Calculating surface tension with the Capillary Equation

AI Thread Summary
The discussion centers on calculating the surface tension of water using the capillary equation while investigating the effects of soap concentrations. The user initially calculated a surface tension value of 2.36E-05, which was significantly lower than the expected value of 0.0644 N/m. After correcting for density and ensuring SI units were used, the user arrived at a new value of 0.02359 kg/s^2, still far from the expected range of 0.073-0.074 N/m. Participants suggested potential issues, such as the contact angle of water in the capillary tube and the accuracy of measurements, particularly regarding the internal diameter of the tube. The user confirmed that they used new tubes for each trial and measured heights accurately, but acknowledged the experimental setup could have introduced errors.
Allen Montgomery
Messages
14
Reaction score
0

Homework Statement


Hello everyone, I am doing an experiment and I've hit a snag with my calculations. I am looking at how concentrations of soap affect surface tension in water. I am have been using the capillary equation and capillary tubes for my calculations. I have practiced this method for 100% water and I have gotten the value of 2.36E-05 as my surface tension of water at 70 degrees F. I would assume that doing this correctly would give me the surface tension for water at 70 degrees F but as I look at other sources the values appear to be .0644 N/M. I am unsure and am looking for help as to where my methodological errors would occur for my value to be this off. I feel like I am missing something but I am unsure as to what it is. (All units should be SI)

Homework Equations


Here is the capillary equation and the variables:

s=(phga)/2

s= surface tension
h= sifference in height between depth of water and the liquid in the capillary tube .0193 m
p= density (of water) =1
g= gravity (9.8) m/s
a= raidus (2.5x10-4) m

The Attempt at a Solution



Raw data (h values)

1.95cm
1.95cm
1.9cm

Av: (1.95+1.95+1.9)/3 = 1.93cm = .0193m = h

h= .0193
p =1
g= 9.8 m/s
a= 2.5x10-4 m

s=((1)(9.8)(2.5x10-4)(.0193))/2

s= 4.7285x10-5
s= 2.364E-5 (final answer)
 
Physics news on Phys.org
What are the units of rho?
 
Chestermiller said:
What are the units of rho?
I am unsure as to what you are referring to, do you mean the radius of the hole? In that case it is m
 
Allen Montgomery said:
I am unsure as to what you are referring to, do you mean the radius of the hole? In that case it is m
The density
 
Should be g/ml
 
You’re using metric units, right? Kg/m^3?
 
I am supposed to be using SI units, did I miss a conversion?
 
Allen Montgomery said:
I am supposed to be using SI units, did I miss a conversion?
What about g?
 
I will have to take a look at that in my calculations, I clearly missed that. I'll post my new results when I get a chance.

Thank you for the help, sir!
 
  • #10
Allen Montgomery said:
I am supposed to be using SI units, did I miss a conversion?
g/ml (grams per millilitre) is not SI. (At least, not since about 1960.)
As Chet suggested, SI would be kg/m3. There's a substantial ratio between the two.
But with that correction, you will still have less than half the value you found elsewhere. And by the way, I think the value you quote is for 70C, not 70F.
 
Last edited:
  • #11
And by the way, I think the value you quote is for 70C, not 70F.[/QUOTE]

Thank you haruspex, you are correct the reference site I used was in F, after some conversion my value should be around .073-.074 N/m range.

I did make the necessary conversion for my density which did solve my issue concerning the "x10^-4" and put my value in the correct range.

However, the new value I have gotten for water is .02359 kg/s^2 which is still off. I am unsure as to where my error has occurred.
g = 9.8m/s^2
h = .0193m
a = 2.5x10^-4
p = 998.2 (more precise value for water than 1)

{(9.8)( .0193)(2.5x10^-4)(998.2)}/2 = .02395 kg/s

Also, Is kg/s^2 equivalent to N/m or is there a conversion I've missed again?

Thank you all for the help so far!
 
Last edited:
  • #12
Allen Montgomery said:
And by the way, I think the value you quote is for 70C, not 70F.

Thank you haruspex, you are correct the reference site I used was in F, after some conversion my value should be around .073-.074 N/m range.

I did make the necessary conversion for my density which did solve my issue concerning the "x10^-4" and put my value in the correct range.

However, the new value I have gotten for water is .02359 kg/s which is still off. I am unsure as to where my error has occurred.
g = 9.8m/s
h = .0193m
a = 2.5x10^-4
p = 998.2 (more precise value for water than 1)

{(9.8)( .0193)(2.5x10^-4)(998.2)}/2 = .02395 kg/s

Also, Is kg/s equivalent to N/m or is there a conversion I've missed again?

Thank you all for the help so far!
The units of surface tension are kg/s^2, which is the same as N/m.
 
  • #13
Ok, that great, and I did accidentally write m/s when it was supposed to be m/s^2.

Any idea on what could have caused my value (.02359 kg/m^s) to be so far off from .073-0.74 N/m that it should be? That is the biggest issue I am having to deal with.

(Also estimated the value for Surface tension of water at 70 degrees F from the second chart in this source https://www.engineeringtoolbox.com/water-surface-tension-d_597.html)
 
  • #14
Allen Montgomery said:
Ok, that great, and I did accidentally write m/s when it was supposed to be m/s^2.

Any idea on what could have caused my value (.02359 kg/m^s) to be so far off from .073-0.74 N/m that it should be? That is the biggest issue I am having to deal with.

(Also estimated the value for Surface tension of water at 70 degrees F from the second chart in this source https://www.engineeringtoolbox.com/water-surface-tension-d_597.html)
Was the tube really only 0.5mm internal dismeter?
Or maybe some detergent residue on the glassware?
 
  • Like
Likes Chestermiller
  • #15
haruspex said:
Was the tube really only 0.5mm internal dismeter?
Or maybe some detergent residue on the glassware?

I used a new capillary tube for every trial so that would eliminate the detergent residue (it was a variable I discussed in my paper).

On the jar of tubes, it read 0.9mm-0.5mm, which I assumed that .5mm referred to the hole and .9mm as the total diameter. Could that be an issue?
I will double check the jar when I have a chance tomorrow though.
 
  • #16
Allen Montgomery said:
I used a new capillary tube for every trial so that would eliminate the detergent residue (it was a variable I discussed in my paper).

On the jar of tubes, it read 0.9mm-0.5mm, which I assumed that .5mm referred to the hole and .9mm as the total diameter. Could that be an issue?
I will double check the jar when I have a chance tomorrow though.
It's not just the tube. Any receptacle the water contacted would need to be clean of surfactants.
If the tubes all had an outside diameter of about 1mm then your interpretation is probably right.
 
  • #17
I see what you mean, I used new plastic cups for each trial but I will have to evaluate my experiment in the near future.

But besides that are there any mathematical errors that could be of an issue?
 
  • #18
Allen Montgomery said:
are there any mathematical errors that could be of an issue?
Not that I can see.
 
  • #19
The equation you used assumes that the contact angle was 90 degrees. Was it?
 
  • #20
Chestermiller said:
The equation you used assumes that the contact angle was 90 degrees. Was it?
Don't you mean zero degrees?
Might be hard to judge without a microscope in a half mm tube!
As I recall, water on glass should have a zero contact angle.
 
  • #21
haruspex said:
Don't you mean zero degrees?
Might be hard to judge without a microscope in a half mm tube!
As I recall, water on glass should have a zero contact angle.
It depends on which angle is being referred to. I have seen it both ways. But, by your definition, it could be as much as 20 degrees on glass. Still, this is not enough to explain the OPs experimental difference. Maybe the height was in inches instead of cm?
 
Last edited:
  • #22
haruspex said:
Don't you mean zero degrees?
Might be hard to judge without a microscope in a half mm tube!
As I recall, water on glass should have a zero contact angle.

I can certainly say that I tried to hold the tube as upright in the experiment but I may have not been that precise with the angle of contact.
 
  • #23
Chestermiller said:
Maybe the height was in inches instead of cm?
That would certainly get close.
 
  • #24
Chestermiller said:
It depends on which angle is being referred to. I have seen it both ways. But, by your definition, it could be as much as 70 degrees on glass. Still, this is not enough to explain the OPs experimental difference. Maybe the height was in inches instead of cm?

I did check but the height was always measured in cm for my raw data then converted to m later on. I will take another look.
 
  • #25
Allen Montgomery said:
may have not been that precise with the angle of contact.
How would you have known? Was there a microscope in the set up?
 
  • #26
haruspex said:
How would you have known? Was there a microscope in the set up?

No there was not, I simply held the tube with my hand and marked the depth and total height with a marker.
 
  • #27
Allen Montgomery said:
No there was not, I simply held the tube with my hand and marked the depth and total height with a marker.
So you would not have been able to judge the contact angle. But I doubt that was a problem.
Holding the tube in your hand sounds like more of a risk. You really should have had it clamped.
Where were you making the marks? On the cup?
 
  • #28
haruspex said:
So you would not have been able to judge the contact angle. But I doubt that was a problem.
Holding the tube in your hand sounds like more of a risk. You really should have had it clamped.
Where were you making the marks? On the cup?

No there were two marks on the capillary tube itself, one for the depth (roughly 2.7cm) and one for total height (4.6, cm on average). The difference between these values should be the h values in the capillary equation.
 
  • #29
Allen Montgomery said:
No there were two marks on the capillary tube itself, one for the depth (roughly 2.7cm) and one for total height (4.6, cm on average). The difference between these values should be the h values in the capillary equation.
So you were making these marks by hand, at some depth within the surrounding cup maybe?, holding the tube with the other hand, while looking down into the cup from above... All sounds a bit awkward.
Presumably the base of the tube was resting on the base of the cup. I don't think that should affect the result, but it does not sound ideal.
 
  • #30
haruspex said:
So you were making these marks by hand, at some depth within the surrounding cup maybe?, holding the tube with the other hand, while looking down into the cup from above... All sounds a bit awkward.
Presumably the base of the tube was resting on the base of the cup. I don't think that should affect the result, but it does not sound ideal.

I was making these markings by hand, although I used a short cup so I could look at it and mark it from the side. But yes it was a bit awkward but I did get mostly consistent results in the experiment

The base of the tube was resting on the base of the cup for every trial as well.
 

Similar threads

Wetting on the outside of a glass cylinder
Replies
12
Views
1K
Calculating Capillary Height with Surface Tension: A Question on Homework
Replies
1
Views
2K
Force acting on one side of a fluid
Replies
22
Views
1K
Measuring Height of Meniscus in Capillary Action
Replies
1
Views
3K
Surface tension and capillary tube
Replies
8
Views
4K
What is the relationship between surface tension and forces?
Replies
2
Views
1K
Surface Tension and Capillary Rise
Replies
1
Views
1K
Surface Tension of needle on water
Replies
4
Views
4K
Capillary Tube and Surface Tension
Replies
7
Views
13K
How to Calculate Capillary Rise on the ISS
Replies
1
Views
2K

Hot Threads

Recent Insights

Back
Top