Which equation to use for hydrostatic pressure?

  • #1

Homework Statement


h = 10 ft
RHO = 10 ppg (or pounds/gallon)

Homework Equations


P = 0.052 * RHO * h
P = RHO * G * h

I am very confused right now as why each equation gives a different answer! What are the correct units for the 2nd equation and how do I convert from second equation to the first one? And where does the g goes?! I looked it up online and but they just say "RHO" takes care of it! What does that mean?!

Thank you!
 

Answers and Replies

  • #2
20,854
4,544

Homework Statement


h = 10 ft
RHO = 10 ppg (or pounds/gallon)

Homework Equations


P = 0.052 * RHO * h
P = RHO * G * h

I am very confused right now as why each equation gives a different answer! What are the correct units for the 2nd equation and how do I convert from second equation to the first one? And where does the g goes?! I looked it up online and but they just say "RHO" takes care of it! What does that mean?!

Thank you!
Let's see your attempt at doing some of the units conversions, starting with the 2nd equation. If RHO is given as 10 ppg, what is density in ##lb_m/ft^3##? What is it in slugs/ft^3?
 
  • #3
So 10 ppg equals 74.8 lb/ft3. What should I use for g in the second equation (in terms of units)?
 
Last edited:
  • #5
I asked for the density in slugs/ft^3.
I never heard about "slug" in my entire life. After googling it, it appears to be 2.33 slug/ft3. But I don't have any feel for the unit or what it represents.
 
  • #6
20,854
4,544
I never heard about "slug" in my entire life. After googling it, it appears to be 2.33 slug/ft3. But I don't have any feel for the unit or what it represents.
A slug is what you use if you are using English units, and you want to write F = ma (without a correction factor), with F in pounds force, mass in slugs, and acceleration in ft/sec^2. So, with g = 32.2 ft/sec^2, what do you get for the pressure using ##P=\rho g h##?
 
  • #7
A slug is what you use if you are using English units, and you want to write F = ma (without a correction factor), with F in pounds force, mass in slugs, and acceleration in ft/sec^2. So, with g = 32.2 ft/sec^2, what do you get for the pressure using ##P=\rho g h##?
I am not sure how to use the slug in P=Rho g h as there is no mass in there!
 
  • #9
You just use rho g h to get the pressure. Slugs IS mass.
So 2.33 * 32.2 *10 = 750 [slug/(ft. sec2)]
What kind of units is this?
 
  • #10
20,854
4,544
So 2.33 * 32.2 *10 = 750 [slug/(ft. sec2)]
What kind of units is this?
It is the same as slug-ft/(ft^2 sec^2). 1 slug-ft/(sec^2) = 1 lb_f. So the unit are lb-f/ft^2=psf. What units did you want the answer in?
 
  • #11
It is the same as slug-ft/(ft^2 sec^2). 1 slug-ft/(sec^2) = 1 lb_f. So the unit are lb-f/ft^2=psf. What units did you want the answer in?
I actually wanted the answer in psi so I can compare it to P = 0.052 * ppg * h = 0.052 * 10 * 10 = 5.2 psi

Edit: So 750/144 = 5.2 psi! Thank you so much.

I still do not understand how we reached this conclusion! :D You kinda took me through it that I don't even understand what was wrong in the first place.
 
  • #12
20,854
4,544
I actually wanted the answer in psi so I can compare it to P = 0.052 * ppg * h = 0.052 * 10 * 10 = 5.2 psi

Edit: So 750/144 = 5.2 psi! Thank you so much.

I still do not understand how we reached this conclusion! :D You kinda took me through it that I don't even understand what was wrong in the first place.
Don't despair. You just gotta get used to manipulating the units. Are you familiar with the use of ##g_c=32.2\ \frac{lb_m\ ft}{lb_f\ sec^2}##?
 

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