Answer:
q = 1.73 W
Explanation:
given data
small end = 5 cm
large end = 10 cm
high = 15 cm
small end is held = 600 K
large end at = 300 K
thermal conductivity of asbestos = 0.173 W/mK
solution
first we will get here side of cross section that is express as
[tex]S = S1 + \frac{S2-S1}{L} x[/tex] ...............1
here x is distance from small end and S1 is side of square at small end
and S2 is side of square of large end and L is length
put here value and we get
S = 5 + [tex]\frac{10-5}{15} x[/tex]
S = [tex]\frac{0.15 + x}{3}[/tex] m
and
now we get here Area of section at distance x is
area A = S² ...............2
area A = [tex](\frac{0.15 + x}{3})^2[/tex] m²
and
now we take here small length dx and temperature difference is dt
so as per fourier law
heat conduction is express as
heat conduction q = [tex]\frac{-k\times A\ dt}{dx}[/tex] ...............3
put here value and we get
heat conduction q = [tex]-k\times (\frac{0.15 + x}{3})^2 \ \frac{dt}{dx}[/tex]
it will be express as
[tex]q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)[/tex]
now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K
[tex]q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}[/tex]
solve it and we get
q (30) = (0.173) × (600 - 300)
q = 1.73 W
