Question

A nuclear power plant operates at 79 percent of its maximum theoretical (Carnot) efficiency between temperatures of 700° and 330°C. If the plant produces electric energy at the rate of 1.3 GW, how much exhaust heat is discharged per hour?

Answer 1

Answer:

Explanation:

T₁ = 700 + 273 = 973 k

T₂ = 330 + 273 = 603 k

Theoretical efficiency = T₁ - T₂ / T₁

= (973 - 603) / 973

= .38 OR 38%

Operating efficiency = .79 x 38

= 30.02 %

Heat input Q₁ , Heat output to sink Q₂ , conversion into power = Q₁ - Q₂

given Q₁ - Q₂ = 1.3 x 10⁹ W

efficiency = Q₁ - Q₂  /  Q₁

Q₁ - Q₂  /  Q₁ = 30.02 / 100

100Q₁ - 100Q₂ = 30.02Q₁

69.98 Q₁  = 100Q₂

Q₁  = 1.429 Q₂

Putting this in the relation

Q₁ - Q₂ = 1.3 x 10⁹ W

1.429Q₂ - Q₂ = 1.3 x 10⁹ W

.429Q₂ = 1.3 x 10⁹

Q₂  = 3.03 x 10⁹W

= 3.03 GW.