Question

Assume that in 2010 the United States will need 2.0×1012 watts of electric power produced by thousands of 1000 MW power plants. These will almost certainly be powered by either coal or nuclear reactors. If all the energy is provided by U-235 burning reactors, how many tonnes of U-235 will be needed annually? NOTE: Rather than using the exact energy from problem 1, please assume that fission of U-235 releases a "rounded" 8.E13 J/kg (Assume a realistic efficiency of 45 percent for the power plant) Do not enter units. If the answer is 1 million tonnes, enter 1.E6. Note that a tonne is a metric ton = 1000 kg)

Answer 1

1752.14 tonnes per year.

Step-by-step explanation:

To solve this exercise it is necessary to apply the concepts related to power consumption and power production.

By conservation of energy we know that:

`\dot{P} = \bar{P}`

Where,

`\dot{P} =` Production of Power

`\bar{P} =` Consumption of power

Where the production of power would be,

`\dot{P} = m \dot{E}\eta`

Where,

m = Total mass required

`\dot{E} =` Energy per Kilogram

`\eta =`Efficiency

The problem gives us the aforementioned values under a production efficiency of 45%, that is,

`\dot{P} = \bar{P}`

`m \dot{E}\eta = \bar{P}`

Replacing the values we have,

`m(8*10^13)(0.45) = 2*10^(12)`

Solving for m,

`m = ( 2*10^(12))/((8*10^13)(0.45))`

`m = 0.0556 (kg)/(s)`

We have the mass in kilograms and the time in seconds, we need to transform this to tons per year, then,

`m = 0.556(kg)/(s)*((3.1536*10^7s)/(1year))((1ton)/(1000kg))`

`m = 1752.14`tonnes per year.