Challenge Problem: An aneurysm is an abnormal enlargement of a blood vessel such as the aorta. Because of an aneurysm, the crosssectional area of the aorta increases by 1.7 times. The speed of the blood (ρ = 1060 kg/m^{3}) through a normal portion of the aorta is 0.40 m/s. Assuming that the aorta is horizontal (the person is lying down), determine the amount by which the pressure P_{2} in the enlarged region exceeds the pressure P_{1} in the normal region.
(move your cursor over the space below to see the solution!)
The Equation of Continuity A_{1}v_{1} = A_{2}v_{2} can be rearranged to v_{2}/v_{1} = A_{1}/A_{2}. Since we know v_{1} and that A_{2} is 1.7 times A_{1} we can solve for v_{2} = 0.4/1.7 = 0.235 m/s. Now we can rearrange Bernoulli's equation for horizontal flow:
P_{2} + ½ρv_{2}^{2} = P_{1} + ½ρv_{1}^{2}
P_{2}  P_{1} = ½ρ[v_{1}^{2}  v_{2}^{2}]
P_{2}  P_{1} = ½(1060)[0.4^{2}  0.235^{2}]
P_{2}  P_{1} = 55 Pa
This positive answer indicates that P_{2} is greater than P_{1}. The excess pressure puts added stress on the already weakened tissue of the arterial wall at the aneurysm.

