(5) The normal inner radius of a large artery is 2 mm. It is 75 cm long, and the

Question

(5) The normal inner radius of a large artery is 2 mm. It is 75 cm long, and the flow through it is 1/200 of the total blood flow. How would the pressure drop across it change if the flow through it were unchanged and there were severe stenosis in the artery (a) across its entire length? or (b) across 15 cm of it? (c) In each case, if the pressure at the beginning of the artery were 85 mmHg, would the pressure drop be severe enough to affect flow in the arterioles and capillaries? (d) In each case, what added pressure would be needed at the beginning of the artery to maintain an unchanged flow in these arterioles and capillaries?

Explanation / Answer

a)Stenosis means narrowing of the inner spaces of the artery. So, the radius of the artery will decrease during stenosis.

Now, the given information is that there is severe stenosis and the flow is unchanged. This means that radius <2mm; and flow = 1/200 of total blood flow.

Apply Bernoulli’s theorem over here:

The pressure will decrease. This is because the increased fluid pressure reduced the internal pressure.

b)

Now, if stenosis occurs across 15cm of the artery, then pressure along this 15cm will be less than the pressure on both sides of the artery.

c)

Let P1 be the pressure in the start; P2 be the pressure of the compressed artery. Then applying Bernoulli’s theorem,

P1 + ½ v12 + gh1 = P2 + ½ v22 + gh2

½ v2 = kinetic energy before and after respectively; and gh = potential energy before and after respectively.

Note that h1 = h2; so, P1 + ½ v12 = P2 + ½ v22

Now, it is given that P1 = P2 = 85mm

So, v12 = v22

So, velocity of flow through the arterioles and artery will become same. So, your answer is ‘yes’. The pressure flow of the arterioles will remain normal.

d)

The pressure required in the beginning is P1:

P1 = P2 + ½ (v22 -v12)

Added pressure = P2 – P1 = ½ (v22 -v12)

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