Differential equation to find activation voltage

I am trying to calculate the activation voltage, with the following code:

i = 100 ;
t = [0 0.2 0.01];

R = 8.314;
G = 9.648*10^4;
T = 303.15; %Conductivity reference temperature
i_0dob = 0.67;
A_m = 232e-4;
b_ca = 0.55 ;
b_an = 1-b_ca ;
C_dl = 3.15e-2*A_m ;

syms u(t)

ode = diff(u,t) == (i-A_m * i_0dob * (exp(((b_an*G)/(R*T))*u)-exp((-(b_ca*G)/(R*T))*u)))/C_dl;
cond = u(0) == 0; 

uSol(t) = dsolve(ode,cond)  %This - is line 19

Warning: Unable to find symbolic solution.

 
uSol(t) =
 
[ empty sym ]
 

I get the warning:

Warning: Unable to find symbolic solution

In dsolve (line 209) %i dont have a line 209??

In Problem43 (line 19)

Can anybody help?

[Sam Chak]

Hi @Anniken Hagen

Please check if the differential equation looks like this in mathematical form:

[Anniken]

Yes! Its this one!

[Sam Chak]

Thanks @Anniken Hagen.

Could you also take a screenshot of the parameters and post it here super quickly? In the problem image, the values of the parameters are not displayed. It would be helpful for us to cross-check because selected values often impact the transient behavior of the system.

Answer 1

[Torsten]

Matlab cannot find an analytical expression for u. You have to use a numerical approach.

I'm not sure about the term exp(b_an*G/(R*T)*u). Shouldn't it be exp(-b_an*G/(R*T)*u) ?

i = 100 ;
R = 8.314;
G = 9.648*10^4;
T = 303.15; %Conductivity reference temperature
i_0dob = 0.67;
A_m = 232e-4;
b_ca = 0.55 ;
b_an = 1-b_ca ;
C_dl = 3.15e-2*A_m ;
fun= @(t,u) (i-A_m * i_0dob * (exp(b_an*G/(R*T)*u)-exp(-b_ca*G/(R*T)*u)))/C_dl;
tspan = [0 0.01];
u0 = 0;
[T,U] = ode15s(fun,tspan,u0);
plot(T,U)

[Anniken]

Thank you! The expression is based off this:

[Torsten]

Ok, then the code should be correct.

[Anniken]

This is the type of graph we should end up with.

[Torsten]

i = [100,10,100];
t = [0,0.2,0.4,0.6];
R = 8.314;
G = 9.648*10^4;
Temp = 303.15; %Conductivity reference temperature
i_0dob = 0.67;
A_m = 232e-4;
b_ca = 0.55 ;
b_an = 1-b_ca ;
C_dl = 3.15e-2*A_m ;
T = [];
U = [];
u0 = 0;
for j = 1:numel(i)
  fun= @(t,u) (i(j)-A_m * i_0dob * (exp(b_an*G/(R*Temp)*u)-exp(-b_ca*G/(R*Temp)*u)))/C_dl;
  tspan = [t(j) t(j+1)];
  [Tstep,Ustep] = ode15s(fun,tspan,u0);
  T = [T;Tstep];
  U = [U;Ustep];
  u0 = Ustep(end,:);
end
plot(T,U)

[Anniken]

Hmm okey i will try! Thank you

[Walter Roberson]

I wonder why was not written in terms of 2 * sinh() instead of in terms of the difference of exp() ?

[Torsten]

@Walter Roberson

b_an does not equal b_ca.

@Anniken Hagen

I included a possible approach to account for a piecewise constant function i.

[Anniken]

Thank you! I will try to interpret the code you have sent and apply it to the code and hopefully it works!

[Walter Roberson]

@Torsten

Ah, you are right, my eyes glazed over that detail.

[Sam Chak]

Hi @Anniken Hagen

I believe I've found the typo in C_dl. However, I'm struggling to plot the piecewise current i and using an anonymous function. Maybe I should ask @Torsten for help.

i       = [100, 10, 100];
t       = [0, 0.2, 0.4, 0.5];
R       = 8.314;        % looks like gas constant (does gas mix with electric?)
G       = 9.648e4;      
Temp    = 303.15;       % Conductivity reference temperature
i_0dob  = 0.67;
A_m     = 232e-4;
b_ca    = 0.55;
b_an    = 1 - b_ca;
C_dl    = (3.15e2)*A_m; % affect the transient

T       = [];
U       = [];
u0      =  0;

for j = 1:numel(i)
    fun   = @(t,u) (i(j) - A_m*i_0dob*(exp(b_an*G/(R*Temp)*u) - exp(- b_ca*G/(R*Temp)*u)))/C_dl;
    tspan = [t(j) t(j+1)];
    [Tstep,Ustep] = ode15s(fun, tspan, u0);
    T     = [T; Tstep];
    U     = [U; Ustep];
    u0    = Ustep(end,:);
end

% Plot Voltage
subplot(211)
plot(T, U, 'linewidth', 1.5), grid on
xlabel('t, [s]')
ylabel('u, [V]')

% Plot Currents
im = A_m*i_0dob*(exp(b_an*G/(R*Temp)*U) - exp(- b_ca*G/(R*Temp)*U));
% i  = 
% id = i - im;
subplot(212)
plot(T, im, 'r'), grid on, ylim([-100 100])
xlabel('t, [s]')
ylabel('i, [A]')

[Anniken]

Yes!Had typo in C_dl and the G and F.... The bug was fixed and also a function was implemented and the resolution was found! Thank you so much for all the help:D