[xiangru119]

Hi, all!
I encountered a difficulty when I was worked out a problem with English unit.
A simple question is like that:
Pressures up to 3000 atm are measured with a dead-weight gauge. The piston diameter is 0.17 (in). What is the approximate mass in (lb_m) of the weights required?
This is what I done:
Abs. pressure = Gauge pressure + Atmospheric pressure
                   = 3000 atm + 1 atm
                   = 3001 atm * 14.696 psi/atm
                   = 44102.696 psi
A = pi*D²/4
   = 0.023 in²
P = F/A
P*A = mg
44102.696*0.023 = m * 32.174 ft/s² (12 in./1 ft)
m= 2.627 lb_m

However, the actual answer is 1000.7 lb_m. So can anyone point out my error?
Any help would be appreciated. Thank you.

[rjb]

Hello,

I'm not 100% sure what the problem you are trying to solve is all about, however, I guess fundamentally the question is how much weight do you have to stick on top of a piston with contents at 3000 atm to meet equilibrium.

Although your units are in old money (non si units), as far as I can tell, your calculations seem to make sense up until you start calculating P*A = mg. This as far as I can tell is unnecessary. You've worked out the pressure exerted as 44102.696 psi, which seems about right, although give some more consideration to the atmospheric pressure addition, I may be wrong, but I'm not sure that makes sense.

The area of the piston is Pi r², which comes out as 0.022698 in². Force required for equilibrium is Pressure X Area which will give you a figure in the ballpark of what you're expecting...

Good Luck

R 

 

[opti384]

Although your units are in old money (non si units), as far as I can tell, your calculations seem to make sense up until you start calculating P*A = mg. This as far as I can tell is unnecessary. You've worked out the pressure exerted as 44102.696 psi, which seems about right, although give some more consideration to the atmospheric pressure addition, I may be wrong, but I'm not sure that makes sense.

The area of the piston is Pi r², which comes out as 0.022698 in². Force required for equilibrium is Pressure X Area which will give you a figure in the ballpark of what you're expecting...

Good Luck

R 

Well pi*D2/4 = Pi r2. xiangru got the area correctly and I think there's no problem with P = F/A , P*A = mg. The result differing from the answer may be because of incorrect coversions of units (I'm also not familiar old money units) or there might be an error in finding the pressure which I'm not 100% sure either. Also, the answer you got m= 2.627 lbm seems to be miscalculate because I get 31.113414 according to your figures.

[xiangru119]

Hi, thanks all for the comments. I'm pretty sure that the steps I done were correct and I just not familiar with the English unit conversion. The pressure gauge gives reading which is the difference between the pressure of interest and the pressure of the surrounding atmsphere. This reading is known as gauge pressure and can be converted to absolute pressure by addition of the barometric pressure. I think there must be an errors in the conversion process. Hopefully someone can work it out and post it, so that I can compare my answer with the correct one. Thank you.

[Borek]

Watch your units. psi is not lb_m per square inch.