91Ó°ÊÓ

First, we need to find the heat transfer rate (Q) required to heat 10,000 kg/h of pressurized water from 35°C to 120°C. We'll use the heat capacity formula: \[Q = m \cdot c_p \cdot \Delta T\] where - \(Q\) = heat transfer rate (W) - \(m\) = mass flow rate (kg/s) - \(c_p\) = specific heat capacity of water (J/kg*K) (We will use 4,184 J/kg*K for water) - \(\Delta T\) = change in temperature (K) First, we need to convert the mass flow rate from kg/h to kg/s by dividing by 3600: \[m = \frac{10000 \mathrm{~kg/h}}{3600 \mathrm{s/h}} = 2.7778 \mathrm{~kg/s}\] Now we can calculate Q: \[Q=2.7778 \mathrm{~kg/s} \times 4,184 \mathrm{~J/kg \cdot K}\times (120-35)^{\circ}\mathrm{C}\]

Now we can calculate the heat transfer rate Q: \[Q = 2.7778 \times 4,184 \times 85 = 985,892 \mathrm{~W}\]

Now we need to find the heat exchanger area (A) using the overall heat transfer coefficient (U) and the heat transfer rate (Q) calculated in the previous step. We can use the formula: \[Q = UA\Delta T_{lm}\] where - \(U\) = overall heat transfer coefficient = 1500 W/m²·K - A = heat exchanger area (m²) - \(\Delta T_{lm}\) = log mean temperature difference (K) But first, we need to calculate the log mean temperature difference (\(\Delta T_{lm}\)). We know the temperatures of the cold stream are 35°C and 120°C, while the hot stream enters at 300°C. We can assume the hot stream exits at 120°C since all its heat is transferred to the cold stream. With this information, we can calculate the temperature difference at both ends of the heat exchanger: \[\Delta T_1 = T_{h_1} - T_{c_1} = 300 - 35 = 265^{\circ}\mathrm{C}\] \[\Delta T_2 = T_{h_2} - T_{c_2} = 120 - 120 = 0^{\circ}\mathrm{C}\] Now we can use these values to calculate the log mean temperature difference: \[\Delta T_{lm} = \frac{\Delta T_1 - \Delta T_2}{\ln{\frac{\Delta T_1}{\Delta T_2}}} = \frac{265 - 0}{\ln{\frac{265}{0}}}\] However, since the temperature difference \(\Delta T_2\) is zero, the logarithm cannot be calculated. In this situation, it is reasonable to assume that the heat exchanger is very efficient, and the temperature difference approaches zero. Hence, we can use \(\Delta T_1\) as the log mean temperature difference: \[\Delta T_{lm} \approx 265^{\circ}\mathrm{C}\] Now we can calculate the heat exchanger area (A): \[A = \frac{Q}{U \Delta T_{lm}} = \frac{985,892 \mathrm{~W}}{1500 \mathrm{~W/m^2 \cdot K} \times 265 \mathrm{~K}}\]

Now we can calculate the heat exchanger area A: \[A = \frac{985,892}{1500 \times 265} \approx 2.480 \mathrm{~m^2}\] So the required heat exchanger area is approximately 2.480 square meters.