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MArishka [77]
3 years ago
13

) If the blood viscosity is 2.7x10-3 Pa.s, length of the blood vessel is 1 m, radius of the blood vessel is 1 mm, calculate the

tubular resistance of the blood vessel (in GPa.S/m3 ). If the blood pressure at the inlet of the above vessel is 43 mm Hg and if the blood pressure at the outlet of the above vessel is 38 mm Hg. Calculate the flow rate (in ml/min).
Engineering
1 answer:
Illusion [34]3 years ago
5 0

Answer:

a) the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b) the flow rate is 5.8 ml/min

Explanation:

Given the data in the question;

Length of blood vessel L = 1 m

radius r = 1 mm = 0.001 m

blood viscosity μ = 2.7 × 10⁻³ pa.s = 2.7 × 10⁻³ × 10⁻⁹ Gpa.s = 2.7 × 10⁻¹² Gpa.s

Now, we know that Resistance = 8μL / πr⁴

so we substitute

Resistance = [8 × (2.7 × 10⁻¹²) × 1] / [π(0.001)⁴]

Resistance = [2.16 × 10⁻¹¹] / [3.14159 × 10⁻¹²]

Resistance = 6.8755 ≈ 6.88 Gpa.s/m³

Therefore, the tubular resistance of the blood vessel is 6.88 Gpa.s/m³.

b)

blood pressure at the inlet of the vessel = 43 mm Hg

blood pressure at the outlet of the vessel = 38 mm Hg

flow rate = ?

we know that;

flow rate Q = ΔP / R

where ΔP is change in pressure and R is resistance.

ΔP = Inlet pressure - Outlet pressure = 43 - 38 = 5 mm Hg =  665 pa

R = 6.8755 Gpa.s/m³ = { 6.8755 × 10⁹ / 60 × 10⁶ } = 114.5916 pa.min.ml⁻¹

so we substitute

Q =  665 pa / 114.5916 pa.m.ml⁻¹

Q = 5.8 ml/min

Therefore, the flow rate is 5.8 ml/min

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We can use the equation

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Q = 62437.833 mm³

we could also use this equation in order to get Q:

Q = (4/3)*(R³-r³)

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V = (83 N/mm²)*(11 mm)*(2615942.11 mm⁴) / (62437.833 mm³)

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