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HOW MUCH EXTRA LIFT
DO WE GET FOR
HOW MUCH EXTRA HEAT?

The basic question arises: if we burn a certain mass of fuel, which generates a certain quantity of heat energy, and we transfer this heat energy to lift gas, how much extra lift do we obtain? To make the answers concrete, we consider three scenarios. Basic information (approximate around 100°C) is that the S.H. of water is 1 cal/gm.°C; the S.H. of steam at constant pressure is 0.5 cal/gm.°C; and the S.H. of air at constant pressure is 0.25 cal/gm.°C. Further, the mass of 1 m3 of air at 15°C is 1.225 kilos, the latent heat of boiling of water is 540 cal/gm, and the molecular weights of water and air are 18 and 29 (avg.) respectively.

SCENARIO ONE - HEATING HOT AIR

A balloon contains a mass of M kilos of hot air at 110°C (say; this is typical). A quantity H kCal of heat is transferred to this air by a burner or the like. The hot air becomes hotter and expands. Question: how much additional lift is generated?

Initially, the volume of the air in m3 is (M/1.225) x (383/288), which equals 1.086M. This displaces 1.330M kilos of air, so the initial lift is 0.330M kilos, or 3.235M newtons.

When H kCal of heat are added to the mass M kilos of air which expands at constant pressure, its temperature rises by H / (0.25 M) degrees to reach (110 + 4H/M) °C, so its volume is now (M/1.225) x (383+(4H/M)/288), which equals 1.086M (as before) + 4H/(1.225 x 288) = 1.086M + H x 0.0113378. (the units have gone for a burton). The additional volume of air displaced is therefore 0.0113378 H, which weighs 0.111 H newtons. Or, to eliminate the decimals, for heating hot air, if 1 mCal of heat is added, the lift increase is 111.2 newton. In joules, addition of 1 mJ of heat gives 26.5 newtons extra lift.

SCENARIO TWO - BOILING WATER TO STEAM

A balloon contains a mass of M kilos of steam at 100°C, and a quantity of water at 100°C is being carried on board. A quantity H kCal of heat is transferred by a burner-boiler to this water, and boils some of it into steam which is injected into the envelope. Of course even though the boiled water (which remains aboard) has changed phase, its weight is still dragging down the balloon; but now the quantity of the steam lift gas is increased. Question: how much additional lift is available?

This heat vaporizes H/540 kilos of the hot water to steam, which has a volume of 0.0031547 H cubic meters. The additional air volume displaced (the water had essentially no volume) is therefore 0.0031547 H m3, which weighs 0.0038613 H kilos, i.e. 0.0378793 H newtons. Or, to eliminate the decimals, for steam being boiled, if 1 mCal of heat is added, the lift increase is 37.9 newtons. In joules, addition of 1 mJ of heat gives 9.0 newtons extra lift.

SCENARIO THREE - SUPERHEATING STEAM

A balloon contains M kilos of steam at 100°C, and it is assumed that there is no water trickling down the inside of the envelope (or that the added heat doesn't reach it within the time frame of this discussion). A quantity H kCal of heat is transferred to this steam by a heating device (superheater). The steam becomes hotter and expands. How much additional lift is generated?

Initially, the volume of the steam in m3 is (M/1.225) x (29/18) x (373/288), which equals 1.703M. This displaces 2.086M kilos of air, so the initial lift is 1.086M kilos, or 10.66M newtons.

When H kCal of heat are added to the mass M kilos of steam which expands at constant pressure, its temperature rises by H / (0.5 M) degrees to reach (100 + 2H/M) °C, so its volume is now (M/1.225) x (29/18) x (373+(2H/M)/288), which equals 2.086M (as before) + 2H x (29/18)/(1.225 x 288) = 2.086M + H x 0.0111849. The additional volume of air displaced is therefore 0.0111849 H, which weighs 0.1097238 H newtons. Or, to eliminate the decimals, for superheating steam, if 1 mCal of heat is added, the lift increase is 109.7 newtons. In joules, addition of 1 mJ of heat gives 26.1 newtons extra lift.

Obviously this figure is essentially the same as in Scenario 1, which indicates that the same mechanism (atmospheric pressure and internal energy) is responsible for the specific heat.... it would be the same for any gas.... fairly obviously, this discussion has plumbed the full depths of my knowledge of thermodynamics and the kinetic theory of gases, and found the bottom....



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