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PassatDream

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## Homework Statement

A pressure of 3000atm is measured with a dead-weight gauge. The piston diameter is 0.17in. What is the mass in pound-mass (lbm)?

## Homework Equations

1) P=F/A

2) F=PA

3) F=mg

4) F/g=m

## The Attempt at a Solution

I first used the given diameter to find the cross-sectional area of the piston:

A = ([tex]\frac{\pi}{4}[/tex])(0.17 in[tex]^{2}[/tex]) = 0.023 in[tex]^{2}[/tex]

Then I plugged the newly-found A into eqn. 2 with the given P value plus a conversion factor (atm->psi).

F = (3000atm)([tex]\frac{14.69595 psi}{1 atm}[/tex])(0.023in[tex]^{2}[/tex])

F= 1014.02 lbf

I then used my F value and plugged it into eqn. 4 and used the standard gravitational value (32.174 ft/s^2)for g.

So,

m = [tex]\frac{1014.02lbf}{32.174\frac{ft lbm}{lbf s^{2}}}[/tex] = 31.52 lbm

However, this is not the answer given in the back of the book, which is 1000.7 lbm. Can someone tell me where I'm going wrong? I'm assuming it's with my application of the grav. constant, since I still dont really understand it's use in the English system. TIA.

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