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Answer:
Explanation:
Given the data in the question;
L = 46 in
Ga = 5 × 10³ ksi
Gs = 11 × 10³ ksi
Outside diameter da = 5 in
ds = 4 in
Tb = 3 kip.in
Now,
Ja = polar moment of Inertia of Aluminum;
Ja ⇒ π/32( 5⁴ - 4⁴ ) = π/32( 625 - 256 ) = π/32( 369 )
Js = polar moment of inertia of steel
Js ⇒ π/32 ds⁴ = π/32( 4⁴ ) = π/32( 256 )
Ta is torque transmitted by Aluminum
Ts is torque transmitted by steel
{composite member }
T = Ta + Ts ------ let this be equation m1
Now, we use the relation;
T/J = G∅/L
JG∅ = TL
∅ = TL/GJ
so, for aluminum rod ∅ = TaLa/GaJa
for steel rod ∅ = TsLs/GsJs
but we know that, ∅a = ∅s = ∅
so
[TaLa/GaJa] = [TsLs/GsJs]
also, we know that, La = Ls = L
∴ [Ta/GaJa] = [Ts/GsJs]
we solve for Ta
TaGsJs = TsGaJa
Ta = TsGaJa / GsJs
we substitute
Ta = [Ts(5 × 10³)( π/32( 369) )] / [ (11 × 10³)( π/32( 256 ) ) ]
Ta = 0.66Ts
now, we substitute 0.66Ts for Ta and 3 for T in equation 1
T = Ta + Ts
3 = 0.66Ts + Ts
3 = 1.66Ts
Ts = 3 / 1.66
Ts = 1.8072 ≈ 1.81 kip-in
so
∅ = TsLs / GsJs
we substitute
∅ = (1.81 × 46 ) / ( 11 × 10³ × π/32( 256 ) )
∅ = 83.26 / 276460.1535
∅ = 0.000301
∅ = 3.01 × 10⁻⁴ rad
so
∅ = ∅
= 3.01 × 10⁻⁴ rad
Therefore, the magnitude of the angle of twist at end B is 3.01 × 10⁻⁴ rad
Answer:
4.8°C
Explanation:
The rate of heat transfer through the wall is given by:
Assumptions:
1) the system is at equilibrium
2) the heat transfer from foam side to interface and interface to block side is equal. There is no heat retention at any point
3) the external surface of the wall (concrete block side) is large enough that all heat is dissipated and there is no increase in temperature of the air on that side
temperature at the interface
Solving for will give the temperature at the interface:
Answer:
The correct answer is "1341.288 W/m".
Explanation:
Given that:
T₁ = 300 K
T₂ = 500 K
Diameter,
d = 0.2 m
Length,
l = 1 m
As we know,
The shape factor will be:
⇒
By putting the value, we get
⇒
⇒
hence,
The heat loss will be:
⇒
This question is incomplete, the missing diagram is uploaded along this answer below.
Answer:
the forces required to keep the artery in place is 1.65 N
Explanation:
Given the data in the question;
Inlet velocity V₁ = 50 cm/s = 0.5 m/s
diameter d₁ = 15 mm = 0.015 m
radius r₁ = 0.0075 m
diameter d₂ = 11 mm = 0.011 m
radius r₂ = 0.0055 m
A₁ = πr² = 3.14( 0.0075 )² = 1.76625 × 10⁻⁴ m²
A₂ = πr² = 3.14( 0.0055 )² = 9.4985 × 10⁻⁵ m²
pressure at inlet P₁ = 110 mm of Hg = 14665.5 pascal
pressure at outlet P₂ = 95 mm of Hg = 12665.6 pascal
Inlet volumetric flowrate = A₁V₁ = 1.76625 × 10⁻⁴ × 0.5 = 8.83125 × 10⁻⁵ m³/s
given that; blood density is 1050 kg/m³
mass going in m' = 8.83125 × 10⁻⁵ m³/s × 1050 kg/m³ = 0.092728 kg/s
Now, using continuity equation
A₁V₁ = A₂V₂
V₂ = A₁V₁ / A₂ = (d₁/d₂)² × V₁
we substitute
V₂ = (0.015 / 0.011 )² × 0.5
V₂ = 0.92975 m/s
from the diagram, force balance in x-direction;
0 - P₂A₂ × cos(60°) + Rₓ = m'( V₂cos(60°) - 0 )
so we substitute in our values
0 - (12665.6 × 9.4985 × 10⁻⁵) × cos(60°) + Rₓ = 0.092728( 0.92975 cos(60°) - 0 )
0 - 0.6014925 + Rₓ = 0.043106929 - 0
Rₓ = 0.043106929 + 0.6014925
Rₓ = 0.6446 N
Also, we do the same force balance in y-direction;
P₁A₁ - P₂A₂ × sin(60°) + R = m'( V₂sin(60°) - 0.5 )
we substitute
⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R = 0.092728( 0.92975sin(60°) - 0.5 )
⇒ 1.5484 + R = 0.092728( 0.305187 )
⇒ 1.5484 + R = 0.028299
R = 0.028299 - 1.5484
R = -1.52 N
Hence reaction force required will be;
R = √( Rₓ² + R² )
we substitute
R = √( (0.6446)² + (-1.52)² )
R = √( 0.41550916 + 2.3104 )
R = √( 2.72590916 )
R = 1.65 N
Therefore, the forces required to keep the artery in place is 1.65 N
