Rewind or Replace a 200 kW Pump Motor
A plant's 200 kW raw-water pump runs 6 000 h/yr at an average load of 75% of rated output. Two options are on the table. Option A: rewind the existing motor for P180 000; its full-load efficiency after rewinding is 90%, and it will need rewinding again every five years. Option B: buy a premium-efficiency motor for P620 000 with 94% efficiency, a 15-year life and a salvage value of P60 000. Energy costs P8.50/kWh, the plant MARR is 12%, and tax and depreciation are to be ignored. Both motors are the same frame size, so installation cost is identical and can be omitted. The plant manager is reluctant to spend P620 000 to 'save a few percent on the power bill'.
1. Compute the annual energy consumption and the annual savings of Option B over Option A.
Read the worked answer
Shaft output = 0.75 × 200 = 150 kW
Option A input = 150/0.90 = 166.67 kW → 166.67 × 6 000 = 1 000 000 kWh/yr → P8 500 000/yr Option B input = 150/0.94 = 159.57 kW → 159.57 × 6 000 = 957 447 kWh/yr → P8 138 298/yr
Annual saving = 42 553 kWh × P8.50 = P361 700/yr
That is 4.3% of the power bill for this motor - small in percentage terms but large in absolute peso terms, which is exactly why a percentage-based judgement ('a few percent') is the wrong lens here.
2. Compare the two options on present worth over 15 years at 12% and make the recommendation.
Read the worked answer
Discount factors at 12%: (P/A,12%,15) = 6.8109; (P/F,12%,5) = 0.5674; (P/F,12%,10) = 0.3220; (P/F,12%,15) = 0.1827.
Incremental analysis (B minus A): Extra first cost: 620 000 − 180 000 = −P440 000 Rewinds avoided in years 5 and 10: +180 000 (0.5674 + 0.3220) = +P160 092 Salvage on B at year 15: +60 000 × 0.1827 = +P10 962 Net incremental investment = 440 000 − 160 092 − 10 962 = P268 946 PV of energy savings = 361 700 × 6.8109 = P2 463 500
Incremental NPW = 2 463 500 − 268 946 = +P2 194 600 → choose Option B.
Simple payback is 440 000/361 700 = 1.2 years, and if the avoided rewinds are credited the effective payback is under a year. A 15-year asset that returns its premium in about 12 months is not a marginal decision; the only real question is whether the plant can fund it.
3. Test the decision against the plant manager's likely objection that run hours may fall. Find the break-even operating hours.
Read the worked answer
At 3 000 h/yr the saving halves to P180 850/yr; PV = 180 850 × 6.8109 = P1 231 700, still far above the P268 946 net investment. NPW ≈ +P963 000 - the decision is unchanged.
Break-even: Required annual saving = 268 946/6.8109 = P39 490/yr Break-even hours = 6 000 × 39 490/361 700 = ≈655 h/yr
So the premium motor is justified as long as the pump runs more than about 655 hours a year - roughly 11 hours a week. Below that, the rewind is the better use of capital.
Caveats to state with the recommendation: (a) the 90% and 94% figures should be guaranteed by the repair shop and the vendor respectively, since a 1-point efficiency error moves the saving by about P90 000/yr; (b) at 75% load, part-load efficiency differs from nameplate, so verify from the actual curves; (c) the five-year rewind cycle for Option A assumes the winding fails from thermal ageing, which is reasonable for a continuously loaded 90% machine but should be checked against the failure history.