Answer:
Modulus of resilience will be
Explanation:
We have given yield strength
Elastic modulus E = 104 GPa
We have to find the modulus
Modulus of resilience is given by
Modulus of resilience , here
is yield strength and E is elastic modulus
Modulus of resilience
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
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
