Electrical Energy Audit in Industry

Energy audit is the systematic study of how energy is used in an industry, to find where it is wasted and to recommend cost-effective measures for saving it. (As per Nepal's Energy Efficiency programme and the Indian Energy Conservation Act, it is "verification, monitoring and analysis of energy use, with a technical report containing recommendations and an action plan to reduce consumption".)

Need of energy audit for an industry

1. Reduce energy cost – energy is a major part (10–40%) of production cost; savings of 10–30% are common.
2. Identify waste – leaks, idle running, oversized motors, poor power factor, poor insulation.
3. Improve productivity and competitiveness – lower unit cost of product.
4. Benchmarking – compare specific energy consumption (kWh/ton) with best practice.
5. Planning – basis for investments in efficient equipment, with payback calculation.
6. Reduce demand charges and penalties – peak load management, pf improvement.
7. Environment – lower fuel use and CO₂ emissions.
8. Equipment reliability – detect overloaded, overheating or unbalanced equipment.
9. Legal and policy compliance – government programmes and standards.

Energy audit technique (methodology)

1. Pre-audit (planning) phase

• Form audit team, meet management, set objectives.
• Collect data: energy bills (12–36 months), production records, single-line diagram, equipment list.
• Walk-through survey to see major energy users.

2. Audit (execution) phase

• Detailed measurements with instruments: power analysers, clamp meters, lux meters, flue gas analysers, thermal cameras, flow and pressure meters.
• Prepare energy balance: where each kWh/litre goes (motors, lighting, compressors, HVAC, furnaces).
• Calculate specific energy consumption; study load profile, pf and losses.
• Identify energy conservation opportunities (ECOs).
• Techno-economic analysis: savings, investment, simple payback = investment / annual savings, NPV/IRR.

3. Post-audit phase

• Prepare report with recommendations ranked as no-cost, low-cost and high-cost.
• Implement measures; monitor and verify savings (M&V); repeat audit periodically.

Pre-audit ─► Data & walk-through ─► Measurements
   ─► Energy balance ─► Find ECOs ─► Cost-benefit
   ─► Report ─► Implement ─► Monitor & verify

Types: preliminary (walk-through) audit – quick, uses existing data; detailed audit – complete measurement and analysis of all systems.

Given: V_L = 415 V, cable rating 235 A, present load 300 A at pf 0.7 (lag). Real power stays the same; capacitors reduce only the reactive part.

Present load

S₁ = √3 × 415 × 300 = 215.64 kVA
P  = S₁ × 0.7       = 150.95 kW
Q₁ = S₁ × sin(cos⁻¹0.7) = 215.64 × 0.7141
   = 154.00 kVAr

Allowed load at rated current

S₂ = √3 × 415 × 235 = 168.92 kVA
Q₂ = √(S₂² − P²) = √(168.92² − 150.95²)
   = 75.82 kVAr
New pf = P / S₂ = 150.95 / 168.92 = 0.894 lag

Capacitor required

Qc = Q₁ − Q₂ = 154.00 − 75.82 = 78.18 kVAr

Answer: about 78.2 kVAr of capacitors (3-phase); the pf rises to about 0.894 lag and the current falls to 235 A. In practice the next standard size, 80 kVAr, is chosen.

Given: average load 1020 kVA at 0.64 pf (lag); transformer 1600 kVA; no-load loss 2.4 kW; full-load (copper) loss 18.57 kW; capacitor cost Rs 800/kVAr; maximum demand charge Rs 190/kVA (taken as per month); energy charge Rs 7.50/kWh; loading 2920 h/year.

i) kVAr required to improve pf to 0.95

P  = 1020 × 0.64          = 652.8 kW
Q₁ = 1020 × sin(cos⁻¹0.64) = 1020 × 0.7684
   = 783.74 kVAr
Q₂ = P × tan(cos⁻¹0.95)   = 652.8 × 0.3287
   = 214.56 kVAr
Qc = Q₁ − Q₂              = 569.18 kVAr
New kVA S₂ = 652.8 / 0.95 = 687.16 kVA

ii) Reduction in transformer loss

No-load loss (2.4 kW) does not change. Copper loss ∝ (S/S_rated)²:

Cu loss before = 18.57 × (1020/1600)²   = 7.547 kW
Cu loss after  = 18.57 × (687.16/1600)² = 3.425 kW
Reduction      = 4.122 kW
Energy saved   = 4.122 × 2920 = 12 035.6 kWh/yr
Cost saved     = 12 035.6 × 7.50 = Rs 90 266.8/yr

iii) Reduction in maximum demand charge

kVA reduction = 1020 − 687.16 = 332.84 kVA
Saving        = 332.84 × 190 = Rs 63 240 per month
              = Rs 758 880 per year

iv) Payback period

Capacitor cost = 569.18 × 800      = Rs 455 342
Annual saving  = 758 880 + 90 266.8 = Rs 849 147
Payback        = 455 342 / 849 147
               = 0.536 year ≈ 6.4 months

Answer: (i) 569.18 kVAr; (ii) 4.12 kW less loss, 12 036 kWh/yr = Rs 90 267/yr; (iii) Rs 63 240/month (Rs 758 880/yr); (iv) payback ≈ 0.54 year (about 6.4 months). (If the demand charge is taken as a one-time yearly figure, annual saving = 63 240 + 90 267 = Rs 153 507 and payback ≈ 2.97 years.)

Load management is the planned control of when and how much electrical load an industry draws, so that the maximum demand and peak-hour consumption are reduced and the load curve is flattened, without lowering production. It lowers demand charges, time-of-day energy cost and losses, and helps the utility too.

Load management techniques

1. Peak clipping (shaving): reduce or switch off non-essential loads during peak hours (e.g. 17:00–23:00 in Nepal's NEA time-of-day tariff) using a maximum demand controller.
2. Load shifting: move flexible loads – water pumping, batch furnaces, charging, grinding – to off-peak or night hours where the tariff is lower.
3. Valley filling: add useful loads in off-peak hours (thermal storage, ice making, battery charging).
4. Staggering / rescheduling of operations: stagger the starting of large motors, shift timings and lunch breaks so that large loads do not run together.
5. Maximum demand controller (MDC): monitors demand continuously and automatically trips selected low-priority loads when demand approaches a set limit.
6. Power factor improvement: capacitor banks reduce kVA demand for the same kW.
7. Use of storage: thermal storage, compressed air receivers, water tanks, batteries to supply part of peak demand.
8. Captive / standby generation during peak hours if cheaper than grid peak tariff.
9. Efficient equipment and VFDs: reduce the load itself (soft start reduces starting peaks).

kW │   ___peak (clip)
   │  /   \        shifted
   │ /     \___    loads
   │/  valley  \__/‾‾‾ (fill)
   └───────────────────── hours

Energy-efficient induction motors (EEMs) are motors designed and built with lower losses, giving 2–6% higher efficiency than standard motors (efficiency classes IE2, IE3, IE4 as per IEC 60034-30).

Technical aspects (how losses are reduced)

1. Stator copper loss (I²R): more copper in the slots, larger conductor area, better slot fill → lower stator resistance.
2. Rotor loss: larger rotor bars and end rings of high-conductivity copper or aluminium → lower rotor resistance and lower slip.
3. Iron (core) loss: thin (0.35–0.5 mm), low-loss, high-grade silicon steel laminations with better insulation; longer core so flux density is lower.
4. Friction and windage loss: high-quality low-friction bearings; smaller, efficient cooling fan.
5. Stray load loss: optimised air gap, slot design and better manufacturing tolerances.
6. Better design optimisation: computer-aided design of magnetic circuit.

Features / benefits

• Higher efficiency and slightly higher power factor at all loads; flatter efficiency curve at part load.
• Runs cooler → longer insulation and bearing life, better overload capacity.
• Lower slip means slightly higher speed – for fans and pumps this can increase load, so check.
• Higher initial cost (15–30% more), but payback usually 1–3 years for motors running long hours.
• Less noise and vibration.

Energy saving opportunities (energy conservation measures) are the changes in equipment, operation and maintenance that reduce energy use without reducing output.

1. Electrical supply system

• Improve power factor with capacitor banks (APFC) – reduces kVA demand, losses and penalties.
• Run transformers near optimum load; switch off lightly loaded transformers.
• Reduce maximum demand through load management and shifting to off-peak hours.
• Balance phase loads; correct voltage level; reduce harmonics with filters.
• Use proper cable sizes to reduce I²R losses.

2. Motors and drives

• Replace old or rewound motors with energy-efficient (IE3/IE4) motors.
• Avoid oversized, under-loaded motors; use star-delta switching for lightly loaded motors.
• Use variable frequency drives (VFDs) on fans, pumps and compressors instead of dampers/throttle valves.
• Use efficient belts (flat/cogged), proper alignment and lubrication.

3. Pumps, fans and blowers

• Correct sizing, trim impellers, avoid throttling; remove unnecessary bends; use VFD.

4. Compressed air system

• Fix air leaks (often 20–30% loss), reduce pressure setting, use intake air from cool place, switch off when not needed.

5. Lighting

• Replace incandescent/CFL/tube lights with LED; use daylight, occupancy sensors, timers, separate switching.

6. Thermal systems (boilers, furnaces)

• Control excess air, insulate pipes and furnaces, recover waste heat, return condensate, fix steam leaks and traps.

7. Refrigeration and HVAC

• Raise chiller set point, clean condensers, use efficient compressors and thermal insulation.

8. Process and management

• Avoid idle running, schedule production, monitor with energy meters, train staff, regular energy audits.

Opportunity        Typical saving
LED lighting       40–60 % of lighting energy
VFD on fan/pump    20–50 % of drive energy
Air leak repair    10–30 % of compressor energy
PF correction      lower kVA demand charge

Energy audit

Energy audit is the verification, monitoring and analysis of energy use in a plant, including submission of a technical report with recommendations for improving energy efficiency, together with cost-benefit analysis and an action plan to reduce energy consumption.

Process of carrying out an energy audit

Phase I – Pre-audit (planning)

1. Meet top management; decide scope, objectives, budget and time.
2. Form the audit team (electrical, mechanical, process staff).
3. Collect historical data: electricity and fuel bills (1–3 years), production data, single-line diagram, list of equipment with ratings.
4. Walk-through survey to understand the process and identify major energy users and obvious wastes.
5. Plan measurements and arrange instruments.

Phase II – Audit (execution) 6. Detailed measurements: power analyser (kW, kVA, pf, harmonics), clamp meters, lux meter, thermal camera, flue gas analyser, flow/pressure/temperature meters, ultrasonic leak detector. 7. Energy and material balance: find how much energy each section and equipment uses and the losses. 8. Compute specific energy consumption (kWh per unit product) and compare with benchmarks. 9. Identify energy conservation opportunities (ECOs): pf correction, efficient motors, VFDs, LED lighting, leak repair, waste heat recovery, load management. 10. Techno-economic analysis of each ECO: annual savings, investment, simple payback = investment/annual saving, NPV or IRR.

Phase III – Post-audit (implementation & follow-up) 11. Prepare the audit report: summary, present consumption, ECOs ranked as no-cost / low-cost / high-investment, action plan. 12. Present to management, implement selected measures. 13. Monitor and verify savings (M&V) and repeat the audit periodically.

 Pre-audit ──► Walk-through ──► Measurement
     │                              │
     ▼                              ▼
 Report ◄── Cost-benefit ◄── Energy balance & ECOs
     │
     └──► Implementation ──► Monitoring & verification

Need for power factor correction

Most industrial loads (induction motors, transformers, welding sets, fluorescent ballasts) draw lagging reactive power. Low pf means:

• Higher current for the same kW → higher I²R losses in cables, transformers and generators.
• Larger kVA rating of generators, transformers and switchgear is needed; less kW can be supplied from a given kVA.
• Larger voltage drop and poor regulation.
• Utility penalties / higher kVA demand charges.

Correcting pf with capacitors supplies the reactive power locally, reducing current, losses and kVA demand, and releasing capacity for extra load.

Numerical

Given: generator 1250 kVA at rated load, pf 0.8; extra load 170 kW at pf 0.85.

Existing load

P₁ = 1250 × 0.8 = 1000 kW
Q₁ = 1250 × 0.6 = 750 kVAr

Additional load

P₂ = 170 kW
Q₂ = 170 × tan(cos⁻¹0.85) = 170 × 0.6197
   = 105.36 kVAr

Total load without capacitor

P = 1000 + 170    = 1170 kW
Q = 750 + 105.36  = 855.36 kVAr
S = √(1170² + 855.36²) = 1449.3 kVA  (> 1250, overload)

Maximum kVAr allowed so that S = 1250 kVA

Q_allowed = √(1250² − 1170²) = √(193 600) = 440.0 kVAr
New pf    = 1170 / 1250 = 0.936 lag

Capacitor required

Qc = 855.36 − 440.0 = 415.36 kVAr

Answer: about 415.4 kVAr of capacitors is needed; the generator then supplies 1170 kW at 0.936 pf, i.e. exactly 1250 kVA.

The execution (audit) phase is the main working stage of an energy audit, where measurements are taken and energy-saving opportunities are found, after the pre-audit planning.

Steps in the execution phase

1. Detailed survey of each section: process, utilities (boilers, compressors, pumps, HVAC, lighting) and electrical distribution.
2. Measurements with portable instruments: power analyser (kW, kVA, pf, harmonics, load profile), clamp-on meters, lux meter, thermal imager, flue gas analyser, tachometer, flow, pressure and temperature meters.
3. Data analysis: load profiles, maximum demand, power factor, transformer and motor loading, losses.
4. Energy and material balance: quantify how much energy each area uses and how much is lost.
5. Specific energy consumption (e.g. kWh/ton) compared with benchmarks and past data.
6. Identify energy conservation opportunities (ECOs): pf correction, efficient motors, VFDs, lighting retrofits, air leak repair, insulation, heat recovery, load shifting.
7. Techno-economic evaluation: estimate savings, investment, payback period, NPV/IRR; rank as no-cost, low-cost and high-cost measures.
8. Discussion with plant staff to check that recommendations are practical and safe.

The output of this phase feeds the post-audit phase (report, implementation and monitoring).

Given: transformer 1000 kVA, average load 800 kVA at 0.6 pf (lag), no-load loss 1.6 kW, full-load loss 13.5 kW, capacitor cost NRs 1500/kVAr, demand charge NRs 210/kVA/month.

kVAr required to improve pf to 0.95

P  = 800 × 0.6               = 480 kW
Q₁ = 800 × 0.8               = 640 kVAr
Q₂ = 480 × tan(cos⁻¹0.95) = 480 × 0.3287
   = 157.77 kVAr
Qc = 640 − 157.77            = 482.23 kVAr

(In practice a standard 500 kVAr APFC panel would be used; the calculation uses 482.23 kVAr.)

Reduction of kVA demand

S₂ = 480 / 0.95   = 505.26 kVA
ΔS = 800 − 505.26 = 294.74 kVA

Reduction in transformer loss

No-load loss (1.6 kW) stays the same; copper loss ∝ (load/rating)²:

Cu loss before = 13.5 × (0.8)²     = 8.640 kW
Cu loss after  = 13.5 × (0.5053)²  = 3.446 kW
Reduction      = 5.194 kW

If the transformer runs all year (8760 h): 5.194 × 8760 = 45 496 kWh saved per year (extra benefit; energy rate not given).

Techno-economic analysis

Cost of capacitor bank = 482.23 × 1500   = NRs 723 347
Demand charge saving   = 294.74 × 210    = NRs 61 895 / month
                       = NRs 742 737 / year
Simple payback         = 723 347 / 61 895
                       = 11.7 months ≈ 0.97 year

The investment is recovered in less than one year from demand charge alone; the loss reduction, lower current, better voltage and released transformer capacity (about 295 kVA) are extra gains. So the pf improvement is technically and economically justified.

Given:

Energy rate Rs 7.5/kWh. Assumption: the transformer is energised for the whole year (8760 h); copper loss depends on loading, so it is worked out for a typical average load of 50% and also for other loadings.

Saving in initial cost

Cost difference = 7,21,577 − 7,00,767 = Rs 20,810 (B cheaper)

Saving in losses with transformer B

Iron-loss difference   = 1500 − 1430  = 70 W
Copper-loss difference = 11 500 − 9845 = 1655 W (at full load)

At 50% average load, Cu loss difference = 0.5² × 1655 = 413.75 W.

Energy saved/yr = (70 + 413.75) × 8760 / 1000
                = 4237.7 kWh
Money saved/yr  = 4237.7 × 7.5 = Rs 31,782

Decision

Transformer B is both cheaper (by Rs 20,810) and has lower losses, so it saves money from the first day: there is no extra investment to pay back. Even at no load it saves Rs 4,599/yr, and at 50% load about Rs 31,782/yr. Over a 25-year life this is several lakh rupees.

Answer: Yes, the second transformer should be purchased (provided its quality, guarantee, impedance and after-sales service are equivalent).

Given: 3-phase load P = 490 kW, pf improved from 0.7 lag to 0.97 lag, delta-connected loss-free capacitors on 11 kV, 50 Hz; cost of var control with switching NRs 2000/kVAr.

kVAr rating of capacitor bank

tan φ₁ = tan(cos⁻¹0.7)  = 1.0202
tan φ₂ = tan(cos⁻¹0.97) = 0.2506
Q₁ = 490 × 1.0202 = 499.90 kVAr
Q₂ = 490 × 0.2506 = 122.81 kVAr
Qc = Q₁ − Q₂      = 377.09 kVAr  (total, 3-phase)

Capacitance per phase (delta)

In delta, each capacitor has full line voltage 11 kV across it.

Qc per phase = 377.09 / 3 = 125.70 kVAr
C = Q_ph / (ω V²)
  = 125 698 / (2π × 50 × 11 000²)
  = 125 698 / (314.16 × 1.21 × 10⁸)
  = 3.307 × 10⁻⁶ F = 3.307 µF
Current per capacitor = 125.70 kVAr / 11 kV = 11.43 A

Cost

Cost = 377.09 × 2000 = NRs 7,54,189

Effect

kVA before = 490 / 0.7  = 700.0 kVA
kVA after  = 490 / 0.97 = 505.15 kVA
Reduction  = 194.85 kVA

Answer: capacitor bank of 377.09 kVAr (125.7 kVAr per phase), C ≈ 3.31 µF per phase in delta, cost ≈ NRs 7.54 lakh; kVA demand falls from 700 to 505.2 kVA.

Load scheduling is the planned arrangement of the operating times of large loads and processes in a plant so that the total demand stays low and smooth, without reducing production. Load management is the wider set of technical and administrative measures used to control the size and shape of the plant's load curve.

Why load scheduling is important

1. Reduces maximum demand (kVA) charges – industrial tariffs in Nepal (NEA) include a demand charge per kVA per month. If large motors, furnaces and compressors all start together, the peak rises and the demand charge is paid on that peak for the whole month.
2. Uses time-of-day (ToD) tariff – energy is cheaper at night (off-peak) and costliest in the evening peak. Moving flexible loads (pumping, batch furnaces, grinding, charging) to off-peak hours lowers the bill.
3. Improves load factor – a flatter load curve means better use of the transformer, cables and DG set; the same equipment serves more kWh.
4. Avoids overloading of the substation transformer, feeders and backup generator, and reduces voltage dips when big motors start.
5. Reduces losses – I²R losses depend on the square of current, so a flat load gives lower losses than a peaky load for the same energy.
6. Helps the utility – lower system peak reduces load shedding and the need for new generation.

kW  Before scheduling        After scheduling
 |      __                    
 |     |  |                   _________
 |  ___|  |___             __|         |__
 |_|          |_          |               |
 +-------------- time     +--------------- time
   high peak, low LF        lower peak, higher LF

Load management in an industrial plant

Load management techniques fall into these groups:

1. Peak clipping – shedding or limiting non-essential loads (air conditioning, water heaters, some lighting) during peak hours, often with a maximum demand controller that trips selected loads when demand approaches a set limit.
2. Valley filling – adding loads in off-peak hours, e.g. running water pumping, ice/thermal storage or battery charging at night.
3. Load shifting – moving batch processes from peak to off-peak periods; staggering shift start times and motor starting.
4. Strategic conservation – reducing total energy by efficient motors, VFDs, efficient lighting and avoiding idle running.
5. Power factor correction – capacitor banks reduce kVA demand for the same kW, directly lowering the kVA charge.
6. Use of captive generation – running a DG set or captive plant during the costliest hours if its cost per kWh is lower.
7. Monitoring and control – energy meters, SCADA or a load management system to record the load curve, set targets and alarm on excess demand.

Steps: record the existing load curve → identify large and flexible loads → prepare a load schedule matching tariff periods and production needs → install demand controller and PF correction → monitor and revise.

Example: a plant with two 100 kW crushers and a 150 kW pump set that run together in the morning may reach 350 kW peak. Running the pumps at night lowers the peak to 200 kW, saving the demand charge on 150 kW every month.

The generator must not exceed 1250 kVA after the extra load is added, so capacitors must supply enough kVAR to keep the total kVA at 1250.

Existing load (rated, pf 0.8):

P1 = 1250 × 0.8          = 1000 kW
Q1 = 1250 × sin(cos⁻¹0.8) = 1250 × 0.6 = 750 kVAR

Additional load (150 kW, pf 0.85):

tan(cos⁻¹0.85) = 0.6197
Q2 = 150 × 0.6197        = 92.96 kVAR

Total load:

P  = 1000 + 150   = 1150 kW
Q  = 750 + 92.96  = 842.96 kVAR

Reactive power allowed on the generator at 1250 kVA:

Q_allowed = √(1250² − 1150²)
          = √(1 562 500 − 1 322 500)
          = √240 000 = 489.90 kVAR

Capacitor kVAR required:

Qc = 842.96 − 489.90 = 353.06 kVAR

New power factor of generator = 1150 / 1250 = 0.92 lagging.

Answer: About 353 kVAR of capacitors is required; the generator then supplies 1150 kW at 0.92 pf, i.e. exactly 1250 kVA (not overloaded).

Note: an induction motor always draws lagging kVAR, so it cannot improve the power factor. The machine that does this while also driving a mechanical load is an over-excited synchronous motor; the solution below treats the motor as such.

Given: load P1 = 500 kW at 0.707 lag; motor load Pm = 100 kW; final pf = 0.95 lag.

Reactive power of the original load:

tan(cos⁻¹0.707) = 1.0003
Q1 = 500 × 1.0003 = 500.15 kVAR (lagging)

Combined load after adding motor:

P  = 500 + 100 = 600 kW
tan(cos⁻¹0.95)  = 0.3287
Q  = 600 × 0.3287 = 197.21 kVAR (lagging)

a) Leading kVAR supplied by motor

Qm = Q1 − Q = 500.15 − 197.21 = 302.94 kVAR (leading)

b) kVA rating of the motor

Sm = √(Pm² + Qm²) = √(100² + 302.94²)
   = √(10 000 + 91 773) = 319.02 kVA

c) Power factor of the motor

cos φm = Pm / Sm = 100 / 319.02 = 0.313 (leading)

Phasor (power triangle) picture:

         Q (kVAR)
 load  ↑ 500.15 lag
         |
 motor   ↓ 302.94 lead
 -----------------------
 total   197.21 lag  → pf 0.95 at 600 kW

Answer: (a) 302.9 kVAR leading, (b) about 319 kVA, (c) about 0.31 leading.

Energy saving comes from the lower input power of the more efficient motor for the same shaft output. Assume the motor runs at full rated output (25 hp) for all 8000 h.

Output power:

Po = 25 hp × 0.746 kW/hp = 18.65 kW

Input power at each efficiency (Pin = Po / η):

At 85 %: Pin1 = 18.65 / 0.85 = 21.941 kW
At 90 %: Pin2 = 18.65 / 0.90 = 20.722 kW

Power saved:

ΔP = 21.941 − 20.722 = 1.219 kW

(Formula form: ΔP = hp × 0.746 × (1/η1 − 1/η2).)

Annual energy saved:

ΔE = 1.219 kW × 8000 h = 9751.6 kWh/year

Annual cost saving:

Saving = 9751.6 × 7.5 = Rs 73 137 per year

If the high-efficiency motor costs, say, Rs 50 000 more, the simple payback is 50 000 / 73 137 ≈ 0.7 year, which shows why energy-efficient motors are recommended in energy audits.

Answer: Annual saving ≈ 9752 kWh, i.e. about Rs 73 140 per year (at full load).

An energy audit is a systematic study of how energy is used in a plant: it measures energy inputs and uses, finds where energy is wasted, and recommends cost-effective measures to reduce consumption and cost without affecting production or safety.

Types

• Preliminary (walk-through) audit – a quick survey using existing bills and a site visit; finds obvious savings in a few days.
• Detailed audit – a full study with instruments, energy balance of each section and detailed cost–benefit of each measure.

Energy audit technique (steps)

1. Pre-audit planning – meet management, form audit team, collect plant layout, single-line diagram, production data and 1–3 years of electricity bills.
2. Data collection – list connected loads (motors, lighting, compressors, furnaces), nameplate ratings and operating hours.
3. Measurement – use portable instruments: power analyser (kW, kVA, pf, harmonics), clamp meters, lux meter, tachometer, thermal camera, flow and pressure meters.
4. Analysis – prepare load curve, energy balance (where each kWh goes), specific energy consumption (kWh per unit product), compare with benchmarks; identify losses.
5. Identify energy saving opportunities (ESOs) and estimate savings.
6. Economic analysis – cost of each measure, annual saving, simple payback, NPV/IRR; rank them (no-cost, low-cost, high-investment).
7. Report and action plan – recommendations with priority and responsibility.
8. Implementation and monitoring – verify savings by measurement, then repeat audit periodically.

Plan → Collect data → Measure → Analyse
  ↑                                 ↓
Monitor ← Implement ← Report ← Find ESOs & costs

Energy saving opportunities in industrial electrical systems

Typical savings from an audit are 10–30 % of the electricity bill, many with payback under two years.

Capacitors connected in delta across the 6.6 kV supply supply the leading kVAR needed to raise the pf from 0.7 to 0.97.

(i) Total kVAR required

Q1 = P tan φ1 = 490 × tan(cos⁻¹0.7)  = 490 × 1.0202 = 499.90 kVAR
Q2 = P tan φ2 = 490 × tan(cos⁻¹0.97) = 490 × 0.2506 = 122.81 kVAR
Qc = Q1 − Q2  = 377.09 kVAR

(ii) Capacitance (delta connection)

In delta, each capacitor sees the line voltage 6.6 kV and supplies one-third of Qc.

Qc per phase = 377.09 / 3 = 125.70 kVAR
Qph = V² ω C   →   C = Qph / (2π f V²)
C = 125 698 / (2π × 50 × 6600²)
  = 125 698 / (314.16 × 43 560 000)
  = 9.19 × 10⁻⁶ F = 9.19 µF per phase

(iii) Payback period

The question does not give the demand tariff; assume the kVA demand charge of Rs 200/kVA/month (as given in the companion version of this problem). Since the capacitors are loss-free, kWh does not change; saving is only in kVA demand.

kVA before = 490 / 0.7  = 700.00 kVA
kVA after  = 490 / 0.97 = 505.15 kVA
Reduction  = 194.85 kVA

Annual saving = 194.85 × 200 × 12 = Rs 467 629
Capital cost  = 377.09 × 1200     = Rs 452 513
Payback = 452 513 / 467 629 = 0.968 year ≈ 11.6 months

Answer: Qc ≈ 377.1 kVAR; C ≈ 9.19 µF per phase (delta); payback ≈ 0.97 year (about 11.6 months) with an assumed Rs 200/kVA/month demand charge.

Loss-free capacitors in delta across the 6.6 kV supply supply leading kVAR. Real power (490 kW) and kWh stay the same; only the kVA demand falls, so the saving is in the kVA demand charge.

a) Total kVAR required

tan(cos⁻¹0.7)  = 1.0202
tan(cos⁻¹0.97) = 0.2506
Q1 = 490 × 1.0202 = 499.90 kVAR
Q2 = 490 × 0.2506 = 122.81 kVAR
Qc = Q1 − Q2      = 377.09 kVAR

b) Capacitance required

For delta connection, each phase capacitor is across the line voltage 6.6 kV.

Qc/phase = 377.09 / 3 = 125.70 kVAR
C = Qph / (ω V²)
  = 125 698 / (2π × 50 × 6600²)
  = 125 698 / (1.3685 × 10¹⁰)
  = 9.19 µF per phase

(Line current of capacitor bank = 377 090 / (√3 × 6600) = 33.0 A.)

c) Payback period

kVA1 = 490 / 0.70 = 700.00 kVA
kVA2 = 490 / 0.97 = 505.15 kVA
ΔkVA = 194.85 kVA

Annual saving in demand charge
  = 194.85 × 200 × 12 = Rs 467 629 /year

Cost of capacitor bank with switching
  = 377.09 × 1200 = Rs 452 513

Payback = 452 513 / 467 629
        = 0.968 year ≈ 11.6 months

Answer: (a) 377.1 kVAR, (b) 9.19 µF per phase in delta, (c) payback ≈ 0.97 year (≈ 11.6 months).

Assumptions: one ballast per tube (electronic ballast cost Rs 300 per tube), lights run 8 h/day for 365 days, and all 8 h fall in the 6 AM – 6 PM band (day shift, Rs 5.75/kWh). The lights are on during the monthly peak, so the kVA saving reduces the demand charge.

Number of tubes:

100 × (1×36 W) = 100 tubes
200 × (2×36 W) = 400 tubes
Total N        = 500 tubes

Power and kVA per tube:

Total load:

Copper:     P = 500 × 44 = 22.0 kW,  S = 44.00 kVA
Electronic: P = 500 × 37 = 18.5 kW,  S = 20.56 kVA
Saving:    ΔP = 3.5 kW,             ΔS = 23.44 kVA

Annual energy cost saving:

ΔE = 3.5 kW × 8 h × 365 = 10 220 kWh/year
Cost saving = 10 220 × 5.75 = Rs 58 765 /year

Annual demand charge saving:

= 23.44 kVA × 210 × 12 = Rs 59 080 /year

Total saving and payback:

Total saving = 58 765 + 59 080 = Rs 117 845 /year
Investment   = 500 × 300       = Rs 150 000
Payback      = 150 000 / 117 845 = 1.27 years ≈ 15.3 months

If the 8 h were instead in the evening peak (Rs 6.55/kWh), the energy saving would be 10 220 × 6.55 = Rs 66 941 and payback ≈ 1.19 year; the conclusion is the same.

Other benefits: lower line current (less cable and transformer loss), no flicker or hum, and longer tube life with electronic ballasts.

Answer: Saving ≈ Rs 1.18 lakh per year (Rs 58 765 energy + Rs 59 080 demand); payback ≈ 1.27 years.

Maximum demand: load factor = average demand / maximum demand, so

Pmax = 50 / 0.5 = 100 kW

The kVA charge applies to this maximum demand.

Most economical pf (loss-free capacitors, kWh unchanged): Let A = annual kVA charge per kVA, B = cost per kVAR of capacitor. The net saving is maximum when

sin φ2 = B / A

No interest/depreciation rate is given, so the full capacitor cost is compared with one year's kVA charge (i.e. capital recovered in one year):

A = 190 × 12 = Rs 2280 /kVA/year
B = Rs 600 /kVAR
sin φ2 = 600 / 2280 = 0.2632
cos φ2 = √(1 − 0.2632²) = 0.9648 lagging

Capacitor size:

Q1 = 100 × tan(cos⁻¹0.75)   = 100 × 0.8819 = 88.19 kVAR
Q2 = 100 × tan(cos⁻¹0.9648) = 100 × 0.2728 = 27.28 kVAR
Qc = 88.19 − 27.28          = 60.91 kVAR

Saving:

kVA1 = 100 / 0.75   = 133.33 kVA
kVA2 = 100 / 0.9648 = 103.65 kVA
ΔkVA = 29.68 kVA
Annual saving in kVA charge = 29.68 × 2280 = Rs 67 670
Capacitor cost = 60.91 × 600 = Rs 36 549
Net saving in first year = 67 670 − 36 549 = Rs 31 121

The energy charge (Rs 7.50/kWh) is the same before and after, since capacitors are loss-free and kW is unchanged.

Payback period:

= 36 549 / 67 670 = 0.54 year ≈ 6.5 months

Note: if instead an annual interest + depreciation of 10 % on the capacitor were assumed (B = Rs 60/kVAR/year), sin φ2 = 60/2280 and the economic pf would be about 0.9997, i.e. practically unity.

Answer: Most economical pf ≈ 0.965 lagging; capacitor ≈ 60.9 kVAR; annual saving ≈ Rs 67 670 (net Rs 31 121 in the first year); payback ≈ 0.54 year.

An energy conservation program is an organised, continuing effort in an industry to reduce energy use per unit of production by eliminating waste and using efficient equipment and practices, while keeping output, quality and safety the same.

Elements / steps of the program

1. Management commitment and energy policy – top management sets targets (e.g. 10 % reduction in kWh/unit in 2 years) and allocates budget.
2. Energy manager and team – appoint a responsible energy manager and a committee from production, maintenance and finance.
3. Energy audit – study present energy use, measure loads, find losses and list energy saving opportunities with costs and payback.
4. Set targets and action plan – rank measures as no-cost, low-cost and capital projects.
5. Implementation – carry out measures such as:
   • load scheduling and shifting to off-peak tariff hours,
   • power factor correction to reduce kVA demand,
   • energy-efficient motors, VFDs on pumps and fans,
   • efficient lighting (LED, electronic ballast, daylight, sensors),
   • fixing compressed air leaks, insulation of hot surfaces,
   • waste heat recovery, good maintenance.
6. Monitoring and targeting – meter each section, track specific energy consumption monthly, compare with targets.
7. Training and awareness – train operators, display posters, reward savings; "switch off when not in use" culture.
8. Review and repeat – evaluate results, report savings and update the program.

Commitment → Audit → Targets → Implement
     ↑                             ↓
   Review  ←  Monitor & train  ←───┘

Benefits: lower production cost, reduced demand on the national grid (less load shedding), lower emissions and better competitiveness.

Data: P = 200 kW at cos φ1 = 0.707; energy charge Rs 10/kWh; demand charge Rs 200/kVA/month; correcting equipment Rs 1200/kVAR.

Annual kVA charge:

A = 200 × 12 = Rs 2400 /kVA/year
B = Rs 1200 /kVAR

(No interest/depreciation rate is given, so the full equipment cost is set against one year's kVA charge.)

a) Optimum power factor

Total annual cost = A·P/cos φ2 + B·P(tan φ1 − tan φ2) is minimum when

sin φ2 = B / A = 1200 / 2400 = 0.5
cos φ2 = √(1 − 0.5²) = 0.866 lagging

b) Capacity of pf correcting equipment

Q1 = 200 × tan(cos⁻¹0.707) = 200 × 1.0003 = 200.06 kVAR
Q2 = 200 × tan(cos⁻¹0.866) = 200 × 0.5774 = 115.47 kVAR
Qc = 200.06 − 115.47 = 84.59 kVAR
Cost = 84.59 × 1200 ≈ Rs 101 508

c) Annual bill (16 h/day, 350 days)

Energy = 200 × 16 × 350        = 1 120 000 kWh
Energy charge = 1 120 000 × 10 = Rs 11 200 000
kVA demand = 200 / 0.866       = 230.94 kVA
Demand charge = 230.94 × 2400  = Rs 554 256
Annual bill = 11 200 000 + 554 256 = Rs 11 754 256

For comparison, without correction the demand is 200/0.707 = 282.89 kVA and the demand charge is Rs 678 925, so the bill would be Rs 11 878 925; correction saves Rs 124 669 per year in demand charge.

Note: if a 10 % annual interest + depreciation on the equipment is assumed instead (B = Rs 120/kVAR/year), sin φ2 = 0.05 and the optimum pf becomes about 0.999.

Answer: (a) optimum pf = 0.866 lagging, (b) ≈ 84.6 kVAR, (c) annual bill ≈ Rs 1.175 crore (Rs 11 754 256).

Power factor is the ratio of real power to apparent power, cos φ = kW/kVA. A pf well below 1 (lagging) is called poor; it means the load draws extra reactive current that does no useful work.

Causes of poor (lagging) power factor

1. Induction motors – the main cause in industry. They need magnetising current; at full load pf is about 0.8–0.85, and at light load it falls to 0.2–0.3. Oversized or lightly loaded motors make it worse.
2. Transformers – magnetising current, especially when lightly loaded or on no load (e.g. at night).
3. Arc and induction furnaces, welding sets – operate at very low pf (about 0.3–0.6).
4. Discharge lamps with magnetic ballasts – fluorescent, mercury and sodium lamps with choke have pf around 0.5.
5. Higher supply voltage than rated increases magnetising current and lowers pf.
6. Harmonics from rectifiers, drives and converters reduce the true (distortion) power factor.
7. Reactors and chokes in the system.

Effects (why it must be improved)

Higher current for the same kW → larger kVA rating of generator, transformer and switchgear, larger cables, higher I²R losses, poorer voltage regulation, and penalty / higher kVA charges in the tariff.

Methods of improvement

Leading kVAR is supplied near the load to cancel lagging kVAR:

1. Static capacitors – most common. Connected in star or delta across the supply, either fixed or automatically switched (APFC panel with relay). Low loss, little maintenance, no moving parts; but life is shortened by over-voltage and harmonics.
2. Synchronous condenser – an over-excited synchronous motor on no load; gives smooth, stepless control of kVAR; used at large substations, costly and needs maintenance.
3. Phase advancers – used with large induction motors; supply exciting ampere-turns to the rotor circuit at slip frequency so the motor runs at higher pf.
4. Synchronous motors driving plant loads, run over-excited to supply leading kVAR.

 Supply ──┬──── Motor load (lagging)
          │
         === Capacitor bank (leading kVAR)
          │

Placement of capacitors: individual (at motor terminals – best for loss reduction), group, or central (at main bus with automatic control).

Good practice: correct motor sizing, avoiding idle running of motors and transformers, and electronic ballasts also help keep pf high. Target pf is usually 0.95–0.99; going to unity is not economical.

Assumptions: the 60 kW and 40 kW are lamp wattages (ballast loss extra); each 200 W incandescent lamp is replaced by one 40 W fluorescent lamp with electronic ballast (similar lumen output); lamp prices are not given, so lamp replacement cost is shown only as number of lamps.

Number of lamps:

Fluorescent  = 60 000 / 40  = 1500 lamps
Incandescent = 40 000 / 200 = 200 lamps

Present load (3500 h/year):

FTL + copper ballast = 1500 × (40 + 8) = 72.0 kW
Incandescent         = 200 × 200       = 40.0 kW
Total                                   = 112.0 kW
Energy = 112 × 3500 = 392 000 kWh/year

Proposed load (hours reduced 10 %: 3150 h/year):

All FTL + electronic ballast = (1500 + 200) × (40 + 1)
                             = 1700 × 41 = 69.7 kW
Energy = 69.7 × 3150 = 219 555 kWh/year

Saving:

Energy saving = 392 000 − 219 555 = 172 445 kWh/year
Cost saving   = 172 445 × 7.50    = Rs 1 293 338 /year

Investment and payback:

Electronic ballasts = 1500 × 500 = Rs 750 000
New FTL fixtures    = 200 × 900  = Rs 180 000
Total investment                 = Rs 930 000
Payback = 930 000 / 1 293 338 = 0.72 year ≈ 8.6 months

Lamp replacement benefit:

Incandescent: 200 × 3500/1000 = 700 lamps/year
FTL (new 200): 200 × 3150/8000 = 78.75 ≈ 79 lamps/year

About 620 fewer lamp replacements per year, plus less maintenance labour and less heat in the building.

Answer: Energy saving ≈ 172 445 kWh/year, cost saving ≈ Rs 12.93 lakh/year, investment Rs 9.3 lakh, payback ≈ 0.72 year.

Importance of load scheduling

Load scheduling means planning when each major load or process runs so that the plant's maximum demand stays low and the load curve is flat.

1. Lower demand (kVA) charges – the monthly demand charge is based on the highest demand; staggering large motors, furnaces and compressors lowers it.
2. Benefit from time-of-day tariff – flexible loads (pumping, batch processes, charging) moved to off-peak night hours use cheaper energy.
3. Higher load factor – better use of transformer, cables and DG set capacity; postpones new investment.
4. Avoids overloading and voltage dips – motors are not started all at once.
5. Lower losses – a flat load has lower I²R loss than a peaky load for the same kWh.
6. Helps the utility reduce system peak and load shedding.

Bus arrangements in industrial distribution

1. Single bus system – all incoming supplies and outgoing feeders on one bus.
   • Simple, cheap, easy to operate.
   • A bus fault or bus maintenance shuts down the whole plant. Least reliable.
2. Single bus with bus sectionaliser – bus split into two sections by a circuit breaker (bus coupler), each fed by its own transformer.
   • A fault on one section is isolated; the other continues. Loads can be shared between sections.
3. Main and transfer bus – a transfer bus with a transfer breaker allows any feeder breaker to be taken out for maintenance without interrupting that feeder.
4. Double bus (double breaker or double bus single breaker) – two main buses; each circuit can be connected to either bus through isolators/breakers.
   • Bus maintenance or bus fault: loads transferred to the other bus.
   • Most flexible and reliable, but costly.
5. Ring main bus – bus forms a closed ring with breakers between circuits; each circuit fed from two sides.

Single bus with sectionaliser
  T1          T2
  |           |
 =X===[CB]===X=   (bus coupler)
  | |       | |
  F1 F2     F3 F4

Double bus
 ==== Bus 1 ===========
 ==== Bus 2 ===========
    |  |   |   |
   CB  CB  CB  CB  (each can select bus 1 or 2)
    F1 F2  F3  F4

Which is more reliable and why

The double bus (especially double bus–double breaker) arrangement is the most reliable, because:

• any bus can be taken out for maintenance or isolated after a fault without loss of supply;
• loads can be divided between buses so a fault affects only part of the plant;
• every circuit has an alternate path.

For most medium industries, a sectionalised single bus fed from two transformers is chosen as a good compromise: it gives continuity for essential loads at much lower cost than a double bus. The single bus without sectionalising is the least reliable.

At a given real power P and voltage V, current and kVA are inversely proportional to power factor:

I   = P / (√3 V cos φ)
kVA = kW / cos φ

So when pf falls, every component carrying this current must be larger, and losses rise.

1. Transformers

• The rating of a transformer is in kVA, limited by winding current (heating) and core flux.
• Lower pf → more kVA for the same kW, so a bigger transformer is needed.
• Example: 800 kW load at pf 1.0 needs 800 kVA; at 0.8 it needs 1000 kVA; at 0.6 it needs 1333 kVA.
• Copper loss ∝ I² ∝ 1/cos²φ – at pf 0.7 the copper loss is about 2 times that at unity pf.
• Voltage regulation becomes poor with lagging pf (reactive voltage drop adds).

2. Generators (alternators, DG sets)

• Rated in kVA at a stated pf (usually 0.8 lag). Below rated pf, the field current needed to hold voltage rises above design, so the alternator must be de-rated or a larger machine bought.
• The prime mover is sized for kW; at low pf the prime mover is under-used while the alternator is at full current – poor utilisation.
• Larger excitation system needed; poorer voltage regulation.

3. Cables and conductors

• Cable size is chosen on current. Low pf increases current, so a larger cross-section is needed:

Cross-section ∝ I ∝ 1/cos φ
Copper volume (for same loss) ∝ 1/cos²φ

• Higher I²R loss and voltage drop along the cable.

4. Switchgear and protective devices

• Breakers, busbars, CTs and fuses must be rated for the higher current, adding to cost.

Summary table

Numerical illustration: a 400 kW, 400 V load at 0.6 pf draws I = 400 000 / (√3 × 400 × 0.6) ≈ 962 A, but at 0.95 pf only ≈ 608 A. The transformer could be 421 kVA instead of 667 kVA.

Hence power factor correction with capacitors (target 0.95–0.98) is used to reduce equipment ratings, losses and kVA charges.