TODAY'S CONCEPT

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.