the value of this drop. If this is true, I would like to add that there are many reasons why a pressure regulator would "droop" other than the velocity head effect. Regulators have springs that compress when the fill valve opens, and there are several friction losses that together comprise the flow performance curve. Many designs have "lock up" that presents a significant offset between the static and flowing conditions. Have I interpreted your question correctly?