Respuesta :
Answer:
The thickness of the insulator is approximately 34.918 mm
Explanation:
From the question, we have;
The outer diameter of the duct, D = 300 mm
The wall thickness of the duct, t₁ = 0.6 mm
The temperature of the inner surface of the duct, [tex]T_C[/tex] = 0°C
The temperature of the outer surface, [tex]T_H[/tex] = 25°C
The thermal conductivity of sheet metal, k₁ = 100 W/m-K
The thermal conductivity of insulation, k₂ = 0.04 W/m-K
Assumed rate of heat transfer through the walls of the refrigerator per second, Q = 30 W = 30 J/s
Therefore, we have;
[tex]Q = \dfrac{T_H - T_C}{R_{total}}[/tex]
[tex]\therefore R_{total} = \dfrac{T_H - T_C}{Q} = \dfrac{25 ^{\circ} - 0^{\circ}}{30 \, W} =\dfrac{5}{6} \ ^{\circ}C/W[/tex]
The outside radius, r₂ = 300 mm/2 = 150 mm
The inner diameter of the pipe, d = D - 2·t₁
∴ d = 300 mm - 2 × 0.6 mm = 298.8 mm
The inside radius, r₁ = d/2 = 298.8mm/2 = 149.4 mm
The heat resistance of the pipe, R₁, is given as follows;
[tex]R_1 = R_{pipe} = \dfrac{ln\left (\dfrac{r_2}{r_1} \right) }{2\cdot \pi \cdot k_1\cdot L}[/tex]
Where;
r₁, r₂, and k₁ are as defined above;
L = The length of the pipe = 1 m
Therefore, we have;
[tex]R_1 = R_{pipe} = \dfrac{ln\left (\dfrac{150}{149.4} \right) }{2\cdot \pi \cdot 100\cdot 1} \approx 6.378964 \times 10^{-6}[/tex]
[tex]R_{total}[/tex] = [tex]R_{insltor}[/tex] + [tex]R_{pipe}[/tex]
∴ [tex]R_{insltor}[/tex] = [tex]R_{total}[/tex] - [tex]R_{pipe}[/tex]
[tex]R_{insltor}[/tex] = 5/6 - 6.378964 × 10⁻⁴ ≈ 0.862695
The heat resistance of the insulator, R₂ = [tex]R_{insltor}[/tex] ≈ 0.862695 °C/W
The heat resistance of the insulator, R₂, is given as follows;
[tex]R_1 = R_{insltor} = \dfrac{ln\left (\dfrac{r_3}{r_2} \right) }{2\cdot \pi \cdot k_2\cdot L}[/tex]
Therefore;
[tex]R_2 = R_{insltor} = 0.862695 = \dfrac{ln\left (\dfrac{r_3}{150} \right) }{2\cdot \pi \times 0.04\times 1}[/tex]
[tex]0.832695 \times 2\times \pi \times 0.04\times 1 = {ln\left (\dfrac{r_3}{150} \right) }{}[/tex]
[tex]0.209279 = {ln\left (\dfrac{r_3}{150} \right) }{}[/tex]
[tex]e^{0.209279} = \dfrac{r_3}{150} \right) }{}[/tex]
r₃ = 150 × [tex]e^{0.209279}[/tex] = 184.918
The outer radius of the insulator, r₃ ≈ 184.918 mm
The thickness of the insulator, t₂ = r₃ - r₂
∴ The thickness of the insulator, t₂ ≈ 184.918 mm - 150 mm = 34.918 mm.
