Power Plant Values Way Out of Whack

[PeterDonis]

This gripe really goes back even to Classic Traveller, but has anyone else noticed that the fuel consumption values for the various ship powerplants seem way out of whack? (The vehicle ones do too, even worse in some cases, but I'll stick to the ship ones for now since they're easier.) Basically, either all these powerplants are horrendously inefficient, or 1 EP equals some prodigious amount of MW of power.

For example, let's look at the fission powerplant, since that is supposed to model something that actually exists so we can look at known numbers. The table in the T20 rulebook says that 1 EP worth of fission powerplant "burns" 1 ton of fuel per month. Here's how I calculate how many MW of direct heat output that equates to:

1 ton = 14 cubic meters, and 1 month = 2.4E6 sec, so the fuel volume flow rate is 5.8E-6 cubic meters per second.

The density of fissionable fuel is about 16000 kg per cubic meter, and the energy released by fission is about 8.1E7 Megajoules per kg (if anyone wants to know where I got these numbers, I have a degree in Nuclear Engineering and I'm taking them from memory and a little bit of back of the envelope calculating, so they may be a little off, but I'm sure they are the right order of magnitude, which is enough for my argument). So the direct heat output in MW is fuel volume flow rate * density * energy per unit mass, or about 7.5E6 MW.

Now this is direct heat output from the power plant; it still has to be converted to usable power. Typical fission plants today convert about 1/3 of their heat output from the fission core into electrical power, so if we use the above calculation to set the value of an "energy point", we find that 1 EP = 2.5E6 MW. This is a *huge* value (the equivalent of 2500 of today's fission power plants, the kind that power cities). Also, if anyone remembers, it is 10,000 times the value that was given to 1 EP in MegaTraveller (250 MW).

Our alternative is to try to calculate the actual thermal efficiency of the above fission power plant by using the "known" value of 1 EP = 250 MW, in which case we come out with the answer that the fission power plant described in T20 is 10,000 times less efficient than today's reactors.

You can calculate numbers for the other types of powerplants (fusion and antimatter), and they all come out with the same general type of answer--either 1 EP is some huge amount of MW, or all these powerplants are horrendously inefficent (the factors range from about 200 to the 10,000 I calculated above).

I suppose that either nobody bothered to check this sort of thing (the numbers were just cribbed out of Classic Traveller), or it wasn't considered a big problem with game design. I'm just wondering if anyone else has noticed this and thought it was an issue worth fixing.

Peter Donis

 

[far-trader]

In CT, way back then, I just figured a large portion of the fuel was actually reaction mass since the magic reactionless thruster hadn't yet been proposed, or at least widely published.

After it became the standard to accept that (most) Traveller used reactionless thrusters I decided all that extra fuel was waste vented as coolant so things didn't melt.

Check the TML for heated (oops, i punned,

) debates on how ships are all glowing puddles of fusion waste or not. If the search engine ever gets fixed that is

So what does that do for your numbers (being a little too lazy to do the math myself). I find my handwaves are good enough for Traveller Tech, imtu, ymmv.

btw welcome aboard

 

[hunter]

Originally posted by PeterDonis:
I suppose that either nobody bothered to check this sort of thing (the numbers were just cribbed out of Classic Traveller), or it wasn't considered a big problem with game design. I'm just wondering if anyone else has noticed this and thought it was an issue worth fixing.

Peter Donis

The power systems are taken directly from CT for compatibility purposes.

Hunter

 

[tjoneslo]

Originally posted by PeterDonis:
You can calculate numbers for the other types of powerplants (fusion and antimatter), and they all come out with the same general type of answer--either 1 EP is some huge amount of MW, or all these powerplants are horrendously inefficent (the factors range from about 200 to the 10,000 I calculated above).

I suppose that either nobody bothered to check this sort of thing (the numbers were just cribbed out of Classic Traveller), or it wasn't considered a big problem with game design. I'm just wondering if anyone else has noticed this and thought it was an issue worth fixing.

The relative inefficency of the CT power plants is a long known issue. If you think the CT plants are bad, do the calculations for the MT ones (derived from Striker). Both TNE and T4 (and GURPS) power plants have much better efficiency.

Here is a counter challenge: Given the optimal efficiency powerplants you propose, how do you get rid of the waste heat? Remember, in space cooling is radiative only, no convection cooling.

 

[PeterDonis]

Originally posted by hunter:
The power systems are taken directly from CT for compatibility purposes.

Hunter [/QB]

That's what I thought but it's nice to have confirmation.

 

[PeterDonis]

Originally posted by tjoneslo:
The relative inefficency of the CT power plants is a long known issue. If you think the CT plants are bad, do the calculations for the MT ones (derived from Striker). Both TNE and T4 (and GURPS) power plants have much better efficiency.

Here is a counter challenge: Given the optimal efficiency powerplants you propose, how do you get rid of the waste heat? Remember, in space cooling is radiative only, no convection cooling. [/QB]

I remember doing similar calculations in the past for MT and T4 (can't remember doing TNE, but IIRC it was similar to T4 numbers-wise) and remarking on the wide disparity in efficiency. I was pretty sure I wasn't the first to notice it, but I hadn't seen any discussion of it before.

Re your counter challenge, most of the waste heat even in a high-efficiency plant would be carried away by the reaction products--which, by the way, you could argue is a form of "convective" cooling, so it's not true that all the waste heat in space has to be radiated away. *All* of the power output from the power plants we're considering starts out as heat--you put in a heat engine to convert as much of that as possible to some other form of usable power (like electricity). The energy that doesn't get converted to usable power stays in the reaction products, so in space all you'd have to do is eject them out a vent somewhere and problem solved. A small amount of heat would be absorbed by the vessel structure, but I don't see any reason why conventional cooling fins or something similar couldn't take care of that.

 

[tjoneslo]

Originally posted by PeterDonis:
I remember doing similar calculations in the past for MT and T4 (can't remember doing TNE, but IIRC it was similar to T4 numbers-wise) and remarking on the wide disparity in efficiency. I was pretty sure I wasn't the first to notice it, but I hadn't seen any discussion of it before.

It's not something that gets hashed out over and over again (like pirates, near-c rocks, nature of the nobility, or any of the other 2E+11 TML flamewar topics). If the efficiency were >100%, then perhaps there would be greater cause for discussion.

Re your counter challenge, most of the waste heat even in a high-efficiency plant would be carried away by the reaction products--which, by the way, you could argue is a form of "convective" cooling, so it's not true that all the waste heat in space has to be radiated away. *All* of the power output from the power plants we're considering starts out as heat--you put in a heat engine to convert as much of that as possible to some other form of usable power (like electricity). The energy that doesn't get converted to usable power stays in the reaction products, so in space all you'd have to do is eject them out a vent somewhere and problem solved. A small amount of heat would be absorbed by the vessel structure, but I don't see any reason why conventional cooling fins or something similar couldn't take care of that.

According to your figures, the 1 EP powerplant uses 5.7kg/sec of fuel, which produces about 4E8 Mj of heat energy. By my calculations (please correct if wrong), the 5.7kg is ejected at about 1.69E11 degrees.

This says to me the powerplant is not running at it's full possible efficiency.

Also: even if the power plant is running at this efficency, since you are dumping some significant fraction out the back, that part becomes irrelevant. We're converting about 1/3 (or some other fraction) of the 4E8 Mj into useful electricity. Which ultimatly turns into heat and needs to be disposed of.

So the handwave becomes that the "fuel" used by the power plant is not all burned/fissioned/fused. Most of it is used to dump waste heat out the back, both the excess heat from the power plant and the heat generated by the electricity produced by the power plant.

 

[PeterDonis]

Originally posted by tjoneslo:
It's not something that gets hashed out over and over again (like pirates, near-c rocks, nature of the nobility, or any of the other 2E+11 TML flamewar topics). If the efficiency were >100%, then perhaps there would be greater cause for discussion.

Yes, I'd be hollering a lot louder if that were the case.

According to your figures, the 1 EP powerplant uses 5.7kg/sec of fuel, which produces about 4E8 Mj of heat energy. By my calculations (please correct if wrong), the 5.7kg is ejected at about 1.69E11 degrees.

Here are the numbers I come up with for a fusion plant: it burns 1 ton of fuel per month per 3 EP produced (let's take the TL 15 plant as our example). This equates to a fuel flow rate of 5.8E-6 cubic meters per second. The density of liquid hydrogen is 70 kg per cubic meter, and the energy released by fusion is 4.7E8 MJ per kg, so we come up with a heat output of 1.9E5 MW. If 1 EP = 250 MW, this is about 800 EP worth of heat, of which only 3 EP becomes useful power out. The rest goes into heating up the reaction products--I haven't calculated the temperature they would reach, but your number is certainly not unreasonable.

This says to me the powerplant is not running at it's full possible efficiency.

Since only 3/800, or 0.375 percent, of the heat output is becoming useful power by the above calculation, this is indeed the case.

Also: even if the power plant is running at this efficency, since you are dumping some significant fraction out the back, that part becomes irrelevant. We're converting about 1/3 (or some other fraction) of the 4E8 Mj into useful electricity. Which ultimatly turns into heat and needs to be disposed of.

So the handwave becomes that the "fuel" used by the power plant is not all burned/fissioned/fused. Most of it is used to dump waste heat out the back, both the excess heat from the power plant and the heat generated by the electricity produced by the power plant.

But you're already going to have to dump your reaction products out the back anyway--by the above calculation, after the 1 ton per month per EP of power output has fused, you have (a) 1 EP of usable power, and (b) 1 ton (less a small amount of mass converted to energy in the fusion process, but it's less than 0.5 percent so I'm ignoring it) of reaction products at a temperature of at least a few billion degrees. You've *got* to dump that stuff out the back, and if you do, it's carrying away your waste heat anyway, so why would you dump a whole mess of unburned fuel out with it?

Another way of putting this is: you are basically saying that you have to make the powerplant inefficient in order to get rid of waste heat. But making the plant more efficient reduces the amount of waste heat, by definition (since more of the energy in the fuel is going into usable power). So by making things more inefficient, you're making the problem worse, not better. The only possible offsetting factor is that dumping out more unburned fuel with the reaction products lowers the temperature of the whole thing, but even with the ratio we calculated above (3 parts reaction products to 800 parts unburned fuel, approximately), the temperature is still going to be in the tens or hundreds of millions of degrees, so it won't be any easier to deal with. Better to make the thing more efficient and reduce the fuel flow rate and the waste heat as much as you can, IMO.

 

[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.

 

[Anthony]

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.

 

[PeterDonis]

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.

 

[PeterDonis]

One other note: my plasma physics profs would have gigged me for this, but in all the above calculations I've totally neglected the electrons, which at these temperatures would form their own population of particles that would soak up some of the heat. So actually in the power plant exhaust (whether it was just reaction products or included unburned fuel), there would be two applicable temperatures, an "electron temperature" and an "ion temperature". I've been calculating the ion temperature above as though it were the only one, but really it isn't. I've left this complication out because it makes no difference to the main conclusion of my argument, but I wanted to be on record in case anyone else noticed.

 

[Anthony]

Per unit volume, water is better (it may have 6x the average atomic mass, but you can fit 14x as much in a given volume); methane is even better (not as dense, but only 3.2x the average atomic mass). Per unit mass, hydrogen is certainly better.

 

[PeterDonis]

Originally posted by Anthony:
Per unit volume, water is better (it may have 6x the average atomic mass, but you can fit 14x as much in a given volume) <snip>

Not once it's a plasma you can't. In fact, not once it's a vapor you can't. All of the density ratios you're talking about only apply when things are liquid or solid, but the whole point is that *nothing* will stay liquid or solid very long under these conditions.

 

[Anthony]

Originally posted by PeterDonis:
Not once it's a plasma you can't. In fact, not once it's a vapor you can't. All of the density ratios you're talking about only apply when things are liquid or solid, but the whole point is that *nothing* will stay liquid or solid very long under these conditions.

Yeah, but it matters in terms of storing the stuff in your fuel tanks. Yes, you have 6x the mass flow rate. The fuel tank still lasts twice as long.

 

[PeterDonis]

Originally posted by Anthony:
Yeah, but it matters in terms of storing the stuff in your fuel tanks. Yes, you have 6x the mass flow rate. The fuel tank still lasts twice as long. [/QB]

Only if you insist on using extra material besides the reaction products to soak up waste heat, and my whole point is, that's pointless because the temperatures involved are still so high that all you can do with the stuff is eject it out into space. Why bother packing a whole extra tank of water just to reduce the temperature of your power plant exhaust from 170 billion degrees to "only" 450 million degrees (or higher if you use water, that number is for hydrogen)? The stuff is still way too hot to do anything useful with.

 

[mike wightman]

I agree that the fusion products will have to be ejected into space.
Just one question how much thrust will this impart on the ship.
My understanding is that if you have a constant flow of fuel and ejected fusion products then you have a fusion drive.
In CT, before the revision of B2 and intro of HG, a PP was only needed for for the M drive, J drives had their own (express boat and annic nova). This implied that the M drive was based on a fusion rocket. My take on this is the PP uses very little of the fuel for the fusion reaction most is used as coolant/drive plasma.
Interestingly TNE and T4 go back to this system, small amount of fuel for the PP per year, the rest is reaction mass.
One final thought, how would military ships mask their radiated heat signatures?

 

[PeterDonis]

Originally posted by Sigg Oddra:
Just one question how much thrust will this impart on the ship.

Not much. I calculate it as follows:

We have 6.2E-15 J per particle (I've translated back to J so I can use straight MKS units throughout this calculation), which, using kinetic energy = mv^2 / 2, gives a particle velocity (for tjoneslo's case where we are dumping extra fuel to absorb waste heat, since that also adds to the reaction mass) of 2.7E6 m/sec. That's less than 1% of the speed of light, so we're safe in ignoring relativistic effects and we can just use straight Newtonian formulas. Multiplying by the mass of each particle (1.7E-27 kg for hydrogen), we get a momentum per particle of 4.5E-21. If we multiply this by the number of particles ejected per second (which is the fuel flow rate of 4.1E-4 kg/s divided by the mass per particle), this gives us the rate of change of momentum of the ship due to the reaction mass being ejected out the back, which is the same as the force imparted to the ship; this comes out to be 1.1E3 Newtons. Dividing by the mass of the ship--we'll assume a 100 disp ton hull and 1 kg/m3 density, which gives 1.4E6 kg--we get the acceleration, which works out to 7.9E-4 m/s^2, or, converting to gees, 8.0E-5 g. As I said, not very much.

If you use my system for reducing power plant fuel consumption instead (don't dump any unburned fuel and slow down the fuel flow rate accordingly), then the math works out to 8.2E-6 g, which is even less.

Note that all of the above only takes into account the T20 numbers for fuel consumption for a power plant; if you really wanted to calculate a fusion rocket, it would be better to use the numbers for a HePlaR thruster in T4 or TNE (I believe both had a version of it) to see if they made sense (those rules required extra fuel consumption for the thruster, over and above what was required for the power plant). I'll leave that as an exercise for the reader.

One final thought, how would military ships mask their radiated heat signatures?

The power radiated is 1500 MW by my earlier calculations, which is of the same order of magnitude as the useful power output of the power plant, so for Traveller purposes, if your rules include an "electromagnetic masking" package, I would assume that would be intended to cover this. In the real world, I'm not sure how you would do it--the rules for the "EMM" package never really explained how it worked. Stealth aircraft today have ways of "masking" their IR signatures, but all that is is controlling in which direction the heat is emitted, so it doesn't get emitted in the direction of the enemy. A starship could do the same thing, but there would always be some direction from which you could see the heat signature. I don't know of a way of masking it completely--at least, not without buying into all the waste heat problems we've worked so hard to avoid.