Engines and Heat Pumps

Engine Cycles

A device which converts heat into useful work is called a heat engine. A heat engine cycles continuously through a series of thermodynamic processes. Two types of engines are shown below.

Four Stroke Petrol Engine

The idealised four stroke petrol engine cycle, also known as the Otto cycle uses a petrol-air mixture as its working substance. The p-V diagram for the cycle is shown below:

petrol_cycle

The net work done by the working substance in one cycle is equal to the
area enclosed by ABCD. This is called an indicator loop.

Idealised Diesel Engine Cycle

The working substance for an idealised diesel engine cycle is diesel fuel. The p-V diagram for this cycle is shown below:

diesel_cycle

Engine Efficiency

The theoretical efficiency of a petrol engine is around 58% while for a diesel engine it is around 65%. However these theoretical efficiencies are not attained due to frictional effects, turbulence and the composition of air in the chamber.

Compression Ratio

A diesel engine has a higher theoretical efficiency due to a greater compression ratio of $1:20$ compared to $1:9$ for a petrol engine. The compression ratio is the ratio of the volume enclosed in the cylinder at the start of the compression stroke to the end of the stroke.

Real Engine Cycles

The p-V diagrams above represent theoretical engine cycles. In real engines:

Input Power

The input power of the engine can be given by:
$$ P_{\text{input}} = \text{calorific value of the fuel} (Jkg^{-1}) \times \text{fuel flow rate} (kgs^{-1}) $$
The calorific value is the energy released by the complete combustion of $1kg$ of fuel. The fuel flow rate depends on the state of the engine, how the car is being driven and the type of engine, among other factors.

Indicated Power

The indicated power of an engine is the theoretical capability of the power an engine, if frictionless motion and no unwanted heat transfer is assumed.

$$ P_{\text{ind}} = \text{area of p-V loop} (J) \times \text{number of cycles per second} (s^{-1}) \\ \times \text{number of cylinders} $$

The indicated power can be determined using an indicator diagram, by dividing the net work done by the time taken for one cycle. It is important to note that the power cycle occurs once every two revolutions.

Output Power

The output power or brake power is the power delivered directly to the crankshaft by the engine. It is also referred to as brake horsepower ($bhp$) where $1~bhp = 746W$. It can be calculated with:
$$ P_{\text{out}} = T\omega $$
Where $T$ is the torque ($Nm$) and $\omega$ is the angular velocity ($rad~s^{-1}$).

Frictional Power

As the indicated power is greater than the output power, the difference between the indicated power and the output power is the power loss due to friction in the engine:

$$ P_{\text{friction}} = \text{indicated power} - \text{output (brake) power} = P_{\text{ind}} - P_{\text{out}} $$

Efficiency of an engine

The mechanical efficiency, $\eta$ of an engine can be calculated by dividing the output power by the indicated power.
$$ \eta = \frac{P_{\text{out}}}{P_{\text{ind}}} $$

The thermal efficiency, $\epsilon$ can be calculated by diving the indicated power by the input power.
$$ \epsilon = \frac{P_{\text{ind}}}{P_{\text{input}}} $$

The overall efficiency of an engine can be calculated by finding the product of the mechanical efficiency and thermal efficiency.

$$ \text{overall efficiency} = \eta \times \epsilon = \frac{P_{\text{out}}}{P_{\text{input}}} $$

Heat Engines

A heat engine converts heat into useful work. The elements of a heat engine are shown below. Heat ($Q_{\text{in}}$) is taken in from the hot reservoir, work $W$ is done and heat ($Q_{\text{out}}$) is released to the cold reservoir.

heat_engine

Second Law of Thermodynamics

The second law of thermodynamics states that the efficiency of any process for converting heat into work cannot approach $100\%$. This means it is impossible for any heat transfer from a high temperature source to produce an equal amount of work.

For a theoretical engine , $\Delta U = 0$ therefore the work done by the heat engine is given by:
$$ W = Q_{\text{in}} - Q_{\text{out}} $$

A reversible engine is one which can also be operated in reverse. The thermal efficiency of a heat engine can be given by the work done in one cycle divided by the heat transferred from the high temperature source:

$$ \epsilon = \frac{W}{Q_{\text{in}}} $$

As the efficiency depends on the temperature of the high temperature source and the temperature of the low temperature sink:

$$ \frac{T_{H}}{T_{C}} = \frac{Q_{\text{in}}}{Q_{\text{out}}} $$

The maximum theoretical efficiency:
$$ \epsilon_{\text{max}} = \frac{T_{H} - T_{C}}{T_{H}} $$
Where $T_{H}$ is the temperature of the hot reservoir ($K$) and $T_{C}$ is the temperature of the cold reservoir ($K$).

As $T_{H} > T_{C}$, the only theoretical way of achieving a maximum theoretical efficiency of $100\%$ is for $T_{C}$ to be at absolute zero or for $T_{H} \rightarrow \infty$ which are both impossible. In order to maximise the efficiency, the temperature of the source must be high as possible and the temperature of the sink as low as possible.

Reversed Heat Engine

A reversed heat engine is one where the work is done on the working substance such that heat is taken in from the low temperature reservoir and rejected at a higher temperature. The diagram below shows this process:

heat_pump

Coefficient of Performance

The effectiveness of heat pumps and refrigerators can be measured using a coefficient of performance ($COP$). The $COP$ shows how well a device can transfer work done into heat transfer.

For a refrigerator, as the useful quantity is the $Q_{\text{out}}$:
$$ COP_{\text{ref}} = \frac{Q_{\text{out}}}
{W} =
\frac{Q_{\text{out}}}{Q_{\text{in}} - Q_{\text{out}}} $$

The maxiumum theoretical COP for a refrigerator is given by:
$$ COP_{\text{max}} = \frac{T_{C}}{T_{H} - T_{C}} $$

For a heat pump, as the useful quantity is the $Q_{\text{in}}$:
$$ COP_{\text{hp}} = \frac{Q_{\text{in}}}
{W} =
\frac{Q_{\text{in}}}{Q_{\text{in}} - Q_{\text{out}}} $$

The maxiumum theoretical COP for a heat pump is given by:
$$ COP_{\text{max}} = \frac{T_{H}}{T_{H} - T_{C}} $$

A heat pump and a refrigerator are identical in principle as they both remove heat from a cold reservoir and provide heat to a hot reservoir. Heat pumps provide a low cost and efficient form of heating as the energy pumped into the building for example, is greater than the work done used to drive the compressor.

© Andrew Deniszczyc, 2018