ERFC

Application:

An application of the complementary error function, ERFC, can be found in the field of heat transfer and diffusion.

Imagine you have a semi-infinite solid, like a very thick metal plate, that is initially at a uniform temperature, T0​. At time t=0, you suddenly change the temperature of the surface of this plate to a new, constant temperature, Ts​. You want to find out how the temperature at any given depth x inside the plate changes over time.

This problem is a classic example of transient heat conduction in a semi-infinite medium. The solution to the one-dimensional heat equation for this scenario is given by:

$\frac{T(x,t)-T_s}{T_0-T_s}=\text{erf}\left(\frac{x}{2\sqrt{\alpha t}}\right)$

This equation is a powerful tool. It tells you the normalized temperature change at any depth x and time t by plugging in the relevant parameters.

Let's use a specific example:

Scenario: A large block of steel (thermal diffusivity α=1.17×10−5 m2/s) is initially at a uniform temperature of 20°C (T0​). At time t=0, one of its surfaces is instantaneously heated to 100°C (Ts​). We want to find the temperature at a depth of x=0.02 m (2 cm) after t=1 hour (3600 s).

First, we calculate the argument for the ERFC function, z:

$z=\frac{x}{2\sqrt{\alpha t}}=\frac{0.02}{2\sqrt{(1.17\times 10^{-5})(3600)}}\approx \frac{0.02}{2\sqrt{0.04212}}\approx \frac{0.02}{0.4105}\approx 0.04872$

Now, we look up the value of erfc(0.04872) from a table or use a calculator. The value is approximately 0.9449.

Using the formula:

$\frac{T(0.02,3600)-20}{100-20}=0.9449$

$T(0.02,3600)-20=80\times 0.9449$

$T(0.02,3600)-20\approx 75.592$

$T(0.02,3600)\approx 95.592^{\circ }\text{C}$

So, after one hour, the temperature at a depth of 2 cm is approximately 95.59°C.

We can create a table to show how the temperature changes over time at the same depth, using the ERFC function: