If the Bi is determined to be sufficiently low in a transient problem, we can use the lumped equation to solve for the temperature evolution.

A good rule of thumb for something being bigger than 1 or smaller than 1 is that it varies by more than an order of magnitude. that is, 0.1 or less is smaller than 1 while 10 or more is greater than 1. In between, we have an "order 1" value.

In this case, we can write that the governing equation is

$\displaystyle{\frac{d(\rho c V T)}{dt} = \rho c V \frac{dT}{dt} = -hA(T-T_\infty)}$

As we will do with many problems moving forward, we would like to make this equation dimensionless prior to solving it. In order to do that we need to identify the relevant time and temperature scales (and length scales in problems where things vary by position).

Recall that for this type of problem we will have **two**
characteristic temperatures, the initial temperature T_{o}
and the bulk fluid temperature $T_\infty$.

For this reason, we use *both* temperatures to scale our
variable by defining:

$\displaystyle{\theta = \frac{T-T_\infty}{T_o-T_\infty}}$

Plugging this into our equation yields

$\displaystyle{\frac{d\theta}{dt} = -\frac{hA}{\rho cV}\theta}$

where $\displaystyle{\frac{\rho c V}{hA}}$ clearly has units of time. Taking this as our characteristic time yields

$\displaystyle{\tau = \frac{thA}{\rho c V}}$

which can be plugged into our equation to give its dimensionless form

$\displaystyle{\frac{d\theta}{d\tau} = -\theta}$

This can be solved providing we recognize that our initial
condition (i.e., T=T_{o} at t=0) has now become $\theta$=1
at $\tau$=0) to yield

$\displaystyle{\theta = e^{-\tau}}$

This solution is much the same for linearized radiation or a combination of linearized radiation and convection as our external heat transfer mechanism(s), provided we replace our h in the $\tau$ definition.

Use the "lumped" equation to solve "1D" transient heat transfer problems