Drawing parabolic pipe flow
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I am trying to plot this on matlab for varying pressures. u_r is just velocity, r radius, and p_1 pressure for the flow through a pipe, but having no luck. Any help would be greatly appreciated!!
回答（1 个）
Star Strider
2017-3-8
I let the Symbolic Math Toolbox solve for ‘p_1(z)’ and ‘u_r’. I am not certain what you want.
syms a p_1(z) r u_r z p_10 dp_10
dp_1dz = diff(p_1,z); % First Derivative
d2p_1dz2 = diff(p_1,z,2); % Second Derivative
Eqn = u_r == (1/a^2) * d2p_1dz2 * (2 - r^2)*r; % Differential Equation
P_1s = dsolve(Eqn, p_1(0) == p_10, dp_1dz(0) == dp_10); % Solve For ‘p_1(z)’
U_r = solve(P_1s == p_1, u_r); % Solve For ‘u_r’
p_1_fcn = matlabFunction(P_1s) % Create Anonymous Function
u_r_fcn = matlabFunction(U_r) % Create Anonymous Function
p_1_fcn = @(a,dp_10,p_10,r,u_r,z) p_10+dp_10.*z-(a.^2.*u_r.*z.^2.*(1.0./2.0))./(r.*(r.^2-2.0));
u_r_fcn = @(a,dp_10,p_10,r,z) 1.0./a.^2.*r.*1.0./z.^2.*(r.^2-2.0).*(p_10-p_1(z)+dp_10.*z).*2.0;
The ‘p_1_fcn’ and ‘u_r_fcn’ are anonymous function versions of those functions that you can use in regular MATLAB code without the Symbolic Math Toolbox. The ‘p_10’ and ‘dp_10’ variables are the initial conditions (values) for ‘p_1(z)’ and ‘dp_1/dz’ respectively.
I leave the rest to you.
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2017-3-8
2017-3-8
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