Gas Cylinder Discharge Time: Solving for V2 Vent

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In summary, the conversation discusses a gas cylinder with two vents, V1 and V2, for discharge. The cylinder empties in 7 minutes when only V1 is open and in 3 minutes when both V1 and V2 are open. The question asks how long it will take to empty the cylinder if only V2 is open. The answer is found by solving the equation 7 * V1 = 3 * (V1 + V2) = T * V2, which gives the value of V2 as 5.25 minutes. The conversation also includes a discussion about the relationship between time and vent size in emptying the cylinder.
  • #1
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
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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.
 
  • #5
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
 
  • #6
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.
 
  • #7
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
 
  • #8
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).
 

1. How do I calculate the discharge time for a gas cylinder?

In order to calculate the discharge time for a gas cylinder, you will need to know the volume of gas in the cylinder (V1), the flow rate of the gas (Q), and the exit pressure of the gas (P2). You can use the equation V2 = V1 - (Q * t / P2) to solve for V2, which is the remaining volume of gas in the cylinder after a given amount of time (t).

2. What is the significance of solving for V2 in gas cylinder discharge time?

Solving for V2 is important because it tells you how much gas is left in the cylinder after a certain amount of time. This information is crucial for determining when a cylinder needs to be refilled or replaced, and for ensuring that the gas supply remains constant and reliable.

3. How does the exit pressure of the gas affect the discharge time?

The exit pressure of the gas has a direct impact on the discharge time. The lower the exit pressure, the longer the discharge time will be, as there is less resistance for the gas to flow out of the cylinder. Conversely, a higher exit pressure will result in a shorter discharge time.

4. Are there any other factors that can affect the discharge time of a gas cylinder?

Yes, there are other factors that can affect the discharge time of a gas cylinder. These include the temperature and composition of the gas, the size and shape of the cylinder, and any obstructions or restrictions in the gas flow path.

5. Can I use the same equation to calculate the discharge time for all types of gases?

The same equation can be used to calculate the discharge time for most gases, as long as the flow rate and exit pressure are known. However, some gases may have different properties that could affect the accuracy of the calculation. It is always best to consult a gas supplier or refer to specific gas cylinder discharge time tables for more precise results.

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