Question 5 of 6: Electrical Services — Insulation Testing, Power Factor, Site Distribution and Transmission

Question 5: Electrical Services — Insulation Testing, Power Factor, Site Distribution and Transmission (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An insulation-resistance test at 500 V DC across a 1.75 MΩ line-to-earth path; and, for the transmission illustration, a load of 10 MW delivered through a line of total resistance 4 Ω at three different transmission voltages.

Find. The leakage current during the insulation test; a definition of power factor with the reason it matters and the value at which correction becomes worthwhile; a diagram of construction-site electrical distribution; and a numerical demonstration of why transmission efficiency rises with voltage.

(1) Leakage current during the insulation-resistance test

Approach. The insulation path is a pure resistance at DC, so Ohm's law applies directly.

1. Apply Ohm's law to the insulation path. With the test voltage $V$ applied across an insulation resistance $R_{ins}$, $$I = \frac{V}{R_{ins}} = \frac{500}{1.75\times 10^{6}}\ \text{A}$$ $$\boxed{I = 2.857\times 10^{-4}\ \text{A} = 285.7\ \mu\text{A}}$$
2. Judge the result against the acceptance criterion. CSA C22.1 and the equipment standards call for a minimum insulation resistance of 1 MΩ on a 500 V test for circuits up to 500 V, so 1.75 MΩ passes — but not comfortably. Healthy new wiring on a small circuit usually reads tens or hundreds of megohms; a reading under about 2 MΩ is a flag for moisture ingress, a damaged conductor, or simply too many parallel circuits and connected loads left in the test. A quarter of a milliampere is far below the 30 mA operating threshold of a residual-current device, so it would not trip anything — which is exactly why insulation testing is done, rather than relying on protective devices to reveal degraded insulation.

(2) Power factor

The power factor of an AC load is the ratio of the real power it converts into useful work or heat to the apparent power the supply must deliver:

where $\phi$ is the phase angle by which the current lags (or leads) the voltage. A purely resistive load — an incandescent lamp, a heating element — draws current exactly in phase with the voltage and has a power factor of 1. Inductive loads store energy in a magnetic field and return it to the supply twice per cycle: motors, transformers, fluorescent and HID ballasts and welding sets all draw a lagging current, and a lightly loaded induction motor can have a power factor as low as 0.3. Where non-linear loads dominate — variable-frequency drives, LED drivers, computer power supplies — the current is distorted rather than merely shifted, and the true power factor includes a distortion component in addition to the displacement $\cos\phi$.

Why it matters. The current a conductor carries is set by the apparent power, not the real power: $I = S/(\sqrt{3}\,V_L)$ for a three-phase system. A 400 kW load at a power factor of 0.75 draws 533 kVA, whereas the same 400 kW at 0.95 draws 421 kVA — a 21% reduction in current. That current difference has four consequences, all of them expensive. Cables, switchgear, transformers and generators must be sized for the kVA, so poor power factor buys larger and more costly equipment for no extra useful output. Distribution losses go as $I^2R$, so the 21% current reduction cuts losses by 38%. Voltage drop along feeders is proportional to current, so poor power factor causes sagging voltage at the far end of long runs, which in turn makes motors run hotter. And utilities bill for it: BC Hydro and most Canadian utilities levy a demand charge on kVA, or apply a power-factor penalty below a threshold, so the customer pays directly.

When to correct. The usual industry threshold is that a power factor below about 0.90 lagging is considered low and worth correcting, and most utility tariffs set their penalty threshold at 0.90 or 0.95. Correction is normally by shunt power-factor-correction capacitors, switched in banks at the main switchboard or fitted locally at large motors, sized to supply the reactive power $Q = P(\tan\phi_1 - \tan\phi_2)$ needed to move from the existing angle to the target. Two cautions belong in the answer: capacitors must not be applied so aggressively that the system goes leading, which raises voltage and can cause self-excitation of motors on shutdown; and on a system with significant harmonic content, plain capacitors can resonate with the supply inductance, so detuned (reactor-connected) banks or active filters are used instead.

(3) Electricity distribution on a construction site

Temporary site distribution is a radial system, deliberately simple, arranged so that every worker-accessible circuit is at the lowest practical voltage and is protected by ground-fault detection. It is governed in Canada by Section 76 of CSA C22.1 (Canadian Electrical Code, Part I) together with Rule 26-700(11) for receptacle ground-fault protection, and by the provincial OH&S regulations (in BC, WorkSafeBC OHS Regulation Part 19).

Reading the diagram from the supply end: power arrives from the utility distribution feeder (or, on remote sites, from a diesel generator, which is bonded and grounded in exactly the same way). It passes through the service entrance — metering, a lockable main disconnect and the main overcurrent device — where the system bonding jumper connects the neutral to the grounding electrode. From the main site distribution board the supply runs radially as sub-mains: dedicated feeders at 600 V three-phase for the heavy fixed plant (tower crane, material and personnel hoists, concrete plant, welding sets), and a feeder to a dry-type step-down transformer that produces 120/208 V for everything a worker will touch. Sub-boards and portable outlet centres distribute from there to final circuits: hand tools, task and egress lighting, temporary heating and the site accommodation. Cables are routed overhead or in protected runs, never where they can be run over or trapped, and every enclosure is weatherproof and rated for the location.

Three protection features are what distinguish a site installation from a permanent one, and all three should appear on the diagram. First, grounding and bonding: one grounding electrode at the service, one system bonding jumper, and a bonding conductor run with every feeder so that every metal enclosure, every board, the crane mast and the hoist structure are all at earth potential. Second, ground-fault protection: Class A GFCIs (5 mA trip) on every 125 V, 15 and 20 A receptacle used by workers, because a construction site is a wet, conductive, damaged-cord environment in which the shock risk is what actually kills. Third, an assured equipment grounding conductor programme — scheduled inspection and continuity testing of every cord set, tool and receptacle, with results recorded — because on a live site the grounding conductor is the component most likely to be broken. Voltage reduction is the other standard control: where practical, hand tools and task lighting are supplied at reduced voltage from a centre-tapped isolating transformer so that no conductor is more than half the nominal voltage above earth.

(4) Why high-voltage transmission is more efficient

Approach. Transmit the same real power at different voltages, note that the current falls in inverse proportion, and evaluate the $I^2R$ loss in the line resistance, which is fixed by the conductor and not by the voltage.

1. Write the current required to deliver a fixed power. For a given real power $P$ delivered at voltage $V$ and power factor $\mathrm{pf}$, $$I = \frac{P}{V\cdot \mathrm{pf}}$$ so at unity power factor the current is simply inversely proportional to the transmission voltage. Doubling the voltage halves the current; raising it tenfold cuts the current to a tenth.
2. Write the loss, which depends only on the current and the conductor. The line loses power by resistive heating: $$P_{loss} = I^2 R_{line} = \left(\frac{P}{V\cdot \mathrm{pf}}\right)^{\!2} R_{line} \;\propto\; \frac{1}{V^2}$$ $$\boxed{P_{loss} \propto \frac{1}{V^{2}}\ \text{for a fixed delivered power and a fixed conductor}}$$ This is the whole answer in one line: the loss falls with the square of the transmission voltage, because the resistance of the copper or aluminium is a property of the conductor and is not changed by raising the voltage.
3. Put numbers to it. Take 10 MW delivered at unity power factor over a line of total resistance 4 Ω. At 10 kV, $$I = \frac{10\times10^{6}}{10\times10^{3}} = 1000\ \text{A}, \qquad P_{loss} = 1000^2\times 4 = 4.00\ \text{MW}$$ At 100 kV the current is 100 A and $P_{loss} = 100^2\times 4 = 40$ kW; at 400 kV the current is 25 A and $P_{loss} = 25^2\times 4 = 2.5$ kW. A tenfold rise in voltage has cut the loss by a factor of one hundred.
4. Express the results as transmission efficiency. Taking efficiency as delivered power over sent power, $\eta = P/(P + P_{loss})$: $$\eta_{10\,\text{kV}} = \frac{10}{14} = 71.4\%, \qquad \eta_{100\,\text{kV}} = \frac{10}{10.04} = 99.60\%, \qquad \eta_{400\,\text{kV}} = \frac{10}{10.0025} = 99.975\%$$ The 10 kV case is not merely inefficient, it is unbuildable: four megawatts dissipated in the line would melt any conductor of that resistance.

Two secondary benefits follow from the same relation and are worth a sentence each. Because the loss for a given acceptable percentage can now be met with a much smaller conductor, high-voltage transmission also saves an enormous amount of aluminium and steel — the capital saving on conductors and towers is often larger than the energy saving. And because voltage drop along the line is $IR$, the smaller current also gives better voltage regulation at the receiving end. The limits on the argument are insulation and clearance: raising the voltage raises the cost of insulators, transformers, switchgear and right-of-way width, and above a few hundred kilovolts corona loss and radio interference start to matter. The optimum transmission voltage is therefore an economic balance, which is why BC Hydro's backbone runs at 500 kV, its regional lines at 230 and 138 kV, distribution at 25 and 12.5 kV, and the service to a building at 600 V or less.