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Energy · Work and power

Power, energy consumption and efficiency

Your electric bill arrives and you wonder why the space heater costs so much more to run than a phone charger left plugged in all week. A kettle and a hot plate both heat water, yet one boils a quart in minutes while the other takes much longer; 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?

A motor lifts a box with an upward force F while the box rises at constant velocity v in the same direction; the power is P = F · v. F force (N) v velocity (m/s) F v = P power (W)

Speeds are often given in km/h, but power calculations need m/s. To convert km/h to m/s, divide by $3.6$ (because $1\ \text{km} = 1000\ \text{m}$ and $1\ \text{h} = 3600\ \text{s}$). For example, $72\ \text{km/h} \div 3.6 = 20\ \text{m/s}$.

Rearranging $P = F \cdot v$ gives $F = P/v$: with the same engine power, a semi truck moving slowly can exert a much larger force (half the speed means twice the force). Rearranging $P = W/t$ gives $t = W/P$, the time a job takes.

In the US, car engines are rated in horsepower: $1\ \text{hp} \approx 746\ \text{W}$. James Watt defined it as the power needed to lift $550\ \text{lb}$ by $1\ \text{ft}$ every second. Some countries use the metric horsepower (CV), the power to lift $75\ \text{kg}$ by $1\ \text{m}$ every second: $W = mgh = 75 \cdot 9.8 \cdot 1 = 735\ \text{J}$ in $1\ \text{s}$, so $P \approx 735\ \text{W}$, very close to $1\ \text{CV} = 736\ \text{W}$. It is nearly the same size as the US horsepower: $1\ \text{CV} \approx 0.986\ \text{hp}$.

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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 $P = W/t$.

  1. a) Calculate the power of the motor, in watts.
  2. b) Express it in kilowatts.
Solve
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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: $1\ \text{CV} = 736\ \text{W}$.

  1. a) Express the car's power in watts.
  2. b) Express it in kilowatts, the unit used by industry today.
Solve
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A passenger car maintains a constant speed of $90\ \text{km/h}$ on a highway. 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 highway. At that speed, air resistance and friction add up to a force of $2000\ \text{N}$ that the engine must overcome.

  1. a) Express the speed in $\text{m/s}$.
  2. 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 $\text{kW}$.
Solve

Energy consumed: $E = P \cdot t$ and the kWh

Power-time graph: a 2 kW heater running for 3 h forms a rectangle of 6 squares of 1 kWh, which represent the energy consumed E = P · t. 1 kWh 1 kWh 1 kWh 1 kWh 1 kWh 1 kWh P (kW) t (h) 1 2 1 2 3 E = P · t = 2 · 3 = 6 kWh P = 2 kW t = 3 h
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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 $E = P \times t$.

  1. a) Calculate the energy consumed, in $\text{kWh}$ (use power in $\text{kW}$ and time in hours).
  2. b) Given that $1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$, express it in joules.
Solve
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Order these three appliances from highest to lowest power: a $350\ \text{W}$ blender, a $2000\ \text{W}$ hair dryer, 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 $2000\ \text{W}$ hair dryer, 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).

Solve
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Unit challenge. Are you more efficient than an electric motor? Design a biomechanical and economic audit of your own motive power at home. 1. Data collection and…

Unit challenge. Are you more efficient than an electric motor? Design a biomechanical and economic audit of your own motive power at home.

  1. Data collection and research: Measure the height of a flight of stairs ($h \pm 0.01\text{ m}$) and the mass of a load ($m \pm 0.1\text{ kg}$) and your own body mass ($M \pm 0.1\text{ kg}$). Look up today's average $\text{kWh}$ price on the U.S. Energy Information Administration (EIA) website.

  2. Experimentation: Carry the load upstairs at four different paces, timing each ascent with its uncertainty ($t \pm 0.2\text{ s}$, performing three trials to find the average).

  3. Data processing: Create a data table showing the useful mechanical work ($W = m \times g \times h$), the power output ($P = W/t$), and their uncertainties. Plot $P$ against $1/t$.

  4. Consumption and efficiency: Assuming a human muscular efficiency of $\eta = 22\,\%$, calculate the actual metabolic energy consumed (in $\text{kJ}$ and $\text{kWh}$), taking into account that you also lift your own body ($E_{\text{metab}} = (m+M) \cdot g \cdot h/\eta$), and compare it with the financial cost if a small domestic motor with $\eta = 85\,\%$ performed the same task (lifting only the load). Write a brief report with your conclusions.

Evaluation criteria:

  • Rigor in estimating uncertainties and quality of the table and graph.

  • Accuracy in applying the concepts of work, power, and efficiency.

  • Critical analysis of efficiency and proper use of EIA data.

Solve

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?

Energy balance diagram of a motor: the total energy going in is split into useful energy that is made use of and energy lost as heat. motor total energy Wtotal, Ptotal useful energy Wuseful, Puseful losses (heat) η = useful energy / total energy
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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 summarizes the energy balance of the motor:

motor 5000 J electrical 4000 J useful losses ? J
  1. a) Calculate the efficiency of the motor and express it as a percentage.
  2. b) How much energy was lost? What do you think it has been transformed into?
Solve
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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 ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).

  1. a) Calculate its efficiency as a percentage.
  2. b) Calculate the energy lost, in $\text{kWh}$ and in joules.
Solve
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The motor of a workshop compressor draws a power of $5\ \text{kW}$ from the power line 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 power line and has an efficiency of $80\ \%$. Efficiency can also be expressed using power: $\eta = \frac{P_{util}}{P_{total}}$.

  1. a) Calculate the useful power of the motor.
  2. b) Calculate the power lost.
  3. c) If the motor operates for $4\ \text{h}$, how much useful energy does it deliver? Express it in $\text{kWh}$ and in joules ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).
Solve

Efficiency chains: several conversions in a row

Diagram of two stages in series, a turbine (η₁ = 90 %) and a generator (η₂ = 80 %), showing the losses of each stage and the overall efficiency as the product of the partial efficiencies. input 100 % turbine η₁ = 90 % 90 % of the input generator η₂ = 80 % useful output 72 % of the input overall η = η₁ · η₂ = 0.9 · 0.8 = 0.72 = 72%
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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:

10 MJ water energy turbine efficiency 90% generator efficiency 80% ? MJ electrical energy
  1. a) Explain why the overall efficiency of the chain is the product of the individual efficiencies and calculate it.
  2. b) If the water supplies $1\times10^{7}\ \text{J}$ of mechanical energy to the power station, how much electrical energy is obtained?
  3. c) Express it in $\text{kWh}$ ($1\ \text{kWh} = 3.6\times10^{6}\ \text{J}$).
Solve
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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 $25\ \text{kW}$.

  1. a) What net resistance force does the engine overcome at that speed? First rearrange for force in the expression $P = F \times v$.
  2. b) If the efficiency of the engine is $25\ \%$, what total power does it consume?
Solve

Energy cost of an installation

An electricity meter measuring energy in kWh is multiplied by the price on a label in ¢/kWh and gives the cost in dollars, shown as coins. 01275 kWh E energy (kWh) ¢/kWh price (¢/kWh) = $ cost (dollars)
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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 bakery oven has a power rating of $2500\ \text{W}$ and operates for $1.5\ \text{h}$ each day. The price of electricity is $0.2\ \text{\$/kWh}$.

  1. a) Calculate the energy consumed by the oven in a month of $30$ days, in $\text{kWh}$.
  2. b) Calculate the monthly electricity cost of the oven.
Solve
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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 $g = 10\ \text{m/s}^2$).

  1. a) Calculate the useful work done by the pump on the water (remember to convert time to seconds).
  2. b) Calculate the useful power, in watts.
  3. c) If the efficiency of the pump is $50\ \%$, what electrical power does it absorb?
  4. d) If it operates for $2\ \text{h}$ a day and electricity costs $0.15\ \text{\$/kWh}$, how much does irrigation cost per day?
Solve

Back to the start: The space heater costs more to run because its power is much higher, so $E = P \cdot t$ is larger for the same time on, and the utility 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.

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