Electrical Machines and Installations · 96 h · Topic 4 of 5

Protection and sizing of installations

A 3 kW circuit 30 m from the panel carries the current with 1.5 mm² cables, but the voltage drops almost 6%. The length decides the cross-section, not the current.

Thermal-magnetic breaker Residual-current device Current-carrying capacity Voltage drop Power factor

01Two risks, two protections

An electrical installation has to protect against two different things. Overcurrent —overload or short circuit— heats the conductors until the insulation is damaged and a fire starts. Electric contact makes current flow through a person: direct, by touching a live part, or indirect, by touching an enclosure that became live because of a fault. Each risk has its own device:

Device	Protects against	How it acts
Thermal-magnetic circuit breaker	Overload and short circuit	A bimetal for slow overloads and an electromagnet for short circuits
Fuse	Overload and short circuit	It melts; high breaking capacity
Residual-current device (RCD)	Direct and indirect contact	Compares the current going out and the current coming back; it trips if more than 30 mA is missing
Grounding	Indirect contact	Gives the fault current a low-resistance path and makes the protection act

A current of about 30 mA through the body, for a fraction of a second, can cause ventricular fibrillation: that is why the 30 mA RCD is mandatory in dwellings. The practical presentation of installations and protections is in the technician program, in electrical installations and electrical protection.

02Switching and protection

Switching devices open and close circuits under normal conditions: switches, disconnectors (which only open without load, to isolate safely), contactors (which switch remotely and thousands of times). Protection devices open on their own on a fault. A thermal-magnetic circuit breaker does both things. Its trip curve (B, C or D) indicates at how many times the rated current the magnetic trip acts: 3 to 5 times for B, 5 to 10 for C, 10 to 20 for D, for loads with high inrush currents such as motors and transformers.

03Conductor sizing

A circuit is sized by chaining three conditions. The design current \( I_B \) comes from the load; the breaker has to have a rated current \( I_n \) greater than it; and the conductor, a current-carrying capacity \( I_z \) greater than the breaker’s, so that the breaker protects it:

\[ I_B \le I_n \le I_z \] \[ I_B = \frac{P}{V \cos\varphi} \] \[ I_B = \frac{P}{\sqrt 3\, V_L \cos\varphi} \] The design current for a single-phase circuit and for a balanced three-phase one. The conductor’s current-carrying capacity depends on the cross-section, the insulation, how it is installed and the ambient temperature, and is given by the tables of the regulations.

The second condition is the voltage drop: even if the conductor withstands the current, if the circuit is long the voltage at the load falls too far. The regulations limit it, typically between 3 and 5% depending on the type of circuit:

\[ \Delta U \approx k\, \frac{\rho\, L\, I_B \cos\varphi}{S} \] With \( k = 2 \) for single-phase and \( \sqrt 3 \) for three-phase, \( \rho \approx 0.0225\ \Omega\,\text{mm}^2/\text{m} \) for copper in service, \( L \) the length and \( S \) the cross-section. The reactance is neglected, valid for small cross-sections. In long circuits, it is the voltage drop and not the current that decides the cross-section.

Lab · sizing a circuit

You choose the load, its power factor and the circuit length. The design current is calculated, along with the standard breaker, the minimum cross-section by current-carrying capacity and the one needed for voltage drop (limit of 3%). The current-carrying capacities are those of IEC 60364-5-52 for copper with PVC in conduit (method B1), at 30 °C, as a reference; for a real installation the AEA (Argentine Electrotechnical Association) regulations apply.

Active power =

Circuit length =

04Power factor correction

An inductive load, such as a motor, takes in addition to the active power a reactive power that goes back and forth without doing work, but that flows through the conductors and heats them. Utilities penalize a low power factor. A capacitor in parallel supplies that reactive power locally:

\[ Q_C = P\,(\tan\varphi_1 - \tan\varphi_2) \] The reactive power of the capacitor to bring the power factor from \( \cos\varphi_1 \) to \( \cos\varphi_2 \). Correcting 15 kW from 0.85 to 0.95 requires 4.4 kvar and lowers the current from 26.8 A to 24.0 A.

The concepts of AC power are in power in AC circuits, in the technician program. With electronic loads the situation changes: the low power factor is due to harmonics and not to phase shift, and a capacitor does not correct it, as seen in power quality.

05In the laboratory

Activity 1 · Design of a circuit

Size the outlet circuit of a workshop with five known loads, choosing the breaker, the cross-section and checking the voltage drop against the regulations in force.

Activity 2 · Testing an RCD

With an RCD tester, measure the trip current and trip time of a 30 mA breaker.

Activity 3 · Correcting a motor

Measure the active and reactive power of a motor at no load and under load, and calculate and connect the capacitor that brings its power factor to 0.95.

06Common mistakes

Choosing the breaker by the cross-section and not by the load. It has to meet both conditions.
Forgetting the voltage drop in long circuits. The conductor withstands the current, but the load receives too little voltage.
Installing an RCD without grounding. It still protects against direct contact, but a fault to the frame stays live until someone touches it.
Overcompensating the power factor. With the load disconnected, the capacitors leave the installation capacitive.

07Self-assessment

What current does a 10 kW three-phase load with cos φ = 0.8 draw at 380 V?

\( 10\,000/(\sqrt3 \cdot 380 \cdot 0.8) \approx 19 \) A.

What drop does a single-phase circuit of 40 m, 10 A, cos φ = 1 and 2.5 mm² have?

\( 2 \cdot 0.0225 \cdot 40 \cdot 10 / 2.5 = 7.2 \) V: 3.3% of 220 V.

What capacitor corrects 10 kW from cos φ 0.7 to 0.95?

\( 10 \cdot (1.020 - 0.329) \approx 6.9 \) kvar.

Why does the RCD not protect against overloads?

Because it only measures the difference between the current going out and the current coming back; an overload flows in full through both conductors.

08Further reading

Argentine Electrotechnical Association (AEA). Reglamentación para la Ejecución de Instalaciones Eléctricas en Inmuebles (AEA 90364) (in Spanish). AEA, current editions. The standard that governs installations in Argentina: protections, cross-sections, voltage drops and grounding.
Marcelo A. Sobrevila. Instalaciones eléctricas (in Spanish). Alsina, various editions. The classic Argentine text on installation design.
Schneider Electric. Electrical Installation Guide (Spanish edition: “Guía de diseño de instalaciones eléctricas”). Schneider Electric, according to IEC. The complete sizing of an installation with the IEC criteria, with examples.