Hybrid &Electrical Vehicles (EX-703 (A)) - Unit 5 Short Notes

UNIT 5: ENERGY AUDIT & MANAGEMENT

1.0 FOUNDATIONS OF ENERGY MANAGEMENT

1.1 Definition, Objectives, and Scope

• Energy Management: The proactive, organized, and systematic coordination of the procurement, conversion, distribution, and use of energy to meet the requirements of an organization with the goal of minimizing energy costs and environmental impact.

• Key Objectives:

  • Reduce energy consumption and costs.

  • Minimize environmental impact (GHG emissions).

  • Enhance energy security and reliability.

  • Comply with regulations and standards.

• Scope: Covers all energy forms (electrical, thermal, fuel) across all departments/processes of an organization.

1.2 General Principles

• Top Management Commitment: Essential for policy and resource allocation.

• Establish Baseline: Measure and document current energy consumption (Energy Baseline).

• Set Targets: Define measurable, time-bound energy reduction goals.

• Implement Action Plans: Deploy energy-efficient technologies and operational practices.

• Monitor & Verify: Continuously track performance against targets.

• Continuous Improvement: Regular review and updation of the management system (Plan-Do-Check-Act cycle).

1.3 Role, Duties, and Responsibilities of an Energy Manager

• Role: A designated person responsible for developing and implementing the organization's energy policy and management plan.

• Key Duties:

  • Conduct energy audits and prepare reports.

  • Analyze energy consumption data.

  • Identify and evaluate Energy Conservation (EC) opportunities.

  • Prepare feasibility reports and investment proposals.

  • Implement and monitor EC projects.

  • Ensure compliance with BEE and other regulations.

  • Promote energy awareness among staff.

  • Maintain records of energy use and savings.

1.4 Barriers to Effective Energy Management & Elimination Strategies

1.5 Environmental Aspects of Energy Consumption

• Non-Renewable Sources (Coal, Oil, Gas):

  • Impacts: GHG emissions (CO₂, CH₄), air pollutants (SOₓ, NOₓ, PM), water pollution, land degradation, thermal pollution.

• Renewable Sources (Solar, Wind, Biomass):

  • Impacts: Land use change, visual impact, noise (wind), embodied energy in manufacturing, waste disposal (biomass ash), intermittent nature requiring backup.

• Elements of Systematic Environmental Assessment:

  1. Goal & Scope Definition: Define purpose and boundaries.

  2. Inventory Analysis: Quantify energy/material inputs and emissions/outputs (Life Cycle Assessment - LCA).

  3. Impact Assessment: Evaluate significance of environmental impacts (global warming, acidification, etc.).

  4. Interpretation & Reporting: Identify hotspots and recommend mitigation.

[!TIP] Exam Focus: Be prepared to contrast environmental impacts of fossil fuels vs. renewables. Link energy conservation directly to reduced environmental footprint.

2.0 ENERGY AUDIT PROCESS & METHODOLOGY

2.1 Definition and Purpose

• Energy Audit: A systematic procedure to quantify energy consumption of a facility, identify areas of energy wastage, and recommend improvement measures with cost-benefit analysis.

• Purpose: To establish energy baseline, identify savings potential, prioritize actions, and provide a roadmap for energy management.

2.2 Types of Energy Audits

2.3 Steps in Conducting an Energy Audit

1. Planning & Organization: Define scope, team, schedule, data requirements.

2. Data Collection:

   • Utility bills (electricity, fuel, water) for 1-3 years.

   • Process flow diagrams, equipment lists, operating schedules.

   • On-site measurements using instruments (power analyzer, flue gas analyzer, etc.).

   • Interviews with operational staff.

3. Data Analysis: Calculate specific energy consumption (SEC), compare with benchmarks, perform load analysis, identify deviations.

4. Energy Conservation Opportunities (ECOs) Identification: For each system (electrical, thermal, mechanical).

5. Technical & Financial Evaluation: Calculate energy savings, cost savings, investment, payback, NPV, IRR.

6. Report Preparation: Executive summary, methodology, findings, recommendations with priority ranking, action plan.

7. Implementation & Follow-up: Assist in project execution and monitor savings realization.

2.4 Manners and Intervals as per BEE Regulations (for Designated Consumers)

• Frequency: Once every 3 years from the date of the previous audit.

• Manner: Must be conducted by ** Accredited Energy Auditors** from the BEE panel.

• Compliance: Audit report must be submitted to the Bureau of Energy Efficiency (BEE) and the State Designated Agency (SDA) within the stipulated time.

• Format: Must follow the BEE-prescribed format (Form 1 to Form 5).

2.5 Equipment and Instruments for Energy Auditing

• Common List: Power Analyzer, Clamp Meter, Lux Meter, Thermometer/Infrared Gun, Flue Gas Analyzer (O₂, CO, CO₂, Stack Temp), Anemometer, Tachometer, Pressure Gauge, Flow Meter, Data Logger.

• Working Principle: Power Analyzer:

  • Measures electrical parameters: Voltage (V), Current (I), Power (W, kW, kVA), Power Factor (PF), Frequency (Hz), Harmonics (THD).

  • Principle: Uses voltage and current transformers to sample waveforms. Digital signal processing calculates RMS values, real/reactive/apparent power, and harmonic distortion by Fourier analysis. Provides real-time data and logs for analysis.

[!TIP] Exam Focus: Know the BEE audit frequency (3 years) and that it's mandatory for Designated Consumers. Be ready to list instruments and explain one (usually Power Analyzer or Flue Gas Analyzer).

3.0 ELECTRICAL SYSTEMS & POWER QUALITY

3.1 Power Factor (PF)

• Definition: Ratio of Real Power (kW) to Apparent Power (kVA). $$\displaystyle PF = \frac{kW}{kVA} = \cos\phi $$.

• Causes of Low PF: Inductive loads (motors, transformers, ballasts) cause current to lag voltage.

• Disadvantages:

  • Increased current for same real power → higher I²R losses.

  • Larger conductor/cable sizing required.

  • Reduced system capacity (kVA limit).

  • Voltage drop, poor voltage regulation.

  • Penalty charges from utilities.

• Benefits of Improvement:

  • Reduced current → lower losses, better voltage regulation.

  • Increased system capacity (same kVA can deliver more kW).

  • Avoidance of PF penalty charges.

  • Reduced cable/transformer sizing for new installations.

• Methods of PF Improvement: Use of Capacitor Banks

  • Capacitors supply leading current to cancel lagging current from inductive loads.

  • Sizing (kVAr): $$\displaystyle Q_c = P (\tan\phi_1 - \tan\phi_2) $$

    • $P$ = Real Power (kW)

    • $$\displaystyle \phi_1 $$ = Initial angle (cos⁻¹ PF₁)

    • $$\displaystyle \phi_2 $$ = Desired angle (cos⁻¹ PF₂)

  • Placement: Best location is as close as possible to the inductive load (at load end or distribution panel). This minimizes line current and losses between capacitor and load.

• Calculation Problem Structure:

  1. Calculate initial kW: $$\displaystyle kW = kVA_{max} \times PF_{initial} $$

  2. Find initial $$\displaystyle \phi_1 $$, desired $$\displaystyle \phi_2 $$.

  3. Required kVAr: $$\displaystyle Q_c = kW (\tan\phi_1 - \tan\phi_2) $$

  4. New kVA: $$\displaystyle kVA_{new} = \frac{kW}{PF_{new}} $$

  5. Check if $$\displaystyle Q_c $$ provided is sufficient; if not, calculate new PF after given kVAr.

[!EXAMPLE] PF Calculation (from Jun 2025 paper):

Given: Max Demand = 800 kVA, PF = 0.80 lag. Install 100 kVAr capacitors. Find new PF & improvement.

Solution:

1. $$\displaystyle kW = 800 \times 0.80 = 640 $$ kW

2. $$\displaystyle \phi_1 = \cos^{-1}(0.80) = 36.87^\circ $$, $$\displaystyle \tan\phi_1 = 0.75 $$

3. $$\displaystyle Q_{initial} = 800 \times \sin(36.87^\circ) = 800 \times 0.6 = 480 $$ kVAr (lag)

4. $$\displaystyle Q_{new} = 480 - 100 = 380 $$ kVAr (lag)

5. $$\displaystyle kVA_{new} = \sqrt{640^2 + 380^2} = \sqrt{409600 + 144400} = \sqrt{554000} = 744.3 $$ kVA

6. $$\displaystyle PF_{new} = \frac{640}{744.3} = 0.86 $$ lag

7. Improvement = 0.86 - 0.80 = 0.06 (or 6%)

8. Penalty: Min PF = 0.90. New PF (0.86) is 4% below min. Penalty = 4 × 20,000 = Rs. 80,000.

3.2 Energy Efficient Motors vs. Standard Motors

• Motor Loading % Calculation (from Jun 2025 paper):

  • Input Power (kW) = $$\displaystyle \frac{\sqrt{3} \times V \times I \times PF}{1000} $$

  • Loading % = $$\displaystyle \frac{\text{Input Power}}{\text{Full Load Rating (kW)}} \times 100\% $$

  • Example: 20 kW motor, V=440V, I=10A, PF=0.78.

    • Input = $$\displaystyle \frac{1.732 \times 440 \times 10 \times 0.78}{1000} = \frac{5956}{1000} = 5.956 $$ kW

    • Loading % = $$\displaystyle \frac{5.956}{20} \times 100 = 29.78\% $$ (Very underloaded. Consider downsizing).

3.3 Harmonics

• Major Problems:

  • Additional Heating: In motors, transformers, cables (due to eddy currents, hysteresis).

  • Torque Pulsations & Cogging: In motors, causing vibration and noise.

  • Neutral Conductor Overloading: In 3-phase 4-wire systems, triplen harmonics (3rd, 9th...) add in neutral.

  • Metering & Relay Malfunction: False readings, nuisance tripping.

  • Capacitor Failure: Resonance with system inductance, leading to overcurrent and failure.

  • Telephone Interference.

• Equipment Contributing to Harmonics:

  • Non-linear Loads: Draw current in non-sinusoidal pulses.

  • Examples: Variable Frequency Drives (VFDs), Uninterruptible Power Supplies (UPS), Switch-mode power supplies (computers, LED drivers), Arc furnaces, Rectifiers (DC drives, battery chargers).

  • Why? They use diodes/thyristors to convert AC to DC. The firing angle control or diode bridge draws current only when voltage exceeds DC bus voltage, creating flat-topped, pulsed current waveforms rich in harmonics.

• Harmonic Distortion Evaluation:

  • Total Harmonic Distortion (THD): $$\displaystyle THD = \frac{\sqrt{\sum_{h=2}^{40} (M_h)^2}}{M_1} \times 100\% $$

    • $$\displaystyle M_h $$ = RMS value of h-th harmonic.

    • $$\displaystyle M_1 $$ = RMS value of fundamental.

  • Total Demand Distortion (TDD): Similar to THD, but denominator is demand current (maximum average current over a period), not fundamental. Used for sizing equipment.

  • Procedure: Use a Power Quality Analyzer to record voltage/current waveforms over 24-48 hours. The analyzer computes THD/TDD for current and voltage at each harmonic order.

3.4 HT vs. LT Systems

3.5 Transformer Losses and Minimization

• Losses:

  • Core (Iron) Losses: Constant, occur whenever energized. Hysteresis + Eddy Current.

  • Copper (I²R) Losses: Vary with load square.

• Minimization Strategies:

  • Select High-Efficiency Transformer: (e.g., ONAN/ONAF, amorphous core).

  • Right-Sizing: Avoid over-sizing; operate near rated load (typically 50-75% for best efficiency).

  • Parallel Operation: Use multiple smaller transformers instead of one large one at low load.

  • Reduce Harmonics: Use K-rated transformers or detuned filters; harmonics increase losses.

  • Proper Maintenance: Tight connections, clean cooling fins.

  • Optimal Loading: For two transformers, compare part-load efficiencies. Run one at near full load if total load is low.

4.0 THERMAL SYSTEMS & BOILERS

4.1 Boiler Efficiency

• Direct Method (Input-Output):

  • $$\displaystyle \eta_{direct} = \frac{\text{Steam Output (kcal/hr)} \times (\text{Enthalpy of Steam} - \text{Enthalpy of Feedwater})}{\text{Fuel Input (kcal/hr)}} \times 100\% $$

  • Advantage: Simple, quick.

  • Disadvantage: Does not identify sources of loss.

• Indirect Method (Loss Method):

  • $$\displaystyle \eta_{indirect} = 100 - (\text{Sum of all % losses}) $$

  • Losses: Flue gas loss, moisture in fuel/air, unburnt carbon (CO, C), radiation/convection loss, blowdown loss.

  • Advantage: Identifies magnitude of each loss → targets for improvement.

• Efficiency on GCV vs. NCV Basis:

  • GCV (Higher Heating Value): Includes latent heat of vaporization in fuel moisture/hydrogen.

  • NCV (Lower Heating Value): Excludes that latent heat (assumes water vapor leaves as vapor).

  • Conversion: $$\displaystyle \eta_{NCV} = \eta_{GCV} \times \frac{GCV}{NCV} $$

  • Example (Dec 2024): Given $$\displaystyle \eta_{GCV}=90\% $$, Fuel: 1.5% moisture, 15% hydrogen. GCV=11500 kcal/kg.

    • NCV ≈ GCV - [9 × H₂% + Moisture% × (587 + 0.45 × T)] (approx. formula)

    • Simplified: NCV ≈ GCV - [0.09 × H₂% + Moisture%] × 587 (if steam at 100°C)

    • Loss due to H₂ = 0.09 × 15 = 1.35 kg/kg fuel → 1.35 × 587 = 792.45 kcal/kg

    • Loss due to moisture = 0.015 × 587 = 8.805 kcal/kg (approx.)

    • Total loss ≈ 801.25 kcal/kg

    • NCV = 11500 - 801.25 = 10698.75 kcal/kg

    • $$\displaystyle \eta_{NCV} = 90 \times \frac{11500}{10698.75} = 90 \times 1.0748 = 96.73\% $$

• Part-Load Efficiency & Boiler Selection (Jun 2025 paper):

  • Principle: Boiler efficiency drops at part load. Two boilers at moderate load may be more efficient than one at full load and one very low.

  • Strategy: Compare total fuel consumption for different operating combinations.

  • Example: Two 15 TPH boilers, full load eff=82%. At 75% load (11.25 TPH), eff=78%. At 45% load (6.75 TPH), eff=66%. Required steam = 20 TPH.

    • Option 1: Both at 10 TPH (66.7% load). Interpolate eff between 45% (66%) and 75% (78%).

      • Eff @ 66.7% ≈ 66 + (66.7-45)/(75-45) × (78-66) = 66 + (21.7/30)×12 = 66 + 8.68 = 74.68%.

      • Total fuel = $$\displaystyle \frac{20 \times 1000}{74.68} = 267.8 $$ kg/h (assuming steam enthalpy constant).

    • Option 2: One at 15 TPH (100% load, 82%), other at 5 TPH (33.3% load). Interpolate for 33.3% load.

      • Eff @ 33.3% ≈ 66 + (33.3-45)/(75-45) × (78-66) = 66 + (-11.7/30)×12 = 66 - 4.68 = 61.32%.

      • Fuel = $$\displaystyle \frac{15}{0.82} + \frac{5}{0.6132} = 18.29 + 8.15 = 26.44 $$ kg/h.

    • Conclusion: Option 1 (both at ~67% load) saves ~1.2% fuel compared to Option 2. Prefer balanced loading.

4.2 Combustion Optimization

• Excess Air: Air supplied beyond theoretical air requirement for complete combustion.

• Importance: Ensures complete combustion (prevents CO, C formation), but too much carries heat away in flue gas.

• Achieving Minimum Excess Air:

  • Use oxygen trim control in boiler control system.

  • Regular tuning of burners.

  • Maintain proper air-to-fuel ratio using flue gas analyzer (target O₂: 3-5% for gas, 4-6% for coal/oil).

  • Seal air leaks in furnace and ducts.

• Why Low-Pressure Steam is More Efficient?

  • For the same quantity of heat, low-pressure steam has higher enthalpy of evaporation (latent heat) than high-pressure steam.

  • Example: At 10 bar, h_fg ≈ 2015 kJ/kg; at 2 bar, h_fg ≈ 2202 kJ/kg.

  • Result: For the same process heat requirement, less steam mass flow is needed at lower pressure → lower boiler fuel consumption, smaller pipe sizes, less heat loss in distribution.

4.3 Steam System

• Steam Distribution Losses:

  • Heat loss from un-insulated pipes.

  • Pressure drop → lower pressure at user → need for initial higher pressure (wasteful).

  • Condensate not recovered → loss of heat and water.

  • Steam leaks.

  • Mitigation: Proper insulation, drip traps, pressure reducing stations at user, condensate recovery, leak detection.

• Steam Traps:

  • Function: Automatically discharge condensate, air, and non-condensable gases while preventing steam leakage.

  • Working Types:

    • Mechanical (Float & Thermostatic): Float responds to condensate level.

    • Thermostatic (Bimetallic, Bellows): Temperature sensing element.

    • Thermodynamic (Disc): Uses kinetic energy difference of steam vs. condensate.

  • Performance Assessment Methods:

    1. Visual/Audio: Listen for continuous flow (steam leak) or no discharge (blockage).

    2. Temperature Measurement: Upstream hot, downstream cold (if discharging condensate).

    3. Ultrasonic Testing: Detects high-frequency sound of steam leakage.

    4. Infrared Thermography: Hot trap = failed open; cold trap = failed closed.

    5. Invasive Testing: Isolate and test in workshop.

• Condensate Recovery Process & Benefits:

  • Process: Collect condensate from steam traps and process equipment → filter → return to boiler feedwater tank (often via pump for pressure).

  • Benefits:

    • Saves heat energy (condensate at 80-100°C).

    • Saves water and associated treatment cost.

    • Reduces boiler blowdown (lower TDS).

    • Improves boiler efficiency.

• Flash Steam Utilization:

  • Concept: When high-pressure condensate is discharged to a lower pressure, a portion of it flashes into steam (flash steam).

  • Example: Condensate at 10 bar (180°C) discharged to a 2 bar deaerator. Enthalpy at 10 bar liquid ≈ 763 kJ/kg, at 2 bar saturated liquid ≈ 504 kJ/kg, saturated vapor ≈ 2736 kJ/kg.

    • Flash steam % = $$\displaystyle \frac{h_{f@10bar} - h_{f@2bar}}{h_{fg@2bar}} \times 100 = \frac{763-504}{2736-504} \times 100 = \frac{259}{2232} \times 100 \approx 11.6\% $$.

  • Utilization: Capture this flash steam and use it for low-pressure applications (e.g., space heating, process at <2 bar) via a flash steam recovery system.

4.4 Thermic Fluid Heating System

• Working Principle:

  1. Heater: Thermic fluid (e.g., Dowtherm, Therminol) is heated in a fired/electrical heater to 300-400°C at near atmospheric pressure.

  2. Circulation: Hot fluid is pumped to heat exchangers (consumers) where it transfers heat to the process.

  3. Return: Cool fluid returns to heater by gravity or pump.

• Advantages over Steam (for specific apps):

  • Operates at high temperature at low pressure → no thick-walled pressure vessels, no licensing, lower safety risks.

  • No phase change → no blowdown, no water treatment, no condensate recovery issues.

  • Precise temperature control.

  • Suitable for temperature-sensitive processes (no steam condensation).

  • Disadvantages: Fluid degradation at high temp, fire risk if leaks, higher initial cost.

4.5 Economic Thickness of Insulation

• Concept: The insulation thickness that minimizes the total annual cost (sum of heat loss cost + annualized insulation cost).

• Process:

  1. Calculate heat loss ($Q$) for varying thickness ($x$): $$\displaystyle Q \propto \frac{1}{x} $$.

  2. Annual heat loss cost = $$\displaystyle Q \times \text{Operating Hours} \times \text{Fuel Cost} / \text{Boiler Efficiency} $$.

  3. Annualized insulation cost = (Insulation Cost) × (Capital Recovery Factor).

  4. Plot Total Annual Cost vs. Thickness. Minimum point = Economic Thickness.

• Factors: Fuel cost, insulation cost, ambient temperature, surface temperature, operating hours.

4.6 Furnaces

• Concept & Classifications:

  • Definition: Enclosed structure for applying heat to materials (metals, ceramics).

  • Classifications:

    • By Heat Source: Oil/gas fired, electric (resistance/arc), coal fired.

    • By Operation: Batch, continuous.

    • By Atmosphere: Air, controlled (inert, reducing).

    • By Temperature: Low (<700°C), medium (700-1200°C), high (>1200°C).

• Role in Steel Industries (with Diagram):

  • Diagram:

    DiagramCANVAS: Show a cross-section of a typical reheating furnace. Label: furnace chamber, burners (on sides/roof), hearth (with moving skids or pusher), flue gas exit, insulation, steel billets/bars being heated.

  • Role: Heat steel billets/slabs to 1200-1300°C for hot rolling. Key requirements: uniform temperature, high throughput, low fuel consumption.

• Advantages of Minimum Excess Air in Furnaces:

  • Reduces flue gas volume → lower sensible heat loss.

  • Reduces radiation loss from flame (less N₂ dilution).

  • Improves heat transfer (higher flame temperature).

  • Reduces NOₓ formation (less N₂ available).

  • How to Achieve: Oxygen trim control, proper burner tuning, regular maintenance, sealed furnace.

5.0 PUMPING, FANNING & REFRIGERATION SYSTEMS

5.1 Pumping Systems

• Factors Affecting Performance:

  • System Curve: Head required = Static Head + Friction Head (∝ Flow²).

  • Pump Curve: Head vs. Flow (decreases with flow). Best Efficiency Point (BEP) is the design point.

  • Affinity Laws (for centrifugal pumps):

    1. $Q \propto N$ (Flow ∝ Speed)

    2. $$\displaystyle H \propto N^2 $$ (Head ∝ Speed²)

    3. $$\displaystyle P \propto N^3 $$ (Power ∝ Speed³)

  • Viscosity, NPSH Available vs. Required.

• Energy Conservation Opportunities:

  • Right-Sizing: Avoid over-pumping; select pump near BEP for typical flow.

  • Throttling Reduction: Use VFDs to control flow by speed, not control valves (saves significant energy).

  • Parallel/Series Operation: Optimize number of pumps running.

  • Reduce System Resistance: Larger pipes, fewer bends, clean filters.

  • Maintenance: Impeller trimming, seal/bearing replacement.

• Significance of Parallel Operation in Centrifugal Pumps:

  • Provides redundancy (one can fail).

  • Allows operation at varying flow demands by switching pumps on/off.

  • Combined Curve: For identical pumps, parallel operation adds flows at same head. System flow increases, but head remains same as single pump at that flow.

  • Caution: Pumps should have steep curves; otherwise, one pump may "hog" flow.

5.2 Fan Systems

• Design & Selection Criteria:

  • Required airflow (CFM) and static pressure (inches w.g.) from system curve.

  • Fan Type: Centrifugal (high pressure) vs. Axial (high flow, low pressure).

  • Efficiency: Select fan with peak efficiency near operating point.

  • Material: Corrosion/erosion resistance.

• Performance Evaluation:

  • Measure: Airflow (anemometer), static pressure (manometer), power input (power analyzer).

  • Calculate: Fan Efficiency $$\displaystyle \eta = \frac{Q \times SP}{P_{input}} $$ (in consistent units).

  • Compare with manufacturer's curve.

• Efficient System Operation:

  • Use inlet guide vanes or VFDs for capacity control (preferred over dampers).

  • Minimize system resistance (smooth ducts, proper sizing).

  • Regular cleaning (dust buildup increases resistance).

  • Ensure proper belt tension (if belt-driven).

5.3 Refrigeration Plants

• Factors Affecting Performance & Energy Efficiency:

  • Evaporating Temperature: Lower evaporating temp → higher compressor work.

  • Condensing Temperature: Higher condensing temp → higher compressor work. Affected by ambient temp, condenser cleanliness.

  • Compressor Efficiency: Volumetric, mechanical, isentropic.

  • Suction/Discharge Pressure Drop: In valves, piping.

  • Subcooling & Superheating: Proper subcooling increases refrigeration effect; optimal superheat protects compressor.

  • Refrigerant Charge: Under/over-charge reduces efficiency.

  • Auxiliary Loads: Fans (condenser, evaporator), pumps (for water-cooled).

  • Maintenance: Clean coils, proper refrigerant charge, oil condition.

6.0 LIGHTING SYSTEMS

6.1 Scope of Energy Conservation

• Largest Opportunity: Lighting accounts for 15-30% of commercial/industrial electricity use.

• Strategies:

  • Technology Upgrade: Replace inefficient lamps (T12, metal halide, CFL) with LEDs.

  • Design Optimization: Use right fixture for task, reduce over-illumination, maximize daylighting.

  • Controls: Occupancy sensors, daylight sensors, timers, dimmers.

  • Maintenance: Regular cleaning, group relamping.

6.2 LED Lighting Technology and Advantages

• Technology: Light Emitting Diode. Semiconductor device that emits light when current flows.

• Advantages over Conventional:

  • High Efficacy: 100-150+ lm/W vs. 60-100 for fluorescent, 15-20 for incandescent.

  • Long Life: 50,000-100,000 hours.

  • Instant On & No UV/IR.

  • Dimmable & Color Tunable (with compatible drivers).

  • Rugged, Low Voltage, Directional.

  • No Mercury (unlike CFL).

6.3 Lighting Economics Calculations

• Energy Savings (Annual): $$\displaystyle E_{save} = (P_{old} - P_{new}) \times H \times N $$

  • $$\displaystyle P_{old}, P_{new} $$ = Power (W) per lamp.

  • $H$ = Operating hours/year.

  • $N$ = Number of lamps.

• Cost Savings (Annual): $$\displaystyle C_{save} = E_{save} \times \text{Electricity Rate} (Rs./kWh) $$

• Simple Payback Period (SPP): $$\displaystyle SPP = \frac{\text{Total Investment Cost}}{C_{save}} $$

• Example (Jun 2025 paper): Replace 500W→350W, 350W→150W, 125W→60W. 4500 hrs/yr, Rs. 5.5/kWh.

  • Assume 1 lamp each for simplicity.

  • Savings per lamp type:

    • Type 1: (500-350)=150W → 150×4500/1000=675 kWh/yr → 675×5.5=Rs. 3712.5

    • Type 2: (350-150)=200W → 200×4500/1000=900 kWh/yr → 900×5.5=Rs. 4950

    • Type 3: (125-60)=65W → 65×4500/1000=292.5 kWh/yr → 292.5×5.5=Rs. 1608.75

  • Total Annual Savings = 3712.5 + 4950 + 1608.75 = Rs. 10,271.25

  • Total Investment = Cost of new lamps (not given, assume provided or calculate if costs given). SPP = Investment / 10271.25.

6.4 Concept of Illumination (Lux)

• Luminous Flux (Lumens, lm): Total light output from source.

• Illuminance (Lux, lx): Luminous flux per unit area. 1 Lux = 1 Lumen/m².

• Measurement: Using a Lux Meter (photocell sensor calibrated to human eye response).

• Design: Determined by task requirements (e.g., office: 300-500 lux, warehouse: 100-200 lux). Calculated using Lumen Method:

  • $$\displaystyle E_{avg} = \frac{N \times \Phi \times CU \times LLF}{A} $$

    • $N$ = No. of luminaires.

    • $\Phi$ = Lumens per lamp.

    • $CU$ = Coefficient of Utilization (fixture efficiency, room reflectance).

    • $LLF$ = Light Loss Factor (maintenance, dirt).

    • $A$ = Area (m²).

[!TIP] Exam Focus: Lighting economics (SPP) is a sure question. Memorize formulas for energy saving, cost saving, SPP. Know LED advantages and lux definition.

7.0 FINANCIAL ANALYSIS & ECONOMICS

7.1 Life Cycle Costing (LCC)

• Definition: Total cost of owning and operating an asset over its entire life, including initial investment, operation, maintenance, and disposal.

• Process:

  1. Identify all cost components (capital, O&M, energy, replacement, salvage).

  2. Convert all costs to present value using discount rate.

  3. Sum all present values → LCC.

• Applications: Compare design alternatives, equipment selection, justify higher upfront cost for efficiency.

• Significance: Avoids "first-cost" bias; reveals true cost-effectiveness over time.

7.2 Investment Evaluation Methods

• Simple Payback Period (SPP):

  • Definition: Time required for cumulative savings to equal initial investment.

  • Calculation: $$\displaystyle SPP = \frac{\text{Initial Investment}}{\text{Annual Net Savings}} $$

  • Significance: Simple, intuitive measure of risk/recovery time.

  • Merits: Easy to calculate, understand. Emphasizes liquidity and short-term risk.

  • Demerits: Ignores cash flows after payback, time value of money, profitability.

• Net Present Value (NPV):

  • Definition: Sum of all future cash flows (savings - costs) discounted to present value.

  • Formula: $$\displaystyle NPV = -I_0 + \sum_{t=1}^{n} \frac{CF_t}{(1+r)^t} $$

    • $$\displaystyle I_0 $$ = Initial Investment.

    • $$\displaystyle CF_t $$ = Net cash flow in year t.

    • $r$ = Discount rate (cost of capital).

    • $n$ = Life of project.

  • Decision Rule: Accept if NPV > 0.

  • Advantages over Payback: Considers time value of money, entire project life, all cash flows.

• Internal Rate of Return (IRR):

  • Concept: The discount rate ($r$) that makes NPV = 0.

  • Decision Rule: Accept if IRR > Required Rate of Return (hurdle rate).

  • Represents the true annualized rate of return on investment.

• Risk Analysis in Payback:

  • Sensitivity Analysis: Vary key assumptions (energy savings, fuel cost, operating hours) to see impact on SPP.

  • Scenario Analysis: Best-case, worst-case, most-likely scenarios.

  • Monte Carlo Simulation: Probabilistic modeling of uncertain variables.

  • Purpose: To assess the robustness of the project and the likelihood of achieving the estimated payback.

7.3 Discounted Cash Flow (DCF) & Discount Period

• DCF: Method of valuing an investment by discounting its expected future cash flows to present value.

• Discount Period: The time period (usually yearly) over which cash flows are discounted. Each year's cash flow is discounted by $$\displaystyle (1+r)^t $$, where t is the year number.

[!EXAMPLE] NPV Calculation (Dec 2024 paper):

Investment: Rs. 2000, Life: 2 years, Savings: Rs. 22000/year, Discount rate: 15%.

Solution:

Year 0: CF = -2000 (PV = -2000)

Year 1: CF = 22000, PV = 22000 / (1.15)¹ = 22000 / 1.15 = 19130.43

Year 2: CF = 22000, PV = 22000 / (1.15)² = 22000 / 1.3225 = 16637.33

NPV = -2000 + 19130.43 + 16637.33 = Rs. 33,767.76 (Positive, so accept).

8.0 ENERGY SERVICE COMPANY (ESCO) CONCEPT & REGULATORY FRAMEWORK

8.1 Basic Concept and Business Model of ESCOs

• Concept: A company that provides comprehensive energy solutions to clients. Its revenue is directly linked to energy savings achieved.

• Business Model (Performance Contracting):

  1. ESCO conducts free detailed energy audit.

  2. Proposes EC measures with guaranteed savings.

  3. Finances, designs, installs, commissions the project.

  4. Monitors & verifies savings for a contract period (3-10 years).

  5. Client pays ESCO from actual energy cost savings.

  6. After contract, all savings accrue to client.

• Key Feature: Guaranteed Savings → client bears no performance risk.

8.2 Role of ESCOs in Promoting Energy Efficiency

• Overcome Barriers: Provide technical expertise and financing that clients lack.

• De-risk Projects: Guarantee savings, assume technical/financial risk.

• One-Stop Solution: Audit → Finance → Implement → Monitor → Maintain.

• Stimulate Market: Create demand for energy-efficient technologies.

• Help Industries: Especially SMEs who cannot afford upfront capital or in-house expertise.

8.3 Bureau of Energy Efficiency (BEE) Regulations

• Manners and Intervals for Energy Audit (Recap from 2.4):

  • Who: Designated Consumers (industries, commercial buildings above certain thresholds).

  • Who Conducts: Accredited Energy Auditors (from BEE panel).

  • Frequency: Once every 3 years from last audit date.

  • Compliance: Submit report to BEE & SDA in prescribed format (Form 1-5) within 4 months of audit completion.

  • Penalty: Non-compliance can lead to fines.

9.0 MISCELLANEOUS & INTEGRATED TOPICS

9.1 Energy Conservation in Various Utility Systems (Integrated View)

• Electrical: PF correction, efficient motors/VFDs, LED lighting, harmonic mitigation, transformer optimization.

• Thermal: Boiler efficiency (excess air, heat recovery), steam system (traps, condensate, insulation), waste heat recovery.

• Mechanical: Pump/fan system optimization (VFDs, right-sizing, parallel operation), HVAC optimization.

• Cross-Cutting: Energy Monitoring & Targeting (M&T) to track performance, Preventive Maintenance, Operator Training, Life Cycle Costing for all investment decisions.

9.2 Common Calculation Problems (Integrated)

1. PF Problem: Given max demand, initial PF, capacitor kVAr → find new PF, new kVA, penalty.

2. Boiler Problem: Given efficiency on GCV, fuel composition → find NCV efficiency. Or, part-load efficiency comparison for multiple boilers.

3. Lighting Problem: Replace lamp types → calculate annual energy/cost savings and SPP.

4. Financial Problem: Given investment, annual savings, life, discount rate → calculate NPV, IRR.

5. Motor Loading Problem: Given voltage, current, PF, full load rating → find loading %.

9.3 Distinction Between Preliminary and Detailed Energy Audit

[!TIP] Final Exam Strategy: For long answers (7 marks), structure as: Definition → Key Components/Steps → Calculations (if applicable) → Advantages/Disadvantages → Conclusion/Recommendation. For numerical problems, show formula → substitution → calculation → boxed answer. Always link recommendations to financial viability (NPV/SPP) and technical feasibility.