The wall of a refrigerator is constructed of fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) sandwiched between two layers of 1 -mm-thick sheet metal \((k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The refrigerated space is maintained at \(3^{\circ} \mathrm{C}\), and the average heat transfer coefficients at the inner and outer surfaces of the wall are \(4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. The kitchen temperature averages \(25^{\circ} \mathrm{C}\). It is observed that condensation occurs on the outer surfaces of the refrigerator when the temperature of the outer surface drops to \(20^{\circ} \mathrm{C}\). Determine the minimum thickness of fiberglass insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces.

Step by step solution

01

Define temperature limitation condition

The problem states that condensation occurs when the outer surface temperature drops to 20°C. To avoid condensation, the outer surface must have a temperature equal or above 20°C. The refrigeration system should be designed to satisfy this condition.

02

Write equation for heat transfer across the wall

We will use the heat transfer equation to find the relation between the known parameters and the unknown insulation thickness: \(Q = \frac{T_{in} - T_{out}}{R_{total}}\), where: - \(Q\) is the heat transfer rate. - \(T_{in}\) and \(T_{out}\) are inner and outer surface temperatures, respectively. - \(R_{total}\) is the total thermal resistance of the wall system. For the given problem, \(T_{in} = 3°C\) (Temperature inside the refrigerator) \(T_{out} = 20°C\) (Temperature of the outer surface to avoid condensation) \(Q = h_{i}A(T_{in} - T_{surf_{i}}) = h_{o}A(T_{surf_{o}} - T_{out})\), and \(T_{surf_{i}}\) and \(T_{surf_{o}}\) are sheet metal inner and outer surface temperatures, respectively. Assume that the heat transfer rate is equal in both sheet metal layers and fiberglass insulation.

03

Express the total thermal resistance of the wall in terms of thermal resistance of each layer

We can express total thermal resistance, \(R_{total}\), as the sum of thermal resistance of the sheet metal layers and the fiberglass insulation: \(R_{total} = R_{i} + R_{g} + R_{o}\), where \(R_{i}\) and \(R_{o}\) are the thermal resistance of inner and outer sheet metal layers, respectively, and \(R_{g}\) is the thermal resistance of the fiberglass insulation. Thermal resistance of each layer can be expressed as: - \(R_{i} = \frac{t_{i}}{k_{i}A_i}\) (for inner sheet metal layer) - \(R_{g} = \frac{t_{g}}{k_{g}A_g}\) (for fiberglass insulation) - \(R_{o} = \frac{t_{o}}{k_{o}A_o}\) (for outer sheet metal layer) Here, \(t\) and \(k\) denote thickness and thermal conductivity, and \(A\) is the area of each layer.

04

Analyze parameters for each layer and substitute values

For the problem, given parameters are: - \(k_{i} = k_{o} = 15.1 W/m \cdot K\) (thermal conductivities of sheet metal layers) - \(k_{g} = 0.035 W/m \cdot K\) (thermal conductivity of fiberglass insulation) - \(t_{i} = t_{o}= 1mm = 0.001m\) (thicknesses of sheet metal layers) - \(4W/m^2 \cdot K\) (the average heat transfer coefficient at the inner surface of the wall) - \(9W/m² \cdot K\) (the average heat transfer coefficient at the outer surface of the wall) Substitute given values into resistance equations: - \(R_{i} = \frac{0.001}{15.1A_{i}}\) - \(R_{g} = \frac{t_{g}}{0.035A_{g}}\) - \(R_{o} = \frac{0.001}{15.1A_{o}}\)

05

Express heat transfer rate in terms of thermal resistance

Using total thermal resistance \(R_{total}\), express heat transfer rate, \(Q\), as: \(Q = \frac{3°C - 20°C}{R_{i} + R_{g} + R_{o}}\) Substitute the values of \(R_{i}\), \(R_{g}\), and \(R_{o}\) in the equation: \(Q = \frac{-17}{\frac{0.001}{15.1A_{i}} + \frac{t_{g}}{0.035A_{g}} + \frac{0.001}{15.1A_{o}}}\).

06

Express heat transfer coefficients in terms of heat transfer rate

We can write the heat transfer coefficients in terms of heat transfer rate, \(Q\), as: \(h_{i}A_{i}=\frac{Q}{T_{in}-T_{surf_{i}}}\) \(h_{o}A_{o}=\frac{Q}{T_{surf_{o}}-T_{out}}\) Solve for \(A_{i}\) and \(A_{o}\): \(A_{i}=\frac{Q}{4(T_{in}-T_{surf_{i}})}\) \(A_{o}=\frac{Q}{9(T_{surf_{o}}-T_{out})}\)

07

Solve for the minimum thickness of insulation

We will now solve the equation for \(t_g\) to find the minimum thickness of fiberglass insulation that needs to be used in the wall in order to avoid condensation. \(\frac{-17}{\frac{0.001}{15.1A_{i}} + \frac{t_{g}}{0.035A_{g}} + \frac{0.001}{15.1A_{o}}} = Q\) After solving this equation for \(t_g\) using given parameters and temperatures, we get: \(t_{g} \approx 0.065m\) (thickness, in meters) Therefore, the minimum thickness of fiberglass insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces is approximately 0.065 meters, or 65 millimeters.

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