Business major.... fluid dynamics question....

[Stephen0311]

Homework Statement

I have a syringe filled with water. This syringe is a cylinder with a length of 3 inches (L), diameter of 1.5 inches (D). Attached to the syringe is a constant force spring, which pushes the plunger into the syringe with a force of 2 pounds (F). Ignore the friction between the plunger and the outside of the syringe.

The syringe is resting on a table, laying horizontally. The needle that the water is trying to escape through has a length of .5 inches (LN), and a diameter of .1mm (LD). If no force is applied by the plunger, the water does not escape through the end of the syringe (due to surface tension I believe).

I'm trying to calculate the amount of time it will take for the syringe to be empty. The goal would be to manipulate F, LN, LD such that it takes 30 minutes for the syringe to empty.

Is this possible to calc? Eg can I set up a spreadsheet so that I can manipulate these variables to see how it effects the amount of time it takes to empty?

Homework Equations

I've been reading up and I think this is related to Bernoulli's equations?

The Attempt at a Solution

I'm about 10 years removed from any calc/physics - so if anyone has the time to walk me through it in detail I would appreciate it a whole bunch. I've been trying to figure this out with a few engineering friends (granted, they aren't fluid dynamics guys), but everyone came up with drastically different solutions.

If you need any more detail, please let me know. I tried to be specific in the initial question, but I may have missed something that you need.

 

[Chestermiller]

The first step is to focus on the needle (which is the most important part of what's happening) and to determine the pressure vs flow rate equation for that. The Bernoulli equation is not the relationship to use for this problem.. The correct equation to use is the Hagen-Poiseulle equation.

 

[Stephen0311]

The first step is to focus on the needle (which is the most important part of what's happening) and to determine the pressure vs flow rate equation for that. The Bernoulli equation is not the relationship to use for this problem.. The correct equation to use is the Hagen-Poiseulle equation.

Really appreciate the response - I'm actually a business major from Michigan so Go Blue back at you!

I watched this video regarding the Hagen-Poiseulle equation and think I have a decent understanding of the different variables - the one outstanding question is how would I calculate the pressure going into that needle vs outside the needle (P1 and P2 in the video).

I'm not really sure if this is a complex question or a simple one... Appreciate any insight you have on this

 

[Stephen0311]

Also, I have it attempted to put it into spreadsheet form.. here is a link to my image... https://imgur.com/a/160Wc

I'm fairly confident that it's completely wrong.. hopefully someone a lot smarter than me can point out my mistakes though.

 

[Chestermiller]

Also, I have it attempted to put it into spreadsheet form.. here is a link to my image... https://imgur.com/a/160Wc

I'm fairly confident that it's completely wrong.. hopefully someone a lot smarter than me can point out my mistakes though.

I think you're solving this backwards. You know the volume rate of flow you want, and you know the geometry, so you can calculate the pressure drop and see if it matches your desired value with the diameter you have specified.

I would solve this using cgs units (dynes, cm, gm, sec). In these units, the viscosity of water at room temperature is 0.01 gm/(cm.sec) and the desired force is 8.9 x 10^5 dynes. The first thing to do is calculate the volume rate of flow in cc/sec. Then, divide that by the cross sectional area of needle to get the average flow velocity, and to see whether that velocity makes sense. For example, if it was coming out of the needle at 600 cm/sec, would that be acceptable.

By the way, the force on the piston is equal to the pressure times the cross sectional area of the syringe and piston, not the cross sectional area of the needle. The axial force on the fluid within the needle is much, much less than 2 lb.

Your original design was for a needle diameter of 0.1 mm, but, in the calculations, you used a radius of 0.1 mm. Which did you mean?

If you specify a geometry, I will run a calculation to get the force and pressure drop to see if we can match results.

 

[Stephen0311]

I think you're solving this backwards. You know the volume rate of flow you want, and you know the geometry, so you can calculate the pressure drop and see if it matches your desired value with the diameter you have specified.

I would solve this using cgs units (dynes, cm, gm, sec). In these units, the viscosity of water at room temperature is 0.01 gm/(cm.sec) and the desired force is 8.9 x 10^5 dynes. The first thing to do is calculate the volume rate of flow in cc/sec. Then, divide that by the cross sectional area of needle to get the average flow velocity, and to see whether that velocity makes sense. For example, if it was coming out of the needle at 600 cm/sec, would that be acceptable.

By the way, the force on the piston is equal to the pressure times the cross sectional area of the syringe and piston, not the cross sectional area of the needle. The axial force on the fluid within the needle is much, much less than 2 lb.

Your original design was for a needle diameter of 0.1 mm, but, in the calculations, you used a radius of 0.1 mm. Which did you mean?

If you specify a geometry, I will run a calculation to get the force and pressure drop to see if we can match results.

Let's go with the diameter is .1mm - I'm flexible on this as one of the variables that can change to make the time equal to 30 minutes. Is there anyway you could throw it into a speadsheet form kind of like I did and show me what you mean? I think I'm following your logic, but it would be super helpful if you could kind of lay the calculation that you are doing step by step so that I can play around with the variables myself and see how that effects the solution.

I really appreciate your help, you've really helped me make more progress in one day than I had in like 2 weeks trying to figure this out by myself

 

[Stephen0311]

Let's go with the diameter is .1mm - I'm flexible on this as one of the variables that can change to make the time equal to 30 minutes. Is there anyway you could throw it into a speadsheet form kind of like I did and show me what you mean? I think I'm following your logic, but it would be super helpful if you could kind of lay the calculation that you are doing step by step so that I can play around with the variables myself and see how that effects the solution.

I really appreciate your help, you've really helped me make more progress in one day than I had in like 2 weeks trying to figure this out by myself

To build off of this - I have attempted to recalc it and have a few questions/comments. I'm hoping it's not overly frustrating trying to teach fluid dynamics to someone that has never practiced it at all before, but I can't reiterate enough how thankful I am.

1) Is there a reason that we care about the velocity of the water exiting the syringe? My understanding was that Q in the Hagen-Poiseuille equation demonstrates the mL/s flow of the water out of the needle/hole. So by adjusting my assumptions (eg varying the diameter of the needle, diameter of syringe, etc) I would be aiming for a Q equal to my syringe volume divided by seconds desired (thus emptying the syringe).

For example, if the volume of my syringe is 100mL, and my desired time is 30 minutes (1800 seconds) then my desired Q would be .056 mL/s. Is this logic correct, or am I missing something?

2) Based on your initial input I attempted to recalc it (and changed the units) - Here is a similar spreadsheet showing my new calculations. Can you look through it and see if you have any feedback? https://imgur.com/a/YXnKr

From this calc, I have a very low velocity and a very low flow. Since I saw this, I recalced it (as shown in comment #3) but adjusted my needle diameter so that my Q was equal to my desired mL/s.

3) Here is the calc in which I simply goalseeked the diameter needed in order to have my Q = .056 mL/s. If you have any comments on #2, no need to repeat them and I'll simply work through solving #2 before redoing the goalseek.
https://imgur.com/a/PJHxb

4) I can upload the spreadsheet somewhere or email it to you if the pictures aren't clear or if you simply would like a copy to play around with.

Thank you for reading this through.

 

[Chestermiller]

SAMPLE CALCULATION

Syringe volume = $\pi\frac{D^2}{4}L=3.14159\frac{(1.5)^2}{4}(3)=5.3\ in^3=5.3(2.54)^3=86.9\ cm^3$

Volume flow rate = $Q = \frac{86.9\ cm^3}{1800\ sec}=0.0483\ \frac{cm^3}{sec}$

Needle diameter d = 0.1 mm = 0.01 cm

Cross sectional area of needle A = $\pi \frac{d^2}{4}=\pi \frac{(0.01)^2}{4}=0.785\times10^{-4}\ cm^2$

Average axial velocity of fluid in needle = $\frac{0.0483}{0.785\times10{-4}}=615\ \frac{cm}{sec}=6.15 \frac{meters}{sec}$

Equation for pressure drop: $\Delta p=\frac{128Q\mu L}{\pi d^4}$ where $\mu=viscosity=0.01\frac{gm}{cm.sec}$ and L = 0.5 inches
$$\Delta p=\frac{128(0.0483)(0.01)(0.5\times2.54)}{(3.14159)(0.01)^4}=2.5\times10^6\ \frac{dynes}{cm^2}$$

Cross sectional area of piston = $\pi \frac{D^2}{4}=(3.14159)\frac{(1.5\times 2.54)^2}{4}=11.4\ cm^2$

Force on piston: $$F = A\Delta p=(11.4)(2.5\times10^6)=2.85\times 10^7 dynes=64.1\ pounds$$

 

[Stephen0311]

SAMPLE CALCULATION

Syringe volume = $\pi\frac{D^2}{4}L=3.14159\frac{(1.5)^2}{4}(3)=5.3\ in^3=5.3(2.54)^3=86.9\ cm^3$

Volume flow rate = $Q = \frac{86.9\ cm^3}{1800\ sec}=0.0483\ \frac{cm^3}{sec}$

Needle diameter d = 0.1 mm = 0.01 cm

Cross sectional area of needle A = $\pi \frac{d^2}{4}=\pi \frac{(0.01)^2}{4}=0.785\times10^{-4}\ cm^2$

Average axial velocity of fluid in needle = $\frac{0.0483}{0.785\times10{-4}}=615\ \frac{cm}{sec}=6.15 \frac{meters}{sec}$

Equation for pressure drop: $\Delta p=\frac{128Q\mu L}{\pi d^4}$ where $\mu=viscosity=0.01\frac{gm}{cm.sec}$ and L = 0.5 inches
$$\Delta p=\frac{128(0.0483)(0.01)(0.5\times2.54)}{(3.14159)(0.01)^4}=2.5\times10^6\ \frac{dynes}{cm^2}$$

Cross sectional area of piston = $\pi \frac{D^2}{4}=(3.14159)\frac{(1.5\times 2.54)^2}{4}=11.4\ cm^2$

Force on piston: $$F = A\Delta p=(11.4)(2.5\times10^6)=2.85\times 10^7 dynes=64.1\ pounds$$

Thank you Chester - This was insanely helpful. I think I'm all set now, I'll let you know if I run into any issues. You're the man! Go Blue!

 

[Chestermiller]

Thank you Chester - This was insanely helpful. I think I'm all set now, I'll let you know if I run into any issues. You're the man! Go Blue!

I would only have done this for a fellow wolverine. Go Blue!