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:

Closed, unsteady:

$ 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 W_{s} 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.

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?