Let's break down the problem and solve it step by step. ## **Given Data** - Water flows in a cylinder at a constant temperature of \( 100^\circ \)C (wall temperature) - Diameter of cylinder, \( D = 50 \) mm \( = .05 \) m - Length of tube, \( L = 6 \) m - Inlet temperature, \( T_i = 15^\circ \)C - Outlet temperature, \( T_o = 57^\circ \)C - Mass flow rate, \( \dot{m} = .25 \) kg/s - Specific heat, \( c_p = 4.178 \) kJ/kg·K \( = 4178 \) J/kg·K ## **Required** Find the **average heat transfer coefficient (\( h \))**. --- ## **Step 1: Calculate Heat Gained by Water** \[ Q = \dot{m} c_p (T_o - T_i) \] \[ Q = .25 \times 4178 \times (57 - 15) \] \[ Q = .25 \times 4178 \times 42 \] \[ Q = .25 \times 175,476 \] \[ Q = 43,869 \text{ W} \] --- ## **Step 2: Log Mean Temperature Difference (LMTD)** \[ \Delta T_1 = T_s - T_i = 100 - 15 = 85^\circ \text{C} \] \[ \Delta T_2 = T_s - T_o = 100 - 57 = 43^\circ \text{C} \] \[ \Delta T_m = \frac{\Delta T_1 - \Delta T_2}{\ln\left(\frac{\Delta T_1}{\Delta T_2}\right)} \] \[ \Delta T_m = \frac{85 - 43}{\ln\left(\frac{85}{43}\right)} \] \[ \Delta T_m = \frac{42}{\ln(1.9767)} \] \[ \ln(1.9767) \approx .681 \] \[ \Delta T_m = \frac{42}{.681} \approx 61.7^\circ \text{C} \] --- ## **Step 3: Surface Area for Heat Transfer** \[ A = \pi D L \] \[ A = \pi \times .05 \times 6 \approx .942 \text{ m}^2 \] --- ## **Step 4: Average Heat Transfer Coefficient (\( h \))** \[ Q = h A \Delta T_m \implies h = \frac{Q}{A \Delta T_m} \] \[ h = \frac{43,869}{.942 \times 61.7} \] \[ h = \frac{43,869}{58.09} \] \[ h \approx 755 \text{ W/m}^2\text{·K} \] --- ## **Correct Option** **The correct answer is:** \[ \boxed{755 \text{ W/m}^2\text{·K}} \] (Corresponds to the option in your image.)