A typical hot air balloon is roughly 55 ft wide and 63 ft hi

A typical hot air balloon is roughly 55 ft wide and 63 ft high, with a total volume of 90000 ft^3. If the balloon, gondola, fuel, passengers and cargo have a total mass of 1000 lbm, what air temperature inside the balloon is required to lift the balloon and its contents? Assume that the ambient air temperature is 20 degree C and pressure is 1 atm. If the surface area of the balloon is 1800 sq yards, estimate how long the balloon can stay aloft at an operating temperature of 100 degree C, with two 20 gallon tanks of propane on board. Note that 1 gallon of propane has a fuel value of approximately 91200 Btu. To estimate the heat transfer coefficient, use the correlation: Nu_D = 2 + 0.589 Ra_D^1/4/[1 + (0.469/Pr)^g/16]^4/9 Note that you can use the width of the balloon as the \"diameter\" of the sphere. Evaluate the properties of air at the film temperature, which is defined as the arithmetic average of the surface temperature and the bulk fluid temperature. The properties of air can be found at http://www.engineeringtoolbox.com/.

Solution

For floating, buoyant force = mass of balloon with all its contents

1 atm = 1.01325 bar

20 deg C = 20+273 K = 293 K

Dia D = 55 ft = 16.764 m

Surface area A = 1800 sq yard = 1505 m^2

Fuel value = 91200 Btu = 96221 kJ

Density of displaced fluid = P / RT

= 1.01325*10^5 / (287*293)

= 1.2 kg/m^3

= 0.075 lb/ft^3

Mass of displaced fluid = 0.075*90000 = 6750 lbm

Mass of air inside balloon = 6750 - 1000 = 5750 lbm

Density of air inside balloon = 5750 / 90000 = 0.06389 lb/ft^3 = 1.022 kg/m^3

P/RT = 1.022

101325 / (287*T) = 1.022

T = 345 K = 72 deg C

Film temperature = (100 + 20)/2 = 60deg C = 273+60 K = 333 K

At 60 deg C for air,

beta = 1/T = 1/333 = 0.003 /K

neu = 1.89*10^-5 m^2/s

alpha = 26.8*10^-6 m^2/s

Pr = 0.699

k = 2.88*10^-5 W/m-K

Ra = g*beta*(Ts - Tamb)*D^3 / (neu*alpha)

= 9.81*0.003*(100-20)*16.764^3 / (1.89*10^-5 * 26.8*10^-6)

= 2.189*10^13

Putting it in expression for Nu, we get

Nu = 983

hD/k = 983

h = 983*2.88*10^-5 / 16.764

h = 0.0017 W/m^2-K

Q = hA(Ts - Tamb)

=0.0017*1505*(100-20)

= 203.3 W

Heat available = 2*20*91200 Btu = 2*20*96221 kJ = 3848840 kJ

Time = 3848840*10^3 / 203.3

= 18931824 s

= 5258.8 hours