The pressure drop of air in the bed is 14.5 kPa.
<u>Explanation:</u>
To calculate Re:
From the tables air property
Ideal gas law is used to calculate the density:
ρ = 
ρ = 1.97 Kg / 
ρ = 
R = 
= 8.2 × 
/ 28.97×
R = 2.83 × atm / K Kg
q is expressed in the unit m/s
q = 1.24 m/s
Re = 
Re = 2278
The Ergun equation is used when Re > 10,
= 4089.748 Pa/m
ΔP = 4089.748 × 3.66
ΔP = 14.5 kPa
Explanation:
thermal expansion ∝L = (δL/δT)÷L ----(1)
δL = L∝L + δT ----(2)
we have δL = 12.5x10⁻⁶
length l = 200mm
δT = 115°c - 15°c = 100°c
putting these values into equation 1, we have
δL = 200*12.5X10⁻⁶x100
= 0.25 MM
L₂ = L + δ L
= 200 + 0.25
L₂ = 200.25mm
12.5X10⁻⁶ *115-15 * 20
= 0.025
20 +0.025
D₂ = 20.025
as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0
Answer:
Some of the internal strain energy is relieved.
There is some reduction in the number of dislocations.
The electrical conductivity is recovered to its precold-worked state.
The thermal conductivity is recovered to its precold-worked state
Explanation:
The process of the recovery of a cold-worked material happens at a very low temperature, this process involves the movement and annihilation of points where there are defects, also there is the annihilation and change in position of dislocation points which leads to forming of the subgrains and the subgrains boundaries such as tilt, twist low angle boundaries.
