Energy Audit & Management (EX-703 (B)) - Unit 3 Short Notes

I. FUNDAMENTALS OF ENERGY AUDIT & MANAGEMENT

Energy Audit: Definition & Purpose

• Definition: Systematic process to determine energy consumption patterns, identify areas of energy wastage, and recommend conservation measures.

• Objectives:

  • Establish baseline energy use.

  • Identify energy-saving opportunities.

  • Reduce energy costs and environmental impact.

  • Ensure compliance with regulations.

• Types of Energy Audit:

• Steps in Conducting Energy Audit:

  1. Planning and organization (team formation, scope definition).

  2. Data collection (energy bills, process data, equipment inventory).

  3. Walk-through inspection (identify obvious wastages).

  4. Detailed measurement and analysis (instrumentation, calculations).

  5. Report preparation (findings, recommendations, economic analysis).

Energy Management: Definition & Principles

• Definition: Application of management techniques to optimize energy use, reduce costs, and minimize environmental impact.

• General Principles:

  • Policy: Clear energy policy from top management.

  • Organization: Dedicated energy manager/team.

  • Planning: Set targets, baseline, action plans.

  • Implementation: Execute conservation measures.

  • Monitoring & Review: Track performance, audit, improve.

• Managerial Functions:

  • Planning, organizing, staffing, directing, controlling.

  • Roles: Develop energy policy, coordinate audits, implement projects, train staff, monitor savings.

• Energy Management in Institutional Organizations:

  1. Form energy committee.

  2. Conduct baseline energy study.

  3. Create awareness among stakeholders.

  4. Implement low-cost/no-cost measures.

  5. Invest in energy-efficient technologies.

  6. Monitor and report regularly.

Regulatory Framework & Procedures (BEE)

• Manners & Intervals:

  • Designated consumers (as per BEE) must conduct energy audit every 3 years.

  • Audit by certified energy auditor.

  • Submit audit report to BEE and designated consumer.

• Duties of Energy Auditor:

  • Assess energy consumption patterns.

  • Identify conservation opportunities.

  • Calculate savings and economics.

  • Prepare audit report as per BEE format.

• Barriers & Elimination:

  • Barriers: Lack of awareness, funds, technical expertise, management commitment, data availability.

  • Elimination: Training programs, financing schemes (ESCO, subsidies), top management support, proper metering.

[!TIP] Exam Focus: Distinguish preliminary vs detailed audit (table format). BEE regulations: 3-year interval, certified auditor. Barriers and solutions are frequently asked.

II. ENVIRONMENTAL ASPECTS OF ENERGY

Environmental Impact of Energy Consumption

• Non-Renewable Sources (coal, oil, gas):

  • Air pollution: SOx, NOx, particulate matter (PM).

  • Greenhouse gas emissions: CO₂, CH₄ (climate change).

  • Water pollution: thermal discharge, contaminants.

  • Land degradation: mining, drilling, waste disposal.

• Renewable Energy Sources (solar, wind, biomass, hydro):

  • Positive: Low operational emissions, sustainable.

  • Negative:

    • Land use (large area for solar/wind farms).

    • Material sourcing (rare earths for wind turbines, PV panels).

    • Intermittency issues (grid stability).

    • Biomass: may cause deforestation if not managed sustainably.

Systematic Environmental Assessment

• Elements:

  1. Screening: Identify potential impacts.

  2. Scoping: Define boundaries and key issues.

  3. Impact Analysis: Quantify and qualify impacts (air, water, noise, ecology).

  4. Mitigation Measures: Propose controls.

  5. Reporting: Environmental Impact Assessment (EIA) report.

  6. Monitoring: Post-implementation checks.

[!TIP] Exam Focus: Compare environmental aspects of non-renewable vs renewable. Systematic assessment steps are often asked in 7-mark questions.

III. ELECTRICAL SYSTEMS & POWER QUALITY

A. Power Factor (PF) Management

Concept & Importance

• Definition: PF = cos φ = Real Power (kW) / Apparent Power (kVA). For three-phase: PF = P / (√3 V I).

• Significance:

  • Reduces current for same real power → lower I²R losses.

  • Improves voltage regulation.

  • Increases system capacity (kVA).

  • Avoids utility penalties.

• Advantages of High PF:

  • Reduced kVA demand → lower maximum demand charges.

  • Lower transmission/distribution losses.

  • Better voltage stability.

  • Improved equipment life.

• Disadvantages of Low PF:

  • Higher current → increased losses, overheating.

  • Voltage drop → poor performance of equipment.

  • Reduced system capacity.

  • Penalty charges from utility.

• Reasons for Low PF:

  • Inductive loads (motors, transformers, reactors).

  • Overloading.

  • Poorly designed systems.

  • Harmonics.

PF Correction Techniques

• Use of Capacitors: Provide leading reactive power to cancel lagging reactive power from inductive loads.

• Best Location for Capacitor Banks:

  • At load terminals (individual correction): Reduces current in entire distribution circuit → maximum energy savings.

  • At distribution boards (group correction): Easier maintenance.

  • At main bus (centralized correction): Simple but less effective for loss reduction.

  • From energy conservation perspective: Install as close to the load as possible.

• HT vs LT Line Connections:

  • LT Side: Capacitors connected on low-voltage side. Advantages: cheaper, easier maintenance, direct reduction in LT kVA demand (billing usually on LT). Disadvantages: higher current in LT cables.

  • HT Side: Capacitors on high-voltage side. Advantages: reduces HT current, less cable losses in HT side. Disadvantages: expensive, complex, may not reduce LT billing demand if transformer is between capacitor and load.

PF Calculation & Penalty Analysis (Problem-Solving Focus)

• Key Formulas:

  • Real Power: \( P = S \times PF \)

  • Reactive Power: \( Q = P \times \tan\phi \)

  • After capacitor installation: \( Q_{\text{new}} = Q - Q_c \)

  • New PF: \( PF_{\text{new}} = \frac{P}{\sqrt{P^2 + Q_{\text{new}}^2}} \)

  • New kVA: \( S_{\text{new}} = \frac{P}{PF_{\text{new}}} \)

• Penalty Computation:

  • Dip in PF = Required PF - Actual PF (if actual < required).

  • Penalty = Dip (%) × Penalty rate per % per month.

Example (Jun 2025):

Max demand = 800 kVA, avg PF = 0.80 lag, min PF = 0.90 lag, penalty = Rs 20,000 per % dip per month. Install 100 kVAr capacitors.

Solution:

1. \( P = 800 \times 0.80 = 640 \text{ kW} \)

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

3. \( Q_1 = P \tan\phi_1 = 640 \times 0.75 = 480 \text{ kVAr} \)

4. \( Q_{\text{new}} = 480 - 100 = 380 \text{ kVAr} \)

5. \( S_{\text{new}} = \sqrt{P^2 + Q_{\text{new}}^2} = \sqrt{640^2 + 380^2} = 744.3 \text{ kVA} \)

6. \( PF_{\text{new}} = \frac{640}{744.3} = 0.8596 \text{ lag} \)

7. Improvement in PF = \( 0.8596 - 0.80 = 0.0596 \)

8. Dip from required PF = \( 0.90 - 0.8596 = 0.0404 \) (4.04%)

9. Penalty = \( 4.04 \times 20000 = \text{Rs } 80,800 \text{ per month} \)

Power Factor in Industrial/Institutional Settings

• Case study: Analyze university/plant data (max demand, PF, penalty structure) to determine capacitor size and savings.

[!TIP] Common Pitfall: Forgetting that PF improvement reduces kVA demand, but penalty is based on dip from required PF. Always calculate new PF after capacitor installation.

B. Harmonics in Power Systems

Sources & Generation

• Equipment Contributing:

  • Rectifiers (AC-DC converters).

  • Variable Frequency Drives (VFDs).

  • Arc furnaces.

  • UPS systems, computers, fluorescent lamps with electronic ballasts.

• How & Why Generated:

  • Non-linear loads draw current in pulses, not sinusoidal.

  • Distorts voltage waveform → harmonic frequencies (multiples of fundamental: 5th, 7th, 11th, etc.).

  • Fourier series analysis shows harmonic components.

Effects & Problems

• Major Problems:

  • Heating: Increased losses in motors, transformers, cables (I²R ∝ (I_h)²).

  • Vibration & Torque Pulsations: In motors and generators.

  • Malfunction: Of protective relays, meters, control systems.

  • Resonance: Between system reactance and capacitor banks → amplification of harmonics.

  • Reduced Efficiency: In electrical machines.

  • Capacitor Failure: Overheating due to harmonic currents.

Assessment & Mitigation

• Harmonic Distortion Evaluation:

  • Measure voltage and current waveforms using power analyzer.

  • Calculate THD (Total Harmonic Distortion):

    \[ \text{THD}_V = \frac{\sqrt{\sum_{h=2}^{\infty} V_h^2}}{V_1} \times 100\%, \quad \text{THD}_I = \frac{\sqrt{\sum_{h=2}^{\infty} I_h^2}}{I_1} \times 100\% \]

    where \(V_1, I_1\) are fundamental components.

  • Compare with standards (IEEE 519: voltage THD <5%, current THD <25% for general systems).

• Methods to Control Harmonics:

  • Passive Filters: Tuned LC circuits to absorb specific harmonics.

  • Active Filters: Inject opposite harmonics to cancel.

  • Detuning Reactors: Series reactors with capacitor banks to avoid resonance.

  • Phase Shifting: Use 12-pulse or 18-pulse converters.

  • Isolation: Separate harmonic-producing loads.

  • Use 6-pulse instead of 3-pulse rectifiers.

C. Energy Efficient Motors

Constructional Differences

Performance & Advantages

• Advantages:

  • Higher efficiency (2-5 percentage points higher).

  • Lower operating temperature → longer insulation life.

  • Lower losses (core, copper, stray).

  • Better power factor.

  • Reduced cooling requirements.

  • Often higher service factor.

• Features:

  • Meets or exceeds IE3/IE4 standards (IEC).

  • Designed for 40°C ambient, 1.0 service factor.

  • Balanced magnetic design.

Motor Performance Evaluation

• Motor Loading at Part Load:

  \[ \text{Loading \%} = \frac{\text{Input Power at Part Load}}{\text{Full Load Input Power}} \times 100 \]

  where Full Load Input = Rated Output / Full Load Efficiency.

• Example (Jun 2025):

  20 kW motor, full load efficiency = 90%. Part load: V=440V, I=10A, PF=0.78. Find loading %.

  Solution:

  1. Full load input = \( \frac{20}{0.90} = 22.22 \text{ kW} \)

  2. Part load input (three-phase) = \( \sqrt{3} \times V \times I \times PF = 1.732 \times 440 \times 10 \times 0.78 = 5.944 \text{ kW} \)

  3. Loading % = \( \frac{5.944}{22.22} \times 100 = 26.76\% \)

  Motor is heavily underloaded → replace with smaller motor or improve load.

• Factors Affecting Performance:

  • Voltage imbalance (>1% reduces efficiency).

  • Ambient temperature.

  • Harmonics.

  • Power quality.

  • Maintenance (bearing lubrication, alignment).

D. Transformers

Losses & Efficiency

• Losses:

  • Core (Iron) Loss: Hysteresis + eddy current → constant, depends on voltage.

  • Copper (I²R) Loss: Depends on load current squared.

  • Stray Loss: Due to leakage flux, proportional to load.

  • Dielectric Loss: In insulation, usually small.

• Efficiency:

  \[ \eta = \frac{\text{Output Power}}{\text{Input Power}} = \frac{\text{Output}}{\text{Output} + \text{Losses}} \]

  Maximum efficiency when core loss = copper loss.

• Minimizing Losses:

  • Use high-grade core material (CRGO silicon steel).

  • Increase conductor cross-section (larger copper windings).

  • Optimize design (reduce flux density, current density).

  • Use amorphous core transformers (very low core loss).

Energy Conservation Opportunities

• Transformer Loading: Operate near rated load for best efficiency (typically 50-75% of rated). Avoid light load (core loss dominates) and overload (copper loss increases).

• Multiple Transformers: For varying loads, use parallel operation with appropriate loading.

• Replace Old Transformers: With high-efficiency (low-loss) models.

• Reduce Harmonics: Use detuning reactors if non-linear loads present.

• Improve Power Factor: Reduces current and copper loss.

[!TIP] Exam Focus: Loss types and minimization. Efficiency maximization at core loss = copper loss. Transformer loading optimization is common.

IV. THERMAL SYSTEMS & BOILERS

A. Boiler Systems

Efficiency Calculations

• Direct Method (Input-Output):

  \[ \eta_{\text{direct}} = \frac{\text{Steam output} \times (h_s - h_w)}{\text{Fuel input} \times \text{GCV}} \times 100\% \]

  where \(h_s\) = enthalpy of steam, \(h_w\) = enthalpy of feedwater.

• Indirect Method (Loss Method):

  \[ \eta_{\text{indirect}} = 100\% - \sum \left( \frac{\text{Losses}}{\text{Fuel input} \times \text{GCV}} \times 100\% \right) \]

  Losses include: stack loss, moisture loss, unburned carbon, radiation, etc.

• Boiler Efficiency on GCV vs NCV:

  • GCV (Gross Calorific Value): Includes latent heat of vaporization in hydrogen and moisture.

  • NCV (Net Calorific Value): Excludes latent heat.

  • Conversion:

    \[ \text{NCV} = \text{GCV} - 0.09 \times (9 \times H_2\% + M\%) \]

    where \(H_2\%\) = hydrogen content, \(M\%\) = moisture content in fuel.

  • Efficiency on NCV basis:

    \[ \eta_{\text{NCV}} = \eta_{\text{GCV}} \times \frac{\text{GCV}}{\text{NCV}} \]

    (since output energy same, denominator smaller → higher efficiency).

Example (Dec 2024):

Efficiency on GCV = 90%, fuel: 1.5% moisture, 15% hydrogen, GCV = 11,500 kcal/kg. Find efficiency on NCV.

Solution:

1. NCV = 11500 - 0.09 × (9×15 + 1.5) = 11500 - 0.09 × (135 + 1.5) = 11500 - 12.285 = 11487.715 kcal/kg.

2. \( \eta_{\text{NCV}} = 0.90 \times \frac{11500}{11487.715} = 0.90096 \approx 90.1\% \).

Part-Load Efficiency

• Efficiency decreases as load decreases due to fixed losses (radiation, standby losses).

• Implication: Running multiple boilers at part load may be less efficient than one boiler at full load.

Combustion & Operation

• Minimum Excess Air:

  • Reduces heat loss in flue gases (lower stack temperature).

  • Improves combustion efficiency.

  • Too little excess air → incomplete combustion, CO formation.

  • Optimal: 10-20% excess air (measure using O₂ analyzer).

• Achieving Optimal Excess Air:

  • Use oxygen trim control.

  • Regular burner tuning.

  • Combustion air flow control.

• Low-Pressure Steam Efficiency:

  • At lower pressure, enthalpy of evaporation (latent heat) is higher.

  • For heating applications, low-pressure steam delivers more latent heat per kg → more efficient.

  • Also reduces piping losses (lower pressure drop).

Boiler Comparison & Optimization (Case Study)

Two identical 15 TPH boilers: full load eff = 82%, part load eff at 75% load = 78%, at 45% load = 66%. Meet 20 TPH requirement.

Option 1: Both at 10 TPH (load = 10/15 = 66.67%).

• Interpolate eff between 75% (78%) and 45% (66%):

\[ > \text{Slope} = \frac{78-66}{75-45} = 0.4 \% \text{ per } \% \text{ load} > \]

Eff at 66.67% = \( 66 + 0.4 \times (66.67 - 45) = 66 + 8.668 = 74.668\% \).

• Total fuel energy for 20 TPH (let Δh = enthalpy rise per kg steam):

\[ > \text{Fuel} = \frac{10}{0.74668} \Delta h + \frac{10}{0.74668} \Delta h = \frac{20}{0.74668} \Delta h = 26.786 \Delta h > \]

Option 2: One at full 15 TPH (eff=82%), other at 5 TPH (load=33.33%).

• For 33.33% load (<45%), extrapolate with same slope:

Eff = \( 66 + 0.4 \times (33.33 - 45) = 66 - 4.668 = 61.332\% \).

• Total fuel = \( \frac{15}{0.82} \Delta h + \frac{5}{0.61332} \Delta h = (18.292 + 8.157) \Delta h = 26.449 \Delta h \).

Comparison: Option 2 uses less fuel. Savings = \( \frac{26.786 - 26.449}{26.786} \times 100\% = 1.26\% \).

Preferred: One boiler at full load, other at 5 TPH.

Note: If "50% capacity" interpreted as 7.5 TPH (50% load), eff at 50% = 66 + 0.4×(50-45)=68%, total steam=22.5 TPH, fuel=29.321Δh for 22.5 TPH → per TPH fuel=1.303Δh → for 20 TPH fuel=26.06Δh, still better than Option 1.

Energy Conservation Opportunities

• Recover heat from flue gases (economizer, air preheater).

• Feedwater heating (using extraction steam).

• Insulation of boiler surfaces.

• Optimize excess air (oxygen trim).

• Blowdown heat recovery (flash steam, heat exchangers).

• Use variable speed drives for FD/ID fans.

• Regular maintenance (soot blowing, burner tuning).

B. Thermic Fluid Heating Systems

Working Principle

• Thermic fluid (e.g., mineral oil, synthetic oil) heated in a furnace or waste heat exchanger.

• Circulated by pump through heat exchangers to provide process heat.

• No phase change → operates at high temperatures (up to 300°C) at low pressure.

• Fluid returns to heater for reheating.

Comparison with Steam Systems

• Preferred Over Steam When:

  • High temperature required (>200°C) without high pressure.

  • No need for steam traps or condensate return.

  • No water treatment required.

  • Simpler operation, no blowdown losses.

  • Safer (no pressure hazards).

• Disadvantages:

  • Lower heat transfer coefficient than steam.

  • Fluid degradation at high temperatures.

  • Higher initial cost.

  • Limited to indirect heating.

C. Steam Systems

Steam Distribution

• Layout: Main steam header → distribution pipes → end users.

• Components: Steam generator, separator, pressure reducing valves, steam traps, condensate return lines.

• Energy Losses:

  • Heat loss from uninsulated pipes.

  • Pressure drop → throttling losses.

  • Condensate drainage without recovery.

  • Leaks.

• Conservation Measures:

  • Insulate all steam lines.

  • Repair leaks promptly.

  • Use steam traps properly.

  • Recover condensate.

  • Optimize steam pressure levels.

Steam Traps

• Working: Automatically discharge condensate and air while retaining steam.

• Types:

  • Mechanical (float): Float mechanism opens with condensate level.

  • Thermostatic (thermostatic bellows): Temperature-sensitive element opens when condensate cools.

  • Thermodynamic (disc): Steam flow creates pressure difference to open/close.

• Performance Assessment:

  • Visual: Observe discharge (steam loss if continuous blowing).

  • Temperature: Upstream and downstream temperature difference.

  • Ultrasonic: Detect steam leakage.

  • Thermal imaging: Identify failed traps.

  • Test: Isolate and measure condensate output.

Condensate & Flash Steam

• Condensate Recovery:

  • Process: Collect hot condensate, filter, pump back to boiler feedwater system.

  • Savings:

    • Energy: Preheats feedwater → reduces fuel needed.

    • Water: Reduces make-up water requirement and treatment costs.

    • Chemical: Less blowdown needed.

• Flash Steam Utilization:

  • When high-pressure condensate is discharged to low pressure,部分 water flashes to steam.

  • Utilization Examples:

    • Use in low-pressure steam applications (e.g., tank heating).

    • De-aerators or low-pressure boilers.

  • Energy Savings: Recover latent heat from flash steam.

D. Furnaces

Concept & Classifications

• Concept: Enclosed chamber for heat treatment/melting.

• Classifications:

  • By Heat Source: Oil-fired, gas-fired, electric, biomass.

  • By Operation: Batch, continuous.

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

Industrial Applications (Steel Industry)

• Role:

  • Blast Furnace: Iron ore reduction.

  • Basic Oxygen Furnace: Steelmaking.

  • Electric Arc Furnace: Scrap melting.

  • Reheating Furnace: Heat steel billets before rolling.

  • Heat Treatment Furnaces: Annealing, hardening.

• Diagram Description:

  • Furnace hearth with burners (top/side).

  • Refractory lining.

  • Flue gas exit.

  • Preheater (if recuperative).

  • Charging/discharging doors.

Factors Affecting Furnace Efficiency

• Combustion efficiency (excess air, burner tuning).

• Heat loss from walls (insulation quality).

• Flue gas temperature (higher → more loss).

• Load factor (operating at design capacity).

• Heat recovery from flue gases (recuperator/regenerator).

• Door openings (heat loss).

E. Thermal Insulation

Economic Thickness

• Concept: Insulation thickness where total annual cost (insulation + energy loss) is minimized.

• Process:

  1. Calculate heat loss \( Q \) for various thicknesses \( x \):

     \[ Q = \frac{A \times (T_1 - T_2)}{\frac{1}{h_i} + \frac{x}{k} + \frac{1}{h_o}} \]

     where \(A\) = area, \(T_1, T_2\) = inner/outer temps, \(h_i, h_o\) = film coefficients, \(k\) = thermal conductivity.

  2. Compute annual energy cost = \( Q \times \text{operating hours} \times \text{fuel cost} \).

  3. Compute annualized insulation cost (capital cost × capital recovery factor).

  4. Sum for each thickness → plot total cost vs thickness.

  5. Minimum point = economic thickness.

• Factors: Insulation cost, fuel cost, operating hours, discount rate, temperature.

[!TIP] Exam Focus: Boiler efficiency direct/indirect methods, GCV/NCV conversion, excess air, economic insulation thickness calculation.

V. MECHANICAL SYSTEMS: PUMPS, FANS & REFRIGERATION

A. Pumping Systems

Performance Factors

• System Curve: Head vs flow, determined by static head + friction head (∝ flow²).

• Pump Curve: Head vs flow (decreasing), efficiency vs flow (bell-shaped).

• Affinity Laws (for constant impeller diameter, speed change):

  \[ \frac{Q_1}{Q_2} = \frac{N_1}{N_2}, \quad \frac{H_1}{H_2} = \left(\frac{N_1}{N_2}\right)^2, \quad \frac{P_1}{P_2} = \left(\frac{N_1}{N_2}\right)^3 \]

• Cavitation: NPSH available < NPSH required → vapor bubbles, damage to impeller. Avoid by ensuring sufficient inlet pressure.

Energy Conservation Opportunities

• Variable Speed Drives (VSD): Adjust speed to match system demand → major savings.

• Impeller Trimming: Reduce impeller diameter for lower flow.

• Parallel Operation: Optimize number of pumps running.

• Reduce Throttling: Avoid control valves; use VSD instead.

• Use High-Efficiency Pumps: Select best efficiency point (BEP) near required flow.

• Maintenance: Clean impellers, seal leaks, align properly.

Parallel Operation

• Significance: Increase flow capacity, provide redundancy.

• System Characteristics:

  • Combined curve is sum of individual pump flows at same head.

  • Efficiency may drop if pumps operate far from BEP.

  • Proper sizing: pumps should have similar curves.

B. Fans & Blowers

Design & Selection Criteria

• Type Selection: Centrifugal (high pressure), axial (high flow).

• Select for Best Efficiency Point (BEP) near required flow and pressure.

• Consider system resistance curve.

• Use backward-curved blades for efficiency and stability.

• Avoid operating at stall or surge regions.

Performance Evaluation

• Measurements: Flow (anemometer, pitot tube), pressure (manometer), power (kW meter).

• Efficiency:

  \[ \eta = \frac{\text{Air flow} \times \text{Pressure}}{\text{Power input} \times \text{conversion factor}} \times 100\% \]

• Efficient Operation:

  • Use VSD for variable flow.

  • Minimize inlet/outlet dampers.

  • Keep blades clean.

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

C. Refrigeration Plants

Performance & Efficiency

• Coefficient of Performance (COP):

  \[ \text{COP} = \frac{\text{Cooling effect (kW)}}{\text{Work input (kW)}} \]

• Factors Affecting COP:

  • Evaporating Temperature: Lower → lower COP.

  • Condensing Temperature: Higher → lower COP.

  • Superheat and subcooling: Optimal values improve COP.

  • Compressor efficiency.

  • Expansion device type (TXV better than capillary).

  • Heat exchanger effectiveness.

• Energy Conservation Opportunities:

  • Optimize evaporating and condensing temperatures.

  • Use variable speed compressors.

  • Heat recovery (e.g., from condenser for hot water).

  • Improve insulation of cold spaces.

  • Regular maintenance (clean coils, check refrigerant charge).

  • Use high-efficiency motors for compressors.

[!TIP] Exam Focus: Affinity laws for pumps/fans. COP definition and factors. Conservation measures: VSD, optimization of temperatures.

VI. LIGHTING SYSTEMS

Lighting Metrics & Design

• Illumination (Lux): Luminous flux per unit area. \( \text{Lux} = \frac{\text{lumens}}{\text{m}^2} \).

• Measurement: Lux meter.

• Design Factors: Task requirements, uniformity, glare control, color rendering.

Energy Conservation in Lighting

• Scope:

  • Replace inefficient lamps (incandescent, halogen) with LEDs.

  • Use occupancy sensors, daylight harvesting controls.

  • Optimize lighting levels (avoid over-illumination).

  • Use reflectors and proper luminaire design.

  • Regular cleaning of fixtures.

• Procedure:

  1. Audit existing lighting (lamp types, wattage, hours of use).

  2. Determine required lux level.

  3. Select energy-efficient alternatives (LEDs, CFLs).

  4. Calculate savings and economics.

  5. Implement and monitor.

LED Lighting Technology

• Advantages:

  • High efficacy (100-150 lm/W vs 15-20 for incandescent).

  • Long life (50,000 hours).

  • Low heat emission.

  • Instant on, dimmable.

  • Eco-friendly (no mercury).

• Scope & Applications:

  • General lighting (offices, homes).

  • Street lighting.

  • Industrial lighting.

  • Automotive, displays.

Lighting Economics

• Energy Savings:

  \[ \text{Energy saving (kWh)} = (\text{Old wattage} - \text{New wattage}) \times \text{hours} \times \text{number of lamps} / 1000 \]

• Cost Savings:

  \[ \text{Cost saving} = \text{Energy saving} \times \text{energy rate} \]

• Simple Payback Period (SPP):

  \[ \text{SPP} = \frac{\text{Investment cost}}{\text{Annual cost saving}} \]

• Example (Jun 2025):

  Replace 500W with 350W, 350W with 150W, 125W with 60W for same light output. Annual operation = 4500 hours, energy rate = Rs 5.5/unit. Calculate savings and payback.

  Note: Number of lamps not given. Assume one lamp each for illustration.

  Solution:

  1. Total wattage reduction = (500-350) + (350-150) + (125-60) = 150 + 200 + 65 = 415 W.

  2. Energy saving = \( 415 \times 4500 / 1000 = 1867.5 \text{ kWh} \).

  3. Cost saving = \( 1867.5 \times 5.5 = \text{Rs } 10,271.25 \).

  4. Payback requires investment cost (cost of new lamps). If given, SPP = Investment / 10271.25.

  In practice, investment includes lamp cost and possibly fixture modification.

[!TIP] Common Pitfall: Forgetting to divide by 1000 for kWh. Payback requires investment cost; if not given, state assumption.

VII. ECONOMIC & FINANCIAL ANALYSIS FOR ENERGY PROJECTS

A. Life Cycle Costing (LCC)

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

• Process:

  1. Identify all cost elements (initial, recurring, salvage).

  2. Estimate costs over project life.

  3. Convert all costs to present value using discount rate:

     \[ PV = \frac{C}{(1+r)^t} \]

     where \(C\) = cost in year \(t\), \(r\) = discount rate.

  4. Sum present values → LCC.

• Applications:

  • Compare equipment alternatives (e.g., efficient motor vs standard).

  • Justify energy efficiency investments.

  • Select building materials, HVAC systems.

• Effects:

  • Considers long-term savings, not just initial cost.

  • Favors high-efficiency options with higher upfront cost but lower operating costs.

B. Payback Period Methods

Simple Payback Period (SPP)

• Definition: Time required to recover initial investment from annual net savings.

  \[ \text{SPP} = \frac{\text{Initial Investment}}{\text{Annual Net Savings}} \]

• Significance:

  • Simple to calculate and understand.

  • Indicates risk (shorter payback → less risk).

• Limitations:

  • Ignores time value of money.

  • Ignores cash flows beyond payback period.

  • No consideration of risk in savings.

Risk Analysis in Payback

• Assess uncertainty in:

  • Energy savings (actual vs predicted).

  • Energy price escalation.

  • Equipment life and performance.

  • Maintenance costs.

• Methods: Sensitivity analysis (vary key parameters), scenario analysis, Monte Carlo simulation.

• Significance: Provides a range of possible payback periods, helps in decision-making under uncertainty.

Return on Investment (ROI)

• Definition: Annual return as percentage of investment.

  \[ \text{ROI} = \frac{\text{Annual Net Savings}}{\text{Initial Investment}} \times 100\% \]

• Example: Investment Rs 100,000, annual saving Rs 20,000 → ROI = 20%.

C. Net Present Value (NPV)

• Definition: Sum of present values of all cash inflows and outflows over project life.

  \[ \text{NPV} = \sum_{t=0}^{n} \frac{C_t}{(1+r)^t} \]

  where \(C_t\) = net cash flow in year \(t\) (negative for outflow), \(r\) = discount rate, \(n\) = life.

• Calculation:

  • Year 0: -Initial investment.

  • Years 1 to n: Annual net savings (or revenue - operating cost).

  • Discount each cash flow to present value.

  • Sum all.

• Decision Rule: Accept if NPV > 0.

• Example (Dec 2024):

  Lamp cost = Rs 2000, annual savings = Rs 22000 for 2 years, discount rate = 15%.

  Solution:

  \[ > \text{NPV} = -2000 + \frac{22000}{(1.15)^1} + \frac{22000}{(1.15)^2} > \]

  \[ > = -2000 + 19130.43 + 16635.59 = 33765.02 > \]

  NPV positive → accept.

• Advantages over Simple Payback:

  • Considers time value of money.

  • Includes all cash flows over project life.

  • Gives absolute monetary value.

  • Aligns with wealth maximization objective.

D. Internal Rate of Return (IRR)

• Definition: Discount rate that makes NPV = 0.

• Significance:

  • Represents true annual return on investment.

  • Compare with hurdle rate (minimum required return).

  • Higher IRR preferred.

• Calculation: Trial-and-error or financial calculator/Excel (IRR function).

E. Energy Service Company (ESCO) Concept

• Definition: Company that provides energy efficiency services and guarantees energy savings to clients.

• Business Model (Performance Contracting):

  1. ESCO conducts energy audit and identifies opportunities.

  2. ESCO finances, designs, installs, and commissions measures.

  3. Client repays ESCO from actual energy savings.

  4. ESCO guarantees savings; if not achieved, ESCO covers shortfall.

  5. Contract period typically 5-10 years.

• Role in Energy Management:

  • Overcome upfront capital barrier.

  • Provide technical expertise.

  • Share performance risk with client.

  • Enable energy efficiency in public/private sectors.

[!TIP] Exam Focus: NPV calculation with discount rate. Compare NPV vs payback. ESCO model and performance contracting. LCC process.

VIII. INSTRUMENTS & MONITORING FOR ENERGY AUDIT

Audit Equipment

Monitoring Systems

• Energy Management System (EMS): Centralized system to monitor, control, and optimize energy use. Includes meters, sensors, communication network, software.

• Sub-metering: Install meters at department/equipment level for detailed tracking.

• SCADA (Supervisory Control and Data Acquisition): For large industrial plants, real-time monitoring and control.

• Data Acquisition: Collect data from instruments, store in database, analyze for trends, anomalies, savings verification.

• Applications: Continuous improvement, verification of savings, predictive maintenance.

[!TIP] Exam Focus: List instruments and working principle of at least one (e.g., power analyzer, thermal imager). Monitoring systems: EMS, sub-metering.

Final Note: This compilation covers all topics from the approved outline and past exam questions. Focus on problem-solving for PF calculations, boiler efficiency, economic methods (NPV, payback), and lighting economics. Use tables for comparisons and box key formulas.