Relationship between Pressure and volume with Air

  • #1

Main Question or Discussion Point

Hey guys!

So the below question might sound like a homework question but in reality it is an oversimplified explanation of what I'm trying to measure

Let's say we have a perfect machine that pushes out 600CFM into a box. On the other end of the box is another small machine that only consumes 300CFM from the box.

What is the pressure in the box? and how do I calculate that? I've looked into Boyles Law and something just doesn't seem right about the results I'm getting so I'm trying to make sure that this is the right law I should be looking into.

If you need area of the box then let's assume it is 360in3

Thanks for lending a hand!
 

Answers and Replies

  • #2
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Hey guys!

So the below question might sound like a homework question but in reality it is an oversimplified explanation of what I'm trying to measure

Let's say we have a perfect machine that pushes out 600CFM into a box. On the other end of the box is another small machine that only consumes 300CFM from the box.

What is the pressure in the box? and how do I calculate that? I've looked into Boyles Law and something just doesn't seem right about the results I'm getting so I'm trying to make sure that this is the right law I should be looking into.

If you need area of the box then let's assume it is 360in3

Thanks for lending a hand!
You will have to mention the temperature as well.
 
  • #3
You will have to mention the temperature as well.
For the sake of this experiment let's assume 310K
 
  • #4
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You are aware that the number of moles of gas in the box is changing with time, correct?
 
  • #5
Hello,

This is a relatively new concept for me so I apologize if I don't understand something. I did not know that the moles of gas would change in time, is this because of the temperature increase that occurs over time? I'm trying to find a good example of how I can measure this so does this mean I would have to provide time? How about we say that this system is running for 5 seconds straight.
 
  • #6
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Hello,

This is a relatively new concept for me so I apologize if I don't understand something. I did not know that the moles of gas would change in time, is this because of the temperature increase that occurs over time?
No. It is because more mass of air is entering than leaving.
I'm trying to find a good example of how I can measure this so does this mean I would have to provide time? How about we say that this system is running for 5 seconds straight.
You're missing much more important information than that. If you don't specify the inlet density of the air, we can't determine the rate of air mass entering. Giving the CFM of air does not specify the mass flow rate.
 
  • #7
For the density of air, let's use 0.069 pound/ft2 of air.

Thanks again!
 
  • #8
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OK. Let's assume that the temperature is constant at 70 F. Then, if p is the pressure in the tank (psi), the number of lb-moles of air in the tank at any time is:
$$m=\frac{pV}{RT}$$where V is the volume of the tank (ft^3), T = 491.7 +70 = 561.7 Rankine, and R is the universal gas constant (10.73 ##\frac{psi\ ft^3}{R\ lb_{mole} }##). The rate of air entering the tank is 41.4 ##lb_m/min##=1.43 ##lb_{mole}/min##. The molar density of the air in the tank is m/V. The rate at which moles of air exit the tank is 300 m/V. So, from a mass balance on the air in the tank, $$\frac{dm}{dt}=1.43-300m/V$$, where t is measured in minutes. So, just solve this differential equation for the moles as a function of time, and then calculate the corresponding pressure.
 
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  • #10
jack action
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You are aware that the number of moles of gas in the box is changing with time, correct?
No. It is because more mass of air is entering than leaving.
I kind of disagree with that. The problem states that more volume of air is entering than leaving, not mass. Therefore we should assume that a steady state will be reached where ##\dot{m}_{in} =\dot{m}_{out}##, or:
[tex]\rho_{in}\dot{V}_{in} = \rho_{box}\dot{V}_{out}[/tex]
And if we assume that the temperature within the box is maintain to a pre-determined value, then:
[tex]p_{box} = \rho_{box}RT_{box} = \rho_{in}\frac{\dot{V}_{in}}{\dot{V}_{out}}RT_{box}[/tex]
With the given data (0.069 lb/ft³ = 1.1053 kg/m³; 600 CFM; 300 CFM; 287 J/kg/K; 310 K), we get a steady-state pressure of 196.7 kPa or 28.5 psi.

Of course, this can become more complex if we assume different conditions, such as no heat input/output from the box (so we have to find the final temperature within the box).
 
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  • #11
I thought about this too but as I said before I'm not expert. I think this is right on the dot on what I'm looking for. Thank you for replying!!

@jack action Can you help me understand the formula a little better? as I said before, this is not a school thing and I'm trying the best I can to remember (and research) this.

To be more specific, how would you order the variables to equate 28.5 PSI? I know you provided the formula but I am unable to connect the variables to it.

Maybe something like:

P=14.7
T=344K

Etc etc
 
Last edited:
  • #12
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I kind of disagree with that. The problem states that more volume of air is entering than leaving, not mass. Therefore we should assume that a steady state will be reached where ##\dot{m}_{in} =\dot{m}_{out}##, or:
[tex]\rho_{in}\dot{V}_{in} = \rho_{box}\dot{V}_{out}[/tex]
And if we assume that the temperature within the box is maintain to a pre-determined value, then:
[tex]p_{box} = \rho_{box}RT_{box} = \rho_{in}\frac{\dot{V}_{in}}{\dot{V}_{out}}RT_{box}[/tex]
With the given data (0.069 lb/ft³ = 1.1053 kg/m³; 600 CFM; 300 CFM; 287 J/kg/K; 310 K), we get a steady-state pressure of 196.7 kPa or 28.5 psi.

Of course, this can become more complex if we assume different conditions, such as no heat input/output from the box (so we have to find the final temperature within the box).
This sounds reasonable, but, of course, as you said, it all depends on the energy balance. My original impression was that this system is significantly underspecified by the OP. That impression hasn't changed.
 
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  • #13
jack action
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To be more specific, how would you order the variables to equate 28.5 PSI?
##\rho_{in}## = 1.1053 kg/m³;
##\dot{V}_{in}## = 600 CFM;
##\dot{V}_{out}## = 300 CFM;
##R## = 287 J/kg/K;
##T_{box}## = 310 K;
Putting those numbers in the equation gives ##p_{box}## = 196 677.082 Pa which you can convert to 196.7 kPa by dividing it by 1000 or to 28.5 psi by dividing it by 6894.757.
 
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  • #14
##\rho_{in}## = 1.1053 kg/m³;
##\dot{V}_{in}## = 600 CFM;
##\dot{V}_{out}## = 300 CFM;
##R## = 287 J/kg/K;
##T_{box}## = 310 K;
Putting those numbers in the equation gives ##p_{box}## = 196 677.082 Pa which you can convert to 196.7 kPa by dividing it by 1000 or to 28.5 psi by dividing it by 6894.757.
You rock!!!!! Thank you so much for this!!!
 
  • #15
@jack action

So I'm trying to plug the numbers in but I might be missing something. Can you check my math below?

X = ρbox * 287 * 310 = 1.1053* (600/300) * 287 * 310

Where do you find the value for ρbox?

Sorry for being so dense.. It's really embarrassing on how much of this stuff I've forgotten!
 
  • #16
Nevermind! I was able to figure it out!
 

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