# BPF: Simplify the appropriate form of the Energy Balance Equation

### Closed Systems:

In a closed system, one typically uses an integral balance, based on internal energy. Furthermore, steady closed balances make no sense (since then nothing at all would happen) so we have the following practical forms:

$U_{final} - U_{initial} + E_{{k}_{final}} - E_{{k}_{initial}}+ E_{{p}_{final}} - E_{{p}_{initial}} = \int_{t_{initial}}^{t_{final}} Q dt - \int_{t_{initial}}^{t_{final}} W dt$

if we consider $\Delta$'s to denote changes from initial to final ([ ]final - [ ]initial) and Q and Ws to be the heat and work over the "life" of the system, we will write this as:

$\Delta U + \Delta E_k + \Delta E_p = Q - W$

There are several "rules" to recall that can help in simplifying this further:

• U depends on chemical composition, phase, and temperature (only a little on pressure). If no reactions, phase changes or temperature changes occur, the change in U is zero! (except a little if the pressure changes a lot)
• Kinetic energy needs an acceleration (positive or negative) to change.
• Potential energy needs a change in position (height).
• If the system and surroundings are at the same T (or if we say that the system is (perfectly) insulated) then Q = 0 (the system is adiabatic).
• Work can only be done by moving parts (a compressive/expansion piston of rotating shaft) or by electrical or radiation energy.

### Open Systems:

In an open system, we can use either an integral (for semi-batch problems) or a differential balance (for continuous problems). For semi-batch problems (by definition unsteady), we must use the general form of the integral balance (see the balances lesson). For continuous problems, we will use a modified form of the general balance which uses enthalpy, rather than energy (so only shaft work is explicitly included):

$\frac{dH}{dt} + \frac{dE_k}{dt} \frac{dE_p}{dt} = \dot H_{in} - \dot H_{out} + \dot E_{{k}_{in}} - \dot E_{{k}_{out}}+ \dot E_{{p}_{in}} - \dot E_{{p}_{out}} + Q - W_s$

For a steady problem (now using $\Delta$'s to denote changes: [ ]out - [ ]in):

$\Delta \dot H + \Delta \dot E_k + \Delta \dot E_p = Q - W_s$

OUTCOME:

Simplify the appropriate form of the Energy Balance Equation

TEST YOURSELF

Write an equation for a steady state, open system that does not change height or speed. What if it is also adiabatic?