# SL: A "mechanical" Energy Balance

### A "mechanical" Energy Balance

Now that we are comfortable with both the first and the second law, it is often required that we combine them. Consider an open system that is at steady state and undergoing a reversible process.

The first law yields:

$0 = -\dot n \left [ d(h + e_k + e_p) \right ] + \delta \dot Q_{sys} + \delta \dot W_s$

The second law yields:

$0 = \dot n ds + \frac{\delta \dot Q_{surr}}{T_{surr}} = \dot n ds - \frac{\delta \dot Q_{sys}}{T_{surr}}$

Eliminating $\dot Q_{sys}$ gives:

$0 = -\dot n \left [ d(h + e_k + e_p) \right ] + \dot n T ds + \delta \dot W_s$

Solving for $\dot W_s$:

$\frac{\dot W_s}{\dot n} = \left [ dh - T ds + de_k + de_p) \right ]$

Noting that for a reversible process:

$du = \delta q_{rev} + \delta w_{rev} = Tds - Pdv$

and that $dh$ can be written as:

$dh = d(u + Pv) = Tds - Pdv + (Pdv + vdP) = Tds +vdP$

combining this with earlier:

$\frac{\dot W_s}{\dot n} = \left [ vdP + de_k + de_p \right ]$

which, for 1in/1out gives:

$\frac{\dot W_s}{\dot n} = \int vdP + (e_{k_2}-e_{k_1}) + + (e_{p_2}-e_{p_1})$

##### Note:

This is related to the Bernoulli equation (when $W_s = 0$).

##### Test Yourself:

Calculate the work extracted from a turbine that is fed an ideal gas at 250 mol/s at 125 bar and a volume of 500 $cm^3$/mol. The final pressure is 8 bar. Take $c_P = \frac{7}{2}R$