91Ó°ÊÓ

(First, we need to find the heat transfer rate for cooling the process fluid. To do this, we will use the formula: $$Q = m_c \times C_{p,c} \times (T_{c1} - T_{c2})$$, where \(m_c\) is the mass flow rate of the process fluid, \(C_{p,c}\) is its specific heat, and \(T_{c1}\) and \(T_{c2}\) are its initial and final temperatures, respectively.) Given: - \(m_c = 2 \frac{\text{kg}}{\text{s}}\) - \(C_{p,c} = 3500 \frac{\text{J}}{\text{kg} \cdot \text{K}}\) - \(T_{c1} = 80^{\circ} \text{C}\) - \(T_{c2} = 50^{\circ} \text{C}\) Using these values, we can calculate the heat transfer rate: $$Q = 2 \cdot 3500 \cdot (80 - 50) = 210000 \ \text{W}$$

(To compute the LMTD for each configuration, we will use the equations established for parallel flow, counterflow, shell-and-tube, and cross-flow exchangers. For each configuration, we need to determine the inlet and outlet temperatures of the hot and cold fluids, and then calculate the LMTD using the respective formula.) For all configurations, the inlet temperature of the process fluid (hot fluid) and chilled water (cold fluid) are the same: - \(T_{h1} = 80^{\circ} \text{C}\) - \(T_{c1} = 15^{\circ} \text{C}\) Since we know that the outlet temperature of the process fluid is \(50^{\circ} \text{C}\), we can determine the outlet temperature of the chilled water using the energy balance equation: $$Q = m_w \times C_{p,w} \times (T_{w2} - T_{w1})$$, where \(m_w\) is the mass flow rate of the chilled water, \(C_{p,w}\) is its specific heat (assumed to be \(4180 \frac{\text{J}}{\text{kg} \cdot \text{K}}\) for water), and \(T_{w1}\) and \(T_{w2}\) are its inlet and outlet temperatures, respectively. Given: - \(m_w = 2.5 \frac{\text{kg}}{\text{s}}\) - \(C_{p,w} = 4180 \frac{\text{J}}{\text{kg} \cdot \text{K}}\) - \(T_{w1} = 15^{\circ} \text{C}\) We can solve for \(T_{w2}\): $$T_{w2} = \frac{Q}{m_w \times C_{p,w}} + T_{w1} = \frac{210000}{2.5 \times 4180} + 15 \approx 35.4^{\circ} \text{C}$$ Now, we can calculate the LMTD for each configuration. 1. Parallel flow: $$\text{LMTD}_{\text{parallel}} = \frac{(80 - 15) - (50 - 35.4)}{\ln \frac{80 - 15}{50 - 35.4}} \approx 42.51 \ \text{K}$$ 2. Counterflow: $$\text{LMTD}_{\text{counter}} = \frac{(80 - 35.4) - (50 - 15)}{\ln \frac{80 - 35.4}{50 - 15}} \approx 66.75 \ \text{K}$$ 3. Shell-and-tube, one shell pass and two tube passes: For this configuration, we will assume the LMTD to be the same as that for the counterflow. 4. Cross-flow, single pass, both fluids unmixed: For this configuration, we can assume the LMTD based on the arithmetic mean of the temperature differences: $$\text{LMTD}_{\text{cross}} \approx \frac{1}{2}[(80 - 15) + (80 - 35.4)] \approx 54.8 \ \text{K}$$

(Now that we have the LMTD for each configuration, we can use the heat exchanger equation to find the required heat transfer area: $$A = \frac{Q}{U \times \text{LMTD}}$$, where \(A\) is the heat transfer area and \(U\) is the overall heat transfer coefficient.) Given: - \(U = 2000 \frac{\text{W}}{\text{m}^2 \cdot \text{K}}\) We can now calculate the required heat transfer area for each configuration: 1. Parallel flow: $$A_{\text{parallel}} = \frac{210000}{2000 \times 42.51} \approx 2.47 \ \text{m}^2$$ 2. Counterflow: $$A_{\text{counter}} = \frac{210000}{2000 \times 66.75} \approx 1.57 \ \text{m}^2$$ 3. Shell-and-tube, one shell pass and two tube passes: $$A_{\text{shell-tube}} \approx A_{\text{counter}} = 1.57 \ \text{m}^2$$ 4. Cross-flow, single pass, both fluids unmixed: $$A_{\text{cross}} = \frac{210000}{2000 \times 54.8} \approx 1.92 \ \text{m}^2$$

(Based on the analysis, the required heat transfer areas for the different exchanger configurations are as follows: 1. Parallel flow: 2.47 m² 2. Counterflow: 1.57 m² 3. Shell-and-tube, one shell pass and two tube passes: 1.57 m² 4. Cross-flow, single pass, both fluids unmixed: 1.92 m² The counterflow and shell-and-tube configurations require the least heat transfer area, making them more efficient choices for this cooling process.)