Question

The composite wall of an oven consists of threematerials, two of which are of known thermal conductivity,k_A = 20 W/m•K and k_C =50 W/(m•K), and known thickness, L_A =0.30m and L_C = 0.15m. The third material,B, which is sandwiched between materials A andC, is known thickness, L_B = 0.15m, butunknown thermal conductivity k_B.



Under steady-state operating conditions, measurementsreveal an outer surface temperature of T_s,o =20ºC, an inner surface temperature of T_s,i= 600ºC, and an oven air temperature ofT_¥=800ºC. The inside convection coefficienth is known to be 25 W/(m²•K). What isthe value of k_B?

Answer given k_B = 1.53W/(m×K)

Answer 1

Concepts and Reason

Plane walls sometimes consist of several layers of different materials. The concept of thermal resistance can be used to determine the rate of steady heat transfer through such ‘composite walls’.

The thermal resistance is analogous to electrical resistance and follows the series and parallel law.

The convective resistance is used at the peripheries of the wall and conductive resistance is used between the walls of similar material.

Fundamentals

The composite wall (consisting of two plane walls in series) and its thermal resistance network for heat transfer are as shown:

The formulae for thermal resistances are:

a. Convective resistance at wall 1:

${R_{conv,1}} = \frac{1}{{A{h_1}}}$

Here, the area of cross section is A and the convective coefficient at wall 1 is ${h_1}$ .

b. Conductive resistance in wall 1:

${R_1} = \frac{{{L_1}}}{{A{k_1}}}$

Here, the length of wall 1 is ${L_1}$ and its thermal conductivity is ${k_1}$ .

c. Conductive resistance in wall 2:

${R_2} = \frac{{{L_2}}}{{A{k_2}}}$

Here, the length of wall 2 is ${L_2}$ and its thermal conductivity is ${k_2}$ .

d. Convective resistance at wall 2:

${R_{conv,2}} = \frac{1}{{A{h_2}}}$

Here, the convective coefficient at wall 2 is ${h_2}$ .

The rate of heat transfer can be calculated as,

$\dot Q = \frac{{{T_{\infty 1}} - {T_{\infty 2}}}}{{{R_{conv,1}} + {R_1} + {R_2} + {R_{conv,2}}}}$

Draw the thermal circuit.

Obtain the thermal resistances per unit area.

${R_{conv,1}} = \frac{1}{h}$

Here, the convective coefficient is h.

${R_A} = \frac{{{L_A}}}{{{k_A}}}$

Here, the length and thermal conductivity of wall A are ${L_A}$ and ${k_A}$ respectively.

${R_B} = \frac{{{L_B}}}{{{k_B}}}$

Here, the length and thermal conductivity of wall B are ${L_B}$ and ${k_B}$ respectively.

${R_C} = \frac{{{L_C}}}{{{k_C}}}$

Here, the length and thermal conductivity of wall C are ${L_C}$ and ${k_C}$ respectively.

Calculate the heat flux.

$\dot q = \frac{{{T_\infty } - {T_{{\rm{S,i}}}}}}{{{R_{conv,1}}}}$

Here, the temperature of air is ${T_\infty }$ and the temperature of the inner wall is ${T_{{\rm{S,i}}}}$ .

Substitute $800^\circ {\rm{C}}$ for ${T_\infty }$ , $\frac{1}{h}$ for ${R_{conv,1}}$ , $25{\rm{ W/}}{{\rm{m}}^2} \cdot {\rm{K}}$ for h, and $600^\circ {\rm{C}}$ for ${T_{{\rm{S,i}}}}$ .

$\begin{array}{c}\\\dot q = \frac{{800 - 600}}{{1/h}}\\\\ = \frac{{800 - 600}}{{1/25}}\\\\ = 5000{\rm{ W/}}{{\rm{m}}^2}\\\end{array}$

Calculate the thermal conductivity of material B.

$\dot q = \frac{{{T_{{\rm{S,i}}}} - {T_{{\rm{S,o}}}}}}{{{R_A} + {R_B} + {R_C}}}$

Here, the outer wall temperature is ${T_{{\rm{S,o}}}}$ .

Substitute $5000{\rm{ W/}}{{\rm{m}}^2}$ for $\dot q$ , $\frac{{{L_A}}}{{{k_A}}}$ for ${R_A}$ , $\frac{{{L_B}}}{{{k_B}}}$ for ${R_B}$ , $\frac{{{L_C}}}{{{k_C}}}$ for ${R_C}$ , $600^\circ {\rm{C}}$ for ${T_{{\rm{S,i}}}}$ , $20^\circ {\rm{C}}$ for ${T_{{\rm{S,o}}}}$ , $20{\rm{ W/m}} \cdot {\rm{K}}$ for ${k_A}$ , $50{\rm{ W/m}} \cdot {\rm{K}}$ for ${k_C}$ , 0.3 m for ${L_A}$ , 0.15 m for ${L_B}$ , and 0.15 m for ${L_C}$ .

$\begin{array}{l}\\5000 = \frac{{600 - 20}}{{\frac{{{L_A}}}{{{k_A}}} + \frac{{{L_B}}}{{{k_B}}} + \frac{{{L_C}}}{{{k_C}}}}}\\\\\frac{{0.3}}{{20}} + \frac{{0.15}}{{{k_B}}} + \frac{{0.15}}{{50}} = 0.116\\\\\frac{{0.15}}{{{k_B}}} = 0.098\\\\{k_B} = 1.53{\rm{ W/m}} \cdot {\rm{K}}\\\end{array}$

Ans:

The thermal conductivity of material B, ${k_B}$ is $1.53{\rm{ W/m}} \cdot {\rm{K}}$ .