Let's solve each part step by step: --- ## **Given Data** - **Pipe Outer Diameter (\(D_o\))** = **3. cm** = .03 m (Radius \(r_o = D_o/2 = .015\,\text{m}\)) - **Pipe Surface Temperature (\(T_s\))** = **120°C** - **Air Temperature (\(T_\infty\))** = **5°C** - **External heat transfer coefficient (\(h\))** = **10 W/m²·K** - **Thermal conductivity of insulation (\(k\))** = **.4 W/m·K** (for part b onward) --- ## **a) Rate of heat loss per unit length (without insulation)** This is an **external convection** problem. **Heat loss per unit length**: \[ \dot{Q}' = \frac{T_s - T_\infty}{R'} \] where \( R' \) is the total thermal resistance per unit length. Since we only have convection from the pipe surface to air (no insulation yet): \[ R' = R'_{\text{conv, air}} = \frac{1}{h \cdot 2 \pi r_o} \] Plug in values: - \( h = 10\, \text{W/m}^2\cdot\text{K} \) - \( r_o = .015\, \text{m} \) - \( T_s = 120^\circ \text{C} \) - \( T_\infty = 5^\circ \text{C} \) \[ R' = \frac{1}{10 \cdot 2 \pi \cdot .015} = \frac{1}{.9425} = 1.061 \,\text{K·m/W} \] \[ \dot{Q}' = \frac{120 - 5}{1.061} = \frac{115}{1.061} = 108.4\, \text{W/m} \] **Final answer (a):** \[ \boxed{\dot{Q}' = 108.4\, \text{W/m}} \] --- ## **b) Critical radius of insulation** For a cylinder: \[ r_{cr} = \frac{k}{h} \] Given: - \( k = .4\, \text{W/m·K} \) - \( h = 10\, \text{W/m}^2\cdot\text{K} \) \[ r_{cr} = \frac{.4}{10} = .04\, \text{m} = 4.\, \text{cm} \] **Final answer (b):** \[ \boxed{r_{cr} = 4.\, \text{cm}} \] --- ## **c) Rate of heat loss per unit length (with insulation, thickness = 1.25 cm)** - **Insulation thickness** = 1.25 cm = .0125 m - **New outer radius** \( r_2 = r_o + \text{thickness} = .015 + .0125 = .0275\, \text{m} \) **Thermal resistances:** 1. **Conduction through insulation:** \[ R'_{\text{ins}} = \frac{\ln(r_2/r_o)}{2\pi k} \] 2. **Convection from insulation surface to air:** \[ R'_{\text{conv}} = \frac{1}{h \cdot 2\pi r_2} \] 3. **Total resistance:** \[ R'_{\text{total}} = R'_{\text{ins}} + R'_{\text{conv}} \] **Calculate each:** - \( \ln(r_2/r_o) = \ln(.0275/.015) = \ln(1.833) = .606 \) \[ R'_{\text{ins}} = \frac{.606}{2 \pi \times .4} = \frac{.606}{2.513} = .241\, \text{K·m/W} \] \[ R'_{\text{conv}} = \frac{1}{10 \cdot 2\pi \cdot .0275} = \frac{1}{1.7279} = .578\, \text{K·m/W} \] \[ R'_{\text{total}} = .241 + .578 = .819\, \text{K·m/W} \] \[ \dot{Q}' = \frac{120 - 5}{.819} = \frac{115}{.819} = 140.4\, \text{W/m} \] **Final answer (c):** \[ \boxed{\dot{Q}' = 140.4\, \text{W/m}} \] --- ## **d) Temperature on the outside surface of the insulation** Let \( T_2 \) = temperature at insulation outer surface. \[ \dot{Q}' = \frac{T_2 - T_\infty}{R'_{\text{conv}}} \] \[ T_2 = \dot{Q}' \cdot R'_{\text{conv}} + T_\infty \] Plug in: - \( \dot{Q}' = 140.4\, \text{W/m} \) - \( R'_{\text{conv}} = .578\, \text{K·m/W} \) - \( T_\infty = 5^\circ \text{C} \) \[ T_2 = 140.4 \times .578 + 5 = 81.2 + 5 = 86.2^\circ \text{C} \] **Final answer (d):** \[ \boxed{T_2 = 86.2^\circ \text{C}} \] --- ### **Summary of Answers** a) \(\boxed{108.4\, \text{W/m}}\) b) \(\boxed{4.\, \text{cm}}\) c) \(\boxed{140.4\, \text{W/m}}\) d) \(\boxed{86.2^\circ \text{C}}\) Let me know if you need a detailed explanation for any step!