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3 years ago

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The aorta is approximately 25 mm in diameter. The mean pressure there is about 100 mmHg and the blood flows through the aorta at

approximately 60 cm/s. Suppose that at a certain point a portion of the aorta is blocked so that the cross-sectional area is reduced to 3/4 of its original area. The density of blood is 1060 kg/m3.
(a) How fast is the blood moving just as it enters the blocked portion of the aorta? (in cm/s)

(b) What is the gauge pressure (in mmHg) of the blood just as it has entered the blocked portion of the aorta? (in mmHg)

1 answer:

Ksenya-84 [330]3 years ago

8 0

Answer:

Explanation:

25 mm diameter

r₁ = 12.5 x 10⁻³ m radius.

cross sectional area = a₁

Pressure P₁ = 100 x 10⁻³ x 13.6 x 9.8 Pa

a )

velocity of blood v₁ = .6 m /s

Cross sectional area at blockade = 3/4 a₁

Velocity at blockade area = v₂

As liquid is in-compressible

a₁v₁ = a₂v₂

a₁ x .6 m /s = 3/4 a₁ v₂

v₂ = .8m/s

b )

Applying Bernauli's theorem formula

P₁ + 1/2 ρv₁² = P₂ + 1/2 ρv₂²

100 x 10⁻³ x 13.6 x10³x 9.8 + 1/2 X 1060 x .6² = P₂ + 1/2x 1060 x .8²

13328 +190.8 = P₂ + 339.2

P₂ = 13179.6 Pa

= 13179 / 13.6 x 10³ x 9.8 m of Hg

P₂ = .09888 m of Hg

98.88 mm of Hg

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