Question

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 1/3 of its original area. The density of blood is 1060 kg/m3.

(a) How fast v ( in cm/s) is the blood moving just as it enters the blocked portion of the aorta?

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

Answer 1

The product of the cross section and the speed remains the same at all points of a tube of flow . This rule is called the equation of continuity .

So,

A1V1 = 4202

(At the point of blockage the cross section area reduces to one third of its original area i.e $A_{2}=\frac{A_{1}}{3}$ ).

  
A1V1 = 41,

  
÷ 23 = 3 = {1

  
U2=3* 60
  (v₁ = 60 cm/s = 0.6 m/s)

  
= 12 = 180
cm/s = 1.8 m/s

b)

According to Bernouli's equation , we can find the pressure difference between the points where there is blockage and where there is no blockage .

Let P₁ =pressure where there is no blockage = 100 mmHg

P₂ = Pressure where there is blockage

P+ pgh+ pu? = constant

Pi + pgh + 5 pui = P2 + pgh + 5 pua

P1+52*0.62 = P2 +50*1.89

Pi-P=

1526.4Newton/m
or
Pascal = 11.45mm Hg

P₂ = 88.55 mmHg