Water enters a tube of diameter, \( D = 60 \, {mm} \) with mass flow rate of 0.01 kg/s\(^{-1}\) as shown in the figure below. The inlet mean temperature is \( T_{{in},i} = 293 \, {K} \) and the uniform heat flux at the surface of the tube is 2000 W/m\(^{-2}\). For the exit mean temperature of \( T_{{m},o} = 353 \, {K} \), the length of the tube, \( L \) is _________m (rounded off to 1 decimal place). Use the specific heat of water as 4181 J kg\(^{-1}\) K\(^{-1}\).
Show Hint
To calculate the length of the tube, use the heat transfer equation that relates heat flux to temperature difference and mass flow rate. Ensure to account for the surface area and specific heat of the fluid.
We are given: Mass flow rate, \( \dot{m} = 0.01 \, {kg/s} \), Inlet temperature, \( T_{{in},i} = 293 \, {K} \), Exit temperature, \( T_{{m},o} = 353 \, {K} \), Heat flux at the surface, \( q = 2000 \, {W/m}^2 \), Specific heat of water, \( C_p = 4181 \, {J/kg·K} \), Diameter of the tube, \( D = 60 \, {mm} = 0.06 \, {m} \). Step 1: Heat Transfer Equation The heat transferred to the water is given by the heat flux at the surface: \[ Q = \dot{m} C_p \Delta T \] where \( \Delta T = T_{{m},o} - T_{{in},i} = 353 - 293 = 60 \, {K} \). Step 2: Heat Flux The heat flux is also given by: \[ Q = q \cdot A \] where \( A \) is the surface area of the tube. For a cylindrical tube: \[ A = \pi D L \] Substituting this into the equation for heat transfer: \[ q \cdot \pi D L = \dot{m} C_p (T_{{m},o} - T_{{in},i}) \] Substituting the given values: \[ 2000 \cdot \pi \cdot 0.06 \cdot L = 0.01 \cdot 4181 \cdot 60 \] Solving for \( L \), we get: \[ L = \frac{0.01 \cdot 4181 \cdot 60}{2000 \cdot \pi \cdot 0.06} \] \[ L \approx 6.8 \, {m} \] Thus, the length of the tube is \( L \approx 6.8 \, {m} \).
