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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.

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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 $\frac{1}{\tau}$.

enter image description here

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  • $\begingroup$ 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! $\endgroup$
    – FEA42
    Apr 12, 2020 at 0:05
  • $\begingroup$ You're welcome! $\endgroup$
    – Gert
    Apr 12, 2020 at 12:13

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