Generating a fourth-degree polynomial curve for the expansion region of a planar nozzle

3 views (last 30 days)
Greetings everyone, I am trying to generate an initial expansion curve of a nozzle using an arc or polynomials. I have attached the screenshot below.
I made the code using the polynomial of 4th degree with a predefined initial expansion height(H_e)[Curve 4 ]. The H_e which I took(1.0834) is based on the use of circular arc to generate the curve. The last ordinate of the curve is H_e, which I then used in this program. However, I am stuck on getting this maximum possible H_e for a given L_e. Is there any way I can get around this? Do we need to evaluate the polynomial again at maximum H_e? Thank you!
G=1.4;
Gp=G+1;
Gm=G-1;
M = 2;
nu = sqrt(Gp/Gm).*atand(sqrt(Gm*(M.^2-1)/Gp))-atand(sqrt(M.^2-1));
ThetaMax = nu/2;
y0 = 1;
H_e = 1.0834;
L_e = 3.155*y0*sind(ThetaMax);
%% Generating the polynomial
% Polynomial coefficients (a4, a3, a2, a1, a0)
syms x a0 a1 a2 a3 a4
% Polynomial of degree 4
Y = a4*x^4 + a3*x^3 + a2*x^2 + a1*x + a0;
% Boundary Conditions
% At O: X = 0, Y = h_h
cond1 = subs(Y, x, 0) == y0;
% At O: dY/dX = 0
dY_dx = diff(Y, x);
cond2 = subs(dY_dx, x, 0) == 0;
% At E: dY/dX = tan(theta_max)
cond3 = subs(dY_dx, x, L_e) == tand(ThetaMax);
% At E: d2Y/dX2 = 0
d2Y_dx2 = diff(dY_dx, x);
cond4 = subs(d2Y_dx2, x, L_e) == 0;
% At E: X = L_e, Y = H_e
cond5 = subs(Y, x, L_e) == H_e;
% Solve for coefficients
coeffs = solve([cond1, cond2, cond3, cond4, cond5], [a0, a1, a2, a3, a4]);
% Display the polynomial coefficients
a0 = coeffs.a0;
a1 = coeffs.a1;
a2 = coeffs.a2;
a3 = coeffs.a3;
a4 = coeffs.a4;
fprintf('Polynomial coefficients:\n');
fprintf('a4 = %f\n', double(a4));
fprintf('a3 = %f\n', double(a3));
fprintf('a2 = %f\n', double(a2));
fprintf('a1 = %f\n', double(a1));
fprintf('a0 = %f\n', double(a0));
%%
% Define the polynomial function using the obtained coefficients
Y_poly = matlabFunction(a4*x.^4 + a3*x.^3 + a2*x.^2 + a1*x + a0);
Plotting the curve:
x_values = linspace(0, L_e, 56 ); % n+1 points from 0 to L_e
% Evaluate the polynomial at each x value
y_values = Y_poly(x_values);
% Plot the polynomial curve
plot(x_values, y_values, 'b', 'LineWidth', 2);
xlabel('X');
ylabel('Y');
title('Polynomial Curve of Degree 4');
grid on;

Accepted Answer

Arnav
Arnav on 19 Aug 2024
From the attached pictures of the equations, the paper simply states that amongst all the curves discussed, the Poly4_Pre curve gives the maximum value of H_e for a given L_e i.e. there is no need to maximize the value of H_e as it is simply the value of the polynomial Poly4_Pre evaluated at x=L_e. Further, this H_e value is used to fit the Bezier curve.
The confusion you had might be due to you assuming the last point At E: X=L_e, Y=H_e as a boundary condition where you assumed the value of H_e to be the value you got from the circular arc.
Omitting this boundary condition, we will get a curve which gives the value of H_e when evaluated at L_e.
This can be done as shown:
Constants
G=1.4;
Gp=G+1;
Gm=G-1;
M = 2;
nu = sqrt(Gp/Gm).*atand(sqrt(Gm*(M.^2-1)/Gp))-atand(sqrt(M.^2-1));
ThetaMax = nu/2;
y0 = 1;
L_e = 3.155*y0*sind(ThetaMax);
% Define x values for plotting
x_values = linspace(0, L_e, 100);
Circular Arc
% Define parameters
y0 = 1; % Initial height, example value
ThetaMax = 15; % Maximum angle in degrees
Le = 3.155 * y0 * sind(ThetaMax); % Length along symmetry axis
% Calculate the radius R based on the slope condition
R = Le / tand(ThetaMax);
% Define the circle center
center_y = y0 + R;
% Calculate corresponding y values for the circular arc
circle_y = center_y - sqrt(R^2 - x_values.^2);
Poly2
syms x a0 a1 a2
% Polynomial of degree 2
Y = a2*x^2 + a1*x + a0;
% Boundary Conditions
% At O: X = 0, Y = h_h
cond1 = subs(Y, x, 0) == y0;
% At O: dY/dX = 0
dY_dx = diff(Y, x);
cond2 = subs(dY_dx, x, 0) == 0;
% At E: dY/dX = tan(theta_max)
cond3 = subs(dY_dx, x, L_e) == tand(ThetaMax);
% Solve for coefficients
coeffs = solve([cond1, cond2, cond3], [a0, a1, a2]);
a0 = coeffs.a0;
a1 = coeffs.a1;
a2 = coeffs.a2;
% Define the polynomial function using the obtained coefficients
Y_poly2 = matlabFunction(a2*x.^2 + a1*x + a0);
poly2_y = Y_poly2(x_values);
Poly3
syms x a0 a1 a2 a3
% Polynomial of degree 4
Y = a3*x^3 + a2*x^2 + a1*x + a0;
% Boundary Conditions
% At O: X = 0, Y = h_h
cond1 = subs(Y, x, 0) == y0;
% At O: dY/dX = 0
dY_dx = diff(Y, x);
cond2 = subs(dY_dx, x, 0) == 0;
% At E: dY/dX = tan(theta_max)
cond3 = subs(dY_dx, x, L_e) == tand(ThetaMax);
% At E: d2Y/dX2 = 0
d2Y_dx2 = diff(dY_dx, x);
cond4 = subs(d2Y_dx2, x, L_e) == 0;
% Solve for coefficients
coeffs = solve([cond1, cond2, cond3, cond4], [a0, a1, a2, a3]);
a0 = coeffs.a0;
a1 = coeffs.a1;
a2 = coeffs.a2;
a3 = coeffs.a3;
% Define the polynomial function using the obtained coefficients
Y_poly3 = matlabFunction(a3*x.^3 + a2*x.^2 + a1*x + a0);
poly3_y = Y_poly3(x_values);
Poly4
syms x a0 a1 a2 a3 a4
% Polynomial of degree 4
Y = a4*x^4 + a3*x^3 + a2*x^2 + a1*x + a0;
% Boundary Conditions
% At O: X = 0, Y = h_h
cond1 = subs(Y, x, 0) == y0;
% At O: dY/dX = 0
dY_dx = diff(Y, x);
cond2 = subs(dY_dx, x, 0) == 0;
% At E: dY/dX = tan(theta_max)
cond3 = subs(dY_dx, x, L_e) == tand(ThetaMax);
% At O: d2Y/dX2 = 0
d2Y_dx2 = diff(dY_dx, x);
cond4 = subs(d2Y_dx2, x, L_e) == 0;
% At E: d2Y/dX2 = 0
cond5 = subs(d2Y_dx2, x, 0) == 0;
% Solve for coefficients
coeffs = solve([cond1, cond2, cond3, cond4, cond5], [a0, a1, a2, a3, a4]);
a0 = coeffs.a0;
a1 = coeffs.a1;
a2 = coeffs.a2;
a3 = coeffs.a3;
a4 = coeffs.a4;
Y_poly4 = matlabFunction(a4*x.^4 + a3*x.^3 + a2*x.^2 + a1*x + a0);
poly4_y = Y_poly4(x_values);
Poly4Pre
syms x a0 a1 a2 a3 a4
% Polynomial of degree 4
Y = a4*x^4 + a3*x^3 + a2*x^2 + a1*x + a0;
% Boundary Conditions
% At O: X = 0, Y = h_h
cond1 = subs(Y, x, 0) == y0;
% At O: dY/dX = 0
dY_dx = diff(Y, x);
cond2 = subs(dY_dx, x, 0) == 0;
% At E: dY/dX = tan(theta_max)
cond3 = subs(dY_dx, x, L_e) == tand(ThetaMax);
% At E: d2Y/dX2 = 0
d2Y_dx2 = diff(dY_dx, x);
cond4 = subs(d2Y_dx2, x, L_e) == 0;
% Solve for coefficients
coeffs = solve([cond1, cond2, cond3, cond4], [a0, a1, a2, a3, a4]);
a0 = coeffs.a0;
a1 = coeffs.a1;
a2 = coeffs.a2;
a3 = coeffs.a3;
a4 = coeffs.a4;
Y_poly4Pre = matlabFunction(a4*x.^4 + a3*x.^3 + a2*x.^2 + a1*x + a0);
poly4Pre_y = Y_poly4Pre(x_values);
H_e = poly4Pre_y(end)
H_e = 1.1286
Bézier
% Control points as symbolic variables
syms x1 y1 x2 y2 real
% Define control points
P0 = [0, y0];
P1 = [x1, y1];
P2 = [x2, y2];
P3 = [L_e, H_e];
% Bézier curve as a symbolic function of t
syms t
B = (1-t)^3 * P0 + 3*(1-t)^2 * t * P1 + 3*(1-t) * t^2 * P2 + t^3 * P3;
% Separate x and y components
Bx = B(1);
By = B(2);
% First derivatives
dBx_dt = diff(Bx, t);
dBy_dt = diff(By, t);
% Slope (dy/dx) at t
slope = dBy_dt / dBx_dt;
% double derivative
DD = diff(slope,t)/dBx_dt;
% Boundary conditions
eq1 = subs(By, t, 0) == y0; % At t = 0, y = y0
eq2 = subs(slope, t, 0) == 0; % At t = 0, dy/dx = 0
eq3 = subs(slope, t, 1) == tand(ThetaMax); % At t = 1, dy/dx = tan(thetaMax)
eq4 = subs(By, t, 1) == H_e; % At t = 1, y = He
eq5 = subs(DD,t,0) == 0;
eq6 = subs(DD,t,1) == 0;
% Solve for the unknowns
sol = solve([eq1, eq2, eq3, eq4, eq5, eq6], [x1, y1, x2, y2]);
% Display the solution
x1_sol = double(sol.x1);
y1_sol = double(sol.y1);
x2_sol = double(sol.x2);
y2_sol = double(sol.y2);
% Substitute the solutions back into the control points
P1_sol = [x1_sol, y1_sol];
P2_sol = [x2_sol, y2_sol];
% Plot the Bézier curve
t_vals = linspace(0, 1, 100);
B_vals = subs(B, {x1, y1, x2, y2}, {x1_sol, y1_sol, x2_sol, y2_sol});
B_vals = subs(B_vals,t,t_vals');
% Extract x and y coordinates
x_vals = B_vals(:, 1);
y_vals = B_vals(:, 2);
Plotting All
figure;
hold on;
plot(x_values, circle_y, 'DisplayName', 'Circular Arc');
plot(x_values, poly2_y, 'DisplayName', 'Poly2');
plot(x_values, poly3_y , 'DisplayName', 'Poly3');
plot(x_values, poly4_y , 'DisplayName', 'Poly4');
plot(x_values, poly4Pre_y , 'DisplayName', 'Poly4Pre');
plot(x_vals, y_vals, 'DisplayName', 'Bezier');
xlabel('X');
ylabel('Y');
title('Comparing Plots');
legend('show', 'Location', 'northwest');
grid on;
hold off;
This graph looks similar to the results shown in the paper which you were referring to.

More Answers (0)

Products


Release

R2024a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!