Answer to Question #249012 in Molecular Physics | Thermodynamics for Prince

Question #249012

.A coal power plant consumes 100000 kg of coal per hour and produces 500 MW of power. If the heat combustion of coal is 30 MJ/kg, what is the efficiency of the power plant?


1
Expert's answer
2021-10-10T16:10:13-0400

Efficiency in terms of power

e=P0Pie=\frac{P_0}{P_i}

The power input by the coal is

Pi=AHPi=(105  kg/hr)(30  Mj/kg)=30×105  mJ/hrPi=[30×105  mJ/hr][1  hr3600  s]=833.3  Mj/s=833.3  MWP_i =AH \\ P_i = (10^5 \;kg/hr)(30\;Mj/kg) \\ = 30 \times 10^5 \;mJ/hr \\ P_i = [30 \times 10^5 \;mJ/hr][\frac{1 \;hr}{3600 \;s}] \\ = 833.3 \;Mj/s \\ = 833.3 \;MW

The efficiency of power plant is

e=P0Pi=500  MW833.3  MW=0.6e =\frac{P_0}{P_i} \\ = \frac{500 \;MW}{833.3 \;MW} \\ = 0.6


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