Understanding Backpressure Effects on Flow Meter Calibration

[ca2n]

Hello all,

I have an in-house designed flow meter with roughly the configuration in the figure. In the setup, a restriction is placed in the flow stream (the rows of horizontal lines in the figure). Pressure measurements upstream and downstream of the restriction, P₁ and P₂, respectively are continuously monitored.

For calibration (determining the relationship between P₁-P₂ and the mass flow rate), the air flowing through the system is supplied to a nozzle with a flow straightener upstream of the nozzle exit. Velocity measurements with hot-wire anemometry allows me to calculate the mass flow rate at the exit of the nozzle at discrete P₁-P₂ values.

So that's how far the calibration goes. Now on to how it's used in experiments.

For some of the experiments that I am conducting there can be a rather significant resistance in the flow (a filter). The backpressure created by this resistance is monitored by a pressure gauge mounted just upstream of P₁ and may reach up to 1 bar (gauge). I am uncertain as to how this backpressure affects the flowmeter calibration. For example:

Let's say from the flow meter calibration, a mass flow rate, m-dot_a is provided by P₁-P₂ = P_a.

If a resistance is fitted downstream, causing backpressure, is m-dot_a still provided by P_a?

Thank you in advanced for any response.

 

[DoItForYourself]

I think that if the difference ΔP is the same, the m-dot_a will almost be the same.

In your case, practically the volumetric flow (Q) depends only on the square root of ΔP. If the density does not change significantly, then the mass flow (m-dot_a), which equals (density)x(volumetric flow), will be the same.

 

[BvU]

If the density does not change significantly

For an ideal gas the density doubles when the pressure goes from 1 Bar to 2 Bar...

You have some $\Delta p$ that follows from $c_d {1\over 2} \rho v^2$ so you can bet that $\Delta p(\dot m)$ is a different function at different $p$ !

 

[DoItForYourself]

I agree with you, the function Δp(m) is not the same at different pressure.

But in the case of bigger pressures (for example from 20 to 21 bars), the difference is not so big.

 

[BvU]

OP is talking about 1 - 2 Bar for the pressure upstream of some filter if I am not mistaken.

If the gas is ideal enough, the $\ \Delta p = c_d {1\over 2} \rho v^2\$ can be worked around using the ideal gas law $\ \rho = {pM\over RT}\$ and $\ v ={ \dot m\over \rho A\ }$ leaving something like $p\,\Delta p \approx {\rm constant} \times\dot m^2$

 

[JRMichler]

Does the flowmeter have turbulent flow, where the pressure drop is proportional to velocity squared? Or is it laminar, where the pressure drop is proportional to velocity?

 

[ca2n]

Does the flowmeter have turbulent flow, where the pressure drop is proportional to velocity squared? Or is it laminar, where the pressure drop is proportional to velocity?

The flow is turbulent.

 

[BvU]

The flow is turbulent.

yes. Perhaps you can find some guidance on the site of a flowmeter company like Brooks. Working principle is a little different but calulations are similar.

 

[JRMichler]

Since you have calibration data, you can fit your data to the following equation: Q = K * (delta P / rho)^0.5, where:
Q is volume flow rate
K is a least squares fit to your test data
Delta P is P1 - P2
rho is the fluid density

Get the density of air from a psychrometric chart. If your flowmeter operates with P2/P1 greater than 0.99, you can ignore the effects of compressibility. A good source on this is Measurement Systems Application and Design, by Doebelin.

 

[ca2n]

Thank you all for the responses. You have been very helpful.

I would like to go back to the original question, if I may:

If a resistance is fitted downstream, causing backpressure, is $\dot{m}_a$ still provided by $P_a$?

I will continue to look deeper into the physics, but is it safe to consider at this point that the answer is no?

 

[BvU]

m-dota still provided by Pa?

No it is not. If that isn't clear from the answers you received, you should seek some assistance with a person who understands these answers.