Answer:
0.00095%
Explanation:
Mass of coal needed every day = Mass needed every hour × 24
[tex]350\times 1000\times 24=8400000\ kg[/tex]
Thermal energy released per hour = Mass × Heat of combustion
[tex]350\times 1000\times 28\times 10^6=9800000000000\ J[/tex]
Power released per second = Energy / 3600 seconds
[tex]P=\frac{E}{t}\\\Rightarrow P=\frac{9800000000000}{3600}\\\Rightarrow P=2722222222.22\ W[/tex]
Thermal efficiency = (Electrical power / Mechanical power) × 100
[tex]\frac{2.6\times 10^6}{2722222222.22}\times 100=0.00095\ \%[/tex]
Thermal efficiency of the plant is 0.00095%