Basil is installing a system in Toronto, Ontario where they are wide temperature variations. What is the TD of the system he’s i
1 answer:
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Answer:
correct option is (A) 0.5
Explanation:
given data
axial column load = 250 kN per meter
footing placed = 0.5 m
cohesion = 25 kPa
internal friction angle = 5°
solution
we know angle of internal friction is 5° that is near to 0°
so it means the soil is almost cohesive soil.
and for a pure cohesive soil
= 0
and we know formula for is
= (Nq - 1 ) × tan(Ф) ..................1
so here Ф is very less should be nearest to zero
and its value can be 0.5
so correct option is (A) 0.5
This question is incomplete, the complete question is;
For a steel alloy it has been determined that a carburizing heat treatment of 11.3 h duration at Temperature T1 will raise the carbon concentration to 0.44 wt% at a point 1.8 mm from the surface. A separate experiment is performed at T2 that doubles the diffusion coefficient for carbon in steel.
Estimate the time necessary to achieve the same concentration at a 4.9 mm position for an identical steel and at the same carburizing temperature T2.
Answer:
the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Explanation:
Given the data in the question;
treatment time t₁ = 11.3 hours
Carbon concentration = 0.444 wt%
thickness at surface x₁ = 1.8 mm = 0.0018 m
thickness at identical steel x₂ = 4.9 mm = 0.0049 m
Now, Using Fick's second law inform of diffusion
/ Dt = constant
where D is constant
then
/ t = constant
/ t₁ =
/ t₂
t₂ = t₁
t₂ = t₁ /
t₂ = ( /
)t₁
t₂ =
/
× t₁
so we substitute
t₂ = 0.0049 / 0.0018
× 11.3 hrs
t₂ = 7.41 × 11.3 hrs
t₂ = 83.733 hrs
Therefore, the required time to achieve the same concentration at a 4.9 is 83.733 hrs
Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e
r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m