Students will learn about Heat Engines, Pressure Volume diagrams and how to calculate ideal efficiencies of heat engines.

### Key Equations

\begin{align*} Q_{\text{in}} = Q_{\text{out}} + W +\Delta U \end{align*}

\begin{align*}U\end{align*} is the internal energy of the gas. (This is the first law of Thermodynamics and applies to all heat engines.)

\begin{align*} e_c = 1-\frac{T_c}{T_h} && \text{Efficiency of a Carnot (ideal) heat engine} \end{align*}

where \begin{align*} T_c \end{align*} and \begin{align*} T_h \end{align*} are the temperatures of the hot and cold reservoirs, respectively.

**adiabatic expansion** is a process occurring without the exchange of heat with the environment

**isothermal expansion** is a process occurring without a change in temperature

**isobaric expansion** is a process occurring without a change in pressure

### Guidance

Heat engines transform input heat into work in accordance with the laws of thermodynamics. For instance, as we learned in the previous Concept, increasing the temperature of a gas at constant volume will increase its pressure. This pressure can be transformed into a force that moves a piston.

The mechanics of various heat engines differ but their fundamentals are quite similar and involve the following steps:

- Heat is supplied to the engine from some source at a higher temperature \begin{align*} (T_h) \end{align*}.
- Some of this heat is transferred into mechanical energy through work done \begin{align*} (W) \end{align*}.
- The rest of the input heat is transferred to some source at a lower temperature \begin{align*} (T_c) \end{align*} until the system is in its original state.

A single cycle of such an engine can be illustrated as follows:

In effect, such an engine allows us to 'siphon off' part of the heat flow between the heat source and the heat sink. The efficiency of such an engine is define as the ratio of net work performed to input heat; this is the fraction of heat energy converted to mechanical energy by the engine: \begin{align*} e = \frac{W}{Q_i} && \text{Efficiency of a heat engine} \end{align*}

If the engine does not lose energy to its surroundings (of course, all real engines do), then this efficiency can be rewritten as

\begin{align*} e = \frac{Q_i-Q_o}{Q_i} && \text{Efficiency of a lossless heat engine} \end{align*}

An ideal engine, the most efficient *theoretically* possible, is called a Carnot Engine. Its efficiency is given by the following formula, where the temperatures are, respectively, the temperature of the exhaust environment and the temperature of the heat input, in Kelvins. In a Carnot engine heat is input and exhausted in isothermal cycles, and the efficiency is \begin{align*} \eta = 1 -\frac{T_{\mathrm{cold}}}{T_\mathrm{hot}} \end{align*}. In all real engines heat is lost to the environment, thus the ideal efficiency is never even close to being obtained.

The Stirling engine is a real life heat engine that has a cycle similar to the theoretical Carnot cycle. The Stirling engine is very efficient compared to a gasoline engine and could become an important player in today's world where green energy and efficiency will reign supreme.

- In a practical heat engine, the change in internal energy must be zero over a complete cycle. Therefore, over a complete cycle \begin{align*}W = \Delta Q\end{align*}.
- The work done by a gas during a portion of a cycle = \begin{align*}P\Delta\!V\end{align*}, note \begin{align*}\Delta\!V\end{align*}can be positive or negative.

- When gas pressure-forces are used to move an object then work is done on the object by the expanding gas. Work can be done on the gas in order to compress it.
- If you plot pressure on the vertical axis and volume on the horizontal axis, the work done in any complete cycle is the area enclosed by the graph. For a partial process, work is the area underneath the curve, or\begin{align*}P\Delta\!V\end{align*}.

#### Example 1

**Question**:A heat engine operates at a temperature of \begin{align*}650\mathrm{K}\end{align*}. The work output is used to drive a pile driver, which is a machine that picks things up and drops them. Heat is then exhausted into the atmosphere, which has a temperature of \begin{align*}300\mathrm{K}\end{align*}.

a) What is the ideal efficiency of this engine?

b) The engine drives a \begin{align*}1200\mathrm{kg}\end{align*} weight by lifting it \begin{align*}50m\end{align*} in \begin{align*}2.5 sec\end{align*}. What is the engine’s power output?

c) If the engine is operating at \begin{align*}50\%\end{align*} of ideal efficiency, how much power is being consumed?

d) How much power is exhausted?

e) The fuel the engine uses is rated at \begin{align*}2.7\times 10^6\mathrm{J/kg}\end{align*}. How many kg of fuel are used in one hour?

**Answer**:

a) We will plug the known values into the formula to get the ideal efficiency.

\begin{align*} \eta = 1 -\frac{T_{\mathrm{cold}}}{T_\mathrm{hot}}=1-\frac{300\mathrm{K}}{650\mathrm{K}}=54\% \end{align*}

b) To find the power of the engine, we will use the power equation and plug in the known values. \begin{align*} P=\frac{W}{t}=\frac{Fd}{t}=\frac{mad}{t}=\frac{1200\mathrm{kg}\times 9.8\mathrm{m/s^2}\times 50\mathrm{m}}{2.5\mathrm{sec}}=240\mathrm{kW} \end{align*}

c) First, we know that it is operating at \begin{align*}50\%\end{align*} of ideal efficiency. We also know that the max efficiency of this engine is \begin{align*}54\%\end{align*}. So the engine is actually operating at \begin{align*} .5\times 54\%=27\% \end{align*} of \begin{align*}100\%\end{align*} efficiency. So \begin{align*}240\mathrm{kW}\end{align*} is \begin{align*}27\%\end{align*} of what? \begin{align*} .27x=240\mathrm{kW} \Rightarrow x=\frac{240\mathrm{kW}}{.27}=890\mathrm{kW} \end{align*}

d) Since we know how much power is being put into the engine and how much energy is actually being used to lift the weight, we can determine how much energy is not actually being used to do work.

\begin{align*} 890\;\text{kW} - 240\;\text{kW}=650\;\text{kW}\\ \end{align*}

e) For this part of the problem, we need to convert the power being put into the engine into the amount of fuel being used(kW or kJ/s \begin{align*}\Rightarrow\end{align*} kg/hr).

\begin{align*} 890\;\text{kW}*\frac{1\;\text{kg}}{2.7*10^6\;\text{J}} * 3600\;\text{s/hr}=1186\;\text{kg/hr}\\ \end{align*}

### Watch this Explanation

### Time for Practice

- Calculate the ideal efficiencies of the following sci-fi heat engines:
- A nuclear power plant on the moon. The ambient temperature on the moon is \begin{align*}15 \;\mathrm{K}\end{align*}. Heat input from radioactive decay heats the working steam to a temperature of \begin{align*}975 \;\mathrm{K}\end{align*}.
- A heat exchanger in a secret underground lake. The exchanger operates between the bottom of a lake, where the temperature is \begin{align*}4 \;\mathrm{C}\end{align*}, and the top, where the temperature is \begin{align*}13 \;\mathrm{C}\end{align*}.
- A refrigerator in your dorm room at Mars University. The interior temperature is \begin{align*}282 \;\mathrm{K}\end{align*}; the back of the fridge heats up to \begin{align*}320 \;\mathrm{K}\end{align*}.

- A heat engine operates through \begin{align*}4\end{align*} cycles according to the \begin{align*}PV\end{align*}diagram sketched below. Starting at the top left vertex they are labeled clockwise as follows: a, b, c, and d.
- From \begin{align*}a-b\end{align*} the work is \begin{align*}75 \;\mathrm{J}\end{align*} and the change in internal energy is \begin{align*}100 \;\mathrm{J}\end{align*}; find the net heat.
- From the a-c the change in internal energy is \begin{align*}-20 \;\mathrm{J}\end{align*}. Find the net heat from b-c.
- From c-d the work is \begin{align*}-40 \;\mathrm{J}\end{align*}. Find the net heat from c-d-a.
- Find the net work over the complete \begin{align*}4\end{align*} cycles.
- The change in internal energy from b-c-d is \begin{align*}-180 \;\mathrm{J}\end{align*}. Find:
- the net heat from c-d
- the change in internal energy from d-a
- the net heat from d-a

#### Answers to Selected Problems

- a. \begin{align*}98\end{align*}% b. \begin{align*}4.0\end{align*}% c. \begin{align*}12\end{align*}%
- a. \begin{align*}1753 \;\mathrm{J}\end{align*} b. \begin{align*}-120 \;\mathrm{J}\end{align*} c. \begin{align*}80 \;\mathrm{J}\end{align*} d. \begin{align*}35 \;\mathrm{J}\end{align*} e. \begin{align*}-100 \;\mathrm{J}, 80 \;\mathrm{J}, 80 \;\mathrm{J}\end{align*}