# Homework Help: Exit velocity of a gas through a hole

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1. Sep 25, 2016

### Peter99

• Moved from a technical forum, so template missing
Hello,

Given: A large pressurized container with a small hole in a side. The sides of the container are VERY thin such that the thickness of the sides can be ignored. The pressure difference between the container and the environment is not large enough to result in choked flow. The outflow is not enough to significantly change the pressure inside the container or the pressure of the surrounding environment.

Question: How does the velocity of the escaping gas relate to the diameter of the hole?

I am not sure even how to approach this. My first impulse is to apply Bernoulli's principal as it basically states that there is a relationship between the pressure of a fluid and the velocity of a fluid.

If I take this approach, simplify things by assuming a non-compressible fluid, a zero elevation difference, and applying the conservation of energy, this yields:

P1 + (1/2) * p * (V1^2) = P2 + (1/2) * p * (V2^2)

where

P1 = pressure inside container
V1 = velocity of gas inside container
P2 = pressure outside container
V2 = exit velocity of gas
p = fluid density

But V1 = 0, and (1/2) * p = constant = C, so:

P1 = P2 + C * V2^2
or
[(P1 - P2)/C] ^ (1/2) = V2

Applying the continuity equation Q = A2 * V2

where

Q = volume flow rate
A2 = area of hole
V2 = velocity of gas

So:

Q/A2 = [(P1 - P2)/C] ^ (1/2)
or
Q = A2 * [(P1 - P2)/C] ^ (1/2)

Which basically says that the volume flow rate increases as the area of the hole increases. But if Q increases then I think I can infer that the linear velocity of the gas also increases.

So the conclusion I get from this is that the exit velocity of the fluid increases as the diameter of the hole increases.

Is this correct??? This result is a bit surprising to me. Why does the velocity increase as the hole diameter increases?

Thanks,
Peter

Last edited: Sep 25, 2016
2. Sep 26, 2016

### The Buttered Cat

I think the Bernoulli equation would work for an incompressible fluid but here you have a gas, so probably not applicable. This has come up before on this forum and here is a link that was referred to: http://www.mcnallyinstitute.com/13-html/13-12.htm

Hopefully that will help.

3. Sep 26, 2016

### Peter99

Hi,

But I'm still not certain I have this conceptually correct and would like confirmation. The relationship Q = AVK is basically the continuity equation (Q = AV) with an adjustment parameter (K). So if I rewrite this as:

V = Q / AK

Then I see that basically the ratio Q / A is what describes the behavior. So Q is a volume rate and A is an area and both Q and A depend on he same linear quantity (the diameter of the hole). So since Q is effectively a cubed quantity and A is a squared quantity, Q will dominate and this ratio will go up with increasing diameters. Thus, the larger the hole the greater the exit velocity. Am I thinking about this correctly?

If so then why is this the case?

I can understand that internal resistance and resistance with the sides of the hole may limit flow for small diameter holes but shouldn't this effect essentially go to zero as the hole diameter gets bigger? But the ratio Q / A does not indicate this. It indicates a linear increase in velocity with increased diameter.

This is counter intuitive to me and so I am not sure I am thinking about this correctly.

Thanks again for the help.

4. Sep 27, 2016

### Peter99

Hi,

OK, I think I am understanding this a little better. So I am writing this to add closure to this conversation.

I believe my "power" analysis in my last comment was incorrect. Yes, the diameter of the hole is in A (the area) and is also in Q (the volume rate), however, the third quantity that is multiplied in Q is NOT the diameter. It comes from the velocity. So the contribution of the area in the equation

V=Q/A

effectively cancels out and one is left with only the contribution of the velocity. So I started to look purely at what would produce the exit velocity of a gas, or more generally a fluid.

Given the initial conditions stated in the original post, and for a non-compressible fluid, the velocity turned out to be

V2 = [(P1 - P2)/C] ^ (1/2)

I also looked at the velocity equation given at the link "The Buttered Cat" responded with and that was

V = (2gh)^(1/2)

where

g = the acceleration due to gravity
h = head across the orifice, or basically the pressure difference from inside the tank to the environment.

Both equations say that the velocity is a function of the pressure difference (Delta P) and not a function of A (the diameter of the hole). This made me think and I realized that given a CONSTANT pressure, the force exerted on an individual element of a fluid is always the same (on average). So it does not matter how large the hole is, the force that is accelerating each individual element of the fluid is the same. The force per unit area is constant.

Given this, the answer to the original question is: The size of the hole does not effect the exit velocity. The exit velocity is determined by the pressure difference between the tank and the environment and also the frictional forces both internal to the fluid and between the fluid and the orifice sides. Ignoring frictional forces and assuming a constant pressure, the exit velocity of a fluid is always the same no matter how big the orifice diameter is.

I think this is correct but if someone would be so inclined to verify my thinking on this it may be helpful to someone else in the future. Thanks to "The Butter Cat" for making me think and not giving a direct answer.

Peter99