i converted this equation before and got this figure.
I forgot how but i lost the file that i wrote the code to get the figure above.
below i am trying to rewrtie it but it is not working. the exponential function is zero or infinity.
Edit!
this is the original equation for the R_e.
i checked the parameters with my report and they are the same.
maybe there is a problem here too?
i did what you asked about the final denominator, still the same results.
clc
Contact_angle = 2.4783666516568;
Volume = 5;
ST_L = 72;
Density = 0.973;
Gravity = 9.807;
Shape_factor = 37.1;
T_zero = 0;
Dynamic_viscosity = 8.9e-4;
R_e = ( 4*Volume / (pi*Contact_angle) )^1/3;
T = 0:0.1:6;
R = R_e*(1 - exp(-(2*ST_L/R_e^12 + Density*Gravity/(9*R_e^10))*(24*Shape_factor*Volume^4*(T+T_zero)/(pi^2*Dynamic_viscosity)))).^(1/6);
plot(T,R);
Thank you All for the help

댓글 수: 8

You need to club the pi^2*neta together in the denominator to correct the definition.
But even after the correction, the result is not similar to the figure above. Are you sure the values are correct?
%Value edited
Contact_angle = 2.4783666516568;%142
Volume = 5;
ST_L = 72;
Density = 0.973;
Gravity = 9.807;
Shape_factor = 37.1;
T_zero = 0;
Dynamic_viscosity = 8.9e-4;
R_e = ( 400*Volume / (pi*Contact_angle))^(1/3);
%T according to the figure
T = 0:6;
R = R_e*(1 - exp(-(2*ST_L/R_e^12 + Density*Gravity/(9*R_e^10))*24*Shape_factor*Volume^4*(T+T_zero)/(pi^2*Dynamic_viscosity))).^(1/6);
plot(T,R)
shammas mohamed
shammas mohamed 2023년 4월 4일
the resutls should be in mm or m. i tried converting all of them but i am still facing the same problem
Dyuman Joshi
Dyuman Joshi 2023년 4월 4일
What are the units of each parameter used?
shammas mohamed
shammas mohamed 2023년 4월 4일
Volume = mm^3;
ST_L = mN/m;
Density = 0.973g/cm^3;
Gravity = m/s^2;
Shape_factor = 1/m;
Dynamic_Viscosity = Pa s
After correcting the units, you get a similar graph -
%Converting to SI units
Contact_angle = 142*pi/180;
Volume = 5 * 1e-9; %m^3;
ST_L = 72 * 1e-3; %N.m
Density = 0.973 * 1e3; %kg/m^3
Gravity = 9.807;
Shape_factor = 37.1;
T_zero = 0;
Dynamic_viscosity = 8.9e-4;
R_e = ( 400*Volume / (pi*Contact_angle))^(1/3);
%T according to the figure
T = 0:6; %seconds
R = R_e*(1 - exp(-(2*ST_L/R_e^12 + Density*Gravity/(9*R_e^10))*24*Shape_factor*Volume^4*(T+T_zero)/(pi^2*Dynamic_viscosity))).^(1/6);
%Convert R to mm
plot(T,R*1e3)
shammas mohamed
shammas mohamed 2023년 4월 4일
thank you so much. its working.
However,
why are you multiplying v by 400 in the R_e equation?
also in my experiment, the final radius reaches a value around 5 just like the original figure
Cris LaPierre
Cris LaPierre 2023년 4월 4일
The 400 is likely because that is what was in your original post before it was edited.
shammas mohamed
shammas mohamed 2023년 4월 4일
yeah sorry its not as the original equation. check the updated question

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답변 (3개)

Sam Chak
Sam Chak 2023년 4월 4일
편집: Sam Chak 2023년 4월 4일

1 개 추천

Edit: There are two correction in the equation. Thanks @Walter Roberson for pointing out.
% Parameters
Contact_angle = 2.4783666516568;
Volume = 5;
ST_L = 72;
Density = 0.973;
Gravity = 9.807;
Shape_factor = 37.1;
T_zero = 0;
Dynamic_viscosity = 8.9e-4;
R_e = ( 4*Volume / (pi*Contact_angle) )^1/3;
% suggest to split up the terms
T = linspace(0,6e-12, 601);
f5 = 24*Shape_factor*Volume^4*(T + T_zero)/((pi^2)*Dynamic_viscosity); % <--- correction
f4 = (Density*Gravity)/(9*(R_e^10));
f3 = (2*ST_L)/(R_e^12);
f2 = f3 + f4;
f1 = - f2*f5;
R = R_e*(1 - exp(f1)).^(1/6); % <--- correction
plot(T, R), grid on, xlabel('T')

댓글 수: 1

shammas mohamed
shammas mohamed 2023년 4월 4일
Thank you for the correction. it gave me a good start on the initial rate that the radius will start increasing exponentially from. however there is still problem that the function is not increasing. (the exponent part is 0).
I dont understand why this is happening. i edited my problem with more information if you need it.

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Walter Roberson
Walter Roberson 2023년 4월 4일

1 개 추천

x^1/3 means (x^1) divided by 3 which is x/3 . If you want to raise to the power of 1/3 you need x^(1/3)
Watch out for ^1/6 for the same issue.

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Dyuman Joshi
Dyuman Joshi 2023년 4월 4일
Walter, I did implement this in the code in my comment, and it still doesn't give any similar result to what OP achieved earlier.

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Cris LaPierre
Cris LaPierre 2023년 4월 4일
편집: Cris LaPierre 2023년 4월 4일

0 개 추천

As written, you need to either include your final denominator in parentheses, or divide by Dynamic_Viscocity.
We have to take the rest of the parameters you have given as correct. One potential error to look into - should your angles be in radians instead of degrees?
Contact_angle = 142;
Volume = 5;
ST_L = 72;
Density = 0.973;
Gravity = 9.807;
Shape_factor = 37.1;
T_zero = 0;
Dynamic_viscosity = 8.9e-4;
R_e = ( 400*Volume / (pi*Contact_angle))^1/3;
T = 0:300;
R = R_e * ( 1 - exp(-(2*ST_L/R_e^12 + (Density*Gravity) / (9*R_e^10))*24*Shape_factor*Volume^4*(T+T_zero)/(pi^2*Dynamic_viscosity))).^(1/6);
plot(T,R);

댓글 수: 6

this is the original equation for the R_e.
i checked the parameters with my report and they are the same.
maybe there is a problem here too?
i did what you asked about the final denominator, still the same results.
clc
Contact_angle = 2.4783666516568;
Volume = 5;
ST_L = 72;
Density = 0.973;
Gravity = 9.807;
Shape_factor = 37.1;
T_zero = 0;
Dynamic_viscosity = 8.9e-4;
R_e = ( 4*Volume / (pi*Contact_angle) )^1/3;
T = 0:0.1:6;
R = R_e*(1 - exp(-(2*ST_L/R_e^12 + Density*Gravity/(9*R_e^10))*(24*Shape_factor*Volume^4*(T+T_zero)/(pi^2*Dynamic_viscosity)))).^(1/6);
plot(T,R);
the value of R_e here is the most accurate to my experiment
Cris LaPierre
Cris LaPierre 2023년 4월 4일
It's likely an issue with inconsistent units. It might be worth doing a dimensional analysis to be sure you are consistent.
shammas mohamed
shammas mohamed 2023년 4월 4일
But why the exponential function becomes zero?
Cris LaPierre
Cris LaPierre 2023년 4월 4일
I also suspect you are using the wrong equation to reproduce the plot you have shared. The equation you have shared is for the partial wetting condition. I wonder if you shouldn't be using the complete wetting situation equation.
See here.
Using that equation, I produced the following curve. I did have to make some assumptions about units, and adjusted the time scale to match the plot you shared.
The exponential function becomes 0 because that's what happens when you raise e to a larger and larger negative number.
exp(0)
ans = 1
exp(-0.1)
ans = 0.9048
exp(-1)
ans = 0.3679
exp(-10)
ans = 4.5400e-05
Because the entire expression 1-exp(...) is multiplied by , as time increases, your result will approach .
If it happens to be doing that quicker or slower than expected, check your units.
shammas mohamed
shammas mohamed 2023년 4월 4일
i understand thanks

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