Let's read the problem and solve step-by-step. --- ### **Problem (from image):** - **Flow in a cylinder:** Water at a constant temperature of 100°C. - **Cylinder diameter (D):** 50 mm = .05 m - **Length of tube (L):** 6 m - **Inlet temperature of water (T₁):** 15°C - **Outlet temperature of water (T₂):** 57°C - **Mass flow rate (ṁ):** .25 kg/s - **Specific heat (cₚ):** 4.18 kJ/kg·K = 418 J/kg·K **Asked:** Find the average heat transfer coefficient (h). --- ## **Step 1: Find Heat Transfer Rate (Q)** \[ Q = \dot{m} c_p (T_2 - T_1) \] \[ Q = .25 \times 418 \times (57 - 15) \] \[ Q = .25 \times 418 \times 42 \] \[ Q = .25 \times 175560 \] \[ Q = 43890 \text{ W} \] --- ## **Step 2: Log Mean Temperature Difference (LMTD)** The surface temperature is constant at 100°C, so: \[ \Delta T_1 = 100 - 15 = 85^\circ C \] \[ \Delta T_2 = 100 - 57 = 43^\circ C \] \[ \Delta T_m = \frac{\Delta T_1 - \Delta T_2}{\ln(\Delta T_1 / \Delta T_2)} \] \[ \Delta T_m = \frac{85 - 43}{\ln(85/43)} \] \[ \Delta T_m = \frac{42}{\ln(1.9767)} \] \[ \ln(1.9767) \approx .681 \] \[ \Delta T_m = \frac{42}{.681} \approx 61.7^\circ C \] --- ## **Step 3: Surface Area (A)** \[ A = \pi D L \] \[ A = \pi \times .05 \times 6 \] \[ A \approx .942 \text{ m}^2 \] --- ## **Step 4: Heat Transfer Coefficient (h)** \[ Q = h A \Delta T_m \] \[ h = \frac{Q}{A \Delta T_m} \] \[ h = \frac{43890}{.942 \times 61.7} \] \[ h = \frac{43890}{58.11} \] \[ h \approx 755 \text{ W/m}^2 \cdot \text{K} \] --- ## **Step 5: Correct Option** **The correct answer is:** **755 W/m²·K** --- ### **Mark:** - [x] 755 W/m²·K --- **Let me know if you need the calculation breakdown or have another question!**