Pressure in a Tank: Questions & Answers

  • Thread starter Thread starter rachit mehra
  • Start date Start date
  • Tags Tags
    Pressure Tank
AI Thread Summary
The discussion centers on calculating the pressure in a tank on an oil tanker during filling. When oil is filled into a 20-meter deep tank with a density of 1.0, the pressure at the bottom valve is influenced by both the liquid column and the inert gas blanket pressure of 400 mm water gauge. At a pressure gauge reading of 1 Bar, the pressure at the bottom line can be determined using the formula for hydrostatic pressure, P_liq = h * d * g. If the tank is filled to a height of 15 meters, the pressure will increase accordingly, factoring in the gas pressure above the liquid. The conversation emphasizes the relationship between gas pressure and liquid height, noting that as the tank fills, the gas pressure will also rise if the gas cannot escape.
rachit mehra
Messages
1
Reaction score
0
Hi all, this will seem a very easy problem for you guys but its really a bit conufusing for me..
I work on an oil tanker and the problem is as follows.
What will be the pressure on the filling line of a tank which is on deck when oil is being filled into the tank. For sake op simplicity let's take the density as 1.0. The tank is 20 metres deep and the valves are at bottom.( The entry point is on the top, and the oil flows down to the bottom of the tank and enters the tank at the bottom) . The tank has an inert gas blanket wih a pressure of about 400 mm water gauge.
Now my questions are-
1) What will be the pressure on the bottom line at the valve when a pressure gauge shows a pressure of 1 Bar.
2) If the tank is filled up to a height of 15 metres what will be the pressure then .

Thanks
 
Physics news on Phys.org
I suppose there will be a gas with oil too. Is it? Let's choose two points in tank. One is in liquid(A) and other is not(B).

Total pressures,

PA = Pgas + P liq
PB = Pgas

P liq = h d g

Also, if gas can not get out, even filling the tank, gas pressure will grow. If this is an ideal gas then,

PV=nRT

and n, R, T is not changing when filling. Thus,

PV= const.

Say, the gas has P pressure in tank and volume is V then, half of tank filled with liquid. Now,

PV = Pf*V/2 (Pf: Final pressure)
Pf=2P

hope this helps. (If i could understand properly)
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top