# High school project -- Finding mass flow rate, air velocity

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1. Jan 17, 2016

### AEGIS

Hi,

First-timer here! I'm designing a physics project that involves pumping a certain pressure of air into an empty bottle.

SETUP: The bottom of the bottle has a small hole through which a thumb tack fits tightly; the top (cap-side) has a valve through which the air is inserted. The bottle is then laid on its side lengthwise, and attached to a small toy car--removing the thumb tack releases a stream of concentrated air from the orifice, which acts as a force of thrust propels the car. All in all, the setup is pretty similar to that seen here.

I'm trying to find the velocity of the air at the orifice as well as within the bottle, assuming bottle pressure po and standard atmospheric pressure outside. I also wanted to determine whether or not the velocity at the orifice would be constant over a period of time given the release of air and loss of pressure. I've researched the topic but what I've found seems to apply more to actual rockets with defined nozzles and throats. What equations would I use?

I'm only in high school, so most of this except the basic motion/forces/thrust equations is entirely new to me. Thanks again for the help!

2. Jan 17, 2016

### billslugg

The general equation relating pressure to velocity of a gas is P=V^2/2g where pressure is in feet of the fluid under consideration, V is in feet per second and g is 32 feet per second squared. For example, suppose you put 10 psi into the bottle, what would the velocity of the air flow coming out of the thumb tack hole?

To put ten pounds of air onto a square inch requires a column of air that is quite high. A pound of air at room temperature occupies about 13 cubic feet. Ten pounds would occupy 130 cubic feet. If you take that 130 cubic feet of air and form it into a column that is one square inch in cross section, it will be 224,640 inches tall or 18,720 feet. Thus the pressure inside the bottle, expressed in terms of feet of the fluid, is 18,720 feet of air.

Take 18,720, multiply by 2 and then by 32 and you get 92,160. Now take the square root of 92,160 and you get 1094 feet per second.

The actual value will be about 80% or 90% of that due to the fact that the orifice is not a smooth transition but jagged and abrupt.

3. Jan 17, 2016

### AEGIS

Okay, that makes sense. Does that mean that the velocity of the air flow hole is gradually decreasing over time as more and more of the air leaves the bottle? (Also, say I put a certain air pressure into the bottle--is it safe to assume that the pressure at the hole as air is escaping is equal to that at any other point within the bottle?)

Also, I'm trying to calculate the mass flow rate, probably using the equation ρ*A*V (ρ = density of gas, A = area of orifice, V = velocity of gas). I thought that I had read somewhere that the density of the air in the bottle is only constant sometimes due to compressibility effects--how can I tell if the density will be constant in this case? I searched it up but did not understand the explanation regarding "Reynolds numbers" and whatnot that was provided.

4. Jan 17, 2016

### billslugg

Yes, the velocity of the air will gradually decrease as the pressure in the bottle gradually decreases. Since the interior of the bottle is large and the flow within the bottle is small, the pressure inside the bottle will be the same everywhere.

The density in the bottle is actually greater than the density of room air. The absolute pressure inside the bottle is 15+10 psi or 25 psi. The density will therefore be 25/15 that of room air. The pressure inside the bottle expressed as feet of the fluid should actually be 15/25 of what I told you in my first reply. That will give you a lower velocity. I forgot to make that correction.

The Reynold's number is not really important here other than to know that the flow inside the bottle is laminar and the flow in the nozzle is fully turbulent. The Reynolds number is used to tell you what flow regime you will be in. The pressure drops are different for different regimes. This really comes into play when calculating losses in long piping systems. For your bottle it is not needed. The basic equation is good enough. I redid it with the higher density in the bottle and got 848 fps for the air flow out of the nozzle. I would say knock that down to about 700 fps to calculate the mass flow rate. Use the density inside the bottle, the 700 fps number and the diameter of the thumb tack and you will get a rough idea of mass flow rate.

5. Jan 20, 2016

### cjl

The actual velocity won't be anything close to that, since you're talking about a compressible flow here (and 1094 ft/s is very nearly supersonic). It's actually a relatively complex problem, and your best bet might be to try to figure out the thrust experimentally.

6. Jan 20, 2016

### AEGIS

Ah. Sorry, but how can you tell that the flow is compressible in this situation? Flow is beyond my high school physics curriculum and I couldn't make sense of the explanations that I found, which used terminology whose definitions didn't make full sense to me.

EDIT: Additionally, what does a situation of compressible flow imply (conversely, what does a situation of non-compressible flow imply)? Are there any additional variables that I then must account for in either of the two scenarios if I need to make calculations of mass flow rate, force of thrust, velocity of air escaping the orifice, etc.? Most importantly, is the velocity of the air escaping the orifice definitely constant in either/both of these scenarios, or if not, what equation can I use to account for the difference? (Unfortunately, I do have to theoretically calculate the force of thrust, mass flow rate, so forth given my teacher's rubric, but could probably explain a significant difference between theoretical/actual as the complexity of the situation/need to make simplifying assumptions. I will also be determining experimental values, however.)

Last edited: Jan 20, 2016
7. Jan 21, 2016

### cjl

I can tell the flow is compressible because at the pressures involved, the density in your storage container will be substantially above ambient density. In addition, any time you run the numbers and find yourself with flows with a mach number in excess of about 0.3 or so, the compressibility contribuition to the fluid dynamics is no longer negligible, and will have a significant effect on the flow. There are some equations to use with compressible flow through an orifice, but I'll have to look them up (I'm a bit busy right now).

8. Jan 21, 2016

Actually the flow problem is still fairly straightforward even for compressible flow under a few reasonable assumptions (assuming I am not brain farting here). It's even easier if the flow is choked, but that determination requires us to know the bottle pressure.

9. Jan 21, 2016

### cjl

I was assuming it wasn't choked (and I honestly don't remember much for a non-choked compressible flow) - you're right that if the flow is choked, it's pretty easy (and that requires what, a bit over 2:1 pressure ratio if I remember right? It's been a while since my last aero class, and compressibility isn't really relevant to my job, sadly)

10. Jan 21, 2016

For air the pressure ratio is 0.528, so just under 2:1.

11. Jan 21, 2016

### cjl

Ahh, that's right. It's really disappointing that I'm forgetting this stuff now that I'm not in grad school anymore (and not using it daily).

12. Jan 21, 2016

Ah, the virtues of remaining in research...

13. Jan 22, 2016

### JBA

14. Jan 22, 2016

### tech99

15. Jan 22, 2016

Assuming it is choked. There is no guarantee.

16. Jan 22, 2016

### JBA

I stand corrected, as for this application there is likely to be subcritical flow as well. In spite of the title the tables show flow at down to 1 psig; and assumed that an atmospheric pressure was the base discharge pressure for the tables. Upon further checking, I have discovered that there is a preceding table specifying the sub- atmospheric outlet pressures required for inlet pressures below the .528 pressure ratio.

17. Jan 22, 2016

At any point, the bottom line is that if the pressure is high enough, the flow will be choked and the mass flow rate coming out of the hole will be linearly proporational to the total pressure inside the bottle. From there you can calculate the flow in time until it is no longer choked via differential equation, at which point you have to use other equations that are a little more complicated. At the end of the day, though, it's all likely above the level of a high school student if they don't know differential equations.

18. Jan 31, 2016

### AEGIS

Sorry to revive a (somewhat) closed thread, but could anyone point me towards the specific choked/non-choked & compressible flow equations describing the velocity and/or mass flow rate of air? I've been having a nightmare trying to find ones for non-choked flow and am not even sure that the choked flow equations I've found are right either. (thankfully, I have at least enough experience with differential equations to get me by for this project.)

If you could provide a citable source as well, that would be much appreciated. Thank you.

19. Jan 31, 2016