Chapter 31, Problem 33P

To determine

To find: Amount of water needed per hour to run 1000 MW reactor.

Answer to Problem 33P

  $5.3\times {10}^3\text{kg}/\text{h}$

Explanation of Solution

Given data:

Power plant output, ${\mathrm{P}}_{\text{output}}=1000\text{W}$

Reactor efficiency, ${\mathrm{\eta }}_{\text{reactor}}=30\mathrm{\%}$

Formula used:

Power output, ${\mathrm{P}}_{\text{output}}={\mathrm{\eta }}_{\text{reactor}}{\mathrm{P}}_{\text{reactor}}$

Calculation:

Water contain two hydrogen molecule, and each hydrogen atom contain 0.000115 d’s

Since, efficiency is 30%, the total power of $\frac{1000\text{MW}}{0.3}=3333.3\text{MW}$ must be obtained from the fusion reactions to provide the required 1000 -MW output.

If the reaction of 4 deuterons, there would be a total of $7.30\text{MeV}$ energy released. So average energy released per deuteron is $1.825\text{MeV}$

Amount of water needed for $3333.3\text{MW}$ fusion reactions

  $\begin{array}{l}=\left[\begin{array}{l}\left(3333.3\times {10}^6\frac{\text{J}}{\text{s}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\left(\frac{1\text{MeV}}{1.60\times {10}^{-13}\text{J}}\right)\left(\frac{1\text{d}}{1.825\text{MeV}}\right)\left(\frac{1\text{H atom }}{0.000115\text{<msup><mo> d</mo><mo>′</mo></msup><mo> s</mo>}}\right)\\ \left(\frac{1{\text{H}}_2\text{O molecule }}{2\text{H atoms }}\right)\left(\frac{0.018{\text{kgH}}_2\text{O}}{6.02\times {10}^{23}\text{molecules}}\right)\end{array}\right]\\ =5342.4\text{kg}/\text{h}\\ \approx 5.3\times {10}^3\text{kg}/\text{h}\end{array}$

Conclusion:

Amount of water required to run the plant is $5.3\times {10}^3\text{kg}/\text{h}$