This was a practice problem for a test review.
A nuclear-fueled electric power plant utilizes a so-called “boiling water reactor”. In this type of reactor, nuclear energy causes water under pressure to boil at 283 °C (the temperature of the hot reservoir). After the steam does the work of turning the turbine of an electric generator, the steam is converted back into water in a condenser at 42.0 °C (the temperature of the cold reservoir). To keep the condenser at 42.0 oC, the rejected heat must be carried away by some means – for example, by water from a river. The plant operates at three-fourths of its Carnot efficiency, and the electrical output power of the plant is 1.20 109 watts. A river with a water flow rate of 9.60 104 kg/s is available to remove the rejected heat from the plant. Find the number of degrees in oC by which the temperature of the river rises.
2 Responses
If I may offer my humble attempt at an answer:
To = 283 C, Tc = 42 C
η = 0.75(ηc) = 0.75(1 – Tc/Th) = 0.75(1 – 42/283) = 0.75*0.8516 = 0.6387
P’ = 1.2×10^9 W,
P = 1.2×10^9/η = 1.2×10^9/0.6387 = 1.879×10^9 W
P – P’ = dQ/dt = (dm/dt)CΔT = (9.6×10^4)4.186[T – 42] = (1.88 – 1.20)x10^6 kJ/s
T = (1.68×10^6) + 0.42*9.6*4.186×10^6)/9.6*4.186×10^4
T = (18.558/40.18)x10² = 46 C
This answer may not be correct since what results is too high of a temp. rise!
Jacy, you have to calculate the Carnot efficiency using the temperatures in Kelvin:
η = 1 – Tc/Th
η = 1 – (42 + 273.15)/(283 + 273.15)
η = 1 – 315.15/556.15
η = 1 – 0.567
η = 0.433
Overall efficiency = 3/4 * 0.433 = 0.325
Now calculate the power wasted as heat to the river:
P = 1.2 * 10^9 / 0.325 * 0.675
P = 2.49 * 10^9 W
Now that we have a known amount of energy per second going into the river, and a known amount of mass flow per second of the river. We can calculate the temperature rise:
Q = mcΔT
2.49 * 10^9 = 9.6 * 10^4 * 4186 * ΔT
2.49 * 10^9 = 401856000 * ΔT
ΔT = 2.49 * 10^9 / 401856000
ΔT = 6.20 K (or C)
And that’s the answer.