- #1
ostrik23
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- 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 [tex]\frac{p_1}{\rho g} = \frac{1}{2} \frac{v_2^2}{2} + \frac{p_2}{\rho g}[/tex] so [tex]\Delta p=\frac{1}{2} \rho v_2^2[/tex]
Then I use that the force must be [tex]F = \textrm{cross section area of piston} \cdot \frac{1}{2} \rho v_2^2[/tex] where [tex]v_2 = \textrm{volume of liquid in syringe/(wanted time to empty \cdot area of barrel cross section)}[/tex]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
[tex]Q=\frac{\pi r^4\Delta p}{8\mu L}[/tex], by recognizing that[tex] Q = V/t[/tex] I solved [tex]\frac{V}{t}=\frac{\pi r^4\Delta p}{8\mu L}[/tex] for t such that [tex]t=\frac{8\mu L V}{\pi r^4 \Delta p}[/tex] 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?
Hence I get [tex]\frac{p_1}{\rho g} = \frac{1}{2} \frac{v_2^2}{2} + \frac{p_2}{\rho g}[/tex] so [tex]\Delta p=\frac{1}{2} \rho v_2^2[/tex]
Then I use that the force must be [tex]F = \textrm{cross section area of piston} \cdot \frac{1}{2} \rho v_2^2[/tex] where [tex]v_2 = \textrm{volume of liquid in syringe/(wanted time to empty \cdot area of barrel cross section)}[/tex]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
[tex]Q=\frac{\pi r^4\Delta p}{8\mu L}[/tex], by recognizing that[tex] Q = V/t[/tex] I solved [tex]\frac{V}{t}=\frac{\pi r^4\Delta p}{8\mu L}[/tex] for t such that [tex]t=\frac{8\mu L V}{\pi r^4 \Delta p}[/tex] 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?
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