Hot water demand is peaky. A hotel uses most of its water 6–9 am and 7–10 pm. A geyser meets a peak with high power. A heat pump meets it with stored hot water, filled slowly over many hours. The tank is a battery of heat.

Three steps:

1. Find daily hot water energy.
2. Choose running hours (typically 8–16 h per day for hot water; more hours means a smaller unit).
3. Size storage to cover the peak that the heat pump cannot meet in real time.

Worked example — 40-room hotel in Pune

Load: 40 rooms × 2 guests × 60 L at 60 °C = 4,800 L/day plus kitchen 1,200 L = 6,000 L/day. Cold water 25 °C (typical Pune winter). ΔT = 35 K. E_day = 6,000 × 4.186 × 35 / 3600 = 244 kWh/day. Morning peak: 50 % of daily use in 3 hours = 122 kWh. Usable fraction of tank (stratification and mixing) η = 0.8 (typical 0.75–0.9).

Running cost is the same for all three if COP is equal: 244 kWh ÷ COP 3.8 = 64 kWh electricity/day × ₹9 = ₹578/day (example rate). Option B is usually best: moderate tank, room to recover if a second peak comes early, and some spare hours for derating. Check: the evening peak must be recovered in the hours between peaks. Option C needs 74 kWh ÷ 16 kW ≈ 4.6 h refill time — fine if peaks are 10 h apart.

Remember

• Energy per day sets total work; running hours set machine size.
• Storage covers what the machine cannot deliver during the peak.
• Leave 20–30 % spare hours for derating and defrost.
• Check recovery time between peaks.