Chapter 9, Problem 61P

(a)

To determine

The required pressure of the fluid in the syringe.

Answer to Problem 61P

The required pressure of the fluid in the syringe is $\underset{\_}{6850\text{Pa}}$.

Explanation of Solution

Given that the internal radius of the syringe is $0.300\text{mm}$, its length is $3.00\text{cm}$, the viscosity of the solution is $2.00\times {10}^{-3}\text{Pa}\cdot \text{s}$, the gauge pressure of the vein is $16.0\text{mm Hg}$, and the rate of injection is $0.250\text{mL/s}$.

Write the expression for the rate of injection.

$\frac{\Delta V}{\Delta t}=\frac{\pi \Delta P{r}^4}{8\eta L}$ (I)

Here, $\Delta V/\Delta t$ is the rate of injection, $\Delta P$ is the change in pressure, $r$ is the radius of the syringe, $\eta$ is the viscosity of the fluid, and $L$ is the length of the syringe.

Use $\Delta P=P_{\text{s}}-P_{\text{v}}$, where $P_{\text{s}}$ is the pressure of the solution in the syringe, and $P_{\text{v}}$ is the pressure in the vein, to modify the equation (I).

$\frac{\Delta V}{\Delta t}=\frac{\pi \left(P_{\text{s}}-P_{\text{v}}\right)r^4}{8\eta L}$

Solve for $P_{\text{s}}$.

$P_{\text{s}}=\frac{8\eta L\left(\frac{\Delta V}{\Delta t}\right)}{\pi r^4}+P_{\text{v}}$ (II)

Conclusion:

Substitute $2.00\times {10}^{-3}\text{Pa}\cdot \text{s}$ for $\eta$, $0.300\text{mm}$ for $r$, $3.00\text{cm}$ for $L$, $16.0\text{mm Hg}$ for $P_{\text{v}}$, and $0.250\text{mL/s}$ for $\Delta V/\Delta t$ in equation (II) to find $P_{\text{s}}$.

$\begin{array}{c}{P}_{\text{s}}=\frac{8\left(2.00\times {10}^{-3}\text{Pa}\cdot \text{s}\right)\left(3.00\text{cm}\right)\left(0.250\text{mL/s}\right)}{\pi {\left(0.300\text{mm}\right)}^4}+16.0\text{mm Hg}\\ =\left(\begin{array}{c}\frac{8\left(2.00\times {10}^{-3}\text{Pa}\cdot \text{s}\right)\left(3.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\left(0.250\text{mL/s}\times \frac{1\text{L}}{1000\text{mL}}\times \frac{1{\text{m}}^3}{1000\text{L}}\right)}{\pi {\left(0.300\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^4}\\ +\left(16.0\text{mm Hg}\times \frac{1.013\times {10}^5\text{Pa}}{760.0\text{mm}\text{Hg}}\right)\end{array}\right)\\ =6850\text{Pa}\end{array}$

Therefore, the required pressure of the fluid in the syringe is $\underset{\_}{6850\text{Pa}}$.

(b)

To determine

The force that must be applied to the plunger.

Answer to Problem 61P

The force that must be applied to the plunger is $\underset{\_}{0.685\text{N}}$.

Explanation of Solution

Given that area of the plunge of the syringe is $1.00{\text{cm}}^2$. It is obtained that the pressure in the syringe is $6850\text{Pa}$.

Write the expression for the force on the plunge of the syringe.

$F=P_{\text{s}}A$ (III)

Here, $F$ is the force, and $A$ is the cross sectional area of the syringe.

Conclusion:

Substitute $6850\text{Pa}$ for $P_{\text{s}}$, and $1.00{\text{cm}}^2$ for $A$ in equation (III) to find the force $F$.

$\begin{array}{c}F=\left(6850\text{Pa}\right)\left(1.00{\text{cm}}^2\right)\\ =\left(6850\text{Pa}\right)\left(1.00{\text{cm}}^2\right)\times {\left(\frac{1\text{m}}{100\text{cm}}\right)}^2\\ =0.685\text{N}\end{array}$

Therefore, the force that must be applied to the plunger is $\underset{\_}{0.685\text{N}}$.