Let's solve the problem step by step: --- # **Step-by-Step Solution** ## **Given Data** - **Sheet Metal Thickness:** \( d_{sm} = 1\ \text{mm} = .001\ \text{m} \) (each side) - **Thermal Conductivity of Sheet Metal (304 SS):** \( k_{sm} = 14.9\ \text{W}/\text{m}\cdot^\circ \text{C} \) - **Thermal Conductivity of Fiberglass Insulation:** \( k_{ins} = .035\ \text{W}/\text{m}\cdot^\circ \text{C} \) - **Inside Heat Transfer Coefficient:** \( h_{in} = 4\ \text{W}/\text{m}^2\cdot^\circ \text{C} \) - **Outside Heat Transfer Coefficient:** \( h_{out} = 9\ \text{W}/\text{m}^2\cdot^\circ \text{C} \) - **Kitchen Temperature:** \( T_{\infty, out} = 25^\circ \text{C} \) - **Refrigerated Space Temperature:** \( T_{\infty, in} = 3^\circ \text{C} \) - **Condensation Point:** Occurs if \( T_{s, out} \leq 10^\circ \text{C} \) - **Insulation Thickness:** \( L = ? \) (to be determined) --- ## **a) Thermal Resistance Network** The wall has the following layers (from inside to outside): 1. **Inside Convective Resistance** (\(R_{conv,in}\)) 2. **Inner Sheet Metal Resistance** (\(R_{sm,in}\)) 3. **Fiberglass Insulation Resistance** (\(R_{ins}\)) 4. **Outer Sheet Metal Resistance** (\(R_{sm,out}\)) 5. **Outside Convective Resistance** (\(R_{conv,out}\)) **Thermal Resistance Network Diagram:** ``` [Inside air] -- [Conv. (h_in)] -- [Sheet metal (1mm)] -- [Insulation (L)] -- [Sheet metal (1mm)] -- [Conv. (h_out)] -- [Outside air] T_inf,in T1 T2 T3 T4 T_inf,out ``` --- ## **b) Calculate Minimum Insulation Thickness** ### **Step 1: Write the Total Thermal Resistance** The total thermal resistance per unit area is: \[ R_{tot} = R_{conv,in} + R_{sm,in} + R_{ins} + R_{sm,out} + R_{conv,out} \] Where: - \( R_{conv,in} = \frac{1}{h_{in}} \) - \( R_{sm,in} = \frac{d_{sm}}{k_{sm}} \) - \( R_{ins} = \frac{L}{k_{ins}} \) - \( R_{sm,out} = \frac{d_{sm}}{k_{sm}} \) - \( R_{conv,out} = \frac{1}{h_{out}} \) --- ### **Step 2: Express the Heat Transfer Rate** The heat transfer per unit area (\( q \)) is: \[ q = \frac{T_{\infty, out} - T_{\infty, in}}{R_{tot}} \] --- ### **Step 3: Find Temperature at the Outer Surface (\(T_{s, out}\))** The temperature at the outer surface (between the outer sheet metal and outside air) is: \[ T_{s, out} = T_{\infty, out} - q \cdot R_{conv,out} \] To **avoid condensation**, \(T_{s, out} \geq 10^\circ \text{C}\). --- ### **Step 4: Substitute and Solve for \(L\)** Let’s put all values and solve for \(L\): - \( h_{in} = 4 \) - \( h_{out} = 9 \) - \( d_{sm} = .001 \) - \( k_{sm} = 14.9 \) - \( k_{ins} = .035 \) - \( T_{\infty, out} = 25 \) - \( T_{\infty, in} = 3 \) #### **Thermal Resistances:** \[ R_{conv,in} = \frac{1}{4} = .25~\text{m}^2\cdot^\circ\text{C}/\text{W} \] \[ R_{sm,in} = R_{sm,out} = \frac{.001}{14.9} = 6.71 \times 10^{-5}~\text{m}^2\cdot^\circ\text{C}/\text{W} \] \[ R_{conv,out} = \frac{1}{9} = .111~\text{m}^2\cdot^\circ\text{C}/\text{W} \] \[ R_{ins} = \frac{L}{.035} \] So, \[ R_{tot} = .25 + 2 \times 6.71 \times 10^{-5} + \frac{L}{.035} + .111 \] \[ R_{tot} \approx .25 + .000134 + \frac{L}{.035} + .111 = .3611 + \frac{L}{.035} \] --- #### **Heat Transfer Rate:** \[ q = \frac{25 - 3}{.3611 + \frac{L}{.035}} \] --- #### **Outer Surface Temperature:** \[ T_{s, out} = T_{\infty, out} - q \cdot R_{conv,out} \] \[ T_{s, out} = 25 - q \cdot .111 \] To avoid condensation: \[ T_{s, out} \geq 10 \] \[ 25 - q \cdot .111 \geq 10 \] \[ q \leq \frac{25 - 10}{.111} \] \[ q \leq 135.135~\text{W}/\text{m}^2 \] But from above: \[ q = \frac{22}{.3611 + \frac{L}{.035}} \] Set equal to upper limit: \[ \frac{22}{.3611 + \frac{L}{.035}} \leq 135.135 \] \[ .3611 + \frac{L}{.035} \geq \frac{22}{135.135} \] \[ .3611 + \frac{L}{.035} \geq .1627 \] \[ \frac{L}{.035} \geq .1627 - .3611 \] \[ \frac{L}{.035} \geq -.1984 \] \[ L \geq -.1984 \times .035 = -.00694~\text{m} \] --- **But a negative thickness is not physical. This means, with the current values, even with no insulation, condensation will not occur!** ### **Check with Zero Insulation (\(L=\))** \[ R_{tot, min} = .3611 \] \[ q_{max} = \frac{22}{.3611} = 60.9~\text{W}/\text{m}^2 \] \[ T_{s, out} = 25 - .111 \cdot 60.9 = 25 - 6.76 = 18.24^\circ \text{C} \] **So, the outer surface with no insulation is at \(18.2^\circ \text{C}\), well above the condensation threshold.** --- ## **Final Answers** ### **a) Thermal Resistance Network** - **Network (in order):** 1. Inside convection 2. Stainless steel sheet (1 mm) 3. Fiberglass insulation (\(L\), to be determined) 4. Stainless steel sheet (1 mm) 5. Outside convection ### **b) Minimum Fiberglass Thickness** - **No insulation is required** to prevent condensation, since even with zero insulation, the outer surface temperature is above the condensation threshold. --- ## **Summary Table** | Layer | Resistance Formula | Value (\(\text{m}^2\cdot^\circ\text{C}/\text{W}\)) | |---------|-------------------------------------|---------------------| | Inside convection | \(1/h_{in}\) | .25 | | Sheet metal (each) | \(d_{sm}/k_{sm}\) | .000067 | | Insulation | \(L/k_{ins}\) | (variable) | | Outside convection | \(1/h_{out}\) | .111 | --- ### **If insulation is required (e.g., for a different dew point):** Use the formula: \[ L_{min} = k_{ins} \left(.1627 - .3611\right) = -.00694~\text{m} \] But for this scenario, **no minimum insulation is needed**. --- **If you have a lower dew point (e.g., 5°C), repeat the calculation with that value.**