.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?
Efficiency in terms of power
e=P0Pie=\frac{P_0}{P_i}e=PiP0
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 \;MWPi=AHPi=(105kg/hr)(30Mj/kg)=30×105mJ/hrPi=[30×105mJ/hr][3600s1hr]=833.3Mj/s=833.3MW
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} \\ = 0.6e=PiP0=833.3MW500MW=0.6
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment