Math Problem Statement

Efficiency derivation

Solution

Efficiency is a widely used concept, particularly in physics, engineering, and economics. Typically, efficiency (η\eta) is expressed as the ratio of useful output to the input energy, work, or power. The basic formula is:

η=Useful OutputInput×100%\eta = \frac{\text{Useful Output}}{\text{Input}} \times 100\%

Let's go through the derivation of this formula for specific systems. Here's a basic derivation using energy:

General Case (Energy Efficiency)

In most cases, efficiency refers to how much of the input energy is converted into useful energy (or work). The system consumes energy (input) and produces useful energy (output). The efficiency is defined as:

η=Useful Energy OutputTotal Energy Input×100\eta = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100

This formula is dimensionless, meaning it is a ratio of two energy quantities, and it yields a percentage.

Derivation Example: Heat Engines

For a heat engine, which converts thermal energy into mechanical work, the input energy is the heat energy supplied QinQ_\text{in}, and the useful output is the mechanical work WW done by the engine.

From the first law of thermodynamics: Qin=W+QoutQ_\text{in} = W + Q_\text{out} where QoutQ_\text{out} is the waste heat expelled by the engine.

The efficiency is given by: η=WQin×100\eta = \frac{W}{Q_\text{in}} \times 100

Now, using the relation W=QinQoutW = Q_\text{in} - Q_\text{out}, the efficiency can also be written as: η=QinQoutQin×100=(1QoutQin)×100\eta = \frac{Q_\text{in} - Q_\text{out}}{Q_\text{in}} \times 100 = \left( 1 - \frac{Q_\text{out}}{Q_\text{in}} \right) \times 100

This shows that efficiency depends on how much of the input heat is lost as waste heat. The closer QoutQ_\text{out} is to zero, the more efficient the system.

Derivation for Electrical Systems

For an electrical system like a motor or generator, the efficiency is the ratio of useful mechanical power output PoutP_\text{out} to the electrical power input PinP_\text{in}.

η=PoutPin×100\eta = \frac{P_\text{out}}{P_\text{in}} \times 100

This can be further detailed if losses such as heat, friction, or electrical resistance are known: Pin=Pout+PlossesP_\text{in} = P_\text{out} + P_\text{losses}

Thus: η=PoutPout+Plosses×100\eta = \frac{P_\text{out}}{P_\text{out} + P_\text{losses}} \times 100

In this form, the efficiency formula shows how system losses reduce the output power.


Would you like more details on this derivation, or do you have any specific system in mind?

Relative Questions:

  1. How does the efficiency of a real system compare to that of an ideal one?
  2. What is Carnot efficiency, and how does it apply to thermodynamic systems?
  3. How can efficiency be improved in an electrical system?
  4. What are the common causes of inefficiency in mechanical systems?
  5. How do friction and resistance impact overall system efficiency?

Tip:

Remember that 100% efficiency is theoretical in most cases because energy losses (e.g., heat, sound) are inevitable.

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Math Problem Analysis

Mathematical Concepts

Efficiency
Energy Conversion
Thermodynamics
Electric Power

Formulas

Efficiency (η) = (Useful Output / Input) × 100%
Heat Engine Efficiency: η = (W / Q_in) × 100
Alternative Form: η = (1 - Q_out / Q_in) × 100
Electrical Efficiency: η = (P_out / P_in) × 100

Theorems

First Law of Thermodynamics

Suitable Grade Level

Grades 10-12