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Amiraneli [1.4K]

2 years ago

13

2. The following segment of carotid artery has an inlet velocity of 50 cm/s (diameter of 15 mm). The outlet has a diameter of 11

mm. The pressure at inlet is 110 mm of Hg and pressure at outlet is 95 mm of Hg. Determine the forces required to keep the artery in place (consider steady state, ignore the mass of blood in the vessel and the mass of blood vessel; blood density is 1050 kg/m3)

1 answer:

ahrayia [7]2 years ago

7 0

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 $_y$ = m'( V₂sin(60°) - 0.5 )

we substitute

⇒ (14665.5 × 1.76625 × 10⁻⁴) - (12665.6 × 9.4985 × 10⁻⁵) × sin(60°) + R $_y$ = 0.092728( 0.92975sin(60°) - 0.5 )

⇒ 1.5484 + R $_y$ = 0.092728( 0.305187 )

⇒ 1.5484 + R $_y$ = 0.028299

R $_y$ = 0.028299 - 1.5484

R $_y$ = -1.52 N

Hence reaction force required will be;

R = √( Rₓ² + R $_y$ ² )

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

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Answer:

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Explanation:

given data

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and we know that is also in term of change in density also

so

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

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$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

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dV % = 2.60 %

so percentage change in volume is 2.60%

and

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now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

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2 years ago

Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 240°C. The water is slowly heated at constan

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Answer:

The final temperature of water is 381.39 °C.

Explanation:

Given that

Mass of water = 5 kg

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$Q=mC_p\Delta T$

We know that for water

$C_p=4.187\ \frac{KJ}{kg.K}$

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Answer:

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Explanation:

Previous concepts

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Applying Newton’s second law to the right hand side of the above equation, we have that r ×ma = r ×F =

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MO = H˙ O

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The equation MO = H˙ O gives us the instantaneous relation between the moment and the time rate of change of angular momentum. Imagine now that the force considered acts on a particle between time t1 and time t2. The equation MO = H˙ O can then be integrated in time to obtain this:

$\int_{t_1}^{t_2}M_O dt = \int_{t_1}^{t_2}H_O dt=H_0t2 -H_0t1$

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$H_Ai +\sum \int_{t_i}^{t_f} M_A dt =H_Af$

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And if we solve for $\omega$ we got:

$\omega =\frac{24}{1.14375}=20.983\frac{rad}{s}$

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Answer:

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