# DBE: General Thermal Energy (Heat) Balance

For our balance, we will choose an arbitrary fixed (in space) control volume. Thus, it is an open system (i.e., one through which mass may flow). Accounting for heat within that control volume (and exchanged with its surroundings), we get:

 rate of accumulation of heat in CV = net rate of heat transport into CV (by flow) + net rate of heat transferred to material in CV + net rate of heat "generation" in CV

It is useful to discuss why we must now include a "generation" term. This term originates from the fact that heat can be "made" from other forms of energy (friction, chemical reactions, electrical resistance, etc.). A very important point therefore is that heat is not conserved!

In writing this mathematically, we use cT to represent the thermal energy per unit mass, where c is the heat capacity of the material. We then get:

$\displaystyle{\frac{\partial}{\partial t}\int\int\int_{CV} \rho c T dV = - \int\int_{CS} \rho c T (\mathbf{v}\cdot \mathbf{n}) dA - \int\int_{CS} (\mathbf{\frac{q}{A}}\cdot \mathbf{n})dA + \int\int\int_{CV} \dot q dV}$

##### NOTATION:
• - Q is the flux of heat to or from the material within the CV (the sign is chosen such that heat flowing into the material is positive.
• $\displaystyle{\dot q}$ is the volumetric rate of thermal energy generation.

Now for a little bit of math wizardry...

There is a theorem attributed to Gauss which states that a surface integral of a quantity can be written as a volume integral of the derivative of that quantity. Using this idea (and rearranging a little), we can write the heat balance instead as:

$\displaystyle{\int\int\int_{CV} \left \{ \frac{\partial}{\partial t}(\rho c T) + \nabla \cdot (\rho c T \mathbf{v}) +\nabla \cdot (\mathbf{\frac{q}{A}}) - \dot q \right \} dV = 0}$

As we chose an arbitrary control volume, in order for this integral to be equal to zero for any choice of CV, the integrand must be zero. So, we now have a differential heat balance equation:

$\displaystyle{ \frac{\partial}{\partial t}(\rho c T) = - \nabla \cdot (\rho c T \mathbf{v}) -\nabla \cdot (\mathbf{\frac{q}{A}}) + \dot q }$

The best way to prove this "zero" thing to you is by contradiction:

### Simplifying the GTEB

At the simplest level there are four assumptions that we can make that effect the terms within the General Thermal Energy Balance (GTEB) Equation:

• Steady State -- this assumption clearly affects the first term on the left-hand side (d/dt = 0); if it is not true we should try the Biot number to see if "lumped" is a more appropriate equation to solve
• No Flow -- this assumption is valid for solids or for truly stagnant fluids, it makes the first term on the right-hand side of the GTEB become zero
• No Generation -- this assumption requires that we have no chemical reactions, electrical current, frictional or mechanical work, etc. that might be turned into thermal energy, if true it makes the last term on the right-hand side become zero
• Constant Properties -- this is the trickiest assumption because it requires that we envision having only a small-to-moderate overall temperature span within our material of interest (particularly difficult to assess when we have generation), if it holds true we pull the density, heat capacity, and conductivity (in the function Q) out of all derivatives
##### OUTCOME:

Simplify the general thermal energy balance equation (identifying assumptions)

##### NOTE:

In order to "set up the equation", one needs to not only write and simplify the appropriate form of this differential equation, but also include the relevant boundary conditions!