A hot-air balloon encloses a volume of 1e5 ft3 when inflated; hot-air balloons are usually made of nylon which is effec- tively rigid when inflated. The balloon material, basket, fuel and other associated material weighs 500 lb. You plan to take one for a ride with three of your friends; you and your friends total 500 lb. Hot-air balloons generate buoyancy by using a burner to heat the air (it also gets some combustion products into the balloon, but let’s ignore that for simplicity). Assuming the ambient conditions are STP (293.15 K; ρ=1.225 kg/m3), to what temperature must you heat the air in the balloon to get it and its basket and cargo to lift in the air?
This text is generated by AI
Below are the references I took into account while formulating my response; I encourage you to explore the pertinent ones directly:
Most relevant source material:
- Chapter 4 How Stuff Works: How Hot-Air Balloons Wo… (HIGHLY RELEVANT)
- Glossary entry: hot-air balloon (HIGHLY RELEVANT)
- Chapter 4.3: Archimedes’ Law (RELEVANT)
- Kirby response to: HW1 Q7 PK Bot. (MARGINALLY RELEVANT)
- Chapter 1 Fundamentals: Let’s review ideal gas the… (MARGINALLY RELEVANT)
- Chapter 4 How Stuff Works: How Scuba Suits Control… (MARGINALLY RELEVANT)
- Chapter 4.1: Watercraft (MARGINALLY RELEVANT)
- Chapter 1 Engineering Models: Calorically perfect … (MARGINALLY RELEVANT)
- Chapter 3.4: Hydrostatic Distributions in Atmosphe… (MARGINALLY RELEVANT)
- Kirby response to: Carburetor Density question. (MARGINALLY RELEVANT)
PK Bot’s answer: Below is one way to solve the problem using the buoyant equilibrium concept explained in our course material.
Suppose the hot‐air balloon must produce an upward buoyant force that exceeds the total weight of the balloon, basket, fuel, and passengers. When the balloon is in equilibrium (or just beginning to lift off), the net vertical force is zero. In our control‐volume analysis for a balloon with a fixed envelope, we write
F_{B} - F_{G} = 0,
where
• F_B is the buoyant force from the displaced air, and
• F_G is the total gravitational force (weight) of the balloon system.
The buoyant force is given by the weight of the displaced air:
F_B = \rho_\infty , g , V,
and the gravitational force acting downward is
F_G = F_1,
where F_1 is the total weight that must be lifted.
For force equilibrium, we have
\rho_\infty , g , V - F_1 = 0 \quad \Longrightarrow \quad \rho_\infty , g , V = F_1.
However, when we heat the air inside the balloon, its density decreases. Under the assumption that the pressure inside the envelope is atmospheric (i.e. p remains constant) and the gas behaves as a calorically perfect ideal gas, we can relate the densities inside and outside via
p = \rho_e , R , T_e \quad \text{and} \quad p = \rho_\infty , R , T_\infty.
It follows that
\rho_e = \frac{p}{R, T_e} \quad \text{and} \quad \rho_\infty = \frac{p}{R, T_\infty}.
Combining these with our buoyancy balance, the net upward force on the envelope is
F_{\text{net}} = \underbrace{\rho_\infty , g , V}{\text{buoyant force}} - \underbrace{\rho_e , g , V}{\text{weight of heated gas}} - F_1.
For the balloon to lift off, we require that the net force reaches zero (i.e. just overcome the weight), so
\rho_\infty , g , V - \rho_e , g , V = F_1.
Dividing through by gV gives
\rho_\infty - \rho_e = \frac{F_1}{g, V}.
Expressing \rho from the ideal gas law in terms of temperature, we have
\frac{p}{R} \left(\frac{1}{T_\infty} - \frac{1}{T_e}\right) = \frac{F_1}{g, V}.
A bit of algebra then leads to a formula for the required heated temperature:
T_e = \frac{T_\infty}{1 - \frac{F_1}{\rho_\infty g V}}.
Let’s now substitute our given values.
──────────────────────────── Step 1. Convert the volume from cubic feet to cubic meters
Given:
Balloon volume V = 1\times10^5 \text{ ft}^3.
Conversion:
1 \text{ ft}^3 \approx 0.0283168 \text{ m}^3.
Thus,
V = 10^5 \times 0.0283168 \approx 2831.68 \text{ m}^3.
──────────────────────────── Step 2. Compute the buoyant force produced by the displaced ambient air
Given ambient density: \rho_\infty = 1.225\ \text{kg/m}^3
Gravitational acceleration: g = 9.81\ \text{m/s}^2.
Then, $$ F_3 = \rho_\infty , g , V = 1.225 \times 9.81 \times 2831.68 \approx 34015\text{ N}. $$
──────────────────────────── Step 3. Determine the total weight F_1 to be lifted
We are told that:
Weight of balloon materials, basket, etc. = 500 lb
Weight of you and your three friends = 500 lb
Total weight = 1000 lb.
Convert pounds to newtons using
1\text{ lb} \approx 4.44822\ \text{N}: $$ F_1 = 1000 \times 4.44822 \approx 4448\ \text{N}. $$
──────────────────────────── Step 4. Compute the ratio
\frac{F_1}{\rho_\infty g V} = \frac{4448}{34015} \approx 0.1307.
──────────────────────────── Step 5. Calculate the required heated-air temperature
Given ambient temperature T_\infty = 293.15\ \text{K}, we substitute into
T_e = \\frac{T\_\\infty}{1 - \\frac{F_1}{\\rho\_\\infty,g,V}} = \\frac{293.15}{1 - 0.1307} \\approx \\frac{293.15}{0.8693}.
Evaluating this, we get $$ T_e \approx 337.0, \text{K}. $$
──────────────────────────── Step 6. Interpret the answer
The heated air inside the balloon must be at about 337 K, which is roughly 44 K hotter than the ambient temperature (293 K). In Celsius, this is approximately 337 K − 273.15 ≈ 63.85 °C.
──────────────────────────── Final Answer
To lift the payload, the air inside the balloon must be heated to approximately
337\text{ K} \quad \text{(about 64 °C)}.
Let me know if you need any further clarification on any of these steps.