$ \frac{dU}{dt} + \frac{dE_{k}}{dt} + \frac{dE_{p}}{dt} = \dot W + \dot Q + $ "flow of energy with material"

$\frac{du}{dt} + \frac{de_{k}}{dt} + \frac{de_{p}}{dt} = \dot w + \dot q + $ "flow of energy with material (per unit material)"

Start with the work terms ...

$ \dot W = \frac{\delta W}{dt} = \frac{-d(PV)}{dt} = -P\frac{Adx}{dt} = -PA\frac{dx}{dt} = -PA\vec{V} = P\dot n v $

Here we are making use of the fact that $A\vec{V}$ is equal to the volumetric flow rate, so that dividing it by the specific volume (which we call $\hat V$ in the scribble notes, but $v$ here) gives us the mass flow rate so $\frac{A\vec{V}}{v}v = \dot n v$.

$ \dot w = \frac{\dot W}{\dot n} = Pv $

This relation holds for *all* input streams and output
streams (with a change in sign).

The way that energy "flows with material" is via that material's internal, kinetic, and/or potential energy (per unit material).

- Differential form of the first law (open system, temporary):

$ \frac{dU}{dt} + \frac{dE_{k}}{dt} + \frac{dE_{p}}{dt}
=$

$\sum_{in} \left [ \dot n \left(u + e_k + e_p \right ) \right
]_{in} - \sum_{out} \left [ \dot n \left(u + e_k + e_p \right
) \right ]_{out} +$

$\dot Q + \left [ \dot W_{shaft} + \sum_{in} \left (\dot n Pv
\right )_{in} - \sum_{out} \left (-\dot n Pv \right )_{out}
\right ]$

$ \frac{du}{dt} + \frac{de_{k}}{dt} + \frac{de_{p}}{dt}
=$

$\sum_{in} \left (u + e_k + e_p \right )_{in} - \sum_{out}
\left (u + e_k + e_p \right )_{out} +$

$\dot q + \left [ \dot w_{shaft} + \sum_{in} \left (Pv \right
)_{in} - \sum_{out} \left (-Pv \right )_{out} \right ]$

The **enthalpy** is defined as follows:

$h \equiv u + Pv$

- Differential form of the first law (open system, real):

$ \frac{dU}{dt} + \frac{dE_{k}}{dt} + \frac{dE_{p}}{dt}
=$

$\sum_{in} \left [ \dot n \left(h + e_k + e_p \right ) \right
]_{in} - \sum_{out} \left [ \dot n \left(h + e_k + e_p \right
) \right ]_{out} +$

$\dot Q + \dot W_{shaft}$

$ \frac{du}{dt} + \frac{de_{k}}{dt} + \frac{de_{p}}{dt}
=$

$\sum_{in} \left (h + e_k + e_p \right )_{in} - \sum_{out}
\left (h + e_k + e_p \right )_{out} +$

$\dot q + \dot w_{shaft}$

Flow work often cannot be "recovered" as useful.

Explain the utility of enthalpy, flow work, and shaft work

For an ideal gas, enthalpy is *also* only a function of
T:

$h = u + Pv = u + RT$

A common form of the open first law is for a steady state process that includes 1 inlet and 1 outlet, which reduces to:

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

Here $\Delta$ refers to $\left [ ()_{out} - ()_{in} \right ]$, rather than final-initial (as with closed systems).