Let's break down the problem step by step: ### Given Data: - Constant temperature of cylinder wall: **100°C** - Diameter of cylinder, \( D = 50 \) mm = **0.05 m** - Length of tube, \( L = 5.6 \) m - Inlet temp of water, \( T_{i} = 15°C \) - Outlet temp of water, \( T_{o} = 37°C \) - Mass flow rate, \( \dot{m} = 0.25 \) kg/s - Specific heat of water, \( c_p = 4178 \) J/kg·K ### To Find: **Average heat transfer coefficient (\( h \)).** --- ### 1. **Calculate heat transfer (\( Q \)):** \[ Q = \dot{m} c_p (T_o - T_i) \] \[ Q = 0.25 \times 4178 \times (37 - 15) \] \[ Q = 0.25 \times 4178 \times 22 \] \[ Q = 0.25 \times 91916 \] \[ Q = 22979 \text{ W} \] --- ### 2. **Log Mean Temperature Difference (LMTD):** \[ \Delta T_1 = T_{wall} - T_{i} = 100 - 15 = 85^\circ C \] \[ \Delta T_2 = T_{wall} - T_{o} = 100 - 37 = 63^\circ C \] \[ \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} \] \[ \Delta T_{lm} = \frac{85 - 63}{\ln(85 / 63)} \] \[ \Delta T_{lm} = \frac{22}{\ln(1.349)} \] \[ \ln(1.349) \approx 0.299 \] \[ \Delta T_{lm} = \frac{22}{0.299} \approx 73.58^\circ C \] --- ### 3. **Surface Area (\( A \)):** \[ A = \pi D L = \pi \times 0.05 \times 5.6 \approx 0.8796 \text{ m}^2 \] --- ### 4. **Heat Transfer Equation:** \[ Q = h A \Delta T_{lm} \] \[ h = \frac{Q}{A \Delta T_{lm}} \] \[ h = \frac{22979}{0.8796 \times 73.58} \] \[ h = \frac{22979}{64.73} \] \[ h \approx 355 \text{ W/m}^2\cdot\text{K} \] --- ### **Closest Answer:** The value is closest to **359 W/m²·K**. --- **Final Answer:** #### **359 W/m²·K** (rounded to the nearest option given)