Water enter a 5-mm-diameter and 13-m-long tube at \(45^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\). The tube is maintained at a constant temperature of \(5^{\circ} \mathrm{C}\). The required length of the tube in order for the water to exit the tube at \(25^{\circ} \mathrm{C}\) is (a) \(1.55 \mathrm{~m}\) (b) \(1.72 \mathrm{~m}\) (c) \(1.99 \mathrm{~m}\) (d) \(2.37 \mathrm{~m}\) (e) \(2.96 \mathrm{~m}\) (For water, use \(k=0.623 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=4.83, v=0.724 \times\) \(10^{-6} \mathrm{~m}^{2} / \mathrm{s}, c_{p}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, \rho=994 \mathrm{~kg} / \mathrm{m}^{3}\).)

The Logarithmic Mean Temperature Difference (LMTD) is a measure of the average temperature difference between the water and the tube over the entire length of the tube. It can be calculated using the formula: \(LMTD = \frac{(T_{in} - T_{out}) - (T_{tube} - T_{in})}{\ln \frac{T_{in} - T_{out}}{T_{tube} - T_{in}}}\) Here, \(T_{in} = 45^{\circ} \mathrm{C}\), \(T_{out} = 25^{\circ} \mathrm{C}\), and \(T_{tube} = 5^{\circ} \mathrm{C}\). Substitute the given values and calculate LMTD: \(LMTD = \frac{(45 - 25) - (5 - 45)}{\ln \frac{45 - 25}{5 - 45}} = \frac{60}{\ln \frac{20}{-40}} = -30 \ln (2) = 20.79\ ^{\circ}\mathrm{C}\)

Now that we have calculated the LMTD, we can determine the heat transfer rate using the given water properties. We first need to determine the heat transfer coefficient (h) and the overall heat transfer area (A). Since we are given the Prandtl number (Pr), thermal conductivity (k), and kinematic viscosity (v), we can find the heat transfer coefficient (h) using the Nusselt number (Nu), as follows: \(Nu = \frac{h \cdot D}{k}\) The Nusselt number can be found using the Dittus-Boelter equation: \(Nu = 0.023 \cdot Re^{0.8} \cdot Pr^{0.3}\) Here, Reynolds number (Re) is given by: \(Re = \frac{VD} {v}\) Using the provided information, we calculate Re, Nu and h: \(Re = \frac{0.3 \cdot 0.005}{0.724 \times 10^{-6}} = 2073.20\) \(Nu = 0.023 \cdot 2073.20^{0.8} \cdot 4.83^{0.3} = 18.97\) \(h = \frac{18.97 \cdot 0.623}{0.005} = 2325.42 \mathrm{~W}/\mathrm{m}^2\cdot \mathrm{K}\) The overall heat transfer area is given by the tube's circumference multiplied by the length of the tube that needs to be determined at the end (L). \(A = \pi \cdot D \cdot L = 0.005\pi \cdot L\)

We can now calculate the heat transfer rate (Q), given the heat transfer coefficient (h), the LMTD, and the overall area of heat transfer (A): \(Q = h \cdot A \cdot LMTD\) We also know that the heat transfer rate is equal to the product of the water's mass flow rate (\(\dot{m}\)), specific heat capacity (\(c_p\)), and the temperature change (\(T_{in} - T_{out}\)): \(Q = \dot{m} \cdot c_p \cdot (T_{in} - T_{out})\) The mass flow rate can be found using the provided water density (\(\rho\)) and volume flow rate (A\(_{flow}\)V): A\(_{flow}\) = \(\frac{\pi D^2}{4} = \frac{\pi (0.005)^2}{4} = 1.963 \times 10^{-5} \mathrm{m}^2\) \(\dot{m} = \rho \cdot A_{flow} \cdot V = 994 \cdot 1.963 \times 10^{-5} \cdot 0.3 = 5.83 \times 10^{-3} \mathrm{kg/s}\) Combining the two equations for heat transfer rate (Q) and solving for tube length (L): \(L = \frac{\dot{m} \cdot c_p \cdot (T_{in} - T_{out})}{h \cdot (\pi \cdot D) \cdot LMTD} = \frac{5.83 \times 10^{-3} \cdot 4178 \cdot (45 - 25)}{2325.42 \cdot (0.005\pi) \cdot 20.79} = 1.99\ \mathrm{m}\) Based on the provided information, the required tube length for the water to exit at \(25^{\circ} \mathrm{C}\) is approximately \(\bold{1.99\ \mathrm{m}}\) (option c).