Calc Air Pressure in Tank with Fluid Level & Height

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SUMMARY

The discussion focuses on calculating the headspace pressure in a closed tank partially filled with fluid, specifically addressing the pressure at the interface between the air and the liquid. The fluid has a specific gravity (s.g.) of 1.2, with a fluid height of 1.7m and a total height of 5m above the measurement point. The correct calculation for the air pressure in the tank's headspace is determined to be 58680 N/m², which is derived using hydrostatic pressure principles rather than the ideal gas law.

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jderulo
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Homework Statement



What is the headspace pressure (air)?

Closed tank part filled with fluid under pressure, there are tubes that indicate the fluid level and pressure. The tube on the left is open to atmosphere.

fluid s.g. 1.2
x = 5m
y= 1.7m

111111111.png

Homework Equations



See below

The Attempt at a Solution



Assuming atmospheric pressure and rho*g*h come into play?
 
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Last edited:
Why? I was wondering whether to use boyles law as the tank full of air at atmospheric will have been compressed as the tank filled?
 
never mind
 
?
 
?
 
Can you please provide an exact statement of the problem.
 
Hi the exact statement contains no further info except a couple of filler words such as the, and etc.

Though it doesn't state about it being heads pace or air pressure it's just the area above the fluid.
 
If atmospheric pressure is present at the top of the left column of water, what is the pressure in the left column of water at the level of the dashed line in the tank?

Chet
 
  • #10
Atmospheric??
 
  • #11
jderulo said:
Atmospheric??

The pressure at the top of the column is atmospheric...The pressure lower in the column is found via Pascals law
 
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  • #12
...
 
Last edited:
  • #13
jderulo said:
1.2*9.81*y ?
Let's check. If that's the pressure at the level of the dotted line, what is the pressure at the bottom of the tank? And also, based on your result, what is the pressure at the top of the left column of fluid?

If you are not able to solve this problem, you need to go back to your textbook and re-read the sections on hydrostatic pressure. If you encounter difficulties with understanding those sections of the textbook, we are prepared to help answer specific questions you may have.

Chet
 
  • #14
I fail to see how I can elaborate further with the info give. I have numerous textbooks, the unconventional setup is throwing me somewhat.
 
  • #15
jderulo said:
I fail to see how I can elaborate further with the info give. I have numerous textbooks, the unconventional setup is throwing me somewhat.

Can you state in your own words what the 'h' in the equation in your OP represents?
If not, can you find a definition in one of your textbook?
 
  • #16
Head of liquid (the x and y values)
 
  • #17
'h' is the height of fluid above the point of measurement.

So what is the height of fluid above the point you are interested in (the dotted line)?
 
  • #18
Interested in the area above the dotted line, 5metres
 
  • #19
Correct.
So what now? Can you reach the answer?
 
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  • #20
No that is where I am stuck, I can calculate the fluid pressure but the air pressure I am lost.

1200kg.m^-3 * 5m * 9.81m.s^-2 = 58680N.m^-2
 
  • #21
jderulo said:
No that is where I am stuck, I can calculate the fluid pressure but the air pressure I am lost.

1200kg.m^-3 * 5m * 9.81m.s^-2 = 58680N.m^-2
Are you aware that pressure is continuous across the interface between the air in the head space and the liquid immediately below?
 
  • #22
Chestermiller said:
Are you aware that pressure is continuous across the interface between the air in the head space and the liquid immediately below?

Hi Chestermiller

Yes, I guess so - but again totally lost as to how I can progress further? Sorry!
 
  • #23
jderulo said:
Hi Chestermiller

Yes, I guess so - but again totally lost as to how I can progress further? Sorry!
At what location(s) do you think your calculated pressure of 58680N.m^-2 is found?

Let's forget about the tank for the moment, and just focus exclusively on the left column of fluid. You already indicated that the pressure at the very top of this column of fluid is atmospheric. As you move downward from the top of this column, does the pressure get higher or lower (think what happens when you go below the surface in a swimming pool)? Suppose you go downward to a point 5 m below the top surface of the liquid in the column. What would the pressure at this location be equal to (over and above atmospheric pressure)?

Chet
 
  • #24
Chet the calculated pressure would be at the bottom of the tank.

The pressure would get higher. But my assumption was to use the ideal gas law to calculate the air pressure but this then doesn't fit the 5m into my equation?. How can I go 5m below the top surface of liquid when there is only 1.7m of liquid?? The 5m is air?
 
  • #25
jderulo said:
Chet the calculated pressure would be at the bottom of the tank.
No. That is not the pressure at the bottom of the tank.
The pressure would get higher. But my assumption was to use the ideal gas law to calculate the air pressure but this then doesn't fit the 5m into my equation?. How can I go 5m below the top surface of liquid when there is only 1.7m of liquid?? The 5m is air?
I told you to forget about the tank, and to focus on the column of liquid on the left. The ideal gas law will not be part of the solution to this problem. Looking at the column of liquid on the left in your figure, did you notice that little triangular "symbol thingie" at the very top of the left column. The left column is full of liquid all the way up to that triangular "symbol thingie." Above that point there is air at atmospheric pressure, but below that point there is liquid. Were you aware of that?

Chet
 
  • #26
Chestermiller said:
No. That is not the pressure at the bottom of the tank.

I told you to forget about the tank, and to focus on the column of liquid on the left. The ideal gas law will not be part of the solution to this problem. Looking at the column of liquid on the left in your figure, did you notice that little triangular "symbol thingie" at the top of the left column. The left column is full of liquid all the way up to that triangular "symbol thingie." Above that point there is air at atmospheric pressure, but below that point there is liquid. Were you aware of that?

Chet

No I wasn't - I had incorrectly assumed the liquid level in the column was the same as in the tank. Thought the triangle symbol was maybe a smudge of a scan. The head is 6.7m?

I am guessing at this point to move onto the tank? If so, a 6.7m head factored into my rho * g * h equation then gives me the pressure at the interface inside the tank?
 
  • #27
jderulo said:
No I wasn't - I had incorrectly assumed the liquid level in the column was the same as in the tank. Thought the triangle symbol was maybe a smudge of a scan. The head is 6.7m?

I am guessing at this point to move onto the tank? If so, a 6.7m head factored into my rho * g * h equation then gives me the pressure at the interface inside the tank?
No. You went downward 6.7 m from the top water surface in the column. This puts you at the bottom of the tank. Now you've got to go back up 1.7 m to get to the water surface in the tank.

Chet
 
  • #28
Chestermiller said:
No. You went downward 6.7 m from the top water surface in the column. This puts you at the bottom of the tank. Now you've got to go back up 1.7 m to get to the water surface in the tank.

Chet

OK - so 9.4m. That makes sense. I didn't realize we added the head backupwards.

I am right in thinking the answer is 110657N.m^-2 ?
 
  • #29
jderulo said:
OK - so 9.4m. That makes sense. I didn't realize we added the head backupwards.

I am right in thinking the answer is 110657N.m^-2 ?
No. If you go back upwards, you have got to subtract, not add.
 
  • #30
Chestermiller said:
No. If you go back upwards, you have got to subtract, not add.

58860N.m^-2 ? Assuming a head of 5m?
 

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