91Ó°ÊÓ

First, we need to determine the heat capacity rates for both fluids, which is given by the product of mass flow rate and specific heat capacity: \( C = m \cdot c_p \) For water, we are given the mass flow rate m_w = 2.5 kg/s. We need to find the specific heat capacity for water, which is typically approximated as c_pw = 4.18 kJ/(kg·K) or 4180 J/(kg·K). Now we can calculate the heat capacity rate for water: \( C_{water} = m_w \cdot c_{pw} = 2.5 \mathrm{~kg/s} \times 4180 \mathrm{~J/(kg·K)} = 10450 \mathrm{~W/K} \) For air, we are given the inlet temperature T_hi = 200°C and the outlet temperature T_ho = 93°C. In general, the specific heat capacity for air can be approximated as c_pa = 1.006 kJ/(kg·K) or 1006 J/(kg·K). To calculate the mass flow rate of air, we should use the energy balance equation: \( Q = m_{air} \cdot c_{pa} \cdot (T_{hi} - T_{ho}) \) Since the heat exchanged between the fluids is the same, we can express Q as: \( Q = C_{water} \cdot (T_{wo} - T_{wi})\) Where T_wi = 35°C and T_wo = 85°C. Now we can calculate the mass flow rate of air: \( m_{air} = \frac{C_{water} \cdot (T_{wo} - T_{wi})}{c_{pa} \cdot (T_{hi} - T_{ho})} = \frac{10450 \mathrm{~W/K} \cdot (85 - 35) \mathrm{~K}}{1006 \mathrm{~J/(kg·K)} \cdot (200 - 93) \mathrm{~K}} = 0.6145 \mathrm{~kg/s}\) Finally, we find the heat capacity rate for air: \( C_{air} = m_{air} \cdot c_{pa} = 0.6145 \mathrm{~kg/s} \times 1006 \mathrm{~J/(kg·K)} = 618.269 \mathrm{~W/K} \)

The effectiveness of the heat exchanger is given by the ratio of the actual heat transfer rate, Q, to the maximum possible heat transfer rate, Q_max: \( \epsilon = \frac{Q}{Q_{max}} \) Since both fluids have the same heat capacity rates, Q_max can be calculated as: \( Q_{max} = \min(C_{water}, C_{air}) \cdot (T_{hi} - T_{wi}) \) In this case: \( Q_{max} = C_{air} \cdot (T_{hi} - T_{wi}) = 618.269 \mathrm{~W/K} \cdot (200 - 35) \mathrm{~K} = 102064.435 \mathrm{~W} \) We have already calculated Q using the energy balance equation between the fluids in step 1: \( Q = C_{water} \cdot (T_{wo} - T_{wi}) = 10450 \mathrm{~W/K} \cdot (85 - 35) \mathrm{~K} = 522500 \mathrm{~W} \) Now we can calculate the effectiveness: \( \epsilon = \frac{522500 \mathrm{~W}}{102064.435 \mathrm{~W}} = 5.12 \)

Using the effectiveness-NTU relationship for a shell-and-tube exchanger with one shell pass and an arbitrary number of tube passes, we have: \( \epsilon = \frac{1 - e^{-NTU(1 - e^{-T})}}{1 + T(1 - e^{-T})} \) Where: - \( \epsilon \) is the effectiveness of the heat exchanger (calculated in step 2). - NTU is the Number of Transfer Units. - T is the ratio of heat capacity rates, given by \( T = \frac{C_{min}}{C_{max}} \). In our case, T is given by: \( T = \frac{C_{air}}{C_{water}} = \frac{618.269 \mathrm{~W/K}}{10450 \mathrm{~W/K}} = 0.0592 \) Now we need to numerically solve the effectiveness-NTU relationship for the NTU value, given \( \epsilon = 5.12 \) and T = 0.0592. Using numerical methods (e.g., Newton-Raphson or bisection method), we find NTU ≈ 0.1659.

Finally, the heat transfer area A can be calculated using the overall heat transfer coefficient U and the NTU value: \( A = \frac{NTU \cdot C_{min}}{U} \) The given overall heat transfer coefficient U = 180 W/(m²·K). Plugging in the values, we get: \( A = \frac{0.1659 \cdot 618.269 \mathrm{~W/K}}{180 \mathrm{~W/(m²·K)}} = 0.57 \mathrm{~m²} \) Therefore, the area of the heat exchanger is approximately 0.57 m².