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Actual load estimation in industrial plants

Short answer

Estimation uses utilization factor ku and simultaneity factor ks. For each load, apparent power = Pn/(η×cosφ). Apply ku (motors 0.75), then ks per group (4 motors: 0.8). Sum at each level, apply main board ks (0.9), add 25% margin. Example: 4×7.5 kW motors, 4 kVA lighting, 10×1 kVA sockets yields 27.06 kVA demand, so a 50 kVA transformer is chosen.

How actual load estimation is done in the industrial plants? Explain with the help of proper examples.

Actual load estimation means finding the maximum power (kVA) that the plant will really draw, not the sum of nameplate ratings. Machines do not run at full load all the time and do not all run together, so the installed power is reduced by suitable factors. The common method (IEC / Schneider Electrical Installation Guide) works level by level.

Steps

  1. List the installed power Pₙ (kW) of each load from nameplates.
  2. Find the installed apparent power of each load:
Pa = Pn / (η × cos φ)    (kVA)

For lamps, add ballast losses.

  1. Apply the factor of maximum utilization ku (≤ 1): the ratio of the actual maximum power a device takes to its rating. Typical: motors 0.75, lighting and heating 1.
  2. Apply the factor of simultaneity ks (≤ 1) for each group, because not all loads in a group run at the same time. Typical for distribution boards (IEC 61439): 2–3 circuits 0.9, 4–5 circuits 0.8, 6–9 circuits 0.7, 10 or more 0.6.
  3. Add up level by level: load → sub-board → main board → transformer. At each level, multiply the sum by that level's ks.
  4. Add a margin for future growth (about 20–25%) and select the next standard transformer size.

Example

A small workshop has:

  • 4 lathe motors, 7.5 kW each, η = 0.88, cos φ = 0.85
  • Lighting load 4 kVA
  • 10 socket outlets of 1 kVA each
Motor:  Pa = 7.5/(0.88×0.85)      = 10.03 kVA
        with ku = 0.75            = 7.52 kVA
        4 motors, ks = 0.8:
        4 × 7.52 × 0.8            = 24.06 kVA
Light:  4 × ku 1 × ks 1           = 4.00 kVA
Socket: 10 × 1 × ks 0.2           = 2.00 kVA
Sum                               = 30.06 kVA
Main board ks = 0.9:
        30.06 × 0.9               = 27.06 kVA
Future margin 25%:
        27.06 × 1.25              = 33.82 kVA

The installed load is 4 × 10.03 + 4 + 10 = 54.1 kVA, but the estimated demand is only about 27 kVA. A 50 kVA transformer (next standard size above 33.8 kVA) is enough, instead of a much larger one sized on connected load.

Why it matters

  • Avoids oversized transformers, cables and switchgear (high cost, poor efficiency at light load).
  • Avoids undersizing, which causes overheating and tripping.
  • Gives the maximum demand used for tariff and contract demand with the utility.

✓ Answer checked by Claude Opus 5.5 (high effort)

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