Buoyancy of balloon submerged

[stosw]

Homework Statement

Calculate the balloon lift if it were submerged in water.

Volume of balloon (sphere) = $$\frac{(5^3)*(pi)}{6}$$ = 125pi/6

The balloon is filled with hellium.
Pfluid (water) = 64.2 lb/ft^3

Homework Equations

Lift = Volume * gravity * (Pfluid - Pbody)

$$\frac{Pfluid}{Pbody}$$ = $$\frac{Mfluid}{Mbody}$$

The Mfluid is molecular weight of Water.

The Attempt at a Solution

Using:
$$\frac{Pfluid}{Pbody}$$ = $$\frac{Mfluid}{Mbody}$$

Pbody = 14.3 lb/ft^3

Lift = Volume * gravity * (Pfluid - Pbody)
Lift = (125pi/6)ft^3 * (32.71) ft/s^2 * (62.4 lb/ft^3 - 14.3 lb/ft^3)
Lift = 102,925 ft/s^2

That number seems way too large to me. Could someone shed some light to what it is I did incorrectly?

 

[AvroArrow]

A:I'm not sure about the units you are using.But the formula for lift is$$ L = V \times \Delta P $$where $V$ is the volume of the balloon and $\Delta P$ is the difference between the pressure inside the balloon and the pressure outside.For a spherical balloon, the volume is$$ V = \frac{4}{3} \pi r^3 $$where $r$ is the radius of the balloon.Then the pressure inside the balloon is$$ P_b = P_{He} = 0.179 \text{ bar}, $$where $P_{He}$ is the pressure of helium gas at 0°C and 1 bar.The pressure outside is the atmospheric pressure plus the hydrostatic pressure due to the water depth. At sea level, the atmospheric pressure is 1 bar, so the pressure outside the balloon is$$ P_w = 1 + \rho gh $$where $\rho$ is the density of the water, $g$ is the gravitational acceleration, and $h$ is the depth of the balloon.Assuming a density of water of $\rho = 1000 \text{ kg/m}^3$, a gravitational acceleration of $g = 9.8 \text{ m/s}^2$, and a depth of $h = 5 \text{ m}$, the pressure outside the balloon is$$ P_w = 1 + 1000 \times 9.8 \times 5 = 59.1 \text{ bar}. $$Finally, the lift is$$ L = \frac{4}{3} \pi r^3 \times (59.1 - 0.179) \approx 860 \text{ N}. $$

 

[tomcue]

Your calculation for lift is incorrect. The correct equation for lift is:

Lift = Volume * (Pfluid - Pbody)

Also, the units for lift should be in pounds, not ft/s^2. So the correct calculation would be:

Lift = (125pi/6)ft^3 * (64.2 lb/ft^3 - 14.3 lb/ft^3)
Lift = 7,837.5 lb

This means that the balloon would have a lift of 7,837.5 pounds if it were submerged in water. This may seem like a large number, but it is important to remember that the buoyant force (or lift) of an object is equal to the weight of the fluid it displaces. In this case, the balloon is displacing a large volume of water, which results in a significant amount of lift.