The Time and Force Required to Empty a Syringe: What Does Science Tell Us?

[ostrik23]

Homework Statement

a) Suppose we have a hypodermic syringe, now without a needle attached. The syringe is filled with 10 ml of water. The inner diameter of the liquid chamber is 15.9mm and the diameter of the barrel is 1.2mm. With how much force must you push the piston for the syringe to be able to empty in 10 seconds

b) Suppose we now put a needle on the syringe. The needle has a length of 25mm and a diameter of 0.210mm. If you now push the piston with a force of 0.10N, how long will it take to empty the syringe given that theviscosity is 0.001 Pascal.

Relevant Equations

Bernoulli's equation
Hagen-Poiseuille equation

Okay, so what I attempted for a) is to use Bernoulli's. The velocity of the fluid in the chamber should be equal to the velocity of the piston, which in comparison to the barrel should be negligible.

Hence I get \frac{p_1}{\rho g} = \frac{1}{2} \frac{v_2^2}{2} + \frac{p_2}{\rho g} so \Delta p=\frac{1}{2} \rho v_2^2

Then I use that the force must be F = \textrm{cross section area of piston} \cdot \frac{1}{2} \rho v_2^2 where v_2 = \textrm{volume of liquid in syringe/(wanted time to empty \cdot area of barrel cross section)}and arrived at about 0.08 Newtons, which didn't seem too unrealistic.

For b) I tried utilising Hagen-poiseuille's equation by saying that

Q=\frac{\pi r^4\Delta p}{8\mu L}, by recognizing thatQ = V/t I solved \frac{V}{t}=\frac{\pi r^4\Delta p}{8\mu L} for t such that t=\frac{8\mu L V}{\pi r^4 \Delta p} but once I plugged in my values I got ridiculous values for time, so I don't understand where I went wrong. Can someone take a look and please help a poor soul out?

 

[Chestermiller]

Homework Statement:: a) Suppose we have a hypodermic syringe, now without a needle attached. The syringe is filled with 10 ml of water. The inner diameter of the liquid chamber is 15.9mm and the diameter of the barrel is 1.2mm. With how much force must you push the piston for the syringe to be able to empty in 10 seconds

b) Suppose we now put a needle on the syringe. The needle has a length of 25mm and a diameter of 0.210mm. If you now push the piston with a force of 0.10N, how long will it take to empty the syringe given that theviscosity is 0.001 Pascal.
Relevant Equations:: Bernoulli's equation
Hagen-Poiseuille equation

Okay, so what I attempted for a) is to use Bernoulli's. The velocity of the fluid in the chamber should be equal to the velocity of the piston, which in comparison to the barrel should be negligible.

Hence I get \frac{p_1}{\rho g} = \frac{1}{2} \frac{v_2^2}{2} + \frac{p_2}{\rho g} so \Delta p=\frac{1}{2} \rho v_2^2

Then I use that the force must be F = \textrm{cross section area of piston} \cdot \frac{1}{2} \rho v_2^2 where v_2 = \textrm{volume of liquid in syringe/(wanted time to empty \cdot area of barrel cross section)}and arrived at about 0.08 Newtons, which didn't seem too unrealistic.

For b) I tried utilising Hagen-poiseuille's equation by saying that

Q=\frac{\pi r^4\Delta p}{8\mu L}, by recognizing thatQ = V/t I solved \frac{V}{t}=\frac{\pi r^4\Delta p}{8\mu L} for t such that t=\frac{8\mu L V}{\pi r^4 \Delta p} but once I plugged in my values I got ridiculous values for time, so I don't understand where I went wrong. Can someone take a look and please help a poor soul out?

Show us the parameter values you used for part b), and your calculation details.

 

[ostrik23]

Okay so for parameter values I used \mu = 10^{-3} \textrm{pascal}

and for radii i used r=\frac{0.210}{2} \cdot 10^{-3} \textrm{m}

For L i used the length of the needle L=25\cdot 10^{-3} \textrm{m}

Where it gets a bit sketchy is in my calculation of \Delta p=P_1-P_2=\frac{F_1}{A_1}-\frac{F_2}{A_2}.

From Pascal's principle

p_1=p_2

Which in theory should give

F_2=\frac{F_1A_2}{A_1}

But that becomes a circular argument when substituting it in the previous equation because then

\Delta p = \frac{F_1}{A_1}-\frac{F_1A_2}{A_1A_2}=0

Which is nonsense. This is specifically where I believe I am stuck.

 

[Chestermiller]

Okay so for parameter values I used \mu = 10^{-3} \textrm{pascal}

and for radii i used r=\frac{0.210}{2} \cdot 10^{-3} \textrm{m}

For L i used the length of the needle L=25\cdot 10^{-3} \textrm{m}

Where it gets a bit sketchy is in my calculation of \Delta p=P_1-P_2=\frac{F_1}{A_1}-\frac{F_2}{A_2}.

From Pascal's principle

p_1=p_2

Which in theory should give

F_2=\frac{F_1A_2}{A_1}

But that becomes a circular argument when substituting it in the previous equation because then

\Delta p = \frac{F_1}{A_1}-\frac{F_1A_2}{A_1A_2}=0

Which is nonsense. This is specifically where I believe I am stuck.

The barrel diameter is 15.9 mm, so the barrel area is 199 mm^2 = $1.99\times 10^{-4}\ m^2$. So the upstream pressure is $0.1/1.99\times 10^{-4}=503\ Pa$. This is the pressure drop $\Delta p$ across the capillary.

The units of viscosity given the the problem statement are wrong. What are the correct units?

 

[ostrik23]

The barrel diameter is 15.9 mm, so the barrel area is 199 mm^2 = $1.99\times 10^{-4}\ m^2$. So the upstream pressure is $0.1/1.99\times 10^{-4}=503\ Pa$. This is the pressure drop $\Delta p$ across the capillary.

The units of viscosity given the the problem statement are wrong. What are the correct units?

I am guessing the correct units are Pascal Seconds? (Pas)

Even so, if I calculate my value for t with the expression I found

t=\frac{8\cdot 10^{-3}\cdot 25\cdot 10^{-3}\cdot 10^{-5}}{\pi \left(\frac{15.9}{2}\cdot 10^{-3}\right)^4\cdot 503}=3.17\cdot 10^{-4} Which is way too fast. I just don't see where I am messing up.

 

[Chestermiller]

I am guessing the correct units are Pascal Seconds? (Pas)

Even so, if I calculate my value for t with the expression I found

t=\frac{8\cdot 10^{-3}\cdot 25\cdot 10^{-3}\cdot 10^{-5}}{\pi \left(\frac{15.9}{2}\cdot 10^{-3}\right)^4\cdot 503}=3.17\cdot 10^{-4} Which is way too fast. I just don't see where I am messing up.

In the evaluation of the volumetric flow rate, you should be using the radius of the capillary in Poiseulle's equation, not the radius of the barrel.

Please do this in two steps. What do you get for the volumetric flow rate?

 

[ostrik23]

Okay, so volumetric flowrate

Q=\frac{\pi r^4 \Delta p}{8\mu L}=\frac{\pi \cdot (0.105\cdot 10^{-3})^4\cdot 503}{8\cdot 10^{-3}\cdot 25\cdot 10^{-3}}=9.603\cdot 10^{-10}

Then t=\frac{V}{9.603\cdot 10^{-10}}=\frac{10^{-5}}{9.603\cdot 10^{-10}}=10413

Which still seems like an unreasonably long time to empty a small 10 millilitre syringe?

 

[Chestermiller]

Okay, so volumetric flowrate

Q=\frac{\pi r^4 \Delta p}{8\mu L}=\frac{\pi \cdot (0.105\cdot 10^{-3})^4\cdot 503}{8\cdot 10^{-3}\cdot 25\cdot 10^{-3}}=9.603\cdot 10^{-10}

Then t=\frac{V}{9.603\cdot 10^{-10}}=\frac{10^{-5}}{9.603\cdot 10^{-10}}=10413

Which still seems like an unreasonably long time to empty a small 10 millilitre syringe?

For the data you gave, this calculation looks OK to me. But that force of 0.1 N = 0.022 lbs seems very low to me.

 

[ostrik23]

I do agree it seems like an awfully small force, but the problem was given as such. As for the result, it heavily makes me doubt my answer in a), given that 0.08 Newtons is able to empty the syringe without a needle in 10 seconds, whereas .1 Newton won't be able to empty the same amount of liquid through a syringe in less than 3 hours no less.

 

[Chestermiller]

I do agree it seems like an awfully small force, but the problem was given as such. As for the result, it heavily makes me doubt my answer in a), given that 0.08 Newtons is able to empty the syringe without a needle in 10 seconds, whereas .1 Newton won't be able to empty the same amount of liquid through a syringe in less than 3 hours no less.

I confirm the 0.08 N. What the second calculation shows is that for the very high viscous resistance to flow in the capillary, the flow rate for the same applied force is enormously lower. That doesn't surprise me at all.