I want to calculate the total pressure exerted by a liquid of density ρ on the walls of a container, say a cylinder, for convenience.

$$P = hρg$$

Therefore integrating from $0$ to $h\; ,$ we get :

$$P = ∫ hρg \\ \implies P = ( h²ρg/2 )$$

Is this way correct?

  • $\begingroup$ However this is dimensionally incorrect and doubtful. Please suggest a suitable way. $\endgroup$ Dec 3 '15 at 13:09

It doesn't really make sense to calculate the total pressure in this manner, because the pressure is the force per unit area, and is different at different heights in the tank as you suggest. You can however calculate the total force on the container from the pressure by integrating the pressure over the height of the container as follows:

$$F = \int PdA$$

In this case, the area element $dA$ will be equal to a differential height element $dh$ times the circumference of the cylinder $2\pi r$, which gives

$$F = 2\pi r\rho g\int_0^h h dh = \pi r \rho g h^2$$

You can then calculate the average pressure by dividing the force by the total area of the cylinder as

$$P_{avg} = \frac{F}{A} = \frac{\pi r \rho g h^2}{2\pi r h} = \frac{\rho g h}{2}$$

This is the long way of doing it just to show you the thought process. You can note that since the pressure is linear in height, you get the same result for the average pressure by taking the pressure to be at the mid-point of the tank.

  • 1
    $\begingroup$ Just a note - it's not clear to me that the "total force" you calculated has any physical significance. That isn't the net force, since you didn't integrate the differential vector forces. $\endgroup$
    – Brionius
    Dec 3 '15 at 13:42
  • $\begingroup$ @Brionius I was using some hydrostatic simplifications which state that the horizontal component of the force on a sample area is the pressure at the center of the area times the area, and the vertical component is related to the volume of fluid above the projected area, which is zero in this case for a cylinder. However, I'm open to alternative approaches. $\endgroup$
    – tmwilson26
    Dec 3 '15 at 13:56
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    $\begingroup$ I think your hydrostatic approximations are fine - that's not what I meant. I mean the force on a differential area due to the pressure is a vector, so the net force would really be $\vec{F_{net}} = \int P d\vec{A}$, which is quite different from $ F= \int P dA$. I can't think of what $F = \int P |d\vec{A}| = \int P dA$ would represent. $\endgroup$
    – Brionius
    Dec 3 '15 at 14:02
  • $\begingroup$ @Brionius I get what you're saying now. It would just be the sum of the forces projected radially outward on the tank, which could be important for determining how well the tank can hold up since it would introduce a certain amount of strain to the tank. However, I'm sure that rating this by the pressure in the tank is a better way to go anyway, so it's something thats going to vary with the height anyway. $\endgroup$
    – tmwilson26
    Dec 3 '15 at 14:10
  • $\begingroup$ I agree with @Brionius. If the directionality of the pressure force is taken into account, then the net force on the wall of the vertical cylinder will be zero. If one is concerned about the stresses in the wall of the tank, one should be looking at the hoop stress. $\endgroup$ Dec 28 '15 at 15:01

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