Electrical Energy Generation, Transmission and Utilization: Applied Calculations
What this domain tests
This companion domain focuses on the applied side of the same syllabus: load estimation, energy auditing, efficiency comparisons and simple payback. It rewards candidates who can connect generation, transmission and end use with numbers rather than descriptions alone.
Core rules
- Load factor = average load/maximum demand; a high load factor spreads fixed costs over more units and improves tariff economics.
- Diversity factor recognises that not all connected loads operate at once; it is used when sizing distribution equipment.
- Energy saved by an efficient device = (input power × hours × efficiency improvement), converted to kWh and then to rupees using the tariff.
- For motors, output power = input power × efficiency; for pumps and fans, shaft power relates to flow and head, and affinity laws show that reducing speed cuts power sharply.
- Simple payback period = extra capital cost/annual saving; it ignores discounting but is widely used in small audits.
Worked example
A 10 kW motor running 8 h a day at 85% efficiency draws 10/0.85 = 11.76 kW input. Replacing it with a 92% efficient motor gives 10/0.92 = 10.87 kW, saving 0.89 kW, or about 7.1 kWh per 8-hour day. At ₹8 per kWh and 300 working days, annual saving is about ₹17,000; if the new motor costs ₹51,000 extra, payback is roughly three years.
Traps that cost marks
Applying efficiency the wrong way round (dividing output by efficiency instead of using it to find input); mixing daily and annual energy; and comparing motors on rated output alone without checking actual loading.