- 2073 Magh · 2+6 marks
Maximum demand and need to calculate it
Short answer
Maximum demand (MD) is the highest average load (kW/kVA) over a 15-30 min period. It sizes transformers, cables, switchgear; sets contract demand and demand charges; guides protection settings; enables load shifting to reduce MD; aids utility planning; and determines capacitor sizing. Example: factory with 80 kW connected load has actual MD 60 kW, leading to 100 kVA transformer and 70 kVA contract demand.
What is Maximum Demand? With suitable example, explain the need to calculate Maximum Demand in an industrial building.
Maximum demand (MD) is the greatest load (kW or kVA) drawn by an installation during a given period. It is usually measured as the highest average demand over a short interval, normally 15 or 30 minutes, by an MD meter. It is always less than or equal to the connected load.
Need to calculate maximum demand in an industrial building
- Sizing the transformer and generator: the substation transformer and standby DG set are chosen to supply the maximum demand plus a margin, not the total connected load. This avoids heavy oversizing.
- Sizing cables, busbars and switchgear: the main incomer, busbars and feeders must carry the maximum demand current without overheating.
- Contract demand and tariff: industrial consumers pay a demand charge (Rs per kVA or kW of MD) plus an energy charge. Knowing the MD lets the plant choose the right contract demand and avoid penalty for exceeding it.
- Protection settings: relay and breaker settings are based on the maximum load current.
- Demand control: knowing when MD occurs lets the plant shift loads (for example run pumps or furnaces off-peak) to lower the MD and the bill.
- Planning: the utility needs the MD to design the feeder and to plan generation.
- Power factor correction: capacitor sizing uses the kW and kVA at maximum demand.
Example
A garment factory has a connected load of:
| Load | Rating |
|---|---|
| 50 sewing machines | 0.4 kW each = 20 kW |
| 2 boilers/irons | 30 kW |
| Lighting and fans | 15 kW |
| Compressor | 15 kW |
| Total connected | 80 kW |
If 80% of the sewing machines, both boilers, all lighting and the compressor were assumed to run together at full rating, the worst-case estimate would be:
0.8×20 + 30 + 15 + 15 = 76 kW
But the boilers and the compressor switch on and off with their thermostats and pressure switches, so they are not all at full load in the same 30 minutes. MD meter records show the actual peak is about 60 kW. Then:
- Demand factor = 60/80 = 0.75.
- A transformer of about 60/0.85 ≈ 71 kVA plus margin → 100 kVA is chosen, not one based on 80 kW plus large margins.
- The contract demand is fixed near 70 kVA, saving demand charges every month.
✓ Answer checked by Claude Opus 5.5 (high effort)
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