# Calculating heat loss to surroundings from a cube suspended in air?

If you have a cube that is suspended in the middle of the air, how would you calculate the heat loss to its surroundings? For this example, I am assuming the following:

Dimensions: 1m×1m×1m

Material: Rough Aluminium

Temperature of Block: 200°C

Temperature of surroundings: 20°C

As this will be a combination of radiation, convection and conduction I thought I could calculate losses due to each individually then sum them?

$$q=\sigma Ae(T_1^4-T_2^4)$$

## Convection - Newton’s law of cooling

$$q=hA\Delta T$$

At this stage I reached a problem, how do I obtain $$h$$? the convective heat transfer coefficient. From my research it appears this isn’t possible to determine analytically as the fluid movement is complex. Instead an empirical approach is required.

Originally, I was going to assume natural convection for the block but it appears that presents a far more complex scenario than forced flow. I am mainly doing these calculations for comparison between different geometries so I could use forced flow conditions (there will be forced flow in the real world application but I haven’t determined the specifics yet and it will vary from face to face, I don’t want to involve these complications at this stage.)

It appears to obtain $$h$$, I might need to use Nusselt’s Number? The ratio of conductive to convective heat transfer

$$Nu=\frac{hL}{k}$$

It appears that $$Nu$$ is a function of geometry and there are tables to look up this value but I am unsure how that applies to a cube?

I know the Nusselt Number is a function of Reynolds and Prandtl but I am not sure how well this applies to an object in atmosphere as opposed to something like pipe flow?

## Conduction

From what I understand heat loss by convection involves some conduction. This is where the air meets the surface creating a no slip condition. On top of this is a boundary layer and the thermal boundary that is created has more heat transfer via conduction than convection.

Is this accounted for in Newton’s law of cooling? Or is there a separate equation or is it insignificant enough to be ignored?

## To summarise:

If I calculate heat loss due to convection, conduction and radiation independently can I sum them to obtain overall heat loss to surrounding?

How do I obtain $$h$$ for this scenario in order to calculate heat loss due to convection?

How do I calculate the heat loss due to conduction?

I’m a design engineering uni student who has only done a basic thermodynamics module and that was a couple of years ago, so I apologise for my lack of knowledge in advance. This is far beyond my scope.

• Is the cube and air in a gravitational field? – cms Jan 3 at 18:32
• @cms Yes they will be, it will be on the earths surface. – FEA42 Jan 3 at 18:57

The first thing to do is to try to get a ballpark estimate of the answer to this problem. According to Bird, Stewart, and Lightfoot, Transport Phenomena, typical values for heat transfer coefficients to gases in natural convection are 3-20 W/m^2K and in forced convection are 10-100 W/m^2K. I would use a value on the order of about 10 W/m^2K to start with in your situation.

The asymptotic (minimum) internal heat transfer coefficient for conductive heat transfer within the cube is going to be on the order of 2k/S, where S is the side of the cube and k is the thermal conductivity of the cube material. For aluminum, the thermal conductivity is about 200 W/mK, so, in the present case, the asymptotic internal heat transfer coefficient will be on the order of 400 W/m^2K. This is much higher than the external convective value, so the main resistance to heat transfer is going to reside in the external convective heat transfer boundary layer. So, to a crude approximation, the internal conductive heat transfer alluded to by Gert can be neglected. And the overall heat transfer coefficient can be taken as roughly 10 W/m^2K. Also, any attempts to improve the accuracy of the calculation should focus on the convective heat transfer at the outside surface of the cube.

I recommend doing the calculation for Newton's cooling with an external heat transfer coefficient of 10 W/m^2K and see what the cooling time come out to be.

Incidentally, in closing, I might mention that conductive and convective heat transfer in the gas surrounding the cube are not separate entities. Convective heat transfer is just conductive heat transfer that is enhanced by fluid flow near a boundary (which acts to increase temperature gradients and heat fluxes in the boundary layer).

• What is the S term? is this the distance from one wall to the opposing one so 1m in this instance? Do you have a reference or name for your 2k/S expression? – FEA42 Jan 4 at 14:33
• S is the side of the cube. It's just a rough order-of-magnitude estimate. You can look up typical asymptotic Nusselt numbers for transient conduction in solids in the literature (e.g., slabs, cylinders, spheres, etc.), as determined by the lowest eigenvalue. They will come out to be on this order of magnitude, or even a little higher. A good reference for transient heat conduction in solids is Carslaw and Jaeger. My objective here was just to establish that the internal conductive resistance in the aluminum cube is going to be negligible compared to the external resistance. – Chet Miller Jan 4 at 14:51
• Yes, the objective makes a lot of sense. That is the same purpose that the Nusselt Number serves isn’t it, showing the ratio of convective to conductive heat transfer. I am still unsure how a value is obtained for S or quite what it is? are there units or is it dimensionless? How would it differ say if the example was a cylinder of height 1m and diameter 1m? – FEA42 Jan 4 at 20:27
• Any tables that I can find only seem to relate Nusselt number to a flow regime and require the Reynolds and Prandtl number to actually obtain a value for Nusselt. I had a look through Carslaw and Jaeger’s conduction of heat in solids textbook but couldn’t seem to find the relevant parts. I can’t find any comparison between geometry and Nusselt, and even if I were to find a value how it is compared to the approximate coefficients of 200W/m^2k for Convection and 10W/m^2k. Would this correspond approximately to a Nusselt of 20? – FEA42 Jan 4 at 20:28
• is S the same as characteristic length? (volume/largest face size) @Chet Miller – FEA42 Jan 9 at 20:45

At this stage I reached a problem, how do I obtain h? the convective heat transfer coefficient. From my research it appears this isn’t possible to determine analytically as the fluid movement is complex. Instead an empirical approach is required.

The heat transfer $$h$$ coefficient is determined empirically (for a given hot object cold object pair). There are plenty of resources on the net that can help you find the value for air to rough aluminium. The $$h$$ value accounts for heat conduction in the transition layer.

Radiative losses only starts to be significant from some threshold temperature of the hot object. If so, then the overall heat transfer can be modeled as:

$$q=hA(T_1-T_2)+\sigma A(T_1^4-T_2^4)\tag{1}$$

where $$A$$ is the total surface area of the cube.

Neat, right?

Unfortunately it's not that simple

The problem is that as the cube sheds heat, its internal temperature $$T_1$$ changes (lowers).

The cube being a large object, heat conduction from the core of the cube to its outer edge really happens.

To account for the conduction, you need to apply Fourier's heat equation:

$$\frac{\partial T}{\partial t}=\alpha \Big( \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} + \frac{\partial^2 T}{\partial y^2}\Big)\tag{2}$$ with $$\alpha$$ the heat diffusivity. Or: $$T_t=\alpha \nabla^2 T$$

So to get the temperature time evolution you would need to solve the PDE $$(1)$$, using $$(2)$$ as a boundary condition and using an initial condition, like:

$$T_1(0)=200°C$$

• Should the emissivity coefficient be included in the Stefan Boltzmann equation? I see you have left it out and I was unsure of it when writing mine. – FEA42 Jan 4 at 14:34
• No, use $\alpha$ only. – Gert Jan 4 at 15:59
• @Gert I believe that FEA42 is right. Eqn 1 requires the emissivity of the Cu and the emissivity of air. The latter is essentially zero. So, the block radiates and the air does not. – Jeffrey J Weimer Jan 5 at 0:28