Question

A steam power plant receives heat from a furnace at a rate of 280 GJ/h. (that is

Answer 1

Answer:

a) $\dot W_{out} = 35.278\,MW$, b) $\eta_{th} = 45.357\,\%$

Explanation:

a) The net power output is:

$\dot W_{out} = \dot Q_{in} - \dot Q_{out} - \dot Q_{loss}$

$\dot W_{out} = \left(280\,\frac{GJ}{h} - 145\,\frac{GJ}{h} - 8\,\frac{GJ}{h}\right)\cdot \left(\frac{1000\,MJ}{1\,GJ}\right)\cdot \left(\frac{1\,h}{3600\,s} \right)$

$\dot W_{out} = 35.278\,MW$

b) The thermal efficiency of the power plant is:

$\eta_{th} = \frac{\dot Q_{in}-\dot Q_{out} - \dot Q_{loss}}{\dot Q_{in}}\times 100\,\%$

$\eta_{th} = \frac{280\,\frac{GJ}{h}-145\,\frac{GJ}{h}-8\,\frac{GJ}{h} }{280\,\frac{GJ}{h} } \times 100\,\%$

$\eta_{th} = 45.357\,\%$