Let's break down the problem and see how to solve it step by step: --- ### **Given:** - Constant surface temperature of the cylinder, \( T_s = 100^\circ C \) - Diameter of cylinder, \( D = 50 \) mm \( = 0.05 \) m - Length of tube, \( L = 6 \) m - Inlet temperature, \( T_{in} = 15^\circ C \) - Outlet temperature, \( T_{out} = 57^\circ C \) - Mass flow rate, \( \dot{m} = 0.25 \) kg/s - Specific heat, \( c_p = 4.18 \) kJ/kg·K \( = 4180 \) J/kg·K --- ### **To Find:** Average heat transfer coefficient, \( h \). --- ### **Step 1: Calculate the Heat Transferred (\( Q \))** \[ Q = \dot{m} c_p (T_{out} - T_{in}) \] \[ Q = 0.25 \times 4180 \times (57 - 15) \] \[ Q = 0.25 \times 4180 \times 42 \] \[ Q = 0.25 \times 175,560 \] \[ Q = 43,890 \text{ W} \] --- ### **Step 2: Log Mean Temperature Difference (LMTD)** \[ \Delta T_1 = T_s - T_{in} = 100 - 15 = 85^\circ C \] \[ \Delta T_2 = T_s - T_{out} = 100 - 57 = 43^\circ C \] \[ \Delta T_m = \frac{\Delta T_1 - \Delta T_2}{\ln{\left(\frac{\Delta T_1}{\Delta T_2}\right)}} \] \[ \Delta T_m = \frac{85 - 43}{\ln{\left(\frac{85}{43}\right)}} \] \[ \Delta T_m = \frac{42}{\ln{(1.9767)}} \] \[ \Delta T_m = \frac{42}{0.681} \] \[ \Delta T_m \approx 61.7^\circ C \] --- ### **Step 3: Surface Area (A)** \[ A = \pi D L = \pi \times 0.05 \times 6 \] \[ A = \pi \times 0.3 = 0.942 \text{ m}^2 \] --- ### **Step 4: Average Heat Transfer Coefficient (h)** \[ Q = h A \Delta T_m \implies h = \frac{Q}{A \Delta T_m} \] \[ h = \frac{43,890}{0.942 \times 61.7} \] \[ h = \frac{43,890}{58.11} \] \[ h \approx 755 \text{ W/m}^2\text{·K} \] --- ### **Final Answer:** The average heat transfer coefficient is **755 W/m²·K**. So the correct answer is the **first option**.