c) Exit temperature of river water:
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:
\(T_{exit} ≈ 37.3^{\circ}C\)
