UNIT 2: ENERGY AUDIT & MANAGEMENT - SHORT NOTES

I. Fundamentals of Energy Audit and Management

Definition and Objectives

• Energy Audit: A systematic process to evaluate energy consumption, identify inefficiencies, and recommend conservation measures. It is the "first step" towards energy management.

• Objectives:

  • Quantify energy use and costs.

  • Identify energy wastage and inefficiencies.

  • Recommend Energy Conservation Measures (ECMs).

  • Establish baseline for future monitoring.

  • Ensure compliance with regulations (e.g., BEE).

Types of Energy Audits

Steps for Energy Audit (Institutional Organization)

1. Planning & Organizing: Form audit team, define scope, secure management commitment.

2. Data Collection: Gather utility bills (electricity, fuel, water) for 1-3 years, building plans, equipment inventory.

3. Walk-through Survey: Visual inspection to identify major areas of consumption/wastage.

4. Detailed Measurement & Verification (M&V): Use instruments to measure actual energy use of key systems (lighting, HVAC, motors).

5. Data Analysis: Calculate specific energy consumption (SEC), benchmark against standards, identify deviations.

6. Identify ECMs & Technical Feasibility: List potential measures (e.g., LED retrofit, VFD installation, boiler tuning).

7. Economic Analysis: Calculate Simple Payback Period (SPP), Net Present Value (NPV), Internal Rate of Return (IRR) for each ECM.

8. Report Preparation & Presentation: Compile findings, recommendations, action plan, and management summary.

Roles and Responsibilities

• Energy Manager:

  • Duties: Implement energy policy, coordinate audits, monitor consumption, promote awareness, track ECM implementation, report to top management.

  • Key Role: In-house champion for continuous energy management.

• Energy Auditor:

  • Duties: Conduct independent/third-party audit, verify data, apply standards/formulas, prepare objective audit report, certify compliance (if designated by BEE).

  • Key Role: External expert providing impartial assessment.

Bureau of Energy Efficiency (BEE) Regulations

• Manners & Intervals (for Designated Consumers - large industries/commercial buildings):

  • Frequency: Every 3 years.

  • Compliance: Must be conducted by a BEE-certified energy auditor.

  • Submission: Audit report (Form 1) must be submitted to BEE and State Designated Agency (SDA) within the stipulated time.

  • Verification: BEE may conduct random verification audits.

Equipment and Instruments for Energy Auditing

Barriers in Energy Auditing & Elimination Strategies

General Principles of Energy Management

1. Management Commitment & Policy: Top-down mandate.

2. Baseline Establishment: Measure and document current energy performance.

3. Monitoring & Targeting (M&T): Regular tracking against benchmarks.

4. Energy Conservation Opportunities (ECOs) Identification: Systematic review of all energy-using systems.

5. Implementation & Verification: Execute ECMs and Measure & Verify (M&V) savings.

6. Continuous Improvement (PDCA Cycle): Plan-Do-Check-Act cycle for ongoing optimization.

7. Awareness & Training: Engage all personnel.

II. Environmental Aspects of Energy

Environmental Impact of Energy Consumption

Elements of Systematic Environmental Assessment

1. Goal & Scope Definition: Define purpose, system boundaries (cradle-to-gate, cradle-to-grave).

2. Inventory Analysis (LCI): Quantify all material/energy inputs and environmental releases (emissions, effluents, solid waste) across the life cycle.

3. Impact Assessment (LCIA): Evaluate potential environmental impacts (global warming, acidification, eutrophication) using inventory data.

4. Interpretation: Summarize results, identify significant issues, draw conclusions, and recommend improvements.

III. Electrical Systems and Power Quality

Power Factor (PF)

• Definition: PF = $$\displaystyle \frac{\text{Real Power (kW)}}{\text{Apparent Power (kVA)}} = \cos\phi $$. It measures effectiveness of power utilization.

• Importance & Benefits of PF Improvement:

  • Reduces kVA demand → lowers electricity bill (if billed on kVA or has PF penalty).

  • Decreases system current → reduces I²R losses in cables/transformers.

  • Improves voltage regulation.

  • Increases system capacity (existing infrastructure can serve more load).

• Causes of Low PF:

  • Inductive loads: Induction motors, transformers, fluorescent ballasts, welding sets.

  • Lightly loaded motors (operate at very low PF).

• Disadvantages of Low PF:

  • Higher current for same kW → larger cable sizes, higher losses.

  • Penalty charges from utility.

  • Poor voltage regulation, overheating of equipment.

PF Improvement using Capacitor Banks

• Principle: Capacitors supply leading kVAr to cancel lagging kVAr from inductive loads.

• Optimal Location:

  1. Individual Motor Compensation: Capacitor bank at motor terminals (best for large, constant-speed motors).

  2. Bus Bar/Common Compensation: At distribution board (good for multiple small loads).

  3. Central/Utility Point Compensation: At main incoming (improves overall PF, but doesn't reduce distribution losses).

  !TIP: For energy & cost saving, locate capacitors as close as possible to the inductive load.

• Sizing Considerations:

  • Calculate existing kVAr demand from kVA, kW, PF.

  • Required capacitor kVAr = kVAr_initial - kVAr_target.

  • Consider harmonics (detuning reactors may be needed).

  • Avoid over-correction (leading PF).

PF Improvement & Penalty Calculation (Example from Past Paper)

Given: Max Demand = 800 kVA, Avg PF = 0.80 lag, Min Required PF = 0.90 lag, Penalty = Rs 20,000 per 1% dip. Capacitor Installed: 100 kVAr.

Step 1: Calculate Initial kVAr & kW

$$ \text{kW} = \text{kVA} \times \text{PF} = 800 \times 0.80 = 640 \text{ kW} $$

$$ \text{Initial kVAr} = \text{kVA} \times \sin\phi = 800 \times \sqrt{1 - 0.80^2} = 800 \times 0.6 = 480 \text{ kVAr} $$

Step 2: Calculate New kVAr after 100 kVAr capacitor

$$ \text{New kVAr} = 480 - 100 = 380 \text{ kVAr} $$

Step 3: Calculate New kVA & PF

$$ \text{New kVA} = \sqrt{\text{kW}^2 + \text{New kVAr}^2} = \sqrt{640^2 + 380^2} = \sqrt{409600 + 144400} = \sqrt{554000} \approx 744.3 \text{ kVA} $$

$$ \text{New PF} = \frac{\text{kW}}{\text{New kVA}} = \frac{640}{744.3} \approx 0.86 \text{ lag} $$

Step 4: Check Penalty

Min Required PF = 0.90. New PF = 0.86 < 0.90. Dip from 0.90 = 0.90 - 0.86 = 0.04 = 4%.

$$ \text{Penalty} = 4 \times 20,000 = \boxed{\text{Rs. 80,000}} $$

!TIP: PF penalty is usually calculated on the dip from the specified minimum PF, not from the original PF.

Harmonics

• Sources/Equipment:

  • Non-linear loads: VFDs (Variable Frequency Drives), UPS systems, computers, LED drivers, SMPS, arc furnaces, rectifiers.

  • Cause: These devices draw non-sinusoidal current (rich in multiples of fundamental frequency, e.g., 5th, 7th, 11th harmonics).

• Effects:

  • Heating: In motors, transformers, cables (core/copper losses increase).

  • Nuisance Tripping: Of circuit breakers, protective relays.

  • Capacitor Failure: Resonance amplification, overheating.

  • Neutral Overload: In 3-phase 4-wire systems (triplen harmonics add in neutral).

  • Metering Inaccuracies.

  • Communication Interference.

• Harmonic Distortion Evaluation:

  1. Measure voltage/current waveforms with power quality analyzer.

  2. Perform FFT (Fast Fourier Transform) to decompose into harmonic components.

  3. Calculate Total Harmonic Distortion (THD):

$$ \text{THD}_V (\%) = \frac{\sqrt{V_2^2 + V_3^2 + ... + V_n^2}}{V_1} \times 100 $$

$$ \text{THD}_I (\%) = \frac{\sqrt{I_2^2 + I_3^2 + ... + I_n^2}}{I_1} \times 100 $$

    (Where subscript 1 = fundamental, 2,3...n = harmonic orders).

4.  Compare with **IEEE 519 standards** for limits.

HT vs. LT Systems

Transformer Losses & Minimization

!TIP: For lightly loaded transformers, core loss dominates. For heavily loaded, copper loss dominates. Select transformer with lowest total loss at typical load factor (often 50-70%).

IV. Thermal Systems and Steam Management

Boiler Efficiency

• On GCV (Gross Calorific Value) Basis:

$$ \eta_{\text{GCV}} = \frac{\text{Steam Output (kg/hr)} \times (h_s - h_w)}{\text{Fuel Input (kg/hr)} \times \text{GCV}} \times 100\% $$

(Where $$\displaystyle h_s $$ = enthalpy of steam, $$\displaystyle h_w $$ = enthalpy of feed water).

• On NCV (Net Calorific Value) Basis:

$$ \eta_{\text{NCV}} = \frac{\text{Steam Output} \times (h_s - h_w)}{\text{Fuel Input} \times \text{NCV}} \times 100\% $$

> **Relationship**: $$\displaystyle \eta_{\text{NCV}} > \eta_{\text{GCV}} $$ because NCV excludes latent heat of vaporization in fuel moisture.

• Direct Method (Input-Output Method):

$$ \eta = \frac{\text{Energy Output (Heat in Steam)}}{\text{Energy Input (Heat in Fuel)}} \times 100\% $$

Simple, measures actual performance.

• Indirect Method (Heat Loss Method):

$$ \eta = 100\% - (\text{Sum of all % losses}) $$

Losses: Stack loss, dry flue gas loss, moisture in fuel/air, unburnt carbon, radiation/convection.

More detailed, identifies *why* efficiency is low.

Part-Load Efficiency & Multiple Boilers Operation

• Part-Load Efficiency: Boiler efficiency decreases at part-load due to:

  • Higher radiation/convection losses (constant).

  • Higher excess air (often not controlled down at low load).

  • Poor combustion stability.

• Multiple Boilers Strategy (Example from Past Paper):

  • Scenario: Two identical 15 TPH boilers (82% full load eff.), part-load eff. at 75% = 78%, at 45% = 66%. Need 20 TPH.

  • Option 1: Both at 10 TPH (66.7% load). Eff. ≈ interpolate between 75% & 45%? Not linear! Usually, efficiency drops sharply below 50%. Assume ~70%? (Given data: 75% load=78%, 45% load=66%. At 66.7%, likely ~72%).

  • Option 2: One at 15 TPH (100% load, 82%), other at 5 TPH (33% load, eff. << 66%, say ~50%).

  • Calculation:

    • Total Fuel for Option 1: (20 TPH steam) / (Avg Eff. ~72%) = ~27.8 TPH fuel equivalent.

    • Total Fuel for Option 2: (15 TPH steam / 82%) + (5 TPH steam / 50%) = 18.29 + 10 = 28.29 TPH fuel equivalent.

  • Conclusion: Running both at higher part-load (Option 1) saves fuel (~1.7% saving). Never run a boiler below ~50% load if another can share.

Steam Distribution Systems & Common Losses

• Distribution: Steam generated → header → distribution pipes → end-use.

• Common Losses:

  1. Radiative/Convective Heat Loss from uninsulated pipes.

  2. Pressure Drop → requires higher generation pressure → lower thermodynamic efficiency.

  3. Condensate Drainage (without recovery) → loss of hot water & heat.

  4. Steam Leaks from joints, valves.

  5. Flash Steam from high-pressure condensate dumped to low-pressure drain.

Steam Traps

• Working Principle: Automatic valve that discharges condensate, air, and non-condensable gases while preventing steam passage.

• Types & Principles:

  | Type | Principle | Typical Application | | :--- | :--- | :--- | | Mechanical (Inverted Bucket, Float & Thermostatic) | Uses buoyancy of condensate (density difference). | General purpose, high capacity. | | Thermostatic (Bimetallic, Bellows) | Uses temperature difference (steam ~hotter than condensate). | Drip legs, tracer lines. | | Thermodynamic (Disc,活塞) | Uses kinetic energy difference of steam vs. condensate. | Main steam lines, high pressure. |

• Performance Assessment Methods:

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

  2. Temperature Measurement: Upstream & downstream temp. A cold trap (blocked) or hot trap (blowing steam) indicates failure.

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

  4. Infrared Thermography: Shows temperature profile; a cold downstream may indicate blockage, hot downstream may indicate live steam loss.

Condensate Recovery & Flash Steam Utilization

• Condensate Recovery:

  • Process: Collect hot condensate from steam traps → return to boiler feed water tank via condensate return lines.

  • Benefits: Saves water, heat (enthalpy), and water treatment chemicals. Improves boiler efficiency.

• Flash Steam Utilization:

  • Process: High-pressure condensate (e.g., from 10 bar process) discharged to a flash vessel at lower pressure. Some condensate flashes into low-pressure steam.

  • Utilization: This flash steam can be used for low-pressure applications (e.g., space heating, domestic hot water, process pre-heat).

  • Benefit: Recovers latent heat that would otherwise be lost.

Thermic Fluid Heating Systems

• Working Principle: Uses a heat transfer oil (thermic fluid) heated in a furnace/coil → circulated by pump → heat exchange with process → returns to heater. Closed loop, no phase change.

• Advantages over Steam:

  • No pressure → no boiler regulations, no steam traps, no condensate recovery needed. Safer.

  • High temperature at low pressure (e.g., 300°C at 5 bar vs. steam needs ~70 bar).

  • Precise temperature control.

  • No scale/rust issues (if oil maintained).

  • Disadvantages: Fire risk (oil leak), higher initial cost, oil degradation over time.

Furnaces

• Concept: Enclosed chamber for direct heat transfer from combustion to material (solid, liquid, gas).

• Classifications:

  • By Heat Source: Oil/gas fired, coal fired, electric.

  • By Material: Metal heating (reheating, annealing), non-metal (glass, ceramic).

  • By Operation: Batch, continuous.

  • By Waste Heat Recovery: Regenerative, recuperative.

• Role in Industries (e.g., Steel): Primary energy consumer for reheating slabs/billets before rolling.

• Energy Efficiency Considerations:

  • Combustion Efficiency: Optimize excess air (minimize stack loss).

  • Heat Recovery: Use recuperators (sensible heat) or regenerators (sensible + latent) from flue gases.

  • Insulation: High-quality lining to minimize skin losses.

  • Furnace Pressure: Slight negative pressure prevents air infiltration (which increases excess air).

  • Soaking Time: Minimize to reduce heat loss.

Insulation: Economic Thickness

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

• Process:

  1. Calculate heat loss per unit area for various thicknesses (using thermal conductivity k).

  2. Calculate total annual heat loss cost = (Heat loss rate) × (Operating hours) × (Fuel cost/unit energy).

  3. Calculate annualized capital cost of insulation (using CRF - Capital Recovery Factor).

  4. Sum (2) + (3) for each thickness.

  5. Choose thickness with minimum total cost.

  !TIP: Beyond economic thickness, savings in heat loss < additional insulation cost.

Combustion Optimization: Minimum Excess Air

• Advantages:

  • Reduces stack loss (less hot flue gas carrying away heat).

  • Reduces fan power consumption (less gas volume to move).

  • Reduces NOₓ formation (less available N₂ at high temp).

• Methods to Achieve:

  1. Regular Tuning: Use flue gas analyzer to measure O₂/CO in flue gas. Adjust air dampers to achieve optimal O₂ level (e.g., 3-5% for gas, 5-8% for coal).

  2. Maintain Burners: Clean nozzles, ensure proper atomization (oil), correct air-fuel mixing.

  3. Control Combustion Air Temperature: Preheat if possible (using waste heat).

  4. Use Oxygen Trim Control: Automated system adjusting air based on real-time O₂ measurement.

Refrigeration Plants: Performance & Efficiency Factors

• Key Performance Indicator: Coefficient of Performance (COP) = $$\displaystyle \frac{\text{Refrigeration Effect (kW)}}{\text{Compressor Input Power (kW)}} $$.

• Factors Affecting COP:

  1. Evaporator Temperature: Lower evaporator temp → lower COP. Operate at highest feasible evaporator temp.

  2. Condenser Temperature: Higher condenser temp (due to fouling, high ambient) → lower COP. Keep condensers clean, use cooling tower optimization.

  3. Compressor Efficiency: Mechanical, volumetric, isentropic efficiency. Regular maintenance.

  4. Expansion Device: Proper sizing (TXV, EEV better than capillary).

  5. Suction/Discharge Pressure Drop: Minimize pipe sizing losses.

  6. Refrigerant Charge: Correct charge level.

  7. Sub-cooling & Superheating: Optimize for system protection, not excessive.

• Energy Conservation Opportunities:

  • Temperature Optimization: Raise evaporator setpoint, lower condenser setpoint (within limits).

  • Preventive Maintenance: Clean coils, check refrigerant charge, replace filters.

  • Variable Speed Drives (VSDs) on compressor motors for part-load.

  • Heat Recovery: Use condenser heat for water heating.

  • Insulation: On cold pipes/equipment.

V. Motors, Pumps, and Fans

Energy-Efficient Motors (Premium Efficiency Motors)

• Construction Differences from Standard Motors:

  • More Copper: Larger cross-section windings → lower I²R loss.

  • Better Steel: Higher grade, thinner laminations (CRGO) → lower core loss.

  • Optimized Design: Longer air gap, improved cooling fan design, better bearings.

  • Tighter Tolerances: Precision manufacturing.

• Advantages:

  • Higher Efficiency (typically 2-5% points higher) across load range.

  • Lower Operating Temperature → longer insulation/bearing life.

  • Lower Energy Costs → higher upfront cost offset by savings.

  • Better Power Factor (often).

  • Smaller Size/Weight for same output (due to better materials).

• Key Features: IE3/IE4 efficiency class (as per IEC/IS), higher service factor (1.15), better starting torque.

Motor Performance Evaluation & Part-Load Efficiency

• Loading Calculation from Input Parameters:

  Given: Voltage (V), Current (I), Power Factor (PF), Rated Output (P_rated).

$$ \text{Input Power (kW)} = \sqrt{3} \times V \times I \times PF \times 10^{-3} \quad (\text{for 3-phase}) $$

$$ \text{Load (\%)} = \frac{\text{Input Power} \times \text{Full Load Efficiency}}{\text{Rated Output (kW)}} \times 100\% $$

> **Note**: Use **nameplate efficiency** to convert input to output. If unknown, assume ~90% for estimation.

• Example from Past Paper:

  Motor: 20 kW, Full Load Eff. = 90%.

  Input: 440V, 10A, PF=0.78.

  Input kW = $$\displaystyle \sqrt{3} \times 0.440 \times 10 \times 0.78 \times 10^{-3} = 1.732 \times 0.440 \times 10 \times 0.78 \times 0.001 \approx 5.95 $$ kW.

  Output kW at this load = $$\displaystyle 5.95 \times 0.90 = 5.36 $$ kW.

  % Loading = $$\displaystyle \frac{5.36}{20} \times 100\% = \boxed{26.8\%} $$.

  !TIP: Motor efficiency at part-load is lower than full-load efficiency. A motor at 26% load may have eff. ~80%, not 90%. So actual loading might be slightly higher than calculated above.

Pumping Systems: Performance & Energy Conservation

• Factors Affecting Pump Performance:

  • System Curve: Static head, friction head (pipe length, fittings, roughness), equipment pressure drop.

  • Pump Curve: Head vs. Flow, efficiency vs. Flow, Best Efficiency Point (BEP).

  • NPSH (Net Positive Suction Head): Cavitation risk if insufficient.

  • Fluid Properties: Viscosity, density.

• Energy Conservation Opportunities:

  1. Operate at BEP: Avoid running far off BEP (low efficiency, cavitation, vibration).

  2. Reduce System Head: Clean pipes, reduce unnecessary fittings, increase pipe diameter (reduces friction loss ∝ 1/D⁵).

  3. Reduce Flow Requirement: Check process need; throttling valves waste energy.

  4. Install Variable Speed Drives (VSDs): Most effective for variable flow systems. Affinity Laws:

$$ Q \propto N, \quad H \propto N^2, \quad P \propto N^3 $$

    (Where Q=flow, H=head, N=speed, P=power). **Small speed reduction saves large power**.

5.  **Parallel Operation**: Use multiple pumps only when needed; **stagger operation** to keep pumps near BEP.

6.  **Replace Inefficient Pumps**: With **high-efficiency models**, correct size.

7.  **Prevent Leakage**: Repair seals, glands.

Fan Systems: Design, Selection & Efficient Operation

• Design & Selection Criteria:

  • Determine system curve (static pressure vs. flow) accurately.

  • Select fan operating point near peak efficiency on its curve.

  • Consider fan type (centrifugal vs. axial) based on pressure/flow requirement.

  • Account for future expansion (margin ~10-15%).

• Performance Evaluation:

  • Measure flow (CFM/m³/s), static pressure (mmWG/Pa), input power (kW).

  • Calculate Fan Efficiency:

$$ \eta_{\text{fan}} = \frac{\text{Air Power (kW)}}{\text{Shaft Power (kW)}} \times 100\% $$

$$ \text{Air Power} = \frac{Q \times SP}{102 \times \eta_{\text{total}}} \quad (\text{in kW, Q in m³/s, SP in mmWG}) $$

• Efficient System Operation:

  • Avoid Throttling/Damper Control: Wastes energy. Use inlet guide vanes or VSDs for flow control.

  • Reduce System Resistance: Clean ducts, remove unnecessary bends/filters.

  • Operate at Highest Efficiency Point: Monitor and adjust.

  • Use Multiple Fans: For variable loads, operate fewer fans at higher efficiency point.

VI. Lighting Systems and Conservation

Scope for Energy Conservation in Lighting

• Replace inefficient lamps: Incandescent → CFL → LED.

• Optimize lighting levels: Use task lighting, reduce overlit areas.

• Improve maintenance: Clean fixtures, lamps, replace old lamps (lumen depreciation).

• Use daylight: Daylight harvesting with photosensors.

• Install occupancy sensors (vacancy sensors) in low-usage areas.

• Use efficient ballasts: Electronic vs. magnetic.

• Select proper luminaires: High reflectance, good optical control.

• Implement lighting control strategies: Zoning, scheduling, dimming.

LED Lighting: Advantages & Applications

• Advantages:

  • Very High Efficacy (100-150+ lm/W vs. 15-60 for incandescent/CFL).

  • Long Life (50,000+ hours).

  • Instant On, no warm-up.

  • Dimmable (with compatible drivers).

  • Rugged, no filament.

  • Directional light (no need for reflectors).

  • Low heat emission.

  • Environmentally friendly (no mercury).

• Applications: General lighting, street lighting, automotive, indicators, displays, horticulture.

Lighting Calculations & Retrofit Analysis

• Illumination Level (Lux): $$\displaystyle E = \frac{\text{Luminous Flux (lm)}}{\text{Area (m²)}} $$.

• Energy & Cost Savings from Retrofit:

  Step 1: Calculate annual energy consumption of existing system.

$$ \text{Energy}_{\text{old}} = \text{No. of Lamps} \times \text{Wattage}_{\text{old}} \times \text{Operating Hours} \times 10^{-3} \text{ (kWh)} $$

**Step 2**: Calculate annual energy of new system.

$$ \text{Energy}_{\text{new}} = \text{No. of Lamps} \times \text{Wattage}_{\text{new}} \times \text{Operating Hours} \times 10^{-3} \text{ (kWh)} $$

**Step 3**: Annual Savings.

$$ \text{Energy Savings} = \text{Energy}_{\text{old}} - \text{Energy}_{\text{new}} \text{ (kWh)} $$

$$ \text{Cost Savings} = \text{Energy Savings} \times \text{Electricity Rate (Rs/kWh)} $$

• Simple Payback Period (SPP):

$$ \text{SPP} = \frac{\text{Total Investment Cost (Rs)}}{\text{Annual Cost Savings (Rs/year)}} \text{ (years)} $$

• Example from Past Paper:

  Replace 500W → 350W, 350W → 150W, 125W → 60W. Same light output. 4500 hrs/yr, Rs 5.5/unit.

  Assumption: 1:1 lamp replacement. Let number of each type = N.

  • Old Total Wattage = N*(500 + 350 + 125) = N*975 W.

  • New Total Wattage = N*(350 + 150 + 60) = N*560 W.

  • Savings per lamp set = 975 - 560 = 415 W = 0.415 kW.

  • Annual Energy Savings per set = 0.415 kW × 4500 hrs = 1867.5 kWh.

  • Annual Cost Savings per set = 1867.5 × 5.5 = Rs. 10,271.25.

  • Investment per set = Cost(350W) + Cost(150W) + Cost(60W) - (Scrap value of old lamps). [Investment cost not given in question, so SPP cannot be computed without it. In exam, investment cost would be provided.]

Procedures for Energy Saving in Lighting Systems

1. Audit: Measure existing lux levels, inventory lamps/luminaires, operating hours.

2. Benchmark: Compare with recommended lux levels (IS/CIE standards).

3. Identify ECMs: Retrofit, delamping, controls, daylighting.

4. Calculate Savings: For each ECM (as above).

5. Economic Evaluation: SPP, NPV, IRR.

6. Implementation Plan: Phasing, procurement, installation.

7. Post-Implementation M&V: Measure new lux, verify energy savings.

VII. Financial and Economic Analysis

Life Cycle Costing (LCC)

• Concept: Total cost of owning and operating an asset over its entire life (investment + O&M + disposal).

• Process:

  1. Identify all cost components: Initial Investment, Installation, Operation, Maintenance, Energy, Replacement, Disposal.

  2. Convert all future costs to Present Worth (PW) using discount rate.

  3. Sum all present worths → LCC.

• Applications: Compare alternative projects (e.g., standard motor vs. premium motor, window vs. split AC), select most economical.

• Effect on Investment Decisions: Reveals true cost. A higher upfront, high-efficiency option often has lower LCC due to reduced energy costs.

Payback Period (PP)

• Definition: Time required for cumulative cash inflows (savings) to equal initial investment.

• Calculation:

$$ \text{Simple Payback Period} = \frac{\text{Initial Investment (Rs)}}{\text{Annual Net Cash Savings (Rs/year)}} $$

**Discounted Payback Period**: Considers time value of money; uses **discounted cash flows**.

• Significance of Risk Analysis:

  • Shorter PP → lower risk (capital recovered faster).

  • Long PP projects are riskier (future savings uncertain, discount rate changes).

  • Sensitivity Analysis: Vary key assumptions (energy cost escalation, savings realization) to see impact on PP.

• Merits: Simple, easy to understand, focuses on liquidity & risk.

• Demerits: Ignores cash flows beyond PP, ignores time value of money (in simple PP), arbitrary cutoff.

Net Present Value (NPV)

• Concept: Sum of all present values of future cash inflows and outflows over project life, discounted at a required rate of return (discount rate).

$$ \text{NPV} = \sum_{t=0}^{n} \frac{CF_t}{(1 + r)^t} $$

Where $$\displaystyle CF_t $$ = net cash flow in year t (t=0 is initial investment, negative), $r$ = discount rate, $n$ = life.

• Decision Rule: Accept if NPV > 0 (project adds value).

• Advantages over Simple PP:

  1. Considers entire project life.

  2. Incorporates time value of money.

  3. Additive (NPV of combined projects = sum of individual NPVs).

  4. Directly measures increase in wealth.

• Example from Past Paper:

  Investment = Rs 2000, Life = 2 years, Discount rate = 15%, Savings = Rs 22000 each year.

  Year 0: CF₀ = -2000 → PV = -2000.

  Year 1: CF₁ = 22000 → PV = 22000 / (1.15)¹ = 22000 / 1.15 ≈ 19130.43.

  Year 2: CF₂ = 22000 → PV = 22000 / (1.15)² = 22000 / 1.3225 ≈ 16635.03.

  NPV = -2000 + 19130.43 + 16635.03 = \boxed{33765.46 Rs}.

  !TIP: Positive NPV indicates project is financially attractive at 15% discount rate.

Internal Rate of Return (IRR)

• Definition: The discount rate (r) that makes the NPV = 0.

$$ 0 = \sum_{t=0}^{n} \frac{CF_t}{(1 + \text{IRR})^t} $$

• Interpretation: The effective annual return on the invested capital.

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

• Merits: Expressed as percentage, easy to compare with other investments, considers time value & all cash flows.

• Demerits: Can have multiple IRRs for non-conventional cash flows, reinvestment assumption (IRR assumes cash flows reinvested at IRR, which may be unrealistic).

Discount Period & Time Value of Money

• Time Value of Money (TVM): A rupee today is worth more than a rupee in the future due to its earning potential (interest, inflation).

• Discounting: Process of converting future cash flows to present value using a discount rate.

• Discount Period: The time interval (usually yearly) over which discounting is applied. The factor is $$\displaystyle \frac{1}{(1+r)^t} $$ for year t.

Energy Service Company (ESCO)

• Concept: A company that provides comprehensive energy solutions to clients: audit, design, finance, implement, and guarantee savings of ECMs.

• Business Models:

  1. Shared Savings: ESCO finances, implements; client pays ESCO a percentage of verified savings over contract period.

  2. Guaranteed Savings: ESCO guarantees a certain level of savings; client pays ESCO a fixed fee from savings.

  3. Energy Supply Contract: ESCO supplies energy (e.g., steam, heat) at a price lower than client's current cost.

• Role in Conservation Projects:

  • Overcomes barrier of high upfront cost (ESCO provides financing).

  • Provides technical expertise.

  • Assumes performance risk (guarantees savings).

  • Enables public sector/institutions with budget constraints to implement projects.

VIII. Additional Cross-Cutting Topics (Integrated Above)

• Parallel Operation of Pumps: Covered in Section V (Pumping Systems). Key: System curve changes, ensure pumps have similar characteristics, avoid one pump running far off BEP.

• Refrigeration Plant Performance Factors: Covered in Section IV (Refrigeration Plants).

• Illumination Lux: Covered in Section VI (Lighting Calculations).