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

q = 1.73 W

Step-by-step 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 + (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 +
`(10-5)/(15) x`

S =
`(0.15 + x)/(3)` m

and

now we get here Area of section at distance x is

area A = S² ...............2

area A =
`((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 =
`(-k* A\ dt)/(dx)` ...............3

put here value and we get

heat conduction q =
`-k* ((0.15 + x)/(3))^2 \ (dt)/(dx)`

it will be express as

`q * (dx)/(((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 {(dx)/(((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