homework and exercises - How fast cool down a tank of water?

I have a tank of water (2000 liter). This is a 1m length 1m width 2m high tank, made of steel. I heat it up with my furnace to e.g. 50 ºC. I would like to know how quick cool down to 20 ºC (this is my environment temperature) without insulation and with insulation. I do not need accurate calculation, an estimated value/method (minute or hour resolution) is enough for me. The type of the insulation should be a parameter in the equation, because I need to test other materials.

Is it possible to calculate this with pure math or I need to collect some sample, or is it an impossible project?

Answer

A conservative (lowest) estimate can be obtained through lumped thermal analysis. It basically assumes that the temperature of the cooling object is uniform (no $x$, $y$ or $z$ temperature gradients) during cooling. The alternative is a mathematical nightmare where I won't take you. I'll also ignore radiative losses, which are minor at these temperatures.

Lumped thermal analysis allows to apply Newton's law of cooling, which defines the rate of cooling as a rate of heat energy ($Q$) loss:

$$\frac{dQ}{dt}=kA[T(t)-T_0]$$

$k$ is a material constant, the so called heat transfer coefficient (see below the fold), depending on the insulation (you'll find estimates on many engineering sites, depending on insulating material), $A$ the total surface area of your tank, $T_0$ the environment temperature and $T(t)$ the temperature (assumed uniform) inside the tank.

During an infinitesimal drop in temperature $dT$ of the tank:

$$dQ=-mcdT(t)$$

Where $m$ is the mass of the tank and $c$ the specific heat capacity of water.

So, dividing both sides by $dt$, we obtain:

$$\frac{dQ}{dt}=-mc\frac{dT(t)}{dt}$$

And with the equations above, we get:

$$-mc\frac{dT(t)}{dt}=kA[T(t)-T_0]$$

This differential equation separates into variables easily:

$$\frac{dT(t)}{T(t)-T_0}=-\alpha dt$$

Where:

$$\alpha=\frac{kA}{mc}$$

Integrating gives:

$$\int_{T_1}^{T_2}\frac{dT(t)}{T(t)-T_0}=-\alpha \int_0^t dt$$

Or:

$$\ln \frac{T_2-T_0}{T_1-T_0}=-\alpha t$$

Where $T_1$ is the starting temperature and $T_2$ the end temperature. Note however that when you set $T_2=T_0$ then $t=+\infty$. This is normal and a consequence of Newton's law of cooling.

So you need to set $T_2$ at a more realistic value, like 95 % of $T_0$, to get some kind of an estimate for your cooling time.

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Heat transfer coefficient $k$:

In the case of an insulated tank, the heat transfer coefficient is calculated from:

$$\frac{1}{k}=\frac{1}{h_{surf}}+\frac{\delta_{ins}}{\lambda_{ins}}$$

Here $h_{surf}$ is the convective heat transfer coefficient at the surface of the insulation and $\lambda_{ins}$ the heat conductivity of the insulation. $\delta_{ins}$ is the latter's thickness.