Question

  Calculate the rate of heat flow per m2 through a furnace wall consisting 200 mm thick inner layer of chrome brick (K= 1.25 W/m2 0C), a centre layer of kaolin brick (K= 0.074 W/m2 0C) 100 mm thick and an outer layer of masonry brick (K= 0.555 W/m2 0C) 100 mm thick. The unit surface conductance at the inner surface is 74 W/m2 0C and the outer surface temperature is 700C. Temperature of gas inside furnace is 16700C. Find temperature of all surfaces.       ​

Answer 1

Answer:

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Answer 2

Given:

Unit surface conductance at the inner surface = 74 W/$m^{2}$ °C

The thickness of the first layer of chrome brick = 200 mm

The conductivity of the first layer = 1.25 W/$m^{2}$ °C

The thickness of the second layer of kaolin brick = 100 mm

The conductivity of the second layer = 0.074 W/$m^{2}$ °C

The thickness of the third layer of masonry brick = 100 mm

The conductivity of the first layer = 0.555 W/$m^{2}$ °C

The temperature inside the furnace = 1670 °C

The temperature outside the furnace = 700 °C

To find:

Rate of heat flow per $m^{2}$

The temperature of all surfaces

Solution:

This is a problem of the type of heat flow through a composite wall. We will use the concept of the overall heat transfer coefficient to find the rate of heat transfer.

The overall heat transfer coefficient will be equal to:

U = $\frac{1}{hf} + \frac{L_{A} }{L_{A} }+ \frac{L_{B} }{L_{B} }+ \frac{L_{C} }{L_{C} }$

Given,

hf = 74 W/$m^{2}$ °C

$L_{A}$ = 200 mm = 0.2 m             $K_{A}$ = 1.25 W/$m^{2}$ °C

$L_{B}$ = 100 mm = 0.1 m               $K_{B}$ = 0.074 W/$m^{2}$ °C

$L_{C}$ = 100 mm = 0.1 m               $K_{C}$ = 0.555 W/$m^{2}$ °C

Putting these values we get

U =  1.7046

Now, we know that

Q = $\frac{A(t_{f} - t_{i}) }{U}$

Where,

Q = Rate of heat transfer

A = surface area

tf =final temperature

ti = initial temperature

U = overall heat transfer coefficient

Now, rate of heat transfer per unit area

q = Q/A =   $\frac{(t_{f} - t_{i}) }{U}$

q = $\frac{1670 - 700 }{1.7046}$

q = 569.048 W/$m^{2}$

The rate of heat flow  $m^{2}$ through the furnace wall will be

569.048 W/$m^{2}$ .

Now,

t1 = temperature on the inner surface of the first layer

q = ti - t1 / hcf

t1 = ti - q/hcf

t1 = 1670 - 7.68

t1 = 1662.3 °C

t2 = temperature between the first and second layer

q = $\frac{t1 - t2}{\frac{lA}{KA} }$

t2 = t1 - q$\frac{LA}{KA}$

t2 = 1662.3 - 91.04

t2 = 1571.25 °C

t3 = temperature between the second and third layer

q = $\frac{t2 - t3}{\frac{lB}{KB} }$

t3 = t2 - q$\frac{LB}{KB}$

t3 = 1571.25 - 768.77

t3 = 802.48 °C

The temperature on the inner surface of the first layer = 1662.3 °C

The temperature between the first and second layer = 1571.25 °C

The temperature between the second and third layer = 802.48 °C