MATLAB: How to run script iteratively

iterationMATLAB

So I just finished a code for a class of mine that works fine. However, I need to be able to change the value of just one variable.

I know I could make that variable equal 2:3:200, but I already have a variable designed in such a way. And if I were to define the variable that I want to change in such a way it wouldn't work because they contain a different number of values. Any Ideas? This is my script:

         clc; clear; close all;
          format long 
          Ma=.84
          Ra=.287
          Rg=.287
          siga=1.4
          sigg=1.3333
          Pa=54.05
          Ta=225.7
          OPR=4
          Va=Ma*sqrt(siga*Ra*Ta)
          Cpoa= (siga*Ra)./(siga - 1)
          TINT = 1100:100:1800
          PRC = sqrt(OPR)
          % Diffuser
          efd=.93
          To1 = Ta*(1+((siga-1)/2)*(Ma^2))
          Toa=To1
          Po1 = Pa*(1 + efd*((siga - 1)/2)*(Ma^2))^(siga/siga-1)
          Poa=Pa*(Toa/Ta)^(siga/siga-1)
          %Compressors  
          efc=.87
          %Low Pressure
          Po2=Po1*(PRC)
          To2 = To1 + ((To1*((PRC.^((siga-1)/siga)-1)))/efc)
          WC_LP = Cpoa*(To1-To2)
          %High Pressure
          To3 = To2 + ((To2.*((PRC.^((siga-1)/siga)-1)))/efc)
          Po3=Po2.*(PRC)
          WC_HP = Cpoa*(To2-To3)
          %Combustion
          efb =.98
          deltPob = .4 
          Po4= (1-deltPob)*Po3
          To4=TINT
          %Combustion Calculations for f
          MC=14.4
          MH=24.9
          MO=0
          Ycc=MC + (MH/4)- (MO/2)
          Tp=To4
          Tr=To3
          To= 298
          HRP_to = -8561991.6
          if Tp <= 1600
          hco2p = 56835 + 66.27*Tp + -11634.0*log(Tp)
          hh2op = 88923 + 49.36*Tp + -7940.8*log(Tp)
          hn2p= 31317 + 37.46*Tp + -4559.3*log(Tp)
          ho2p = 43388 + 42.27*Tp + -6635.4*log(Tp)
          else
          hco2p = 93048 + 68.58*Tp + -16979.0*log(Tp)
          hh2op = 154670 + 60.43*Tp + -19212.0*log(Tp)
          hn2p = 44639 + 39.32*Tp + -6753.4*log(Tp)
          ho2p =127010 + 46.25*Tp + -18798*log(Tp)   
          end
          if Tr <= 1600
          hco2r = 56835 + 66.27*Tr + -11634.0*log(Tr)
          hh2or = 88923 + 49.36*Tr + -7940.8*log(Tr)
          hn2r= 31317 + 37.46*Tr + -4559.3*log(Tr)
          ho2r = 43388 + 42.27*Tr + -6635.4*log(Tr)
          else
          hco2r = 93048 + 68.58*Tr + -16979.0*log(Tr)
          hh2or = 154670 + 60.43*Tr + -19212.0*log(Tr)
          hn2r = 44639 + 39.32*Tr + -6753.4*log(Tr)
          ho2r =127010 + 46.25*Tr + -18798*log(Tr)
          end
          hco2 = hco2p-hco2r
          hh2o = hh2op-hh2or
          hn2 = hn2p-hn2r
          ho2 =ho2p-ho2r
          h2oa=30.5
          h2ob=.0103
          co2a=28.8
          co2b=.028
          o2a=27
          o2b=.0079
          fuela= 38.1
          fuelb=.656
          Delt_HRP = (MH/2)*(h2oa*Tr +.5*h2ob*Tr.^2) - (h2oa*To +.5*h2ob*To.^2)+MC*(co2a*Tr +.5*co2b*Tr.^2) - (co2a*To +.5*co2b*To.^2)- Ycc*(o2a*Tr +.5*o2b*Tr.^2) - (o2a*To +.5*o2b*To.^2)-(fuela*Tr +.5*fuelb*Tr.^2) - (fuela*To +.5*fuelb*To.^2)
          HRP =HRP_to + Delt_HRP
          y = (-HRP - MC*hco2 - (MH/2).*hh2o + Ycc.*ho2)./(3.76.*hn2+ho2)
          fideal=197.7./(4.76*y*28.97)
          f=fideal*efb
          %Turbines | HIgh and Low have equal efficiencies 
          eft=.90
          efm=.99
          Cpog = (sigg*Rg)/(sigg-1)
          % High Pressure
          To5= To4 + (WC_HP./(efm*(1+f)*Cpog))
          Po5 = Po4*(1-((1-(To5/To4))/eft))^(sigg/(sigg-1))
          % Low Pressure
          To6= To5 + (WC_LP./(efm*(1+f)*Cpog))
          Po6 = Po5*(1-((1-(To6/To5))/eft))^(sigg/(sigg-1))
          %Nozzle
          %Choke Test
          efn = .95
          P_Po6 = (1-((1/efn)*(1-(2/(sigg+1)))))^(sigg/(sigg-1))
          R=Pa./Po6
          Me=1
          Pe=Po6*(P_Po6)
          Te=To6*(2/(sigg+1))
          roe= Pe./(Rg*Te)
          Ve= Me*sqrt(sigg*Rg*1000*Te)
          % Fs and T.S.F.C Calculations
          Fs = ((1+f).*Ve-Va) + (Pe-Pa).*((1+f)./(roe.*Ve))
          tsfc= (f./Fs).*3600

OPR is the variable I'm interested in

Best Answer

 clc; clear; close all;
          format long 
          Ma=.84
          Ra=.287
          Rg=.287
          siga=1.4
          sigg=1.3333
          Pa=54.05
          Ta=225.7
          OPR=2:3:200
          tsfc=cell(1,67) ;      %PRE-ALLOCATION
          for i = 1:numel(OPR)
          Va=Ma*sqrt(siga*Ra*Ta)
          Cpoa= (siga*Ra)./(siga - 1)
          TINT = 1100:100:1800
          PRC = sqrt(OPR(i))
          % Diffuser
          efd=.93
          To1 = Ta*(1+((siga-1)/2)*(Ma^2))
          Toa=To1
          Po1 = Pa*(1 + efd*((siga - 1)/2)*(Ma^2))^(siga/siga-1)
          Poa=Pa*(Toa/Ta)^(siga/siga-1)
          %Compressors  
          efc=.87
          %Low Pressure
          Po2=Po1.*(PRC)
          To2 = To1 + ((To1*((PRC.^((siga-1)/siga)-1)))/efc)
          WC_LP = Cpoa*(To1-To2)
          %High Pressure
          To3 = To2 + ((To2.*((PRC.^((siga-1)/siga)-1)))/efc)
          Po3=Po2.*(PRC)
          WC_HP = Cpoa*(To2-To3)
          %Combustion
          efb =.98
          deltPob = .4 
          Po4= (1-deltPob)*Po3
          To4=TINT
          %Combustion Calculations for f
          MC=14.4
          MH=24.9
          MO=0
          Ycc=MC + (MH/4)- (MO/2)
          Tp=To4
          Tr=To3
          To= 298
          HRP_to = -8561991.6
          if Tp <= 1600
          hco2p = 56835 + 66.27*Tp + -11634.0*log(Tp)
          hh2op = 88923 + 49.36*Tp + -7940.8*log(Tp)
          hn2p= 31317 + 37.46*Tp + -4559.3*log(Tp)
          ho2p = 43388 + 42.27*Tp + -6635.4*log(Tp)
          else
          hco2p = 93048 + 68.58*Tp + -16979.0*log(Tp)
          hh2op = 154670 + 60.43*Tp + -19212.0*log(Tp)
          hn2p = 44639 + 39.32*Tp + -6753.4*log(Tp)
          ho2p =127010 + 46.25*Tp + -18798*log(Tp)   
          end
          if Tr <= 1600
          hco2r = 56835 + 66.27*Tr + -11634.0*log(Tr)
          hh2or = 88923 + 49.36*Tr + -7940.8*log(Tr)
          hn2r= 31317 + 37.46*Tr + -4559.3*log(Tr)
          ho2r = 43388 + 42.27*Tr + -6635.4*log(Tr)
          else
          hco2r = 93048 + 68.58*Tr + -16979.0*log(Tr)
          hh2or = 154670 + 60.43*Tr + -19212.0*log(Tr)
          hn2r = 44639 + 39.32*Tr + -6753.4*log(Tr)
          ho2r =127010 + 46.25*Tr + -18798*log(Tr)
          end
          hco2 = hco2p-hco2r
          hh2o = hh2op-hh2or
          hn2 = hn2p-hn2r
          ho2 =ho2p-ho2r
          h2oa=30.5
          h2ob=.0103
          co2a=28.8
          co2b=.028
          o2a=27
          o2b=.0079
          fuela= 38.1
          fuelb=.656
          Delt_HRP = (MH/2)*(h2oa*Tr +.5*h2ob*Tr.^2) - (h2oa*To +.5*h2ob*To.^2)+MC*(co2a*Tr +.5*co2b*Tr.^2) - (co2a*To +.5*co2b*To.^2)- Ycc*(o2a*Tr +.5*o2b*Tr.^2) - (o2a*To +.5*o2b*To.^2)-(fuela*Tr +.5*fuelb*Tr.^2) - (fuela*To +.5*fuelb*To.^2)
          HRP =HRP_to + Delt_HRP
          y = (-HRP - MC*hco2 - (MH/2).*hh2o + Ycc.*ho2)./(3.76.*hn2+ho2)
          fideal=197.7./(4.76*y*28.97)
          f=fideal*efb
          %Turbines | HIgh and Low have equal efficiencies 
          eft=.90
          efm=.99
          Cpog = (sigg*Rg)/(sigg-1)
          % High Pressure
          To5= To4 + (WC_HP./(efm*(1+f)*Cpog))
          Po5 = Po4*(1-((1-(To5/To4))/eft))^(sigg/(sigg-1))
          % Low Pressure
          To6= To5 + (WC_LP./(efm*(1+f)*Cpog))
          Po6 = Po5*(1-((1-(To6/To5))/eft))^(sigg/(sigg-1))
          %Nozzle
          %Choke Test
          efn = .95
          P_Po6 = (1-((1/efn)*(1-(2/(sigg+1)))))^(sigg/(sigg-1))
          R=Pa./Po6
          Me=1
          Pe=Po6*(P_Po6)
          Te=To6*(2/(sigg+1))
          roe= Pe./(Rg*Te)
          Ve= Me*sqrt(sigg*Rg*1000*Te)
          % Fs and T.S.F.C Calculations
          Fs = ((1+f).*Ve-Va) + (Pe-Pa).*((1+f)./(roe.*Ve))
          tsfc{i}= (f./Fs).*3600;
          end
          celldisp(tsfc)        %each subcell represents each value of OPR
tsfc = vertcat(tsfc{:})    %double array

Related Solutions

MATLAB: Can’t figure out how to finalize the code

%V1=volume of first reactor in L
%V2=volume of second reactor in L
%V3=volume of third reactor in L
%V4=volume of forth reactor in L
 
%k1=first order reaction rate in units of (L^3)(M^-1)(T^-1)
%k2=second order reaction rate in units of (L^3)(M^-1)(T^-1)
%Each reactor has different reaction rate
 
%M=mass in units of M
%Qin=inflow in units of (L^3)(T^-1)
%Qout=outflow in units of (L^3)(T^-1)
%Qxy=the flow from reactor x to reactor y (x&y can be 1,2,3..)
%UNITS
%The unit for V1, V2, V3 & V4 is L
%The unit for k1 is hr^-1
%The unit for k2 is L/(mg*hr)
%The unit for M is mg
%The unit for Qin, Qout,Q12, Q23, Q34, Q32, Q43 & Qxy is L/hr
%Given and calculated values  
Qin=10;
Q12=10;
Q23=10+5; %Q23=Q12+Q32
Q34=15-5+3; %Q34=Q23-Q32+Q43
Qout=10;
Q32=5;
Q43=3;
 
Cain=1;
Cbin=0;
 
V1=25;
V2=75;
V3=100;
V4=25;
 
k11=0.05;
k12=0.1;
k13=0.5;
k14=0.1;
 
k21=0.04;
k22=0.08;
k23=0.4;
k24=0.08;
 
%% PARTA
 
%Mass Balance Eqns for Chemical Species A
 
%For reactor 1

%(Qin*Cain)-(Q12*Ca1)-(k11*Cain*V1)==0  %for first order reaction
%(Qin*Cain)-(Q12*Ca1)-(k21*(Cain)^2*V1)==0  %for second oreder reaction
 
%For reactor 2

%(Q12*Ca1)+(Q32*Ca3)-(Q23*Ca2)-(k12*Ca2*V2)==0 %for first order reaction
%(Q12*Ca1)+(Q32*Ca3)-(Q23*Ca2)-(k22*(Ca2)^2*(V2))==0 %for second oreder reaction
 
%For reactor 3

%(Q23*Ca2)+(Q43*Ca4)-(Q32*Ca3)-(Q34*Ca3)-(k13*Ca3*V3)==0  %for first order reaction
%(Q23*Ca2)+(Q43*Ca4)-(Q32*Ca3)-(Q34*Ca3)-(k23*(Ca3)^2*(V3))==0  %for second order reaction
 
%For reactor 4

%(Q34*Ca3)-(Q43*Ca4)-(Qout*Ca4)-(k14*Ca4*V4)==0  %for first order reaction
%(Q34*Ca,3)-(Q43*Ca,4)-(Qout*Ca,4)-(k24*(Ca4)^2*V4)==0  %for second order reaction
 
%Mass Balance Eqns for Chemical Species B
 
%For reactor 1
%(Qin*Cbin)-(Q12*Cb1)+(k11*Ca1*V1)==0  %for first order reaction
%(Qin*Cbin)-(Q12*Cb1)+(k21*(Ca1)^2*V1)==0  %for second oreder reaction
 
%For reactor 2
%(Q12*Cb1)+(Q32*Cb3)-(Q23*Cb2)+(k12*Ca2*V2)==0 %for first order reaction
%(Q12*Cb1)+(Q32*Cb3)-(Q23*Cb2)+(k22*(Ca2)^2*(V2))==0 %for second oreder reaction
 
%For reactor 3
%(Q23*Cb2)+(Q43*Cb4)-(Q32*Cb3)-(Q34*Cb3)+(k13*Ca3*V3)==0  %for first order reaction
%(Q23*Cb2)+(Q43*Cb4)-(Q32*Cb3)-(Q34*Cb3)+(k23*(Ca3)^2*(V3))==0  %for second order reaction
 
%For reactor 4
%(Q34*Cb3)-(Q43*Cb4)-(Qout*Cb4)+(k14*Ca4*V4)==0  %for first order reaction
%(Q34*Cb3)-(Q43*Cb4)-(Qout*Cb4)+(k24*(Ca4)^2*V4)==0  %for second order reaction
 
syms Ca1 Ca2 Ca3 Ca4 Cb1 Cb2 Cb3 Cb4
Eqns_1st_Order=[(Qin*Cain)-(Q12*Ca1)-(k11*Cain*V1)%=0















(Q12*Ca1)+(Q32*Ca3)-(Q23*Ca2)-(k12*Ca2*V2)%=0
(Q23*Ca2)+(Q43*Ca4)-(Q32*Ca3)-(Q34*Ca3)-(k13*Ca3*V3)%=0
(Q34*Ca3)-(Q43*Ca4)-(Qout*Ca4)-(k14*Ca4*V4)%=0
(Qin*Cbin)-(Q12*Cb1)+(k11*Cain*V1)%=0
(Q12*Cb1)+(Q32*Cb3)-(Q23*Cb2)+(k12*Ca2*V2)%=0
(Q23*Cb2)+(Q43*Cb4)-(Q32*Cb3)-(Q34*Cb3)+(k13*Ca3*V3)%=0
(Q34*Cb3)-(Q43*Cb4)-(Qout*Cb4)+(k14*Ca4*V4)]%=0
Eqns_1st_Order = 
 
Eqns_2nd_Order=[(Qin*Cain)-(Q12*Ca1)-(k21*(Cain)^2*V1)%=0
(Q12*Ca1)+(Q32*Ca3)-(Q23*Ca2)-(k22*(Ca2)^2*(V2))%=0
(Q23*Ca2)+(Q43*Ca4)-(Q32*Ca3)-(Q34*Ca3)-(k23*(Ca3)^2*(V3))%=0
(Q34*Ca3)-(Q43*Ca4)-(Qout*Ca4)-(k24*(Ca4)^2*V4)%=0
(Qin*Cbin)-(Q12*Cb1)+(k21*(Cain)^2*V1)%=0
(Q12*Cb1)+(Q32*Cb3)-(Q23*Cb2)+(k22*(Ca2)^2*(V2))%=0
(Q23*Cb2)+(Q43*Cb4)-(Q32*Cb3)-(Q34*Cb3)+(k23*(Ca3)^2*(V3))%=0
(Q34*Cb3)-(Q43*Cb4)-(Qout*Cb4)+(k24*(Ca4)^2*V4)]%=0
Eqns_2nd_Order = 
sol = solve(Eqns_2nd_Order, [Ca1 Ca2 Ca3 Ca4 Cb1 Cb2 Cb3 Cb4])
sol = struct with fields:
    Ca1: [8×1 sym]
    Ca2: [8×1 sym]
    Ca3: [8×1 sym]
    Ca4: [8×1 sym]
    Cb1: [8×1 sym]
    Cb2: [8×1 sym]
    Cb3: [8×1 sym]
    Cb4: [8×1 sym]
sol.Ca1
ans = 
sol.Ca2
ans = 
vpa(sol.Ca2, 3)
ans =

MATLAB: Loop to create matrix with 2 inputs and 2 outputs

Colby - you almost have it. Your code for the two for loops is as follows

     for TINT = 1000:1700
         TINT = TINT + 100;
         for OPR = 2:28
             OPR = OPR + 2;

Since TINT and OPR are your indexing variables, you don't need to increment either through code. This will happen automatically so long as you set your step value correctly in the for loop. The above can be rewritten as

     for TINT = 1100:100:1800
         for OPR = 4:2:30
             % etc.
         end
     end

So on the first iteration of the outer for loop, TINT will be assigned a value on 1100, and subsequent iterations of this loop will assign values of 1200, 1300, 1400,...,1800 to TINT.

The inner for loop will initialize *OPR to 4 with subsequent iterations assigning it values of 6,8,10,...,30.

Best Answer

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MATLAB: Can’t figure out how to finalize the code

MATLAB: Loop to create matrix with 2 inputs and 2 outputs

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