Question

5. Find the heat flow rate

through the composite wall as shown in

Fig. 15.12. Assume one dimensional flow.

kA = 150 W/m°C,

kB = 30 W/m°C,

kC = 65 W/m°C and

kD = 50 W/m°C.

Answer 1

The thermal circuit for heat flow in the given composite system (shown in Fig. 15.12) has been illustrated in Fig. 15.13.

Thickness :

$L_A= 3\; cm= 0.03 \;m \\ L_B=L_C=8 \;cm = 0.08 \;m \\ L_D= 5 \;cm = 0.05 \;m$

Areas :

$A_A= 0.1 \times 0.1 = 0.01 \;m^2 \\ A_B= 0.1 \times 0.03 = 0.003 \;m^2 \\ A_C= 0.1 \times 0.07 = 0.007 \;m^2 \\ A_D= 0.1 \times 0.1 = 0.01 \;m^2$

Heat flow rate, Q :

The thermal resistances are given by

$R_{th-A}= \frac{L_A}{k_AA_A}= \frac{0.03}{150 \times 0.01}= 0.02 \\ R_{th-B}= \frac{L_B}{k_BA_B}= \frac{0.08}{30 \times 0.003}= 0.89 \\ R_{th-C}= \frac{L_C}{k_CA_C}= \frac{0.08}{65 \times 0.007}= 0.176 \\ R_{th-D}= \frac{L_D}{k_DA_D}= \frac{0.05}{50 \times 0.01}= 0.1$

The equivalent thermal resistance for the parallel thermal resistances $R_{th–B}$ and $R_{th–C}$ is given by:

$\frac{1}{(R_{th})_{eq.}} = \frac{1}{R_{th-B}} + \frac{1}{R_{th-C}} = \frac{1}{0.89} + \frac{1}{0.176} = 6.805 \\ (R_{th})_{eq.}= \frac{1}{6.805}= 0.147$

Now, the total thermal resistance is given by

$(R_{th})_{total}= R_{th-A} + (R_{th})_{eq.} + R_{th-D} \\ = 0.02+ 0.147+0.1=0.267 \\ Q= \frac{(∆t)_{overall}}{(R_{th})_{total}}= \frac{400-60}{0.267}= 1273.4 \;W$