Archimedes principle - special case

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Homework Help Overview

The discussion revolves around a physics problem involving a Chinese lantern balloon, focusing on the application of Archimedes' principle to determine how much additional weight can be added before the balloon can no longer take off. The problem includes parameters such as the balloon's dimensions, mass, and the temperature of the surrounding air.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the volume of the balloon and the buoyant forces acting on it, but expresses uncertainty about the temperature of the hot air and its effect on density. Participants discuss the potential need for assumptions regarding the energy content of the fuel and the specific heat capacity of air to estimate the temperature of the air in the balloon after combustion.

Discussion Status

Participants are exploring various assumptions and considerations that affect the calculations. Some suggest focusing on the moment when the fuel is nearly consumed to simplify the analysis, while others raise concerns about the accuracy of temperature estimates and the composition of gases in the balloon.

Contextual Notes

There are several omitted factors that could influence the outcome, such as the temperature of the hot air, the volume of combustion products, and the effects of heat loss. The original poster acknowledges these uncertainties and seeks further guidance.

Lucky7
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Homework Statement


We have a chinese lantern (balloon) made of paper, cylinder shaped with the following sizes:
base diameter - 45 cm, height - 70 cm. Mass of the balloon is 57 g (21 g from this is mass of "fuel" - the fuel is wax paper!). Fuel is then ignited at the centre of the base, which is open. Therefore the balloon will soar. The question is, how much additional weight can be hung to the balloon and the balloon still takes off. Outside temperature is 5 °C (Let us denote it as cold air).

Homework Equations


Just see the text.

The Attempt at a Solution


First I can compute volume of the balloon, which is of course V = π*r2*h = π*22.52*70 cm3 = 111330.189662 cm3≈0.11133m3.
Now the gravitational force of air pressed up by the balloon gives the magnitude of buoyancy:
Gcold_aircold_air*V*g = 1.2697*0.11133*g≈1.38723 N.
Now we should determine the volume of the fuel, which is
Vfuel= mfuelfuel=0.021/650 = 0.00003 m3.
Now we will compute Ghot_airhot_air*(V-Vfuel)*g = 1.1277*0.11130*9.871373 ≈ 1.23175 N. Now we can use Archimedes principle:
Gcold_air = Ghot_air + m*g + M*g, and we want to solve this equation for M, which is our burden, that can be carried:
M = (Gcold_air-Ghot_air-m*g)/g, which gives us M≈-0.04116 kg, which is rubbish...

There are just too many determinants, that were omitted:
1) temperature of hot air in the ballon, I chose 40 °C, but how can I know? What if it is 60 °C, then the density would be higher, but how can I know?
2) It would be best if all fuel would just burn out, then the temperature would be the highest and the mass the lowest, so should I wait till the fuel is burn out and only after that I bind the burden?
3) I omitted the air pressed up by the wax paper and the burden.
4) ... inf)
There are just too many...
I'd be so much grateful for your help!
 
Last edited:
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Do you know the energy content of the fuel and the specific heat capacity of air? That may allow you to calculate the temperature of the air in the balloon after the fuel is burnt. Some assumptions may have to be made!
 
This is all I know, we should probably make some of our own assumptions, yeah...
 
Since you only need it to take off eventually, yes, you can restrict attention to the time at which the fuel is almost gone. I suspect that assuming all the heat stays in the balloon will give a significant overestimate of the temperature. You could calculate the volume of gas from the combustion products, assume that mixes uniformly with the original air, find the resulting expansion, and hence find how much heat is lost by spillage out of the envelope. (There'll also be radiative losses, but they're probably a lot less.)
Of course, the gases in the balloon will no longer be just hot air - there'll be a lot of CO2, affecting the density.
 

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