Pressure to Force to Mass with English units

  • #1

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.
 
Last edited:

Answers and Replies

  • #2
136
0
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.
 
  • #3
tiny-tim
Science Advisor
Homework Helper
25,832
251
slugs! :yuck:
 

Related Threads on Pressure to Force to Mass with English units

  • Last Post
Replies
5
Views
13K
Replies
9
Views
13K
Replies
0
Views
2K
Replies
1
Views
10K
  • Last Post
Replies
2
Views
490
  • Last Post
Replies
4
Views
9K
Replies
5
Views
2K
  • Last Post
Replies
7
Views
20K
  • Last Post
Replies
2
Views
613
  • Last Post
Replies
1
Views
8K
Top