Chapter 3.8, Problem 27E

Refer to the law of laminar flow given in Example 7. Consider a blood vessel with radius 0.01 cm, length 3 cm, pressure difference 3000 dynes/cm², and viscosity η = 0.027.

(a) Find the velocity of the blood along the center-line 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?

EXAMPLE 7

To determine

To find: The velocity of the blood along the center line $r=0$, at radius $r=0.005\text{cm}$ and at the wall $r=R=0.01\text{cm}$.

Answer to Problem 27E

Velocity of the blood along the center line is $v\left(0\right)=0.\overline{925}\text{cm/s}$.

Velocity of the blood at 0.005cm as radius is $v\left(0.005\right)=0.69\overline{4}\text{cm/s}$.

Velocity of the blood at the wall is $v\left(0.01\right)=0$.

Explanation of Solution

Given:

The law of laminar flow is as given below.

$v\left(r\right)=\frac{P}{4\eta l}\left(R^2-r^2\right)$ (1)

The length of the blood vessel is given below.

$l=3\text{cm}$

The pressure difference between the ends of the vessel is given below.

$P=3000{\text{dynes/cm}}^{\text{2}}$

The viscosity of the blood is given below.

$\eta =0.027$

Calculation:

Calculate the velocity of the blood along the center line using the equation (1).

$v\left(r\right)=\frac{P}{4\eta l}\left(R^2-r^2\right)$

Substitute 3cm for $l$, 0 for r, $3000{\text{dynes/cm}}^{\text{2}}$ for $P$, 0.027 for $\eta$ and 0.01cm for $R$ in the equation (1).

$\begin{array}{c}v\left(0\right)=\frac{3000}{4\times 0.027\times 3}\left({0.01}^2-0\right)\\ =9259.25\times 0.0001\end{array}$

$v\left(0\right)=0.925\text{cm/s}$

Thus, the Velocity of the blood at $r=0\text{cm}$ is $0.925\text{cm/s}$.

Calculate the velocity of the blood at $r=0.005\text{cm}$ using the equation (1).

$v\left(r\right)=\frac{P}{4\eta l}\left(R^2-r^2\right)$

Substitute 3 cm for $l$, 0.005 cm for r, $3000{\text{dynes/cm}}^{\text{2}}$ for $P$, 0.027 for $\eta$ and 0.01 cm for $R$ in the equation (1).

$\begin{array}{c}v\left(0.005\right)=\frac{3000}{4\times 0.027\times 3}\left({0.01}^2-{0.005}^2\right)\\ =9259.259\times 0.000075\\ =0.694\text{cm/s}\end{array}$

$v\left(0\right)=0.694\text{cm/s}$

Thus, the Velocity of the blood at $r=0.005\text{cm}$ is $0.694\text{cm/s}$.

Calculate the velocity of the blood at the wall using the equation (1).

$v\left(r\right)=\frac{P}{4\eta l}\left(R^2-r^2\right)$

Substitute 3 cm for $l$, 0.01 cm for r, $3000{\text{dynes/cm}}^{\text{2}}$ for $P$, 0.027 for $\eta$ and 0.01 cm for $R$ in the equation (1).

$\begin{array}{c}v\left(0.01\right)=\frac{3000}{4\times 0.027\times 3}\left({0.01}^2-{0.01}^2\right)\\ =0\end{array}$

$v\left(0.01\right)=0$

Thus, the Velocity of the blood at $r=0.01\text{cm}$ is $0$.

To determine

To find: The velocity gradient at $r=0$, $r=0.005$ and $r=0.01$.

Answer to Problem 27E

Velocity gradient at $r=0$ is $v\text{'}\left(0\right)=0$.

Velocity gradient at $r=0.005$ is $v\text{'}\left(0.005\right)=-92.\overline{592}\left(\text{cm/s}\right)\text{/cm}$.

Velocity gradient at $r=0.01$ is $v\text{'}\left(0.01\right)=-185.\overline{185}\left(\text{cm/s}\right)\text{/cm}$.

Explanation of Solution

Calculate the velocity gradient at center line $r=0$.

Differentiate equation (1) with respect to $r$.

$\begin{array}{c}v\text{'}\left(r\right)=\frac{P}{4\eta l}\left(-2r\right)\\ =\frac{-rP}{2\eta l}\end{array}$

$v\text{'}\left(r\right)=\frac{-rP}{2\eta l}$ (2)

Substitute 3 cm for $l$, 0 for r, $3000{\text{dynes/cm}}^{\text{2}}$ for $P$, 0.027 for $\eta$ in the equation (1).

$\begin{array}{c}v\text{'}\left(0\right)=\frac{-rP}{2\eta l}\\ v\left(0\right)=0\end{array}$

Thus, the Velocity of the blood at radius $r=0$ is $0$.

Calculate the velocity gradient at $r=0.005$ using the equation (2).

$v\text{'}\left(r\right)=\frac{-rP}{2\eta l}$

Substitute 3 cm for $l$, 0.005 for r, $3000{\text{dynes/cm}}^{\text{2}}$ for $P$, 0.027 for $\eta$ in the equation (1).

$\begin{array}{c}v\text{'}\left(0.005\right)=\frac{-\left(0.005\right)\times 3000}{2\times 0.027\times 3}\\ =\frac{-15}{0.162}\\ v\text{'}\left(0.005\right)=-92.592\left(\text{cm/s}\right)\text{/cm}\end{array}$

The velocity gradient at $r=0.005$ is $-92.592\left(\text{cm/s}\right)\text{/cm}$.

Calculate the velocity gradient at the wall edge $r=0.01$ using the equation (2).

$v\text{'}\left(r\right)=\frac{-rP}{2\eta l}$

Substitute 3 cm for $l$, 0.01 for r, $3000{\text{dynes/cm}}^{\text{2}}$ for $P$, 0.027 for $\eta$ in the equation (1).

$\begin{array}{c}v\text{'}\left(0.01\right)=\frac{-\left(0.01\right)\times 3000}{2\times 0.027\times 3}\\ =\frac{-30}{0.162}\\ v\text{'}\left(0.01\right)=-185.185\left(\text{cm/s}\right)\text{/cm}\end{array}$

The velocity gradient at $r=0.01$ is $v\text{'}\left(0.01\right)=-185.185\left(\text{cm/s}\right)\text{/cm}$.

To determine

To find: Where is the velocity the greatest and where it changes the most.

Answer to Problem 27E

The velocity is greatest where $r=0$ and the velocity is changing most where $r=R=0.01\text{cm}$

Explanation of Solution

The velocity is greatest where $r=0$(at the center) and the velocity is changing most where $r=R=0.01\text{cm}$ (at the edge), which means the velocity gradient is highest at the edge.