Flow rate of air through constriction

I've been trying to find the formula(s) to determine the flow rate of air (and properties) in a pipe of varying diameters connecting two theoretical reservoirs of infinite size. The pipe is short in length so friction can be neglected. Reservoir 1 is always a higher pressure. There is no heat transfer to or from the air.

If I have the:
pressure of reservoir 1
pressure of reservoir 2
Inside diameter of pipe
Temperature of air in reservoir 1

How can I find:
Mass flow rate of air flowing through pipe
Pressure of air flowing through pipe
Velocity of air flowing through pipe
Temperature of air flowing through pipe

I'm aware of choked flow conditions which seems to simplify this problem somewhat, however, I'm also interested in being able to determine the properties of the gas flow in subsonic conditions. I've been studying Bernoulli's principle for compressible flow, but am having trouble determining the initial flow conditions based only on pressures. I could use some guidance from an expert on the subject. Any help would be appreciated.

This is too broad as asked. Which flow regime are you interested in, or what are the pressures and temperatures?

russ_watters said:

This is too broad as asked. Which flow regime are you interested in, or what are the pressures and temperatures?

Does the flow regime matter when the pipe length is negligible? I'm imagining a 1" long pipe connecting two reservoirs. I'm interested in all pressures and temperatures, but if I had to choose an example, 320 psi at 100 C to 300 psi.

Maybe use the "modified" (for compressible flow) Darcy formula as shown in Crane 410, Eq 3-20:

$$W=1891~ Y ~d^2 \sqrt{ \frac {\Delta P} {K~\bar{V_{1}}}}$$

Imperial units
W is flow rate in pounds per hour
Y compressibility factor is found via charts, but I believe for short pipe would be 0.6 for air flow.
1891 is a units conversion
K factor fL/D
##\bar{V_{1}}## is inlet specific volume (ft^3 / pound)

Interesting. Looking into it. Thanks!

MysticDream said:

Does the flow regime matter when the pipe length is negligible?

Yes. Making the pipe shorter/ssmaller cannot, for example, turn subsonic flow into choked/supersonic flow.

MysticDream said:

I'm imagining a 1" long pipe connecting two reservoirs. I'm interested in all pressures and temperatures, but if I had to choose an example, 320 psi at 100 C to 300 psi.

Those pressures are outside of my recent experience so let's give others a chance to weigh in, but my first thought is that with such a low relative pressure differential you are barely into compressible subsonic flow. Compressible Bernoulli's might be accurate enough there.

gmax137 said:

Y compressibility factor is found via charts, but I believe for short pipe would be 0.6 for air flow.

Oops, I was way off - for your 320/300 psi (are those gauge or absolute pressures?) Y is going to be only slightly less than 1.0, say 0.98.

I would probably use K=1.5 for a sharp edged entrance & exit. What is the opening diameter? How accurately are you trying to predict? These kinds of formulas can approximate reality but for critical applications you have to do some testing to get more reliable values.

I was wondering if there was a formula that could be used to determine these properties for most conditions. I'm aware that Bernoulli's equation could be used for velocities under Mach .3, but for any velocities above that, I would assume there would be a useful formula, if even an approximation.

The 320/300 psi would be absolute. The opening diameter could be .5". The accuracy just has to be reasonably close, not exact.

russ_watters said:

Yes. Making the pipe shorter/ssmaller cannot, for example, turn subsonic flow into choked/supersonic flow.Those pressures are outside of my recent experience so let's give others a chance to weigh in, but my first thought is that with such a low relative pressure differential you are barely into compressible subsonic flow. Compressible Bernoulli's might be accurate enough there.

Thanks.

MysticDream said:

The opening diameter could be .5".

Then using
$$W=1891~ Y ~d^2 \sqrt{ \frac {\Delta P} {K~\bar{V_{1}}}}$$
W is flow rate in pounds per hour
Y compressibility factor for your 320/300 psi Y ~ 0.98.
K factor fL/D = 1.5 for a sharp edged entrance & exit
##\bar{V_{1}}## is inlet specific volume (ft^3 / pound)

$$W=1891~ (0.98) ~0.5^2 \sqrt{ \frac {20} {(1.5)~(0.7)}} = 2000 \frac {lbm}{hr}$$
for opening diameter of 1/2 inch the flow area is 0.00136 ft^2, then the velocity would be
$$0.7~\frac{ft^3} {lbm}~2000 \frac {lbm} {hr}~\frac {1} {0.00136 ft^2}~ \frac {hr}{3600~sec} = 285~ \frac{ft}{sec}$$

As always, check my arithmetic!

Also might be a good idea to bounce this off of @boneh3ad

gmax137 said:

Then using
$$W=1891~ Y ~d^2 \sqrt{ \frac {\Delta P} {K~\bar{V_{1}}}}$$
W is flow rate in pounds per hour
Y compressibility factor for your 320/300 psi Y ~ 0.98.
K factor fL/D = 1.5 for a sharp edged entrance & exit
##\bar{V_{1}}## is inlet specific volume (ft^3 / pound)

$$W=1891~ (0.98) ~0.5^2 \sqrt{ \frac {20} {(1.5)~(0.7)}} = 2000 \frac {lbm}{hr}$$
for opening diameter of 1/2 inch the flow area is 0.00136 ft^2, then the velocity would be
$$0.7~\frac{ft^3} {lbm}~2000 \frac {lbm} {hr}~\frac {1} {0.00136 ft^2}~ \frac {hr}{3600~sec} = 285~ \frac{ft}{sec}$$

As always, check my arithmetic!

Also might be a good idea to bounce this off of @boneh3ad

Sorry for the late reply. I'm going to have a look at this. Thanks a lot for the help.