# SIB: Energy 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 energy within that volume (and transfer with surroundings) we get:

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

We denote e as the total energy per unit mass, and note that energy can be transferred to the material in the CV through both heat (thermal energy) flow and work. Therefore, we can write this in equation form as:

$\displaystyle{\frac{d}{dt}\int\int\int_{CV} e \rho dV = -\int\int_{CS} e \rho \left(\mathbf{v}\cdot\mathbf{n}\right ) dA +q +W}$

##### NOTATION:

- e, the total energy per unit mass, is comprised of mechanical (kinetic and potential) as well as internal (thermal) energy

- q represents the heat that is transferred to the material within the volume

- W represents any work done on or by the system (the chosen sign convention is that work done on the system is positive).

We note that the energy per unit mass may have an internal energy component as well as a kinetic and potential energy component, so that we can write it as:

$\displaystyle{e = \hat U + \frac{1}{2}v^2 + gz}$

We further note that the work can have several components:

$\displaystyle{W = W_s + W_\sigma + W_\tau}$

where Ws is the shaft work and the other two components ($\displaystyle{ W_\sigma}$ and $\displaystyle{W_\tau}$) correspond to stresses at the boundaries (both normal and shear). If we extract the pressure from the normal stresses we can write our work in the form:

$\displaystyle{W = W_s - \int\int_{CS}P\left(\mathbf{v}\cdot \mathbf{n}\right ) dA + W_\mu}$

where the pressure stresses now make up what is typically called the "flow work" term, and remaining viscous stresses are lumped into the term $\displaystyle{W_\mu}$, which is only non-zero if we have a non-zero shearing velocity on the control surfaces.

Combining these observations into a final form of the macroscopic energy balance equation, we get

$\displaystyle{\frac{d}{dt}\int\int\int_{CV} \rho\left(\hat U +\frac{1}{2}v^2 + gz\right ) dV = -\int\int_{CS} \rho\left(\hat U +\frac{1}{2}v^2 + gz + \frac{P}{\rho}\right ) \left(\mathbf{v}\cdot\mathbf{n}\right ) dA +q +W_s + W_\mu}$

##### OUTCOME:

Derive and use a macroscopic energy balance