Nominal Linear Velocity Gas flow

In summary, the gas flow is currently at 7.5 ml/s and the desired linear velocity is 350 m/s. The gas is flowing through a 100 mm doughnut-shaped tube with a 363 um diameter center and a 508 um total diameter. The flow is at 450 ml/min using a gas flow meter. The velocity of the gas is currently 76 m/s. The continuity equation for fluid flow is Q = A*V, where Q is the flow rate, A is the cross sectional area of the conduit, and V is the flow velocity. To calculate the velocity of the gas, use the area of the annulus instead of the area of the tube.
  • #1
Jamesan89
6
0
Hello, thanks in advance for your help,

I am not a physicist, however have a fairly good grasp on mathematics, also this isn't homework or an assignment but a real world problem.

Currently trying to create a system with a nebulising gas at a linear velocity of 350 m/s. The calculations confusing me as the system is complex. The gas flow is through a 100 mm, doughnut shaped tube with an 363 um diameter centre and a 508 um total diameter. The gas flow is currently at 450 ml/min (using a gas flow meter).

My solution would be to convert the flow rate to ml/s (7.5 ml/s) and take the volume of the tube in m^3 by Pi*R^2*h and divide these values, however the output value is clearly not correct. A point in the right direction would be greatly appreciated .
 

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  • #2
Please explain configuration a bit more . Also confirm the 350 m/s speed .
 
  • #3
The nominal linear velocity I aim to achieve is 350 m/s, I am trying to work out if I am close to achieving this speed for the gas flow.

As for the configuration the gas is a sheath gas flow down the outer tube with a liquid flowing down the inner tube, therefore I really need to know the gas flow required (currently 450 ml/min) to achieve the desired linear velocity.
 
  • #4
350 metres/sec is around speed of sound for many common gasses so there could be problems in achieving this speed .
 
  • #5
We can start in a different place and derive an achievable velocity . What is pressure at inlet and outlet of annular tube and what is actual gas in use ?
 
  • #6
The gas is nitrogen and the pressure at the inlet is 7 bar, I am unsure however of the pressure at the outlet of the tube. The nitrogen gas is at the highest pressure achievable using the labs equipment.

After doing some maths and thinking, would I be correct in saying the velocity of the gas currently is 76 m/s. I am starting to wonder whether 350 m/s is an error if its around that speed!
 
  • #7
I will attach a paper and put this more into context, the linear velocity value of 350 m/s is found at the bottom right hand corner of the page
 

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  • #8
See message .
 
  • #9
Jamesan89 said:
Hello, thanks in advance for your help,

I am not a physicist, however have a fairly good grasp on mathematics, also this isn't homework or an assignment but a real world problem.

Currently trying to create a system with a nebulising gas at a linear velocity of 350 m/s. The calculations confusing me as the system is complex. The gas flow is through a 100 mm, doughnut shaped tube with an 363 um diameter centre and a 508 um total diameter. The gas flow is currently at 450 ml/min (using a gas flow meter).

My solution would be to convert the flow rate to ml/s (7.5 ml/s) and take the volume of the tube in m^3 by Pi*R^2*h and divide these values, however the output value is clearly not correct. A point in the right direction would be greatly appreciated .
It's not clear what you are doing here.

The continuity equation for fluid flow is Q = A*V, where Q is the flow rate, A is the cross sectional area of the conduit, and V is the flow velocity.

For tube ID = 0.363×10-3 m, A = 1.03×10-7 m2
flow rate Q = 7.5×10-3 l/s = 7.5×10-6 m3/s

flow velocity V = Q / A = 72.5 m/s

Once the velocity of a gas exceeds about 0.3 Mach (0.3 times the speed of sound), compressibility effects become significant.

The speed of sound in nitrogen at standard conditions is about 349 m/s. Therefore, your flow velocity has a Mach No. = 72.5 / 349 = 0.21, so the flow can be treated as incompressible for all practical purposes.
 
  • #10
Just a quick observation, according to the drawing the gas is flowing through the annulus rather than the internal tube and it appears to me you have used the tube's ID in the above calc.
 
  • #11
JBA said:
Just a quick observation, according to the drawing the gas is flowing through the annulus rather than the internal tube and it appears to me you have used the tube's ID in the above calc.
Then use the area of the annulus to calculate the velocity instead of the area of the tube.

The point is, Q = A*V is the correct formula to use, rather than what was described in the OP.

For the annulus, the ID = 0.363×10-3 m, OD = .508×10-3 m,
Annular area, A = 9.92×10-8 m2
flow rate Q = 7.5×10-3 l/s = 7.5×10-6 m3/s

flow velocity V = Q / A = 75.6 m/s

Mach No. = 75.6 / 349 = 0.22 which is still less than M = 0.3, so the flow can be assumed to be incompressible.
 
  • #12
Thanks for your messages and help with simplifying the calculations, I think the author of the paper must have made a mistake or have been referring to the liquid flow through the internal tube. The Gas flow is indeed going through the annulus.

From your reply the maximum flow I could expect to be achievable for my system would be
M = 0.3(max) so V = 349 * 0.3 = 104.7 m/s
Q = V * A s so Q = 104.7 m/s * 9.92x10^-8 m^2 = 1.04x10^-5 m^3/s or 624ml/min.
 
  • #13
Jamesan89 said:
Thanks for your messages and help with simplifying the calculations, I think the author of the paper must have made a mistake or have been referring to the liquid flow through the internal tube. The Gas flow is indeed going through the annulus.

From your reply the maximum flow I could expect to be achievable for my system would be
M = 0.3(max) so V = 349 * 0.3 = 104.7 m/s
Q = V * A s so Q = 104.7 m/s * 9.92x10^-8 m^2 = 1.04x10^-5 m^3/s or 624ml/min.

You can put more gas thru the tube, it's just that the calculations get more complicated due to the effects of compressibility. When the outlet velocity reaches M = 1, then the flow becomes choked, and the maximum mass flow is obtained. The length of the tube and the ratio of the inlet and outlet pressures all factor into what this maximum mass flow number actually is.
 
  • #14
Tech Notes
 

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Related to Nominal Linear Velocity Gas flow

1. What is nominal linear velocity gas flow?

Nominal linear velocity gas flow is a measurement of the speed at which gas flows through a given area. It is typically expressed in units of meters per second (m/s) or feet per second (ft/s).

2. How is nominal linear velocity gas flow calculated?

Nominal linear velocity gas flow is calculated by dividing the volumetric flow rate of gas by the cross-sectional area through which it is flowing. This can be represented by the equation V = Q/A, where V is the velocity, Q is the volumetric flow rate, and A is the cross-sectional area.

3. What factors can affect nominal linear velocity gas flow?

Several factors can affect nominal linear velocity gas flow, including the pressure and temperature of the gas, the viscosity and density of the gas, and the size and shape of the conduit through which the gas is flowing.

4. Why is nominal linear velocity gas flow an important measurement?

Nominal linear velocity gas flow is an important measurement because it helps to determine the efficiency and performance of gas systems and equipment. It can also be used to ensure that gas is flowing at a safe and optimal rate for a given application.

5. How is nominal linear velocity gas flow used in scientific research?

Nominal linear velocity gas flow is commonly used in scientific research to study the properties and behavior of gases. It can also be used to design and optimize gas systems for various industrial and scientific applications.

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