## College Physics (4th Edition)

The pressure drop across the aorta is $26.7~kPa$
We can use Poiseuille's law to find an expression for the flow rate: $Q = \frac{\pi~\Delta P~r^4}{8~\eta~L}$ $Q$ is the flow rate ($m^3~s^{-1}$) $\Delta P$ is the pressure difference ($Pa$) $r$ is the radius ($m$) $\eta$ is the fluid viscosity $L$ is the length of the tube ($m$) We can find the pressure drop $\Delta P$: $\Delta P = \frac{8~Q~\eta~L}{\pi~r^4}$ $\Delta P = \frac{(8)~(4.1\times 10^{-3}~m^3/s)~(4.0\times 10^{-3}~Pa~s)~(0.40~m)}{(\pi)~(5.0\times 10^{-3}~m)^4}$ $\Delta P = 26.7~kPa$ The pressure drop across the aorta is $26.7~kPa$.