## College Physics (4th Edition)

The blood pressure is $75.7~Pa$ higher in the aneurysm then in the unenlarged part of the aorta.
Let $v_1$ be the speed in the aorta. Let $v_2$ be the speed in the aneurysm. The cross-sectional area in the aneurysm $A_2$ is 9 times larger than the cross-sectional area in the aorta. We can find an expression for $v_2$ in terms of $v_1$: $A_2~v_2 = A_1~v_1$ $(9A_1)~v_2 = A_1~v_1$ $v_2 = \frac{v_1}{9}$ We can find the speed in the aorta: $A_1~v_1 = (120 \times 10^{-6})~m/s$ $v_1 = \frac{(120 \times 10^{-6})~m/s}{A_1}$ $v_1 = \frac{(120 \times 10^{-6})~m/s}{\pi~r_1^2}$ $v_1 = \frac{(120 \times 10^{-6})~m/s}{(\pi)~(0.010~m)^2}$ $v_1 = 0.382~m/s$ We can use Bernoulli's equation to find the pressure difference: $P_2+\frac{1}{2}\rho~v_2^2 = P_1+\frac{1}{2}\rho~v_1^2$ $P_2-P_1 = \frac{1}{2}\rho~(v_1^2-v_2^2)$ $P_2-P_1 = \frac{1}{2}\rho~(v_1^2-(\frac{v_1}{9})^2)$ $P_2-P_1 = \frac{1}{2}\rho~\frac{80~v_1^2}{81}$ $P_2-P_1 = \frac{1}{2}(1050~kg/m^3)~\frac{80~(0.382~m/s)^2}{81}$ $P_2-P_1 = 75.7~Pa$ The blood pressure is $75.7~Pa$ higher in the aneurysm then in the unenlarged part of the aorta.