Question

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 11mm. 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)

Answer 1

To find the forces required to maintain the position of a carotid artery segment with given inlet and outlet pressures and velocities, one must use the continuity equation to relate areas and velocities and Bernoulli's equation to relate pressures and velocities. The net force is then calculated from the pressure difference and cross-sectional area.

Step-by-step explanation:

The student's question involves applying principles of fluid mechanics, specifically the continuity equation and Bernoulli's equation, to calculate the forces required to keep a segment of the carotid artery in place given the inlet and outlet velocities and diameters, as well as the inlet and outlet pressures. The blood density is provided as well.

To solve this problem, we start by employing the continuity equation (A1v1 = A2v2) to find the relation between inlet and outlet velocities and cross-sectional areas. Then, Bernoulli's equation can be applied to relate pressures and velocities at the inlet and outlet.

Once we have all the necessary parameters, we can use the difference in pressures along with the cross-sectional area to calculate the net force (F = ΔP x A) needed to keep the artery in place. This force is the result of the pressure difference across the artery section and is necessary for the maintenance of a steady state in the absence of other forces such as the vessel's mass or the gravitational pull on the blood.

Answer 2

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

Step-by-step 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