SUMMARY
The efficiency of a steam-electric power plant delivering 900 MW of electric power is calculated to be approximately 25%. This conclusion is derived from the heat loss through cooling water, which has a mass flow rate of 629,000 kg/s and a temperature increase of 1.03 °C. The calculations involve using the specific heat capacity of water (4184 J/kg·K) to determine the total heat loss, allowing for an accurate efficiency assessment. The initial miscalculation of 19% was corrected by factoring in the specific heat and total output of the plant.
PREREQUISITES
- Understanding of thermodynamic efficiency calculations
- Knowledge of specific heat capacity, particularly for water (4184 J/kg·K)
- Familiarity with power units, specifically megawatts (MW) and joules per second (J/s)
- Basic principles of heat transfer and energy conservation
NEXT STEPS
- Study thermodynamic efficiency formulas, specifically Efficiency = W/Qh
- Learn about heat transfer principles in power plants
- Explore the impact of cooling systems on power plant efficiency
- Investigate advanced calculations for thermal efficiency in steam-electric plants
USEFUL FOR
Engineering students, energy analysts, and professionals involved in power generation and thermal efficiency optimization will benefit from this discussion.