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syms D g H Do

tspan = [0 120];

mgiren=0

Do=3;

D=2/10;

h0=h;

g=9.81;

y0 = 2;

ySol= ode45(@(t,h)(mgiren-(pi()*D^2/4*(2*g*h)^0.5)/(pi()*Do^2/4)), tspan, y0)

for t=linspace(0,100,11)

fprintf('%15.5f',t,deval(ySol,t)),;fprintf('\n')

end

My differantial code is here, dt/dh=(mgiren-(pi()*D^2/4*(2*g*h)^0.5)/(pi()*Do^2/4)),h(0)=2, h(tx)=1, how can i find tx, is there anyway to find tx, i can find it with interpolate but is there any easier way.

Alan Stevens
on 21 Aug 2020

Edited: Alan Stevens
on 21 Aug 2020

Something like this perhaps:

tspan = 0:120;

h0=2;

[t, h] = ode45(@rate, tspan, h0);

tx = interp1(h,t,1);

plot(t,h,tx,1,'o'), grid

text(tx+5,1,['tx = ' num2str(tx)])

xlabel('t'),ylabel('h')

function dhdt = rate(~, h)

mgiren=0;

Do=3;

D=2/10;

g=9.81;

dhdt = (mgiren-(pi*D^2/4*(2*g*h)^0.5)/(pi*Do^2/4));

end

Ok, I guess this still uses interpolation!

You could use fzero to find it, but, for this curve I think interpolation is far simpler.

Alan Stevens
on 21 Aug 2020

Tht's because you can't get to zero with the data specified. There is an analytical solution to your equation, which is most easily expressed with t as a function of h - see below. You'll notice there is a logarithmic term, which tries to calculate log(0) when both mgiren and h are zero.

tspan = 0:120;

h0=2;

mgiren=0;

Do=3;

D=2/10;

g=9.81;

a = mgiren/(pi*Do^2/4);

b = (D/Do)^2*sqrt(2*g);

T = @(h) -2*(b*sqrt(h) + a*log(a-b*sqrt(h)))/b^2 ...

+ 2*(b*sqrt(h0) + a*log(a-b*sqrt(h0)))/b^2;

h = h0:-0.01:0.01;

t = T(h);

plot(t,h), grid

xlabel('t'), ylabel('h')

Alan Stevens
on 21 Aug 2020

Hmm. Thinking further, the log term is zeroed all the way through because it's multiplied by a (which is zero)., so, basically, you just have a square root relationship (when a is zero).

You get NaN if you try T(0).

T(h = 0) can be found from 2*sqrt(h0)/b;

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