Answer:
pneumatic power system
Explanation:
pneumatic power can be used to quietly operate power windows, door locks, power mirrors, and much much more, also negative pressure pneumatics (vacuum) is used to control many engine and fuel systems
Determine if the fluid is satisfied
1 answer:
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Answer:
33.429 N-m
Explanation:
Given :
Inclination angle of two shaft, α = 20°
Speed of shaft A, = 1000 rpm
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Now we know that for maximum velocity,
= 1064.1 rpm
Now we know
Mass of flywheel, m = 30 kg
Radius of Gyration, k =100 mm
= 0.1 m
Therefore moment of inertia of flywheel, I = m.
=30 X
= 0.3 kg-
Now torque on the output shaft
T₂ = I x ω
= 0.3 X 1064.2 rpm
=
= 33.429 N-m
Torque on the Shaft B is 33.429 N-m
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
