Let's break down the problem and solve it step by step: ### **Given Data** - Cylinder temperature (\(T_s\)): \(100^\circ C\) - Diameter of cylinder (\(D\)): \(50 \text{ mm} = .05 \text{ m}\) - Length of tube (\(L\)): \(6 \text{ m}\) - Inlet temperature (\(T_{m,i}\)): \(15^\circ C\) - Outlet temperature (\(T_{m,o}\)): \(57^\circ C\) - Mass flow rate (\(\dot{m}\)): \(.25 \text{ kg/s}\) - Specific heat (\(c_p\)): \(418 \text{ J/kg·K}\) ### **To Find** Average heat transfer coefficient (\(h\)). --- ## **Step 1: Calculate Heat Transferred (\(\dot{Q}\))** \[ \dot{Q} = \dot{m} c_p (T_{m,o} - T_{m,i}) \] \[ \dot{Q} = .25 \times 418 \times (57-15) \] \[ = .25 \times 418 \times 42 \] \[ = .25 \times 175560 \] \[ = 43890 \text{ W} \] --- ## **Step 2: Log Mean Temperature Difference (\(\Delta T_{lm}\))** \[ \Delta T_{1} = T_s - T_{m,o} = 100 - 57 = 43^\circ C \] \[ \Delta T_{2} = T_s - T_{m,i} = 100 - 15 = 85^\circ C \] \[ \Delta T_{lm} = \frac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2 / \Delta T_1)} \] \[ = \frac{85 - 43}{\ln(85/43)} \] \[ = \frac{42}{\ln(1.9767)} \] \[ = \frac{42}{.681} \] \[ \approx 61.65^\circ C \] --- ## **Step 3: Surface Area (\(A\))** \[ A = \pi D L = \pi \times .05 \times 6 = .942 \text{ m}^2 \] --- ## **Step 4: Heat Transfer Equation** \[ \dot{Q} = h A \Delta T_{lm} \] \[ h = \frac{\dot{Q}}{A \Delta T_{lm}} \] \[ h = \frac{43890}{.942 \times 61.65} \] \[ h = \frac{43890}{58.07} \] \[ h \approx 755.6 \text{ W/m}^2\cdot\text{K} \] --- ### **Final Answer** **The correct option is:** \[ \boxed{755 \text{ W/m}^2\cdot\text{K}} \]