Convective heat transfer coeficient equation

Question

I am attempting to model a situation, in which a sphere of some material of identifiable thermal properties is receiving heat at a constant rate. And that material is coming into equilibrium with the surrounding fluid, such that the heat it is receiving is equivalent to the heat that is going out via convection.

Almost everywhere that I have looked has given the same simple formula that the rate of heat transfer

$$\frac{\mathrm{d}Q}{\mathrm{d}t} = hA(T_1 - T_2)$$

with $A$ being the surface area in contact, $T_1$ being the temperature of the object and $T_2$ being the ambient fluid temperature. Then comes the heat transfer coefficient $h$, which I have not found any equation for calculating. This seems like it should be a very simple thing and I apologize for my lack of knowledge about this, but any response is appreciated.

Answer 1

The equation you state is a very general expression related to heat transfer, and basically everything goes into that constant. Convection of course is one thing, but what about radiative cooling (often important), diffusive cooling (might be important), and heat resistance, since the temperature of your object is not uniform.

All these contributions can be summed up into one heat transfer coefficients. This is very similar to summation of resistances in an electrical circuit. (tempature difference <-> voltage, heat flow <-> current and resistance <-> $1/h$)

You are specifically referring to convective cooling. When looking at convective cooling, you cannot live without Nusselt numbers

$$Nu=\dfrac{h L}{k}$$ where $L$ is a typical length scale and $k$ the thermal conductivity of the fluid.

Why you cannot live without them, is that all fluids in essence behave in the same way, also with respect to cooling. Therefore, the Nusselt number is a property of your geometry, and studied for a lot of geometries. They are specified in Nusselt-relations, normally as a function of the Reynolds number and Prandtl number. You calculate the Nusselt number, which gives you a good approximation of the heat transfer coefficient $h$. Here you have to realize that there is a difference between forced convection (fluid flow is driven by some external factor) and natural convection (i.e. buoyancy driven flow caused by the temperature difference itself).

Answer 2

For simple conductive heat transfer, h is $\kappa$, the thermal conductivity, divided by the length over which the temperature gradient exists. You can look this up for a given material. For convective heat transfer, this constant will depend on the details of your problem, including the dynamics of the liquid in question (can't simply look it up, you'd have to model the flow to determine how much heat it can carry away).

There are two important material constants for heat transfer. These are the thermal conductivity ($\kappa$) and the heat capacity ($c_p$). These are the values that you'd typically look up for a material. There is one derived parameter, called the thermal diffusivity ($\chi$), which is defined as the ratio $\kappa / c_p$.

What you're asking about follows from Fourier's Law:

$$q = -\kappa \nabla T$$

Here, $q$ is the rate of heat transfer ($W/m^2$). The power flowing through this surface ($Q$) is just the surface integral:

$$\int{q\cdot dA} = \dfrac{dQ}{dt} = -\kappa \nabla T A$$

Then, assume that $\nabla T$ is small such that it can be approximated as $\left(T_1 - T_2\right) / l$. Substitute this in the above equation and you get:

$$\dfrac{dQ}{dt} = \dfrac{\kappa A}{l}(T_1-T_2)$$

Adding to what's already been said about convective cooling, the Nusselt number is the ratio of convective heat transfer to conductive heat transfer. If the Nusselt number is << 1 in your system you can ignore convection. Otherwise you can look up the Nusselt number if it's previously been computed for a geometry, calculate the conductive heat transfer and use the two to get convective heat transfer if that's what you're really interested in.

Answer 3

Convection is very hard to model. It most likely changes with increasing temperature difference. (With very a low temperature differences there maybe no convection.) The best link I've found is here,

http://www.thermopedia.com/content/660/?tid=110&sn=7 (or go to Thermopedia and look under convection.)

Answer 4

Transport Phenomena by Bird, Stewart, and Lightfoot provides extensive information on correlations to determine the convective heat transfer coefficient in various situations. Convective heat transport in flow past a sphere is specifically covered. This is a very well-known situation, and has been analyzed extensively.