Let's break down the problem step-by-step: **Given:** - Cylinder temperature, \( T_s = 100^\circ C \) - Diameter of cylinder, \( D = 50 \text{ mm} = 0.05 \text{ m} \) - Length of tube, \( L = 6 \text{ m} \) - Inlet temperature, \( T_{i} = 15^\circ C \) - Outlet temperature, \( T_{o} = 57^\circ C \) - Mass flow rate, \( \dot{m} = 0.25 \text{ kg/s} \) - Specific heat of water, \( c_p = 4.178 \text{ kJ/kg K} = 4178 \text{ J/kg K} \) **To Find:** - Average heat transfer coefficient, \( h \) --- ### 1. **Calculate the heat transfer rate, \( Q \):** \[ Q = \dot{m} \cdot c_p \cdot (T_o - T_i) \] \[ Q = 0.25 \times 4178 \times (57 - 15) \] \[ Q = 0.25 \times 4178 \times 42 \] \[ Q = 0.25 \times 175,476 \] \[ Q = 43,869 \text{ W} \] --- ### 2. **Log Mean Temperature Difference (LMTD):** Since the wall temperature is constant, we use: \[ \Delta T_1 = T_s - T_i = 100 - 15 = 85^\circ C \] \[ \Delta T_2 = T_s - T_o = 100 - 57 = 43^\circ C \] \[ \Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} \] \[ \Delta T_{lm} = \frac{85 - 43}{\ln(85/43)} \] \[ \Delta T_{lm} = \frac{42}{\ln(1.9767)} \] \[ \ln(1.9767) \approx 0.681 \] \[ \Delta T_{lm} = \frac{42}{0.681} \approx 61.7^\circ C \] --- ### 3. **Area for heat transfer:** \[ A = \pi D L \] \[ A = \pi \times 0.05 \times 6 \] \[ A = \pi \times 0.3 \] \[ A \approx 0.942 \text{ m}^2 \] --- ### 4. **Calculate Heat Transfer Coefficient \( h \):** \[ Q = h \cdot A \cdot \Delta T_{lm} \] \[ h = \frac{Q}{A \cdot \Delta T_{lm}} \] \[ h = \frac{43,869}{0.942 \times 61.7} \] \[ h = \frac{43,869}{58.13} \] \[ h \approx 755 \text{ W/m}^2\text{K} \] --- ### **Final Answer:** \[ \boxed{755 \text{ W/m}^2\text{K}} \] So, the correct option is the last one: **755 W/m²·K**.