Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the \(2.5\)-cm-internaldiameter tube of a double-pipe counter-flow heat exchanger at \(17^{\circ} \mathrm{C}\) at a rate of \(1.8 \mathrm{~kg} / \mathrm{s}\). Water is heated by steam condensing at \(120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)\) in the shell. If the overall heat transfer coefficient of the heat exchanger is \(700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the length of the tube required in order to heat the water to \(80^{\circ} \mathrm{C}\) using ( \(a\) ) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

Since the mass flow rate of water is given (\(1.8\,\text{kg/s}\)), no calculations are required for this step.

We need to find out the heat transfer required to heat the water from \(17^\circ\text{C}\) to \(80^\circ\text{C}\). We can calculate this using the formula: \(Q = m \cdot c_p \cdot \Delta T\), where \(Q\) is the heat transfer, \(m\) is the mass flow rate, \(c_p\) is the specific heat capacity of the water, and \(\Delta T\) is the temperature difference. \(\Delta T = 80^\circ\text{C} - 17^\circ\text{C} = 63^\circ\text{C}\) \(Q = 1.8\,\text{kg/s} \cdot 4180\,\text{J/kg}\cdot\text{K} \cdot 63\,\text{K} = 473076\,\text{W}\) The required heat transfer is \(473076\,\text{W}\).

To evaluate the LMTD, the hot and cold outlet temperatures must be estimated. The steam condenses at a constant temperature of \(120^\circ\text{C}\). Assuming the steam temperature (\(T_h\)) remains constant throughout the process, the temperature difference at the inlet (\(\Delta T_1\)) and outlet (\(\Delta T_2\)) can be calculated as follows: \(\Delta T_1 = T_h - T_{c1} = 120^\circ\text{C} - 17^\circ\text{C} = 103^\circ\text{C}\) \(\Delta T_2 = T_h - T_{c2} = 120^\circ\text{C} - 80^\circ\text{C} = 40^\circ\text{C}\)

Now, we can find the Logarithmic Mean Temperature Difference (LMTD) using the formula: \(\mathrm{LMTD} = \frac{\Delta T_1 - \Delta T_2}{\ln \left(\frac{\Delta T_1}{\Delta T_2}\right)}\) \(\mathrm{LMTD} = \frac{103^\circ\text{C} - 40^\circ\text{C}}{\ln \left(\frac{103^\circ\text{C}}{40^\circ\text{C}}\right)} = 63.76^\circ\text{C}\) The heat transfer area \(A\) can be calculated using the formula: \(Q = U \cdot A \cdot \mathrm{LMTD}\), where \(U\) is the overall heat transfer coefficient. Rearranging the formula to calculate \(A\), we get: \(A = \frac{Q}{U \cdot \mathrm{LMTD}} = \frac{473076\,\text{W}}{700\,\text{W/m}^2\cdot\text{K} \cdot 63.76\,\text{K}} = 1.067\,\text{m}^2\) The internal diameter of the tube is given. To find the length of the tube, we use the formula for the surface area of the tube: \(A = \pi d L\), where \(d\) is the diameter, and \(L\) is the length. We can solve for \(L\): \(L = \frac{A}{\pi d} = \frac{1.067\,\text{m}^2}{\pi \cdot 0.025\,\text{m}} \approx 13.56\,\text{m}\) So, the length of the tube required using the LMTD method is approximately \(13.56\,\text{m}\).

The number of transfer units (\(\mathrm{NTU}\)) can be calculated with the formula: \(\mathrm{NTU} = \frac{UA}{mc_p}\), where \(U\), \(A\), \(m\), and \(c_p\) are the overall heat transfer coefficient, heat transfer area, mass flow rate, and specific heat capacity of water, respectively. Using the values from the LMTD section, we get: \(\mathrm{NTU} = \frac{700\,\text{W/m}^2\cdot\text{K} \cdot 1.067\,\text{m}^2}{1.8\,\text{kg/s} \cdot 4180\,\text{J/kg}\cdot\text{K}} \approx 0.056\)

The effectiveness (\(\varepsilon\)) can be calculated using the formula: \(\varepsilon = \frac{Q_{actual}}{Q_{max}}\), where \(Q_{actual}\) and \(Q_{max}\) are the actual and maximum possible heat transfers, respectively. Since the outlet coolant temperature is given, we can calculate the actual heat transfer: \(Q_{actual} = m c_p (T_{c2} - T_{c1}) = 1.8\,\text{kg/s} \cdot 4180\,\text{J/kg}\cdot\text{K} \cdot (80^\circ\text{C} - 17^\circ\text{C})= 473076\,\text{W}\) The maximum possible heat transfer occurs when the coolant reaches the temperature of steam: \(Q_{max} = m c_p (T_h - T_{c1}) = 1.8\,\text{kg/s} \cdot 4180\,\text{J/kg}\cdot\text{K} \cdot (120^\circ\text{C} - 17^\circ\text{C})= 627096\,\text{W}\) \(\varepsilon = \frac{473076\,\text{W}}{627096\,\text{W}} \approx 0.754\) We can use the effectiveness-NTU relation to check the NTU value: \(\varepsilon = 1 - \exp\left(-\frac{1}{c} \cdot \left(1 - \exp(-c\,\mathrm{NTU})\right)\right)\), where \(c = \frac{mc_p}{U\cdot A}\) Using the given values, \(c = \frac{1.8\,\text{kg/s}\cdot 4180\,\text{J/kg}\cdot\text{K}}{700\,\text{W/m}^2\cdot\text{K} \cdot 1.067\,\text{m}^2} \approx 17.77\) Solving for NTU, we obtain \(\mathrm{NTU} \approx 0.056\), which agrees with the value we calculated earlier. Now, we can calculate the heat transfer area using the NTU and ε values: \(A = \frac{m c_p}{U \cdot \mathrm{NTU}} = \frac{1.8\,\text{kg/s} \cdot 4180\,\text{J/kg}\cdot\text{K}}{700\,\text{W/m}^2\cdot\text{K} \cdot 0.056} \approx 1.067\,\text{m}^2\) Finally, the tube length can be found using the same formula as in the LMTD method: \(L = \frac{A}{\pi d} = \frac{1.067\,\text{m}^2}{\pi \cdot 0.025\,\text{m}} \approx 13.56\,\text{m}\) So, the length of the tube required using the ε-NTU method is approximately \(13.56\,\text{m}\).