A mass of 8000 kg of slightly enriched uranium (2% U-235, 98% U-238) is exposed for 30 days in a reactor operating at (6.18) hea
t power 2000 MW. Neglecting consumption of U-238, what is the final fuel enrichment?
1 answer:
You might be interested in
Answer:
F = 8849 N
Explanation:
Given:
Load at a given point = F = 4250 N
Support span = L = 44 mm
Radius = R = 5.6 mm
length thickness of tested material = 12 mm
First compute the flexural strength for circular cross section using the formula below:
σ
σ = FL / π R³
Putting the given values in the above formula:
σ = 4250 ( 44 x 10⁻³ ) / π ( 5.6 x 10⁻³ ) ³
= 4250 ( 44 x 10⁻³ ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 4250 ( 11 / 250 ) / 3.141593 ( 5.6 x 10⁻³ ) ³
= 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³
= 187 / 3.141593 (0.0056)³
= 338943767.745358
= 338.943768 x 10⁶
σ = 338 x 10⁶ N/m²
Now we compute the load i.e. F from the following formula:
= 2 σ
d³/3 L
F = 2σd³/3L
= 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)
= 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)
= 676000000 ( 12 x 1/1000 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 3 / 250 )³ / 3 ( 44 x 10⁻³)
= 676000000 ( 27 / 15625000 ) / 3 ( 44 x 10⁻³)
= 146016 / 125 / 3 ( 44 x 1 / 1000 )
= ( 146016 / 125 ) / (3 ( 11 / 250 ))
= 97344 / 11
F = 8849 N
Answer:
Rate of heat transfer to river=1200MW
So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes
Explanation:
In order to find the actual heat transfer rate is lower or higher than its value we will first find the rate of heat transfer to power plant:
From First law of thermodynamics:
Rate of heat transfer to river=heat transfer to power plant-work done
Rate of heat transfer to river=2000-800
Rate of heat transfer to river=1200MW
So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes.