A coal-burning power plant produces \(3000 .\) MW of thermal energy, which is used to boil water and produce supersaturated steam at \(300 .{ }^{\circ} \mathrm{C}\). This high-pressure steam turns a turbine producing \(1000 .\) MW of electrical power. At the end of the process, the steam is cooled to \(30.0^{\circ} \mathrm{C}\) and recycled. a) What is the maximum possible efficiency of the plant? b) What is the actual efficiency of the plant? c) To cool the steam, river water runs through a condenser at a rate of \(4.00 \cdot 10^{7} \mathrm{gal} / \mathrm{h}\). If the water enters the condenser at \(20.0^{\circ} \mathrm{C}\), what is its exit temperature?

To find the maximum possible efficiency for the plant, we can use the Carnot efficiency formula: Carnot efficiency = \(1 - \frac{T_C}{T_H}\) where \(T_C\) is the temperature of the cold reservoir and \(T_H\) is the temperature of the hot reservoir. Now, we must convert the temperatures to Kelvin: \(T_C = 30 + 273.15 = 303.15\) K \(T_H = 300 + 273.15 = 573.15\) K Plugging these values into the Carnot efficiency formula, we get: Carnot efficiency = \(1 - \frac{303.15}{573.15} ≈ 0.471\) or \(47.1\%\)

To find the actual efficiency of the plant, we can use the following formula: Actual efficiency = \(\frac{Electrical \ power \ produced}{Thermal \ power \ supplied}\) Given that the electrical power produced is \(1000\) MW and the thermal power supplied is \(3000\) MW, the actual efficiency is: Actual efficiency = \(\frac{1000}{3000} = 0.333\) or \(33.3\%\)

To find the exit temperature of the river water, we need to equate the heat removed from the steam during the cooling process to the heat gained by the river water: \(Q_{steam} = Q_{water}\) where \(Q_{steam}\) refers to the heat removed from the steam, and \(Q_{water}\) refers to the heat absorbed by the river water. For both steam and water, specific heat capacities are needed, so assume: \(c_{steam} = 4.18 \, kJ/kgK (specific \, heat \, capacity \, of \, water)\) \(c_{water} = 4.18 \, kJ/kgK (specific \, heat \, capacity \, of \, water)\) The mass of river water (\(m_{water}\)) can be found using the given volume flow rate and density of water: \(4.00 \cdot 10^{7} \, gal/h \times \frac{3.78541 \, L}{1 \, gal} \times \frac{1\, kg}{1\, L} \times \frac{1\, h}{3600 \, s} = 42220.22 \, kg/s\) Now, we can equate the heat removed from the steam to the heat absorbed by the river water: \((m_{steam} c_{steam} \Delta T_{steam}) = (m_{water} c_{water} \Delta T_{water})\) Since the steam is cooled from \(300^{\circ}\)C to \(30^{\circ}\)C, and the river water enters at \(20^{\circ}\)С, we have: \(\Delta T_{steam} = 270\) K \(\Delta T_{water} = (T_{exit} - 20)\) K, where \(T_{exit}\) is the exit temperature of the river water in Celsius. To find the mass of steam (\(m_{steam}\)), we use the electrical energy output and actual efficiency: \(P_{thermal} = 3000 \, MW = 3000 \cdot 10^6 \, W\) \(P_{thermal} = m_{steam} c_{steam} \Delta T_{steam}\) Thus, \(m_{steam} = \frac{P_{thermal}}{c_{steam} \Delta T_{steam}}\) Now we can substitute everything into the equation for the heat balance: \((\frac{P_{thermal}}{c_{steam} \Delta T_{steam}} c_{steam} 270) = (42220.22 \, kg/s \cdot c_{water} (T_{exit} - 20))\) Solving for \(T_{exit}\), we find: \(T_{exit} ≈ 37.3^{\circ}C\)