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?morty92222 said: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!
I asked for the density in slugs/ft^3.morty92222 said:So 10 ppg equals 74.8 lb/ft3. What should I use for g in the second equation (in terms of units)?
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.Chestermiller said:I asked for the density in slugs/ft^3.
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##?morty92222 said: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.
Chestermiller said: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##?
You just use rho g h to get the pressure. Slugs IS mass.morty92222 said:I am not sure how to use the slug in P=Rho g h as there is no mass in there!
Chestermiller said:You just use rho g h to get the pressure. Slugs IS mass.
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?morty92222 said:So 2.33 * 32.2 *10 = 750 [slug/(ft. sec2)]
What kind of units is this?
Chestermiller said: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?
Don't despair. You just got to 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}##?morty92222 said: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.
The formula for calculating hydrostatic pressure is P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.
The density of the fluid can be determined by using a hydrometer or by looking up the density of the fluid in a table. It is typically measured in units of kilograms per cubic meter (kg/m³).
Yes, the hydrostatic pressure equation can be used for any type of fluid, as long as the density of the fluid and the acceleration due to gravity are known.
The unit of measurement for hydrostatic pressure is typically Pascal (Pa), which is equivalent to Newtons per square meter (N/m²). However, other units such as bar or psi may also be used.
The hydrostatic pressure is directly proportional to the height of the fluid column. This means that as the height increases, the pressure also increases. For example, doubling the height of the fluid column will result in double the hydrostatic pressure.