Homework Help: Hypodermic Needle Fluid Dynamics

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1. Aug 24, 2015

A hypodermic syringe contains a medicine with the density of water (figure below). The barrel of the syringe has a cross-sectional area of 2.18 10-5 m2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm.
A force of magnitude 1.78 N is exerted on the plunger, making medicine squirt from the needle. Determine the medicine's flow speed through the needle.
Assume the pressure in the needle remains equal to 1.00 atm and that the syringe is horizontal.

I assume the Bernoulli equation is needed
pv1^2/2 + p1 = pv2^2/2 + p2 = constant
pgh would be irrelevant as the needle is horizontal, hence can be removed from both sides of the equation.

So far I have P1 = 101 325 + 1.78/2.18*10^-5 = 182976.38 (2dp)

where do I go from here? Any help would be awesome.

Last edited by a moderator: Apr 14, 2017
2. Aug 24, 2015

Nidum

Need the needle bore .

3. Aug 24, 2015

I wasnt given any other measurements, how can I solve for the area of the needle?

4. Aug 24, 2015

Nidum

Bore of needle is essential to calculate flow rate . If you don't actually know then choose some possible bores and calculate flow for each one to get ball park answers .

You can do it backwards and specify a suitable flow rate and then solve for needle bore if this is easier .

If this is a medical syringe then needle sizes for different purposes are probably available from manufacturers data .

5. Aug 24, 2015

If it says that the pressure in the needle remains at 1atm, even though liquid is being pushed through, would the force through the first part be equal to the force equal through the needle, and hence p1a1=p2a1?
Would this give me a correct area value?

6. Aug 24, 2015

Nidum

Is this a real problem to be solved or a study/homework task ?

7. Aug 24, 2015

for study

8. Aug 24, 2015

Since they give you all the pressures, you actually don't need the bore of the needle if you make a few simplifying assumptions such as Bernoulli's equation being valid, a constant flow rate, and the syringe being much larger than the needle. In that case, you know everything except for the dynamic pressure in the syringe and can solve for the velocity that way.

9. Aug 24, 2015

uh, how? I really don't know...

10. Aug 24, 2015

So then let's think about what you do know: baseline pressure in both the needle and syringe (which is equal to atmospheric pressure) and a force applied to a plunger of known cross-sectional area. Is there any way that you can use this information to eliminate all of the other variables from Bernoulli's equation besides the outgoing pressure, particularly in light of the simplifying assumptions I mentioned earlier?

11. Aug 26, 2015

Yes, I have P1, cannot find P2, I can eliminate pgh from both sides of the equation.
I am now left with P1 = 182976.376, A1 = 2.18*10^-5, and the rest of the equation
p1-p2 = 1/2rho (v^2 - v^2)

12. Aug 26, 2015

But you are given $p_2$.

13. Aug 26, 2015

p2 doesnt increase from 1atm when the plunger is pushed?

14. Aug 26, 2015

if not then i have (182976 - 101325)/.5*1000 = v2^2-v1^2

15. Aug 26, 2015

In the problem statement you gave originally, it said to "assume the pressure in the needle remains equal to 1.00 atm and that the syringe is horizontal."

The above approximation isn't so great in real life for something as narrow as a needle, but the advantage of something like a narrow needle is that you can make use of another approximation that I mentioned before: namely that the cross section of the syringe is much, much larger than that of the needle.

16. Aug 26, 2015

I'm failing to see how this helps me, I can't just say that the cross sectional area of the needle is 0, nor can I say that the velocity is extremely large. (as per a1v1=a2v2, I don't even have a velocity for the tube.)

17. Aug 26, 2015

Ah, but you were almost there. You have just said $v_1 A_1 = v_2 A_2$, with the 1 being the syringe and 2 being the needle. You can safely assume that $A_1 \gg A_2$, so try looking at it in the form
$$\dfrac{A_2}{A_1} = \dfrac{v_1}{v_2}.$$

18. Aug 26, 2015

hence,
$$\dfrac{A_1}{2.18*10^{-5}} = \dfrac{v_1}{v_2}$$
where A1 would be negligible?
hence
$$\dfrac{v_1}{v_2} = 2.18*10^{-5}$$

19. Aug 26, 2015

This could be expressed as
$${v_2} = {2.18*10^{-5}}*{v_1}$$
If this is true then
$${182976}-{101325} = {500}*(({2.18*10^{-5}*v_1})-{v_1})$$

20. Aug 26, 2015

True, but you are overcomplicating matters. If $A_2/A_1$ is negligibly small, then $v_1/v_2$ must be equally small, and you could say that $v_1 \ll v_2$, which is essentially what you did when you just estimated it as five orders of magnitude smaller (though you had your ratios reversed). For all intents and purposes, then, you could say $v_1 \approx 0$.

21. Aug 26, 2015

By solving this I get a negative velocity value, something tells me this means that it's incorrect.

22. Aug 26, 2015

this would essentially mean that $${v_2} = 12.779 m.s^{-1}$$

23. Aug 26, 2015

hence
$${2.14*10^{-5}}*{v_1}={A_2}*{12.78}$$

24. Aug 26, 2015

Right, as I previously pointed out, you mixed up your ratios. It is $v_1$ that should be substantially less than $v_2$, not the other way around.

25. Aug 26, 2015

okay so,
$${A_1}{V_1} = {A_2}{V_2}$$
$${A_1} = {2.14*10^{-5}}$$
$$\dfrac{A_1}{A_2} = \dfrac{V_2}{V_1}$$
$$\dfrac{2.14*10^{-5}}{A_2} = \dfrac{V_2}{V_1}$$

what do I do from here, please no more riddles, struggling to understand this as it is.

Last edited: Aug 26, 2015