Here’s the text from the image: --- **Question 16** Consider a flow of water in a cylinder maintained at constant temperature of 100°C, the diameter of the cylinder is 50 mm and the length of the tube is 6 m. The inlet and the outlet temperature of the water is \( T_i = 15^\circ C \) and \( T_o = 57^\circ C \). Find the average heat transfer coefficient associated with the flow of water. Mass flow rate is 0.25 kg/s and \( c_p = 4.178 \) kJ/kg·K. --- Let's solve the problem step by step: ### **Given:** - Cylinder temperature, \( T_s = 100^\circ C \) - Diameter, \( D = 50\, \text{mm} = 0.05\, \text{m} \) - Length, \( L = 6\, \text{m} \) - Inlet temp, \( T_i = 15^\circ C \) - Outlet temp, \( T_o = 57^\circ C \) - Mass flow rate, \( \dot{m} = 0.25\, \text{kg/s} \) - Specific heat, \( c_p = 4.178\, \text{kJ/kg·K} = 4178\, \text{J/kg·K} \) --- ### **Step 1: Calculate the heat absorbed by the water (\( Q \)):** \[ Q = \dot{m} c_p (T_o - T_i) \] \[ Q = 0.25 \times 4178 \times (57 - 15) \] \[ Q = 0.25 \times 4178 \times 42 \] \[ Q = 0.25 \times 175,476 \] \[ Q = 43,869 \text{ W} \] --- ### **Step 2: Calculate the log mean temperature difference (\( \Delta T_{lm} \)):** \[ \Delta T_1 = T_s - T_o = 100 - 57 = 43^\circ C \] \[ \Delta T_2 = T_s - T_i = 100 - 15 = 85^\circ C \] \[ \Delta T_{lm} = \frac{\Delta T_2 - \Delta T_1}{\ln(\Delta T_2 / \Delta T_1)} \] \[ \Delta T_{lm} = \frac{85 - 43}{\ln(85 / 43)} \] \[ \Delta T_{lm} = \frac{42}{\ln(1.9767)} \] \[ \ln(1.9767) \approx 0.681 \] \[ \Delta T_{lm} = \frac{42}{0.681} \approx 61.7^\circ C \] --- ### **Step 3: Calculate the heat transfer area (\( A \)):** \[ A = \pi D L = \pi \times 0.05 \times 6 = 0.942 \text{ m}^2 \] --- ### **Step 4: Find the average heat transfer coefficient (\( h \)):** \[ Q = h A \Delta T_{lm} \] \[ h = \frac{Q}{A \Delta T_{lm}} \] \[ h = \frac{43,869}{0.942 \times 61.7} \] \[ h = \frac{43,869}{58.1} \] \[ h \approx 755 \text{ W/m}^2\cdot\text{K} \] --- ### **Final Answer:** The average heat transfer coefficient associated with the flow of water is: \[ \boxed{755 \text{ W/m}^2\cdot\text{K}} \]