Question

A $2.0mL$syringe has an inner diameter of $6.0mm$, a needle inner diameter of 0.25mm, and a plunger pad diameter (where you place your finger) of $1.2cm$. A nurse uses the syringe to inject medicine into a patient whose blood pressure is $140/100$.

A. What is the minimum force the nurse needs to apply to the syringe?

B. The nurse empties the syringe in $2.0s$. What is the flow speed of the medicine through the needle?

Short answer

The required velocity is $20.4m/s$.

Step-by-step solution

Introduction

The expression for the force is,

$F=PA$

Here, $P$ is the pressure and A is the area.

The area of the syringe(part A)

(A)The area of the syringe is,

$A={\mathrm{\pi r}}^2$

Here, $r$is the radius of the syringe

Substitute $\frac{d}{2}$for $r$.

$A=\pi {\left(\frac{d}{2}\right)}^2$

Here, $d$is the diameter.

Substitute $6mm$for $d$.

$A=\pi {\left(\frac{6\mathrm{mm}\left(\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)}{2}\right)}^2$

$=2.83\times {10}^{-5}{\mathrm{m}}^2$

Convert the pressure from mmHg to Pa

$P=\left(100\mathrm{mmHg}\right)\left(\frac{1\mathrm{atm}}{760\mathrm{mmHg}}\right)\left(\frac{1.013\times {10}^5\mathrm{Pa}}{1\mathrm{atm}}\right)$

$=13.32\times {10}^3\mathrm{Pa}$

So, the minimum force is,

$F=PA$

Substitute $2.83\times {10}^{-5}{\mathrm{m}}^2$for $A$and $13.32\times {10}^3\mathrm{Pa}$for $P$.

$F=\left(13.32\times {10}^3\mathrm{Pa}\right)\left(2.83\times {10}^{-5}{\mathrm{m}}^2\right)$

$=0.38N$

Hence, the required force is $0.38N$

The volume flow rate(part B)

(B)The volume flow rate is,

$Q=\frac{V}{t}$

Here, $V$is the volume and $t$is the time.

Substitute $2mL$for $2s$and for $t$.

$Q=\frac{2\mathrm{mL}\left(\frac{{10}^{-6}{\mathrm{m}}^3}{1\mathrm{mL}}\right)}{2\mathrm{s}}$

$=1\times {10}^{-6}{\mathrm{m}}^3/\mathrm{s}$

The area of the needle is,

Substitute $\frac{d}{2}$for r

$A=\pi {\left(\frac{d}{2}\right)}^2$

Substitute $0.125mm$for $d$

$A=\pi {\left(\frac{0.125\mathrm{mm}\left(\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)}{2}\right)}^2$

$=4.91\times {10}^{-8}{\mathrm{m}}^2$

The flow rate is,

$Q=Av$

$v=\frac{Q}{A}$

Substitute $4.91\times {10}^{-8}{\mathrm{m}}^2$for $A$and$1\times {10}^{-6}{\mathrm{m}}^3/\mathrm{s}$for $Q$.

$v=\frac{1\times 10{\mathrm{m}}^3/\mathrm{s}}{4.91\times {10}^{-8}{\mathrm{m}}^2}$

$=20.4\mathrm{m}/\mathrm{s}$

As a result, the necessary speed is $20.4m/s$.