# How does a radiator work fast enough on a space craft?

Basically, after some heat exchanger, a fluid is pumped to a radiator and then releases its heat to the void of space. But I'm having a hard time seeing how that is done fast enough. The fluid should radiate its heat immediately, so some heat is put back into the spaceship before it can reach the outer tubes of the radiator.

So now I'm going to provide my own "research" about this, because for one thing, I want to know if my simple understanding of thermodynamics is correct.

I understand heat is a scalar, and power is a vector. Heat is measured in joules and power is measured in joules/second (the watt). It's very annoying tho because there are many types of energy, so I will refer to them as heat energy/power or electic energy/power.

There are two types of thermal management. One is internal heat moving/rearranging, and the other is radiation.

Ultimately all heat must be lost as radiation, because space is a vacuum. (I know there are some evaporative cooling things even in space, but that expends a liquid and I'd rather stick with pure electrically powered thermal management).

Very simple example. Spherical spacecraft has a wall and inner air. Wall mass is 200 kg. Air mass is 2 kg. This craft could have a uniform temperature of 400 K. Or the walls could have a temperature of 401 K and the air is 300 K.

This would have the same total heat energy because 400x202 = 80,800, and 401x200 + 300x2 = 80,800. (I actually don't understand why heat energy, the joule, is not measured in K x kg for this purpose.)

Correct me if I'm wrong. If this is wrong so far then I'll always be totally lost unless it's corrected. I know total heat energy is measured in joules but I'd prefer to know how to calculate that based on mass and temperature.

Anyway, so those two examples were to illustrate that you can "rearrange" internal heat but you haven't actually reduced the total heat of the spacecraft. I made that example because I wanted something where the outer walls act like a hot radiator, but the internal air is cool and comfortable for people. (Ignoring that the walls will radiate inwards as well as outwards, for now.)

So now imagine a nuclear power plant inside the space ship lol. That's a lot of heat generation. If you are generating 1 GW of electrical power then you are also generating more than 1 GW of heat power (because all things have inefficiency).

Btw, it's been done before. The RORSATs. It's not 1 GW, but it is a nuclear power plant on an unmanned spacecraft.

I understand you can use heat exchangers, which then pump the fluid thru some insulated pipe, and this pipe leads to a radiator. The problem is, the fluid should be radiating away heat all the time.

So after the heat exchanger, the fluid is say 600 K. It starts radiating away heat immediately, into the surrounding pipe, which warms up and then transfers that heat to the surrounding air of the cabin.

Then the fluid reaches the radiator, where it is maybe 400 K. It's still radiating away heat, but slower now because of lower temperature.

So then what? Does it just sit there until it reaches some cold temperature like 250 K, then it comes back to help cool the overall temperature of the spacecraft? That sounds like it would take a long time. And the fluid needs to be pumped at some constant rate anyway.

I also understand the Stefan–Boltzmann law, where radiated joules per second is equal to $$\sigma T^4$$. To me this seems like you want the fluid to be as hot as possible, so it radiates heat away as quickly as possible.

But I don't understand how you can pump heat to concentrate heat somewhere. The heat exchanger simply "equalizes temperatures" so it's not going to be hotter than the initial ambient temp of whatever it's cooling. So if the steam is 600 K, the fluid in your heat exchanger will be 600 K or a little less after it passes thru the steam.

What I'd like to do is somehow pump heat out of something at any temperature. So the air might be 100 kg at 310 K, but I'd like to take 10 K out of it and into some small amount of fluid at some rly rly hot temperature like 1200 K. But a heat exchanger will never do that. The fluid in its pipe will reach 310 K at the most, and never get hotter.

Anyway, I'm lost now. I have looked at similar questions and answers (link, link), but they never seem to describe exactly how the radiator works or how heat transfer actually works. I feel like if I don't understand these at the component level, I'll never really understand it at all.

• @OrganicMarble 600 K object is going to radiate heat, whether it is inside or outside the spacecraft. When inside, the heat radiates into the spacecraft, first to its pipe. Apr 17 '21 at 21:14
• @OrganicMarble Undesirable, but will. We want radiation happening outside and away from the ship. That is desirable. But it will start radiating inside the ship immediately. That is undesirable. Apr 17 '21 at 21:24
• You misunderstand how heat exchangers work, for one thing. Your last point about the fluid only reaching 310k is incorrect. The heat exchanger has two closed loops, and changes pressure with a compressor. A refridgerator is a simple example, it moves heat from the inside to the outside, but the liquid temp is not directly related to the ambient air or inner temp, due to it changing pressure. Apr 17 '21 at 21:59
• @Innovine At the risk of embarassing myself further, i think you are confusing two devices. The vapor-compression cycle uses a compressor, and then the hotter fluid goes to a radiator. They are two different things. HX, i thought it was just a closed circuit running thru some air or steam (or whatever ambient), so it picks up that heat and "equalizes" temperature thru pure conduction. If there is some compression involved in a HX then i really need to see diagrams. (there is a pump for fluid flow, and some ppl argue about whether a pump primarily creates flow or pressure). Apr 17 '21 at 22:13
• heatpumpingtechnologies.org/market-technology/heat-pump-work the heat exchange part is performed by conduction, yes, but the difference in temperature in the two loops is maintained by the compressor/expansion cycle. So a "heat pump" has extra parts over a "heat exchanger". Apr 18 '21 at 8:12

You asked for corrections, and indeed I see three immediately. They involve: a material's emissivity; its specific heat; and the concept of back radiation.

The full-up expression of the Stefan-Boltzmann law for calculating radiated power is P = A$${\epsilon}{\sigma}T^4$$, where P is the radiated power, A is the effective radiating area, and $$\epsilon$$ is the material's emissivity. No materials have $$\epsilon$$ exactly 1 over all wavelengths but some get close, especially in the infrared. This is of relatively minor importance for this particular discussion.

In your example where you calculate the equivalence of total energy for two situations, you omitted consideration of the specific heats of the two different materials. Specific heat is defined as the amount of energy it takes to change a unit mass of a material by 1 K, and this varies hugely from material to material, not only among solids, liquids, and gases, but also among different solids, different liquids, etc. An example: the specific heat of liquid water is ~4200 J/kg-K, while for solid aluminum it is 900 J/kg-K. As a result, the total energies of the two situations are not equal. This is why heat energy is not measured in K-kg: multiplying K-kg by the specific heat converts it to Joules. An added complication: for gases, the specific heat at constant pressure is different from the specific heat at constant volume. The ratio of those two is important in rocket engine dynamics. For your example, it isn't possible to calculate the total heat energy without knowing which metal constitutes the chamber wall and which gases constitute the "air".

OK, back radiation. It is true that for a given area of a given material at a given temperature, the surroundings will not affect the amount of power radiated (as electromagnetic radiation) into those surroundings. But the amount of power radiated by those surroundings back at the radiator is critical to calculating the net power dissipated by the radiator. If the surroundings are uniformly at 0 K, then the net radiated power is equal to the power indicated by Stefan-Boltzmann. But if those surroundings are not uniformly at 0 K, then they also emit radiation, and some or most (or even all, in the academic case of $$\epsilon$$ = 1) of the part falling onto the radiator will be absorbed by the radiator. The net power dissipated by the radiator is the forward-radiated power minus the back-radiated power.

An example: a sphere of 1 m^2 surface area, with $$\epsilon$$ = 1 over all wavelengths, at 400 K, with surroundings at a uniform 0 K. This object radiates 1452 W, mostly in the IR. With zero radiation coming back from the surroundings, the net effective radiated power is just that 1452 W. Now let the surroundings be at a uniform temperature of 336.6 K over 4$$\pi$$ steradians. Those surroundings will radiate back at the radiator, with the radiator absorbing a total power of 726 W. The net effective power dissipated by the radiator will be 1452 W – 726 W = 726 W; the "warm" surroundings have reduced the net power dissipated by one half.

The practical situation is more complex. For instance, in LEO you often have Earth radiating at you from nearly 2$$\pi$$ steradians at its effective temperature of ~255 K; and the sun, occupying only ~6 x 10^-5 steradians of the sky but radiating at you at 5700 K — take that to the fourth power! Also you often have other parts of the spacecraft radiating back at the radiators, and the temperatures of those parts are affected by the heat the radiator is radiating. The back radiation can cause the temperature of the radiator to increase, which increases its net radiative power, but usually will also increase the temperature of the heat exchanger, which will increase the temperature of the other equipment upstream, and so on.

By the way, the variation of $$\epsilon$$ with wavelength is why radiators for space applications are often white: in the infrared their $$\epsilon$$ is fairly close to unity, but in the visible it is much lower, reflecting rather than absorbing most incident sunlight.

Oh, yeah: "How does a radiator work fast enough...?" Given the materials available and the anticipated surrounding environment (Earth, sun, etc.) you design the operating temperature, orientation, and area of the radiator to yield the desired net dissipated power.

• Another by-the-way: when you have a pipe inside a spacecraft at some temperature significantly higher than the desired temperature of the pipe's surroundings, you wrap the pipe with insulation, preventing radiative, convective, and most conductive heat transfer. Apr 18 '21 at 21:42
• Very nice explanations. A lot of it is more clear now but still one thing bothering me. Insulation protects against thermal conductivity...but not radiation, correct? What im worried about is the hot fluid in the pipe radiating energy into the pipe/insulation. If it's inside the spacecraft, that energy will make its way into the ambient conditions somehow. So even if u use a high heat capacity material, i think it will still eventually become as hot as the fluid? (second caveat: can u insulate for radiation too by internal lining of silver/white paint?) Apr 18 '21 at 21:50
• The insulation absorbs all the radiative heat load from the pipe in a very thin layer next to the pipe. It also has heat conducted into it directly from the pipe. The sum of those two (plus a very minor amount of convective heat transfer) give the total heat load in the insulator. The insulator is designed such that it has a very low thermal conductivity, so the amount of heat transmitted across its thickness is quite low. The small amount of heat arriving at the insulator's outer surface will increase the temperature of that surface just a bit, so it radiates away what little heat arrives. Apr 18 '21 at 21:56
• Remember that most materials — such as pipe materials, insulators, etc. — are essentially opaque to electromagnetic radiation in the IR and visible bands. Radiation from the fluid within a pipe is totally absorbed by the pipe's inner wall, arriving at the outer wall by conduction. Convection and conduction also deliver heat to the pipe's inner wall, and it makes it to the outer wall only by conduction. Apr 18 '21 at 21:58
• For back radiation, do we always assume e = 1? So a sphere of 1 m2, the sun effective temp is 5700 K, and the area is (6 x 10^-5)/4pi = 4.77 x 10^-6 square meters. Therefore P = 4.77x10^-6 x 1 x 5.67 x 5700^4 = 28.5 GW...gigawatts. What am i doing wrong? Apr 18 '21 at 22:34

I find it useful to start with the basics: steady state. In real practical systems, there's all sorts of transients and dynamics to deal with, but the fundamentals can be understood in the steady state case.

At the absolute basics of this, you have $$S=I-O$$. Storage rate is the difference between the input rate and the output rate. This is a very general equation that works for all sorts of things, like adding water into a bucket with a hole in it. For a satellite, the processes that add heat are any electrical activity (or nuclear activity, in your example) which generates thermal energy from electrical/nuclear potential energy, and the radiation of the sun warming you. Heat energy sinks (outputs) are simpler: all of the heat energy must eventually be radiated out into the cold of space. It may take a few steps, but in the end, that's where all of the heat goes.

For simplicity, I'll ignore the heating/cooling effects of the Earth. It will play a part, given its angular size and temperature, but whether its an input or an output depend on how hot or cold you are. We'll stick to the cold of space, which is obviously an output of heat, and the heat of the sun, which is obviously an input. Adding Earth effects in is left as an exercise for the reader.

So the first question is "is there a steady state?" and the answer is "yes!" For simplicity, consider an example with a steady-state electrical/nuclear heat load and a steady-state solar heat load. Thus the inputs are constant (it just makes our life easier, and avoids dynamics).

Let's say there's not enough radiator, which means the heat output of the radiator is less than that of the heat input from the sun/electrical/nuclear. This means the storage rate is positive, which is a fancy phrasing for "heating up." Now at first glance, it might heat up forever, but a reality eventually sets in. As you noted, the heat output of the radiator into the cold of space is proportional to $$T^4$$, the temperature of the radiating surface to the fourth power. Thus, as the radiator heats up, the radiated power grows quartically. This means we have more heat output into the cold of space, and less storage. Eventually we find a radiator temperature at which the outputs balance the inputs, and we achieve a steady state temperature.

Likewise, if the radiators are radiating too much energy, the outputs are more than the inputs and the storage rate is negative. This means the craft is cooling down.
Eventually this will cool the radiators down, mostly through conduction/convection. This causes the output to drop off. This causes the negative storage rate to move closer to zero. Like before, we achieve a steady state.

Indeed, if we look at the steady state case, the equations become simple. You have the inputs from the internal power sources, the input radiation from the sun, and the output radiation to space. The input radiation from the sun doesn't change much with temperature (it changes a little), and we've already stated that we're looking at the constant internal heat load case, so we can treat the inputs as constants. The output is a function of the temperature of the radiating surfaces to the 4th power, so we have $$S=I-O$$ $$0=i-kT^4$$ $$T = \sqrt[4]{i/k}$$

This is a key equation. Some of the answers will provide equations involving things like emissivities with which we can calculate numbers for $$i$$ and $$k$$. However, even without those calculations, we see the cold hard reality: Given any heat load ($$i$$) and radiator design ($$k$$), the temperature will approach a steady state radiator temperature.

I'm pedantic about calling it a radiator temperature, rather than just "temperature" in general because what matters is the temperature of the radiating surfaces, like the skin of the satellite and the surfaces of the radiators themselves (if any). The temperature of any other component isn't important for this steady state balance. What matters is the heat load of the system, and the temperature of the radiating surfaces. We can talk about the temperatures of the heating elements and fluids, and pipes all we want later. The rubber-meets-the-road temperature is the one of the radiators.

Now if all of your components can operate at the temperature of your radiators, its just a matter of trying to get the temperatures even. Using thick (heavy!) conducting components or liquids makes it easier to transfer heat via conduction to the radiating surfaces. Yes, the pipes will radiate into the body of the craft as well, but you don't mind, because they're at a temperature that your components can work with!

If that radiator temperature is too warm for your components, then you have a problem. Now you can't just conduct the heat away. We just showed that the temperature will increase until that steady-state temperature. We have to do something clever. You need a device which actively tries to maintain a difference in temperature between components.

These devices are very common planetside. Air conditioners and refrigerators do exactly that ! They expend power to make one side of the device colder than the other. A refrigerator expends power to make the inside colder than the outside. An air conditioner expends power to make the refrigerant liquid on one side colder than the refrigerant on the other side (refrigerators and air conditioners are basically the same).

So now, if your radiators have to be at a temperature t, the rest of your components can be "steady state" at some $$t-x$$. This is generally accomplished with a heat exchanger, like the one in an air conditioner. The air conditioner makes the refrigerant liquid cold (much colder than the air in the house), and then passes it through a heat exchanger, which is just a device that's designed to make heat transfer as fast as possible. This typically means lots of contact between the cold bits and the warm bits. It lets you leverage the coldness that your refrigerator/air conditioner was able to cause.

Now, there are inefficiencies here. The hot-side tubes do put heat energy into the cold side of the system. Insulation helps here, as you just want to slow the rate of heat transfer. The compressors/peltiers/etc do consume electric energy, which means the input heat load is higher. However, from a practical design perspective, we tend to be able to design things to radiate much more heat than they generate. Its easier for refrigerators and air conditoners, which can sink heat into the air via convection, but the same process works for sinking heat into the cold of space via radiation.

There's plenty of complications of course. The systems are certainly more dynamic than I made them sound. And while air conditioners and refrigerators are in principle the same as heat management in a satellite, the devil is in the details. However, I like using them as visualizations because they are conceptually the same. If a refrigerator bothers you, you can study how they work on the ground, without having to deal with all the strangeness of space, then look at how it works in space.

This is a partial answer hoping to address some points in the question that may help the OP to formulate it better.

I understand heat is a scalar, and power is a vector. Heat is measured in joules and power is measured in joules/second (the watt).

This is close, but power doesn't have to be a vector by itself; a 100 W light bulb (remember those) doesn't have much of a vector associated with it for example.

To get started working with heat flow, the analogies of electricity and and water flowing through permeable medium are helpful. The heat(power)/current/water are constantly moving from one end to the other of some conducting segment being studied.

First let's consider a long narrow conductors of various types where the flow is parallel. We know the direction so we don't have to think about vectors for the moment.

The temperature/voltage/pressure drop is equal to the heat(power)/current/water divided by the thermal/electrical/flow resistance.

Heat flow $$q$$ is given by a simplified form of Fourier's law:

$$q = -k \frac{\Delta T}{L}.$$

If the heat flow (power) $$q$$ through the long conductor of length $$L$$ is in watts/second and the temperature difference between the input and output end is $$\Delta T$$, then $$k$$ is the material's thermal conductivity.

The fact that the right side $$-k$$ is negative tells us that this is really a vector equation in disguise. We should really write this as

$$\mathbf{q} = -k \ \nabla T$$

where the boldface $$\mathbf{q}$$ is now the heat flow (power) vector in watts per square meter at some single point inside any shaped material structure, and $$\nabla T$$ is the gradient of the temperature at that point, which is also a vector.

In electricity we could write:

$$I = -\rho \frac{\Delta V}{L}$$

where $$rho$$ is the bulk electrical resistivity of the material, $$\Delta V$$ is the voltage drop (rather than temperature drop) and $$L$$ is still the length, and

$$\mathbf{j} = -\rho \ \nabla \phi$$

where $$\mathbf{j}$$ is now current density and $$\nabla \phi$$ is now the gradient of the electrical potential, also know as the electric field vector $$\mathbf{E}$$.

Because laminar flow of liquid interacts with the walls of its tube strongly compared to how we normally experience heat and electrical flow in everyday examples, the analogy is harder to extend, but for water flow through a permeable material we can use Darcy's law

$$Q = k \frac{A}{\mu L} \Delta p$$

where $$k$$ is the permeability of the material, $$\mu$$ is the viscosity of the liquid and $$\Delta p$$ is the pressure drop from one end to the other. In this case they drop the minus sign by convention, we're supposed to remember that the pressure gradient and the flow point in opposite directions. When they write the differential form the minus sign is intact:

$$\mathbf{q} = -k \frac{\nabla p}{\mu}.$$