1. Jul 28, 2011

### tenbee

1. The problem statement, all variables and given/known data

Please help me understand when I should use $\rho$gh vs. $\rho$gy.

2. Relevant equations

P = $\rho$gy
P =$\rho$gh

3. The attempt at a solution
(In a uniformly dense solution)

$\rho$gh is essentially the gravitational potential energy at a specified point in a fluid column. So "h" is taken from the bottom of the fluid to that specified point. It seems like this version of pressure isn't really pressure at all because pressure is dependent on the density (and thus mass) of a fluid.

$\rho$gy is an actual measure of pressure because it takes into account the mass (and thus the density) of all the fluid above some arbitrary point in the fluid.

Is my understanding correct?
...BTW I don't know calculus...

2. Jul 28, 2011

### Pengwuino

What are 'y' and 'h' that supposedly make them different?

Last edited: Jul 28, 2011
3. Jul 28, 2011

### tenbee

"you definitely can't mix these up - one is gravitational potential energy, and the other is the potential energy due to pressure, which is a result of the weight of fluid, air, etc. pushing down from above.

imagine that you were a water molecule in a bucket, diffusing around. at the top of the bucket, you would have more gravitational potential energy (ρgh), and at the bottom of the bucket, you would have more potential energy from pressure (ρgy)

in bernoulli's equation:
pressure + ½ρv² + ρgh = constant, the point is that energy can be converted between pressure, kinetic, and gravitational potential, but stays the same everywhere.

ρgh by itself cannot be used to calculate pressure, since this is a measure of gravitational potential energy, as compared to some arbitrary reference point. For pressure, you need to know the weight of all of the matter above the point pushing down (think about this: what would the pressure be at the 1 ft dept of a swimming pool when it is filled to 10 ft vs when it is filled to 2 ft - it would be much higher, even though h at that point is the same)

remember that density = mass / volume (ρ = m / V)

ρgh is more-or-less the same thing as potential energy = mgh (that you use for problems on land), and as you should know in those problems, h could be the distance to the floor in the room, the ground outside, sea level, etc. h = 0 only because that it the height defined in the problem as being the bottom point."