I've seen that it's defined as $h_0 = h + \frac{v^2}{2}$, where $h_0$ is the total enthalpy per unit mass. I can see that the $\frac{v^2}{2}$ term probably has something to do with kinetic energy, but I am still really confused. Doesn't $h$ already account for flow work? And also, how can both sides of the equation have the same units if $h_0$ is "per unit mass"?

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    $\begingroup$ In chemistry, we don't usually throw things around, but if we will, they'll surely have an additional energy of $mv^2\over2$. $\endgroup$ – Ivan Neretin Aug 13 '19 at 10:05

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