A typical volume of a modern hot air balloon is 2500 cubic metres, and a typical maximum temperature of the hot air is 120 degrees Celsius. Given these figures, and an outside air temperature and density of 15 ∘C and 1.225 kg/m3 respectively, compute the maximum mass (in kilograms) of the balloon, basket and payload.
The only parameters we need to convert are the temperature difference (ΔT=120−15=105) and the outside air temperature in Kelvin (T = 288.15 K).
Now the lift of this hot air balloon is found using:
L=ρVg(ΔTT+ΔT)
L=1.225⋅2500⋅9.81(105288.15+105)=mg
m⋅9.81=8023.727
m=818kg
818.225
Archimedes says:
volume of fluid displaced * density of fluid * g = upward buoyant force = weight is neutrally buoyant
mass = weight/g = volume*density
2500 * 1.225 = 3063 kg up
now subtract the mass of air in the balloon
2500 * rho
what is rho, the density of air inside at 120 C ?
approximate as perfect gas ?
P V = n R T
P is close to the same in and out or balloon would burst
V the same
so
n T is about constant
density proportional to mass in the balloon volume
n hot T hot = n cold Tcold
n hot /n cold = Tcold/Thot
Tcold = 273 + 15 = 288
T hot = 273 + 120 = 393
so density hot/density cold = 288/393 = .733
.733 * 1.225 = .8977 kg/m^3
so mass of air in balloon = .8977*2500 = 2244 kg
then finally
mass of other stuff = mass of cold air displaced - mass of hot air = 3063 -2244 = 819 kg
Well, let's do some "hot" air math!
First, let's find the mass of the hot air inside the balloon. We'll use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
So, we need to convert the temperatures to Kelvin. 120 degrees Celsius is 393 Kelvin, and 15 degrees Celsius is 288 Kelvin. Now, let's plug in the values:
(P_hot_air)(V_hot_air) = (n)(R)(T_hot_air)
(P_cold_air)(V_hot_air) = (n)(R)(T_cold_air)
The pressure of the cold air cancels out, so we're left with:
V_hot_air = (T_hot_air / T_cold_air) x V_cold_air
V_cold_air is 2500 cubic meters, so:
V_hot_air = (393 / 288) x 2500 = 3409.03 cubic meters
Now we need to find the mass of the hot air inside the balloon. We'll use the ideal gas law again, but this time we'll solve for the number of moles (n):
(P_hot_air)(V_hot_air) = (n)(R)(T_hot_air)
n = (P_hot_air)(V_hot_air) / (R)(T_hot_air)
Let's plug in the values: P_hot_air is the same as the pressure of the cold air, which is 1.225 kg/m^3. V_hot_air is 3409.03 cubic meters. R is the gas constant, which is 8.3145 J/(mol K), and T_hot_air is 393 Kelvin.
n = (1.223 kg/m^3)(3409.03 cubic meters) / (8.3145 J/(mol K))(393 K)
n = 1660.66 moles
Now, let's calculate the mass of the hot air inside the balloon. We'll use the molar mass of air, which is approximately 28.97 g/mol:
mass_hot_air = n x molar mass of air
mass_hot_air = 1660.66 moles x 28.97 g/mol
mass_hot_air = 48,033.28 grams
To convert grams to kilograms, we divide by 1000:
mass_hot_air = 48.033 kilograms
Now, let's add the mass of the balloon, basket, and payload. Assuming they all weigh zero (since they are being lifted by the hot air), the maximum mass of the balloon, basket, and payload is simply the mass of the hot air:
maximum mass = mass_hot_air
maximum mass = 48.033 kilograms
So, the maximum mass of the balloon, basket, and payload is 48.033 kilograms. Keep in mind that this calculation does not take into account any additional weight, such as fuel, equipment, or passengers. Just something to keep you "air-ware"!
To compute the maximum mass of the balloon, basket, and payload, we need to determine the mass of the hot air inside the balloon and account for the mass of the envelope, basket, and payload.
First, let's calculate the mass of the hot air inside the balloon. We can use the ideal gas law equation:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature
Rearranging the equation, we get:
n = PV / RT
We can calculate the number of moles of hot air in the balloon using the given volume, temperature, and outside air pressure. The outside air pressure is not explicitly mentioned, but assuming it is at sea level, we can use the standard atmospheric pressure of approximately 101325 Pa.
Next, we utilize the molar mass of air, which is approximately 28.97 grams per mole, to determine the mass of the hot air:
Mass of hot air = number of moles of hot air * molar mass of air
After obtaining the mass of hot air, we can add the masses of the envelope, basket, and payload to find the maximum total mass.
Note: For simplicity, we'll assume the envelope, basket, and payload have negligible volume compared to the hot air.
Now let's calculate it step by step:
Step 1: Calculate the number of moles of hot air.
Given:
Volume of the balloon, V = 2500 m^3
Temperature of the hot air, T = 120 °C = 393.15 K (convert to Kelvin)
Using the ideal gas law equation:
n = PV / RT
P = outside air pressure = 101325 Pa
R = ideal gas constant = 8.314 J/(mol·K)
n = (P * V) / (R * T)
Step 2: Calculate the mass of the hot air.
Given:
Molar mass of air (M) = 28.97 g/mol
Mass of hot air = number of moles of hot air * M
Convert the mass of hot air to kilograms (1 kg = 1000 g).
Step 3: Calculate the total mass.
Given:
Mass of envelope = assume negligible
Mass of basket = assume negligible
Mass of payload = assume negligible
Total mass = mass of hot air + mass of envelope + mass of basket + mass of payload
Keep in mind that this calculation assumes the envelope, basket, and payload are negligible compared to the mass of the hot air. If any of these components have significant mass, they should be included in the total mass calculation.
Following these steps, you can calculate the maximum mass of the balloon, basket, and payload.