Talking of heat transfer, He Who Shall Not Be Named has recently presented Rene's calculations that 'show' the PLSS needed to contain 48 kg of water to cool the astronauts 'on a J-type mission.'
You're kidding me. Do the terms "heat of vaporization" appear anywhere? If you can merely get the correct figure for water, its relatively high evaporative cooling potential becomes rather obvious.
Or is he claiming the suit isn't as well insulated as it is?
Or are the astronauts each operating at a 10 kW metabolic rate?
Inquiring (and morbid) minds what to know: how does someone get such bogus results time after time without simply running away with his tail between his legs as any normal human would do when he's been as soundly and repeatedly thrashed in public as he has?
This is the excerpt from NASA Mooned America. It's clear he has no understanding of heat exchangers. Some of the assumptions are staggering.
According to the authors of First On The Moon each PLSS was built to catch and disperse metabolic heat generated by the astronaut at an average rate of sixteen hundred British Thermal Units an hour.
Since a BTU equals .2928 watts we have a total of 368 Watts (René's arithmetic error.) This should be added to the Sun's heat value for a total heat input of 571 watts. However we should calculate the heat radiated by the shady side of the suit. Before proceeding we must determine a temperature for the air in the suit. The higher the temperature, the easier it is for the air cooler to do the job. Let's assume that their suits stayed at 100°F. Looking back to the Temperature Conversion chart we see that this temperature is 311° Kelvin which we need to know in order to use the Stefan-Boltzmann radiation equation. We must invert the original formula to look like this.
I (watts) = K
4 x ( A x e x a)
Thus we find that there are 80 watts being radiated. This must be subtracted from the 571 total watts, which leaves us with 491 watts.
To round out the numbers we add 9 watts for radios, pump heat, etc. for a total of 500 watts.
Since there are 860 calories per watt and, assuming we can work at 100% efficiency we must make enough ice to carry off 430,000 calories per hour. " In 4 hours that adds up to 1,720,000 calories
To lower the temperature of one gram of water one degree C requires the loss of one calorie of heat. Upon the formation of ice, a gram of water loses 80 calories. Therefore a temperature drop from 100° F (38° C) down to freezing (0° C) entails the transfer of 38 calories, and when that gram freezes it absorbs another 80 calories for a total of 118 calories per gram vented out the blowhole. If we divide that 1,720,000 calories by 118 we get 14,576 gms of water that we must eject. This is 14.6 liters, which equals .514 cubic feet. That would take up 1/4 of the PLSS's volume. The weight of this is 32 pounds on Earth, which is or 38 % of the total claimed weight.
Using an efficiency of 40 %, which is still high compared to most mechanisms, and a suit temperature of 80° F, we find that 23.78 liters of throw away water is needed. This is 52.3 pounds on Earth, 62 % of the PLSS's total weight and .839 cubic feet which is 40 % of the unit's volume.