Hello, I'm trying to solve Rayleigh–Plesset equation to compute bubble radius at certain time instants.
The image above shows the equation. I've chosen fluid density, kg/m³, Pascal. I have to vary the Pressure at infinity () as a function of time. As the paper shows (from which I have taken the values of P at infinity and time range), the bubble radius must increase first and then it should collapse. But I have been getting the radius increment out of bound. What would be the solution to this problem? Am I missing something obvious?
The pulse code (varying P at infinity) is here:
function z = pulse(t)
z = (-.0002*t^3) + (0.0851*t^2) - (8.7073*t) + 52.853; % code
end
The differential equation code is here:
function f = RP(t, y)
p_ref = 3200; % reference pressure (Pa)
rho = 1e3; % density of the surrounding fluid (kg/m3)
f = [y(2);
(p_ref - pulse(t))/(rho*y(1)) - 1.5*(y(2)^2)/y(1)];
end
I'm taking the initial radius as 50 micrometer. The plotting code is here:
[t, y] = ode45(@RP, [0, 0.3], [50e-6, 0]);
plot(t, y(:,1)), grid
xlabel('time'),
ylabel('$R, \;(\rm{m})$', 'interpreter', 'latex')
title('Radius of the spherical bubble')
for i = 1:length(t)
z(i) = pulse(t(i));
end
plot(t, z), grid
xlabel('time'),
ylabel('$P_{\infty}, \;(\rm{Pa})$', 'interpreter', 'latex')
title('External pressure infinitely far from the bubble')
Any help with the code will be appreciated. Thanks.

回答 (2 件)

Sulekha
Sulekha 2018 年 2 月 16 日

0 投票

Were you able to find a solution to this issue? I am also facing the same problem.
Jhonath wensenber
Jhonath wensenber 2025 年 11 月 21 日
編集済み: Jhonath wensenber 2025 年 11 月 21 日

0 投票

Lo que sucede es que el término de (PR-PL)/rho debe tener unidades de m2/s2 para ello modifica la expresión a: g(PR-PL)/(g*rho), de tal modo que, [g]= m/s2, [PR-PL]=N/m2 y [g*rho]=9810 N/m3.

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