Questions: Uma parede é composta por 5 diferentes materiais e com L1 = 2cm, L2 = 6 cm e L3=4 cm, conforme a figura abaixo. Sabe-se que a temperatura externa do material A é 370°C enquanto TE=60°C, kA=175 W / m°C, kB=35 W / m°C, kC=60 W / m°C, kD=50 W / m°C e kE=80 W / m°C. As áreas (transversais ao fluxo de calor) dos materiais A e E valem 0,3 m² e as áreas dos materiais B, C e D são iguais a 0,1 m². Desenhe o esquema de resistências e calcule o fluxo de calor transferido considerando esse fluxo unidimensional.

Solution Steps

First, we need to calculate the thermal resistances of each material. The thermal resistance \( R \) is given by:

\[ R = \frac{L}{k \cdot A} \]

where \( L \) is the thickness, \( k \) is the thermal conductivity, and \( A \) is the cross-sectional area.

For material A: \[ R_A = \frac{L_1}{k_A \cdot A_A} = \frac{0.02 \, \text{m}}{175 \, \text{W/m}^\circ\text{C} \cdot 0.3 \, \text{m}^2} = 3.81 \times 10^{-4} \, \text{m}^2\text{C/W} \]

For material B: \[ R_B = \frac{L_2}{k_B \cdot A_B} = \frac{0.06 \, \text{m}}{35 \, \text{W/m}^\circ\text{C} \cdot 0.1 \, \text{m}^2} = 0.0171 \, \text{m}^2\text{C/W} \]

For material C: \[ R_C = \frac{L_2}{k_C \cdot A_C} = \frac{0.06 \, \text{m}}{60 \, \text{W/m}^\circ\text{C} \cdot 0.1 \, \text{m}^2} = 0.01 \, \text{m}^2\text{C/W} \]

For material D: \[ R_D = \frac{L_2}{k_D \cdot A_D} = \frac{0.06 \, \text{m}}{50 \, \text{W/m}^\circ\text{C} \cdot 0.1 \, \text{m}^2} = 0.012 \, \text{m}^2\text{C/W} \]

For material E: \[ R_E = \frac{L_3}{k_E \cdot A_E} = \frac{0.04 \, \text{m}}{80 \, \text{W/m}^\circ\text{C} \cdot 0.3 \, \text{m}^2} = 1.67 \times 10^{-3} \, \text{m}^2\text{C/W} \]

Next, we calculate the equivalent thermal resistance of the system. Materials B, C, and D are in parallel, so their combined resistance \( R_{BCD} \) is:

\[ \frac{1}{R_{BCD}} = \frac{1}{R_B} + \frac{1}{R_C} + \frac{1}{R_D} \]

\[ \frac{1}{R_{BCD}} = \frac{1}{0.0171} + \frac{1}{0.01} + \frac{1}{0.012} \]

\[ R_{BCD} = 0.0036 \, \text{m}^2\text{C/W} \]

The total resistance \( R_{total} \) is the sum of the resistances of materials A, BCD, and E:

\[ R_{total} = R_A + R_{BCD} + R_E \]

\[ R_{total} = 3.81 \times 10^{-4} + 0.0036 + 1.67 \times 10^{-3} \]

\[ R_{total} = 0.00565 \, \text{m}^2\text{C/W} \]

Finally, we calculate the heat flux \( q \) using the temperature difference and the total thermal resistance:

\[ q = \frac{T_A - T_E}{R_{total}} \]

\[ q = \frac{370^\circ\text{C} - 60^\circ\text{C}}{0.00565 \, \text{m}^2\text{C/W}} \]

\[ q = \frac{310^\circ\text{C}}{0.00565 \, \text{m}^2\text{C/W}} \]

\[ q = 54867 \, \text{W/m}^2 \]

Final Answer