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Lesson 12 of 16

Work, Energy and Power · Lesson 12 of 16

Power, Efficiency and Energy Resources

“Apply energy rates to elevators, pumps, wind, solar panels and human mechanical work.”

Learning Objectives

• Separate useful output from input power. • Use mass flow to calculate lifting and wind power. • Check units, efficiencies and physical assumptions.

Efficiency connects two energy accounts

Useful output and inputLaTeX
Use the power ratio for the same interval or steady operation. Convert percentages into fractions.

An efficiency of 30% means 0.30 of the input becomes the specified useful output; the rest reaches other forms or destinations. Dividing useful power by 0.30 finds the input, which must be greater than the useful output. Do not multiply output by efficiency to find input. What counts as useful must be named, such as gravitational-energy gain of pumped water.

Elevator power at constant speed

Lifting while overcoming resistance

Problem
An elevator plus passengers has mass 1800 kg, rises at 2 m/s and faces 4000 N resistance. Use g = 10 m/s². Find the mechanical motor output.

  1. 1.At constant speed, acceleration is zero, so upward motor force is mg+f = 18000+4000 = 22000 N.
  2. 2.Mechanical output power is Fv = 22000×2 = 44000 W = 44.0 kW.
  3. 3.In horsepower this is 44000/746 ≈ 59.0 hp. This is delivered mechanical power; electrical input is higher if motor efficiency is below 100%.
  4. 4.Zero net force and zero net kinetic-energy change do not mean zero motor power: 36 kW raises gravitational energy and 8 kW compensates resistance.

Water pumps and mass flow

Steady lifting powerLaTeX
V with calligraphic style denotes volume here; mass flow is density times volume flow. Ignore extra outlet kinetic energy unless specified.
Pump input requirement

Problem
A pump fills a 30 m³ tank 40 m above ground in 15 min. Water density is 1000 kg/m³, g = 9.8 m/s² and efficiency is 30%. Find electrical input.

  1. 1.The lifted mass is 30000 kg and time is 900 s. Useful energy is mgh = 11760000 J.
  2. 2.Useful power is 11760000/900 ≈ 13066.7 W.
  3. 3.Input power is 13066.7/0.30 ≈ 43555.6 W = 43.6 kW.

Wind carries kinetic-energy flow

Imagine a cylinder of air passing through a rotor area A. In time t, a uniform perpendicular wind of speed v advances length vt. Its volume is Avt and its mass is ρAvt. Multiply by v²/2 to obtain its kinetic energy. Divide by time to obtain the available kinetic power. This explains the strong dependence on wind speed. A real turbine cannot take all this energy while still allowing air to flow through; use the stated overall conversion fraction.

Wind energy fluxLaTeX
Use upstream wind perpendicular to the swept area and a uniform-flow approximation.
A small windmill

Problem
A = 30 m², v = 36 km/h, ρ = 1.2 kg/m³ and η = 0.25. Find electric power.

  1. 1.Convert wind speed to 10 m/s.
  2. 2.Incident kinetic power is ½×1.2×30×10³ = 18000 W.
  3. 3.Electric output is 0.25×18000 = 4500 W = 4.5 kW. Doubling speed would multiply the ideal incident kinetic power by eight at fixed density and area.

Solar collection and physiological energy

Area needed for a household

Problem
A household needs an average 8 kW. Sunlight averages 200 W/m² on the surface and conversion efficiency is 20%. Find collecting area.

  1. 1.Useful average output per square metre is 200×0.20 = 40 W/m².
  2. 2.Area is 8000/40 = 200 m², approximately a 14.1 m by 14.1 m square.
  3. 3.This is the energy-rate estimate under the stated averages. Actual roof orientation, shading, available area and storage for time mismatch require a separate design.
Mechanical work and food energy

Problem
A person lifts a 10 kg mass 1000 times through 0.5 m, with lowering energy dissipated. Use g = 9.8 m/s², 20% conversion efficiency and fat energy 3.8×10⁷ J/kg. Find positive lifting work and the ideal fat equivalent.

  1. 1.Positive lifting work is 1000×10×9.8×0.5 = 49000 J.
  2. 2.Required chemical input for that lifting work is 49000/0.20 = 245000 J.
  3. 3.The ideal fat equivalent is 245000/(3.8×10⁷) ≈ 0.00645 kg = 6.45 g.
  4. 4.This is a simplified mechanical-energy exercise, not a prediction of actual body-fat loss. Lowering, resting metabolism and biological fuel use are separate processes.

Check your understanding

Quiz

Quick check

Useful output 100 W at 25% efficiency needs input

Explanation: Input=output/0.25.

Quick check

At constant elevator speed, motor power can be nonzero because

Explanation: Zero ΔK does not eliminate other energy transfers.

Quick check

Wind kinetic power scales with speed as

Explanation: Mass flow adds one power of v to the v² kinetic energy per mass.

Quick check

A 20% solar conversion of 200 W/m² gives useful flux

Explanation: Multiply incident flux by efficiency.

Quick check

Before using 15 min in a watt calculation, convert it to

Explanation: Watts are joules per second.

Key Takeaways

Key Takeaways

• Name the useful output and divide by efficiency for input. • Use consistent SI units and mass flow. • Device power and energy resource estimates follow the same energy-rate balance.

Previous · Lesson 11

Power and Time-Dependent Energy Transfer

Next · Lesson 13

Collisions, Impulse and Energy Changes