ElectricityDesign and Technical Drawing III (year 6)

The industrial electrical project

In a house a calculation error is paid for with a warm cable; in a plant it is paid for with a production stoppage, a burnt transformer or a switchboard that blows up when a breaker is closed. The industrial project follows the same logic as a house —survey, calculate, verify, draw— with three differences: the powers are hundreds of times larger, the short circuit stops being a hypothesis, and the protective devices have to work as a team. This page follows a single example, a metalworking shop fed by a 315 kVA transformer, from the load schedule to the capacitor bank.

Maximum demand Transformer Voltage drop Short circuit Selectivity cos φ

01Power demand and the transformer

Every project starts with the question “how much power will this plant ask for?”. The answer is not the sum of the motor nameplates: not all equipment is installed at full load, and not all of it runs at the same time. That is why two powers are distinguished:

Installed power

The sum of the electrical power of everything connected. It tells what was bought, not how to size the supply.

Maximum demand

The greatest power the plant draws at some real moment of its operation, usually averaged over 15 minutes. It is the one that sizes the transformer, busbars and feeders.

Two factors take you from one to the other. The utilization factor f_u tells what fraction of its power each load really uses when it runs (an overhead crane moves little; a furnace runs almost all the time). The diversity (simultaneity) factor f_s tells what fraction of the total runs at once: the probability that all ten machines start in the same minute is low. The values come from experience with the type of industry, from manufacturers’ guides and, in an existing plant, from measuring them (see lab 1). Active and reactive power are calculated separately:

P_{\max} = f_{s} \cdot \sum (P_{i} \cdot f_{u,i}) Q_{\max} = f_{s} \cdot \sum (Q_{i} \cdot f_{u,i})

Pᵢ: installed electrical active power of each load · fᵤ: utilization factor of each load · fₛ: diversity (simultaneity) factor of the group · Qᵢ = Pᵢ · tan φᵢ

The example that runs through the page

A metalworking shop is supplied at medium voltage (13.2 kV) from the utility grid and has its own substation. It has a production hall 120 m from the substation, a furnace and pump area next to it and an office building. It is organized in a main distribution board (MDB) and three sub-distribution boards (SDB): SDB1 (production), SDB2 (furnaces, pumps and extraction) and SDB3 (lighting and offices). The table is the load survey, already solved. The installed power of a motor is what it draws from the grid, $P = P_{n} / η$ , and its reactive power comes from its working cos φ.

Load	Pn (shaft)	P installed	cos φ	f_u	P dem. (kW)	Q dem. (kvar)	Board
Screw compressors	2 × 37 kW	78.7 kW	0.84	0.75	59.0	38.1	TS1
Machine tools	10 × 7.5 kW	85.2 kW	0.74	0.60	51.1	46.4	TS1
Overhead crane	1 × 22 kW	25.0 kW	0.70	0.30	7.5	7.7	TS1
Process pumps	3 × 15 kW	49.5 kW	0.80	0.70	34.6	25.9	TS2
Extraction and ventilation	4 × 5.5 kW	25.3 kW	0.76	0.80	20.2	17.3	TS2
Resistance furnaces	2 × 30 kW	60.0 kW	1.00	0.70	42.0	0.0	TS2
Lighting	—	20.0 kW	0.92	1.00	20.0	8.5	TS3
Sockets and offices	—	24.0 kW	0.85	0.50	12.0	7.4	TS3
Sum of the demands		367.7 kW			246.4	151.3
Maximum demand of the plant			0.85	f_s = 0.85	209.4	128.6

Each row was calculated like this: installed power is $n \cdot P_{n} / η$ (compressors: 2 · 37 / 0.94 = 78.7 kW), active demand is that power times f_u, and reactive demand is the active one times tan φ. With the totals, f_s = 0.85 is applied:

S_{\max} = \sqrt{{P_{\max}}^{2} + {Q_{\max}}^{2}} = \sqrt{{209.4}^{2} + {128.6}^{2}} kVA = 245.8 kVA

\cos φ = \frac{P_{\max}}{S_{\max}} = \frac{209.4}{245.8} = 0.85

Add kW and kvar, not kVA

The apparent power of each load has its own angle, so it cannot be added arithmetically. Active powers (kW) and reactive powers (kvar) are added separately and only then is the apparent power found with Pythagoras. Adding kVA gives a number larger than the real one and the transformer ends up oversized (see Power in alternating current).

Example · Choosing the transformer

Maximum demand is 245.8 kVA with cos φ = 0.85. A transformer is not chosen exactly for that power: a reserve of 25 % is left for expansion and the next standard size is adopted (160, 200, 250, 315, 400, 500, 630 kVA).

S_{req} = 1.25 \cdot 245.8 kVA = 307.2 kVA \to S_{n} = 315 kVA

The rated current on the low-voltage side and the loading are:

I_{n} = \frac{S_{n}}{\sqrt{3} \cdot U_{n}}

rated current of the transformer secondary

I_{n} = \frac{315 000 VA}{\sqrt{3} \cdot 400 V} = 454.7 A \frac{S_{\max}}{S_{n}} = \frac{245.8}{315} = 78 %

At peak the transformer runs at 78 % of its rating: between 60 and 80 % loading it has good efficiency and still leaves margin. A 315 kVA, 13.2/0.4 kV, Dyn11 transformer with short-circuit voltage u_k = 4 % is adopted. Cross-check: at 315 kVA and 245.8 kVA of demand, 69 kVA are left over, 22 % of the rating, which covers the planned reserve.

The transformer, its connection and its cooling are studied in Transformers; the complete transformer station (MV cubicles, earthing, protections) in Substations and distribution lines, and its regulation (AEA 95401) in Regulations for electrical installations.

02The single-line diagram, lines, circuits and the environment

The single-line diagram is the map of the installation: one line stands for the three phases and shows, from source to load, which transformer, breaker, busbar and cable is in each stretch. It is the drawing that is calculated first and the one consulted by anyone who operates or repairs the plant. Figure 1 is the diagram of the example; its four states show, on the same drawing, the design currents, the protections, the conductors and the short-circuit currents.

Figure 1. Single-line diagram of the metalworking shop, animated. One drawing and four readings: the current in each stretch, the breaker that protects it, the cable that connects it and the fault current at each busbar. The numbers are obtained in sections 01 to 06.

Lines and circuits

In an industrial project a line (or feeder) is the conductor that carries power from one board to another, and a circuit the one that leaves a board towards a load: a motor, a bank of luminaires, a group of sockets. The criterion for splitting the plant into sub-boards is always the same:

By zone, so that a feeder does not cross the whole plant and the voltage drop stays reasonable (SDB1 is in the hall, next to the machines).
By type of load: heavy-starting motors, lighting and offices do not share a board, because their disturbances and criticality differ.
By criticality: what cannot stop (emergency lighting, control) has its own circuit so that a fault in another area does not take it out.
With a reserve: each board leaves 20 to 25 % free space to expand without rebuilding it, and each feeder is calculated with that expansion in mind.

Control and signalling circuits (contactor coils, push buttons, pilot lights, sensors) are drawn separately, in the functional diagram: they are supplied by a control transformer or a 24 V supply, with their own protection. The project of a house, with its outlets, sockets and bill of quantities, is in Installation project; the general rules for raceways, boards and earthing in Electrical installations. What changes in industry is the calculation, which follows.

Environmental conditions decide the material

A board that works in a clean office is destroyed in a hall with metal dust and cutting oil. Before choosing enclosures, cables and apparatus the environmental conditions of each room are surveyed: maximum and minimum temperature, dust (conductive or not), humidity and water splashes, chemical or corrosive agents, risk of impact and of explosion. The protection degree of an enclosure is given by the IP code (IEC 60529): two digits, the first against solids (0 to 6) and the second against liquids (0 to 8).

Code	What it protects against	Where it is used in the example plant
IP 20	Fingers and objects larger than 12 mm; nothing against water.	Boards inside the clean, dry electrical room of the substation.
IP 31	Objects over 2.5 mm; vertically dripping water.	Office boards with a ceiling, without significant dripping risk.
IP 54	Dust (limited ingress, not enough to cause harm); splashing from any direction.	SDB1 and push-button stations in the hall: machining dust and coolant mist.
IP 55	Same dust protection; low-pressure water jets.	SDB2, next to the furnaces and pumps, where the floor is hosed down.
IP 65	Dust-tight; water jets.	Push-button stations and terminal boxes in washdown areas or outdoors.
IP 66 / IP 67	Dust-tight; powerful jets (66) or temporary immersion (67).	Limit switches and luminaires in intense washdown areas or with flood risk.

Temperature

Cables and breakers are rated at a reference temperature. If the room or the inside of the board is hotter, they carry less current (factor k_T in section 03). The room is ventilated, the board moved away from the furnaces, or the equipment oversized.

Dust and humidity

Conductive dust (filings) short-circuits terminals; humidity condenses and lowers insulation. They are handled with the right IP, anti-condensation heaters and, in corrosive atmospheres, polyester or stainless-steel enclosures.

Impact and explosion

The IK code rates impact resistance. In rooms with flammable gases, vapours or dusts a specific explosive-atmosphere regulation applies (IEC 60079 series), with certified equipment; it is not improvised.

03Conductors: current and voltage drop

Each feeder must pass the same checks as in a house (see Installation project): current-carrying capacity, voltage drop, short circuit and coordination with the protective device. The largest of the resulting sizes is adopted. What is new in industry is that current capacity is corrected by two factors that hardly matter in a house.

Current capacity with correction factors

The current a cable can carry without exceeding its insulation temperature depends on how it is installed (on a tray, in conduit, buried), on the ambient temperature and on how many neighbouring cables also heat up. The regulation (AEA 90364) and the manufacturer publish tables with the current capacity for a reference installation and correction factors to depart from it. The values on this page are typical of the IEC 60364-5-52 series, on which those tables are based; in a real project they are always taken from the current table and the manufacturer’s datasheet.

Ambient temperature	PVC (70 °C)	XLPE (90 °C)	Cables grouped on a perforated tray	k_G
30 °C (reference)	1.00	1.00	1 circuit	1.00
35 °C	0.94	0.96	2 circuits	0.88
40 °C	0.87	0.91	3 circuits	0.82
45 °C	0.79	0.87	4 circuits	≈ 0.80
50 °C	0.71	0.82	6 circuits	0.73

The condition is that the chosen size, once corrected, carries at least the rating of the protective device (the device is what protects the cable, so the cable must withstand what the device lets through):

I_{table} \geq \frac{I_{n}}{k_{T} \cdot k_{G}}

Iₙ: protective device rating · kₜ: ambient-temperature factor · k_G: grouping factor (values come from the tables of the regulation and of the manufacturer)

Example · The feeder to the hall (MDB → SDB1)

SDB1 demands 149.4 kVA with cos φ = 0.787, i.e. a design current I_b = 215.7 A. It is protected by a 250 A breaker (the standard value above I_b). The four feeders share a perforated tray along part of the route and the hall can reach 40 °C in summer: k_T = 0.91 (XLPE) and k_G = 0.80.

I_{table} \geq \frac{250 A}{0.91 \cdot 0.80} = \frac{250 A}{0.728} = 343 A

The table for copper cables with XLPE insulation, three loaded conductors, on a perforated tray, gives 352 A for 95 mm²: it is enough (352 ≥ 343). By current, 95 mm² is enough. Now the voltage drop, at 120 m, with the three-phase formula:

Δ V = \sqrt{3} \cdot I \cdot L \cdot (r \cdot \cos φ + x \cdot \sin φ)

r = ρ / S: resistance per metre (ρ ≈ 0.022 Ω·mm²/m for copper at 90 °C) · x ≈ 0.08 mΩ/m: reactance per metre of a multicore cable · L in metres

r = \frac{0.022 Ω·mm²/m}{95 mm²} = 0.232 mΩ/m Δ V = \sqrt{3} \cdot 215.7 A \cdot 120 m \cdot (0.232 mΩ/m \cdot 0.787 + 0.08 mΩ/m \cdot 0.617)

Δ V = 10.4 V \to \frac{10.4 V}{400 V} \cdot 100 % = 2.60 % > 2 %

With 95 mm² the drop is 2.60 % and the 2 % limit assigned to the feeder is exceeded. With 150 mm²:

Δ V = \sqrt{3} \cdot 215.7 A \cdot 120 m \cdot (0.147 mΩ/m \cdot 0.787 + 0.08 mΩ/m \cdot 0.617) = 7.4 V = 1.85 %

It complies with 1.85 %. 4 × 150 mm² Cu/XLPE is adopted: voltage drop, not current, defined the size. In long runs this happens almost always.

The total admissible voltage drop, from the origin of the installation to the farthest appliance, is shared between the feeder, the sub-board and the final circuit. As a design criterion about 3 % is used for lighting and 5 % for power (check the current value in AEA 90364 and in the client’s specification); here 2 % is assigned to the feeder, 0.5 % to the busway and the rest to the final circuit. The motor must also be able to start: during starting the drop can reach several percent without being a defect, but not so much that the contactor drops out or the motor cannot accelerate.

The other two feeders are solved the same way, and show the two possible paths:

Feeder	I_b (A)	I_n (A)	I_tab req. (A)	By current	ΔV with that size	Adopted	Final ΔV
MDB → SDB1 · production · 120 m	215.7	250	343	95 mm²	2.60 %	4 × 150 mm² (ΔV)	1.85 %
MDB → SDB2 · furnaces and pumps · 35 m	153.0	160	220	50 mm²	1.01 %	4 × 50 mm²	1.01 %
MDB → SDB3 · lighting and offices · 80 m	51.6	63	87	16 mm²	2.26 %	4 × 25 mm² (ΔV)	1.47 %

SDB2 is close and current governs; SDB3 is far and voltage drop governs (16 mm² would carry the current, but with 2.26 % it exceeds 2 %). Neutral and protective (PE) conductor sizes are adopted separately, from the regulation tables; with non-linear loads the neutral can carry as much as a phase (see Power quality and machine protection).

04Short-circuit currents, busbars and busway

A short circuit is not a rare failure: it is the design case that decides whether the board resists or blows up. It must be known as two values. The maximum current (three-phase fault right at the board) sizes the breaking capacity of the breakers and the stresses on the busbars. The minimum current (fault at the farthest end of the line) checks that the protective device notices the fault and clears it in time. The general method is IEC 60909; what follows is the impedance method, which with a calculator gives the maximum value with good approximation.

The idea: from the grid to the fault point there is a chain of impedances in series (the grid, the transformer, each stretch of cable). They are added —resistances with resistances and reactances with reactances— and the current is the voltage divided by the total impedance. It is calculated at 400 V, the line voltage on the low-voltage side.

Figure 2. Impedance chain up to the fault, animated. At point 1 the fault current is the largest possible; every metre of cable away from the transformer adds impedance and lowers it.

X_{grid} = \frac{{U_{n}}^{2}}{S_{sc,grid}}

equivalent impedance of the medium-voltage grid referred to 400 V; it is almost purely reactive

Z_{T} = \frac{u_{k}}{\cdot} R_{T} = \frac{P_{Cu}}{\cdot} X_{T} = \sqrt{{Z_{T}}^{2} - {R_{T}}^{2}}

uₖ: short-circuit voltage (per unit) · P_Cu: copper losses at full load · Sₙ in VA

Z_{T} = \frac{0.04}{\cdot} = 20.3 mΩ R_{T} = \frac{4 500 W}{\cdot} = 7.3 mΩ

X_{T} = \sqrt{{20.3}^{2} - {7.3}^{2}} mΩ = 19.0 mΩ X_{grid} = \frac{{400 V}^{2}}{250 000 000 VA} = 0.6 mΩ

The reactance of the medium-voltage grid is calculated from its short-circuit power (data supplied by the utility; here 250 MVA), and the transformer data from its nameplate or catalogue (u_k and copper losses; typical values: the real ones are on the nameplate). The fault at the low-voltage terminals of the transformer and the fault at the end of the hall feeder give:

Element	R (mΩ)	X (mΩ)	Z (mΩ)	I_cc3 (kA)
MV grid (250 MVA)	—	0.6	0.6
Transformer 315 kVA, 4 %	7.3	19.0	20.3
Point 1 · LV terminals of the transformer	7.3	19.6	20.9	11.0
Cable MDB → SDB1: 4 × 150 mm², 120 m	17.6	9.6	20.0
Point 2 · busbars of board SDB1	24.9	29.2	38.4	6.0

I_{sc3} = \frac{U_{n}}{\sqrt{3} \cdot Z}

Z = √(R² + X²): total impedance from the source to the fault point

Example · The two fault currents

Point 1, at the transformer terminals (where the main breaker sits and the feeders start):

I_{sc3} = \frac{400 V}{\sqrt{3} \cdot 20.9 mΩ} = 11.0 kA

Point 2, at the SDB1 busbars, 120 m away (adds the 4 × 150 mm² cable: R = 17.6 mΩ, X = 9.6 mΩ). Besides the three-phase fault, the two-phase fault is estimated; it is about 87 % of the three-phase one and the smallest of the phase-to-phase faults:

I_{sc3} = \frac{400 V}{\sqrt{3} \cdot 38.4 mΩ} = 6.0 kA I_{sc2} \approx \frac{\sqrt{3}}{2} \cdot 6.0 kA = 5.2 kA

Cross-check: with an infinite-power grid, the fault at the terminals would be I_n/u_k = 11.4 kA; the reactance of the real grid lowers it to 11.0 kA, as expected.

\frac{I_{n}}{u_{k}} = \frac{454.7 A}{0.04} = 11.4 kA

The fault gets smaller the farther from the transformer, but it is still large: 6.0 kA at the end of the feeder, enough to melt an unprotected cable. A formal calculation also applies a voltage factor c = 1.05 (IEC 60909) and would give about 5 % more; that is why, when choosing breakers, one standard step of margin is always left.

The phase-to-earth or phase-to-neutral fault depends on the earthing system (TT, TN, IT) and on the loop impedance; it is measured on site with a loop tester and is the value used to check that the protection trips in time (see Electrical protection). For the 250 A breaker of SDB1, whose magnetic trip is around 10 · I_n = 2,500 A, the minimum two-phase fault of 5.2 kA exceeds it by a wide margin: the breaker sees the fault at the end of the line.

Busbars

The copper busbars of the MDB carry the current of all the sub-boards. They are pre-sized by current density —about 1.25 A/mm² as a conservative criterion in a naturally ventilated board— and checked against the manufacturer’s table, which accounts for shape, paint, bar stacking and the temperature inside the board:

Example · Busbars of the main board

Current to carry: the main breaker of 500 A (sized for the transformer, which delivers up to 454.7 A). Required cross-section: $500 / 1.25 = 400 mm²$ . A 40 × 10 mm copper bar (400 mm²) per phase is adopted.

But the bar must withstand the fault. Two stresses: the dynamic one, because the opposing currents of two parallel bars repel each other with a force proportional to the square of the peak current, and the thermal one, because all the energy of the fault turns into heat before the breaker opens. The peak of the first half-wave exceeds the rms value by the factor κ, which depends on R/X:

κ = 1.02 + 0.98 \cdot e^{- 3 R / X} I_{p} = κ \cdot \sqrt{2} \cdot I_{sc3}

Iₚ: peak value of the first half-wave, which produces the mechanical stress

κ = 1.02 + 0.98 \cdot e^{- 3 \cdot 0.37} = 1.34 I_{p} = 1.34 \cdot \sqrt{2} \cdot 11.0 kA = 21.0 kA

With insulating supports every 0.5 m and phases 0.1 m apart (conservative formula, for two long parallel conductors):

F = 2 \cdot 10^{- 7} \cdot {I_{p}}^{2} \cdot \frac{l}{a}

F in newtons per span · Iₚ in amperes · l: distance between supports · a: distance between phases (both in metres)

F = 2 \cdot 10^{- 7} \cdot {20 971 A}^{2} \cdot \frac{0.5 m}{0.1 m} = 440 N

Each support and insulator must withstand those 440 N (about 45 kgf) with margin: the bending-strength data comes from the insulator manufacturer. Thermally, with the main breaker delayed 0.2 s (next section):

S_{\min} \geq \frac{I_{sc3}}{\cdot}

k ≈ 140 A·s½/mm² for the bare copper of a busbar (manufacturer value) · t: time taken to clear the fault

S_{\min} \geq \frac{11 041 A}{\cdot} = 35 mm²

The chosen 400 mm² far exceeds the 35 mm² required. Cross-check: a 400 mm² bar withstands, at 140 A/mm², 56 kA for 1 s, or 125 kA for 0.2 s: well above 11.0 kA.

Busway

A busway (busbar trunking system, IEC 61439-6) is a metal enclosure holding insulated copper or aluminium bars, with straight sections, bends and tap-off boxes that plug in at any point. It replaces cable in two cases: when the power is so high that many cables in parallel would be needed, and when the loads are spread along a line and a tap must be added or moved without touching the rest. It is the typical solution for a production hall.

In the example, inside the hall, SDB1 feeds a 250 A, 40 m busway, with a tap-off box every few metres for the machines (the design current of SDB1 is 215.7 A). It is checked like a feeder, with three differences:

Voltage drop is calculated with the resistance and reactance per metre from the catalogue; with illustrative values of 0.20 and 0.08 mΩ/m, at full current over the whole length it comes to 0.77 %.
Since the loads are spread along the line, the current falls stretch by stretch and the real drop is about half: 0.39 %. With the 1.85 % of the feeder, the total up to the machines is 2.24 %, within the plan.
Its short-time withstand current (I_cw) and the peak it withstands must be checked: nameplate data that must exceed the fault at its point of installation (6.0 kA at SDB1).

Busways are also corrected for the environment

Their rated current is given for a specific ambient temperature and horizontal mounting. They are corrected for temperature like cables and, where they cross fire-rated walls or floors, the manufacturer’s fire-barrier sections must be used. And like any enclosure they are chosen with the IP suited to the room (IP 54 or 55 in the example hall).

05Switching, protection and selectivity

Each circuit needs three functions, which may be in one device or several: isolating (to work without voltage), switching (making and breaking in service) and protecting against overload and short circuit. The general protection criteria are in Electrical protection; here they are applied to the example.

The circuit breaker and its breaking capacity

A circuit breaker has a thermal trip (overload, with a delay that depends on how much the current is exceeded) and a magnetic trip (short circuit, instantaneous). In power breakers both are adjustable: I_r (long delay), I_sd (short delay, with its time) and I_i (instantaneous). It also has two breaking capacities:

I_cu · ultimate breaking capacity

The greatest current it can interrupt, once. Afterwards it may be unusable. It must be greater than or equal to the short-circuit current at the point where it is installed.

I_cs · service breaking capacity

A percentage of I_cu (25, 50, 75 or 100 %) after which the breaker is still fit for service. Where an outage cannot be tolerated, a high value is required.

The key point is where the current is taken: the breaker sees the fault at its own terminals. So the MDB breakers (main and feeders) face 11.0 kA, and those inside SDB1, 6.0 kA:

Breaker	Location	I_sc at its terminals	I_n	I_cu chosen
Main	MDB	11.0 kA	500 A	25 kA
Feeder SDB1	MDB	11.0 kA	250 A	25 kA
Feeder SDB2 / SDB3	MDB	11.0 kA	160 A / 63 A	25 kA
SDB1 incomer	SDB1	6.0 kA	250 A	≥ 10 kA

The standard I_cu value immediately above 11.0 kA is 16 kA; 25 kA is adopted to leave margin for a larger transformer. In SDB1 it is possible to take advantage of the current limitation of the upstream breaker (back-up protection or cascading): the manufacturer guarantees, with tables, that two breakers in series withstand a fault larger than one alone. It is an economical solution, but it must be backed by the manufacturer’s table, not by an assumption.

The motor circuit

A motor needs a set of devices with different functions: short-circuit protection (aM fuses or a magnetic breaker), switching (contactor) and overload protection (thermal relay). Motor details are in Single-phase and three-phase AC motors, and those of the devices in Maintenance of electrical components. With the 37 kW compressor of the example:

I_{n} = \frac{P_{shaft}}{\sqrt{3} \cdot U_{n} \cdot η \cdot \cos φ}

rated current of a three-phase motor from its shaft power

I_{n} = \frac{37 000 W}{\sqrt{3} \cdot 400 V \cdot 0.94 \cdot 0.84} = 67.6 A I_{start} \approx 7 \cdot 67.6 A \approx 473 A

Element	Function	Choice for the compressor
aM fuses	Short circuit. They withstand starting without blowing.	80 A (a rating that does not act on the starting peak; the manufacturer’s coordination table confirms it).
Contactor	Motor switching, utilization category AC-3 (squirrel cage).	80 A in AC-3 (standard step ≥ 67.6 A), with control-voltage coil.
Thermal relay	Overload and phase loss.	Set at 67.6 A, class 10 (short start) or 20 (long start).

Coordination among these three elements is type 1 (after a severe fault the starter may be damaged, though without risk) or type 2 (it remains usable, only slight contact welding allowed). For a compressor that cannot stay down, type 2 is preferred. And there is an industry-specific check: the start. Direct on line, the compressor would draw about 473 A, practically the rated current of the whole transformer (454.7 A): it would cause a voltage dip across the plant. With star-delta starting the current drops to a third, about 158 A, and with a soft starter or a drive (see Variable-speed drives) it is even better controlled.

Coordinating protections: selectivity

A well-protected installation is not the one with many breakers, but the one in which, on a fault, only the breaker immediately upstream of the fault opens. This is called selectivity (or discrimination). If the main breaker opens because one machine in the hall failed, the whole plant goes dark. There are three ways to achieve it:

Current-based: the upstream device has a magnetic trip threshold higher than the maximum downstream fault. Simple, but it only works where the fault falls with distance.
Time-based: the upstream device waits an adjustable instant (short-circuit delay) to give the downstream one time to clear the fault. It requires the upstream device to withstand the short circuit during that time.
By manufacturer tables: for combinations of breakers of the same range, the manufacturer guarantees selectivity up to a certain current. It is the usual approach in low voltage.

Figure 3 shows the curves of two breakers of the example, the 500 A main of the MDB and the 250 A one that protects SDB1. They are typical curves of the breaker type (not of a brand), drawn on a logarithmic scale: current on the horizontal axis; on the vertical axis, the time each takes to trip.

Figure 3. Selectivity between the main breaker (500 A) and the SDB1 feeder breaker (250 A), animated. With the 0.2 s delay on the main, the 250 A always clears first; if the main had only instantaneous trip, on a strong fault both would open and the plant would lose power.

Example · Who opens on each fault?

The main is set to I_r = 500 A, I_sd = 6 · I_r = 3,000 A with a 0.2 s delay, and no instantaneous trip; the SDB1 breaker (250 A) has a magnetic trip at 10 · I_n = 2,500 A, acting in under 0.01 s.

700 A overload in SDB1: the 250 A trips thermally in about 55 s; the main would take about 440 s. Only the 250 A opens.
6.0 kA short circuit at the end of the feeder: the 250 A clears in under 0.01 s; the main is still waiting its 0.2 s. Only the 250 A opens.
11.0 kA short circuit right at the output of the 250 A breaker: same, as long as the main withstands 11.0 kA for 0.2 s (its short-time withstand current I_cw must be greater).
Without delay (if the main tripped instantaneously at 5 kA): at 6.0 kA both would open and selectivity is lost.

There is a price: while the main waits 0.2 s the fault keeps flowing. The cable and the busbars must take that energy. For the 25 mm² cable of SDB3, protected at its own outlet, the maximum time with the fault at the MDB is:

t \leq {(\frac{k}{\cdot})}^{2}

k = 143 for XLPE-insulated copper · S: conductor cross-section in mm² · I: fault current in A

t \leq {(\frac{143}{\cdot})}^{2} = 0.105 s

The SDB3 cable depends on its own (instantaneous) breaker clearing in a fraction of a cycle: if it depended on the main, its 0.2 s would destroy it. That is why coordination is always checked against the cable and the busbars, not only between breakers.

06Reactive power and power-factor correction

Motors, transformers and lamps with ballasts need reactive power to magnetize: it goes back and forth between the grid and the load doing no work, but it flows through the cables and the transformer as if it did (see Power in alternating current and Three-phase circuits). The example plant has cos φ = 0.85: out of every 100 amperes flowing, only 85 do work. Correcting it has three benefits:

Less current through the transformer and feeders: lower losses, less voltage drop and more power available for expansion.
No penalty: utilities surcharge the bill below a reference cos φ (of the order of 0.85; the value and the surcharge appear in the current tariff schedule approved by the provincial regulator).
Better use of the investment in copper and in the transformer.

Reactive power is compensated with capacitors, which supply it where it is consumed. The bank power is calculated from the angles before and after:

Q_{c} = P \cdot (\tan φ_{1} - \tan φ_{2})

P: active demand power · φ₁: angle before correction · φ₂: target angle (cos φ₂ = 0.95)

Example · The capacitor bank

The plant demands P = 209.4 kW and Q = 128.6 kvar (tan φ₁ = 0.614, cos φ₁ = 0.85). The aim is to bring cos φ to 0.95 (tan φ₂ = 0.329):

Q_{c} = 209.4 kW \cdot (0.614 - 0.329) = 209.4 kW \cdot 0.285 = 59.8 kvar \to 60 kvar

An automatic 60 kvar bank in steps (for example 6 × 10 kvar) is adopted, with a controller that connects and disconnects capacitors according to the measured cos φ. Result:

Q_{end} = 128.6 - 60 = 68.6 kvar S_{end} = \sqrt{{209.4}^{2} + {68.6}^{2}} kVA = 220.4 kVA \cos φ = \frac{209.4}{220.4} = 0.95

The low-voltage current drops from 354.7 A to 318.1 A (10 % less) and the transformer runs at 70 % of its rating. The bank data:

C = \frac{Q_{c}}{3 \cdot ω \cdot {U_{n}}^{2}}

capacitance of each phase with the capacitors connected in delta (U = line voltage)

C = \frac{60 000 var}{3 \cdot 314.2 rad/s \cdot {400 V}^{2}} = 398 µF

I_{c} = \frac{60 000 var}{\sqrt{3} \cdot 400 V} = 86.6 A \to 1.43 \cdot 86.6 A = 123.8 A \to 125 A

The bank breaker is chosen at no less than 1.43 times the rated capacitor current, because capacitors withstand continuously up to 30 % extra current from harmonics and overvoltage, and their real capacitance may exceed the nominal by 10 % (1.3 · 1.1 = 1.43); 125 A is the standard step. 35 mm² cable (table: 185 A · 0.728 = 135 A ≥ 125 A).

Figure 4. Power triangle before and after correction, animated. The capacitor bank removes reactive power from the grid; active power does not change, and apparent power (what the transformer must deliver) decreases.

Where to install it

Fixed compensation at each large motor: a capacitor in parallel with the motor, switched with it. It corrects only what that motor needs and avoids carrying the reactive power along the line.
Centralized compensation at the MDB: an automatic bank that follows demand. This is the case of the example.
Mixed: fixed at the large motors and automatic at the MDB for the rest.

Harmonics: the capacitor’s enemy

A capacitor presents an impedance that falls as frequency rises; with drives, rectifiers and electronic luminaires it can enter resonance with the transformer, amplify a harmonic and burn out. If more than roughly 20–25 % of the load is non-linear, the bank is installed with a detuning reactor (for example 7 %) in series with each step; see Power quality and machine protection.

A disconnected capacitor is still charged

A capacitor keeps its charge after the breaker opens. The bank has discharge resistors, but before touching it you must wait the time the manufacturer states (one to several minutes), verify the absence of voltage with an instrument at the terminals and only then short-circuit and earth them. A capacitor is never short-circuited “just in case” without having opened first. See Occupational health and safety.

The project is one chained calculation

Demand fixes the transformer; the transformer fixes the short-circuit current; that fixes the breaking capacity and the strength of busbars and cables; cos φ comes back to the demand. Changing one datum (another transformer, another length, one more motor) forces you to go through the chain again. That is why it is done in a spreadsheet and documented in a calculation report that accompanies the drawings (see Technical drawing and computer-aided design).

07In the workshop

Lab 1 · Survey the loads of a board and measure their factors

Materials: a sub-board of the school, true-rms clamp meter (an analyzer if available), spreadsheet. Steps: with the installation in service, record for each outgoing way the line current at three moments of the day, the voltage and the cos φ; fill in the sheet with P, Q and S of each load. What is measured: real current against the rated current of each load (f_u) and the feeder current against the sum of the outgoing currents (f_s). Acceptance criterion: the vector sum of P and Q of the outgoing ways matches the one measured at the feeder to within 10 %; the f_u and f_s obtained are recorded for use in future projects.

Lab 2 · The example calculation in a spreadsheet

Materials: spreadsheet and the manufacturer’s cable tables. Steps: enter the load table of this page (with formulas, not numbers), obtain the demand and the transformer; then the three feeders with current and voltage drop; then I_sc at the two points; then the bank. Change the SDB1 length from 120 to 60 m and the target cos φ to 0.98. Acceptance criterion: the sheet reproduces the values in the text (315 kVA, 4 × 150 mm², 11.0 and 6.0 kA, 60 kvar) and, with the changes, the results move in the expected direction, which the student explains in writing.

Lab 3 · Measure the loop impedance and estimate the fault current

Materials: loop impedance tester (or earth tester with that function), training sub-board. Steps: with the teacher and with the circuit in normal service, measure at the farthest terminal the phase-neutral and phase-earth loop impedance; calculate $I = U / Z$ ; compare it with the value the project calculation predicts. What is measured: loop impedance and the estimated fault current. Acceptance criterion: the estimated current greatly exceeds the magnetic threshold of the breaker, so that it would trip instantaneously. An insufficient value points to a cable that is too long or a loose connection.

Lab 4 · Draw the single-line diagram and the load schedule

Materials: CAD software and the table of this page. Steps: draw the single-line diagram of the example with standard symbols, labelling in each stretch the cross-section, length, protection and calculated currents; add the load schedule and the title block with the revision. Acceptance criterion: a classmate who has not seen the calculation can reproduce the SDB1 cross-section using only the drawing (see Technical drawing and computer-aided design).

Measuring on a board is not just looking

Any measurement with the board energized is made by a trained and authorized person, with dielectric footwear and gloves, face protection and insulated tools; an instrument with a suitable measurement category (CAT III or IV) is used, and nothing is opened or touched unless the teacher has said so. Live work follows Resolution SRT 592/2004 and Law 19,587 with its Decree 351/79.

08Common mistakes

Mistake	Consequence
Sizing with installed power or adding kVA arithmetically	Oversized transformer, busbars and feeders, with the cost that implies. Demand is calculated with f_u, f_s and adding kW and kvar separately.
Adopting f_u and f_s arbitrarily, “to make it work”	An undersized plant is discovered when the main breaker trips at peak. If in doubt, measure in a similar installation.
Choosing the cable by current alone in a long run	Voltage falls at the far end: motors start poorly and overheat. The drop is always checked.
Forgetting the temperature and grouping factors	The cable heats beyond what its insulation withstands; cable life shortens or a fire starts.
Choosing the breaking capacity with the current at the end of the line	A breaker is sized with the fault at its own terminals. With the end value it falls short and may blow up while clearing.
Taking the I_sc of one board and using it everywhere	Fault current changes with distance from the transformer: each board has its own.
Installing “higher-ampere” breakers so they do not trip	Cable protection and selectivity are lost: on a fault the main opens and the whole plant loses power.
Ignoring the starting of large motors	A direct start of 37 kW draws as much as the whole transformer: voltage dip across the plant. Star-delta, soft starter or drive is used.
Installing capacitors without knowing the harmonics	Resonance with the transformer and burnt capacitors. With non-linear loads, the bank has detuning reactors.
An IP 20 enclosure in an atmosphere with conductive dust or humidity	Short circuits and insulation failures within months.

09Self-assessment

Why is maximum demand lower than installed power, and how is it obtained?

Because not all loads are at full load or run at the same time. The utilization factor f_u is applied to each load, kW and kvar are added separately, and the total gets the diversity factor f_s. Apparent power then follows from Pythagoras. In the example, 367.7 kW installed give 209.4 kW of demand.

A 90 m feeder carries 120 A at 400 V three-phase, with cos φ = 0.85 (sin φ = 0.527). Does 35 mm² comply if the limit is 2 %? (r = 0.022/S, x = 0.08 mΩ/m)

r = 0.022/35 = 0.629 mΩ/m → R = 56.6 mΩ; X = 7.2 mΩ. ΔV = √3 · 120 · (0.0566 · 0.85 + 0.0072 · 0.527) = 207.8 · (0.0481 + 0.0038) = 10.8 V, i.e. 2.7 %: it does not comply. With 50 mm²: R = 39.6 mΩ, ΔV = 207.8 · (0.0337 + 0.0038) = 7.8 V = 1.95 %: it complies. Even if 35 mm² may be enough for current, voltage drop defines the size.

Why is a breaker’s breaking capacity chosen with the short-circuit current at its terminals and not with that at the end of the line it protects?

Because the worst case for the breaker is a fault right after it, where the downstream impedance is minimal and the current maximal. The fault at the far end is smaller and serves another check: that the magnetic trip detects it.

A 400 kVA, 4 % transformer fed from an infinite-power grid. What short-circuit current is there at the 400 V terminals?

I_n = 400,000/(√3 · 400) = 577.4 A; I_sc = I_n/u_k = 577.4/0.04 = 14.4 kA. With a real grid, a little less; and in practice the next standard breaking-capacity step is chosen.

What does it mean for two breakers to be selective, and how is it achieved by delay?

That on a fault only the breaker immediately upstream of it opens. With delay, the upstream breaker waits an adjustable time (for example 0.2 s) before tripping on short circuit, giving the downstream one, which is instantaneous, time to clear. You must check that it withstands the fault during that time (I_cw) and that the cable and busbars can take it.

A plant draws P = 150 kW at cos φ = 0.80. What capacitor bank brings it to cos φ = 0.95?

tan φ₁ = 0.75 (cos 0.80) and tan φ₂ = 0.329 (cos 0.95). Q_c = 150 · (0.75 − 0.329) = 150 · 0.421 = 63 kvar; a 60 or 70 kvar bank is adopted. Current falls from 150/(√3 · 0.4 · 0.80) = 270.6 A to 150/(√3 · 0.4 · 0.95) = 227.9 A.

Which environmental conditions are surveyed before choosing the board, and what is done with them?

Temperature (corrects the current capacity of cables and devices), dust (IP of the enclosure, first digit), humidity and water (second digit, anti-condensation heaters), chemical agents (enclosure material), impact (IK) and explosion risk (specific regulation). In a hall with machining dust and coolant, IP 54 or 55 is adopted.

← Previous topic

Electricity Topic Map

Next topic →

Electrical protection