Balloon lifting force
Online calculator for solving problems related to balloon lifting force. It allows you to find the mass of the balloon shell, or the mass of the payload, or the required hot air temperature, or the volume of the balloon, from other known values.
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Let's say right away that we are talking about balloons where hot air is used as the lifting gas, i.e. the gas that creates buoyancy. If helium is used as the lifting gas, the calculation changes slightly, see Lifting force of a helium-filled balloon.
There are a set of balloon parameters, such as the volume of the balloon, the mass of its shell, the mass of the payload - the basket and the balloonists, and the temperature of the air in the balloon. In addition, the lifting force also depends on the environmental conditions - the temperature and density of the surrounding air. Problems related to the balloon are referred to problems on molecular-kinetic theory. For example, the problem may ask to what minimum temperature the air in the balloon should be heated to so that the balloon takes off with the cargo, given the known mass of the shell, cargo, volume of the balloon, temperature and density of the surrounding air. Or they can ask what load the balloon can lift, given a known volume, mass of the shell, temperature of the air in the balloon, temperature and density of the surrounding air. The basic equations are the same, only the unknown quantity changes. The calculator below allows you to solve such problems.
It is worth noting that the density of the surrounding air is almost always specified in the problems. If you are not solving problems on molecular kinetic theory, but, for example, you want to calculate the parameters of a balloon for aircraft modeling, you may not know the density of the air. Then you can use a calculator to get its value Density of air as a function of pressure and temperature.
As usual, the theory and calculation formulas are listed below the calculator.
Calculation of balloon parameters
A balloon is lifted by a buoyant force, or Archimedes' force, because Archimedes' law applies not only to liquids but also to gases. Accordingly, for the balloon to begin to rise, the buoyant force acting on the balloon must exceed the force of gravity. To solve problems, we usually find a boundary condition (minimum temperature, minimum volume, etc.), which allows us to equate the force of gravity and the buoyant force. That is, a little more - and the ball starts to rise. The equation of equilibrium looks like this:
,
where
M is the mass of the shell,
m is the mass of the load,
mₐ is the mass of heated air in the balloon,
mₑ is the mass of the surrounding air displaced by the balloon,
g is the acceleration of free fall.
Let's replace the air mass by the product of the air density by the volume of the balloon
Reducing g and transferring the mass of heated air to the right-hand side, we get
Let's deal with the density of a heated gas. When air is heated in a balloon, pressure and volume do not change - we have the same inextensible shell with pressure equal to atmospheric pressure. At the initial moment, the temperature of the air inside the shell is equal to the temperature of the surrounding air. We can use the Mendeleev-Clapeyron equation:
In order to compensate for the increase in temperature with heating, the mass of air in the balloon must decrease, and hence the density. Substituting the mass, we get
After reduction and transposition, we get the following expression for the density of heated air (we just learned that the density of a gas is inversely proportional to its temperature):
.
By the way, the air temperature inside the shell can reach 100-120°C.
Substituting ρₐ into the equality above, we get the final formula that relates all the parameters of the balloon and the environment:
From this equality we can obtain formulas to calculate the desired unknown.
The mass of the shell:
The mass of the load:
Volume of the ball:
The temperature of the heated air:
It is worth noting that since the shell cannot be heated above a certain temperature, the warmer the surrounding air, the smaller the density difference, and the worse the balloon's lifting force.
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