Answer:
"13.48 Kwhr" is the right solution.
Explanation:
The given values are:
Average rate of heat energy,
= 0.063 W/m²
Diameter,
= 8m
Efficiency of conversion,
= 50%
Now,
The area of hotspot will be:
⇒ [tex]A=\frac{\pi}{4} d^2[/tex]
On substituting the values, we get
⇒ [tex]=\frac{3.14}{4} (8)^2[/tex]
⇒ [tex]=0.785\times 64[/tex]
⇒ [tex]=50.24 \ m^2[/tex]
Total heat generation rate will be:
⇒ [tex]Q=q\times A[/tex]
[tex]=0.063\times 50.24[/tex]
[tex]=3.16 \ W[/tex]
hence,
The electricity generation capacity will be:
⇒ [tex]P=\eta Q t[/tex]
On substituting the values, we get
⇒ [tex]=0.5\times 3.16\times 8760[/tex]
⇒ [tex]=13840.8 \ Whr[/tex]
On converting into Kwhr, we get
⇒ [tex]=13.84 \ Kwhr[/tex]
