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Chapter 9: Problem 88

In an aortic aneurysm, a bulge forms where the walls of the aorta are weakened. If blood flowing through the aorta (radius \(1.0 \mathrm{cm}\) ) enters an aneurysm with a radius of \(3.0 \mathrm{cm},\) 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 $120 \mathrm{cm}^{3} / \mathrm{s} .$ Assume the blood is non-viscous and the patient is lying down so there is no change in height.

Short Answer

Expert verified
Answer: The average blood pressure difference between the normal aorta and the aneurysm is approximately 5738 Pa.

Step by step solution

01

Understand the Problem and Define the Variables

Let's denote the pressure in the normal aorta as \(P_1\), the pressure in the aneurysm as \(P_2\), the radius of the aorta as \(r_1\), the radius of the aneurysm as \(r_2\), and the flow rate as \(Q\). We have \(r_1 = 1.0 \:cm\), \(r_2 = 3.0\: cm\), and \(Q = 120\: cm^3/s\). We need to find the pressure difference, \((P_2 - P_1)\).
02

Apply the Bernoulli's Equation

Because the patient is lying down and the blood is assumed to be non-viscous, the difference in potential energy and viscosity can be ignored. So, Bernoulli's equation simplified for this case is: \[P_1 + \frac{1}{2} \rho v_1^2 = P_2 + \frac{1}{2} \rho v_2^2\] where \(\rho\) is the density of the fluid, and \(v_1\) and \(v_2\) are the velocities of the fluid in the normal aorta and aneurysm, respectively.
03

Relate Flow Rate, Velocity, and Area

The flow rate is given by the product of the cross-sectional area and the velocity of the fluid, i.e. \(Q = Av\). So, we have: \[v_1 = \frac{Q}{\pi r_1^2}\] \[v_2 = \frac{Q}{\pi r_2^2}\]
04

Substitute in Bernoulli's Equation

Substitute the expressions for \(v_1\) and \(v_2\) in the Bernoulli's equation: \[P_1 + \frac{1}{2} \rho \left(\frac{Q}{\pi r_1^2}\right)^2 = P_2 + \frac{1}{2} \rho \left(\frac{Q}{\pi r_2^2}\right)^2\]
05

Solve for the Pressure Difference

Rearrange the equation to solve for pressure difference \[(P_2 - P_1) = \frac{1}{2} \rho \left[\left(\frac{Q}{\pi r_1^2}\right)^2 - \left(\frac{Q}{\pi r_2^2}\right)^2\right]\]
06

Insert Known Values and Calculate

Now plug in the known values \(r_1 = 1.0 \: cm\), \(r_2 = 3.0\: cm\), \(Q = 120\: cm^3/s\), and \(\rho \approx 1060\: kg/m^3\) for the density of blood (convert the units of the measurements to be consistent): \[(P_2 - P_1) = \frac{1}{2} \times 1060 \: kg/m^3 \left[\left(\frac{120 \times 10^{-6} m^3/s}{\pi (0.01\: m)^2}\right)^2 - \left(\frac{120 \times 10^{-6} m^3/s}{\pi (0.03\: m)^2}\right)^2\right]\] After calculating the above expression, we obtain the pressure difference: \[(P_2 - P_1) \approx 5738\: Pa\] Thus, the average blood pressure inside the aneurysm is approximately \(5738 \: Pa\) higher than the pressure in the unenlarged part of the aorta.

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