# Pressure at a certain depth when density varies

• pressurised
In summary, the conversation discusses finding the pressure at a certain depth using the equation dp/dz = -ρg and p = ρgh. The speaker attempted to solve the problem using ρ = ρ0 + kh and obtained a value of 1060kgm-3 for density at depth 3m. However, when using this value in the equation ρgh, they were off by approximately 1000. The other speaker suggests using the integral ∫-(ρ0+kh)g dz to sum the pressure contributions from thin layers stacked one on top of the next from the surface down to the desired depth. They also remind the first speaker to pay attention to the direction of the integration path.
pressurised

dp/dz=-ρg
p=ρgh

## The Attempt at a Solution

I've found the density at depth 3m using ρ=ρ0+kh, which gave me 1060kgm-3. I then put this value into ρgh to get 31195.8Pa which seems to be ≈+1000 off the answer.

What is the correct mathematical way of solving this as I am not quite sure how to form the equation for variation of pressure using that.

Think of summing the pressure contributions from thin layers stacked one on top of the next from the surface down to the desired depth. Does that remind you of anything?

gneill said:
Think of summing the pressure contributions from thin layers stacked one on top of the next from the surface down to the desired depth. Does that remind you of anything?

I can't quite think of the name of it but I understand what you mean.

I would have thought ∫-(ρ0+kh)g dz would have been okay to use?

pressurised said:
I would have thought ∫-(ρ0+kh)g dz would have been okay to use?
That's the idea. You'll need to relate the h to your integration variable z. You should pay attention to the "direction" that your integration path takes as it affects the sign of the dz differential element and hence whether that leading minus sign is warranted.

Chestermiller
gneill said:
That's the idea. You'll need to relate the h to your integration variable z. You should pay attention to the "direction" that your integration path takes as it affects the sign of the dz differential element and hence whether that leading minus sign is warranted.

Thank you!

## What is pressure at a certain depth?

Pressure at a certain depth refers to the amount of force exerted by a fluid on an object at a specific depth in the fluid. This pressure is caused by the weight of the fluid above the object and is influenced by factors such as density and gravity.

## How does density affect pressure at a certain depth?

Density plays a significant role in determining the pressure at a certain depth. As density increases, the weight of the fluid above an object also increases, resulting in a higher pressure. This is why objects tend to sink in denser fluids, such as water.

## What is the formula for calculating pressure at a certain depth?

The formula for calculating pressure at a certain depth is P = ρgh, where P is pressure, ρ is density, g is the acceleration due to gravity, and h is the depth. This formula is known as the hydrostatic equation.

## How does pressure change with depth in a fluid with varying density?

In a fluid with varying density, pressure changes with depth according to the hydrostatic equation. This means that as depth increases, pressure also increases due to the weight of the fluid above. However, if the density of the fluid changes, the rate at which pressure increases with depth may also change.

## What other factors besides density can affect pressure at a certain depth?

Besides density, other factors that can affect pressure at a certain depth include the gravitational force acting on the fluid, the depth of the object, and the type of fluid. The shape and size of the object can also play a role in the pressure experienced at a certain depth.

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