Question

please screenshot the MATlab code if possible

The following equation is available for the saturation pressure of air as a function of temperature: In (psat/Pc) = C1 x + C2 x2 + C3 In (Tsat/Tc), where x = 1/Tc - 1/Tsat Pc =3.77 MPa is the critical pressure, and Tc = 132.5 K is the critical temperature. The following data are available for the saturation pressure of air as a function of temperature: Taat (K) psat (MPa) Tsat (K) paat (MPa) 76.5 80.4 83.3 85.7 88.1 90.2 92.0 93.6 95.1 96.5 98.9 0.08 0.12 0.16 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.60 101 103 106.3 109.2 111.8 114.2 116.4 118.4 122.9 127.0 0.70 0.80 1.00 1.20 1.40 1.60 1.80 2.00 2.50 3.00 Obtain the values of ci, C2 and c from a least squares fit of the data. Analyze your regression fit by the computation of the standard error of the estimate, the coefficient of determination and the correlation coefficient.

Need ASAP

it is for a course that uses MATlab to solve for missing variables and such.

Answer 1

Least Square method is regression method which is used to fit data (calculate coefficients of polynomial to be fitted)

It is done by finding A and B matrices. Summation is taken here is over given number of data points,

Procedure is given below

For simplification some substitution are done.

Given eq in (96%) = 4,4+** Shu (test) where that and Pc = 3.77 MPa, To= 182.5k Let Z = ln en lehet egy Y = ln (Teat) z = Gatan²+ GzY Using hear regression. We can find a , C2. & C3. Take summation on both sides Şzi = 9EX + cz 3x: + ez EY: (2) (1) by 2 Multiply na = qa²+ Gn² + BAY ΣχZ = GIR + C₂ {X² + Eat (i) 22 Multiply (1) by x²z cx + x + Cz x²Y Exz- , {a² ta ex & Co Ex²7 (in)

C Σα + C2 Σ'+ G, ΣΥΞ ΣΖ ΕΙΣ? + C, Σ2 3 4. C3 ΣυY = ΣκZ 6, Ση - , Σ+ Σκάγο Σηθ 2. Σ Ση2 EX Ο, Σ? Σχ ΣκY Enz ζω και Ση 3 Ση Enz E2²7 A [phi phi - inv(A) XB

Standard enor of estimate Se Jo  [ Y: - ce+b+c)] 2 given value fitted value Co-relation coefficient, 8= n-1 (4:- =) (9:-5) 5 = mean of x values on sy ỹ - mean of y values (fitted values) standard deviation coefficient of determination R R = 82

MATLAB CODE FOR regression fit

leastSqRegression.m

clc;
%MALAB CODE for finding least square regression
% data given Tsat and Psat
T=[76.5 80.4 83.3 85.7 88.1 90.2 92 93.6 95.1 96.5 98.9 101 ...
103 106.3 109.2 111.8 114.2 116.4 118.4 122.9 127];
P=[0.08 0.12 0.16 0.2 0.25 0.3 0.35 0.4 0.45 0.5 0.6 ...
0.7 0.8 1 1.2 1.4 1.6 1.8 2 2.5 3];
% Constant values Critical Pressure and Critical Temperature
Pc=3.77;
Tc=132.5;
%define x
x=(1/Tc)-(1./T);
%define Y as ln(Tsat/Tc)
Y=log(T/Tc);
%define Z as ln(Psat/Pc)
Z=log(P/Pc);

% Coefficient Matrix A
A=[sum(x) sum(x.^2) sum(Y);sum(x.^2) sum(x.^3) sum(x.*Y);sum(x.^3) sum(x.^4) sum(x.^2.*Y) ];
B=[sum(Z);sum(x.*Z);sum(x.^2.*Z) ];
phi=A\B;
c1=phi(1);
fprintf('Value of C1 is = %f ',c1);
c2=phi(2);
fprintf('\nValue of C2 is = %f ',c2);
c3=phi(3);
fprintf('\nValue of C3 is = %f ',c3);

% Standard error of estimation
x_1=(1/Tc)-(1./T);
Z_new=c1*x_1+c2*x_1.^2+c3*log(T/Tc);
P_Cal=exp(Z_new)*Pc;
n=length(T);
Se=sqrt(sum((P-P_Cal).^2)/(n-2));
fprintf('\nStandard error of estimate = %f ',Se);

%Corelation coefficient
A=sum((T-mean(T)).*(P_Cal-mean(P_Cal)))/(n-1);
r=A/(std(T)*std(P_Cal));
fprintf('\nCorelation Coeficient = %f ',r);

% Coefficient of Determination
Coe_det=r^2;
fprintf('\nCoeficient of Determination = %f ',Coe_det);
plot(T,P,'dr'); hold on

T_finer=76.5:0.1:127;
x_1=(1/Tc)-(1./T_finer);
Z_new=c1*x_1+c2*x_1.^2+c3*log(T_finer/Tc);
P_finer=exp(Z_new)*Pc;
plot(T_finer,P_finer,'-b','LineWidth',1.5);
legend('Data Points','Regression Fit');
xlabel('Tsat (K)');
ylabel('Psat (MPa)');
title('Psat vs Tsat');

Psat vs Tsat 3.5 Data Points Regression Fit 3 2.5 1 0.5 - 0 70 80 90 100 110 120 130 Tsat (K)

MATLAB R2015a HOME PLOTS APPS EDITOR PUBLISH VIEW EDIT RUN 38 II - Find Files Insert fx Fli Run Section Compare Go To Comment %%% New Open Save Breakpoints Run Run and Advance Run and Print Find Indent 目 Advance Time FILE NAVIGATE BREAKPOINTS 0: Program Files ▸ MATLAB MATLAB Production Server R2015a bin Current Folder Editor - C:\Program Files\MATLAB\MATLAB Production Server 2015a\bin\leastSqRegression.m Name run Mass Spring.m xfunMass SpringProblem.mx leastSqRegression.mx + worker.bat A=sum (1-mean (1) TF_car-mean (F_Lal)))/n-1); 39 volumeOfBox.m r=A/ (3td (T)*3td (P_Cal)); voltagePlot.m 40 fprintf("\nCorelation Coeficient = $f',r); viscosityInterpolation Proble... 41 vibration Problem.m 42 Coefficient of Determination Vibrational_Response.xlsx 43 Coe_det=r^2; Vibration.m 44 fprintf("\nCoeficient of Determination = $f ', Coe_det); VelTrevor.m 45 plot (T, P, 'dr'); hold on velocity.jpg 46 untitled1.jpg 47 - T_finer=76.5:0.1:127; Underdamped.m x_1=(1/Tc)-(1./T_finer); Undamped.m 49 - Z_new=c1*x_1+c2*x_1.^2+c3*log (T_finer/Tc); umeanCal.m 50 P_finer=exp (2_new) *Pc; tutSubplot.m 51 plot (T_finer, P_finer,'-b', 'LineWidth', 1.5); trapelntegral.m 52 - legend ("Data Points', 'Regression Fit'); toughnessProblem.m 53 xlabel('Tsat (K)'); test.dat 54 - Details ylabel('Pat (MPa)'); 55 - title('Psat vs Tsat'); Workspace 56 Name Value Command Window [953.5950;6.3061e+10;-13 Value of ci is = 1728.519839 A 11.4468 Value of c2 is = 45245.015991 ans 0.9074 Value of C3 is = -7.855230 B [-38.8106;0.1310;-5.0761€ Standard error of estimate = 0.002044 c1 1.7285e+03 Corelation Coeficient = 0.952 600 c2 4.5245e+04 fx Coeficient of Determination = 0.907446 >> a im HH 70552 TI