Bro. Daines got tired of trying to teach his students Fluid Mechanics and really

Question

Bro. Daines got tired of trying to teach his students Fluid Mechanics and really wants to connect with his alien abductors. He wants to be abducted again and plans to float up and wait for a passing flying saucer to notice him so he is going to build himself a brightly-colored hot-air balloon. Bro. Daines and the stuff he want to show to the aliens (along with the balloon basket) have a combined mass of 130kg. Bro. Daines knows it may take a while to attract the attention of a passiong saucer, so he wants to have a "loiter time" at altitude as laong as possible (the balloon has a 20 gallon propane tank just for loitering with a burner capable of 12 times 10^6 Btu/hr. The energy content of propane is 91.690 Btu/gallon.) Since Bro. Daines teaches heat transfer, he knows the balloon will lose energy at a rate given by j q = U A (Tb = Tamb) where q is the heat transfer rate in watts, U is the overall heat transfer coefficient, about 18 W/m^2 K for this situation, A is the surface area of the balloon, Tb is the ari temperature in the balloon, and Tamb is the ambient temperature outside the balloon. Bro. Daines is also cheap and wants to buy as little fabric as possible. (He does know that this fabrichas a maximum working temperature of 120 C and a mass of 63.6 g/m^2.) The preferred altitude for alien abductions, he happens to know, is 5000 m. This is a warm summer day in Rexburg (altitude: 1600 m) and the ground temperature is 30 C, so Bro. Daines figures the temperature at 5000 m is about 5 C. What balloon diameter and balloon air temperature do you recommend he use? What would hi sloiter time be? How much fabric will he need to pruchase? As a team, you will submit a filled- in spreadsheet form (downloaded from I-Learn) with the 3 key numbers above, lplus a 1-paragraph explanation of how you came up with your optimum value and a graph or table that supports your paragraph.

Explanation / Answer

The basic principle behind hot air balloon physics is the use of hot air to create buoyancy, which generates lift.A burner sits in the basket and is used to heat the air inside the envelope through an opening. This heated air generates lift by way of a buoyant force.

The hot air inside the envelope is less dense than the surrounding (cooler) air. This difference in density causes the hot air balloon to be lifted off the ground due to the buoyant force created by the surrounding air. The principle behind this lift is called Archimedes' principle.

For a hot air balloon, the upward buoyant force acting on it is equal to the weight of the cooler surrounding air displaced by the hot air balloon. Since the air inside the envelope is heated it is less dense than the surrounding air, which means that the buoyant force due to the cooler surrounding air is greater than the weight of the heated air inside the envelope. And for lift to be generated, this buoyant force must exceed the weight of the heated air, plus the weight of the envelope, plus the weight of passengers and equipment on board.Mass of Bro Daines combined with his stuff=130 kg.So weight of Bro Daines with his stuff=1300

So, total weight of the hot air balloon=1300 + 0.636y where y= surface area of the fabric

Therefore,net buoyant force pushing upwards on the heated air inside the envelope should be

Fb >1300 + 0.636y

Now, let us divert our attention to the air at Rexburg.

Temperature=30 C

Altitude=1600 m.

Pressure at sea level is 1 atm=101325 Pa approx.Air pressure above sea level can be measured by P = 101325 (1 - 2.25577 10-5 h)5.25588.This derivation is done elsewhere in standard thermodynamics books.We shall use the formula for our calculation of Pressure at 1600m. at Rexburg.

By the ideal gas law,we can find out the density of air at these conditions.

P=dRT

where d=density of air,

R=Gas constant of air= 287 J/kgK

T=temperature = 303 K

P=Pressure at 1600m at 30 C= 84340 Pa

From this,substituting in the ideal gas equation,we get,

d=0.969860 kg/m3 = density of air at Rexburg

For creating a buoyant force , we need a density of air inside the heated envelope to be lesser than that of the outside air.That is, density of air inside heated envelope=d'<0.969860

Since we know, that the fabric can have maximum temperature of 120 C,let us assume we heat air inside the envelope at 110 C.

The heated air inside the envelope is at roughly the same pressure as the outside air. With this in mind we can calculate the density of the heated air at a given temperature, using the Ideal gas law, as follows:

P=d'RT'

where d' and T' are density of hot air and temperature of hot air respectively

T=temperature = 383 K

P=84340 Pa

R=Gas constant of air= 287 J/kgK

We get, d'=0.7672 kg/m3

The net buoyant force is defined here as the difference in density between the surrounding air and the heated air, multiplied by the envelope volume. Thus,

Fb =(0.9698-0.7672) x V where V=volume of the hot air balloon

We take the hot air balloon to be approximately spherical.Actually,the hot air balloon shape is distinctive and neither a sphere nor a cylinder, so we have to make certain approximation which we make of a spherical shape here.

V=4/3 x Pi x r3 =4.1866 r3

Fb =(0.9698-0.7672) x 4.1866 r3
We know,from previous explanations,

Fb >1300 + 0.636y where y=surface area of the balloon.

y=4 x Pi x r2 =12.56 r2

Fb >1300 + 7.98816r2

Equating the two relations for Fb will give us the threshold condition of floating of the balloon where the net buyant force is equal to the gravitational force acting downwards.

1300+7.98816r2=(0.9698-0.7672) x4.1866 r3

=> 1300+7.98816r2= 0.8478 x r3

Solving for r, we get r=15.66 m. approximately

This r is for the threshold condition of floating of the balloon Fb =1300 + 0.636y where y=surface area of the balloon.Which implies that,for upward acceleration,the buoyant force must be greater than the force acting downwards. So, r>15.66 m. Because that is when the buoyant force will cause an upward acceleration.

Let us then take r to be in the range 16 to 17 m. (say 16.5 m).> 15.66 m.

Area of fabric= 4 x 3.14 x 272.25 m2 =3419.46 m2

To Summarize, Bro Daines needs a fabric of area 3419.46 m2 for a balloon of radius 16.5 m with a temperature of 110 C kept inside the balloon.With these conditions, the buoyant force will exceed the gravitational force and set the ballon into upward motion.

The hot air balloon loses heat at the rate of UA(Tb - Tambient)=18 x 3419.46 x (383-278)=6461910 J/s =6125 Btu/sec since 1 Btu=1055 J

Now,the burner is working upto 12 x 106 Btu/hr. = 3333.33Btu/sec.

We see that the loss is more than the heat produced by the burner.As we have taken temperature to be almost maximum (110 C is very near to 120 C)there is no further scope for reducing the radius of the ballon.And hence, no scope for reducing the heat loss as heat loss is proportional to the surface area of the balloon.

With the current burner,the loiter time is zero.And at current capacity,even maximizing the efficiency by smallest possible radius and highest temperature inside,the balloon will not reach 5000 m. It will start going down at a height where the loss of heat=UA(Tb-Tambient) becomes more than the heat produced by burner (3333.3 Btu/sec).

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