Calculating Absolute and Gauge Pressure in a Gasoline-Filled Container

  • Thread starter Thread starter JWHooper
  • Start date Start date
  • Tags Tags
    Pressure
AI Thread Summary
To calculate the absolute and gauge pressure at the bottom of a 2.5 m deep gasoline-filled container, the gauge pressure is determined to be 16.9 kPa, and the absolute pressure is 118 kPa. The formula used is P = P_0 + Dhg, where P_0 represents atmospheric pressure. The discussion confirms that absolute pressure equals atmospheric pressure plus gauge pressure. Clarifications about the nature of gauge pressure and its potential to be negative when submerged are also noted.
JWHooper

Homework Statement


Calculate the absolute and gauge pressure at the bottom of a 2.5 m deep container, if it is full of gasoline (density = 690 kg/m^3). Assume the container is open at the sea level.


Homework Equations


P = P_0 + Dhg



The Attempt at a Solution


P = (690 kg/m^3)(2.5m)(9.8 m/s^2) = (1.69 x 10^4 Pa)(1 kPa/10^3Pa) = 1.69 x 10 kPa (gauge pressure)
(1.69 x 10^4 Pa) + (1.01 x 10^5 Pa) = 1.18 x 10^5 Pa (absolute pressure)

My questions (important!):
1. Did I do these right?
2. Also, what is P_0 in the equation P = P_0 + Dhg?
3. gauge pressure + 14.7 lb/in^2 = absolute pressure ----> true or false? If not, please explain.


Thanks,

J.
 
Physics news on Phys.org
2. Also, what is P_0 in the equation P = P_0 + Dhg?
I think you've answered your own question with the formula

P = P_0 + D.h.g

absolute pressure = atm. pressure + gauge pressure.

Interesting to speculate whether, if it was all underwater, the gauge pressure would be negative ?
 
Mentz114 said:
I think you've answered your own question with the formula

P = P_0 + D.h.g

absolute pressure = atm. pressure + gauge pressure.

Interesting to speculate whether, if it was all underwater, the gauge pressure would be negative ?

So, P is the absolute pressure, P_0 is the atm. pressure, and Dhg is the gauge pressure?
 
Okay. Now, disregarding P_0, did I do the solution correctly?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top