Originally posted by Anthony:

** </font><blockquote>quote:</font><hr />Originally posted by tjoneslo:**

** Do this calculation the other way. Assume the power plant works at 100% efficency. It uses 1.27 kg of hydrogen per month to produce 1EP. It uses the remainder of hydrogen (980 kg) as a heat sink then dumped. Which comes out at about 6 million degrees. Of course, Hydrogen is about the worst product you could use for a heat sink like this. A dton of Iron with 600,000,000J heats about 14 degrees. In a month. **

**Um...1 dton of iron = ~110 tons.**

Specific heat of iron = 0.44J/gK

Energy to heat 1 dton by 1K: 49 MJ

1 EP = 250 MJ/sec, so the iron will heat by 5 degrees per second. It will be vapor in minutes.

Hydrogen, because it has a low atomic mass, is actually a decent choice if you want to avoid high temperatures, though water, at 6x the average atomic mass and 14x the density, is better (iron, at 56x the average atomic mass and 56x the density, is not better). On the other hand, hydrogen may be relatively easy to control or something. </font>[/QUOTE]First of all, tjoneslo, you're right that I did the calculation the wrong way--I didn't take into account your suggestion about using extra unburned fuel to soak up waste heat. Serves me right for trying to post too late at night.

Anthony's value for the specific heat of iron only applies until it vaporizes; then it becomes lower, but it only takes a few more minutes for it to become plasma, where the idea of specific heat as you apply it at room temperature is really meaningless. When you're dealing with energies this high, it's better just to think of temperature as average kinetic energy per particle. So the best "heat sink" at these temperatures is the one that has the maximum number of particles per unit mass--i.e., hydrogen. At lower temperatures, when things are liquid or solid, intermolecular bonds come into play and make the calculation of heat capacity per unit mass much more complex, but at millions of degrees none of that stuff matters.

So here's what I come up with for average energy per particle, calculating the right way. First of all, as I said before, if you assume that the power plant is 100% efficient, then you have *zero* waste heat--that's the definition of waste heat. If the power plant is 100% efficient, then every MW it produces goes directly to useful work somewhere. So we have to assume that the portion of the fuel that does get "burned" does not get converted to useful work with 100% efficiency.

Also, I misspoke earlier, I said I was using the TL 15 power plant, but I quoted 3 EP per ton of power plant when it should have been 2. So let's start fresh:

1 ton of power plant, assumed to use 1 ton of fuel per month to produce 2 EP of useful power output. Let's suppose that the power plant is 25% efficient at converting the energy of the fuel it burns into useful power output; that means that in order to produce 2 EP of output, it has to burn enough fuel to produce 8 EP, or 2000 MW, of heat energy from the fuel. That tells us how much fuel actually has to be "burned" to supply power. We'll assume the rest of the fuel gets dumped, unburned, to help soak up waste heat.

How much waste heat will there be? We're supplying 2000 MW of heat to produce 500 MW (2 EP) of useful power, so we have 1500 MW left over to dispose of as waste heat. The numbers I gave earlier are enough to calculate that this requires 4.2E-6 kg per second of fuel to be burned in the power plant, out of a total of 4.1E-4 kg per second total fuel flow (that's the number that equates to 1 ton per month). So the amount of fuel burned is negligible and we'll ignore it and just assume that 4.1E-4 kg per second of hydrogen is available to soak up 1500 MJ per second of waste heat. Hydrogen has 6.0E26 atoms per kg, so our number for average energy per particle (per atom) comes out to be 6.2E-21 MJ. Converting this to a temperature using Boltzmann's constant, which is 1.4E-29 MJ per Kelvin, we come up with a temperature of 4.5E8 degrees K (450 million degrees).

We can compare this to the number we would get if we did things my way and slowed down the fuel flow so we are only supplying enough fuel for fusion and it's all getting burned. The fuel flow in this case we already know, 4.2E-6 kg/sec, and now the heat is being soaked up by helium, which only has 1.5E26 atoms per kg, so the final number turns out to be 1.7E11 K, or 170 billion degrees.

Now it's true that the "slow fuel rate" temperature is several orders of magnitude higher than the "fast fuel rate, use excess fuel to soak up waste heat" temperature, but both of them are so high that I don't think it matters much. Either way practically all of that heat is going out into space, and the actual amount of power being rejected (1500 MW in each case) is small enough that it's reasonable to assume it could be dealt with in this fashion. In short, no matter what you do, you are going to eject out of your ship whatever ends up holding the waste heat from your fusion power plant, because it's literally too hot to handle any other way, so why waste extra fuel when the temperature is still going to be far too extreme to do anything else?

Also, this calculation should make it clear why iron wouldn't be a good idea; the number of atoms per kg of iron is 14 times smaller than for helium (56 times smaller than for hydrogen), so your temperatures would be proportionately higher still. Water wouldn't be as bad as iron, but still not as good as hydrogen or helium--at these temperatures it would be a mixture of hydrogen and oxygen, and the oxygen's larger atomic mass (i.e., smaller number of atoms per kg) wouldn't help.