Power, Energy Consumption, and Efficiency
Two robots weld exactly the same part, but one takes half as long as the other; two motors use the same amount of electrical energy, but one makes much better use of it than the other. In both cases the work can be the same and yet one machine is better than the other. Power and efficiency tell apart "how much work is done" from "how fast and how well it is done." At the end of the page we will come back to these two cases.
Power: how fast energy is transferred
Two cranes lift a $500\ \text{kg}$ bundle to a height of $10\ \text{m}$: both do the same work, $49\,000\ \text{J}$, but one takes $10\ \text{s}$ and the other $40\ \text{s}$. What quantity measures how fast that same energy is transferred?
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.
The driving force of a car engine moving at a constant speed of $20\ \text{m/s}$ is equal to $1500\ \text{N}$. When the force and velocity have the same direction and…
The driving force of a car engine moving at a constant speed of $20\ \text{m/s}$ is equal to $1500\ \text{N}$. When the force and velocity have the same direction and sense, power can be calculated as $P = F \times v$. Calculate the power developed by the engine, 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 $\text{kW}$.
Horsepower (hp)
On the spec sheet of a car or a jet ski you will still see power given in "horses." How many watts is that?
Where does the 736 come from? Lifting $75\ \text{kg}$ by $1\ \text{m}$ takes work $W = m\,g\,h = 75 \times 9.8 \times 1 = 735\ \text{J}$, done in $1\ \text{s}$: $P = 735\ \text{W}$. The small difference from $736\ \text{W}$ comes from the fact that $g$ is exactly $9.80665\ \text{m/s}^2$ and not $9.8$.
In the technical specifications of a car, a power rating of $100\ \text{CV}$ is listed. The metric horsepower is a historical unit of power:…
In the technical specifications of a car, a power rating of $100\ \text{CV}$ is listed. The metric horsepower is a historical unit of power: $1\ \text{CV} = 736\ \text{W}$.
- a) Express the car's power in watts.
- b) Express it in kilowatts, the unit used by industry today.
James Watt defined horsepower ($\text{CV}$) as the power required to lift $75\ \text{kg}$ at a rate of $1\ \text{m}$ every second, at constant speed (take…
James Watt defined horsepower ($\text{CV}$) as the power required to lift $75\ \text{kg}$ at a rate of $1\ \text{m}$ every second, at constant speed (take $g = 9.8\ \text{m/s}^2$).
- a) Calculate the work done against weight in $1\ \text{s}$ under these conditions and the corresponding power.
- b) Compare the result with today's equivalence, $1\ \text{CV} = 736\ \text{W}$, and comment on the reason for the small difference.
- c) A well-trained cyclist can maintain around $300\ \text{W}$ for one hour: approximately how many $\text{CV}$ is this equivalent to? Comment on the result.
Power when climbing a slope
A truck climbs slowly up a hill on the federal highway to Oaxaca, while on flat ground it speeds up without any trouble. What extra force does the engine have to overcome, and how much power does it need?
If the vehicle climbs at constant velocity, the net force is zero. The engine must balance two things: the component of the weight parallel to the road, $m\,g\,\sin\alpha$, and the friction plus air resistance, $F_r$.
A passenger car of $1200\ \text{kg}$ drives up a mountain pass with an $8\ \%$ gradient (for every $100\ \text{m}$ traveled horizontally, it rises by $8\ \text{m}$; you…
A passenger car of $1200\ \text{kg}$ drives up a mountain pass with an $8\ \%$ gradient (for every $100\ \text{m}$ traveled horizontally, it rises by $8\ \text{m}$; you may take $\sen \alpha \approx 0.08$) at a constant speed of $20\ \text{m/s}$. Frictional and aerodynamic resistance forces add up to $600\ \text{N}$ (take $g = 10\ \text{m/s}^2$), as shown in the figure.
- a) Calculate the component of weight parallel to the road.
- b) Calculate the total force that the engine must develop to maintain speed.
- c) Calculate the useful power of the engine, in $\text{kW}$.
- d) If the efficiency of the engine is $30\ \%$, what power does it consume? Comment on why consumption on an uphill stretch is much higher than on flat ground.
A high-speed train develops a power of $8\ \text{MW}$ when travelling at $288\ \text{km/h}$. a) Express the speed in $\text{m/s}$. b) Calculate the total resistance…
A high-speed train develops a power of $8\ \text{MW}$ when travelling at $288\ \text{km/h}$.
- a) Express the speed in $\text{m/s}$.
- b) Calculate the total resistance force that the train overcomes at that speed, rearranging from $P = F \times v$.
- c) If the train travelled at half speed while maintaining the same power, what maximum force could it overcome? Reason your answer and comment on why freight locomotives, which haul huge loads, travel slowly.
Energy consumed: $E = P \times t$ 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 versus total power
When $100\ \text{J}$ of electrical energy is supplied to an incandescent bulb, only about $5\ \text{J}$ comes out as light and the remaining $95\ \text{J}$ heats the glass. How do we measure what fraction of the energy an appliance actually puts to use?
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
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?
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 $\text{kWh}$ ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).
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 $4\ \text{\$/kWh}…
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 $4\ \text{\$/kWh}$.
- 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 $3\ \text{\$/kWh}$, how much does irrigation cost per day?
Back to the beginning: the robot that takes half the time develops twice the power ($P = W/t$) for the same work, and the motor that makes better use of the energy has a higher efficiency ($\eta = P_{\text{useful}}/P_{\text{total}}$) and dissipates less heat.