- #1
Peter99
- 14
- 3
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
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
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