Posted by **trace** on Thursday, January 27, 2011 at 11:16pm.

In an aortic aneurysm, a bulge forms where the walls of the aorta are

weakened. If blood flowing through the aorta (radius 1.0 cm) enters an aneurysm with a radius of 2.5 cm, how much on average is the blood pressure higher inside the aneurysm than the pressure in the unenlarged part of the aorta? The average flow rate through the aorta is 120 cm3/s. Assume the blood is nonviscous and the patient is lying down so there is no change in height.

A. 150 kPa B. 75Pa C. 75 kPa D. 62 Pa E. 750 Pa

So I thought I was on the right track, I found the area of both the normal blood vessel, and where the aneurysm is using A=pi*r^2 for A1= 3.142 and A2= 19.63, then I used rate flow=A1v1=A2v2 to find the velocity in both. I got v1= 38.20 and v2= 6.116.

I know I should somehow use Bernoulli's equation, or I think I should, but I have no idea what to do now.

Any help would be appreciated!

Thanks

- physics, I'm stuck -
**drwls**, Friday, January 28, 2011 at 1:06am
The dimensions of your velocities are cm/s and you have used the continuity equation correctly to get the two velocities.

You are correct that the Bernoulli equation is what you need to compute the pressure difference. It is only valid if viscous effects can be neglected, which is OK over such a short distance. With no difference in height,

(p2 - p1) = (1/2)(density)(V1^2 - V2^2)

Use 1.06 g/cm^3 for the density of blood.

p2 - p1 = (0.5)(1.06)[38.2^2 - 6.1^2] = 746 dyne/cm^2 (round to 750)

= 7.5*10^4 Pa = 75 kPa

That is quite high, about 3/4 atmosphere. No wonder people die of ruptured aneurysyms

- P.S. thanks -
**drwls**, Friday, January 28, 2011 at 1:10am
.. for showing your own work. We encourage students to do that here, but very few do.

- physics, I'm stuck -
**trace**, Friday, January 28, 2011 at 1:35am
thanks!

- physics, I'm stuck -
**trace**, Friday, January 28, 2011 at 1:39am
thanks!

but when i checked the conversion online, 750 dyne converts to 75 Pa, not 75kPa, which conversion is correct?

Is there a reason you have a different conversion, that I'm not seeing?

- physics, I'm stuck -
**drwls**, Friday, January 28, 2011 at 1:55am
1 Pascal = 1 N/m^2

= 10^5 dyne/10^4 cm^2

= 10 dyne/cm^2

You are quite right. My conversion was wrong.

- physics, I'm stuck -
**trace**, Friday, January 28, 2011 at 1:56am
thank you so much!

Have a good night

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