Catto / Topic Map · Industrial Installations Year 6
Industrial Installations · 168 h · Topic 2 of 7

Electrical materials

An installation is, materially, a selection of materials: copper, insulation, sheet metal, plastic and iron. Choosing any of them badly is not noticed on the first day; it is noticed two years later, when something heats up, cracks or fills with water.

Conductors Insulation IP rating Ferromagnetic materials

01Conductors

MaterialResistivity at 20 °CWhen it is used
Copper0.0172 Ω·mm²/mPractically everything: installations, boards, motors. The best common conductor, ductile and easy to connect.
Aluminum0.0282 Ω·mm²/mOverhead lines and large-cross-section feeders: it weighs a third as much and costs less. It requires specific terminals and pastes.
Silver0.0159 Ω·mm²/mOnly in contacts: it is the best conductor, but its price rules it out as a line conductor.
Case: single-phase 220 V circuit, 30 m long, 20 A loadconductor cross-sectionCurrent-carrying capacity1.515 A2.521 A4.028 A6.036 A10.050 Aload: 20 A1.5 mm2Voltage drop13.8 V= 6.3 %CheckfailsOverloaded: 15 A allowable for a 20 A load, and also a 6.9 V drop.2.5 mm2Voltage drop8.3 V= 3.8 %CheckborderlineIt carries the current, but the voltage drop sits at the 3 % limit.4.0 mm2Voltage drop5.2 V= 2.3 %CheckpassesPasses both checks with a reasonable margin.6.0 mm2Voltage drop3.4 V= 1.6 %CheckpassesComfortable on both checks: the typical choice if the run may grow.10.0 mm2Voltage drop2.1 V= 0.9 %CheckpassesVery comfortable. Justified only if the run is much longer or the load will increase.The cross-section is not chosen by current alone: on long runs, voltage drop is what governs.
Figure 1. Cross-section, current and voltage drop, animated. As the cross-section increases, the resistance falls, and with it the heating and the voltage drop. The cross-section is not chosen only by current: also by the length of the run.
R=ρLSΔU=2·ρL·IS In a single-phase circuit the current flows out and back: hence the factor 2. In a balanced three-phase circuit the factor is √3.
Worked example · Is 2.5 mm² enough?

A single-phase circuit of 2.5 mm² supplies a 16 A load at 28 m from the distribution board, at 220 V.

ΔU = 2 × 0.0172 × 28 × 16 / 2.5 = 6.2 V, that is 2.8 %. Just inside the allowable 3 %, but with no margin.

With 4 mm²: ΔU = 3.85 V, or 1.8 %. The current-carrying capacity was already fine with 2.5 mm²; what forces the move to a larger cross-section is the distance. It is the most frequent calculation mistake on long runs: the current is checked and the drop is forgotten.

InsulationMaximum temperatureCharacteristics
PVC70 °CThe standard in home installations. Inexpensive. Gives off dense, corrosive smoke when burning.
Halogen-free (LSZH)70 to 90 °CMandatory in places with many people: schools, hospitals, subways. It gives off no toxic or corrosive gases.
XLPE / EPR90 °CCross-linked polyethylene and ethylene-propylene rubber. They carry more current at the same cross-section and withstand short circuits better.
Silicone180 °COvens, boilers, motors. Very flexible and very expensive.
Current-carrying capacity is not a fixed number

It depends on the cross-section, the insulation, and above all on how the cable is installed: in free air, embedded in conduit, buried, or bundled with others. The same 2.5 mm² conductor carries much less current in a conduit next to five others. The regulation tables include those correction factors, and skipping them is what produces warm cables inside walls.

02Insulators and enclosures

EnclosureIP20Dry indoorFirst digit · solidsprotected against fingers (12 mm)Second digit · waterno protection against waterBoards inside a home or an office.IP44SplashingFirst digit · solidssolid objects larger than 1 mmSecond digit · watersplashing from any directionBathrooms, laundry rooms and covered porches.IP55Dust and hoseFirst digit · solidsdust-protectedSecond digit · waterlow-pressure water jetsDusty industry with hose-down cleaning.IP65OutdoorsFirst digit · solidsdust-tightSecond digit · waterpressurized water jetsOutdoor enclosures.IP68ImmersionFirst digit · solidsdust-tightSecond digit · watercontinuous immersionManholes and submersible pumps.Drilling an IP65 enclosure without a cable gland ends its IP65 rating: the whole installation sets it.
Figure 2. The IP rating, animated. The first digit indicates protection against solid objects and the second against water. IP65 and IP20 look almost the same from the outside: what changes is where they can be installed.
Insulating materials
  • Thermoplastics (PVC, polyethylene, polyamide): they soften with heat and are easy to mold. Switch bodies, boxes, raceways.
  • Thermosets (Bakelite, epoxy resins): they do not soften, and resist temperature and arcing. Lamp holder bases, insulators, encapsulated windings.
  • Ceramics and glass: line insulators, fuse holders, power resistors. They withstand weather and temperature for decades.
  • Impregnated paper and varnishes: insulation between turns of transformers and motors.
What defines an insulator
  • Dielectric strength: how many kV per millimeter it withstands before breaking down.
  • Operating temperature: above it, it ages quickly and becomes brittle.
  • Insulation resistance: measured with a megohmmeter, and the basic acceptance test of an installation.
  • Fire behavior: whether it spreads flame and what gases it emits.
RatingWhere it applies
IP20Boards in dry indoor locations. Protects against fingers, not water.
IP44Bathrooms, laundry rooms, covered porches: splashing.
IP55Dusty industry with hose-down cleaning.
IP65Outdoors, water jets. Dust-tight.
IP67 / IP68Temporary or continuous immersion: manholes, submersible pumps.
The rating is defined by the installation, not the enclosure

An IP65 enclosure that gets a hole drilled with a screwdriver to pass a cable is no longer IP65. The rating is maintained only with cable glands, blanking plugs and gaskets in good condition. It is by far the most common cause of “sealed” enclosures full of water.

03Ferromagnetic materials

Everything that transforms or converts energy —transformers, motors, contactors, reactors— needs a magnetic circuit, and that circuit is as important as the electrical one.

MaterialUse and reason
Silicon-iron sheetCores of transformers and motors. Silicon increases resistivity and reduces eddy currents.
Grain-oriented sheetPower transformers: the crystals are aligned with the direction of the flux and losses drop considerably.
FerriteHigh frequency: switching power supplies, filters, pulse transformers. Extremely high resistivity, almost no eddy currents.
Permanent magnetsFerrite for low cost; neodymium for compact, high-efficiency motors.
Why the core is made of laminations

The varying flux induces currents in the iron itself —eddy currents, or Foucault currents— that close in circles and heat the core without doing any work. By cutting the core into thin laminations insulated from one another, those circuits are interrupted and the losses fall drastically. On top of that come the hysteresis losses, which depend on the loop area of the material and on the frequency.

Both are studied in more detail in transformers and in magnetism and electromagnetism.

04Losses, efficiency and thermography

Where energy is lost
  • In the copper: I²R. It grows with the square of the current, so doubling the load quadruples the loss.
  • In the iron: hysteresis and eddy currents. They are almost constant: they exist even if the equipment is unloaded.
  • In the contacts: a loose or corroded terminal behaves like a series resistance that dissipates power and heats up.
  • Mechanical: friction and ventilation in rotating machines.
Thermography

An infrared camera shows where there is heat without touching anything and with the equipment in service. It is the most efficient diagnostic tool in distribution boards: a loose terminal shows up as a hot spot long before it fails.

The temperature of the point is always compared with that of an equivalent point —the same phase at another terminal— and with the ambient temperature. The difference is what matters, not the absolute value.

Example · The cost of a loose terminal

A terminal with 10 mΩ of contact resistance carrying 30 A dissipates P = I²R = 900 × 0.01 = 9 W at a point the size of a fingernail. Those 9 W concentrated raise the temperature of the terminal by tens of degrees, the heat degrades the insulation, the resistance increases, and the process feeds on itself until it ends in fire. That is why periodic re-tightening of terminals is not a detail: it is essential maintenance.

05In the lab

Lab 1 · Measuring resistivity

With a micro-ohmmeter or the four-point method, measure the resistance of known lengths of conductor of different cross-sections and calculate the resistivity. Compare copper with aluminum and with the table. Verify the increase with temperature by heating a sample.

Lab 2 · Real voltage drop

Build a circuit with 30 m of 1.5 mm² cable and a resistive load. Measure the voltage at the source and at the load with different currents, and compare with the calculation. Repeat with 2.5 mm²: the difference shows on the instrument and on the thermometer.

Lab 3 · Insulation

Measure with a megohmmeter the insulation resistance of new cables, of used cables and of a cable deliberately damaged or wetted. Record the values and establish an acceptance criterion.

Lab 4 · Thermography of a distribution board

With a thermal camera —or a phone attachment— survey a board in service. Identify the hot spots, compare with the corresponding terminal, re-tighten and measure again. Document the before and the after: it is a real maintenance report.

06Common mistakes

MistakeConsequence
Choosing the cross-section only by currentOn long runs the voltage drop goes above 3 % and the equipment works at reduced voltage.
Ignoring grouping factorsSix cables in one conduit do not carry the same as a single one: the whole bundle heats up.
Joining copper and aluminum directlyGalvanic couple: the contact corrodes, the resistance increases and it ends up overheating. Bimetallic terminals exist for this.
Drilling a sealed enclosure without a cable glandThe IP rating is lost and the enclosure fills with water or dust.
Using PVC where halogen-free is requiredIn a fire, the smoke is toxic and corrosive exactly where people are being evacuated.
Solid core instead of laminatedHuge eddy currents: the iron heats up and the efficiency collapses.
Not re-tightening terminalsContact resistance grows, the point heats up and ends in a fault or a fire.

07Self-assessment

Why is aluminum used in overhead lines if it conducts worse than copper?

Because it weighs about a third as much and costs less. For the same current more cross-section is needed, but the whole is still lighter and cheaper, and in an overhead line weight is decisive.

Calculate the voltage drop over 20 m of 1.5 mm² with 10 A, single-phase.

ΔU = 2 × 0.0172 × 20 × 10 / 1.5 = 4.6 V, or 2.1 % of 220 V. Acceptable, but at 30 m it would already exceed the limit.

What does the current-carrying capacity of a conductor depend on?

On the cross-section, the type of insulation and the installation conditions: in free air, in conduit, buried, bundled with others, and at what ambient temperature.

What does IP44 mean and where does it apply?

Protection against solid objects larger than 1 mm and against water splashing from any direction. It applies in bathrooms, laundry rooms and covered porches.

When is halogen-free cable mandatory?

In places where people gather or must be evacuated: schools, hospitals, cinemas, subways. In a fire, PVC gives off dense, toxic and corrosive smoke.

Why is a transformer core made of insulated laminations?

To interrupt the eddy currents that would be induced in a solid core and heat it without producing useful work.

What is the difference between copper losses and iron losses?

Copper losses are I²R and depend on the load —they grow with the square of the current—; iron losses are hysteresis and eddy currents, and are practically constant even if the equipment is unloaded.

What is compared in a thermographic inspection?

The temperature of the suspect point with that of an equivalent point —the same phase at another terminal— and with the ambient temperature. What matters is the difference.

What happens when copper is joined directly to aluminum?

A galvanic couple forms: the contact corrodes, the resistance increases and the point heats up. It is solved with bimetallic terminals and specific pastes.

How much power does a 5 mΩ terminal dissipate at 40 A?

P = I²R = 1,600 × 0.005 = 8 W concentrated in a tiny spot. Enough to degrade the insulation and make the problem get worse by itself.

Development of the topic “Electrical materials technology” of Industrial Installations (Year 6), based on the “Curriculum Proposal – Second Cycle of the Technical-Vocational Track, Secondary Education – Electronics,” Ministry of Education of the Province of Córdoba, DGETyFP. Back to the Topic Map · catto.ar