Choosing a Hostel Water Pump
A hostel committee must replace a worn-out water pump that lifts water from a ground tank to a rooftop tank 20 m above. The hostel needs 50,000 litres per day. Two pumps are quoted, both rated at 0.75 kW and suitable for a 20 m head. Pump A delivers 100 litres per minute and costs NPR 18,000; Pump B delivers 150 litres per minute and costs NPR 26,000. Electricity costs NPR 13 per kWh. Pump A is expected to need NPR 3,000 per year in maintenance, while Pump B needs NPR 4,000 per year. The committee can run the pump only during off-peak hours, but the rooftop tank holds only 10,000 litres, so the pump cannot run continuously for more than about two hours without the tank overflowing unless the daily demand is drawn down concurrently. The committee wants a payback-period comparison and must also consider noise, voltage fluctuation and spare-part availability. Assume 365 operating days and ignore motor efficiency differences beyond the quoted ratings.
1. Calculate each pump's daily operating time, daily energy use and annual electricity cost. Use 365 days.
Read the worked answer
Pump A: daily volume 50,000 L at 100 L/min gives 500 min = 8.33 h. Energy = 0.75 kW × 8.33 h = 6.25 kWh/day. Annual energy = 6.25 × 365 = 2,281 kWh. Annual electricity cost = 2,281 × NPR 13 ≈ NPR 29,656. Pump B: 50,000 L at 150 L/min gives 333.3 min = 5.56 h. Energy = 0.75 kW × 5.56 h = 4.17 kWh/day. Annual energy = 4.17 × 365 = 1,521 kWh. Annual electricity cost = 1,521 × NPR 13 ≈ NPR 19,771. The higher flow of Pump B shortens its running time enough to reduce electricity use despite the same rated power.
2. Find the simple payback period for choosing Pump B, including maintenance. Then check whether each pump can meet the daily demand within a 6-hour off-peak window, given the 10,000-litre tank.
Read the worked answer
Annual operating cost of A = 29,656 + 3,000 = NPR 32,656. Annual operating cost of B = 19,771 + 4,000 = NPR 23,771. Annual saving from B = 32,656 - 23,771 = NPR 8,885. Extra capital cost of B = 26,000 - 18,000 = NPR 8,000. Simple payback = 8,000/8,885 ≈ 0.90 year, or about 10.8 months. Off-peak check: in 6 h = 360 min, Pump A can deliver 100 × 360 = 36,000 L, less than the 50,000 L daily demand; Pump B can deliver 150 × 360 = 54,000 L, so it can meet the demand hydraulically. However, the 10,000-litre tank is only 20% of daily demand, so neither pump can simply fill the tank once and stop. The tank must be drawn down while pumping, or filled in several cycles; if no concurrent draw is allowed, both options fail the storage constraint and a larger tank or daytime pumping is needed.
3. Recommend a pump. Address at least two non-economic factors and state one condition under which your recommendation would change.
Read the worked answer
Recommend Pump B if its quoted 150 L/min is achievable at the actual 20 m head. It pays back its extra capital in under a year, and it is the only one that can move 50,000 L within a 6-hour off-peak window at the rated flow. Pump A would need 8.33 h, so it would force some pumping into peak tariff periods or require a larger tank. Non-economic factors: Pump B's faster fill means more frequent start-stop cycles, so the committee should check whether a level controller or small pressure tank is needed to avoid motor overheating; spare-part availability matters because an imported pump with no local service could cause long outages even if its energy cost is lower; noise and voltage fluctuation should be checked at the actual installation because both pumps are single-phase 0.75 kW units and weak wiring could cause voltage dips. Recommendation would change if Pump B's local spare-part support is poor, if its real flow at 20 m is much below the quoted value, or if the off-peak window is removed so that the shorter runtime no longer gives a tariff advantage.