# Difference of pressure of fluids in containers.

1. Aug 1, 2004

### izytang

Hello everyone,

I have a question that pertains to the physics of fluids and how the pressure of which changes in different given containers, which contains water. Basically, here is what I’m trying to do:

The shape of the container is that defined by a the function:
$$f(x) = \frac{5}{8} x$$
Which is rotated about the x-axis from the limits 1 to 8, with a circular hole in the smaller part of the cone. So, essentially it is a cut-off cone with a height of 7 and large radius of 5 (from top). The cone container is flipped so that the cone is “upside-down” (or that the larger radius is upward), and water will then flow out at an unknown rate. The following link gives a diagram I drew out to better illustrate what is happening:

http://www.angelfire.com/fang/ca2/waterTankEx.jpg
(Copy and paste in new browser window, since angelfire doesn’t allow external linking)

The essential problem is that I need to find a function in time of the water’s rate of the change of the container’s volume with pressure taken into account, so that I could simply find it’s instantaneous change of the Volume, height, or radius. My thought process has gone as follows:

First, I needed to write a formula for the volume of the cut-off cone. And since the cone is defined by a function, I used calculus to get:
$$\int_{1}^{x} pi \left \frac{5}{8} x \right ^2$$
or
$$\int_{1}^{x} pi \frac{25}{64} x^2 dx$$
With $x$ being the cone’s height from “0” (although the cone does not start until x=1).

My next step was to find out the container’s rate of flow, or pressure by calculating it initially at the point of which the tank is full, then writing a formula representing the rate of flow in terms of time, which is what I am ultimately looking for. I started by using the Pressure Force relation:
$$P = \frac{F}{A}$$
or
$$P = \frac{ma}{A}$$
Where F is force which equals mass*acceleration, and A is the area, in meters squared. I had then calculated the volume of the container when it is full (purely as a test), and got 209.03 meters cubed, which equals 209,030 Kg in mass. And because we are dealing with water, the volume density relation is 1cm cubed = 1 gram.

I was just about to calculate the pressure, but it suddenly hit me that the shape of the container might possibly affect the pressure of the fluid. In others words, there might be more pressure from fluids in a container that is a cylinder as opposed to a cone, even if the volume of the fluids are the same. If there is such a relation, it is very important to take that into account for calculating a rate of flow from the container, but unfortunately, I do not know where to start to find this relation, if any exists at all. Essentially, all I need to know is how pressures of fluids are affected by the shape of the container of which those fluids are in. But, if anyone would like to double check all my previous work and concepts, I’d gladly appreciate it. Of course, any help is appreciated.

-Josh

2. Aug 2, 2004

### ArmoSkater87

Hi, i cant help much because I know NO calculus, i have yet to take it when school starts. But i can tell you that im pretty sure the shape doesnt make a difference in the pressure, its only the height that matters. Let me show you why...

$$P = F/A$$
$$P = ma/A$$
$$P = DVg/A$$ D is the density of the fluid, in your case its water. V is the volume of fluid you have, D times V gives you mass as you can see. You replace a with g.
$$P = D(Ah)g/A$$ The volume can be expresed as the area times height
$$P = Dgh$$
So, height is the dominant factor in pressure, not the volume of fluid...which would be determined by the shape of what you are using.

3. Aug 2, 2004

### Staff: Mentor

Pressure of a liquid inside a container depends only on the density of the liquid (usually assumed to be constant) and the height of the fluid. Thus it is a linear relationship between height (depth) and pressure.