Hot Air Balloon Energy Estimate Comparison

[Eigenentity]

I'm playing around with the physics of hot air balloons, and I've come to two different estimates of the energy required to get a hot air balloon off the ground. The two estimates make wildly different predictions and I'm hoping to get some insight on which one is wrong.

Okay, so first from thermodynamics. Let's assume we're going to heat our balloon (volume $V$, basket mass $m$) up to some temperature $T_{\mathrm{in}}$. The ambient temperature is $T_0$, and the condition for liftoff is:

$$V(\rho_0 - \rho_{\mathrm{in}}) \ge m$$

Some faffing around with the ideal gas law results in:

$$V \ge \frac{m \bar{R}}{p_0 (\frac{1}{T_0} - \frac{1}{T_{\mathrm{in}}})}$$

where $\bar{R}$ is the specific gas constant ($R / M_{r,\mathrm{air}}$).

So that gives us the volume of balloon we need given a specified temperature difference between the inside and outside of the balloon. Using this expression we find that we can lift off a basket of 300 kg with a 2500 m³ balloon heated to 60 deg C. That sounds reasonable.

Ok, so now, making the (extreme and unrealistic) assumption that we can do the heating adiabatically, and ignoring the gas lost out of the balloon on heating, we can guess the heat required from the burner:

$$\mathrm{d}Q = C \mathrm{d}T = \frac{7}{2}\frac{pV}{T} \mathrm{d}T \Rightarrow \Delta Q = \frac{7}{2} p V \ln{\frac{T_{\mathrm{in}}}{T_0}}$$

This gives numbers of order 100 MJ.

Another way of guessing is to say that the "buoyancy energy" must balance the gravitational potential at the surface of the earth, or:

$$\Delta Q \ge m g r_{\mathrm{Earth}}$$

This gives an estimate of order 20 GJ.

Which of these estimates is wrong? I think it must be the latter, not least because that sounds like an enormous amount of energy, but I can't work why it doesn't work.

 

[omoplata]

I think the second estimate would have been right if the burner somehow pushed the balloon from the center of the Earth up to $r_{Earth}$. But no mechanical motion happened here. The energy $\Delta Q$ was used only to heat the gas in the balloon from $T_{0}$ to $T_{in}$.

 

[Eigenentity]

I'm not sure it's as simple as that. Sure, the work done by the burner goes into heat, but that heat is clearly doing mechanical work. Think of what happens if I run the burner a little longer: I lift the entire balloon and its payload off the ground. I'm definitely doing real work.

One way of thinking of it, I guess, is that the density difference between the inside and outside of the balloon serves as an additional (effective) potential. Certainly, when the liftoff condition is satisfied, the balloon is in some nearly-flat potential, as I can walk up to it and poke it and it will fly up or down or sideways.

I think perhaps the reason the second estimate is wrong is that I'm tacitly assuming a cold balloon (with contents of density $\rho_0$) would sink to the center of the Earth in the absence of any rock in the way. This clearly isn't true, as any hypothetical atmosphere would get denser, and the buoyancy forces would balance at some $r \neq 0$.

Edit: I still don't understand how this can be wrong. At $T_0$ the gas inside is in equilibrium with the atmosphere, so the only contribution to the gravitational PE is the payload. The PE of the balloon at $T_{\mathrm{in}}$ must be be at least $m g r_{\mathrm{Earth}}$ greater than at $T_0$, otherwise it would still be subject to net downward forces. Isn't this reasoning correct?

 

[omoplata]

OK, let me write down what I've been thinking. It might help someone who is reading this tread to arrive at a solution.

I was wrong when I said "second estimate would have been right if the burner somehow pushed the balloon from the center of the Earth up to $r_{Earth}$". Assume no frictional or electromagnetic forces. If we assumed the mass distribution of Earth to be concentrated at a point at the center, the gravitational force on the balloon would be proportional to $1/r^{2}$ inside the earth. The gravitational force infinitesimally near the point mass would be infinite. Then it would take infinite work to get from that infinite point mass to the surface of the earth. The general formula for the gravitational potential of the Earth is $U = -\frac{G M m}{r} + K$, where $K$ is a constant of integration (http://en.wikipedia.org/wiki/Gravitational_potential_energy#General_formula", the gravitational force on the balloon would be proportional to $r$ inside the earth. So only a finite amount of work would have to be done for the balloon to be brought from the center of the Earth to the surface. So for different mass distributions of the Earth inside the surface, the gravitational work done for the balloon to be brought from the center of the Earth to the surface would be different. But the gravitational force above the Earth would be the same according to the shell theorem, and the balloon will behave the same way. This suggests that the movement that compensates for the energy supplied to the balloon would be outside the surface of the Earth and not within.

Let's say the gravitational potential field is $\phi_{g} (r)$, and the buoyancy potential field is $\phi_{b} (r)$. Then the gravitational force is $\vec{\nabla} \phi_{g}$ and the buoyancy force is $\vec{\nabla} \phi_{b}$. When the density inside the balloon is equal to the density outside the balloon, $\phi_{b}$ is at a minimum and there is no buoyancy force. So by heating the gas we change $\phi_{b}$ so the minimum shifts to another place at a larger $r$. But $\phi_{g}$ does not change at all during this process.

The condition for the balloon being at a state where it's just about to liftoff is $\vec{\nabla}\phi_{g} (r_{Earth}) + \vec{\nabla}\phi_{b} (r_{Earth}) = \vec{0}$. So we have to change $\phi_{b}$ by heating the gas until that condition is satisfied.

Let's name the buoyancy potential field before the gas is heated $\phi_{b1}$ and the buoyancy potential field after the gas is heated $\phi_{b2}$. So $\vec{\nabla} \phi_{b1} (r_{Earth}) = \vec{0}$ and $\vec{\nabla}\phi_{g} (r_{Earth}) + \vec{\nabla}\phi_{b2} (r_{Earth}) = \vec{0}$. The amount of potential change, or the energy used to heat the gas will be $\Delta Q = \phi_{b2} (r_{Earth}) - \phi_{b1} (r_{Earth})$.

Let $h$ be the altitude at which, after the gas has been heated, the buoyancy force is zero, i.e. $\vec{\nabla} \phi_{b2} (r_{Earth} + h) = \vec{0}$. Then, if $\phi_{b2} (r_{Earth} + h) = \phi_{b1} (r_{Earth})$, we can find the energy used to heat the gas by $\Delta Q = \phi_{b2} (r_{Earth}) - \phi_{b2} (r_{Earth} + h)$. But we need an equation for $\phi_{b2} (r)$ to actually calculate this.

Also, I don't know whether the assumption $\phi_{b2} (r_{Earth} + h) = \phi_{b1} (r_{Earth})$ is correct. That means the minimum value for $\phi_{b}$ did not change as it was shifted. That looks like to be saying that the gravitational and buoyancy potentials are independent of each-other. But I guess that approximation may be made for small distance changes?

 

[paranormal]

I would first like to commend you for exploring the physics of hot air balloons and coming up with these estimates. It is always important to question and verify our assumptions and calculations.

After reviewing the two estimates you have provided, I believe that the second estimate using the "buoyancy energy" is incorrect. While it is true that the buoyancy energy must balance the gravitational potential at the surface of the Earth, this calculation does not take into account the fact that the hot air balloon is constantly losing gas as it rises due to the decrease in atmospheric pressure.

This loss of gas, along with other factors such as air resistance and heat loss, significantly reduces the amount of energy required to lift the balloon off the ground. Therefore, the first estimate using thermodynamics and the ideal gas law is more accurate in predicting the energy required for liftoff.

Additionally, the assumption of adiabatic heating and ignoring gas lost during heating is unrealistic and would not accurately reflect the actual energy required in a hot air balloon. It is important to consider all factors and assumptions when making these calculations.

In conclusion, I believe that the first estimate using thermodynamics is more accurate in predicting the energy required for liftoff of a hot air balloon. However, further research and experimentation may be needed to refine and improve these estimates.