Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 25

The equation $\displaystyle \nu = \frac{P}{4\eta \ell}(R^2 - r^2)$ represents the law of Laminar flow.
Where $\eta$ is the viscosity of the blood and $P$ is the pressure difference between the ends of the tube with length $\ell$ and radius $R$. While $r$ is the distance from the axis. Consider a blood vessel with radius 0.01cm, length 3cm, pressure difference $\displaystyle 3000 \frac{\text{dynes}}{\text{cm}^2}$ and viscosity $\eta = 0.027$
a.) Determine the velocity of blood along the centerline $r = 0$, at radius $r = 0.005$cm, and at the wall $r = R = 0.01$cm.
b.) Find the velocity gradient at $r=0$, $r=0.005$, and $r=0.01$
c.) Where is the velocity the greatest? Where is the velocity changing most?


a.) @ $r = 0$


$
\begin{equation}
\begin{aligned}
\nu(r) &= \frac{P}{4\eta \ell} (R^2 - r^2)\\
\\
\nu(0) &= \frac{3000}{4(0.027)(3)}\left( (0.01)^2 - (0)^2\right)\\
\\
\nu(0) &= 0.9259 \frac{cm}{s}
\end{aligned}
\end{equation}
$


@ $r = 0.005$cm


$
\begin{equation}
\begin{aligned}
\nu(0.005) &= \frac{3000}{4(0.027)(3)} \left[ (0.01)^2 - (0.005)^2 \right]\\
\\
\nu(0.005) &= 0.6944 \frac{\text{cm}}{s}
\end{aligned}
\end{equation}
$


@ $r=0.01$cm


$
\begin{equation}
\begin{aligned}
\nu(0.01) &= \frac{3000}{4(0.027)(3)} \left[ (0.01)^2 - (0.01)^2 \right]\\
\\
\nu(0.01) &= 0 \frac{\text{cm}}{s}
\end{aligned}
\end{equation}
$


b.) The velocity gradient $\displaystyle = \frac{d\nu}{dr}$

$
\begin{equation}
\begin{aligned}
\nu &= \frac{P}{4\eta \ell} \left( R^2 - r^2 \right)\\
\\
\nu &= \frac{PR^2}{4 \eta \ell} - \frac{Pr^2}{4 \eta \ell}\\
\\
\frac{d \nu}{dr} &= \frac{d}{dr} \left( \frac{PR^2}{4 \eta \ell}\right) - \frac{P}{4 \eta \ell} \frac{d}{dr} (r^2)\\
\\
\frac{d \nu}{dr} &= 0 - \frac{P}{4\eta \ell} (2r)\\
\\
\frac{d \nu}{dr} &= \frac{-Pr}{ 2 \eta \ell}
\end{aligned}
\end{equation}
$

Velocity gradient at $r=0$

$
\begin{equation}
\begin{aligned}
\frac{d \nu}{dr} &= \frac{-3000 (0)}{2(0.027)(3)}\\
\\
\frac{d \nu}{dr} &= 0 \frac{\frac{\text{cm}}{s}}{\text{cm}}
\end{aligned}
\end{equation}
$

Velocity gradient at $r = 0.005$

$
\begin{equation}
\begin{aligned}
\frac{d \nu}{dr} &= \frac{-3000(0)}{2(0.027)(3)}\\
\\
\frac{d \nu}{dr} &= -92.5926 \frac{\text{cm}/s}{\text{cm}}
\end{aligned}
\end{equation}
$

The velocity gradient at $r = 0.01$

$
\begin{equation}
\begin{aligned}
\frac{d \nu}{dr} &= \frac{-3000(0.01)}{(0.0027)(3)}\\
\\
\frac{d \nu}{dr} &= -185.1852 \frac{\text{cm}/s}{\text{cm}}
\end{aligned}
\end{equation}
$


c.) The velocity is greatest at $r = 0$. While the velocity is changing most (decreasing) at when $r = R = 0.01$cm.