Question

Consider a plane composite wall that is composed of two materials of thermal conductivities kA 0.1 W/mK and kB 0.04 W/mK and thicknesses LA 10 mm and LB 20 mm. The contact resistance at the interface between the two materials is known to be 0.30 m2 K/W. Material A adjoins a fluid at 200°C for which h = 10 W/m2 K, and material B adjoins a fluid at 40°C for which h = 20 W/m2 K.
a)What is the rate of heat transfer through a wall that is 2.5 meters high by 2 meters side?


b)What is the temperature at the exposed surface of material A?


c)What is the temperature at the exposed surface of material B?


d)Sketch the temperature distribution.

Answer 1

Answer:

(a)  761.9 W

(b) 184.762 °C  

(c) 55.238 °C

(d) see figure

Explanation:

Data

$k_A = 0.1 W/mK$

$k_B = 0.04 W/mK$

$L_A = 0.010 m$

$L_B = 0.020 m$

resistance, $R = 0.30 (m^2 K)/W$

$T_1 = 200 C$

$h_1 = 10 W/m^2 K$

$T_2 = 40 C$

$h_2 = 20 W/m^2 K$

area, $A = 2.5 m \times 2 m = 5 m^2$

(a)

The rate of heat transfer is calculated as

$Q = A \, \frac{1}{R_t} \, (T_1 - T_2) (1)$

Total flux resistance is

$R_t = 1/h_1 + 1/h_2 + L_A/k_A + L_B/k_B + R$

$R_t = (m^2 K)/10 W + (m^2 K)/20 W + 0.010 m (mK)/0.1 W+ 0.020 m (mK)/0.04 W + 0.30 (m^2 K)/W$

$R_t = 1.05 (m^2 K)/W$

From equation 1

$Q = 5 m^2 \, \frac{1}{1.05 (m^2 K)/W} \, (200 - 40) K$

$Q = 761.9 W$

(b)

Between ambient next to material A and material A heat flux is

$Q = A \, h_1 \, (T_1 - T_A)$

$T_A = T_1 - \frac{Q}{A \, h_1}$

$T_A = 200 C - \frac{761.9 W}{5 m^2 \, 10 W/m^2 C}$

$T_A = 184.762 C$

(c)

Between material B and ambient next to material B heat flux is

$Q = A \, h_2 \, (T_B - T_2)$

$T_B = \frac{Q}{A \, h_2}+ T_2$

$T_B = \frac{761.9 W}{5 m^2 \, 20 W/m^2 C} + 40 C$

$T_B = 55.238 C$

(d)

See figured attached