### 2018-03-13

CHEMICAL ENGINEERING : HEAT TRANSFER

• Q:1
• As asbestos pad, square in cross-section, measure 0.05 m on a side and increases linearly to 0.1 m on the other side. The lenght  of the pad is 0.15 m. If the small end is held at 600 K the larger end at 300 K, what will be the heat flow if the other four sides are insulated? Assume one dimensional heat flow. Thermal conductivity of asbestos is 0.173 W/mK.

• Solution :
• Heat flow rate,  $$\hat {Q} = -kA \frac {dT}{dx}$$  ..............(1)
In this probllem, area is the variable and it varies with length.

Let a be the side of the square face at any lenght. The side of the square can be represented as the function of length (x) of the pad as below:

$$a = a_0 \space + \space bx$$$$here \space a_0 = 0.05 m$$

We need to find b, so taking a = 0.1 m and l = 0.15 m we get:

$$=> 0.1 = 0.05 \space + \space b \times 0.15$$

$$=> b = \frac{1}{3}$$

So, area equation becomes $$a= 0.05 \space + \space \frac{1}{3}x$$       ..............(2)

From equation (1) we can write:

$$\hat {Q} \frac{dx}{A} = -kdT$$  $$=> \hat{Q} \frac{dx}{(0.05 \space + \space \frac{1}{3}x)^2} = -kdT$$
Integrating both sides with proper limits:

$$=> \hat{Q} \int ^{x_1}_{x_2}\frac{dx}{(0.05 \space + \space \frac{1}{3}x)^2} =\int^{300}_{600} -kdT$$

$$=> 3\hat{Q} \int ^{x_1}_{x_2}\frac{\frac{1}{3}dx}{(0.05 \space + \space \frac{1}{3}x)^2} = k(600 \space - \space 300)$$   ........(3)

Let l = 0.05 + (1/3)x, differentiating both sides we get:

dl = (1/3)dx

Substituting the above value in (3) we get:

$$=> 3\hat{Q} \int ^{0.15}_{0}\frac{dl}{(l)^2} = 300 \times 0.173$$

$$=> 3\hat{Q} [\frac{-1}{0.05 \space + \space \frac{1}{3}x}] ^{0.15}_{0} = 51.9$$

$$=> \hat {Q} [\frac{-1}{1/10} \space + \space [\frac{1}{0.05}]] = 51.9$$

$$=> \hat{Q} = {51.9\over 10} = 5.19$$

Therefore, heat flow rate $$\hat{Q}$$ = 51.9 W.