Electrification of Industrial Building

The cable is first chosen by current rating from the given table, then checked for voltage drop; the larger of the two sizes is selected.

Assumptions: power factor 0.8 (not given), allowed voltage drop 2.5 % of 400 V = 10 V for the MDB–SDB feeder.

Formulas and assumptions

Line current   I  = P / (√3 · V_L · cosφ)
Voltage drop   ΔV = √3 · I · (ρ·L/A) · cosφ
               (reactance neglected)
Min. area      A  = √3 · I · ρ · L · cosφ / ΔV_allowed
ρ (copper)     = 0.0178 Ω·mm²/m

MDB to SDB1 (80 kW, 10 m)

I  = 80 000 / (√3 × 400 × 0.8) = 144.34 A
From table: 50 mm² → 145 A ≥ 144.34 A
ΔV = √3 × 144.34 × 0.0178 × 10 × 0.8 / 50
   = 0.71 V  (0.18 %)  < 10 V  OK

50 mm² is just enough by current; since it runs at almost 100 % of its rating, 70 mm² (175 A) is the safer practical choice.

MDB to SDB2 (10 kW, 100 m)

I  = 10 000 / (√3 × 400 × 0.8) = 18.04 A
From table: 1.5 mm² → 21 A ≥ 18.04 A
ΔV (1.5 mm²) = √3 × 18.04 × 0.0178 × 100 × 0.8 / 1.5
             = 29.67 V (7.42 %)  > 10 V  NOT OK
A_min = √3 × 18.04 × 0.0178 × 100 × 0.8 / 10
      = 4.45 mm²  → next size 6 mm² (45 A)
ΔV (6 mm²) = 7.42 V (1.85 %)  OK

Answer: MDB–SDB1: 3½ core 50 mm² Cu (70 mm² recommended for margin); MDB–SDB2: 3½ core 6 mm² Cu. The short heavy feeder is decided by current rating; the long light feeder is decided by voltage drop.

The rating of a panel board (MDB) or distribution board (SDB) means the rating of its incomer (MCCB/MCB), busbar and incoming cable. It is fixed from the design current of the loads it feeds, after applying utilization and simultaneity (diversity) factors and a safety margin.

Procedure

1. List the loads on each DB: kW, pf, efficiency, single or three phase. Balance single-phase loads on R, Y, B.
2. Find the current of each outgoing circuit: I = P/(√3·V·η·cosφ) for 3-phase, I = P/(V·cosφ) for single phase.
3. Select outgoing MCBs: In ≥ 1.25 × Ib (next standard size: 6, 10, 16, 20, 25, 32, 40, 50, 63 A); C curve for general loads, D curve for motors.
4. SDB demand: add circuit currents and multiply by the simultaneity factor Ks (0.9 for 2–3 circuits, 0.8 for 4–5, 0.7 for 6–9, 0.6 for 10 or more).
5. SDB incomer: In ≥ 1.25 × SDB demand; MCB up to 63/100 A, MCCB above.
6. MDB demand: add the SDB demands, apply Ks of the MDB, add spare for future.
7. MDB incomer: MCCB/ACB ≥ 1.25 × MDB demand; check that it does not exceed the transformer LT full-load current, and its breaking capacity ≥ prospective fault current.
8. Busbar: rating ≥ incomer rating (copper about 1.2–1.6 A/mm²). Keep 20 % spare ways.
9. Coordination: Ib ≤ In ≤ Iz (cable rating), and upstream device larger than downstream for discrimination.

Example

A factory, 400 V, 3-phase: SDB-1 feeds 3 motors of 7.5 kW (η = 0.88, pf = 0.85); SDB-2 feeds 15 kW balanced lighting and sockets at pf 0.9.

Motor current = 7500 / (√3 × 400 × 0.88 × 0.85)
              = 14.47 A
Motor MCB     = 1.25 × 14.47 = 18.09 A → 20 A (D curve)

SDB-1 demand  = 3 × 14.47 × Ks 0.9 = 39.08 A
SDB-1 incomer = 1.25 × 39.08 = 48.84 A → 50 A TP MCB

SDB-2 current = 15000 / (√3 × 400 × 0.9) = 24.06 A
SDB-2 incomer = 1.25 × 24.06 = 30.07 A → 32 A TPN MCB

MDB demand = (39.08 + 24.06) × Ks 0.9 = 56.82 A
MDB incomer = 1.25 × 56.82 = 71.02 A → 80 A TPN MCCB

Cables are then chosen so that their rating is above the protective device rating (e.g. MDB–SDB-1 with 50 A MCB → 10 mm² Cu), and checked for voltage drop.

Electrical installation (wiring) is the system of cables, conduits, accessories and protective devices that carries power from the distribution board to the points of use. The type is chosen according to the building use, environment, safety, appearance, cost and need for future changes.

Types of electrical installation

1. Cleat wiring: PVC cables held on porcelain or plastic cleats on walls. Cheap and quick, but poor appearance and protection. Use: temporary installations, construction sites.
2. Batten (CTS/TRS) wiring: sheathed cables clipped on wooden battens. Cheap and easy to inspect; not for damp or high-temperature areas. Use: low-cost residential buildings (now less common).
3. Casing and capping: cables in PVC or wooden channels with a cover. Neat, cheap, easy to alter. Use: offices, shops, rented buildings.
4. Surface conduit wiring: PVC or GI conduit fixed on walls/ceilings. Good mechanical protection, easy to modify. Use: workshops, factories, plant rooms.
5. Concealed conduit wiring: conduits buried in walls/slabs before plastering. Best appearance, safe, long life, but costly and hard to alter. Use: residential, hotels, hospitals, offices.
6. Cable tray / ladder installation: armoured or unarmoured cables on perforated trays or ladders. Good cooling and easy to add circuits. Use: industries, substations, plant rooms, data centres.
7. Trunking and busbar trunking (bus duct): metal or PVC ducts; busbar trunking with tap-off boxes. Flexible for changing machine layouts. Use: workshops with many machines, rising mains in high-rise buildings.
8. Underground cable installation: armoured cables direct in ground or in ducts. Use: feeders between buildings, external lighting.
9. Special installations: flameproof (explosion-proof) wiring with sealed fittings in hazardous areas (fuel depots, paint shops, mines); fire-resistant cables for fire pumps and emergency lighting; waterproof installations for wet areas.

Which installation for which function

The choice is a balance between safety, durability, appearance, flexibility and cost.

A cable is sized so that it carries the design current without overheating, keeps voltage drop within limits and withstands short-circuit current. The same method is used for each stage, but the governing condition changes from stage to stage.

Transformer to MDB

• Design current = full-load LT current of the transformer: I = kVA × 1000 / (√3 × 400). Example: 250 kVA → 360.8 A.
• Cable (or parallel runs/bus duct) must carry this current after derating for grouping and temperature; it must also withstand the transformer LT fault current for the clearing time (A ≥ I_sc·√t / k).
• Run is short, so current rating and short-circuit withstand govern; voltage drop is usually small (aim < 1 %).

MDB to SDB

• Design current = maximum demand of the SDB (connected load × Ku × Ks), plus 20–25 % future margin.
• Cable rating ≥ rating of the protective MCCB/MCB at the MDB.
• Feeder lengths are moderate to long, so check voltage drop (about 1.5–2.5 %); long feeders often need a larger size than the current alone requires.

SDB to end use

• Design current of each final circuit: motor full-load current (I = P/(√3·V·η·pf)), lighting or socket circuit current.
• Cable rating ≥ MCB rating; minimum 1.5 mm² Cu for lighting, 2.5–4 mm² for power sockets.
• For motors, check voltage drop during starting (5–7 × FLC) and running.
• Total drop from transformer to the farthest load should be within about 5 %.

Factors on which selection depends

1. Load current (design current) and type of load.
2. Current-carrying capacity from the manufacturer's table.
3. Derating factors: ambient temperature, grouping, method of laying (air, duct, ground), depth and soil resistivity.
4. Permissible voltage drop and cable length.
5. Short-circuit current and fault clearing time.
6. Rating of the protective device (coordination Ib ≤ In ≤ Iz).
7. Conductor material (Cu/Al), insulation (PVC/XLPE) and armouring.
8. Future load growth and economics (cost of losses).

Here 5 kW and 10 kW are motor output ratings, so input power = output / η. A safety factor of 1.25 on full-load current is used for MCB selection, and the cable is chosen so that its rating is not less than the MCB rating (Ib ≤ In ≤ Iz).

Input power  P_in = P_out / η
Line current I    = P_in / (√3 × V × pf)

Motor (i): 5 kW, η = pf = 0.8

P_in = 5000 / 0.8 = 6250 W
I1   = 6250 / (√3 × 415 × 0.8) = 10.87 A
1.25 × I1 = 13.59 A → MCB 16 A
Cable: rating ≥ 16 A → 1.5 mm² (21 A)

Motor (ii): 10 kW, η = pf = 0.85

P_in = 10000 / 0.85 = 11764.71 W
I2   = 11764.71 / (√3 × 415 × 0.85) = 19.26 A
1.25 × I2 = 24.07 A → MCB 32 A
Cable: rating ≥ 32 A → 4 mm² (36 A)

MDB to SDB

I_total = I1 + I2 = 10.87 + 19.26 = 30.12 A
1.25 × 30.12 = 37.66 A → MCB 40 A
Cable: rating ≥ 40 A → 6 mm² (45 A)

(Arithmetic sum of currents is taken, which is slightly conservative since the power factors differ.)

Voltage drop check (10 m, ρ = 0.0178 Ω·mm²/m)

All drops are well below 2.5 %, so the current rating decides the size.

Answer

Use D-curve (motor) MCBs so that starting current does not cause nuisance tripping; overload protection of the motor is given by a thermal overload relay in the starter.

Efficiency is not given, so the 5 kW and 10 kW ratings are taken as the electrical input power of the motors (η = 1). MCBs are chosen with a 1.25 safety factor on full-load current, and cables so that their rating is not less than the MCB rating (Ib ≤ In ≤ Iz).

I = P / (√3 × V_L × pf)

Motor (a): 5 kW, pf 0.8

Ia = 5000 / (√3 × 415 × 0.8) = 8.70 A
1.25 × 8.70 = 10.87 A → MCB 16 A
Cable rating ≥ 16 A → 1.5 mm² (21 A)

Motor (b): 10 kW, pf 0.85

Ib = 10000 / (√3 × 415 × 0.85) = 16.37 A
1.25 × 16.37 = 20.46 A → MCB 32 A
Cable rating ≥ 32 A → 4 mm² (36 A)

MDB to SDB

I = 8.70 + 16.37 = 25.06 A
1.25 × 25.06 = 31.33 A → 32 A would be enough,
but for discrimination with the 32 A motor MCB
the next size is taken → MCB 40 A
Cable rating ≥ 40 A → 6 mm² (45 A)

Voltage drop check (10 m, Cu, ρ = 0.0178 Ω·mm²/m)

ΔV = √3 × I × ρ × L × cosφ / A
Motor (a): √3 × 8.70 × 0.0178 × 10 × 0.8 / 1.5
         = 1.43 V (0.34 %)
Motor (b): √3 × 16.37 × 0.0178 × 10 × 0.85 / 4
         = 1.07 V (0.26 %)
MDB–SDB : about 1.1 V (0.26 %)

All are far below 2.5 %, so the current rating governs.

Answer

If the 5 kW and 10 kW are output ratings, the currents rise by the efficiency factor; with a typical η of about 0.85 the same MCB and cable sizes still apply.

The lift motor must raise the unbalanced load (car + passengers − counterweight) at rated speed, allowing for the efficiency of the gear, ropes and motor.

Method

1. Rated load: number of persons × 68 kg (standard value per person, IS 14665 / EN 81).
2. Car weight: from the manufacturer (depends on size and finish).
3. Counterweight = car weight + 40–50 % of rated load, so the motor works against the smaller unbalanced weight in both directions.
4. Unbalanced mass (worst case, full car going up): m = (rated load + car weight) − counterweight
5. Power: P = m × g × v / η, where v = rated speed (m/s), η = overall efficiency (about 0.45–0.6 for geared, 0.7–0.85 for gearless drives).
6. Select the next higher standard motor rating; check starting duty (number of starts per hour, about 120–180), and use VVVF drive for smooth starting.

Example: 10-person lift in an industry office block

Rated load      = 10 × 68 = 680 kg
Car weight      = 700 kg (assumed)
Counterweight   = 700 + 0.45 × 680 = 1006 kg
Unbalanced mass = (680 + 700) − 1006 = 374 kg
Speed v         = 1.5 m/s
Overall η       = 0.5 (geared machine)

P = m × g × v / η
  = 374 × 9.81 × 1.5 / 0.5
  = 11 007 W ≈ 11.01 kW

Selected motor: 15 kW (next standard size above 11.01 kW, giving margin for starting and frequent duty), 3-phase, 400 V, with VVVF drive.

Other points in selection

• Motor duty is intermittent (S4/S5), so a lift-duty motor with high starting torque is used.
• Cable and MCCB are sized on the drive input current; standby (DG) supply is required for at least one lift for evacuation.
• For goods lifts in industry, rated load is the heaviest material load plus trolley, and speed is lower (0.5–1 m/s).

The cable must carry the total current from the supply end and keep the voltage drop at the farthest point within 2 %. Since the loads are distributed along the run, the drop is found by the current-moment method.

Assumptions: 10 HP is the motor output with η taken as 1 (not given); heaters are at unity pf; ρ (Cu) = 0.0178 Ω·mm²/m; drop in both go and return conductors (2-wire circuit), reactance neglected.

Load currents

10 kW heater (75 m): I = 10000 / 230       = 43.48 A
12 kW heater (35 m): I = 12000 / 230       = 52.17 A
10 HP motor  (70 m): P = 10 × 746 = 7460 W
                     I = 7460 / (230 × 0.8) = 40.54 A

Total current at supply end

I = 43.48 + 52.17 + 40.54 (0.8 − j0.6)
  = 128.08 − j24.33
|I| = 130.38 A   (arithmetic sum 136.20 A)

The largest cable in the table (50 mm², 90 A) cannot carry 130.38 A, so one cable from the table is not enough by current rating.

Voltage drop condition

Allowed drop = 2 % of 230 = 4.60 V
Σ I·cosφ·L = 52.17×35 + 40.54×0.8×70 + 43.48×75
           = 1826.09 + 2270.43 + 3260.87
           = 7357.39 A·m
ΔV = 2 ρ Σ(I cosφ L) / A ≤ 4.60
A ≥ 2 × 0.0178 × 7357.39 / 4.60 = 56.94 mm²

Selection

Answer: required area ≥ 56.94 mm² and current ≈ 130 A; using the given table, use two 50 mm² Cu cables in parallel per conductor (100 mm² total, 180 A, drop 2.62 V = 1.14 %). A single 70 mm² or 95 mm² Cu cable could be used if its rating (from a fuller table) exceeds 130 A.

The cable is selected by current rating and then checked for voltage drop; the larger size governs. Power factor is not given, so cosφ = 1 is assumed (largest resistive drop, conservative). ρ (Cu) = 0.0178 Ω·mm²/m.

By current rating

I = 60 A → from the table, 10 mm² carries 60 A (exactly), so 10 mm² is the minimum by current.

By voltage drop

Allowed drop = 2 % of 400 V = 8 V
ΔV = √3 × I × ρ × L / A
For 10 mm²: ΔV = √3 × 60 × 0.0178 × 200 / 10
              = 36.99 V (9.25 %)  → NOT OK
A_min = √3 × 60 × 0.0178 × 200 / 8
      = 46.25 mm² → next standard size 50 mm²
Check 50 mm²: ΔV = 369.97 / 50 = 7.40 V (1.85 %) → OK

Answer: use 3½ core, 50 mm² copper cable (rating 145 A, voltage drop 7.40 V = 1.85 % < 2 %). The long length makes voltage drop, not current, the deciding factor. (With pf 0.8 the minimum area would be 37 mm², still giving 50 mm².)

Each feeder is sized by current rating from the table and checked for voltage drop.

Assumptions: pf = 0.8 (not given), allowed drop 2.5 % of 400 V = 10 V, ρ (Cu) = 0.0178 Ω·mm²/m.

I  = P / (√3 × 400 × 0.8)
ΔV = √3 × I × ρ × L × cosφ / A

MDB to SDB1 (70 kW, 30 m)

I = 70000 / (√3 × 400 × 0.8) = 126.30 A
By current: 50 mm² (145 A)
ΔV = √3 × 126.30 × 0.0178 × 30 × 0.8 / 50
   = 1.87 V (0.47 %)  OK

MDB to SDB2 (90 kW, 120 m)

I = 90000 / (√3 × 400 × 0.8) = 162.38 A
By current: 70 mm² (175 A)
ΔV = √3 × 162.38 × 0.0178 × 120 × 0.8 / 70
   = 6.87 V (1.72 %)  OK

Answer: MDB–SDB1: 3½ core 50 mm² Cu; MDB–SDB2: 3½ core 70 mm² Cu.

Each feeder is sized by current rating from the given table and checked for voltage drop.

Assumptions: pf = 0.8 (not given), allowed drop 2.5 % of 400 V = 10 V, ρ (Cu) = 0.0178 Ω·mm²/m.

I  = P / (√3 × V_L × cosφ)
ΔV = √3 × I × ρ × L × cosφ / A

MDB to SDB-1 (60 kW, 80 m)

I = 60000 / (√3 × 400 × 0.8) = 108.25 A
By current: 35 mm² (120 A)
ΔV = √3 × 108.25 × 0.0178 × 80 × 0.8 / 35
   = 6.10 V (1.53 %)  OK

MDB to SDB-2 (75 kW, 125 m)

I = 75000 / (√3 × 400 × 0.8) = 135.32 A
By current: 50 mm² (145 A)
ΔV = √3 × 135.32 × 0.0178 × 125 × 0.8 / 50
   = 8.34 V (2.09 %)  OK (< 2.5 %)

Answer: MDB–SDB-1: 3½ core 35 mm² Cu; MDB–SDB-2: 3½ core 50 mm² Cu. If a stricter 2 % limit is used, SDB-2 needs 70 mm² (drop 5.96 V = 1.49 %).

The number of lifts is found from the population to be moved, the time allowed, and the round trip time of one car.

Interpretation: travel height is taken as 20 floors × 3.5 m = 70 m; one stop every 10 m on a trip; lift speed 135 m/min both ways.

Step 1: population

Persons = rental area / area per person
        = 20000 / 10 = 2000 persons

Step 2: round trip time (RTT) of one car

Travel height       H = 20 × 3.5 = 70 m
Running time (up + down) = 2 × 70 / 135 min
                         = 1.037 min = 62.22 s
Number of stops     = 70 / 10 = 7 stops
Stop time           = 7 × 10 s = 70 s
Ground floor time   = 30 s
RTT = 62.22 + 70 + 30 = 162.22 s

Step 3: handling capacity of one lift

Trips per hour   = 3600 / 162.22 = 22.19 trips
Persons per hour = 22.19 × 10 = 221.92 persons

Step 4: number of lifts

Persons to move in 60 min = 2000
N = 2000 / 221.92 = 9.01  → round up

Answer: 10 lifts (each 10-person, 135 m/min) are needed to empty the building in 60 minutes.

Power system vs lighting system

The two are kept on separate circuits (often separate DBs) so that a fault on a machine does not black out the lighting, and so that energy can be metered separately.

Factors in determining cable size

1. Load (design) current and type of load.
2. Current-carrying capacity of the cable from tables.
3. Derating for ambient temperature, grouping and laying method.
4. Permissible voltage drop and length of run.
5. Short-circuit current and fault duration.
6. Rating of the protective device (Ib ≤ In ≤ Iz).
7. Conductor material and insulation (Cu/Al, PVC/XLPE).
8. Future expansion and economy (energy loss cost).

A vertical rising main is a system of busbars (in a metal enclosure, called a rising main or busbar riser) or heavy cables running vertically through an electrical shaft of a multi-storey building, from the main LT panel at the bottom to the top floor, with tap-off points at each floor that feed the floor distribution boards.

  Top floor   ---[tap-off]--> floor DB
              |
  Floor 3     ---[tap-off]--> floor DB
              |  rising main
  Floor 2     ---[tap-off]--> floor DB
              |  (4 busbars R Y B N
  Floor 1     ---[tap-off]--> floor DB
              |   + earth)
  Ground      [MDB / LT panel]

Benefits

• One riser replaces many long individual cables from MDB to each floor, saving space, cable and cost.
• Easy and quick tap-off at any floor; new loads can be added without new cables from the basement.
• Low voltage drop and good current balance due to large busbars.
• Neat, safe, enclosed installation in the shaft; fire barriers at each slab.
• Easier fault finding and maintenance; each floor is isolated by its own tap-off switch-fuse or MCCB.

Uses

• High-rise apartments, commercial complexes, hotels, hospitals and office buildings.
• Multi-storey industrial buildings where each floor has its own DB.
• Separate rising mains may be used for normal and emergency (DG-backed) supply.

Types of cable used in industrial installations

• By voltage: LT 1.1 kV cables for 400/230 V; HT 11 kV XLPE cables for incoming supply.
• By insulation: PVC (70 °C) and XLPE (90 °C) cables; rubber/EPR flexible cables.
• By protection: armoured (direct burial, exposed routes) and unarmoured (in conduit, trays, trunking).
• By cores: single, 2, 3, 3½ and 4-core; multi-core control cables.
• By conductor: copper (motor and building circuits) and aluminium (large feeders).
• Special: flexible trailing cables, fire-resistant (FRLS) cables, screened instrumentation cables.

Cable size for the 15 hp motor

Assumption: allowed voltage drop 2.5 % of 400 V = 10 V; ρ (Cu) = 0.0178 Ω·mm²/m.

Output  = 15 × 746 = 11190 W
Input   = 11190 / 0.85 = 13164.71 W
I = 13164.71 / (√3 × 400 × 0.8) = 23.75 A
By current: 2.5 mm² (27 A)

Voltage drop check over 79 m:

ΔV = √3 × I × ρ × L × cosφ / A
2.5 mm²: √3 × 23.75 × 0.0178 × 79 × 0.8 / 2.5
       = 18.51 V (4.63 %)  NOT OK
4 mm²  : 11.57 V (2.89 %)  NOT OK
6 mm²  : 7.71 V (1.93 %)   OK
A_min = √3 × 23.75 × 0.0178 × 79 × 0.8 / 10
      = 4.63 mm² → 6 mm²

Answer: use 3½ core (unarmoured) 6 mm² copper cable (rating 45 A, full-load current 23.75 A, drop 7.71 V = 1.93 %). If a 5 % drop were allowed, 2.5 mm² would be acceptable by drop but leaves little margin for starting current.

Planning an industrial electrical installation means arranging supply, distribution, protection and wiring so that power reaches every machine safely, reliably and economically, with room for growth.

Fundamental considerations

1. Load assessment: list all machines, lighting, HVAC and utilities with ratings; estimate maximum demand using utilization (Ku) and simultaneity (Ks) factors.
2. Supply and substation: supply voltage (11 kV/400 V), transformer size and number, substation location at the load centre.
3. Distribution scheme: radial or ring; MDB and SDB locations near load groups; bus trunking for machine shops; separate lighting and power circuits.
4. Safety and protection: correct MCCB/MCB/fuse ratings, earth-fault protection (RCCB/ELCB), proper earthing, coordination and discrimination; compliance with codes (IEC/IS, NEA rules).
5. Cable selection and routing: size by current, voltage drop and short circuit; routes in trays/trenches away from heat and traffic.
6. Reliability and continuity: standby DG set, UPS for critical loads, duplicate feeders for essential processes.
7. Power quality and efficiency: power factor correction, harmonic control, efficient motors and lighting, metering for energy audit.
8. Environment: dusty, wet, corrosive or hazardous areas need suitable enclosures (IP rating) and flameproof equipment.
9. Flexibility and expansion: spare capacity (20–25 %) and spare ways in DBs.
10. Maintenance and economy: easy access to equipment, labelling, minimum initial plus running cost.

The cross-section is found from the permissible voltage drop of the go-and-return (2-wire) circuit.

Constants: resistivity of aluminium ρ = 0.0283 Ω·mm²/m (2.83 × 10⁻⁸ Ω·m); the whole I·R drop is taken (conservative, reactance neglected).

Load current I = P / (V × cosφ)
               = 25000 / (220 × 0.7) = 162.34 A
Allowed drop   = 3 % of 220 = 6.60 V
Loop length    = 2 × 40 = 80 m
ΔV = I × R = I × ρ × 2L / A ≤ 6.60
A  = ρ × 2L × I / ΔV
   = 0.0283 × 80 × 162.34 / 6.60
   = 55.69 mm²

Next standard size: 70 mm².

Check: ΔV = 0.0283 × 80 × 162.34 / 70
          = 5.25 V (2.39 %) < 6.60 V  OK

Answer: minimum area = 55.69 mm²; use a 70 mm² aluminium cable (drop 5.25 V). If only the in-phase component of drop (I·R·cosφ) is considered, A = 38.98 mm² and a 50 mm² cable would just satisfy the drop limit; the current rating of the chosen size must also be checked from the manufacturer's table for 162 A.

The size (capacity and speed) and number of lifts for an office building are chosen so that the peak traffic (morning arrival or evening departure) is handled within an acceptable time and waiting interval.

Steps

1. Population: net rentable area ÷ area per person (about 8–12 m²/person in offices).
2. Peak demand: persons to be moved in the critical 5 minutes (handling capacity 12–15 % of population in 5 min for offices), or the whole population in a given time (e.g. emptying in 60 min).
3. Choose car capacity and speed: capacity 8–20 persons (68 kg each); speed rises with building height (about 1 m/s up to 5 floors, 1.5–2.5 m/s for 10–20 floors, higher for taller).
4. Round trip time (RTT) = running time up and down (2H/v) + door and stop time (number of probable stops × time per stop) + loading time at ground floor.
5. Handling capacity per car = car capacity × (time period / RTT).
6. Number of lifts = required persons per period ÷ capacity of one car; round up.
7. Interval (waiting time) = RTT / number of lifts; should be about 25–40 s for offices.
8. Motor size of each lift from P = (unbalanced mass) × g × v / η.
9. Provide at least one fire/service lift, and connect lifts to standby supply.

Example in brief: 2000 persons, 10-person cars, RTT = 162 s → each car moves 10 × 3600/162 ≈ 222 persons/h; to move 2000 in an hour, 2000/222 = 9.01 → 10 lifts.

MCCBs (moulded case circuit breakers, about 16–1600 A, adjustable trips, high breaking capacity) are used as MDB incomers and feeders; MCBs (up to 63–125 A, fixed trips, 6–10 kA) are used in SDBs for final circuits. Both are selected by rated current, breaking capacity, number of poles and tripping characteristic.

Procedure

1. Load current of each final circuit (Ib): I = P/(√3·V·η·pf) for 3-phase, P/(V·pf) for 1-phase.
2. MCB for each outgoing circuit: In ≥ 1.25 × Ib, next standard rating (6, 10, 16, 20, 25, 32, 40, 50, 63 A). Curve: B for lighting, C for mixed/socket, D for motors and transformers.
3. SDB incomer: total SDB current × simultaneity factor Ks, × 1.25 → MCB (up to 63 A) or MCCB.
4. MDB outgoing feeders: MCCB ≥ SDB incomer rating (one size higher for discrimination).
5. MDB incomer: sum of feeders × Ks of MDB × 1.25 → MCCB/ACB; it should match the transformer LT full-load current (I = kVA·1000/(√3·400)).
6. Breaking capacity: Icu ≥ prospective fault current at that point (e.g. near a 250 kVA, 4.5 % impedance transformer, Isc ≈ 361/0.045 ≈ 8 kA → 10 kA MCB/MCCB at SDB, 25–36 kA MCCB at MDB).
7. Poles: SP for single-phase, TP for motors, TPN/4P for 3-phase 4-wire incomers.
8. Coordination: Ib ≤ In ≤ Iz (cable rating); upstream larger than downstream for selectivity.
9. Earth leakage: RCCB (30 mA) for socket circuits, 100–300 mA at SDB incomer.

Example (400 V, 3-phase)

• SDB-1: three motors, 7.5 kW each, η = 0.88, pf = 0.85.
• SDB-2: 15 kW lighting and sockets, pf 0.9, balanced.

Motor Ib = 7500 / (√3 × 400 × 0.88 × 0.85) = 14.47 A
Motor MCB: 1.25 × 14.47 = 18.09 → 20 A TP, D curve

SDB-1: 3 × 14.47 × 0.9 = 39.08 A
       1.25 × 39.08 = 48.84 → 50 A TP MCB incomer
SDB-2: 15000 / (√3 × 400 × 0.9) = 24.06 A
       1.25 × 24.06 = 30.07 → 32 A TPN MCB incomer

MDB feeders: 63 A MCCB (to SDB-1), 40 A MCCB (to SDB-2)
MDB demand  = (39.08 + 24.06) × 0.9 = 56.82 A
MDB incomer = 1.25 × 56.82 = 71.02 → 80 A TPN MCCB, 25 kA

Electrical installation in a factory is the complete system that takes power from the supply authority and delivers it safely to every machine, lamp and socket, with proper protection, earthing and control.

Main parts

11 kV → substation (TR) → MDB → SDBs → machines/lights
                           |
                 DG set + ATS, APFC panel

1. Substation: 11/0.4 kV transformer with LA, isolator, HT fuse/VCB, placed near the load centre.
2. Main distribution board (MDB): ACB/MCCB incomer, busbars, MCCB feeders, metering, APFC panel and DG changeover.
3. Sub-distribution boards (SDBs): one per shop or bay; MCB/MCCB outgoing to machines, lighting and sockets.
4. Cabling: armoured XLPE cables in trenches or on trays for feeders; conduit or trunking for final circuits; bus trunking for machine rows.
5. Motor circuits: MCB/MPCB, contactor and overload relay (DOL, star-delta or VFD), isolator near machine.
6. Lighting: high-bay LED/HID luminaires designed for 200–500 lux, on separate circuits; emergency lighting.
7. Earthing and safety: equipment and neutral earthing, RCCBs, lightning protection, flameproof fittings in hazardous zones.
8. Standby supply: DG set and UPS for critical loads and fire pumps.

Design steps

1. Prepare the plant layout and load list.
2. Estimate maximum demand (Ku, Ks) and size the transformer.
3. Locate substation, MDB and SDBs near load centres.
4. Size cables (current, voltage drop, short circuit) and protective devices.
5. Design lighting, earthing and power factor correction.
6. Prepare SLD, layout drawings, BOQ and cost estimate; test (insulation, earth resistance, polarity) before commissioning.

The installation must follow IEC/IS standards and NEA regulations, and keep 20–25 % spare capacity for expansion.

Light and power sub-circuits

A sub-circuit is a branch circuit taken from a distribution board (DB) through its own fuse/MCB to feed a group of points. In building wiring the points are split into two separate kinds of sub-circuit:

• Light sub-circuit: feeds lamps, fans and 6 A (5 A) socket points. As per IS 732 / NEA practice it carries at most 10 points or 800 W, is wired in 1.5 mm² Cu and is protected by a 6 A (or 10 A) MCB.
• Power sub-circuit: feeds 16 A (15 A) power sockets for heaters, irons, AC, machines, etc. It carries at most 2 power sockets (about 3000 W), is wired in 2.5–4 mm² Cu and protected by a 16–20 A MCB.
• The two are kept on separate circuits (often separate DBs) so that a fault or overload on a power point does not plunge the area into darkness, and so that each can be sized for its own current.

Cable size from MDB to SDB1 and SDB2

Assumptions: power factor 0.8 lagging; Cu resistivity 0.0178 Ω·mm²/m; permissible voltage drop 2.5 % of 400 V (= 10 V) on a sub-main.

Current: I = P / (√3 × V_L × cos φ)

SDB1: I1 = 50000 / (√3 × 400 × 0.8) = 90.21 A
SDB2: I2 = 60000 / (√3 × 400 × 0.8) = 108.25 A

From the table (3.5-core Cu, laid in ground):

• SDB1: 90.21 A → 25 mm² (rating 99 A)
• SDB2: 108.25 A → 35 mm² (rating 120 A)

Voltage drop check: ΔV = √3 × I × R × cos φ, with R = ρL/A (reactance neglected).

SDB1: R  = 0.0178 × 60 / 25  = 0.0427 Ω
      ΔV = √3 × 90.21 × 0.0427 × 0.8 = 5.34 V
         = 5.34/400 = 1.33 %   < 2.5 %  OK
SDB2: R  = 0.0178 × 100 / 35 = 0.0509 Ω
      ΔV = √3 × 108.25 × 0.0509 × 0.8 = 7.63 V
         = 7.63/400 = 1.91 %   < 2.5 %  OK

Answer: MDB → SDB1: 3.5-core 25 mm² Cu cable; MDB → SDB2: 3.5-core 35 mm² Cu cable.

Note: if a 25 % margin for future load is added (1.25 × I = 112.76 A and 135.32 A), the sizes become 35 mm² and 50 mm² respectively.

The power and lighting system of an industry must be safe, reliable, economical and flexible. The main factors to consider are:

1. Load assessment: list of machines, their kW/kVA, power factor, duty cycle and starting current; lighting load from the required lux levels. Demand factor and diversity factor are applied to get the maximum demand, which fixes transformer, cable and switchgear sizes.
2. Supply system and voltage: HT or LT supply, number of phases, 400/230 V or 11 kV distribution, and whether an in-plant substation is needed. Large motors may need 11 kV or 3.3 kV supply.
3. Safety: correct earthing (equipment and system), protection against shock (RCCB/ELCB), short circuit and overload protection, insulation suited to the environment, and compliance with the Electricity Regulations and IS/IEC codes.
4. Reliability and continuity: critical processes need ring or duplicate feeders, standby DG sets, UPS and emergency lighting so that a single fault does not stop production.
5. Voltage regulation: cable sizes chosen so that the voltage drop stays within limits (about 5 % for power, 3 % for lighting), because low voltage reduces motor torque and lamp output.
6. Environment of the area: dust, moisture, corrosive fumes, high temperature or explosive gases decide the type of wiring (conduit, armoured cable), luminaire (dust-proof, flameproof) and enclosure (IP rating).
7. Illumination requirements: required lux for each task (IS 3646), uniformity, glare control, colour rendering, and avoiding stroboscopic effect near rotating machines.
8. Energy efficiency and power factor: efficient motors and LED lamps, capacitor banks for pf correction, and good control (switching, sensors) to cut running cost.
9. Flexibility and future expansion: spare ways in DBs, spare capacity in cables and transformer (about 20–25 %), and bus-duct systems where machine layout may change.
10. Economy: balance of initial cost (cable, switchgear) against running cost (losses, maintenance) over the life of the plant.
11. Ease of maintenance and operation: accessible DBs, clear labelling, isolation of each machine, and standard equipment ratings to reduce spares.

An electric drive is a system that uses an electric motor as the prime mover to control the speed, torque and direction of a mechanical load. It consists of the power source, a power modulator/controller (starter, converter), the motor, the transmission (shaft, gears, belt) and the driven load, with sensing and control units.

Supply → Power modulator → Motor → Transmission → Load
              ↑                          |
          Control unit  ←── sensors ──────┘

Factors affecting the selection of motor

1. Nature of supply: AC or DC, single or three phase, voltage and frequency available. Most industries use 3-phase AC induction motors because the supply is AC.
2. Load characteristics:
   • Speed–torque characteristic of the load (constant torque like conveyors, torque ∝ speed² like fans and pumps, constant power like winders).
   • Starting torque required (cranes and compressors need high starting torque; fans need little).
   • Speed range and need for speed control (wide control → DC motor or VFD-fed induction motor).
3. Electrical characteristics of the motor: starting current and its effect on supply voltage, running characteristics, speed regulation, braking (plugging, rheostatic, regenerative), power factor and efficiency.
4. Mechanical considerations: type of enclosure (open, drip-proof, totally enclosed fan-cooled, flameproof) suited to dust, moisture or explosive atmosphere; type of bearings; mounting (foot, flange); transmission (direct, belt, gear); noise and vibration.
5. Size and rating: rating chosen from the duty cycle: continuous, short-time or intermittent periodic duty. Under-rating causes overheating; over-rating gives poor pf and efficiency.
6. Service conditions: ambient temperature, altitude, humidity, frequency of starting and reversal.
7. Cost: initial cost of motor and control gear, running cost (efficiency, pf), and maintenance cost. Squirrel-cage motors are cheapest and almost maintenance-free.
8. Availability and standardisation: standard frame sizes and ratings make spares and replacement easy.

Example: a lift needs high starting torque, frequent starts, reversal and smooth speed control, so a VFD-fed induction motor (or DC motor) is chosen; a ceiling fan works well with a simple single-phase capacitor motor.

Assumptions: the strip radiates from both faces (edges neglected); power per phase is supplied at phase voltage of the star connection; Stefan's law for heat dissipation:

H = 5.72 × 10⁴ × e × k × [(T₁/1000)⁴ − (T₂/1000)⁴] W/m²

where e = emissivity, k = radiating efficiency, T₁ and T₂ are absolute temperatures of strip and charge.

Electrical data per phase

Power per phase P   = 30/3       = 10 kW
Phase voltage Vph   = 400/√3     = 230.94 V
Resistance per phase R = Vph²/P  = 230.94²/10000 = 5.333 Ω

Heat dissipated per m² of strip surface

T₁ = 1100 + 273 = 1373 K,  T₂ = 700 + 273 = 973 K
(1.373)⁴ − (0.973)⁴ = 3.5537 − 0.8963 = 2.6574
H = 5.72 × 10⁴ × 0.9 × 0.6 × 2.6574
  = 82,082 W/m²

Equations for width w and length l (thickness t = 0.25 cm)

Resistance: R = ρl/(w t) → l/w = R t/ρ

l/w = 5.333 × 0.0025 / 1.03×10⁻⁶ = 12,945        ...(1)

Heat balance (two faces): P = H × 2 w l

w × l = 10000 / (2 × 82082) = 0.06091 m²           ...(2)

Multiplying and dividing (1) and (2):

w² = 0.06091 / 12945  → w = 2.17 × 10⁻³ m = 2.17 mm
l  = 12945 × w        → l = 28.08 m

Answer: width ≈ 2.17 mm and length ≈ 28.08 m of strip per phase (with t = 0.25 cm as given).

Note: a width smaller than the thickness is not a practical strip, which suggests the intended thickness is 0.25 mm (0.025 cm). With t = 0.25 mm: l/w = 1294.5, w = √(0.06091/1294.5) = 6.86 mm and l = 8.88 m per phase.

Method: the wire must (a) have the resistance that draws 15 kW at 220 V and (b) have enough surface to radiate 15 kW at the given temperature. Heat dissipated per unit surface (Stefan's law):

H = 5.72 × 10⁴ × e × k × [(T₁/1000)⁴ − (T₂/1000)⁴] W/m²

Assumption: single-phase supply, one continuous wire element.

Step 1: Resistance

R = V²/P = 220²/15000 = 3.2267 Ω

Step 2: Heat dissipation per m²

T₁ = 1000 + 273 = 1273 K,  T₂ = 600 + 273 = 873 K
(1.273)⁴ − (0.873)⁴ = 2.6261 − 0.5808 = 2.0453
H = 5.72 × 10⁴ × 0.9 × 0.6 × 2.0453 = 63,174 W/m²

Step 3: Two equations in d and l

R = ρl/(πd²/4)  → l/d² = πR/(4ρ)
              = π×3.2267/(4×1.016×10⁻⁶) = 2.4943×10⁶  ...(1)
P = H × πdl     → l × d = 15000/(π × 63174)
              = 0.075579 m²                 ...(2)

Step 4: Solve

Dividing (2) by (1): d³ = 0.075579 / (2.4943 × 10⁶)

d³ = 3.030 × 10⁻⁸ m³
d  = 3.118 × 10⁻³ m = 3.12 mm
l  = 0.075579 / 0.003118 = 24.24 m

Check: ρl/(πd²/4) = 1.016×10⁻⁶ × 24.24/(7.634×10⁻⁶) = 3.227 Ω ✓

Answer: diameter d ≈ 3.12 mm, length l ≈ 24.24 m of nichrome wire.

Vapour compression refrigeration is the cycle used in refrigerators, water coolers, cold stores and air conditioners. A refrigerant (R-134a, R-410A, R-600a, ammonia in industry) absorbs heat when it evaporates at low pressure inside the cold space and rejects that heat when it condenses at high pressure outside. An electric motor-driven compressor supplies the work needed to move heat from cold to hot.

Sketch of the cycle

        Heat out (to room/air)
             ↑  ↑  ↑
      +----[ CONDENSER ]----+
      |   high P, high T    |
   (2)|                     |(3) liquid
      |                     ↓
 [COMPRESSOR]        [EXPANSION VALVE]
   ↑  driven by             |
   |  motor M               | (4) low P,
(1)|  low P vapour          |  cold mix
      +----[ EVAPORATOR ]---+
             ↓  ↓  ↓
        Heat in (from food / room)

Working (four processes)

1. Compression (1→2): the compressor draws low-pressure, low-temperature vapour from the evaporator and compresses it. Its pressure and temperature rise well above room temperature (superheated vapour).
2. Condensation (2→3): hot vapour passes through the condenser coils (fins at the back of a fridge, or outdoor unit of an AC). Heat flows to the surrounding air, and the refrigerant condenses into high-pressure liquid.
3. Expansion (3→4): the liquid passes through the expansion valve or capillary tube. Its pressure drops suddenly; part of it flashes to vapour and its temperature falls far below the space temperature.
4. Evaporation (4→1): the cold liquid–vapour mixture flows through the evaporator inside the cabinet/room. It takes latent heat from the food or air and boils into vapour. This heat absorption is the cooling (refrigerating) effect. The vapour returns to the compressor and the cycle repeats.

Electrical circuit of a domestic refrigerator

 L ──[Thermostat]──[OLP]──┬──[Relay coil]── Run winding ──┐
                          │                               │
                          └─[Relay contact]─ Start winding┤
 N ───────────────────────────────────────────────────────┘
 Door switch ── Lamp (in parallel across L–N)

• Thermostat switches the compressor motor on when the cabinet warms and off when the set temperature is reached.
• Starting relay (current relay or PTC) connects the start winding for a moment at starting; overload protector (OLP) trips the motor on overcurrent or overheating.
• The compressor motor is a hermetically sealed single-phase induction motor.

Coefficient of performance: COP = refrigerating effect / work input = (h₁ − h₄)/(h₂ − h₁), typically 2–4, i.e. 1 kW of electric input removes 2–4 kW of heat.

Dielectric heating is the heating of an electrically non-conducting (insulating) material by placing it between two metal plates (electrodes) connected to a high-frequency AC voltage (about 10–50 MHz, a few kV). The alternating field makes the molecules of the dielectric flip and rub repeatedly (dielectric hysteresis loss), so heat is generated uniformly throughout the material itself.

   HF oscillator (10–50 MHz)
      |                |
   ===+=== electrode   |
   [  material (charge) ]   ← heat produced inside
   ===+=== electrode   |
      |________________|

Heat produced: P = 2π f C V² tan δ ≈ 2π f C V² cos φ, so heat rises with frequency f and voltage V. The material behaves like a lossy capacitor (loss angle δ).

Difference from induction heating

Applications of dielectric heating

• Drying of timber, paper, textiles and tobacco.
• Gluing of plywood and laminated wood (heat cures glue uniformly inside).
• Welding of PVC and plastic sheets (raincoats, bags).
• Pre-heating of plastic moulding powder and vulcanising rubber.
• Food processing: baking, dehydration, sterilisation; the microwave oven works on the same principle at 2.45 GHz.
• Medical diathermy (heating body tissue).

Method: the wire must give the right resistance and also radiate 4.5 kW from its surface at the given temperatures. Stefan's law gives heat dissipated per m²:

H = 5.72 × 10⁴ × e × k × [(T₁/1000)⁴ − (T₂/1000)⁴] W/m²

Data: P = 4.5 kW, V = 200 V, ρ = 42.5 μΩ-cm = 42.5 × 10⁻⁸ Ω-m, k = 1, e = 0.9.

Step 1: Resistance

R = V²/P = 200²/4500 = 8.889 Ω

Step 2: Heat dissipation per m²

T₁ = 1000 + 273 = 1273 K,  T₂ = 500 + 273 = 773 K
(1.273)⁴ − (0.773)⁴ = 2.6261 − 0.3570 = 2.2691
H = 5.72 × 10⁴ × 0.9 × 1 × 2.2691 = 116,812 W/m²

Step 3: Equations

R = ρl/(πd²/4) → l/d² = πR/(4ρ)
             = π×8.889/(4×42.5×10⁻⁸) = 1.6427×10⁷  ...(1)
P = H × πdl    → l d = 4500/(π × 116812)
             = 0.012262 m²                ...(2)

Step 4: Solve

d³ = 0.012262 / 1.6427×10⁷ = 7.465 × 10⁻¹⁰ m³
d  = 9.07 × 10⁻⁴ m = 0.907 mm
l  = 0.012262 / 9.071×10⁻⁴ = 13.52 m

Check: ρl/(πd²/4) = 42.5×10⁻⁸ × 13.52 / (6.463×10⁻⁷) = 8.89 Ω ✓

Answer: diameter ≈ 0.907 mm and length ≈ 13.52 m of nichrome wire.

Why electric drive is preferred over mechanical drive

A mechanical drive uses a steam engine, diesel engine or water wheel and a system of shafts, belts and gears; an electric drive uses an electric motor as the prime mover. Electric drives are preferred because:

1. Easy and wide speed control: speed, torque and direction can be changed smoothly using starters, VFDs or converters.
2. Simple starting and stopping: a push button or remote signal starts the motor at once; no warm-up time as in engines.
3. Clean and quiet: no smoke, exhaust or fuel storage; low noise.
4. High efficiency and low running cost: motors are 85–95 % efficient; losses in long line shafts and belts are avoided.
5. Compact and flexible layout: the motor can be mounted directly on the machine, so machines can be placed anywhere in the shop.
6. Wide range of ratings and characteristics: from fractional watts to many MW, with motors matching any load characteristic.
7. Electric braking: plugging, dynamic and regenerative braking are available; regeneration returns energy to the supply.
8. Automation and remote control: easy to interlock, automate and control with PLCs and sensors.
9. Long life and low maintenance: squirrel-cage motors need little attention.
10. Overload capacity and protection: short-time overloads are allowed, and overload/short-circuit relays protect the motor.

The only limits are dependence on a reliable supply and loss of drive during power failure.

Individual drive versus group drive

• Group drive: one large motor drives a line shaft, and several machines take power from it by belts and pulleys.
• Individual drive: each machine (or each operation of a machine) has its own motor.

Justification: although group drive has lower first cost, individual drive gives better continuity of production, higher efficiency, independent speed control, safer working and easy expansion. Over the life of the plant, these savings outweigh the higher initial cost, so individual drive is the better choice for modern industries.

An electric wiring system is the method of running insulated conductors from the distribution board to the points (lights, fans, sockets) in a building. The common types are:

1. Cleat wiring: PVC or VIR cables are held on porcelain/plastic cleats screwed to walls and ceilings. Cheap and quick; used for temporary installations (exhibitions, construction sites). Untidy and unprotected.
2. Casing and capping (wooden) wiring: cables run inside a wooden casing with grooves, covered by a wooden capping. Older method; now replaced by PVC casing (PVC channel) wiring, which is cheaper and resists moisture better.
3. Batten (TRS/CTS) wiring: tough-rubber-sheathed or PVC-sheathed cables are fixed on a straight teak-wood batten with brass/aluminium link clips. Neat and cheap; used in residences and offices for indoor dry places.
4. Lead-sheathed (metal-sheathed) wiring: cables with an outer lead/aluminium sheath fixed on battens with link clips; the sheath is earthed. Good mechanical and moisture protection but costly and the sheath can corrode.
5. Conduit wiring: PVC cables are drawn through steel or PVC conduit pipes.
   • Surface conduit: pipes fixed on the wall surface with saddles (factories, workshops).
   • Concealed conduit: pipes buried in walls/slabs during construction (modern buildings). Best appearance and protection; rewiring is easy by pulling wires through the pipe.

Comparison

Choice: concealed PVC conduit wiring for modern residential and commercial buildings; surface steel conduit or armoured cable in factories with risk of mechanical damage and fire; cleat wiring only for temporary work.

Given: P_out = 15 hp, V = 400 V (3-phase), η = 85 %, pf = 0.8, length SDB → motor = 79 m.

Assumptions: 1 hp = 746 W; cable chosen for 125 % of motor full-load current (motor circuit rule of IS 732/NEC); Cu resistivity 0.0178 Ω·mm²/m; permissible voltage drop 5 % (20 V). Table ratings are used for the unarmoured multicore Cu cable.

Step 1: Full-load current

Output P_out = 15 × 746        = 11,190 W
Input  P_in  = 11190 / 0.85    = 13,164.7 W
I_FL = P_in / (√3 × V × pf)
     = 13164.7 / (√3 × 400 × 0.8)
     = 23.75 A

Step 2: Design current and cable from table

Design current = 1.25 × 23.75 = 29.69 A

From the table, the next rating above 29.69 A is 31 A → 4 mm².

Step 3: Voltage drop check

R  = ρL/A = 0.0178 × 79 / 4 = 0.3515 Ω
ΔV = √3 × I × R × cos φ
   = √3 × 23.75 × 0.3515 × 0.8 = 11.57 V
%ΔV = 11.57/400 × 100 = 2.89 %  (< 5 %, OK)

Answer: 4 mm² Cu unarmoured (3.5/4-core) cable — full-load current 23.75 A, voltage drop 11.57 V (2.89 %). Protect the motor with a 32 A TP MCB (D-curve) or MPCB and a thermal overload relay set at about 24 A.

Electric building (space) heating converts electrical energy into heat to keep rooms at a comfortable temperature. It is clean, needs no fuel storage or chimney, is easy to control with thermostats, and has nearly 100 % conversion efficiency at the point of use (heat pumps give even more heat than the electricity used). The main methods are:

1. Direct resistance heaters

Current through a resistance element produces heat (I²R).

• Convection heaters (convectors): an element at the bottom heats air, which rises and circulates through the room by natural convection. Oil-filled radiators work the same way using oil as a heat store.
• Fan heaters / forced-air heaters: a fan blows air over the hot element; heats a room quickly.
• Radiant heaters: a red-hot element with a reflector (bar heater, quartz/infrared lamp) radiates heat directly onto people and objects; good for large halls and outdoor areas.

2. Embedded (panel) heating

Heating cables or panels are buried in the floor, ceiling or walls. The large warm surface heats the room by low-temperature radiation and convection. Underfloor heating gives uniform, comfortable heat and no visible equipment.

3. Storage heating

Heaters with a block of high-heat-capacity material (bricks, ceramic) are charged at night with cheap off-peak electricity and release heat during the day, controlled by dampers. Flattens the utility load curve.

4. Central electric heating

An electric boiler (resistance or electrode type) heats water or air, which is circulated through radiators or ducts to all rooms. Suits large buildings.

5. Heat pump heating

A vapour-compression heat pump (reverse-cycle air conditioner) extracts heat from outside air or ground and delivers it indoors. It gives about 3 units of heat per unit of electricity (COP ≈ 3), so it is the most energy-efficient electric method, and the same unit cools in summer.

6. Infrared heating

Infrared lamps or panels heat people and surfaces directly, not the air; used in factory work bays, warehouses with high ceilings and loading docks.

Control: thermostats, timers and zoning switch the heaters on only where and when needed, which saves energy.