Waste Heat Management - sswelm/KSPInterstellar GitHub Wiki

In KSP Interstellar, waste heat is a very important mechanic. All of the reactors will produce very large amounts of waste heat, and even solar panels will generate some. If your waste heat level climbs too high, many components will shut off and become non-functional. It is therefor necessary to ensure you have enough radiators to dissipate your craft's waste heat.

Determining the Number of Radiators

How many radiators DO you need to cool your ship in space? If you want to be safe, you can add up the thermal power of all your reactors and put enough radiators to dissipate all of that. A 3 GW 3.75m fission reactor will require 10 huge unupgraded radiators in order to dissipate that much heat. In practice, however, you will almost never need that many radiators.

Cooling Thermal Rockets

If you are running thermal rockets, your reactor's waste heat will be dumped out the back along with the propellant, so you never need to worry about it as long as your engine is on. If you are running a fission or fusion reactor, when you shut the engine off the reactor will idle at 30%, and that heat must be dissipated in it's entirety. Fission is a pain to shut down and restart, so it's usually best to keep at least enough radiators to dissipate that much heat. For a 3 GW reactor, that equates to 900 MW, or only 3 huge radiators(4 recommended to prevent accidental shutdowns). That's a big difference!

If you are using fusion, it's even better. Fusion can be shut off when you don't need it as long as you have enough power to restart it. For fusion reactors, when run in D/T mode (which is the best mode for thermal rockets), they generate 80% thermal power, which is consumed by the thermal rocket, and 20% charged particles, which either gets consumed by a generator or accumulates and creates waste heat. 20% of 525 MW is only 105 MW, which can be dissipated by two medium unupgraded radiators. However, since it will only operate at that level of power while running the rocket and you can shut it off when you don't need it, you can actually get away with even less radiators. The exact amount depends on how long you want to run your thermal rocket and how long you give it to cool off in between burns. A handful of small radiators should be enough if you are only making short burns with long cool-down periods in between.

For antimatter thermal rockets, in theory you don't actually need radiators at all! Since they automatically throttle themselves to match the requirement of the thermal rocket nozzle, they will never produce any excess heat, except in the very beginning when they first fill up the thermal power. However, in practice, you will need some radiators, as you will need to power the antimatter containment devices. A small 62.5 cm or 1.25m fission reactor should suffice to power them, and you will only have to worry about dissipating the heat they create.

Cooling Electrical Generators

When you have an electrical generator, you will need at least one radiator for the generator to function. The efficiency of a generator is dependent on the temperature of the attached reactor and the temperature of the radiators. In short, the cooler your radiators, the more efficiency you get out of your generators.

If you want to figure out exactly how many radiators you need for a given efficiency, you will need to use several formulas. The efficiency of your generators is calculated with the formula E = (1 - Tc/Th) * Et. E is your actual efficiency, Tc is your radiator temperature, Th is your reactor temperature, and Et is your theoretical maximum efficiency.

To figure out how many radiators you need, first you need to figure out what temperature they should be at, which you can get by solving for Tc: Tc = (1 - E/Et) * Th. Once you have the temperature, you can use the Stefan-Boltzmann law, solved for area, to figure out how much radiator area you need: A = P / (o * e * T^4). P is power (measured in watts), and should be equal to your reactor's output multiplied by one minus the desired efficiency of the radiator (this is the amount of thermal power that becomes waste heat). A is the radiator area in m^2, which you are solving for. o is the Stefan-Boltzmann constant, and is equal to 5.6704e-8. e is the emissivity of your radiators, which in KSPI is always 1. T is the temperature that your radiator is at, in K.

Example: You have an unupgraded 1.25 fission reactor (40 MW @ 1674 K) and you would like to get at least 8 MW (20% efficiency, 32 MW of waste heat) of electricity out of it using an unupgraded generator (Theoretical 31% maximum efficiency).

Tc = (1 - E/Et) * Th

Tc = (1 - 0.2/0.31) * 1674

Tc = 594

You will need a radiator temperature of 594 K or lower. To achieve that temperature, you will need enough radiator area to dissipate your waste heat at that temperature:

A = P / (o * e * T^4)

A = 32,000,000 / (5.6704e-8 * 1 * 594^4)

A = 4,533.05

You will need 4,533.05 m^2 of radiator area. In order to achieve this, you will need 3 huge radiators (1600 m^2 area each for a total of 4800 m^2) or 12 medium radiators (400 m^2 each for a total of 4800 m^2)

Notice that it is difficult to get close to you theoretical efficiency using unupgraded parts. This is because unupgraded reactors have low core temperatures (necessitating lower radiator temperatures) and generators have low theoretical efficiency (resulting in more waste heat that needs to be dissipated). It should also be pointed out that these equations give the same result regardless of whether your radiators are upgraded. This is because upgrades only increase the maximum radiator temperature and nothing else. Upgraded radiators will reduce the amount you need to keep an idle reactor cool, but will not increase the efficiency of your electrical generators.