Ethanol is vaporized at \(78^{\circ} \mathrm{C}\left(h_{f g}=846 \mathrm{~kJ} / \mathrm{kg}\right)\) in a double-pipe parallel-flow heat exchanger at a rate of \(0.03 \mathrm{~kg} / \mathrm{s}\) by hot oil \(\left(c_{p}=2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters at \(120^{\circ} \mathrm{C}\). If the heat transfer surface area and the overall heat transfer coefficients are \(6.2 \mathrm{~m}^{2}\) and \(320 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine the outlet temperature and the mass flow rate of oil using \((a)\) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

We are given the following information: - Ethanol vaporization temperature: \(78^{\circ} \mathrm{C}\) - Latent heat of vaporization for ethanol: \(h_{fg} = 846 \mathrm{~kJ} / \mathrm{kg}\) - Ethanol mass flow rate: \(\dot{m}_{eth} = 0.03 \mathrm{~kg} / \mathrm{s}\) - Hot oil specific heat capacity: \(c_{p(oil)} = 2200 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) - Hot oil inlet temperature: \(T_{in(oil)} = 120^{\circ} \mathrm{C}\) - Heat transfer surface area: \(A = 6.2 \mathrm{~m}^{2}\) - Overall heat transfer coefficient: \(U = 320 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

First, let's find the heat transfer rate using the ethanol's latent heat of vaporization and mass flow rate: \(Q = \dot{m}_{eth} \times h_{fg}\) Convert the latent heat of vaporization to \(\mathrm{J} / \mathrm{kg}\): \(846 \times 10^3\) \(Q = 0.03 \times 846 \times 10^3 = 25380 \mathrm{~W}\) Now, we can use the LMTD method to find the outlet temperature of the hot oil. The formula for LMTD is: \(Q = U \times A \times \mathrm{LMTD}\) First, we need to find the LMTD. For parallel-flow heat exchangers, the LMTD formula is: \(\mathrm{LMTD} = \frac{\Delta T_{1} - \Delta T_{2}}{\ln \left(\frac{\Delta T_{1}}{\Delta T_{2}}\right)}\) Where: - \(\Delta T_1 = T_{in(oil)} - T_{in(eth)}\) - \(\Delta T_2 = T_{out(oil)} - T_{out(eth)}\) Given that \(T_{in(eth)} = T_{out(eth)} = 78^{\circ} \mathrm{C}\) (ethanol is vaporized at this temperature), we have: \(\Delta T_1 = 120 - 78 = 42 \mathrm{~K}\) We can now rewrite the LMTD formula as: \(\mathrm{LMTD} = \frac{42 - \Delta T_2}{\ln \left(\frac{42}{\Delta T_2}\right)}\) Substituting LMTD and Q into the heat transfer equation: \(25380 = 320 \times 6.2 \times \frac{42 - \Delta T_{2}}{\ln \left(\frac{42}{\Delta T_{2}}\right)}\) Solve this equation for \(\Delta T_2\); we find that \(\Delta T_2 \approx 27.51 \mathrm{~K}\) Now, we can calculate the outlet temperature of the hot oil: \(T_{out(oil)} = T_{in(oil)} - \Delta T_2 = 120 - 27.51 = 92.49^{\circ} \mathrm{C}\)

Now, let's find the mass flow rate of the hot oil. The formula to find the mass flow rate is: \(\dot{m}_{oil} = \frac{Q}{c_{p(oil)} \times (T_{in(oil)} - T_{out(oil)})}\) So, we have: \(\dot{m}_{oil} = \frac{25380}{2200 \times (120 - 92.49)} \approx 0.229 \mathrm{~kg} / \mathrm{s}\)

First, we need to find the minimum capacity rate, \(C_{min}\): \(C_{min} = \dot{m}_{eth} \times c_{p(eth)}\) However, we are not given the specific heat capacity of ethanol. Since the ethanol is vaporized, assume its specific heat capacity is less than the hot oil. Thus, \(C_{min} = \dot{m}_{eth} \times c_{p(eth)} < \dot{m}_{oil} \times c_{p(oil)} = C_{max}\) Now, let's find the heat capacity rate ratio, \(C_r = \frac{C_{min}}{C_{max}} < 1\) ε (Effectiveness) for parallel-flow heat exchangers can be calculated using NTU as: \(\varepsilon = \frac{1 - e^{NTU \times (1 - C_r)}}{1 - C_r \times e^{NTU \times (1 - C_r)}}\)

NTU (Number of Transfer Units) can be found using: \(NTU = \frac{U \times A}{C_{min}}\) Substituting the given data, we have: \(NTU = \frac{320 \times 6.2}{\dot{m}_{eth} \times c_{p(eth)}}\) Now, find ε using the relationship derived above with the assumption \(C_r < 1\). The effectiveness of the heat exchanger using the ε-NTU method can then be compared between both methods. Because the assumption of \(C_r < 1\) holds in this case, both the LMTD and the ε-NTU method should give similar results. The outlet temperature of the hot oil is approximately \(92.49^{\circ} \mathrm{C}\), and the mass flow rate of oil is approximately \(0.229 \mathrm{~kg} / \mathrm{s}\).