Calculating the He Needed for Balloon Lift: An Analysis with Example Calculations

In this article, we delve into the fascinating world of balloon physics and explore how to calculate the amount of helium (He) required for a balloon to lift a specific mass. We will compare helium with hydrogen, discuss the buoyant force, and provide detailed calculations and real-world examples.

The Basics of Balloon Physics

When a balloon is filled with a lighter-than-air gas like helium, it experiences a buoyant force that can lift it and its payload. This phenomenon is explained by Archimedes' principle, which states that the buoyant force on a submerged object is equal to the weight of the fluid displaced by the object. In the case of a balloon, the fluid is the surrounding air.

Example Calculations with Helium

Let's consider a scenario where we need to lift a specific mass using helium. For demonstration purposes, we take an example of a pound of helium, which has a known volume in air. We will compare this with hydrogen and then move to a more mathematical model for a more accurate calculation.

Comparison of Helium and Hydrogen

Helium (He) is a noble gas with a specific gravity of about 0.1785, which means it is less dense than air (which has a density of about 1.293 kg/m3). Conversely, hydrogen (H?) has a specific gravity of only 0.0708, making it even lighter. The density of air (Da) at sea level is approximately 1.29 kg/m3, while the density of helium (Dh) is approximately 0.1785 kg/m3.

Example Calculation: 20 Pound Helium Balloon

If we fill a balloon with 20 pounds of helium, the balloon and the payload it lifts will be significantly heavier. To understand this, let's break it down step by step:

Pure helium occupies about 7 pounds 8 ounces of air per pound of helium. For 20 pounds of helium, the displaced air volume is 20 pounds x 7.5 pounds of air/pound of helium. The total weight lifted is 130 pounds (20 pounds of helium 110 pounds of displaced air). Similarly, if hydrogen is used, the same volume can lift 140 pounds (10 pounds of hydrogen 130 pounds of displaced air).

Mathematical Model for Balloon Lift

To perform a more accurate calculation, we can use a mathematical model based on the principles of buoyancy and gravity. The buoyant force on the balloon is equal to the weight of the displaced air, while the gravitational force is the weight of the balloon, payload, and the gas used for inflation.

Key Assumptions

Here are the key assumptions for our model:

The mass 'M' includes the mass of the uninflated balloon and payload, but not the gas used to inflate it. The volume of the uninflated balloon and payload is negligible compared to the volume of the gas. The density of air (Da) is approximately 1.29 kg/m3, and the density of helium (Dh) is approximately 0.18 kg/m3. Gravity 'g' is 9.8 m/s2.

Formulas and Calculations

Let's define:

Dh the density of helium (approximately 0.1785 kg/m3) Da the density of air (approximately 1.29 kg/m3)

The upward buoyant force (Fbuoyant) is given by:

Fbuoyant Da * V * g, where V is the volume of the inflated balloon.

The downward gravitational force (Fgravity) is given by:

Fgravity (M Dh * V) * g, where M is the mass of the uninflated balloon and payload.

Setting the buoyant force equal to the gravitational force gives us the volume needed to lift a specific mass:

Da * V * g (M Dh * V) * g

Solving for V:

V M / (Da - Dh)

Example of a Simple Model

For a very approximate and initial sizing, we can use the rule of thumb that a balloon volume of 1 m3 is needed to lift approximately 1 kilogram of payload at sea level. This is a practical and easy way to get an initial estimate without going into complex calculations.

Conclusion

In conclusion, the amount of helium needed to lift a specific mass can be calculated using a simple mathematical model. Helium and hydrogen both offer the opportunity to lift payloads, but helium is safer and more reliable due to its flammability issues with hydrogen.

If you are interested in more detailed calculations or specific examples, the model and principles discussed here can be extended with further precision.