# Nominal Linear Velocity Gas flow

Tags:
1. Sep 22, 2015

### Jamesan89

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 .

#### Attached Files:

• ###### Snapshot.jpg
File size:
9.4 KB
Views:
117
2. Sep 22, 2015

### Nidum

Please explain configuration a bit more . Also confirm the 350 m/s speed .

3. Sep 22, 2015

### Jamesan89

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. Sep 22, 2015

### Nidum

350 metres/sec is around speed of sound for many common gasses so there could be problems in achieving this speed .

5. Sep 22, 2015

### Nidum

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. Sep 22, 2015

### Jamesan89

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. Sep 22, 2015

### Jamesan89

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

#### Attached Files:

• ###### DESI.pdf
File size:
158.6 KB
Views:
125
8. Sep 22, 2015

### Nidum

See message .

9. Sep 22, 2015

### SteamKing

Staff Emeritus
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. Sep 22, 2015

### JBA

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. Sep 22, 2015

### SteamKing

Staff Emeritus
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. Sep 23, 2015

### Jamesan89

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 refering 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. Sep 23, 2015

### SteamKing

Staff Emeritus
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. Sep 23, 2015

Tech Notes

File size:
437.5 KB
Views:
188