# Calculating heat transfer coefficient from cooling rate of a block?

If I have a cube of metal with known surface area and time to cool from $$T_2$$ to $$T_1$$, determined experimentally. Can I calculate the convective heat transfer coefficient, h of this situation from this?

Seen as h is the rate of transfer per unit area per unit temperature difference it seems like it should be possible? However, I can't find an appropriate equation.

Can I calculate the convective heat transfer coefficient, $$h$$ of this situation from this?

It depends. If your material has a sufficiently high thermal conductivity $$k$$ then we can assume temperature gradients in the $$x$$, $$y$$ and $$z$$ directions to be negligible. Put slightly pedantically:

$$\frac{\partial T}{\partial x}=\frac{\partial T}{\partial y}=\frac{\partial T}{\partial y}\approx 0$$

In thermal analysis we often use the so-called Biot dimensionless number, $$\text{Bi}$$:

$$\text{Bi}=\frac{h L}{k}$$

where $$L$$ is a characteristic length (for a cube typically the length of the sides)

If $$k$$ is large then $$\text{Bi}\ll1$$. In those circumstances we can apply lumped thermal analysis, based on Newton's Law of Cooling:

$$\frac{\text{d}Q}{\text{d}t}=mc_p\frac{\text{d}T(t)}{\text{d}t}=-hA[T(t)-T_{env}]$$

Without going into details, this is a differential equation that solves to:

$$\ln\Theta=-\frac{t}{\tau}\tag{1}$$

where:

$$\Theta =\frac{T(t)-T_{env}}{T_0-T_{env}}\text{ and } \frac{1}{\tau}=\frac{hA}{mc_p}$$

$$\tau$$ is called the characteristic time. Here $$A$$ is the total surface area exposed to the environment, $$m$$ the total mass of the object and $$c_p$$ the specific heat capacity of the material.

By empirically plotting the LHS of $$(1)$$ versus time $$t$$, we can determine the gradient $$\frac{1}{\tau}$$, and thus $$h$$ (see illustration below). Use linear regression on a minimum of 3 data points (the more the merrier) to estimate $$\frac{1}{\tau}$$. • That's brilliant thanks, exactly what I needed. I calculated the k value both by your described method of plotting the graph and also by substituting values straight into the equations and both were nearly identical. The h value I obtained in the end seems reasonable, your help is greatly appreciated! Apr 12, 2020 at 0:05
• You're welcome!
– Gert
Apr 12, 2020 at 12:13