6.2 For a typical human aorta, the diameter of the lumen is 30 mm and the thickn

Question

6.2 For a typical human aorta, the diameter of the lumen is 30 mm and the thickness of the wall is 4 mm. Assuming a blood density of 1.056 g/cm and a viscosity of 0.035 P, calculate the Womersley number if the heart rate is 72 bpm Diameters for the carotid and femoral arteries of a tvpical human are about 0.8 and 0.5 cm respectively. Compute the Womersley numbers for each of these arterial segments for the same heart rate. 6.3In an experiment on an exposed abdominal aorta of a dog, the pulse wave speed is determined to be 1.5 m/s. The wall thickness of the artery was measured to be 5% of the diameter. Estimate the Young's modulus for the aorta. In another segment of the same artery, the pulse wave speed was double the previous value. What can you conclude from this?

Explanation / Answer

The Womersley number, usually denoted , is defined by the relation

where L is an appropriate length scale (for example the radius of a pipe), is the angular frequency of the oscillations, and, , are the kinematic viscosity, density, and dynamic viscosity of the fluid, respectively.[2] The Womersley number is normally written in the powerless form

Given values from problem :

Diameter of lumen , D = 30 mm = 30 * 10-3 m

So radus is = 15 * 10-3 m

Thickness of the wall : 4 mm = 4 * 10-3 m

Blood density : 1.056 g/cm3 = 1.056 * (10-3/ 10-6) = 1.056 * 103 Kg/m3

Viscosity = 0.035 P

Heart rate : 72 per minute

So Frequency = 72/60 = 1.2 Hz

Therefore,

Womersley number = = L ( / )(1/2)

= (15 * 10-3 )((1.2 * 1.056 * 103 ) /0.035 )(1/2)

     = 2854.16 * 10-3N_w

Now Diameter of carotid is = 0.8 cm = 0.8 * 10-2 m

Radius = 0.4 * 10-2 m

   Diameter of Femoral is = 0.5 cm = 0.5 * 10-2 m

Radius = 0.25 * 10-2 m

Therefore,

Womersley number = = L ( / )(1/2)

For carotid,

= (0.4 * 10-2) * ((1.2 * 1.056 * 103 ) / 0.035)(1/2)

= (0.4 * 10-2) * 190.27

    = 76.11 * 10-2N_w

For Femoral,

= (0.25 * 10-2) * ((1.2 * 1.056 * 103 ) / 0.035)(1/2)

= (0.25 * 10-2) * 190.27

    = 47.56 * 10-2N_w

6.3 ) Speed = (Y/ )(1/2)

Young’s modulus is,

Y = Speed2 * density

Blood density : 1.056 g/cm3 = 1.056 * (10-3/ 10-6) = 1.056 * 103 Kg/m3

Therefore,

Y = (1.5)2 * 1.056 * 103 = 2376 N/m2

If speed is doubled,

Young’s modulus is,

Y = (3)2 * 1.056 * 103 = 9504 N/m2

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