Thermodynamics - Hot-air balloon

[Feodalherren]

Homework Statement

The mass of a hot-air balloon and its cargo (not including the air inside) is 200kg. The air outside is at 10C and 101kPa. The volume of the balloon is 400m^3. To what temperature must the air in the balloon be warmed before the balloon will lift off? (The density of air a 10C is 1.25kg/m^3).

Homework Equations

PV=nRT
and maybe some other basic thermodynamics.

The Attempt at a Solution

\sum F = _{}\rho_{inside}Vg - _{}\rho_{outside}Vg - mg = 0
Simplifies to
m= V(_{}\rho_{inside} - {}\rho_{outside})

Assuming the volume is the same... But I can't really wrap my head around it. Would the volume be the same? I have no idea what to do next and I'm not even sure that I'm doing it right.

 

[ehild]

Homework Statement

The mass of a hot-air balloon and its cargo (not including the air inside) is 200kg. The air outside is at 10C and 101kPa. The volume of the balloon is 400m^3. To what temperature must the air in the balloon be warmed before the balloon will lift off? (The density of air a 10C is 1.25kg/m^3).

Homework Equations

PV=nRT
and maybe some other basic thermodynamics.

The Attempt at a Solution

\sum F = _{}\rho_{inside}Vg - _{}\rho_{outside}Vg - mg = 0
Simplifies to
m= V(_{}\rho_{inside} - {}\rho_{outside})

Assuming the volume is the same... But I can't really wrap my head around it. Would the volume be the same? I have no idea what to do next and I'm not even sure that I'm doing it right.

The volume of the balloon is given, 400m³.

You have to sum all forces acting on the balloon and its cargo, as you did, but check the signs. Which force acts downward and which one acts upward?


ehild

 

[Andrew Mason]

Homework Statement

The mass of a hot-air balloon and its cargo (not including the air inside) is 200kg. The air outside is at 10C and 101kPa. The volume of the balloon is 400m^3. To what temperature must the air in the balloon be warmed before the balloon will lift off? (The density of air a 10C is 1.25kg/m^3).

Homework Equations

PV=nRT
and maybe some other basic thermodynamics.

The Attempt at a Solution

\sum F = _{}\rho_{inside}Vg - _{}\rho_{outside}Vg - mg = 0
Simplifies to
m= V(_{}\rho_{inside} - {}\rho_{outside})

Assuming the volume is the same... But I can't really wrap my head around it. Would the volume be the same? I have no idea what to do next and I'm not even sure that I'm doing it right.

The condition for lift is net upward force > 0, which means:

m < (ρ_outside - ρ_inside)V

You cannot change m (the mass of the balloon itself), V or ρ_outside or P. So how do you lower ρ_inside? (hint: express PV= nRT in terms of density).

AM

 

[Feodalherren]

I still don't understand how the volume can be the same on the inside and the outside... Wouldn't the volume of the gas outside be all of the air above the balloon?

P(m/ρ)=nRT...?

 

[Andrew Mason]

I still don't understand how the volume can be the same on the inside and the outside... Wouldn't the volume of the gas outside be all of the air above the balloon?

P(m/ρ)=nRT...?

V is the volume of air displaced by the balloon i.e. the volume of the balloon (you are to assume that the volume of the balloon apparatus apart from the balloon itself has negligible volume). Buoyancy requires the balloon displacing a mass of air that is greater than its mass, which is comprised of the mass of the balloon apparatus and the air inside the balloon.

AM

 

[Feodalherren]

Well wouldn't that only make the initial volume of the balloon Vi=400
Then using

PiVi=PfVf I could find some final volume? Maybe?

 

[Andrew Mason]

Well wouldn't that only make the initial volume of the balloon Vi=400
Then using

PiVi=PfVf I could find some final volume? Maybe?

Assume that the balloon does not expand so the volume is constant. You would you reduce the mass of the inflated balloon? (hint: you can't reduce the mass of the balloon apparatus).

Am

 

[ehild]

Well wouldn't that only make the initial volume of the balloon Vi=400
Then using

PiVi=PfVf I could find some final volume? Maybe?

You wrote the equation for an isothermal process.

Have you seen a hot-air balloon launching? (watch from 15 minutes. )

The gas in the balloon is heated by burning gas. The balloon is open, so the pressure is the same inside and outside. The air expands while heated and some of it leaves the balloon. But the volume of the balloon does not change. The mass and the density of air inside the balloon changes with the temperature - how?

ehild

 

[Feodalherren]

If the balloon doesn't expand... I guess heating the gas is the only option but that only means that the pressure inside increases. But if the volume can't expand I don't see how that is of any significance to the problem.

ehild, I have no idea... It doesn't many any intuitive sense to me at all

ρ=m/V

PV=nRT

ρ=mnRT/P

 

[Andrew Mason]

If the balloon doesn't expand... I guess heating the gas is the only option but that only means that the pressure inside increases. But if the volume can't expand I don't see how that is of any significance to the problem.

ehild, I have no idea... It doesn't many any intuitive sense to me at all

ρ=m/V

PV=nRT

ρ=mnRT/P

You are trying to reduce the mass of the air inside the balloon! What has to happen if the pressure and volume cannot increase and the temperature increases? Hint: the balloon is open at the bottom...

AM

 

[Feodalherren]

I guess some of the air molecules have to escape from the bottom then.

 

[ehild]

Very good... If some molecules escape, less stay inside the balloon. T increases, P and V are constant, so what else changes in the equation PV=nRT?

ehild

 

[Feodalherren]

n decreases.

 

[ehild]

n is the number of moles in the balloon. If n decreases the mass of air inside the balloon also decreases. How much should be the mass of air inside the balloon, so that the weight of the air in the balloon + the weight of the balloon and cargo is equal or less than the buoyant force?

ehild