Balloon Thermo/buoyancy question

[efekwulsemmay]

Homework Statement

Hot air balloons use a burner (typically around 2 MW of power output) to heat the air in
the envelope (the balloon part of the hot air balloon). Consider a typical envelope with a
volume of V = 2800 m³. The weight of the balloon (envelope, basket and passengers, but not including the air in the envelope) is 400kg. Assume that all the heat from the burner goes into heating the air in the envelope and that there is no conductive loss to the environment. (Note that as the air in the balloon expands, hot air will move out of the envelope; i.e., there will be convective loss of heat.) Treat the air as a diatomic ideal gas with P = 1atm and molar mass M = 28 g/mol. Use an external air temperature of 298K.

Using that the buoyant force on the envelope needs to support the
weight of the balloon plus the weight of the air in the envelope, determine the
minimum temperature that the air in the envelope must be to lift the balloon.

Homework Equations

Density: $\rho=MP/RT$ (derived from the Ideal Gas Law)

The Attempt at a Solution

Ok, so I know from Wikipedia that buoyancy can be defined as:

$\frac{density of object}{density of fluid}=\frac{weight (of object)}{weight of fluid}$

and I've gotten to here:

$\frac{density of air envelope}{density of air environmental}=\frac{weight of balloon and air envelope}{weight of air displaced by balloon}$

Just from thinking about the problem this is where I am at but I don't know where to go from here. Any ideas/suggestions?

Thanks

 

[mfb]

What do you know about the total balloon system in equilibrium (=the minimal temperature to stay in the air)? Can you relate this to the mass of the total system?

 

[efekwulsemmay]

Wait so could the answer be related to the ratio of the densities? The weight over weight portion of the equation, when I compute it, will just end up being a constant with the information I am given, so could that mean that the density of the air in the envelope has to be greater than the density of the air in the environment?

Like this? :

$density of air envelope = \frac{weight of balloon and air envelope}{weight of air displaced by balloon} \cdot density of air environment$

 

[mfb]

I don't think those ratios will help you at the current stage of the solving process. Absolute numbers are easier to understand.

weightofballoonandairenvelope/weightofairdisplacedbyballoon is an interesting quantity, however. What happens if it is smaller/equal to/larger than 1?

so could that mean that the density of the air in the envelope has to be greater than the density of the air in the environment?

That would not give lift.

 

[Nickie]

for providing the problem statement and your attempt at a solution. I would approach this problem by first defining the variables and equations that are relevant to the situation.

First, let's define the variables:

V = volume of the envelope (2800 m3)
m = mass of the balloon (400 kg)
P = pressure (1 atm)
T = temperature
M = molar mass of air (28 g/mol)
R = gas constant (8.314 J/mol·K)
g = acceleration due to gravity (9.8 m/s2)

Now, let's consider the equations that are relevant to the situation:

1. Ideal Gas Law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

2. Density: ρ = m/V, where ρ is density, m is mass, and V is volume.

3. Buoyancy: Fb = ρVg, where Fb is the buoyant force, ρ is density, V is volume, and g is acceleration due to gravity.

Using these equations, we can solve for the minimum temperature needed to lift the balloon:

1. From the Ideal Gas Law, we can solve for n (the number of moles of air in the envelope): n = PV/RT.

2. Using the density equation, we can calculate the density of the air in the envelope: ρ = m/V = (400 kg)/(2800 m3) = 0.143 kg/m3.

3. Now, we can substitute this value for density into the buoyancy equation to solve for the buoyant force: Fb = ρVg = (0.143 kg/m3)(2800 m3)(9.8 m/s2) = 398.32 N.

4. We know that the buoyant force must be equal to the weight of the balloon and the air in the envelope (m), so we can set these two forces equal to each other: Fb = m = 398.32 N.

5. Rearranging the Ideal Gas Law equation to solve for temperature (T), we get T = PV/nR.

6. Substituting the values we know into this equation, we get T = (1 atm)(2800 m3)/[(398.32 N)(8.314 J/mol·K)(28