Unit Conversion: What went wrong?

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Homework Help Overview

The problem involves calculating the mass of weights required for a dead-weight gauge measuring pressures up to 3000 atm, with a given piston diameter. The original poster attempts to convert pressures and areas to find the mass in pounds.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the relationship between mass and weight, questioning whether the calculated value represents mass (lbm) or force (lbf). There are attempts to clarify the role of gravitational acceleration in the calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the results and the necessary conversions. Some guidance has been offered regarding the definitions of mass and weight, but no consensus has been reached on the correct interpretation of the calculations.

Contextual Notes

Participants note the importance of distinguishing between lbm and lbf in the context of the problem, as well as the implications of gravitational acceleration in the calculations. There is an acknowledgment of communication barriers affecting the discussion.

xiangru119
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Homework Statement


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 (lbm) of the weights required?

The Attempt at a Solution


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 lbm

4. The actual answer: 1000.7 lbm. So can anyone point out my error?
Any help would be appreciated. Thank you.
 
Physics news on Phys.org
A mass of 1 lbm weighs 1 lbf in standard gravitational field.

<br /> 4.41 \times 10^{4} \, \frac{\mathrm{lb}}{\mathrm{in}^{2}} \times (2.3 \times 10^{-2} \, \mathrm{in}^{2}) = 1.01 \times 10^{3} \, \mathrm{lb}<br />
 
Dickfore said:
A mass of 1 lbm weighs 1 lbf in standard gravitational field.

<br /> 4.41 \times 10^{4} \, \frac{\mathrm{lb}}{\mathrm{in}^{2}} \times (2.3 \times 10^{-2} \, \mathrm{in}^{2}) = 1.01 \times 10^{3} \, \mathrm{lb}<br />

Hi, the answer obtained from this should be equivalent to pound force. How about the accelerate force of gravity? Thanks for your comment.
 
xiangru119 said:
Hi, the answer obtained from this should be equivalent to pound force. How about the accelerate force of gravity? Thanks for your comment.

I don't know what you are talking about. I was not supposed to solve even this much I think. Please use proper English to communicate your ideas.
 
I apologise for my broken English. I mean the answer obtained (1.01x10^3 lb) might be lbf instead of lbm. I think acceleration of gravity needed to be taken into account in order to get lbm.
 
But, it is taken into account. Did you read this:
Dickfore said:
A mass of 1 lbm weighs 1 lbf in standard gravitational field.
 
Ok, I will think about it. Thanks
 

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