Question

An asbestos pad is square in cross section, measuring 5 cm on a side at its small end increasing linearly to 10 cm on a side at the large end. The pad is 15 cm high. If the small end is held at 600 K and the large end at 300 K, what heat-ow rate will be obtained if the four sides are insulated? Assume onedimensional heat conduction. The thermal conductivity of asbestos may be taken as 0:173W/mK:

Answer 1

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

$S = S1 + \frac{S2-S1}{L} x$     ...............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 + $\frac{10-5}{15} x$

S = $\frac{0.15 + x}{3}$    m

and  

now we get here Area of section at distance x is

area A = S²    ...............2

area A = $(\frac{0.15 + x}{3})^2$    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 = $\frac{-k\times A\ dt}{dx}$      ...............3

put here value and we get

heat conduction q = $-k\times (\frac{0.15 + x}{3})^2 \ \frac{dt}{dx}$  

it will be express as

$q \times \frac{dx}{(\frac{0.15 + x}{3})^2} = -k (dt)$  

now we intergrate it with limit 0 to 0.15 and take temp 600 to 300 K

$q \int\limits^{0.15}_0 {\frac{dx}{(\frac{0.15 + x}{3})^2 } = -0.173 \int\limits^{300}_{600} {dt}$          

solve it and we get

q (30)  = (0.173) × (600 - 300)

q = 1.73 W