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:
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Answer:
a). 8.67 x m
b).0.3011 m
c).0.0719 m
d).0.2137 N
e).1.792 N
Explanation:
Given :
Temperature of air, T = 293 K
Air Velocity, U = 5 m/s
Length of the plate is L = 6 m
Width of the plate is b = 5 m
Therefore Dynamic viscosity of air at temperature 293 K is, μ = 1.822 X Pa-s
We know density of air is ρ = 1.21 kg /
Now we can find the Reyonld no at x = 1 m from the leading edge
Re =
Re =
Re = 332052.6
Therefore the flow is laminar.
Hence boundary layer thickness is
δ =
=
= 8.67 x m
a). Boundary layer thickness at x = 1 is δ = 8.67 X m
b). Given Re = 100000
Therefore the critical distance from the leading edge can be found by,
Re =
100000 =
x = 0.3011 m
c). Given x = 3 m from the leading edge
The Reyonld no at x = 3 m from the leading edge
Re =
Re =
Re = 996158.06
Therefore the flow is turbulent.
Therefore for a turbulent flow, boundary layer thickness is
δ =
=
= 0.0719 m
d). Distance from the leading edge upto which the flow will be laminar,
Re =
5 X =
x = 1.505 m
We know that the force acting on the plate is
=
and at x= 1.505 for a laminar flow is =
=
= 1.878 x
Therefore, =
=
= 0.2137 N
e). The flow is turbulent at the end of the plate.
Re =
=
= 1992316
Therefore =
=
= 3.95 x
Therefore =
=
= 1.792 N
Answer:
(b). T = 22.55 ⁰C
(c). q = 557.8 W
Explanation:
we take follow a step by step process to solving this problem.
from the question, we have that
The two glass pieces is separated by a 1.8 cm distance layer of air.
the thickness of glass piece is 1 cm
width = 4 m
the height = 3 m
(a). the sketch of the thermal circuit is uploaded in the picture below.
(b). the thermal resistance due to the conduction in the first glass plane is given thus;
R₁ = Lg / Kg A ................(1)
given that Kg rep. the thermal conductivity of the glass plane
A = conduction surface area
Lg = Thickness of glass plane4
taking the thermal conductivity of glass plane as Kg = 0.78 w/mk
inputting values into equation (1) we have,
R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]
R₁ = 1.068 ˣ 10 ⁻³ k/w
Being that we have same thermal resistance in the first and second plane,
therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w
⇒ Also the thermal resistance between air and glass as a result of the conduction by the layer is given thus
R₂ = La/KaA .....................(2)
given Ka = thermal conductivity of air
A = surface area
La = thickness of air
substituting values into the equation we have
R₂ = [1.8 (cm) ˣ 1 (m)/100 (cm)] / [(0.0262 w/mk)(4m ˣ 3m)]
R₂ = 5.73 ˣ 10⁻² k/w
Given the thermal resistance on the outer surface due to convection, we have
R₄ = 1/hA
inputting value gives R₄ = 1 / (12 w/m² ˣ 12m) = 6.94 ˣ 10⁻³k/w
R₄ = 6.94 ˣ 10⁻³k/w
Finally the sum total of thermal resistance = R₁ + R₂ + R₃ + R₄
R-total = 0.0663 kw
From this we can calculate the rate of heat loss
using q = Ti - To / R-total ..............(3)
given Ti and To is the inside and outside temperature i.e. 27⁰C and -10⁰C
from equation (3),
q = 27- (-10) / 0.0063 = 557.8 W
q = 557.8 W
⇒ Applying the heat transfer formula for inside surface glass temperature gives;
q = Ti - T₂ / R₃ + R₄
T₂ = Ti - q (R₃ + R₄)
T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C
T₂ = 22.55°C
cheers i hope this helps
