Question

-The Unit 1 reactor was a “boiling water” type reactor. These reactors generally work under high pressure and produce steam at temperatures around 380C. The power plant is a “heat engine”, so some of this thermal energy is released to the atmosphere and some is converted to electrical power. Use the efficiency equation from chapter 2 (page 60) to estimate the maximum possible efficiency for this reactor.

[The formula is Efficiency is less than or equal to 1 - (Tcold/Thot)]

-The Unit 1 reactor was capable of producing 460MW of electrical power. Assuming that the plant’s efficiency was about 10% less than the optimal efficiency you found in the previous question, what was the thermal power output (in units of MW) of the reactor?

- The reactor was shut down quickly after the earthquake struck, and hence the nuclear chain reaction was halted. However, the highly radioactive fission fragments within the reactor continued to decay and thus produce heat (the so-called “decay heat”). Combine your estimate of the thermal output during normal operation with the information in your textbook (see text above Fig. 5.13) to estimate the thermal output of the reactor just after the shutdown and one year after the shutdown.

-Heat can be removed from the reactor complex by allowing it to boil water. The process of boiling off a gallon (=about 4 liters) of water that is initially at 20C takes approximately 9.8×106 Joules (9.8MJ) . Combine this information with your previous work to compute the rate that water must be supplied (in gallons per second) to carry away the heat so the reactor does not heat up and melt. Remember that a Watt is a Joule per second, or 1W = 1J/s.

Answer 1

Maximum efficiency,

Efficiency = 10% less than maximum efficiency = 0.496 = Work / Thermal input = 460 / Thermal power. So, thermal power output = 927.3MW.

Wigner-Way's formula: power output at t sec after shutdown, where, P₀ is the thermal power output before the shutdown (927.3 MW here), t is the time elapsed since shut down (2 cases here: 0 sec and 1year), t₀ is the time from the setup of reactor to its shutdown. Since t₀ is not given, we cannot calculate P_d (0) or P_d(1year).

The maximum thermal power is given off immediately after the shutdown. So the maximum water flow required is calculated as shown. Let P_d(0) = x MW, i.e., it is giving out x MJ of thermal power every second. 1 gallon of water takes 9.8 MJ of energy, so x MJ of energy requires x/9.8 gallons. Since x MJ is produced in 1 second, we need (x/9.8) gallons/ sec to carry the heat away. This flow rate can be reduced as the decay heat reduces with time.