Introduction

An electric load centre is the point in an industrial plant from which electric power is distributed to a group of loads, located as close as possible to the "centre of gravity" of those loads. A unit substation or main distribution board placed at this point is called a load-centre substation.

Locating the load centre

If loads P₁, P₂, … Pₙ are at coordinates (x₁, y₁), (x₂, y₂), … on the plant layout, the load centre is:

X = Σ(Pᵢ·xᵢ) / ΣPᵢ
Y = Σ(Pᵢ·yᵢ) / ΣPᵢ

The transformer or main panel is placed at (X, Y), or at the nearest practical spot (near a wall, away from heat, dust and traffic).

 Incoming 11 kV
       |
  [Load-centre S/S] 11 kV/400 V
   /     |      \
 MCC-1  MCC-2   LDB
 (machines)    (lighting)

High voltage is carried close to the loads and only short low-voltage feeders go out from the load centre.

Advantages

1. Lower voltage drop: the LV feeders are short, so the voltage at the machines stays close to rated. Motors give their full torque and lamps their full output.
2. Lower I²R losses: power travels most of the way at HV and low current, so copper losses fall and energy cost goes down.
3. Saving in cable/conductor: short LV runs mean smaller and cheaper cables. Large LV cables are costly, and HV cables carry far less current.
4. Better voltage regulation: each load centre feeds a small area, so the voltage is easier to hold within limits.
5. Lower fault level on LV side: several smaller transformers give lower short-circuit currents than one big transformer, so cheaper switchgear can be used.
6. Higher reliability: a fault in one load centre affects only its own area. The rest of the plant keeps running.
7. Easy expansion: a new load centre can be added for a new production line without redesigning the whole system.
8. Better power factor control: capacitor banks can be placed at each load centre, close to the inductive loads.

Example: A factory has a 300 kW machine shop 200 m from the main gate. Feeding it at 400 V from the gate would need very heavy cable and give a large voltage drop. Taking 11 kV to a 400 kVA load-centre substation inside the shop and running short 400 V feeders to the machines reduces cable cost, losses and voltage drop.

Electrical installations must follow the rules of the regulator and the standards of good practice. In Nepal these come from the Electricity Regulations, 2050, NEA's connection requirements and NBC 207 (Nepal National Building Code, electrical design), which are largely based on the Indian Electricity Rules 1956 and IS 732. The main general rules are:

1. Approved materials: all wires, cables, switches and fittings must meet the relevant standard (NS/IS/IEC) and be rated for the circuit voltage and current.
2. Switches and fuses in the phase conductor: single-pole switches, fuses and MCBs go in the phase (live) wire only, never in the neutral, so that opening the switch makes the equipment dead.
3. Main switch and cut-out: every consumer installation must have a main switch (or MCB/MCCB) and a cut-out near the point of supply, easy to reach and clearly marked.
4. Earthing: the metal body of every appliance, the conduit, metal enclosures and the neutral of the supply transformer must be properly earthed. Earth resistance is kept low (about 1 Ω for large substations, a few ohms for installations). Three-pin sockets must have the earth pin connected.
5. Protection of every circuit: each sub-circuit is protected by a fuse/MCB rated for the cable. Earth-leakage protection (RCCB/ELCB) is used where shock risk is high (wet areas, sockets).
6. Separate circuits: lighting and power circuits are kept separate. A lighting sub-circuit normally has no more than about 10 points or 800 W; a power sub-circuit has no more than 2 power sockets (about 3 kW).
7. Polarity and colour code: phases R-Y-B (red, yellow, blue), neutral black and earth green (or green/yellow) must be used consistently.
8. Clearances: overhead lines and service mains must keep minimum clearances from ground, buildings and other lines as given in the regulations.
9. Mechanical protection: cables are run in conduit, trunking or trays, or are armoured, where they may be damaged.
10. Danger notices: HV equipment and substations must carry danger boards in Nepali and English.
11. Testing before energising: insulation resistance, earth continuity, polarity and earth resistance tests must be done and the results recorded before the supply is connected. Insulation resistance must not be less than the value required by IS 732/regulations (about 1 MΩ for new LV wiring).
12. Licensed persons: design, installation and testing must be done or supervised by a licensed electrician/engineer.
13. Accessibility: distribution boards and switchgear must be accessible, with enough working space in front, and not placed in wet or hazardous locations.

The supply (distribution) system of an industrial plant is the arrangement of transformers, switchgear and feeders that carries power from the incoming utility point to the loads. It is chosen on the basis of reliability needed, cost, plant size and ease of maintenance.

Main types

1. Simple radial system: one incoming line and one transformer feed all loads through radial feeders. Cheapest and simplest, but any fault in the line or transformer stops the whole plant.
2. Expanded radial system: the HV is taken to several load-centre transformers, each feeding its own area radially. Lower losses and voltage drop than simple radial.
3. Primary selective system: each load-centre transformer can be switched to either of two HV feeders. If one feeder fails, the transformer is switched to the other.
4. Primary loop (ring main) system: HV feeders form a loop through all load-centre substations, so each one is fed from two directions. A faulty section can be isolated and supply restored.
5. Secondary selective system: two transformers feed two LV buses joined by a normally open tie breaker. If one transformer fails, the tie closes and the other carries the essential load.
6. Secondary spot network: two or more transformers fed from different HV feeders are paralleled on the LV side through network protectors. Highest reliability, highest cost.

Radial system in detail

 Utility 11 kV
      |
    [CB]  incoming breaker
      |
    (TR)  11 kV / 0.4 kV
      |
    [ACB] LV main breaker
      |
 ===== LV main bus 400 V =====
   |       |       |       |
 [MCCB] [MCCB]  [MCCB]  [MCCB]
   |       |       |       |
 MCC-1   MCC-2    PDB     LDB
 motors  motors  sockets  lights

Working: power comes from one source over one path to each load. The incoming breaker and transformer feed the LV main bus. From the bus, separate outgoing feeders, each with its own MCCB, go to motor control centres, power distribution boards and lighting distribution boards. Power flows in one direction only, so protection is simple: overcurrent relays and MCCBs graded from load to source.

Advantages:

• Lowest first cost and simple layout.
• Simple protection and easy operation.
• Easy to expand by adding feeders.

Disadvantages:

• Poor reliability: a fault on the incoming line, transformer or main bus shuts the whole plant.
• Maintenance of the transformer needs a full shutdown.
• Voltage drop at far ends if feeders are long.

Use: small and medium plants where a short interruption is acceptable, or as the base layout that is later expanded to a selective or loop system.

Actual load estimation means finding the maximum power (kVA) that the plant will really draw, not the sum of nameplate ratings. Machines do not run at full load all the time and do not all run together, so the installed power is reduced by suitable factors. The common method (IEC / Schneider Electrical Installation Guide) works level by level.

Steps

1. List the installed power Pₙ (kW) of each load from nameplates.
2. Find the installed apparent power of each load:

Pa = Pn / (η × cos φ)    (kVA)

For lamps, add ballast losses.

3. Apply the factor of maximum utilization ku (≤ 1): the ratio of the actual maximum power a device takes to its rating. Typical: motors 0.75, lighting and heating 1.
4. Apply the factor of simultaneity ks (≤ 1) for each group, because not all loads in a group run at the same time. Typical for distribution boards (IEC 61439): 2–3 circuits 0.9, 4–5 circuits 0.8, 6–9 circuits 0.7, 10 or more 0.6.
5. Add up level by level: load → sub-board → main board → transformer. At each level, multiply the sum by that level's ks.
6. Add a margin for future growth (about 20–25%) and select the next standard transformer size.

Example

A small workshop has:

• 4 lathe motors, 7.5 kW each, η = 0.88, cos φ = 0.85
• Lighting load 4 kVA
• 10 socket outlets of 1 kVA each

Motor:  Pa = 7.5/(0.88×0.85)      = 10.03 kVA
        with ku = 0.75            = 7.52 kVA
        4 motors, ks = 0.8:
        4 × 7.52 × 0.8            = 24.06 kVA
Light:  4 × ku 1 × ks 1           = 4.00 kVA
Socket: 10 × 1 × ks 0.2           = 2.00 kVA
Sum                               = 30.06 kVA
Main board ks = 0.9:
        30.06 × 0.9               = 27.06 kVA
Future margin 25%:
        27.06 × 1.25              = 33.82 kVA

The installed load is 4 × 10.03 + 4 + 10 = 54.1 kVA, but the estimated demand is only about 27 kVA. A 50 kVA transformer (next standard size above 33.8 kVA) is enough, instead of a much larger one sized on connected load.

Why it matters

• Avoids oversized transformers, cables and switchgear (high cost, poor efficiency at light load).
• Avoids undersizing, which causes overheating and tripping.
• Gives the maximum demand used for tariff and contract demand with the utility.

Electrical rules (such as Nepal's Electricity Regulations, 2050 and NBC 207) and standards (IS/IEC/NS) set the minimum requirements for a safe and reliable installation. Following them during installation and testing is important for these reasons:

During installation

1. Safety of people: correct earthing, use of RCCBs, proper insulation and switching in the phase conductor protect workers and users from electric shock.
2. Fire prevention: correct cable sizing and matching fuses/MCBs stop overheating and short-circuit fires, which are a main cause of building fires in Nepal.
3. Correct ratings of material: standards make sure cables, switchgear and fittings are of proper quality and rating for the voltage, current and fault level.
4. Uniformity: standard colour codes, symbols and layouts let any electrician understand, operate and maintain the system safely later.
5. Legal approval: the utility (NEA) gives a connection only if the installation meets the rules. Insurance claims and liability also depend on compliance.

During testing

1. Proves the installation is safe before energising: insulation resistance, earth continuity, earth electrode resistance, polarity and RCD tests find faults while the system is dead.
2. Gives reference values: recorded test results are the baseline for future maintenance and fault finding.
3. Checks protection works: tests confirm that fuses, MCBs and relays will clear a fault in the required time (correct earth-fault loop impedance).
4. Protects equipment: finding wrong connections or low insulation early prevents damage to costly motors and transformers on first switch-on.

Example: if an installer skips the polarity test and a switch ends up in the neutral, a lamp holder stays live even when the switch is off. A person changing the lamp can get a shock. A standard test catches this before the supply is connected.

Load factor is the ratio of the average load to the maximum demand over a given period (day, month or year):

Load factor = Average load / Maximum demand
            = Energy consumed / (Max demand × hours)

It is always ≤ 1 and shows how evenly the load is used.

Calculation (period = 1 day = 24 h)

Each consumer's maximum demand is its constant load.

Consumer A:
 Energy  = 20 kW × 5 h       = 100 kWh
 Avg load = 100 / 24         = 4.167 kW
 LF(A)   = 4.167 / 20        = 0.2083

Consumer B:
 Energy  = 5 kW × 20 h       = 100 kWh
 Avg load = 100 / 24         = 4.167 kW
 LF(B)   = 4.167 / 5         = 0.8333

Answer: Load factor of A = 0.208 (20.8%); load factor of B = 0.833 (83.3%).

Significance

Both consumers use the same energy, 100 kWh per day, but:

• Utility point of view: to supply A, the utility must keep 20 kW of generation, transformer and line capacity ready, which is used only 5 hours a day. For B only 5 kW of capacity is needed and it is used 20 hours. The fixed cost of plant per kWh sold is four times larger for A. So a high load factor (B) is better for the utility: plant is used well, cost per unit falls.
• Energy expenditure (consumer) point of view: under a two-part tariff (demand charge per kW + energy charge per kWh), A pays the same energy charge as B but four times the demand charge (20 kW vs 5 kW). A's cost per kWh is therefore higher. Improving load factor by spreading loads over time lowers the electricity bill.
• That is why utilities give lower rates to high load factor consumers and encourage off-peak use.

Importance of electrical load estimate

Load estimation finds the actual maximum demand of the installation from the connected load using demand and diversity factors (or ku and ks). It is important because:

• It fixes the size of transformer, generator, cables, busbars and switchgear. Over-estimate wastes money and runs equipment at low efficiency; under-estimate causes overheating and tripping.
• It sets the contract demand and tariff with the utility (NEA).
• It is needed for voltage drop, fault level and protection calculations.
• It allows planning for future expansion with a known margin.

Key features of the supply system to industry

1. Reliability and continuity: process loads need supply with few interruptions; alternative sources or ring/selective schemes are used.
2. Good voltage regulation: voltage within about ±5% of rated, so motors give rated torque.
3. Correct frequency (50 Hz in Nepal) and a sinusoidal waveform with low harmonics.
4. Adequate capacity for present load plus future expansion.
5. Safety: proper earthing, protection and isolation.
6. Flexibility: easy to extend and to isolate parts for maintenance.
7. Economy: low first cost, low losses, good power factor.
8. Supply voltage matched to the load size: 400/230 V for small plants, 11 kV or 33 kV for medium and large plants, 66 kV or 132 kV for very large plants.

Sources of supply to an industrial plant

1. Utility grid (NEA): the main source for most plants. Small industries get 400 V three-phase; larger ones take 11 kV or 33 kV and have their own substation. Cheapest and most reliable for normal running, but subject to load shedding and outages.
2. Captive power plant: the plant's own generator, such as a small hydro plant (common in Nepal for remote industries), a diesel generator set or a steam turbine. Used where grid supply is absent or unreliable.
3. Cogeneration: industries like sugar, paper and cement use process steam or waste heat (bagasse) to make electricity and heat together, and may sell surplus power to the grid.
4. Standby/emergency sources: diesel generator sets with automatic transfer switches, UPS and battery banks for critical loads (control systems, emergency lighting, computers).
5. Renewable sources: rooftop solar PV (grid-tied or with batteries), wind, biogas. They reduce the energy bill and dependence on the grid.
6. Dual utility feeders: two incoming feeders from different substations for high reliability.

Formulas used

Demand factor = Maximum demand / Connected load
Load factor   = Average load / Maximum demand
Average load  = Energy per day / 24 h

Connected load

Lamps   : 12 × 80 W  =  960 W
Heaters :  4 × 100 W =  400 W
Connected load       = 1360 W

Daily energy and average load

Lamps   : 8 × 80 W × 6 h  = 3840 Wh
Heaters : 4 × 100 W × 4 h = 1600 Wh
Energy per day            = 5440 Wh
Average load = 5440 / 24  = 226.67 W

Factors

Demand factor = 1500 / 1360   = 1.103
Load factor   = 226.67 / 1500 = 0.1511

Answer (data as printed): load factor = 0.151 (15.1%), demand factor = 1.10.

Note on the data: a demand factor greater than 1 is not possible, because the maximum demand cannot exceed the connected load. The heater rating is almost certainly 1000 W (the standard textbook version of this problem). With 4 heaters of 1000 W:

Connected load = 960 + 4000           = 4960 W
Energy/day = 3840 + 4×1000×4          = 19840 Wh
Average load = 19840 / 24             = 826.67 W
Demand factor = 1500 / 4960           = 0.3024
Load factor   = 826.67 / 1500         = 0.5511

Answer (heaters 1000 W): demand factor = 0.302, load factor = 0.551 (55.1%).

Maximum demand (MD) is the greatest load (kW or kVA) drawn by an installation during a given period. It is usually measured as the highest average demand over a short interval, normally 15 or 30 minutes, by an MD meter. It is always less than or equal to the connected load.

Need to calculate maximum demand in an industrial building

1. Sizing the transformer and generator: the substation transformer and standby DG set are chosen to supply the maximum demand plus a margin, not the total connected load. This avoids heavy oversizing.
2. Sizing cables, busbars and switchgear: the main incomer, busbars and feeders must carry the maximum demand current without overheating.
3. Contract demand and tariff: industrial consumers pay a demand charge (Rs per kVA or kW of MD) plus an energy charge. Knowing the MD lets the plant choose the right contract demand and avoid penalty for exceeding it.
4. Protection settings: relay and breaker settings are based on the maximum load current.
5. Demand control: knowing when MD occurs lets the plant shift loads (for example run pumps or furnaces off-peak) to lower the MD and the bill.
6. Planning: the utility needs the MD to design the feeder and to plan generation.
7. Power factor correction: capacitor sizing uses the kW and kVA at maximum demand.

Example

A garment factory has a connected load of:

If 80% of the sewing machines, both boilers, all lighting and the compressor were assumed to run together at full rating, the worst-case estimate would be:

0.8×20 + 30 + 15 + 15 = 76 kW

But the boilers and the compressor switch on and off with their thermostats and pressure switches, so they are not all at full load in the same 30 minutes. MD meter records show the actual peak is about 60 kW. Then:

• Demand factor = 60/80 = 0.75.
• A transformer of about 60/0.85 ≈ 71 kVA plus margin → 100 kVA is chosen, not one based on 80 kW plus large margins.
• The contract demand is fixed near 70 kVA, saving demand charges every month.

Electric load centre

An electric load centre is the point from which power is distributed to a group of loads, placed at or near the "centre of gravity" of those loads so that low-voltage feeders are as short as possible. Its position is found from:

X = Σ(Pᵢ·xᵢ)/ΣPᵢ ,   Y = Σ(Pᵢ·yᵢ)/ΣPᵢ

The substation or main distribution board placed there reduces voltage drop, losses and cable cost and makes the system more reliable.

(i) Connected load

Connected load is the sum of the nameplate (rated) ratings of all equipment connected to the supply system of a consumer, in kW or kVA.

Example: a workshop has 4 motors of 10 kW, 20 lamps of 40 W and 2 heaters of 2 kW.

Connected load = 4×10 + 20×0.04 + 2×2
               = 40 + 0.8 + 4 = 44.8 kW

(ii) Maximum demand

Maximum demand is the greatest load actually drawn by the consumer at any time during a period, usually measured as the highest 15- or 30-minute average. Because all equipment never runs at full load together, maximum demand is less than connected load.

Example: in the workshop above, the highest load recorded by the MD meter in a month is 30 kW (3 motors near full load plus lights and one heater). So:

Maximum demand = 30 kW
Demand factor  = 30 / 44.8 = 0.67

Maximum demand decides the transformer, cable and switchgear size and the demand charge on the bill.

The factor of maximum utilization (ku) is the ratio of the maximum power actually taken by a load in normal operation to its rated (installed) power:

ku = Actual maximum power used / Rated power   (ku ≤ 1)

It is used in industrial load estimation (IEC method) because equipment rarely runs at full rating. For example, a motor is chosen with a margin and usually runs at about 75% of its rating.

Typical values:

• Motors in industry: ku ≈ 0.75
• Incandescent lighting and heating: ku = 1
• Socket outlets: depends on what is plugged in

Example: a 10 kVA motor with ku = 0.75 is counted as 7.5 kVA when the demand of its distribution board is worked out. ku is applied to each individual load, before the factor of simultaneity ks is applied to groups.

The factor of simultaneity (ks), also called the coincidence factor, accounts for the fact that all loads of a group do not run at the same time. It is the ratio of the maximum demand of a group of loads to the sum of their individual maximum demands:

ks = Max demand of group / Σ(individual MDs)
(ks ≤ 1)

It is the reciprocal of the diversity factor. ks is applied at each level of distribution (sub-board, main board, transformer) after ku has been applied to each load.

Typical values (IEC 61439 for distribution boards):

By function: lighting 1, heating and air conditioning 1, socket outlets 0.1–0.2.

Example: a board feeds 5 motors each with a demand of 7.5 kVA. With ks = 0.8, the board demand is 5 × 7.5 × 0.8 = 30 kVA instead of 37.5 kVA.

The fifth column repeats "(12:00–16:00)"; it is taken as a separate 4-hour interval as printed, giving 7 intervals of data.

Step 1: Combined load in each interval

Step 2: Capacity of substation

The substation must supply the highest combined (simultaneous) load:

Max demand of complex = 73 kVA

Answer (i): Required substation capacity = 73 kVA (in practice the next standard transformer, 100 kVA, would be installed).

Step 3: Diversity factor

Individual maximum demands:

Block A max = 22 kVA
Block B max = 34 kVA
Block C max = 40 kVA
Sum         = 96 kVA

Diversity factor
 = Σ individual MDs / Max demand of system
 = 96 / 73
 = 1.315

Answer (ii): Diversity factor = 1.32.

Because the blocks reach their peaks at different times, the substation needs only 73 kVA instead of 96 kVA.

These terms describe how much of the installed equipment is actually used and when. One example workshop is used for all of them.

Example workshop: 5 motors of 10 kW, lighting 5 kW and heaters 15 kW.

i) Connected load

The sum of the rated (nameplate) powers of all equipment connected to the supply.

Connected load = 5×10 + 5 + 15 = 70 kW

ii) Maximum demand

The greatest load drawn at any time in a period (usually the highest 15- or 30-minute average, read by an MD meter). Not all equipment runs at full load at once, so MD < connected load.

Recorded maximum demand = 42 kW

iii) Demand factor

The ratio of maximum demand to connected load (always ≤ 1).

Demand factor = MD / Connected load = 42/70 = 0.6

It is used to find the MD of a new installation from its connected load.

iv) Diversity factor

The ratio of the sum of individual maximum demands of parts of a system to the maximum demand of the whole system (always ≥ 1).

If the workshop and two other workshops have individual MDs of 42 kW, 30 kW and 28 kW, and the feeder's recorded MD is 80 kW:

Diversity factor = (42 + 30 + 28) / 80 = 100/80 = 1.25

A high diversity factor means smaller transformers and feeders can supply many loads that peak at different times.

v) Load factor

The ratio of average load to maximum demand over a period.

If the workshop uses 504 kWh in a day:

Average load = 504 / 24 = 21 kW
Load factor  = 21 / 42  = 0.5

A high load factor means the plant capacity is used well and the cost per kWh is lower.

Electric load centre

An electric load centre is the point from which power is distributed to a group of loads, located at the centre of gravity of those loads (X = ΣPx/ΣP, Y = ΣPy/ΣP). Placing the substation there keeps LV feeders short, reducing voltage drop, losses and cable cost.

Demand factor of Block-1 and Block-2

Demand factor = Maximum demand / Connected load

Block-1 connected load
 = 70 + 90 + 20 + 50 + 10 + 20 = 260 kW
DF(1) = 200 / 260 = 0.769

Block-2 connected load
 = 60 + 40 + 40 + 70 + 30 = 240 kW
DF(2) = 160 / 240 = 0.667

Diversity factor of the substation

Sum of individual maximum demands
 = 200 + 160 + 150 + 200 = 710 kW
Maximum demand of substation = 600 kW

Diversity factor = 710 / 600 = 1.183

Answer: Demand factor of Block-1 = 0.769, of Block-2 = 0.667; diversity factor of the substation = 1.18.

The blocks do not reach their peaks at the same time, so the substation needs to be sized for 600 kW rather than 710 kW.

Connected load is the sum of the rated (nameplate) powers of all electrical equipment connected to the supply system of an industry, expressed in kW or kVA.

In an industry it includes:

• Motors of machines, pumps, fans, compressors, cranes and conveyors
• Heating loads such as furnaces, ovens, boilers and welding sets
• Lighting (lamps with ballast/driver losses) and fans
• Socket outlets and office equipment
• Air conditioning and other building services

Points to note:

• Connected load is not the load the plant actually draws. Machines run below full rating and not all at once, so the actual maximum demand is less (demand factor = MD / connected load, typically 0.5–0.8 in industry).
• It is the starting point for load estimation: maximum demand is found from it using demand and diversity factors (or ku and ks).
• The utility uses it to classify the consumer and to approve the connection.

Example: a small factory has 10 motors of 7.5 kW, 2 furnaces of 20 kW, 60 lamps of 40 W and 4 kW of office load.

Connected load = 10×7.5 + 2×20 + 60×0.04 + 4
               = 75 + 40 + 2.4 + 4
               = 121.4 kW

These four terms describe how the electrical capacity of an installation is used. A single example plant is used to show each.

Example plant: 6 motors of 15 kW, furnace 30 kW, lighting 10 kW.

Connected load

The sum of the rated (nameplate) powers of all equipment connected to the supply. It is the maximum the plant could draw if everything ran at full load together.

Connected load = 6×15 + 30 + 10 = 130 kW

Maximum demand

The greatest load drawn during a period, measured as the highest average over a short interval (15 or 30 minutes) by an MD meter. Since machines run below rating and not all together, MD is less than the connected load. It decides the size of transformer, cables and switchgear, and the demand charge on the bill.

Maximum demand recorded = 78 kW

Demand factor

The ratio of maximum demand to connected load. It is always ≤ 1 and is used to estimate MD at the design stage.

Demand factor = 78 / 130 = 0.6

Typical values: motors 0.6–0.8, lighting about 1.

Load factor

The ratio of average load to maximum demand over a period (day, month, year). It shows how steadily the plant uses power.

Energy used in a day = 936 kWh
Average load = 936 / 24 = 39 kW
Load factor  = 39 / 78 = 0.5

A high load factor means equipment is used well and the cost per unit of energy is low. Running some loads (pumps, furnaces) in off-peak hours raises the load factor.