91Ó°ÊÓ

First, we'll find the length of the heat exchanger using the given data. For this, we'll use the following heat transfer equation: \(Q = U * A * \Delta T_{lm}\) where: \(Q\) = total heat transfer rate (3000 W), \(U\) = overall heat transfer coefficient (500 W/m²·K), \(A\) = heat transfer area, \(\Delta T_{lm}\) = log mean temperature difference. To find the log mean temperature difference we will use: \(\Delta T_{lm} = \frac{(\Delta T_2 - \Delta T_1)}{ln(\Delta T_2 / \Delta T_1)}\) where: \(\Delta T_1\) = temperature difference at the inlet, \(\Delta T_2\) = temperature difference at the outlet. \(\Delta T_1 = T_{i_o} - T_{i_w} = 160 - 20 = 140^{\circ}C\) \(\Delta T_2 = T_{o_o} - T_{o_w} = 140 - 80 = 60^{\circ}C\) Now, finding the log mean temperature difference: \(\Delta T_{lm} = \frac{(60 - 140)}{ln(60 / 140)} = -94.47^{\circ}C\) The minus sign is due to the countercurrent flow; the actual temperature difference is positive. Now, we have: \(Q = U *\ A *\ \Delta T_{lm}\) Since the exchanger is a concentric tube type, the heat transfer area per unit length can be expressed as: \(A_{exchanger} = 2 \pi * D_{i} * L\) where \(D_{i}\) = diameter of the inner tube (0.02 m), \(L\) = length of the exchanger. Putting the values into the equation and solving for length, we get: \(3000 = 500 * 2 \pi * 0.02 * L * 94.47\) \(L = \frac{3000}{500 * 2 \pi * 0.02 * 94.47} \approx 1.59 m\) So, the length of the heat exchanger is approximately 1.59 m.

Let's use the data after degradation due to fouling and find the corresponding values. Given the following conditions after degradation: \(T_{o_w}^{'} = 65^{\circ}C\) (new outlet temperature of water) Now, let's calculate the new values of heat transfer rate, ΔT1', and ΔT2': \(Q^{'} = 500 * 2 \pi * 0.02 * L * \Delta T_{lm}^{'}\) So, we need to find \(\Delta T_{lm}^{'}\) for the degraded case: \( \Delta T_1^{'} = T_{i_o} - T_{i_w} = 160 - 20 = 140^{\circ}C\) \(\Delta T_2^{'}\) = temperature difference at the outlet for the degraded case = \(T_{o_o}^{'} - T_{o_w}^{'}\) To calculate this, we need to know \(T_{o_o}^{'}\), the new oil outlet temperature. Let's first find \(Q^{'}\) value by using the given water outlet temperature, \(T_{o_w}^{'}\). \(Q^{'} = m_w * c_w * (T_{o_w}^{'} - T_{i_w})\) \(Q^{'} = m_o * c_o * (T_{i_o} - T_{o_o}^{'})\) We know the initial condition of heat transfer rate, \(Q = 3000 W\), and we can deduce: \(m_w * c_w * (80 - 20) = m_o * c_o * (160 - 140)\) Thus, the ratio \(\frac{m_w * c_w}{m_o * c_o} = \frac{160 - 140}{80 - 20} = 0.5\) Now let's use it to find \(T_{o_o}^{'}\): \(Q^{'} = 0.5 * (T_{i_o} - T_{o_o}^{'}) * (T_{o_w}^{'} - T_{i_w})\) \(Q^{'} = 0.5 * (160 - T_{o_o}^{'}) * (65 - 20)\) Now, solving for the new heat transfer rate and the new oil outlet temperature, we get: \(Q^{'} \approx 1687.5 W\) \(T_{o_o}^{'} \approx 149.55^{\circ}C\) Next, we find \(\Delta T_2^{'}\): \( \Delta T_2^{'} = T_{o_o}^{'} - T_{o_w}^{'} = 149.55 - 65 = 84.55^{\circ}C\) Now, find \(\Delta T_{lm}^{'}\): \(\Delta T_{lm}^{'} = \frac{(\Delta T_2^{'} - \Delta T_1^{'})}{ln(\Delta T_2^{'} / \Delta T_1^{'})} = \frac{(84.55 - 140)}{ln(84.55 / 140)} \approx -107.5^{\circ}C\) Now, let's find the new overall heat transfer coefficient, \(U^{'}\): \(Q^{'} = U^{'} * A * \Delta T_{lm}^{'}\) \(U^{'} = \frac{Q^{'}}{A * \Delta T_{lm}^{'}} = \frac{1687.5}{2 \pi * 0.02 * 1.59 * 107.5} \approx 262.6 W/m^2 K\) The water-side fouling factor, \(R_{f}^{n}\), can be found using the relation: \( \frac{1}{U^{'}} = \frac{1}{U} + R_{f}^{n}\) \( R_{f}^{n} = \frac{1}{U^{'}} - \frac{1}{U} \approx \frac{1}{262.6} - \frac{1}{500} = 0.0006438 m^2 K/W\) The corresponding values are: 1. Heat transfer rate: \(Q^{'} \approx 1687.5 W\) 2. Outlet temperature of the oil: \(T_{o_o}^{'} \approx 149.55^{\circ}C\) 3. Overall heat transfer coefficient: \(U^{'} \approx 262.6 W/m^2 K\) 4. Water-side fouling factor: \(R_{f}^{n} \approx 0.0006438 m^2 K/W\)