--- title: "Break-even hours for two pumps" subject: "Engineering Economics" chapter: 1 key: "semester_6/engineering_economics:breakeven-hours-khasa-sarvo-pump" source: https://www.bctengineer.com/q/semester_6/engineering_economics/breakeven-hours-khasa-sarvo-pump site: BCT Engineering --- # Break-even hours for two pumps Engineering Economics · Chapter 1: Introduction - Asked in: 2070 Magh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/engineering_economics/breakeven-hours-khasa-sarvo-pump - Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-1 ## Question Calculate break-even hours of operation per year to become cost equal and recommend the economic pump if it is to be operated 5 hours daily at full load. | | KHASA Pump | SARVO Pump | |---|---|---| | Capacity | 100 hp | 100 hp | | Purchase cost (Rs.) | 5,00,000 | 10,00,000 | | Tax per year (Rs.) | 10,000 | 15,000 | | Maintenance cost per year (Rs.) | 36,500 | 29,200 | | Efficiency | 80% | 90% | | Life, year | 5 | 5 | | Salvage value | 20% of purchase cost for both | | | MARR | 20% per year | | | Electricity cost | Rs. 16/kwh | | ## Short answer Fixed costs: Khasa Rs 200,252, Sarvo Rs 351,704; energy cost/hour: Rs 1,492 and Rs 1,326.22. Equating: 200252+1492h = 351704+1326.22h gives break-even 914 hours/year. At 5h/day (1,825 h) Sarvo saves ~Rs 1.51 lakh/year and is recommended. ## Answer #### Approach Compare the annual cost of each pump as a function of operating hours h per year: Annual cost = fixed annual cost + (energy cost per hour) × h. Set them equal to get the break-even hours. #### Assumptions 1 hp = 0.746 kW; the motor draws power = rated output ÷ efficiency (full load); interest rate = MARR = 20%; capital cost is converted to an annual figure by capital recovery with salvage. Capital recovery (CR) = (P − S)(A/P, 20%, 5) + S(0.20), where (A/P, 20%, 5) = 0.33438 #### Annual fixed cost | Item | KHASA | SARVO | |---|---|---| | P (Rs) | 500,000 | 1,000,000 | | S = 20% of P (Rs) | 100,000 | 200,000 | | CR (Rs/yr) | 400,000×0.33438 + 100,000×0.2 = 153,752 | 800,000×0.33438 + 200,000×0.2 = 307,504 | | Tax (Rs/yr) | 10,000 | 15,000 | | Maintenance (Rs/yr) | 36,500 | 29,200 | | **Fixed cost (Rs/yr)** | **200,252** | **351,704** | #### Energy cost per hour - KHASA: 100 × 0.746 / 0.80 = 93.25 kW; × Rs 16 = **Rs 1,492 per hour** - SARVO: 100 × 0.746 / 0.90 = 82.89 kW; × Rs 16 = **Rs 1,326.22 per hour** #### Break-even hours 200,252 + 1,492 h = 351,704 + 1,326.22 h 165.78 h = 151,452 **h = 913.6 hours per year** #### Decision for 5 hours daily Hours = 5 × 365 = 1,825 h per year (> 913.6 h). - KHASA: 200,252 + 1,492 × 1,825 = Rs 2,923,152 per year - SARVO: 351,704 + 1,326.22 × 1,825 = Rs 2,772,059 per year **Answer: Break-even = about 914 hours per year. At 1,825 hours the SARVO pump is cheaper (Rs 2.77 million vs 2.92 million per year, saving about Rs 151,000), so SARVO is recommended.** SARVO is better whenever it runs more than about 914 hours per year. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The approach used for break-even hours is: A. Equate purchase costs B. Equate efficiencies C. Equate salvage values D. Equate annual cost = fixed annual cost + hourly energy cost x h for both pumps Answer: D. The two annual cost lines cross at the break-even hours. (Key point: Approach) 2. Energy cost per hour for KHASA is: A. 100 x 0.746 / 0.80 x 16 = Rs 1,492 B. 100 x 0.746 x 0.80 x 16 = Rs 954 C. 100 x 16 = Rs 1,600 D. 100 x 0.746 / 0.90 x 16 = Rs 1,326 Answer: A. Input power = output / efficiency. (Key point: Energy cost) 3. Annual capital cost uses: A. P x 0.20 only B. CR = (P - S)(A/P, 20%, 5) + S(0.20) C. (P + S)/5 D. P/5 Answer: B. Capital recovery with salvage at MARR 20%. (Key point: Capital recovery) 4. At 5 h/day (1,825 h/yr), which is cheaper and why? A. Both equal B. KHASA, since its fixed cost is lower C. SARVO, since 1,825 h exceeds the break-even of about 914 h D. KHASA, since 1,825 h exceeds break-even Answer: C. Above break-even hours, the lower running cost (higher efficiency) pump wins. 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