Six animated diagrams for the concepts that carry the 100 descriptive marks in Paper-3. Each shows what the formula actually does when a value changes — which is the part that makes a numerical obvious instead of memorised. Every one is grounded in the 2014 BEE guidebook with its section reference.
kW is the work you actually get and it never moves. Adding a capacitor supplies the magnetising kVAr locally, so the vertical side shrinks — and the kVA hypotenuse collapses toward kW. Same work done, smaller kVA, smaller current, lower I²R losses and a smaller maximum-demand bill.
Mark-losing trap: kVAr is not kVA − kW. Use the triangle: kVAr = √(kVA² − kW²), and for correction kVAr = kW(tanφ₁ − tanφ₂) where φ = cos⁻¹(PF).
2 · Transformer losses vs load
📖 Book-3 §1.6 — Transformer losses and efficiency
Total loss = P_iron + (load fraction)² × P_copper(full load)
η = output / (output + losses) η_max when iron loss = copper loss
Iron (no-load) loss is there the moment you energise the transformer and never changes. Copper loss follows the square of load — at half load it is only a quarter of its full-load value. Efficiency peaks exactly where the two lines cross, which for most distribution transformers is around 40–60% load, not at 100%.
Mark-losing trap: at 50% load copper loss is 0.5² = 0.25× the full-load figure, not 0.5×. Squaring the load fraction is the single most common slip in this numerical.
3 · Vapour-compression cycle & COP
📖 Book-3 §4.2–4.3 — Refrigeration cycle, COP and kW/TR
Four components, one loop. The evaporator takes heat in (that is the cooling you are paid for), the compressor puts work in, and the condenser throws the sum of both out. COP is the ratio of what you get to what you pay. kW/TR is the same story upside-down — lower kW/TR is better, higher COP is better.
Mark-losing trap: Carnot COP uses kelvin, never °C — COP = T_evap / (T_cond − T_evap) with T in K. Forgetting the +273 destroys the answer.
📖 Book-3 §7.2 — Range, approach and tower effectiveness
Range = hot water − cold water
Approach = cold water − wet bulb
Effectiveness % = Range / (Range + Approach) × 100
The wet bulb temperature is a wall the water can never cross. Range is how far you actually cooled it; Approach is how much you left on the table. A smaller approach means a better tower — and effectiveness just expresses that as a percentage. Here: 12/(12+5) = 70.6%.
Mark-losing trap: Range uses two water temperatures. Approach compares water to air (wet bulb). Swapping them is the classic error — and examiners set both in the same question.
One speed change, three different exponents. Power is the cube, which is why a small speed reduction produces a large energy saving — and why VFDs pay back so fast on pumps and fans running below design flow.
Mark-losing trap: "Speed doubles, power doubles" is wrong — it is 2³ = 8×. If speed triples, power is 27×. Examiners set exactly this and put the wrong-exponent answer in the options.
Shut the plant down so nothing legitimately consumes air, then time how long the compressor spends loaded (T) versus unloaded (t) over several cycles. Any air it still has to make is escaping through leaks. Book benchmark: leakage above 10% needs attention.
Mark-losing trap: the denominator is T + t, the full cycle — not just t. And the test is only valid with the plant genuinely idle.