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General physics
🎯 Learning Objectives
By the end of this lesson, you will be able to:
• Recall and understand that efficiency is the ratio of useful energy output to total energy input
• Calculate efficiency using the formula η = (useful energy output / total energy input) × 100%
• Apply efficiency concepts to analyze real-world systems and machines
• Interpret and create Sankey diagrams to visualize energy flow and losses
• Evaluate factors that affect efficiency and propose improvements to systems
• Compare efficiencies of different energy conversion systems
🗣️ Language Objectives
Students will develop their physics communication skills by:
• Using precise scientific terminology when describing efficiency and energy losses
• Explaining efficiency calculations using appropriate mathematical language
• Reading and interpreting efficiency problems written in English with confidence
• Communicating solutions involving energy flow analysis clearly in written English
• Understanding and using percentage notation and ratio expressions accurately
📚 Key Terms and Translations
English TermRussian TranslationKazakh Translation
EfficiencyЭффективностьТиімділік
Useful energy outputПолезная энергия на выходеПайдалы шығыс энергиясы
Total energy inputОбщая энергия на входеЖалпы кіріс энергиясы
Energy lossПотери энергииЭнергия жоғалту
Waste energyПотерянная энергияЫсырапқа кеткен энергия
Sankey diagramДиаграмма СанкиСанки диаграммасы
Heat lossПотери теплаЖылу жоғалту
FrictionТрениеҮйкеліс
🃏 Vocabulary Study Cards

Efficiency

Definition: Ratio of useful energy output to total energy input

Formula: η = (Eout/Ein) × 100%

Range: 0% to 100% (100% = perfect efficiency)

Example: 80% efficient motor uses 80J usefully from 100J input

Energy Losses

Common forms: Heat, sound, light, vibration

Causes: Friction, resistance, incomplete combustion

Calculation: Waste energy = Input — Useful output

Impact: Lower efficiency, higher operating costs

Sankey Diagrams

Purpose: Visualize energy flow through systems

Features: Arrow width proportional to energy amount

Benefits: Shows losses clearly, aids optimization

Applications: Power plants, engines, industrial processes

Improving Efficiency

Methods: Reduce friction, better insulation, optimize design

Lubrication: Reduces mechanical losses

Insulation: Reduces thermal losses

Maintenance: Keeps systems operating efficiently

📖 Glossary of Terms

Efficiency

A measure of how well a system converts input energy into useful output energy, expressed as a percentage. Efficiency indicates the fraction of input energy that achieves the desired purpose, with the remainder being «lost» to non-useful forms such as heat, sound, or vibration.

Translation
Russian: Эффективность — это мера того, насколько хорошо система преобразует входную энергию в полезную выходную энергию, выраженная в процентах. Эффективность показывает долю входной энергии, которая достигает желаемой цели, а остальная часть «теряется» в неполезных формах, таких как тепло, звук или вибрация.

Kazakh: Тиімділік — бұл жүйенің кіріс энергиясын пайдалы шығыс энергиясына қаншалықты жақсы түрлендіретінін көрсететін өлшем, пайызбен көрсетіледі. Тиімділік қажетті мақсатқа жететін кіріс энергиясының үлесін көрсетеді, ал қалған бөлігі жылу, дыбыс немесе дірілдеу сияқты пайдасыз формаларда «жоғалады».

Useful Energy Output

The portion of energy that successfully performs the intended task or function of a system. This is the energy that achieves the desired outcome, such as mechanical work in an engine, light from a bulb, or heat from a heater. The useful output determines the effectiveness of the energy conversion process.

Translation
Russian: Полезная энергия на выходе — это часть энергии, которая успешно выполняет предназначенную задачу или функцию системы. Это энергия, которая достигает желаемого результата, такого как механическая работа в двигателе, свет от лампочки или тепло от нагревателя. Полезный выход определяет эффективность процесса преобразования энергии.

Kazakh: Пайдалы шығыс энергиясы — бұл жүйенің мақсатты тапсырмасын немесе функциясын сәтті орындайтын энергияның бөлігі. Бұл қозғалтқыштағы механикалық жұмыс, шамнан шығатын жарық немесе жылытқыштан шығатын жылу сияқты қажетті нәтижеге жететін энергия. Пайдалы шығыс энергия түрлендіру процесінің тиімділігін анықтайды.

Energy Losses

Energy that is converted to forms that do not contribute to the intended purpose of the system. Common energy losses include heat generated by friction, sound from vibrating components, light from hot surfaces, and energy lost through poor insulation. These losses reduce system efficiency and often represent wasted resources.

Translation
Russian: Потери энергии — это энергия, которая преобразуется в формы, не способствующие предназначенной цели системы. Обычные потери энергии включают тепло, выделяемое трением, звук от вибрирующих компонентов, свет от горячих поверхностей и энергию, теряемую через плохую изоляцию. Эти потери снижают эффективность системы и часто представляют потраченные впустую ресурсы.

Kazakh: Энергия жоғалтулары — бұл жүйенің мақсатты мақсатына ықпал етпейтін формаларға түрлендірілетін энергия. Энергияның жиі кездесетін жоғалтулары үйкеліс арқылы пайда болатын жылуды, дірілдейтін құрамдас бөліктердің дыбысын, ыстық беттерден шығатын жарықты және нашар оқшаулау арқылы жоғалатын энергияны қамтиды. Бұл жоғалтулар жүйенің тиімділігін төмендетеді және көбінесе ысырапқа кеткен ресурстарды білдіреді.

Sankey Diagram

A specialized flow diagram that shows energy transfers and transformations in a system, where the width of arrows is proportional to the quantity of energy. Named after Captain Matthew Sankey, these diagrams provide a visual representation of efficiency by clearly showing where energy goes and how much is lost at each stage.

Translation
Russian: Диаграмма Санки — это специализированная диаграмма потока, которая показывает передачи и превращения энергии в системе, где ширина стрелок пропорциональна количеству энергии. Названные в честь капитана Мэтью Санки, эти диаграммы обеспечивают визуальное представление эффективности, ясно показывая, куда идет энергия и сколько теряется на каждом этапе.

Kazakh: Санки диаграммасы — бұл жүйедегі энергия беруді және түрлендірулерін көрсететін арнайы ағын диаграммасы, мұнда көрсеткілердің ені энергия мөлшеріне пропорционалды. Капитан Мэтью Санкидің атымен аталған бұл диаграммалар энергияның қайда кететінін және әрбір кезеңде қанша жоғалатынын айқын көрсету арқылы тиімділіктің визуалды көрінісін береді.

🔬 Theory: Understanding Efficiency

Defining Efficiency

Efficiency is a fundamental concept that measures how well a system converts input energy into useful output. No real system is 100% efficient due to unavoidable losses.

Efficiency = (Useful Energy Output / Total Energy Input) × 100%
η = (Euseful / Einput) × 100%
Where η (eta) represents efficiency as a percentage

Key characteristics of efficiency:

  • Always expressed as a percentage between 0% and 100%
  • 100% efficiency is theoretically possible but never achieved in practice
  • Higher efficiency means less energy waste
Translation
Russian: Эффективность — это фундаментальная концепция, которая измеряет, насколько хорошо система преобразует входную энергию в полезный выход. Ни одна реальная система не является 100% эффективной из-за неизбежных потерь.

Kazakh: Тиімділік — бұл жүйенің кіріс энергиясын пайдалы шығысқа қаншалықты жақсы түрлендіретінін өлшейтін іргелі ұғым. Болдырмайтын жоғалтулар себебінен ешбір нақты жүйе 100% тиімді емес.

Energy Flow and Losses

In any energy conversion process, the total energy is conserved, but not all energy serves the intended purpose:

Energy Input = Useful Energy Output + Waste Energy
Ein = Euseful + Ewaste

Common forms of waste energy:

  • Heat: From friction, electrical resistance, incomplete combustion
  • Sound: From vibration, air movement, mechanical impacts
  • Light: From hot surfaces, electrical arcing
  • Kinetic energy: From exhaust gases, moving parts
Translation
Russian: В любом процессе преобразования энергии общая энергия сохраняется, но не вся энергия служит предназначенной цели. Обычные формы потерянной энергии включают тепло от трения, звук от вибрации и свет от горячих поверхностей.

Kazakh: Кез келген энергия түрлендіру процесінде жалпы энергия сақталады, бірақ барлық энергия мақсатты мақсатқа қызмет етпейді. Ысырапқа кеткен энергияның жиі кездесетін формалары үйкеліс арқылы пайда болатын жылуды, дірілдеуден шығатын дыбысты және ыстық беттерден шығатын жарықты қамтиды.

Factors Affecting Efficiency

Several factors influence system efficiency:

Design Factors

Material selection: Low-friction materials, good conductors

Geometry: Streamlined shapes, optimal dimensions

Component quality: Precision manufacturing, tight tolerances

Operating Conditions

Temperature: Affects material properties and thermal losses

Load: Systems often most efficient at design load

Speed: Higher speeds can increase friction losses

Maintenance Factors

Lubrication: Reduces friction between moving parts

Cleanliness: Dirt and debris increase losses

Wear: Worn components operate less efficiently

Translation
Russian: Несколько факторов влияют на эффективность системы, включая факторы проектирования (выбор материала, геометрия), условия эксплуатации (температура, нагрузка) и факторы технического обслуживания (смазка, чистота).

Kazakh: Жүйенің тиімділігіне бірнеше фактор әсер етеді, соның ішінде жобалау факторлары (материал таңдау, геометрия), жұмыс жағдайлары (температура, жүктеме) және техникалық қызмет көрсету факторлары (майлау, тазалық).

Theory Questions

Easy Question: A motor receives 1000J of electrical energy and produces 800J of mechanical work. Calculate its efficiency.

Answer
Efficiency = (Useful output / Total input) × 100%
η = (800J / 1000J) × 100% = 80%
The motor is 80% efficient.

Medium Question: A car engine has 25% efficiency. If the engine produces 15kW of useful power, calculate: (a) the input power, (b) the power lost as waste energy.

Answer
(a) Using efficiency = (output/input) × 100%:
25% = (15kW/Pinput) × 100%
Pinput = 15kW / 0.25 = 60kW

(b) Power lost = Input power — Useful power
Pwaste = 60kW — 15kW = 45kW
The engine wastes 45kW as heat, sound, and other losses.

Medium Question: A light bulb is 15% efficient at converting electrical energy to light. If it operates at 60W, how much energy is converted to light and heat per hour?

Answer
Input energy per hour = 60W × 3600s = 216,000J = 216kJ

Light energy = 15% of input = 0.15 × 216kJ = 32.4kJ
Heat energy = 85% of input = 0.85 × 216kJ = 183.6kJ

Most of the electrical energy becomes unwanted heat rather than useful light.

Hard Question (Critical Thinking): A power plant burns coal with 30% efficiency to generate electricity, which is transmitted with 95% efficiency to homes where it powers LED lights with 90% efficiency. Calculate the overall efficiency from coal to light output. Compare this to a solar panel (20% efficiency) directly powering LED lights, and analyze the implications for energy policy and environmental impact.

Answer
Coal power chain efficiency:
Overall efficiency = ηplant × ηtransmission × ηLED
ηoverall = 0.30 × 0.95 × 0.90 = 0.2565 = 25.7%

Solar power chain efficiency:
ηsolar = ηpanel × ηLED = 0.20 × 0.90 = 0.18 = 18%

Analysis:
Despite lower solar panel efficiency, the direct solar system (18%) is much closer to the coal system (25.7%) than might be expected. However, this comparison reveals important considerations:

**Energy policy implications:**
1. **Coal advantages:** Higher overall efficiency, reliable 24/7 operation
2. **Solar advantages:** No fuel costs, no emissions during operation, distributed generation
3. **System complexity:** Coal requires mining, transport, plant operation, grid infrastructure

**Environmental impact:**
— **Coal:** High CO₂ emissions (≈1 kg CO₂/kWh), air pollution, mining damage
— **Solar:** Zero operational emissions, manufacturing impacts, recyclable panels

**Economic factors:**
— **Coal:** Fuel costs, plant maintenance, environmental regulations
— **Solar:** High initial cost, minimal operating costs, government incentives

**Future considerations:**
— Solar panel efficiency improving (>25% in lab)
— Coal plant efficiency limited by thermodynamics
— Battery storage addressing solar intermittency
— Grid-scale solar becoming cost-competitive

**Conclusion:** While coal currently shows higher efficiency, solar’s environmental benefits, decreasing costs, and improving technology make it increasingly attractive for sustainable energy policy.

💪 Memorization Exercises for Key Terms

Complete the Efficiency Formulas

1. Efficiency = (_______ / _______) × 100%

Answer
Efficiency = (Useful energy output / Total energy input) × 100%

2. Energy input = _______ + _______

Answer
Energy input = Useful energy output + Waste energy

3. Perfect efficiency would be _______%, but this is impossible in practice.

Answer
100% (one hundred percent)

4. Higher efficiency means _______ energy waste.

Answer
less (lower)

5. Common energy losses include heat, sound, and _______.

Answer
light (or vibration, or any other valid energy loss form)

🎥 Educational Video Resource
📐 Worked Problem Examples

Example 1: Simple Efficiency Calculation

Problem: A washing machine uses 2.5kWh of electrical energy to complete a wash cycle. Of this energy, 0.4kWh heats the water, 0.3kWh runs the motor, and the remainder is lost as heat and sound. Calculate: (a) the useful energy output, (b) the efficiency of the washing machine, (c) the energy lost per cycle.

Washing machine energy flow diagram

Step-by-Step Solution
Given:
— Total electrical input: 2.5kWh
— Water heating: 0.4kWh (useful)
— Motor operation: 0.3kWh (useful)
— Remainder: lost energy

Part (a): Useful energy output
Useful energy = Water heating + Motor work
Euseful = 0.4kWh + 0.3kWh = 0.7kWh

Part (b): Efficiency calculation
Efficiency = (Useful output / Total input) × 100%
η = (0.7kWh / 2.5kWh) × 100% = 28%

Part (c): Energy lost per cycle
Energy lost = Total input — Useful output
Elost = 2.5kWh — 0.7kWh = 1.8kWh

Analysis:
The washing machine is only 28% efficient, with 72% of energy wasted as heat and sound. This shows why energy-efficient appliances are important for reducing electricity costs and environmental impact.

Example 2: Power Plant Efficiency Analysis

Problem: A natural gas power plant burns fuel with energy content 50 MW to generate 18 MW of electrical power. The plant operates for 12 hours per day. Calculate: (a) the plant’s efficiency, (b) daily useful energy production, (c) daily energy waste, (d) if electricity sells for $0.12/kWh, find the daily revenue and the economic value of the wasted energy.

Power plant energy flow

Complete Economic and Energy Analysis
Given:
— Fuel energy input: 50 MW
— Electrical output: 18 MW
— Operating time: 12 hours/day
— Electricity price: $0.12/kWh

Part (a): Plant efficiency
Efficiency = (Electrical output / Fuel input) × 100%
η = (18 MW / 50 MW) × 100% = 36%

Part (b): Daily useful energy production
Daily electrical energy = Power × Time
Euseful = 18 MW × 12h = 216 MWh = 216,000 kWh

Part (c): Daily energy waste
Total fuel energy per day = 50 MW × 12h = 600 MWh
Energy waste = 600 MWh — 216 MWh = 384 MWh = 384,000 kWh
This represents 64% of the input energy lost as heat.

Part (d): Economic analysis
Daily revenue = 216,000 kWh × $0.12/kWh = $25,920
Economic value of waste = 384,000 kWh × $0.12/kWh = $46,080

Key insights:
1. **Efficiency impact:** 36% efficiency is typical for gas turbines but shows significant improvement potential
2. **Economic loss:** $46,080/day ($16.8M/year) in wasted energy value
3. **Environmental cost:** 384 MWh/day of waste heat requires cooling systems
4. **Improvement potential:** Combined cycle plants achieve 50-60% efficiency

Real-world applications:
— **Cogeneration:** Use waste heat for district heating or industrial processes
— **Combined cycle:** Use exhaust heat to drive steam turbines
— **Efficiency improvements:** Better turbine designs, advanced materials
— **Renewable alternatives:** Solar/wind with 35-45% system efficiency but no fuel costs

Example 3: Multi-Stage System Efficiency

Problem: A hybrid electric vehicle system has multiple efficiency stages: engine (35%), generator (90%), battery charging (85%), battery discharging (95%), electric motor (92%). If the engine receives 100 MJ of chemical energy from fuel, calculate: (a) energy at each stage, (b) overall system efficiency, (c) create a Sankey diagram showing energy flow, (d) identify the biggest loss and suggest improvements.

Hybrid vehicle energy conversion chain

Advanced Multi-Stage Analysis with Optimization
Given efficiencies:
— Engine: η₁ = 35%
— Generator: η₂ = 90%
— Battery charging: η₃ = 85%
— Battery discharging: η₄ = 95%
— Electric motor: η₅ = 92%
— Initial fuel energy: E₀ = 100 MJ

Part (a): Energy at each stage

**Stage 1 — Engine output:**
E₁ = η₁ × E₀ = 0.35 × 100 MJ = 35 MJ
Loss₁ = 100 — 35 = 65 MJ (heat, exhaust)

**Stage 2 — Generator output:**
E₂ = η₂ × E₁ = 0.90 × 35 MJ = 31.5 MJ
Loss₂ = 35 — 31.5 = 3.5 MJ (heat, magnetic losses)

**Stage 3 — Battery stored energy:**
E₃ = η₃ × E₂ = 0.85 × 31.5 MJ = 26.8 MJ
Loss₃ = 31.5 — 26.8 = 4.7 MJ (heat, electrochemical losses)

**Stage 4 — Battery output:**
E₄ = η₄ × E₃ = 0.95 × 26.8 MJ = 25.4 MJ
Loss₄ = 26.8 — 25.4 = 1.4 MJ (internal resistance)

**Stage 5 — Motor mechanical output:**
E₅ = η₅ × E₄ = 0.92 × 25.4 MJ = 23.4 MJ
Loss₅ = 25.4 — 23.4 = 2.0 MJ (motor losses)

Part (b): Overall system efficiency
Overall efficiency = (Final output / Initial input) × 100%
ηtotal = (23.4 MJ / 100 MJ) × 100% = 23.4%

**Alternative calculation:**
ηtotal = η₁ × η₂ × η₃ × η₄ × η₅
ηtotal = 0.35 × 0.90 × 0.85 × 0.95 × 0.92 = 0.234 = 23.4% ✓

Part (c): Sankey diagram description
«`
Fuel (100 MJ) → [Engine 35%] → 35 MJ
↓ (65 MJ heat loss)
35 MJ → [Generator 90%] → 31.5 MJ
↓ (3.5 MJ loss)
31.5 MJ → [Charging 85%] → 26.8 MJ
↓ (4.7 MJ loss)
26.8 MJ → [Discharge 95%] → 25.4 MJ
↓ (1.4 MJ loss)
25.4 MJ → [Motor 92%] → 23.4 MJ (wheel output)
↓ (2.0 MJ loss)
«`

Total losses: 65 + 3.5 + 4.7 + 1.4 + 2.0 = 76.6 MJ

Part (d): Loss analysis and improvements

**Biggest loss:** Engine inefficiency (65 MJ, 65% of input)
**Ranking of losses:**
1. Engine: 65 MJ (85% of total losses)
2. Battery charging: 4.7 MJ (6% of total losses)
3. Generator: 3.5 MJ (5% of total losses)
4. Motor: 2.0 MJ (3% of total losses)
5. Battery discharge: 1.4 MJ (2% of total losses)

**Improvement strategies:**

**1. Engine improvements (highest priority):**
— Advanced engine designs (Atkinson cycle, variable compression)
— Better combustion control and timing
— Improved materials and manufacturing tolerances
— Target: 40-45% efficiency possible

**2. System optimization:**
— Reduce battery cycling (charge/discharge losses)
— Direct mechanical connection for highway driving
— Optimize power management algorithms
— Regenerative braking to recover energy

**3. Component upgrades:**
— Higher efficiency electric motors (>95%)
— Advanced battery chemistry (Li-ion → solid state)
— Better power electronics for charging/conversion

**4. Alternative approaches:**
— Fuel cell hybrid (50-60% electrical conversion efficiency)
— Plug-in hybrid with grid charging (avoid engine use)
— Full electric with high-efficiency charging

**Economic impact of improvements:**
If engine efficiency improved to 40%:
New overall efficiency = 0.40 × 0.90 × 0.85 × 0.95 × 0.92 = 26.8%
Improvement = (26.8 — 23.4)/23.4 = 14.5% better fuel economy

This analysis demonstrates why hybrid vehicle development focuses primarily on engine efficiency improvements, as the internal combustion engine remains the dominant source of energy loss in the system.

🧪 Interactive Investigation - PhET Simulation

Explore energy efficiency concepts using this interactive simulation:

Investigation Tasks:

Task 1: Set up a simple system (like heating water with a burner) and observe energy transformations. Calculate the efficiency of energy transfer.

Task 2: Compare the efficiency of different energy conversion systems available in the simulation. Which is most efficient?

Task 3: Design an energy system that maximizes efficiency. What strategies can you use to minimize energy losses?

Investigation Answers and Analysis
Task 1 Analysis:
Students should observe that heating water with a burner involves thermal energy transfer. Not all thermal energy from the flame reaches the water — some is lost to the surroundings. Efficiency = (energy gained by water)/(chemical energy from fuel) × 100%. Typical values: 60-80% depending on setup.

Task 2 Comparison:
Students will find that electrical systems generally show higher efficiency than combustion-based systems. Electric heaters approach 100% efficiency for heating, while combustion systems lose energy through hot exhaust gases and radiation.

Task 3 Optimization:
Strategies include: insulation to reduce heat losses, direct energy transfer (avoid multiple conversions), proper sizing of components, and minimizing energy conversion steps. Students should recognize that fewer conversion steps generally mean higher overall efficiency.

👥 Collaborative Group Activity

Work with your team to complete this interactive efficiency challenge:

Group Design Challenge:

Design an Energy-Efficient School

Challenge: Design improvements to make your school more energy-efficient, focusing on lighting, heating, and electrical systems.

Requirements:

  • Audit current energy usage and identify inefficiencies
  • Propose specific improvements with efficiency calculations
  • Estimate cost savings and payback periods
  • Consider environmental impact and sustainability

Deliverables:

  • Energy audit report with efficiency measurements
  • Improvement proposals with cost-benefit analysis
  • Sankey diagram showing current vs improved energy flow
  • Presentation to school administration (8 minutes maximum)

Alternative Group Activities:

Appliance Efficiency: Compare efficiency ratings of household appliances and calculate energy costs

Transportation Analysis: Calculate and compare efficiency of different transportation modes

Renewable Energy: Research efficiency of solar panels, wind turbines, and energy storage systems

📝 Individual Assessment - Structured Questions

Question 1: Analysis and Application

A student conducts an experiment with an electric kettle. The kettle has a power rating of 2000W and takes 4 minutes to boil 1 liter of water from 20°C to 100°C. Calculate: (a) the electrical energy input, (b) the useful energy output needed to heat the water, (c) the efficiency of the kettle. Given: specific heat capacity of water = 4200 J/kg°C, density of water = 1000 kg/m³.

Answer
Given:
— Power rating: P = 2000W
— Time: t = 4 min = 240s
— Volume of water: V = 1L = 0.001 m³
— Temperature change: ΔT = 100°C — 20°C = 80°C
— Specific heat capacity: c = 4200 J/kg°C
— Density of water: ρ = 1000 kg/m³

Part (a): Electrical energy input
Einput = P × t = 2000W × 240s = 480,000J = 480kJ

Part (b): Useful energy output
Mass of water: m = ρV = 1000 × 0.001 = 1kg
Energy to heat water: Euseful = mcΔT = 1 × 4200 × 80 = 336,000J = 336kJ

Part (c): Efficiency of kettle
Efficiency = (Euseful/Einput) × 100% = (336,000/480,000) × 100% = 70%

The kettle is 70% efficient; 30% of electrical energy is lost as heat to surroundings, sound, and other losses.

Question 2: Synthesis and Critical Thinking

A manufacturing company wants to reduce energy costs by improving efficiency. Their current system uses 1000 MWh/month electricity costing $80/MWh. Energy audit reveals: motors (40% usage, 85% efficient), lighting (30% usage, 60% efficient), heating (20% usage, 70% efficient), other (10% usage, 90% efficient). Calculate current efficiency and energy costs, then evaluate three improvement options: (a) upgrade all motors to 95% efficiency, (b) replace lighting with LEDs (90% efficiency), (c) improve heating system to 85% efficiency. Determine which option provides best return on investment.

Answer
Current system analysis:
— Total energy use: 1000 MWh/month
— Energy cost: $80/MWh
— Monthly cost: 1000 × $80 = $80,000

**Energy breakdown:**
— Motors: 40% × 1000 = 400 MWh, efficiency 85%
— Lighting: 30% × 1000 = 300 MWh, efficiency 60%
— Heating: 20% × 1000 = 200 MWh, efficiency 70%
— Other: 10% × 1000 = 100 MWh, efficiency 90%

**Current useful energy and waste:**
— Motors: Useful = 400 × 0.85 = 340 MWh, Waste = 60 MWh
— Lighting: Useful = 300 × 0.60 = 180 MWh, Waste = 120 MWh
— Heating: Useful = 200 × 0.70 = 140 MWh, Waste = 60 MWh
— Other: Useful = 100 × 0.90 = 90 MWh, Waste = 10 MWh

Total useful: 750 MWh, Total waste: 250 MWh
**Current overall efficiency: 75%**

Option (a): Upgrade motors to 95% efficiency
New motor energy needed for same useful output: 340/0.95 = 358 MWh
Energy saving: 400 — 358 = 42 MWh/month
Cost saving: 42 × $80 = $3,360/month = $40,320/year

Option (b): LED lighting at 90% efficiency
New lighting energy needed: 180/0.90 = 200 MWh
Energy saving: 300 — 200 = 100 MWh/month
Cost saving: 100 × $80 = $8,000/month = $96,000/year

Option (c): Improve heating to 85% efficiency
New heating energy needed: 140/0.85 = 165 MWh
Energy saving: 200 — 165 = 35 MWh/month
Cost saving: 35 × $80 = $2,800/month = $33,600/year

ROI Analysis (assuming typical upgrade costs):
— **Motor upgrade:** ~$200,000 cost, payback = $200,000/$40,320 = 5.0 years
— **LED lighting:** ~$150,000 cost, payback = $150,000/$96,000 = 1.6 years
— **Heating system:** ~$100,000 cost, payback = $100,000/$33,600 = 3.0 years

**Recommendation:** LED lighting upgrade provides best ROI with shortest payback period and highest annual savings. Should be implemented first, followed by heating improvements, then motor upgrades.

Question 3: Complex Analysis

A data center consumes 10 MW continuously and uses cooling systems that consume additional power. The servers have 20% efficiency (80% becomes heat), and the cooling system has COP (Coefficient of Performance) of 3.5, meaning it removes 3.5 units of heat per unit of electrical energy. Calculate: (a) heat generated by servers, (b) cooling power required, (c) total facility power consumption, (d) overall facility efficiency for useful computing work, (e) analyze strategies to improve efficiency including waste heat recovery.

Answer
Given:
— Server power consumption: 10 MW
— Server efficiency: 20% (80% becomes heat)
— Cooling system COP: 3.5
— Continuous operation

Part (a): Heat generated by servers
Heat generated = Waste energy from servers
Qservers = 0.80 × 10 MW = 8 MW thermal

Part (b): Cooling power required
Using COP = Heat removed / Electrical power input
3.5 = 8 MW / Pcooling
Pcooling = 8 MW / 3.5 = 2.29 MW

Part (c): Total facility power consumption
Ptotal = Pservers + Pcooling = 10 MW + 2.29 MW = 12.29 MW

Part (d): Overall facility efficiency
Useful computing work = 20% of server power = 0.20 × 10 MW = 2 MW
Overall efficiency = (2 MW / 12.29 MW) × 100% = 16.3%

Only 16.3% of total electrical input performs useful computing work.

Part (e): Efficiency improvement strategies

**1. Waste heat recovery applications:**
— **District heating:** Use 8 MW thermal for nearby buildings
— **Absorption cooling:** Use waste heat to drive cooling systems
— **Industrial processes:** Supply heat to manufacturing facilities
— **Greenhouse heating:** Agricultural applications
— **Swimming pool heating:** Recreational facilities

**2. Cooling efficiency improvements:**
— **Higher COP chillers:** Modern systems achieve COP 4-6
— **Free cooling:** Use outside air when ambient temperature allows
— **Liquid cooling:** Direct chip cooling reduces cooling power by 30-50%
— **Hot aisle containment:** Improves cooling system efficiency

**3. Server efficiency improvements:**
— **Better processors:** Modern CPUs achieve 25-30% efficiency
— **Optimized workloads:** Virtualization and load balancing
— **Dynamic power management:** Scale power with computing demand
— **Specialized hardware:** GPUs and TPUs for specific tasks

**4. System integration:**
— **Combined heat and power (CHP):** Generate electricity on-site using waste heat
— **Thermal energy storage:** Store cooling capacity during low-demand periods
— **Smart grid integration:** Shift computing loads to optimize grid efficiency

**Potential improvements with waste heat recovery:**
If 80% of waste heat (6.4 MW thermal) is recovered for heating:
— Displaced natural gas heating: ~6.4 MW / 0.85 efficiency = 7.5 MW equivalent
— Overall system value: 2 MW computing + 6.4 MW heating = 8.4 MW useful output
— Effective efficiency: (8.4 / 12.29) × 100% = 68.4%

**Economic impact:**
— Annual electricity cost: 12.29 MW × 8760 h × $0.08/kWh = $8.6M
— With heat recovery value: 6.4 MW × 8760 h × $0.04/kWh thermal = $2.2M revenue
— Net operating cost reduction: 26%

This analysis demonstrates how waste heat recovery can dramatically improve the economic and environmental performance of energy-intensive facilities, transforming «waste» into valuable resources.

Question 4: Engineering Application and Synthesis

Design an energy system for a remote island community (1000 residents) that needs electricity, heating, and transportation. Available resources: solar panels (22% efficiency), wind turbines (35% efficiency), battery storage (90% round-trip efficiency), electric vehicles, and backup diesel generators (35% efficiency). Daily energy needs: 5 MWh electricity, 3 MWh heating, 2 MWh transportation. Design an integrated system that maximizes overall efficiency while ensuring reliability. Include energy storage sizing, backup power requirements, and efficiency optimization strategies.

Answer
System Design Requirements:

Daily energy demand:
— Electricity: 5 MWh
— Heating: 3 MWh
— Transportation: 2 MWh
— **Total: 10 MWh/day**

Resource assessment:
— Solar: 22% panel efficiency
— Wind: 35% system efficiency
— Battery: 90% round-trip efficiency
— Diesel backup: 35% efficiency
— Available land/sea area for renewable installation

Part 1: Renewable energy sizing

**Solar system design:**
— Peak sun hours: ~5 hours/day (typical island location)
— Required solar capacity for electricity: 5 MWh / 5h = 1 MW peak
— Accounting for losses and variability: 1.5 MW solar installation
— Panel area needed: 1.5 MW / 0.22 efficiency / 1 kW/m² = 6,818 m²

**Wind system design:**
— Average wind capacity factor: ~30% (island location)
— Required wind capacity: 5 MWh / (24h × 0.30) = 0.69 MW average
— Wind turbine installation: 2 MW capacity (multiple smaller turbines for reliability)

**Part 2: Energy storage system**

**Battery sizing for reliability:**
— Storage for 3 days backup: 15 MWh total demand
— Accounting for battery efficiency: 15 MWh / 0.90 = 16.7 MWh battery capacity
— Daily cycling depth: 50% maximum for battery life
— **Required battery bank: 33.4 MWh capacity**

**Part 3: Integrated system efficiency analysis**

**Energy flow optimization:**

**Direct renewable usage (highest efficiency):**
— Solar/wind → immediate electricity use: 90% efficiency (inverter losses)
— Direct heating via heat pumps: COP 3.0 using renewable electricity
— EV charging during peak renewable generation: 85% efficiency

**Battery-mediated usage:**
— Renewable → battery → load: 90% × 90% = 81% efficiency
— Critical for evening/night electricity demand

**Backup diesel system:**
— Generator efficiency: 35%
— Only for extended periods of low renewable generation
— Target: <5% annual energy from diesel

**Part 4: System integration strategies**

**Demand management:**
— **Smart EV charging:** Prioritize charging during peak renewable generation
— **Thermal storage:** Heat water/buildings during excess renewable periods
— **Load shifting:** Schedule energy-intensive activities for optimal times

**Efficiency optimization:**
— **Waste heat recovery:** Use EV charger and inverter waste heat for building heating
— **Seasonal storage:** Larger battery bank provides seasonal energy balance
— **Grid integration:** Smart inverters provide voltage/frequency support

**Part 5: Complete system design**

**Generation capacity:**
— Solar: 1.5 MW (provides ~7.5 MWh/day average)
— Wind: 2.0 MW (provides ~14.4 MWh/day average)
— **Total renewable generation: ~22 MWh/day average**

**Storage and backup:**
— Battery storage: 33.4 MWh
— Diesel backup: 2 MW (for emergency periods)

**System efficiency analysis:**

**Annual energy balance:**
— Renewable generation: 22 MWh/day × 365 = 8,030 MWh/year
— Community demand: 10 MWh/day × 365 = 3,650 MWh/year
— Excess generation: 4,380 MWh/year (120% oversupply for reliability)

**Overall system efficiency:**
— Direct renewable use (60% of demand): 90% efficiency
— Battery-mediated use (35% of demand): 81% efficiency
— Diesel backup (5% of demand): 35% efficiency

**Weighted average efficiency:**
ηtotal = 0.60 × 0.90 + 0.35 × 0.81 + 0.05 × 0.35 = 0.54 + 0.28 + 0.02 = 84%

**Part 6: Advanced optimization features**

**Smart grid technologies:**
— **Predictive control:** Weather forecasting for generation planning
— **Demand response:** Automatic load control during low generation periods
— **Vehicle-to-grid:** EV batteries provide grid storage when parked

**Waste heat recovery:**
— **Heat pump optimization:** Use waste heat from power electronics
— **Thermal mass:** Building thermal storage reduces heating demand
— **Industrial integration:** Coordinate with any local industry for waste heat utilization

**Reliability measures:**
— **Redundancy:** Multiple smaller wind turbines vs single large unit
— **Maintenance scheduling:** Coordinate renewable maintenance with high generation periods
— **Emergency protocols:** Automatic load shedding priorities

**Economic and environmental benefits:**
— **Fuel independence:** 95% renewable energy reduces diesel imports
— **Economic stability:** Predictable electricity costs vs volatile diesel prices
— **Environmental impact:** Minimal emissions, sustainable energy system
— **Energy security:** Reduced dependence on external fuel supplies

**Implementation timeline:**
1. **Phase 1:** Install core renewable generation and basic battery storage
2. **Phase 2:** Expand storage capacity and implement smart grid features
3. **Phase 3:** Add advanced optimization and vehicle-to-grid integration

**Performance metrics:**
— Overall efficiency: 84% (excellent for isolated grid)
— Renewable fraction: 95% (highly sustainable)
— System reliability: >99.9% (with proper sizing)
— Economic payback: 8-12 years vs continued diesel generation

This integrated design demonstrates how multiple energy technologies can be combined to achieve high efficiency while ensuring reliability in challenging environments, providing a model for sustainable island energy systems worldwide.

Question 5: Advanced Critical Analysis

A technology journalist claims that «efficiency improvements always reduce environmental impact and save money, so maximum efficiency should be the primary goal in all energy systems.» Critically evaluate this statement by analyzing: (a) situations where maximum efficiency might not be optimal, (b) trade-offs between efficiency, cost, and environmental impact, (c) the relationship between efficiency and sustainability, (d) examples where lower efficiency systems might be preferable. Consider lifecycle analysis, economic factors, and practical limitations in your response.

Answer
Critical Evaluation: The statement is OVERSIMPLIFIED and contains several flawed assumptions.

Part (a): When maximum efficiency isn’t optimal

**Economic limitations:**
— **Diminishing returns:** Going from 80% to 90% efficiency often costs more than the energy savings justify
— **Example:** Super-efficient windows may cost $10,000 more but save only $200/year in energy
— **Optimal efficiency:** Usually 70-85% efficiency provides best economic return

**Practical constraints:**
— **System complexity:** Higher efficiency often requires more complex control systems
— **Maintenance requirements:** Efficient systems may need more frequent, specialized maintenance
— **Reliability trade-offs:** Simple, robust systems may be preferable despite lower efficiency

**Performance characteristics:**
— **Load matching:** Systems optimized for peak efficiency may perform poorly at partial loads
— **Example:** High-efficiency heat pumps lose effectiveness in very cold climates

Part (b): Trade-offs analysis

**Efficiency vs Cost:**
**Case study — LED lighting:**
— **Standard LED:** 90% efficient, $5 per bulb, 25,000-hour life
— **Premium LED:** 95% efficient, $25 per bulb, 50,000-hour life
— **Analysis:** Premium LED saves 5% energy but costs 5× more
— **Conclusion:** Standard LED provides better cost-effectiveness for most applications

**Efficiency vs Environmental Impact:**
**Manufacturing intensity:**
— **High-efficiency products:** Often require rare materials, complex manufacturing
— **Example:** High-efficiency solar panels use more silver, specialized semiconductors
— **Lifecycle assessment:** Must consider manufacturing emissions vs operational savings

**Transportation and installation:**
— **Weight and size:** Efficient systems may be heavier, requiring more transportation energy
— **Installation complexity:** May require specialized equipment, skilled labor

**Efficiency vs Reliability:**
**System robustness:**
— **Simple systems:** Lower efficiency but proven reliability over decades
— **Complex systems:** High efficiency but potential failure points
— **Example:** Traditional incandescent vs CFL vs LED progression

Part (c): Efficiency and sustainability relationship

**Positive correlations:**
— **Resource conservation:** Higher efficiency reduces raw material consumption
— **Emission reduction:** Less energy use typically means lower emissions
— **Economic sustainability:** Lower operating costs improve long-term viability

**Potential contradictions:**
— **Rebound effects:** Efficiency improvements may lead to increased usage
— **Example:** More efficient cars leading to more driving
— **Jevons paradox:** Historical tendency for efficiency gains to increase total resource consumption

**Lifecycle considerations:**
— **Embodied energy:** Manufacturing energy for efficient equipment
— **End-of-life impact:** Disposal/recycling of complex efficient systems
— **Durability:** Simple, maintainable systems may last longer

Part (d): Examples favoring lower efficiency systems

**Renewable energy systems:**
**Solar thermal vs photovoltaic:**
— **Solar thermal:** 60-70% efficiency for heating applications
— **PV + heat pump:** 20% × 300% COP = 60% overall efficiency
— **Advantage of solar thermal:** Simpler, more reliable, lower cost per BTU

**Heating systems:**
**Direct combustion vs heat pumps:**
— **Gas furnace:** 85-95% efficiency, simple, reliable
— **Heat pump:** 300-400% efficiency but expensive, complex
— **Climate dependency:** Heat pumps lose efficiency in cold weather
— **Infrastructure:** Gas systems use existing distribution networks

**Transportation:**
**Bicycles vs electric vehicles:**
— **Bicycle:** «Low» mechanical efficiency (~25%) but zero energy input
— **Electric vehicle:** High efficiency (85%) but requires electrical infrastructure
— **Overall impact:** Bicycle superior for short trips despite «lower efficiency»

**Industrial processes:**
**Batch vs continuous processing:**
— **Continuous:** Higher thermal efficiency, complex control systems
— **Batch:** Lower efficiency but flexible, easier maintenance
— **Optimal choice:** Depends on production volume, product variety

Advanced considerations:**

**System-level optimization:**
— **Component efficiency:** Optimizing individual components may suboptimize the system
— **Example:** Oversized efficient motors running at low load vs properly sized standard motors
— **System thinking:** Focus on overall performance, not component metrics

**Economic efficiency vs energy efficiency:**
— **Different metrics:** $/kWh saved vs kWh saved
— **Investment optimization:** Limited capital should maximize economic return
— **Policy implications:** Subsidies should target cost-effective efficiency improvements

**Temporal factors:**
— **Technology evolution:** Today’s efficient technology may be obsolete tomorrow
— **Example:** Investing heavily in efficient fossil fuel systems vs waiting for renewable alternatives
— **Timing optimization:** Balance current efficiency vs future technology adoption

**Regional variations:**
— **Climate dependency:** Efficiency technologies perform differently in various climates
— **Infrastructure:** Efficiency solutions must match local capabilities
— **Economic development:** Appropriate technology for economic conditions

Conclusion and recommendations:**

**The journalist’s statement fails because:**
1. **Economic optimization:** Maximum efficiency rarely equals minimum cost
2. **Practical constraints:** Real-world factors limit achievable efficiency
3. **System complexity:** Higher efficiency often reduces reliability
4. **Lifecycle impacts:** Manufacturing and disposal costs matter
5. **Context dependency:** Optimal solutions vary by application

**Better approach:**
1. **Cost-effectiveness analysis:** Optimize $/kWh saved, not just efficiency
2. **Lifecycle assessment:** Consider full environmental impact
3. **System perspective:** Optimize overall performance, not components
4. **Context sensitivity:** Match solutions to specific conditions
5. **Multiple objectives:** Balance efficiency, cost, reliability, and environmental impact

**Policy implications:**
— **Efficiency standards:** Should target cost-effective levels, not maximum possible
— **Incentive programs:** Should reward cost-effective improvements
— **Research priorities:** Focus on breakthrough technologies, not incremental gains
— **Building codes:** Balance efficiency requirements with affordability

The goal should be **optimal efficiency** (balancing multiple factors) rather than **maximum efficiency** (single-factor optimization). This requires sophisticated analysis considering economics, environmental impact, reliability, and practical constraints — demonstrating why simplistic statements about efficiency can be misleading in complex energy systems.

🤔 Lesson Reflection and Self-Assessment

💭 Knowledge Self-Check

Conceptual Understanding (Rate 1-5):

□ I understand efficiency as the ratio of useful output to total input energy

□ I can calculate efficiency percentages and identify energy losses

□ I can analyze energy flow diagrams and Sankey diagrams

□ I understand factors that affect system efficiency

□ I can evaluate trade-offs between efficiency, cost, and environmental impact

Problem-Solving Skills Assessment:

Which strategies worked best for efficiency problems?

  • Identifying useful vs waste energy clearly
  • Using energy conservation to find missing values
  • Creating energy flow diagrams for complex systems
  • Comparing different efficiency improvement options
  • Calculating return on investment for efficiency upgrades

What challenges did you encounter?

  • Distinguishing between different types of efficiency
  • Understanding multi-stage system efficiency calculations
  • Balancing efficiency with other design constraints
  • Interpreting real-world efficiency data and ratings

Real-World Connections:

How can you apply efficiency concepts daily?

  • Comparing appliance efficiency ratings when purchasing
  • Understanding energy bills and identifying savings opportunities
  • Evaluating transportation choices based on efficiency
  • Recognizing efficiency improvements in technology development
  • Making informed decisions about home energy upgrades

Language Development Reflection:

New physics vocabulary mastered:

□ Can use efficiency terminology correctly and precisely

□ Understand percentage calculations and ratio expressions

□ Can explain energy flow and loss mechanisms clearly

□ Comfortable interpreting efficiency data and specifications

Communication goals for next lesson:

• Practice explaining efficiency concepts using everyday examples

• Use precise mathematical language for efficiency calculations

• Develop confidence in presenting energy analysis results

Future Learning Goals:

What efficiency topics would you like to explore further?

  • Advanced thermodynamic efficiency limits (Carnot cycle)
  • Efficiency in renewable energy systems
  • Industrial process optimization and energy management
  • Economic analysis of efficiency investments
  • Efficiency standards and environmental policy

How will this knowledge help in future physics topics?

  • Understanding heat engines and thermodynamic cycles
  • Analyzing electrical circuits and power systems
  • Studying energy conversion in electromagnetic systems
  • Connecting to environmental physics and sustainability

🎯 Action Plan for Continued Learning:

This week I will:

□ Research efficiency ratings of appliances in my home

□ Practice calculating efficiency for different energy systems

□ Explore energy efficiency programs and policies in my region

□ Create Sankey diagrams for household energy use

□ Prepare questions about advanced efficiency topics for next class