Let's break down the solution step by step: --- # **Given Data** - **Tube material:** Type-302 stainless steel - **Inner diameter, \( D_i \):** 22 mm = .022 m - **Outer diameter, \( D_o \):** 27 mm = .027 m - **Fouling factors:** - Inner surface, \( R_{fi}'' = .0004 \) m²·K/W - Outer surface, \( R_{fo}'' = .0002 \) m²·K/W --- ## **(a) Determine the Overall Heat Transfer Coefficient (\( U_o \)) for Air Outside** ### **Step 1: Thermal Resistance Network** The total resistance per unit length (\( L \)) is the sum of: - Inner convection (\( R_{conv, i} \)) - Inner fouling (\( R_{f,i} \)) - Tube wall conduction (\( R_{wall} \)) - Outer fouling (\( R_{f,o} \)) - Outer convection (\( R_{conv, o} \)) \[ R_{total} = R_{conv,i} + R_{f,i} + R_{wall} + R_{f,o} + R_{conv,o} \] --- ### **Step 2: Individual Resistances** #### **1. Inner Convection** \[ R_{conv,i} = \frac{1}{h_i A_i} \] - \( h_i \): Inner convection coefficient - \( A_i = \pi D_i L \) #### **2. Inner Fouling** \[ R_{f,i} = \frac{R_{fi}''}{A_i} \] #### **3. Tube Wall (Cylindrical Conduction)** \[ R_{wall} = \frac{\ln(D_o/D_i)}{2\pi k L} \] - \( k \): Thermal conductivity of SS302 (approx. 16.3 W/m·K) #### **4. Outer Fouling** \[ R_{f,o} = \frac{R_{fo}''}{A_o} \] - \( A_o = \pi D_o L \) #### **5. Outer Convection** \[ R_{conv,o} = \frac{1}{h_o A_o} \] - \( h_o \): Outer convection coefficient --- ### **Step 3: Overall Heat Transfer Coefficient (\( U_o \))** \[ U_o = \frac{1}{R_{total, o} \cdot A_o} \] Where \( R_{total, o} \) is the total resistance expressed on the **outside area** basis. #### **Express All Resistances in Terms of \( A_o \):** - \( \frac{A_o}{A_i} = \frac{D_o}{D_i} \) - \( R_{conv,i} = \frac{1}{h_i A_i} = \frac{D_i}{h_i D_o A_o} \) - \( R_{f,i} = \frac{R_{fi}''}{A_i} = \frac{R_{fi}'' D_i}{D_o A_o} \) - \( R_{wall} = \frac{\ln(D_o/D_i)}{2\pi k L} \) (already per length) - \( R_{f,o} = \frac{R_{fo}''}{A_o} \) - \( R_{conv,o} = \frac{1}{h_o A_o} \) --- ### **Step 4: Assign Typical Values** #### **For water inside:** - \( h_i \approx 230 \) W/m²·K (turbulent, typical for water) #### **For air outside:** - \( h_o \approx 50 \) W/m²·K (cross-flow air, typical) #### **Tube wall:** - \( k = 16.3 \) W/m·K --- ### **Step 5: Calculate Each Resistance Per Meter Length (\( L = 1 \) m)** - \( D_i = .022 \) m, \( D_o = .027 \) m - \( A_i = \pi (.022)(1) = .0691 \) m² - \( A_o = \pi (.027)(1) = .0848 \) m² #### **1. Inner Convection:** \[ R_{conv,i} = \frac{1}{230 \times .0691} = .0063 \ \text{K/W} \] #### **2. Inner Fouling:** \[ R_{f,i} = \frac{.0004}{.0691} = .0058 \ \text{K/W} \] #### **3. Tube Wall:** \[ R_{wall} = \frac{\ln(.027/.022)}{2\pi \times 16.3 \times 1} = \frac{\ln(1.227)}{102.4} = \frac{.204}{102.4} = .002 \ \text{K/W} \] #### **4. Outer Fouling:** \[ R_{f,o} = \frac{.0002}{.0848} = .0024 \ \text{K/W} \] #### **5. Outer Convection:** \[ R_{conv,o} = \frac{1}{50 \times .0848} = .236 \ \text{K/W} \] --- ### **Step 6: Convert All to Outside Area Basis** - For \( R_{conv,i} \) and \( R_{f,i} \), multiply by \( \frac{A_o}{A_i} = \frac{.0848}{.0691} = 1.227 \): \[ R_{conv,i, o\ basis} = .0063 \times 1.227 = .0077 \ \text{K/W} \] \[ R_{f,i, o\ basis} = .0058 \times 1.227 = .0071 \ \text{K/W} \] --- ### **Step 7: Total Resistance** \[ R_{total} = .0077 + .0071 + .002 + .0024 + .236 = .255 \ \text{K/W} \] --- ### **Step 8: Find \( U_o \)** \[ U_o = \frac{1}{R_{total} \cdot A_o} = \frac{1}{.255 \times .0848} = \frac{1}{.0216} = 46.3 \ \text{W/m}^2\cdot\text{K} \] --- ### **Step 9: Compare Resistances** - **Dominant resistance:** Outer convection (\( .236 \) K/W) - Others are much smaller. --- ## **(b) Outside Fluid is Water (\( V_o = 1 \) m/s, \( T = 15^\circ C \))** - For water, **outside convection \( h_o \approx 300 \) W/m²·K** (turbulent) #### **Outer convection resistance:** \[ R_{conv,o,water} = \frac{1}{300 \times .0848} = .0039 \ \text{K/W} \] --- ### **New Total Resistance** \[ R_{total,water} = .0077 + .0071 + .002 + .0024 + .0039 = .0231 \ \text{K/W} \] \[ U_{o,water} = \frac{1}{.0231 \times .0848} = \frac{1}{.00196} = 510 \ \text{W/m}^2\cdot\text{K} \] --- ### **Compare Thermal Resistances** - All resistances are now of similar order, with convection and fouling being equally important. --- # **Final Answers** --- ## **(a) For air outside:** - **Overall heat transfer coefficient:** \( \boxed{U_o = 46 \ \text{W/m}^2\cdot\text{K}} \) - **Dominant resistance:** Outer convection (air) is the limiting factor. --- ## **(b) For water outside:** - **Overall heat transfer coefficient:** \( \boxed{U_o = 510 \ \text{W/m}^2\cdot\text{K}} \) - **Dominant resistances:** All resistances (convection, wall, fouling) are significant and similar. --- **If you need detailed calculations for any step or property correlations, let me know!**