为什么我的matlab图填写?

时间:2014-03-05 00:10:21

标签: matlab plot

我正在尝试使用Matlab对气体进行一些模拟,我希望绘制从Matrix M中获取的预测数据。这些数据应该显示为单行,但它看起来像是在两个值之间绘制和填充。我哪里错了? 提前致谢

%Basic One Dimensional Simulation

%Set Parameters
Y = 1.32;   %spec heat ratio
po = 103200; %pressure at outlet
pi = 300000; %pressure at inlet
dx = 0.3; %length of control volume
cbar = 340.29; %average speed of sound, m/s
Fr = 4; %Frequency of wave
Cd = 0.7; %Discharge coefficient
Tu = 1800; %Rough temperature of LNG combustion
Am = 1; %amplitude of wave
A = 1; %Cross sectional area of exhaust chamber
Ao = 0.4; %Cross sectional area of exhaust exit
Ai = 0.3; %Cross sectional area of exhaust inlet

%Define simulation setting time constants
Ts = 0.0001;  %Sampling interval, sec
T = 1;      %simulation time, sec
mt = T/Ts;   %Number of iterations


%Initial values
%Mass flow rates
Mg = 0.0110352; %Mass flow rate of gas
Ma = 0.189189; %Mass flow rate of air
Mi = Mg + Ma; %Total Mass flow rate at inlet
M4 = Mi;
M3 = M4;
M2 = M3;
M1 = M2;

p1 = po; %Pressure, Pa
p2 = p1;
p3 = p2;
p4 = p3;

%Simulation    
M = zeros(mt);
N = zeros(mt);
for j = 0:1:mt
    t=j*Ts;

Mi = Mg + Ma; %Mass flow rate based on input flow rates
a = (Ma/0.172);
Rexhaust = (84373 + 70841*a)/(290.67 + 210.67*a);

%Predicting Mass Flow Rate at exit
Cm1 = (((2*Y)/(Rexhaust*(Y-1)))*(((po/p1)^(2/Y))*((po/p1)^((Y+1)/Y))))^(1/2);
Cm2 = ((2/(Y+1))^(1/(Y-1)))*(2*Y/Rexhaust*(Y+1))^(1/2);

Pratio = (2/(Y+1))^(Y/(Y-1));

if po/p1 > Pratio;
Mo = Cd*Cm1*Ao*p1/(Tu)^(1/2);
else
Mo = Cd*Cm2*Ao*p1/(Tu)^(1/2);
end

%Position 1
dpdt1 = ((Rexhaust*Tu)/(dx*A))*(Mo-M1);
drhodt1 = dpdt1/(Rexhaust*Tu);
p1 = dpdt1*Ts;
M1 = (dpdt1*A*dx)/(Rexhaust*Tu);


%Position 2
dpdt2 = ((Rexhaust*Tu)/(dx*A))*(M1-M2);
drhodt2 = dpdt2/(Rexhaust*Tu);
p2 = dpdt2*Ts;
M2 = (dpdt2*A*dx)/(Rexhaust*Tu);


%Position 3
dpdt3 = ((Rexhaust*Tu)/(dx*A))*(M2-M3);
drhodt3 = dpdt3/(Rexhaust*Tu);
p3 = dpdt3*Ts;
M3 = (dpdt3*A*dx)/(Rexhaust*Tu);


%Position 4
%Predicting Mass Flow Rate at inlet
Cm1 = (((2*Y)/(Rexhaust*(Y-1)))*(((p4/pi)^(2/Y))*((p4/pi)^((Y+1)/Y))))^(1/2);
Cm2 = ((2/(Y+1))^(1/(Y-1)))*(2*Y/Rexhaust*(Y+1))^(1/2);

Pratio = (2/(Y+1))^(Y/(Y-1));

if p4/pi > Pratio;
M4 = Cd*Cm1*Ai*pi/(Tu)^(1/2);
else
M4 = Cd*Cm2*Ai*pi/(Tu)^(1/2);
end
dpdt4 = ((Rexhaust*Tu)/(dx*A))*(M3-M4);
drhodt4 = dpdt4/(Rexhaust*Tu);
p4 = dpdt4*Ts;
M4 = (dpdt4*A*dx)/(Rexhaust*Tu);    


%Save current values to Matrix M
d = [t Mo M1 M2 M3 M4 Mi dpdt1 dpdt2 dpdt3 dpdt4 drhodt1 drhodt2 drhodt3 drhodt4];
for i = 1:1:15
    M(j+1,i) = d(i);
end

%Save current p  to Matrix N
e = [t po p1 p2 p3 p4];               
for k=1:1:6
   N(j+1,k) = e(k);
end


end


%FIGURES
F1 = figure(1);
subplot(3,1,1);
plot(M(:, 1), M(:, 2), 'y'); %outlet
hold on
plot(M(:, 1), M(:, 3), 'g'); %position 1
plot(M(:, 1), M(:, 4), 'r'); %position 2
plot(M(:, 1), M(:, 5), 'b'); %position 3
plot(M(:, 1), M(:, 6), 'm'); %position 4
plot(M(:, 1), M(:, 7), 'c'); %inlet
grid on
hold off
legend('outlet','M1','M2','M3','M4','Mi');
title('MASS FLOW RATES');

subplot(3,1,2);
plot(M(:, 1), M(:, 8), 'g'); %position 1
hold on
plot(M(:, 1), M(:, 9), 'r'); %position 2
plot(M(:, 1), M(:, 10), 'b'); %position 3
plot(M(:, 1), M(:, 11), 'm'); %position 4
grid on
hold off
legend('dpdt1','dpdt2','dpdt3','dpdt4');
title('RATE OF CHANGE OF PRESSURES');

subplot(3,1,3);
plot(M(:, 1), M(:, 12), 'g'); %position 1
hold on
plot(M(:, 1), M(:, 13), 'r'); %position 2
plot(M(:, 1), M(:, 14), 'b'); %position 3
plot(M(:, 1), M(:, 15), 'm'); %position 4
grid on
hold off
legend('drhodt1','drhodt2','drhodt3','drhodt4');
title('RATE OF CHANGE OF DENSITIES');


F2 = figure(2);
plot(N(:, 1), N(:, 2), 'g');
hold on
plot(N(:, 1), N(:, 2), 'r');
plot(N(:, 1), N(:, 2), 'b');
plot(N(:, 1), N(:, 2), 'm');
hold off`enter code here`
legend('p1','p2','p3','p4');
title('PRESSURES');

1 个答案:

答案 0 :(得分:0)

填充区域是由于数据的性质(即高频分量)。下图显示了相同的数据plot(M(:, 1), M(:, 6), 'm');,但是在不同的比例上(右侧缩放)。

enter image description here

如果您选择M(1:2:end, 1)M(2:2:end, 1),则会有本地极值。