Power, energy consumption and efficiency
Your ESB bill arrives and you wonder why the immersion costs so much more to run than a phone charger left plugged in all week. A kettle and an immersion both heat water, yet one boils a litre in minutes while the other takes hours; and two motors can draw the same electrical energy while one wastes far more of it as heat. The work done may be the same, and yet one machine is quicker, or cheaper to run, than the other. Power and efficiency are the two quantities that tell apart "how much work is done" from "how well and how fast it is done".
Power: how quickly energy is transferred
Two cranes lift a pallet of $500\ \text{kg}$ to a height of $10\ \text{m}$: both do exactly the same work of $49{,}000\ \text{J}$, but one takes $10\ \text{s}$ and the other $40\ \text{s}$. Which quantity measures how quickly that same energy is transferred?
Rearranging $P = F \cdot v$ gives $F = P/v$: with the same engine power, a lorry moving slowly can exert a much larger force. Rearranging $P = W/t$ gives $t = W/P$, the time a job takes.
The motor of a garage door does $6000\ \text{J}$ of work in $30\ \text{s}$ when lifting it. The average power is calculated as $P = W/t$. a) Calculate the power of the…
The motor of a garage door does $6000\ \text{J}$ of work in $30\ \text{s}$ when lifting it. The average power is calculated as $P = W/t$.
- a) Calculate the power of the motor, in watts.
- b) Express it in kilowatts.
A construction tower crane lifts a $400\ \text{kg}$ load of blocks to a height of $20\ \text{m}$, at constant speed, in $40\ \text{s}$ (take $g = 10\ \text{m/s}^2$). a)…
A construction tower crane lifts a $400\ \text{kg}$ load of blocks to a height of $20\ \text{m}$, at constant speed, in $40\ \text{s}$ (take $g = 10\ \text{m/s}^2$).
- a) Calculate the work done by the crane against weight.
- b) Calculate the useful power it develops, in watts and in kilowatts.
A passenger car maintains a constant speed of $90\ \text{km/h}$ on a motorway. At that speed, air resistance and friction add up to a force of $2000\ \text{N}$ that the…
A passenger car maintains a constant speed of $90\ \text{km/h}$ on a motorway. At that speed, air resistance and friction add up to a force of $2000\ \text{N}$ that the engine must overcome.
- a) Express the speed in $\text{m/s}$.
- b) Since the speed is constant, the driving force equals the total resistance: calculate the power developed by the engine using $P = F \times v$, in $\text{kW}$.
Energy consumed: $E = P \cdot t$ and the kWh
An electric heater has a power rating of $2000\ \text{W}$ and operates for $3\ \text{h}$. The energy consumed is calculated as $E = P \times t$. a) Calculate the energy…
An electric heater has a power rating of $2000\ \text{W}$ and operates for $3\ \text{h}$. The energy consumed is calculated as $E = P \times t$.
- a) Calculate the energy consumed, in $\text{kWh}$ (use power in $\text{kW}$ and time in hours).
- b) Given that $1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$, express it in joules.
Order these three appliances from highest to lowest power: a $350\ \text{W}$ blender, a $2000\ \text{W}$ hairdryer, and a $9\ \text{W}$ LED light bulb. Then, calculate…
Order these three appliances from highest to lowest power: a $350\ \text{W}$ blender, a $2000\ \text{W}$ hairdryer, and a $9\ \text{W}$ LED light bulb. Then, calculate the energy consumed by the most powerful appliance operating for $10\ \text{min}$, expressed in joules (remember to convert time to seconds).
Efficiency: useful power compared with total power
When $100\ \text{J}$ of electrical energy is supplied to a traditional light bulb, only $5\ \text{J}$ is given off as visible light and the remaining $95\ \text{J}$ heats the glass. How do we measure quantitatively the fraction of energy that a device actually makes use of compared with the fraction it wastes?
The motor of a Ferris wheel consumes $5000\ \text{J}$ of electrical energy and delivers $4000\ \text{J}$ of useful work in raising the cabins. Efficiency is defined as…
The motor of a Ferris wheel consumes $5000\ \text{J}$ of electrical energy and delivers $4000\ \text{J}$ of useful work in raising the cabins. Efficiency is defined as $\eta = \frac{W_{util}}{W_{total}}$. The diagram summarises the energy balance of the motor:
- a) Calculate the efficiency of the motor and express it as a percentage.
- b) How much energy was lost? What do you think it has been transformed into?
An industrial electric motor consumes $8\ \text{kWh}$ of electrical energy during its shift and delivers $6\ \text{kWh}$ of useful work (…
An industrial electric motor consumes $8\ \text{kWh}$ of electrical energy during its shift and delivers $6\ \text{kWh}$ of useful work ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).
- a) Calculate its efficiency as a percentage.
- b) Calculate the energy lost, in $\text{kWh}$ and in joules.
The motor of a workshop compressor draws a power of $5\ \text{kW}$ from the mains supply and has an efficiency of $80\ \%$. Efficiency can also be expressed using power:…
The motor of a workshop compressor draws a power of $5\ \text{kW}$ from the mains supply and has an efficiency of $80\ \%$. Efficiency can also be expressed using power: $\eta = \frac{P_{util}}{P_{total}}$.
- a) Calculate the useful power of the motor.
- b) Calculate the power lost.
- c) If the motor operates for $4\ \text{h}$, how much useful energy does it deliver? Express it in $\text{kWh}$ and in joules ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).
Efficiency chains: several conversions in a row
In a mini hydroelectric power station, the turbine has an efficiency of $90\ \%$ and the generator coupled to it has an efficiency of $80\ \%$. The diagram shows the…
In a mini hydroelectric power station, the turbine has an efficiency of $90\ \%$ and the generator coupled to it has an efficiency of $80\ \%$. The diagram shows the conversion chain:
- a) Reason why the overall efficiency of the chain is the product of the individual efficiencies and calculate it.
- b) If the water supplies $1\times10^{7}\ \text{J}$ of mechanical energy to the power station, how much electrical energy is obtained?
- c) Express it in $\text{kWh}$ ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).
A jet ski travels at a constant speed of $10\ \text{m/s}$ and its engine develops a useful power of $25\ \text{kW}$. a) What net resistance force does the engine…
A jet ski travels at a constant speed of $10\ \text{m/s}$ and its engine develops a useful power of $25\ \text{kW}$.
- a) What net resistance force does the engine overcome at that speed? First rearrange for force in the expression $P = F \times v$.
- b) If the efficiency of the engine is $25\ \%$, what total power does it consume?
Energy cost of an installation
A bakery oven has a power rating of $2500\ \text{W}$ and operates for $1.5\ \text{h}$ each day. The price of electricity is $0.2\ \text{€/kWh}$. a) Calculate the energy…
A bakery oven has a power rating of $2500\ \text{W}$ and operates for $1.5\ \text{h}$ each day. The price of electricity is $0.2\ \text{€/kWh}$.
- a) Calculate the energy consumed by the oven in a month of $30$ days, in $\text{kWh}$.
- b) Calculate the monthly electricity cost of the oven.
A solar irrigation pump raises $200\ \text{kg}$ of water (remember: $1\ \text{L}$ of water has a mass of $1\ \text{kg}$) from a tank to a terraced field located…
A solar irrigation pump raises $200\ \text{kg}$ of water (remember: $1\ \text{L}$ of water has a mass of $1\ \text{kg}$) from a tank to a terraced field located $15\ \text{m}$ higher, in $5\ \text{min}$ (take $g = 10\ \text{m/s}^2$).
- a) Calculate the useful work done by the pump on the water (remember to convert time to seconds).
- b) Calculate the useful power, in watts.
- c) If the efficiency of the pump is $50\ \%$, what electrical power does it absorb?
- d) If it operates for $2\ \text{h}$ a day and electricity costs $0.15\ \text{€/kWh}$, how much does irrigation cost per day?
Back to the start: The immersion costs more to run because its power is much higher, so $E = P \cdot t$ is larger for the same time on, and the ESB meter charges for every kWh drawn. A kettle with the same work to do but twice the power finishes in half the time ($P = W/t$), while the motor that makes better use of the energy has a higher efficiency ($\eta = P_{\text{useful}}/P_{\text{total}}$), so less is dissipated as heat.