Gas Cylinder Discharge Time: Solving for V2 Vent

[nampe]

hi :

i have this question, which it reads pretty simple but I am trying to see if someone can walk me through the answer...
A gas cylinder has two vents, V1 and V2, for
discharge. The cylinder empties in 7 minutes when
only vent V1 is open. The cylinder empties in 3
minutes when both vent V1 and vent V2 are open. About
how many minutes will the cylinder take to empty if
only vent V2 is open?

the answer is in actual minutes...
thanks
pedro

 

[arcnets]

nampe,
obviously it doesn't matter what the vessel's volume is. So you could just assume it's 100 litres or something, and calculate how many litres per minute go thru each vent.

 

[turin]

You have to know how (why) it empties for the conditions of the vents being open. Is there a piston pushing the stuff out? Is it in an infinite negative pressure reservoir?

 

[Doc Al]

turin is right, but my guess is that this is meant as an algebra problem, not a physics problem. Is that correct?

In that case, the rates are:

V₁ = 1/7 (tanks per minute)
V₁ + V₂ = 1/3

Solve for V₂ and invert to get the time for Valve 2 alone.

 

[nampe]

thanks every one for your replies... i found the answer and here it is...Think about it like this. For a given cylinder, the product (time to empty) x (total vent size) should be constant. Basically, if you open two vents of the same size, it should take half the time to empty. So let's say V1 and V2 are the sizes (areas) of vents 1 and 2. Then,

7 * V1 = 3 * (V1 + V2) = T * V2

From the first part of the equation, you can solve for V1 in terms of V2. That will then allow you to solve the second half of the equation for T, which is your answer.
so using 7 * V1 = 3 * (V1 + V2) = T * V2, i am solving for the first part of the equation...
7 * V1 = 3 * (V1 + V2)... 7V1=3V1+3V2...V1=3V2/4...now u got to place 7 * V1= T * V2 to obtain T value
7*3V2/4= T*V2...V2=7*3V2/4/V2 which V2=7*3/4=5.25

this was a gre practice question... pretty interesting ehhh?!
thanks again

 

[turin]

Originally posted by nampe
For a given cylinder, the product (time to empty) x (total vent size) should be constant.

Why? Did you find this in the solution guide or something? The rate that the cylinder empties is a lot more intimately related to why it empties, than how it empties.

 

[Jess]

Originally posted by turin
Why? Did you find this in the solution guide or something? The rate that the cylinder empties is a lot more intimately related to why it empties, than how it empties.

I assume that the tank empties because the gas inside is at greater than atmospheric pressure. When the valves are opened the tank pressure changes until the pressure in the tank is the same as the pressure outside the tank. This way it doesn't matter how the tank empties as long as the starting tank pressure is the same each time.

Jess

 

[turin]

If inactive pressure equalization is what causes the tank to empty, then it will never be empty (unless you assume that the ambient can be at a negative absolute pressure). Even if we assume that what is meant is pressure equalization with the ambient, then this will still take an infinite amount of time according to the proposed mechanism of strict baromotivation (I think it is a decaying exponential).