Pressure to Force to Mass with English units

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SUMMARY

The discussion revolves around converting pressure measurements to mass using English units, specifically addressing a problem involving a pressure of 3000 atm and a piston diameter of 0.17 inches. The user calculated the force using the equations P=F/A and F=PA, resulting in a force of 1014.02 lbf. However, the user encountered a discrepancy when converting this force to mass, arriving at 31.52 lbm instead of the expected 1000.7 lbm. The confusion stemmed from the misunderstanding of the gravitational constant in the English system and the distinction between pound-force (lbf) and pound-mass (lbm).

PREREQUISITES
  • Understanding of pressure calculations using P=F/A
  • Familiarity with unit conversions, particularly between atm and psi
  • Knowledge of the relationship between force, mass, and gravity in the English system
  • Basic grasp of piston mechanics and cross-sectional area calculations
NEXT STEPS
  • Study the concept of gravitational constants in the English system, particularly 32.174 ft/s²
  • Learn about the differences between pound-force (lbf) and pound-mass (lbm)
  • Practice unit conversions between different pressure units, such as atm and psi
  • Explore the implications of rounding errors in engineering calculations
USEFUL FOR

Students in engineering or physics, particularly those dealing with fluid mechanics and pressure calculations, as well as professionals needing to convert between force and mass in the English unit system.

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 = (\frac{\pi}{4})(0.17 in^{2}) = 0.023 in^{2}

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

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

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 = \frac{1014.02lbf}{32.174\frac{ft lbm}{lbf s^{2}}} = 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 don't really understand it's use in the English system. TIA.
 
Last edited:
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You converted your answer to slugs, not lbm. Personally, I loathe the idea of lbm. In practical use - although not entirely correct - lbf is equivalent to lbm. So, your original numerical answer of 1014 is correct - just different due to rounding in between steps.
 
slugs!
 

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