# I Efficiency of engine when non-boundary work done is done?

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1. Jul 9, 2016

### Soren4

The thermal engine efficiency is defined as
$$\eta = \frac{W_{\mathrm{produced}}}{Q_{\mathrm{absorbed}}}\tag{1}$$
A more general definition of efficiency (not only for thermal machines is) $$\eta = \frac{\mathrm{Work \, produced}}{\mathrm{Energy \, absorbed}}\tag{2}$$
But suppose that, in one of the processes that takes place in the cycle of an engine that works with an ideal gas, work is supplied to the gas, for istance with a fan delivering a certain power. That work is not a "boundary work" but it is supplied to the system by the fan.

That work of the fan (say $W_{\mathrm{fan}}$) is a form of energy that must be supplied in order to make the engine work, so should it be taken into account in the efficiency? That is should I write, in that case, $$\eta = \frac{W_{\mathrm{produced}}}{Q_{\mathrm{absorbed}}+W_{\mathrm{fan}}}\,?\tag{3}$$

It makes sense but often the definition of efficiency is just $(1)$, the only thing in denominator is heat and not work. So is $(3)$ the correct form for efficiency for such situations?

2. Jul 12, 2016

### jambaugh

I think such auxiliary work components are taken out of the W in the numerator. They would be subtracted out of production work because they are a form of non-entropy energy or free energy.

I mean imagine the heat just passed on through and you hooked the fan motor to the heat engine's output shaft. You're getting gross output work and you can scale it up as high as you like using a beefier fan motor so the ratio as you write it can be arbitrarily close to unity. But if you subtract out the input work you get the net work of zero which I think better describes this extreme case.