JimO

2003-Jan-25, 04:11 PM

In reviewing the aspects of HB hoaxing that can be verified in High Schools (and at home), I'm looking at the thermodynamics offered by 'self-taught physicist' Ralph Rene, and wanted to ask some basic questions.

In "NASA Mooned America" chapter 11 ("The LEM's Problem"), page 90-91 in my edition, he offers to calculate the equilibrium temperature of the LM on the lunar surface.

Solar heating is 1353 watts per square meter.

He assumes the LM sunlit area is 18 square meters.

"I have chosen an emissivity factor of 0.5 simply because that lies halfway between a perfect mirror and a perfect black body".

Thus he computes "12,168 watts per hour" (he frequently shows confusion over units).

Then he offers to compute the temperature at which the LM would be emitting this much heat (plus some from the crewmen and electronics). Using "the Stefan-Bolzmann formula", he computes a temperature of 120 deg C for the LM, at equilibrium.

"Did I miss something?" he asks. To start with, I'd suggest his absorption number (he states that absorption and emissivity values are the 'same number') of 0.5 is way too high.

What's the true 'equilibrium temperature' of the LM in space? Assume lighting angles as on the Moon.

In chapter 12 ("Blowholes of Sea & Space"), page 101, he gets even more amusing by calculating how the cooling water for the spacesuit sublimator was totally inadequate. He states that heat is rejected from the suit BY COOLING AND FREEZING WATER, so that each gram of water can account for so many calories from dropping to 0 deg C and then 80 calories for freezing (the 'heat of fusion', a term he does not use). He writes, "When that gram freezes it absorbs another 80 calories..."

Now, the physics I learned in High School is that water freezing 'gives off' heat to that amount (it resists cooling further until it is frozen solid). Rene seems to have it totally backwards -- the heat that a gram of water can dispose of is the calories absorbed when it rises from ambient to 100degC plus the heat of vaporization when it boils off. That value is about seven times the 'heat of fusion', so naturally a gram of water can cool a whole lot more than in the upside-down computations of our home-schooled 'physicist'.

Sure there is ice, since some of the water is 'over-cooled' by evaporation, but it then sublimates and absorbs heat just like it was supposed to. Any flecks of ice that get ejected with the water vapro stream are 'wasted' because they don't serve to cool the syste, -- sublimator or flash evaporator or water spray boiler, whatever it's called ("swamp cooler" in New Mexico).

I'm counting on High School physics teachers to be able to run through these numbers in a one-hour lecture, and to point to Rene's claims as a good example of 'intellectual over-reaching' that deserve laughs and giggles rather than credulity.

But first I wanted to run the numbers here to see if I've forgotten anything.

In "NASA Mooned America" chapter 11 ("The LEM's Problem"), page 90-91 in my edition, he offers to calculate the equilibrium temperature of the LM on the lunar surface.

Solar heating is 1353 watts per square meter.

He assumes the LM sunlit area is 18 square meters.

"I have chosen an emissivity factor of 0.5 simply because that lies halfway between a perfect mirror and a perfect black body".

Thus he computes "12,168 watts per hour" (he frequently shows confusion over units).

Then he offers to compute the temperature at which the LM would be emitting this much heat (plus some from the crewmen and electronics). Using "the Stefan-Bolzmann formula", he computes a temperature of 120 deg C for the LM, at equilibrium.

"Did I miss something?" he asks. To start with, I'd suggest his absorption number (he states that absorption and emissivity values are the 'same number') of 0.5 is way too high.

What's the true 'equilibrium temperature' of the LM in space? Assume lighting angles as on the Moon.

In chapter 12 ("Blowholes of Sea & Space"), page 101, he gets even more amusing by calculating how the cooling water for the spacesuit sublimator was totally inadequate. He states that heat is rejected from the suit BY COOLING AND FREEZING WATER, so that each gram of water can account for so many calories from dropping to 0 deg C and then 80 calories for freezing (the 'heat of fusion', a term he does not use). He writes, "When that gram freezes it absorbs another 80 calories..."

Now, the physics I learned in High School is that water freezing 'gives off' heat to that amount (it resists cooling further until it is frozen solid). Rene seems to have it totally backwards -- the heat that a gram of water can dispose of is the calories absorbed when it rises from ambient to 100degC plus the heat of vaporization when it boils off. That value is about seven times the 'heat of fusion', so naturally a gram of water can cool a whole lot more than in the upside-down computations of our home-schooled 'physicist'.

Sure there is ice, since some of the water is 'over-cooled' by evaporation, but it then sublimates and absorbs heat just like it was supposed to. Any flecks of ice that get ejected with the water vapro stream are 'wasted' because they don't serve to cool the syste, -- sublimator or flash evaporator or water spray boiler, whatever it's called ("swamp cooler" in New Mexico).

I'm counting on High School physics teachers to be able to run through these numbers in a one-hour lecture, and to point to Rene's claims as a good example of 'intellectual over-reaching' that deserve laughs and giggles rather than credulity.

But first I wanted to run the numbers here to see if I've forgotten anything.