# Heat engines

Vast majority of electricity generated like this:

• chemical bonds are rearranged (burned), or nuclear bonds are rearranged (nuclear reactor) to generate heat
• heat is used to boil water and generate steam
• ...to drive an electric generator

• chemical bonds are rearranged to generate heat
• to generate heat
• heat is used to expand a gas; push a piston

• combustion (heat): coal/natural gas

Heat is *not* an intermediate stage with a minority of industrial processes:

• Photovoltaic panels
• electric motors use DC

What about trains? nuclear submarines? electric vehicles? the human body?]

It turns out that there are fundamental limits to how much energy can be converted from heat energy to mechanical energy (work).

### Changing heat $\Leftrightarrow$ work

The heat engine is a useful abstraction consisting of...

• a system,
• that does mechanical work,
• can exchange heat with constant temperature heat reservoirs.

In what sense is a gasoline engine a "heat engine"?

The simplest possible heat engine (shown schematically as a machine by the M) on the left has work done on it, and converts all the work to heat, which is dumped into a heat reservoir.

The process on the right involves a machine which absorbs heat and converts it completely into work. Such a heat engine...

• is not prohibited by the laws of mechanics,
• is not prohibited by the first law (conservation of energy),
• but has never been found to exist, and
• turns out to be prohibited by the second law of thermodynamics.

A more realistic heat engine that converts some of the heat it absorbs to work can be depicted as shown here. It absorbs heat from one reservoir at $T_2$, performs an amount of work $W$, and dumps (rejects) heat to a reservoir at a different temperature $T_1$, where $T_2 > T_1$.

The fraction of the input energy (only heat $Q_2$) that is converted $\to$ work $$\eta = \frac{W}{Q_2}.$$ This is called 'efficiency' and $\eta$ is pronounced 'eta'.

Using the first law, for a system that makes a complete cycle, $$\Delta U = 0 = Q_1 + Q_2 - W.$$

So the efficiency can also be written... $$\eta = \frac{Q_1+Q_2}{Q_2} = 1+\frac{Q_1}{Q_2}=1-\frac{|Q_1|}{|Q_2|}.$$ ...remember that the heats entering the system are $Q_2 > 0$, but $Q_1 < 0$.