Notes · PhysicsUK · A-Levels

Engineering physics (optional)

This is one of the five optional A-Level modules (AQA Paper 3, Section B); a student studies just one option. Engineering physics develops rotational dynamics - moment of inertia, torque and angular momentum - as the rotational analogue of linear mechanics, then applies thermodynamics to heat engines: the first and second laws, the work done by a gas, engine cycles and their indicator diagrams, and the efficiency of engines and heat pumps.

5 sections~14 min reading time3 competenciesLevel Standard 1 · Advanced 4

T·111111 / 13

Exam profile

AO1 · Define moment of inertia, angular momentum and the laws of thermodynamics and describe engine cycles (optional module)AO2 · Calculate rotational energy, torque, angular momentum, work done by a gas, engine power and efficienciesAO3 · Interpret indicator (p-V) diagrams, compare theoretical and real cycles and evaluate engine and heat-pump performance

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basic level

This is an optional A2 module examined in Paper 3 Section B; it is studied only if chosen as the school's option. It requires rotational dynamics, the thermodynamic laws and the engine cycles.

higher level

The full option demands the rotational analogues of the linear equations, angular-momentum conservation, indicator-diagram analysis and quantitative engine and heat-pump efficiency.

Depth

Text

Contents · 5 sections

Engineering physics (optional)

Rotational dynamics and moment of inertia#

StandardAQA 7408 3.11.1DfE GCE Physics - engineering physics option

Torque on a rotating disc

Fig. 1A tangential force

F

applied at the rim of a disc, a distance

r

from the axis, produces a torque

T = Fr

that angularly accelerates it.

Rotational motion mirrors linear motion, with each linear quantity having a rotational analogue: displacement becomes angle

\theta

, velocity becomes angular velocity

\omega

, acceleration becomes angular acceleration

\alpha

. For constant angular acceleration the rotational equations of motion take exactly the form of the suvat equations, for example

\omega_2 = \omega_1 + \alpha t

and

\theta = \omega_1 t + \tfrac{1}{2}\alpha t^2

The rotational equivalent of mass is the moment of inertia

I

, which measures how the mass of a body is distributed about the axis of rotation. For a set of point masses it is

I = \sum m r^2

; the further the mass from the axis, the larger the moment of inertia and the harder the body is to spin up or slow down. Standard shapes have standard formulae (a solid disc, a ring, a rod), always with the units

\text{kg m}^2

A rotating body stores rotational kinetic energy

E_k = \tfrac{1}{2}I\omega^2

, the analogue of

\tfrac{1}{2}mv^2

. A flywheel exploits this: a heavy rim (large moment of inertia) spun to high angular velocity stores substantial energy that can be released smoothly, evening out the power of an engine or storing energy for later use.

Because the moment of inertia depends on where the mass sits, redistributing mass changes how a body rotates without changing its total mass. A tightrope walker's long pole and a diver's tuck both work by adjusting the moment of inertia - a theme completed by angular momentum in the next section.

I = \sum m r^2

Moment of inertia

Depends on the distribution of mass about the axis.

E_k = \tfrac{1}{2}I\omega^2

Rotational kinetic energy

The rotational analogue of $\tfrac{1}{2}mv^2$ .

Worked example

Energy stored in a flywheel

A flywheel has moment of inertia $1.2\,\text{kg m}^2$ and rotates at $120\,\text{rad s}^{-1}$ . Find its rotational kinetic energy.

Use the rotational KE formula
$E_k = \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}(1.2)(120)^2$ .
Evaluate
$120^2 = 14400$ , so $E_k = 0.5 \times 1.2 \times 14400$ .
$E_k = 8.6\times10^{3}\,\text{J}$

Result: The flywheel stores about $8.6\,\text{kJ}$ of rotational kinetic energy.

Typical mistakes

Forgetting that moment of inertia depends on the distribution of mass, not just its amount.
Using degrees instead of radians for angle and angular velocity in the rotational equations.

Active revision

A flywheel of moment of inertia $0.80\,\text{kg m}^2$ spins at $150\,\text{rad s}^{-1}$ . Calculate its rotational kinetic energy.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Physics 7408 specification (AQA)

Angular momentum, torque, work and power#

AdvancedAQA 7408 3.11.1DfE GCE Physics - angular momentum

A torque is the rotational analogue of force - the turning effect that produces angular acceleration. Newton's second law in rotational form is

T = I\alpha

, the torque equals the moment of inertia times the angular acceleration. This is the equation of motion for rotating machinery: a given torque spins up a large moment of inertia only slowly.

Angular momentum is

L = I\omega

, the rotational analogue of linear momentum

mv

. The principle of conservation of angular momentum states that, in the absence of an external resultant torque, the total angular momentum of a system is constant. This is why a spinning ice skater who pulls in their arms (reducing

I

) spins faster (increasing

\omega

), keeping

I\omega

constant, and why a diver tucks to somersault faster.

Work and power have rotational forms too. The work done by a constant torque turning a body through an angle

\theta

W = T\theta

, and the power delivered is

P = T\omega

- the analogue of

P = Fv

. These give the output of a rotating shaft: an engine's power is its torque multiplied by its angular speed, which is why gearboxes trade torque for speed.

The rotational quantities complete a full parallel with linear mechanics:

T = I\alpha

mirrors

F = ma

L = I\omega

mirrors

p = mv

W = T\theta

mirrors

W = Fs

, and conservation of angular momentum mirrors conservation of linear momentum. Recognising which analogue applies turns a rotational problem into a familiar linear one.

T = I\alpha

Rotational Newton's second law

Torque equals moment of inertia times angular acceleration.

L = I\omega

Angular momentum

Conserved when there is no external torque.

P = T\omega

Rotational power

The analogue of $P = Fv$ .

Worked example

Conservation of angular momentum

A turntable of moment of inertia $0.40\,\text{kg m}^2$ rotates freely at $8.0\,\text{rad s}^{-1}$ . A ring is lowered onto it, raising the total moment of inertia to $0.60\,\text{kg m}^2$ . Find the new angular velocity.

No external torque - conserve angular momentum
$I_1\omega_1 = I_2\omega_2$ .
Solve for the new angular velocity
$\omega_2 = \dfrac{I_1\omega_1}{I_2} = \dfrac{0.40 \times 8.0}{0.60}$ .
$\omega_2 = \dfrac{3.2}{0.60} = 5.3\,\text{rad s}^{-1}$

Result: The turntable slows to $5.3\,\text{rad s}^{-1}$ as its moment of inertia rises - angular momentum is conserved.

Typical mistakes

Forgetting that angular momentum is conserved only when there is no external torque.
Confusing the rotational work $T\theta$ with the rotational kinetic energy $\tfrac{1}{2}I\omega^2$ .

Active revision

A turntable of moment of inertia $0.50\,\text{kg m}^2$ rotates freely at $6.0\,\text{rad s}^{-1}$ . A lump of clay is dropped on, raising the moment of inertia to $0.80\,\text{kg m}^2$ . Find the new angular velocity.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

The first law of thermodynamics and p-V work#

AdvancedAQA 7408 3.11.2DfE GCE Physics - first law of thermodynamics

Work done as the area under a p-V curve

Fig. 2As a gas expands, the work it does is the area under the pressure-volume curve; here the shaded area is the work done between the two volumes.

The first law of thermodynamics is the principle of conservation of energy applied to a gas: the heat supplied to a system equals the increase in its internal energy plus the work it does,

Q = \Delta U + W

. The sign convention matters:

Q

is positive when heat is added,

W

is positive when the gas does work on its surroundings (expands), and

\Delta U

is the change in the internal energy, which for an ideal gas depends only on temperature.

When a gas expands at constant pressure, the work it does is

W = p\,\Delta V

- the force on the piston (

pA

) times the distance it moves (

\Delta V/A

). More generally the work done in any process is the area under the curve on a pressure-volume (indicator) diagram. Expansion (increasing volume) does work on the surroundings; compression does work on the gas.

Four standard processes recur. An isothermal change keeps temperature (and so internal energy) constant, so all the heat supplied becomes work:

Q = W

. An adiabatic change happens with no heat exchange (

Q = 0

), so any work done changes the internal energy and hence the temperature - rapid compression heats a gas, rapid expansion cools it. A constant-pressure (isobaric) change does work

p\,\Delta V

; a constant-volume (isochoric) change does no work at all, so all the heat changes the internal energy.

These processes are the building blocks of engine cycles. The area enclosed by a closed loop on a p-V diagram is the net work done by the gas per cycle, and the shape of the loop - which processes make it up - determines both the work output and the efficiency, developed in the next section.

Q = \Delta U + W

First law of thermodynamics

Heat added equals the rise in internal energy plus work done by the gas.

W = p\,\Delta V

Work at constant pressure

The area under a constant-pressure line on a p-V diagram.

Worked example

Work done by an expanding gas

A gas at constant pressure $1.5\times10^{5}\,\text{Pa}$ expands from $1.0\times10^{-3}\,\text{m}^3$ to $4.0\times10^{-3}\,\text{m}^3$ . Find the work done by the gas.

Change in volume
$\Delta V = (4.0 - 1.0)\times10^{-3} = 3.0\times10^{-3}\,\text{m}^3$ .
Work at constant pressure
$W = p\,\Delta V = 1.5\times10^{5} \times 3.0\times10^{-3}$ .
$W = 4.5\times10^{2}\,\text{J}$

Result: The gas does $450\,\text{J}$ of work on its surroundings as it expands.

Typical mistakes

Getting the sign convention wrong - work done BY the gas is positive; work done ON the gas is negative.
Assuming an adiabatic change involves no temperature change - with no heat flow, the work changes the internal energy and so the temperature.

Active revision

A gas at a constant pressure of $1.2\times10^{5}\,\text{Pa}$ expands from $2.0\times10^{-3}\,\text{m}^3$ to $3.5\times10^{-3}\,\text{m}^3$ . Find the work done by the gas.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Engine cycles and indicator diagrams#

AdvancedAQA 7408 3.11.3DfE GCE Physics - engine cycles

Idealised Otto-cycle indicator diagram

Fig. 3A schematic pressure-volume loop for the Otto cycle: 1 to 2 adiabatic compression, 2 to 3 heat added at constant volume, 3 to 4 adiabatic power stroke, 4 to 1 heat rejected at constant volume. The enclosed area is the net work per cycle.

A four-stroke petrol engine works in a repeating cycle of four piston strokes: induction (fuel-air mixture drawn in), compression (mixture squeezed and heated), power (mixture ignited, the hot gas expands and drives the piston), and exhaust (burnt gas expelled). The idealised thermodynamic version is the Otto cycle: two adiabatic strokes (compression and power) and two constant-volume processes (heat added at ignition, heat rejected at exhaust).

The cycle is represented on an indicator diagram - a plot of pressure against volume. The idealised Otto cycle is a closed loop with two curved adiabatics and two vertical constant-volume lines. The diesel cycle is similar but adds its heat at constant pressure (no spark; the fuel ignites on injection into air already hot from a higher compression), giving a slightly different loop.

The area enclosed by the loop on the indicator diagram is the net work done by the gas in one cycle. The real indicator diagram, measured from a running engine, is a rounded, smaller version of the theoretical one, because valves open and close over finite times, combustion is not instantaneous, and heat is lost - so the real work per cycle is less than the ideal.

The power output follows from the work per cycle. The indicated power is the work per cycle multiplied by the number of cycles per second (for a four-stroke engine there is one power cycle every two revolutions, and each cylinder contributes). The useful output at the crankshaft, the brake power, is less than the indicated power because some is lost to friction: friction power

=

indicated power

-

brake power.

\text{net work per cycle} = \text{area enclosed by the indicator loop}

Work per cycle

The area of the closed loop on the p-V diagram.

\text{friction power} = \text{indicated power} - \text{brake power}

Power balance

Friction losses account for the difference.

The four-stroke cycle

Fig. 4The repeating sequence of the four strokes of a petrol engine: induction, compression, power and exhaust.

Worked example

Indicated power of an engine

A single-cylinder four-stroke engine does $480\,\text{J}$ of net work per cycle and runs at $3600\,\text{rpm}$ . Find the indicated power. (A four-stroke engine fires once every two revolutions.)

Revolutions per second
$3600\,\text{rpm} = 3600/60 = 60\,\text{rev s}^{-1}$ .
Cycles per second
One power cycle every two revolutions: $60/2 = 30$ cycles per second.
Indicated power
Work per cycle times cycles per second: $480 \times 30$ .
$P = 480 \times 30 = 1.4\times10^{4}\,\text{W}$

Result: The indicated power is about $14\,\text{kW}$ .

Typical mistakes

Confusing the four mechanical strokes with the thermodynamic processes of the Otto cycle.
Forgetting that a four-stroke engine fires only once every two revolutions when converting to power.

Active revision

A single-cylinder four-stroke engine does $520\,\text{J}$ of net work per cycle and runs at $3000\,\text{rpm}$ . Calculate the indicated power.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA)

Engine performance, the second law and heat pumps#

AdvancedAQA 7408 3.11.4DfE GCE Physics - second law and heat engines

Energy flow through an engine

Fig. 5The fuel energy input divides into useful work and losses (exhaust, cooling and friction); the width of each flow suggests its share.

The input power of an engine is set by its fuel: input power

=

calorific value of the fuel

\times

the fuel flow rate. The overall (thermal) efficiency is the useful output power divided by this input power, and the mechanical efficiency is the brake power divided by the indicated power. Because energy is lost to exhaust gases, cooling and friction, real engines are typically only

20

40\%

efficient.

The second law of thermodynamics places a fundamental limit on efficiency: no heat engine can be perfectly efficient, because some heat must always be rejected to a cold reservoir. A heat engine takes heat

Q_H

from a hot source, converts part to work

W

, and rejects the rest

Q_C

to a cold sink, so its efficiency is

\eta = \dfrac{W}{Q_H} = \dfrac{Q_H - Q_C}{Q_H}

The maximum possible efficiency of any engine working between temperatures

T_H

and

T_C

(in kelvin) is the Carnot efficiency,

\eta_{\max} = 1 - \dfrac{T_C}{T_H}

. This is why engines are designed to run as hot as materials allow and to reject heat to as cold a sink as possible - a larger temperature difference permits a higher efficiency, though the Carnot limit can never quite be reached.

Run in reverse, a heat engine becomes a heat pump or refrigerator: work is done to move heat from a cold region to a hot one. A refrigerator's performance is judged by how much heat it removes per unit work, its coefficient of performance

\dfrac{Q_C}{W}

; a heat pump's by the heat it delivers per unit work,

\dfrac{Q_H}{W}

. Because they move rather than generate heat, these coefficients can exceed one - a heat pump can deliver several joules of heat to a house for each joule of electrical work.

\eta = \dfrac{W}{Q_H} = \dfrac{Q_H - Q_C}{Q_H}

Efficiency of a heat engine

Work done divided by heat taken from the hot source.

\eta_{\max} = 1 - \dfrac{T_C}{T_H}

Carnot (maximum) efficiency

The upper limit set by the second law; temperatures in kelvin.

Worked example

Maximum efficiency of a heat engine

A steam turbine takes heat at $800\,\text{K}$ and rejects it at $320\,\text{K}$ . Find the maximum possible efficiency, and state the useful power output if it takes in heat at $50\,\text{kW}$ and achieves $80\%$ of this maximum.

Carnot efficiency
$\eta_{\max} = 1 - T_C/T_H = 1 - 320/800 = 1 - 0.40 = 0.60$ .
Actual efficiency
$80\%$ of the maximum: $0.80 \times 0.60 = 0.48$ .
Useful power
$P_{\text{out}} = 0.48 \times 50\,\text{kW}$ .
$P_{\text{out}} = 0.48 \times 50 = 24\,\text{kW}$

Result: The maximum efficiency is $60\%$ ; at $80\%$ of this the turbine delivers about $24\,\text{kW}$ .

Typical mistakes

Using Celsius instead of kelvin in the Carnot efficiency.
Treating the coefficient of performance like an efficiency that must be less than one - it can exceed one for a heat pump.

Active revision

A heat engine operates between $600\,\text{K}$ and $300\,\text{K}$ . Find its maximum possible efficiency, and comment on how a real engine compares.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Physics 7408 specification (AQA) · GCE AS and A level subject content for the sciences (Department for Education)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Physics 7408 specification

Engineering physics (optional)

5 sections~14 min reading time3 competenciesLevel Standard 1 · Advanced 4

Engineering physics (optional)

Key points

Energy stored in a flywheel

Key points

Conservation of angular momentum

Key points

Work done by an expanding gas

Key points

Indicated power of an engine

Key points

Maximum efficiency of a heat engine

Sources

Engineering physics (optional)

Key points

Energy stored in a flywheel

Key points

Conservation of angular momentum

Key points

Work done by an expanding gas

Key points

Indicated power of an engine

Key points

Maximum efficiency of a heat engine

Sources