Help me convert this Equation
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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
8 Comments
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
on 4 Apr 2023
Dyuman Joshi
on 4 Apr 2023
What are the units of each parameter used?
shammas mohamed
on 4 Apr 2023
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
on 4 Apr 2023
Cris LaPierre
on 4 Apr 2023
The 400 is likely because that is what was in your original post before it was edited.
shammas mohamed
on 4 Apr 2023
Answers (3)
% 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')
1 Comment
shammas mohamed
on 4 Apr 2023
Walter Roberson
on 4 Apr 2023
1 vote
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.
1 Comment
Dyuman Joshi
on 4 Apr 2023
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.
Cris LaPierre
on 4 Apr 2023
Edited: Cris LaPierre
on 4 Apr 2023
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);
6 Comments
shammas mohamed
on 4 Apr 2023
Cris LaPierre
on 4 Apr 2023
It's likely an issue with inconsistent units. It might be worth doing a dimensional analysis to be sure you are consistent.
shammas mohamed
on 4 Apr 2023
Cris LaPierre
on 4 Apr 2023
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.
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.

Cris LaPierre
on 4 Apr 2023
Edited: Cris LaPierre
on 4 Apr 2023
The exponential function becomes 0 because that's what happens when you raise e to a larger and larger negative number.
exp(0)
exp(-0.1)
exp(-1)
exp(-10)
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
on 4 Apr 2023
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