B Archimedes' Principle for gases - derivation?

[Amaterasu21]

Hi all,

I understand where Archimedes' Principle comes from in liquids:

If we imagine a cylinder immersed in a liquid of density ρ whose cross-sectional area is A and whose top is at depth h₁ and whose bottom is at depth h₂:

Force_{(top of cylinder)} F_T = ρgh₁A
Force_{(bottom of cylinder)} F_B = ρgh₂A
Buoyant force = ρgA(h₂-h₁)
= ρA(h₂-h₁)g = ρV_cylinderg = m_{displaced liquid}g = weight of displaced liquid
therefore buoyant force = weight of displaced liquid.

I also understand that this works even if the object immersed in liquid isn't a cylinder because pressure at any depth is constant, so the horizontal components of pressure on any surface of the object cancel out, and the vertical components produce the same buoyant force as the cylinder.

But what I don't understand is why this is true for gases as well. I've been told that the buoyant force in gases is equal to the weight of displaced gas. However, gases unlike liquids are compressible and a large mass of gas in a gravitational field (e.g. the atmosphere) will be far denser at the bottom than at the top. Therefore while ρ is constant during the derivation for liquids, it would certainly NOT be constant across the height of the cylinder in a gas!

Force_{(top of cylinder)} F_T = ρ₁gh₁A
Force_{(bottom of cylinder)} F_B = ρ₂gh₂A
Buoyant force = gA(ρ₂h₂ - ρ₁h₁)
...and I don't see how we get from here to "= weight of displaced gas!"

Any help would be appreciated!

 

[Orodruin]

Force(top of cylinder) FT = ρ1gh1A
Force(bottom of cylinder) FB = ρ2gh2A

This is not correct, the pressure a priori not dependent on the density in the way you describe. What is true is that the pressure differential is given by $dp = \rho g \,dh$. In the case where $\rho$ is constant, this implies that $p = p_0 + \rho g h$, where $h$ is the vertical distance to some reference level where the pressure is $p_0$.

The pressure force acting on your cylinder is given by the pressure difference between the top and bottom. This pressure difference is given by
$$
\Delta p = \int_{h_1}^{h_2} dp = \int_{h_1}^{h_2} \rho g \, dh = g \int_{h_1}^{h_2} \rho\, dh.
$$
If you have a cylinder with a cross-sectional area $A$, this implies
$$
\Delta p = g \int_{h_1} ^{h_2} A \rho \, dh = g M,
$$
where $M$ is the mass of the displaced fluid (since it is the volume integral of the corresponding density).

 

[Amaterasu21]

Thank you, this explains it! I should have realized we'd need to integrate since we're dealing with a constantly changing density with height.

This is not correct, the pressure a priori not dependent on the density in the way you describe. What is true is that the pressure differential is given by $dp = \rho g \,dh$. In the case where $\rho$ is constant, this implies that $p = p_0 + \rho g h$, where $h$ is the vertical distance to some reference level where the pressure is $p_0$.

The pressure force acting on your cylinder is given by the pressure difference between the top and bottom. This pressure difference is given by
$$
\Delta p = \int_{h_1}^{h_2} dp = \int_{h_1}^{h_2} \rho g \, dh = g \int_{h_1}^{h_2} \rho\, dh.
$$
If you have a cylinder with a cross-sectional area $A$, this implies
$$
\Delta p = g \int_{h_1} ^{h_2} A \rho \, dh = g M,
$$
where $M$ is the mass of the displaced fluid (since it is the volume integral of the corresponding density).

 

[sophiecentaur]

I've been told

Were you given any reference for this information? I appreciate that you may have reservations because it looks 'too simple'.
If you stratify your volume into sufficiently shallow layers to consider the density to be constant over the thickness and equal to the mean value then the buoyant force for that slice would in fact be the weight of the slice displaced. IMO, that should be sufficient.

 

[Orodruin]

If you stratify your volume into sufficiently shallow layers to consider the density to be constant over the thickness and equal to the mean value then the buoyant force for that slice would in fact be the weight of the slice displaced. IMO, that should be sufficient.

It should be pointed out that, in essence, this is nothing but the typical Riemann sum, which in the end results in the integral outlined in #2.