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//Chapter 7 //Example 7_8 //Page 153 clear;clc; mw=5; kv=33; pf=0.8; cost=4; id=0.1; p=1e-6; //cost = Rs (25000*a + 2500) ra=p*1e5; i=mw*1e6/sqrt(3)/kv/1000/pf; e=3*i^2*ra*8760/1000; ac=cost*e/100; cc=25000; vac=id*cc; a=sqrt(ac/vac); printf("Resistance of one conductor = %.3f/a ohm \n\n", ra); printf("Line current = %.2f A \n\n", i); printf("Energy lost per annum = %.1f/a kWh \n\n", e); printf("Annual cost of energy lost = Rs. %d/a \n\n", ac); printf("Capital cost is given to be Rs.20*a per metre. Threfore for 1km cable = Rs. %d*a \n\n", cc); printf("Variable annual charge = Rs. %d*a \n\n", vac); printf("Area of cross section = %.2f cm^2 \n\n", a);
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//laplace// syms t s; y=laplace('5+6*t^2+3*%e^(-2*t)',t,s); disp(y,"ans=")
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/////////////////////////////////////////////////////////////////////// /////////////// IF STATEMENTS /////////////// /////////////////////////////////////////////////////////////////////// let condition = yes // all examples print yes if condition, print(yes) else print(no) if condition, print("Block of code") print(yes) else print("Block of code") print(no) end if condition, print("Block of code") print(yes) else print(no) if condition, print(yes) else print("Block of code") print(no) end
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// Ejemplo de una ecuaci�n diferencial parcial en 1D // Autor: Antonio Carrillo Ledesma // -Uxx-k^2U=0 // 0<=U<=1 // U(0)=1 y Ux(1)=ikU(1) TEST = 1; // (0) Diferencias finitas, (1) Diferencias finitas exactas segun Yau Shu Wong y Guangrui Li function y=LadoDerecho(x) y=0.0; endfunction function y=SolucionAnalitica(x, k) //y=cos(k*x)+%i*sin(k*x); y=exp(%i*k*x); endfunction K = 150; KK = K*K; a=0; // Inicio dominio c=1; // Fin dominio M=300; // Partici�n N=M-1; // Nodos interiores h=(c-a)/(M-1); // Incremento en la malla Y0=1; // Condici�n Dirchlet inicial en el inicio del dominio Y1=%i*K; // Condici�n Neumann inicial en el fin del dominio A=zeros(N,N); // Matriz A b=zeros(N); // Vector b if TEST = 0 then R=-1/(h^2); P=2/(h^2)-KK; Q=-1/(h^2); else R=-1/(h^2); P=(2*cos(K*h)+(K*h)^2)/(h^2) - KK; Q=-1/(h^2); end // Primer renglon de la matriz A y vector b A(1,1)=P; A(1,2)=Q; b(1)=LadoDerecho(a)-Y0*R; // Frontera dirichlet // Renglones intermedios de la matriz A y vector b for i=2:N-1 A(i,i-1)=R; A(i,i)=P; A(i,i+1)=Q; b(i)=LadoDerecho(a+h*(i-1)); end // Relglon final de la matriz A y vector b if TEST = 0 then A(N,N-1)=1/(h^2); A(N,N)=-1/(h^2)+ Y1/h; b(N)=LadoDerecho(c)/2; else A(N,N-1)=1/(h^2); A(N,N)=-1/(h^2)+ %i*sin(K*h)/(h^2); b(N)=LadoDerecho(c)/2; end // Resuleve el sistema lineal Ax=b x=inv(A)*b; ESC = 5; xxx=zeros(M*ESC,1); zzz=zeros(M*ESC,1); for i=1:M*ESC xxx(i)=a+h/ESC*(i-1); zzz(i)=SolucionAnalitica(xxx(i),K); end // Prepara la graficaci�n xx=zeros(M,1); zz=zeros(M,1); for i=1:M xx(i)=a+h*(i-1); zz(i)=SolucionAnalitica(xx(i),K); end yy=zeros(M,1); yy(1)=Y0; // Condici�n inicial for i=1:N yy(i+1)=x(i); end // Grafica la soluci�n de la Ecuaci�n Diferencial Parcial en 1D plot2d(xx,yy,15) plot2d(xxx,zzz)
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10000 ~~~~~~~~~~~~~~~~~~~~~~~~~~ 104715764
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// ex.6.3 // solvent enters to the tank with flow of 0.1 L/s // A + B -> C, k // accu = in - out + prod function dy = model(t, y) // instantaneous values: V = y(1) nA = y(2) nB = y(3) nC = y(4) // instant concentrations A = nA/V B = nB/V C = nC/V //rates: dVdt = 0.1 // 0.1 L/s dnAdt = -k*A*B*V dnBdt = -k*A*B*V dnCdt = k*A*B*V dy=[dVdt, dnAdt, dnBdt, dnCdt] endfunction // initial conditions V0 = 10 //L nA0 = 2.0 // mol nB0 = 1.5 // mol nC0 = 0 // mol k = 1e-2 y0 = [V0; nA0; nB0; nC0] t0 = 0 t = linspace(0,3600) // time in seconds y = ode(y0, t0, t, model) //V(t) total profile V=y(1,:) nA=y(2,:) nB=y(3,:) nC=y(4,:) A = nA ./ V // ./ ==> element-by-element division B = nB ./ V // ./ ==> element-by-element division C = nC ./ V // ./ ==> element-by-element division clf plot(t, nA, '-or') plot(t, nB, '-ob') plot(t, nC, '-og') legend(['A']) xlabel('Time, s') ylabel('Concentration, mol/L')
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//EXAMPLE 9-38 PG NO=624-625 C=1/9; //CAPACITOR X=2; //R/L=X Y=6-X; //G/C G=4*C; disp('i) G (G) = '+string (G)+' ohm') L=0.9; R=1.8;
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clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 3.1 Page 104 \n') //Example 3.1 // Find Skin Temperature & Aerogel Insulation Thickness A=1.8; // [m^2] Area for Heat transfer i.e. both surfaces Ti = 35+273; //[K] - Inside Surface Temperature of Body Tsurr = 10+273; //[K] - Temperature of surrounding Tf = 283; //[K] - Temperature of Fluid Flow e=.95; // Emissivity of Surface Lst=.003; //[m] - Thickness of Skin kst=.3; // [W/m.K] Effective Thermal Conductivity of Body kins = .014; // [W/m.K] Effective Thermal Conductivity of Aerogel Insulation hr = 5.9; //[W/m^2.k] - Natural Thermal Convectivity from body to air stfncnstt=5.67*10^(-8); // [W/m^2.K^4] - Stefan Boltzmann Constant q = 100; //[W] Given Heat rate //Using Conducion Basic Eq 3.19 Rtot = (Ti-Tsurr)/q; //Also //Rtot=Lst/(kst*A) + Lins/(kins*A)+(h*A + hr*A)^-1 //Rtot = 1/A*(Lst/kst + Lins/kins +(1/(h+hr))) //Thus //For Air, h=2; //[W/m^2.k] - Natural Thermal Convectivity from body to air Lins1 = kins * (A*Rtot - Lst/kst - 1/(h+hr)); //For Water, h=200; //[W/m^2.k] - Natural Thermal Convectivity from body to air Lins2 = kins * (A*Rtot - Lst/kst - 1/(h+hr)); Tsa=305; //[K] Body Temperature Assumed //Temperature of Skin is same in both cases as Heat Rate is same //q=(kst*A*(Ti-Ts))/Lst Ts = Ti - q*Lst/(kst*A); //Also from eqn of effective resistance Rtot F printf("\n\n (I) In presence of Air, Insulation Thickness = %.1f mm",Lins1*1000) printf("\n (II) In presence of Water, Insulation Thickness = %.1f mm",Lins2*1000); printf("\n\n Temperature of Skin = %.2f degC",Ts-273); //END
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function t = expo(l,n) u = rand(n, 1); t = u t = -log(1-t)/l endfunction histplot(20,expo(1,100000)) a=0:0.01:12; lambda=1; b=lambda*exp(-lambda*a); plot2d2(a,b,style=1) legend("Simulation de la loi exponentielle","Densité de la loi exponentielle")
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//Simple if-else-end flag=input ("Enter the value of flag:"); if flag then disp("True"); else disp("False"); end //Ladder if-elseif-else-end example a=input ("Enter marks:") if a >=75 then disp("Honors"); elseif a>=60, //then replaced by a comma disp ("First division"); elseif a>=45 // then replaced by a space disp ("Second division"); elseif a>=35; // then replaced by a semicolon disp ("Third division"); else disp("Fail"); end
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//CHAPTER 8- DIRECT CURRENT MACHINES //Example 38 clc; disp("CHAPTER 8"); disp("EXAMPLE 38"); //VARIABLE INITIALIZATION v_t=215; //in Volts r_a=0.4; //in Ohms p=5*1000; //in Watts N_g=1000; //speed as generator in rpm ratio=1.1; //according to the solution, Φ_b:Φ_a=1.1 //SOLUTION //As generator I_ag=p/v_t; E_a=v_t+(I_ag*r_a); //As motor I_am=p/v_t; E_b=v_t-(I_am*r_a); N_m=(1/ratio)*N_g*(E_b/E_a); N_m=round(N_m); //to round off the value disp(sprintf("The speed of the machine as motor is %d rpm",N_m)); //END
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//signals and systems //time domain analysis of discreet time systems //iterative solution clear; close; clc; n=(-2:10)'; y=[1;2;zeros(length(n)-2,1)]; x=[0;0;n(3:length(n))]; for k=1:length(n)-2 y(k+2)=y(k+1)-0.24*y(k)+x(k+2)-2*x(k+1); end; clf; plot2d3(n,y); disp([msprintf([n,y])]);
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clc a=40*%pi/180; //Mach angle in radians y=1.4; R=287; //J/kg K T=288; //K C=sqrt(y*R*T); V=C/sin(a); disp("Velocity of bullet =") disp(V) disp("m/s")
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clear //Given Ev=200 //V I0=0.9 //A f=50 //Hz //Calculation // E0=sqrt(2)*Ev Xl=E0/I0 L=Xl/(2*%pi*f) //Result printf("\n The value of inductance is %0.0f H",L)
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clc; clear all; disp("Laminar flow over plate") L=5;//m plate length w=2.5;//m plate width x=1.2;//m distance from the leading edge of plate v=15.4*10^(-6);//m^2/s kinematic viscocity U=4;//m/s velocity of air rho=1.208;//kg/m^3 density of air v=1.47*10^(-5);//m^2/s kinematic viscosity of air Re=5*10^5;// Reynold's number x=Re*v/U;// length of plste over which boundary layer is laminar disp("m",x,"length of plste over which boundary layer is laminar =") delta=5*x*1000/(Re)^0.5;//mm disp("mm",delta,"thickness of boundary layer =") Cfx=0.664/(Re)^0.5; disp(Cfx,"drag coefficient =") tau=Cfx*0.5*rho*U^2;// shear stress disp("N/m^2",tau,"Shear stress =") Cf=1.328/(Re)^0.5; A=x*w;//m^2 area of plate Fd=2*Cf*0.5*rho*A*U^2; disp("N",Fd,"Total drag force on both sides of plate, =")
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// Example 4_9 clc;funcprot(0); // Given data R=5;// The radius of a jar in cm n=33;// tThe turntable has been revolving at a steady speed in rpm g=9.807;// The acceleration due to gravity in m/s^2 // Calculation omega=(2*%pi*n)/60;// Acceleration h=(omega*R*10^-2)^2/(2*g);// The height h in m printf("\nThe height,h=%1.3e m",h);
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//Example 4_4 clc;funcprot(0); //Given data P=40000;//kW N=500;//r.p.m H=240;// m h=30;// m SG=1.1;// Specific gravity of water q=150;// litres/sec q=q*SG;// kg/sec n_m=0.88;// The over all efficiency //Calculation w=1000*9.81;// N p=(q*w*h*n_m)/(1000*1000);// kW //d_r=D/d; d_r=sqrt(P/p)*(h/H)^0.75; n=N*d_r*sqrt(h/H);// r.p.m n_s=(n*sqrt(p))/h^(5/4); N_s=(N*sqrt(P))/(H)^(5/4); printf('\n(a)The design speed for a turbie,n=%0.0f r.p.m',n); printf('\nThe runner is of Francis type'); // The answer provided in the textbook is wrong
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clear; clc; printf("\t Example 2.5.a\n"); //position 1 moles molefraction // air 80 0.8 // water 20 0.2 //position 2 moles molefraction // air 10 0.1 // water 90 0.9 ya1=0.8; ya2=0.1; T=(273+35); //temperature in kelvin pt=1*1.013*10^5; //total pressure in pascal z=0.3*10^-3; //gas film thickness in m Dab=.18*10^-4; //diffusion coefficient in m^2/s R=8314; //universal gas constant Na=Dab*pt*(ya1-ya2)/(z*R*T) //diffusion flux in kmol/m^2*s rate=Na*100*10^-4*3600*46; //since molecular weight of mixture is 46 printf("\n rate of diffusion of alcohol-water vapour :%f kg/hr ",rate); //end
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clc r1=2 //Assigning values to parameters r2=0.02 wi=412 pf=0.8 x=1 kva=50 e1=2300 e2=230 i2=kva*1000/e2 i1=kva*1000/e1 wcf=(i1*i1*r1)+(i2*i2*r2) n1=x*kva*pf*100/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001)) x=0.5 n2=x*kva*pf*100/((x*kva*pf)+(wi*0.001)+(x*x*wcf*0.001)) disp("Percent",n1,"Efficiency at full node 0.8pf is") disp("Percent",n2,"Efficiency at half full node 0.8pf is")
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clc //initialisation of variables k= 1.667 L= 0.5 //ft g= 32.2 //ft/sec^2 //CALCULATIONS Cd= (k/L)/(sqrt(2*g)*(2/3)) //RESULTS printf (' Coefficiant of discharge = %.3f ',Cd)
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clc //Initialization of variables T1=77.32 //K P=1 //atm T2=126 //K Pc=33.5 //atm //calculations dS=27/32 *1.987*P/Pc *(T2/T1)^3 //results printf("Change in entropy = %.2f eu/mol",dS)
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clc //initialisation of variables M= 25.01 //gms n= 1.0053 //moles n1= 6.6*10^-5 //moles e= 1.350*10^-3 //coloumbs //CALCULATIONS x= M/n y= n1*x nm= y*10^3+e*10^3-(x/10) t= nm/(e*10^3) //CALCULATIONS printf (' transference number = %.3f ',t)
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function [vec]=setparammmat3d(name, var, ni,nj,nk,nr,nc,elist) //AddMetadata(name, property, port) nargin=length(elist); if nargin>0 then server=elist(1); if nargin>1 then port=elist(2); if nargin>2 then id=elist(3); else id=0; end else port=8080; end else server='localhost'; port=8080; id=0; end try ind=1; for i1=1:ni for i2=1:nj for i3=1:nk for i=1:nr for j=1:nc tmat(ind)=var(i1,i2,i3,i,j); ind=ind+1; end end end end end mstring=vectostring(tmat, ni*nj*nk*nr*nc,','); //put double quotes around the vec string so that it is //passed into unix shell script as a single variable umstring=sprintf('""%s""',mstring); scommand=sprintf("iogs setparam mmat3d %s %s %d %d %d %d %d %d %d %s", name, umstring, ni,nj,nk,nr,nc, id,port,server); vec=unix_g(scommand); catch disp('Setmmat3dParam Error!'); vec=-1; end endfunction
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<?xml version="1.0" encoding="utf-8"?> <test> <description>StdProject1D Segment Modified basis P=6 Q=7</description> <executable>StdProject</executable> <parameters>-s segment -b Modified_A -o 6 -p 7</parameters> <metrics> <metric type="L2" id="1"> <value tolerance="1e-12">5.07435e-15</value> </metric> <metric type="Linf" id="2"> <value tolerance="1e-12">5.10703e-15</value> </metric> </metrics> </test>
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//example 12.2 clc; funcprot(0); Ap=%pi/4*1.75^2; q=135.6; w=0.83; FS=4; phi=36; Nq=0.21*exp(0.17*phi); Qp=Ap*q*(w*Nq-1); Qpall=Qp/FS; disp(Qpall,"allowed load in kN");
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function [y,y1]=convol(h,x,y0) // convol - convolution (overlap add method) //%CALLING SEQUENCE // [y,y1]=convol(h,x,y0) // [y]=convol(h,x) //%PARAMETERS // x,h :input sequences (h is a "short" sequence, x a "long" one) // y0 : old piece to overlap add // y : output of convolution // y1 : new piece to overlap add //%DESCRIPTION // calculates the convolution y= h*x of two discrete sequences by the overlap // add method (fft). //%EXAMPLE // For x=[x1,x2,...,] // First call : [y,y1]=convol(h,x1) // Subsequent calls : [y,y1]=convol(h,xk,y1). //! [lhs,rhs]=argn(0) n=prod(size(x)) m=prod(size(h)) m1=n+m-1 x(m1)=0;h(m1)=0 if norm(imag(x))==0&norm(imag(h))==0 then y=real(fft(fft(matrix(x,1,m1),-1).*fft(matrix(h,1,m1),-1),1)) else y=fft(fft(matrix(x,1,m1),-1).*fft(matrix(h,1,m1),-1),1) end if lhs+rhs=5 then, y0(n)=0;//update carried from left to right y1=y(n+1:n+m-1) y=y(1:n)+y0 elseif lhs+rhs=4 then if rhs=2 then y1=y(n+1:n+m-1) y=y(1:n) //initial update else y0(n+m-1)=0 //final update y=y(1:n+m-1)+y0 end, else y=y(1:n+m-1) //no update end
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clc;funcprot(0); //Example 5.7 //Initializing the variables Vj = 5*10^6; //Velocity of Jet Mr = 150000; // Mass of Rocket Mf0 = 300000; // Mass of initial fuel Vr = 3000; //Velocity of jet relative to rocket g = 9.81; // Acceleration due to gravity //Calculations m = Vj/Vr; //Rate of fuel consumption T = Mf0/m; // Burning time function[DVt]=f(t) DVt = m*Vr /(Mr + Mf0 - m*t) - g; endfunction function[V]=h(t) V = -g*t - Vr*log(1 - t/269.95); endfunction Vt = intg(0, 180 ,f); Z1=intg(0,180,h); Z2 = Vt^2/(2*g); disp(T, "(a)Burning time (s): "); disp(Vt,"(b)Speed of rocket when all fuel is burned (m/s):"); disp((Z2+Z1)/1000,"(c)Maximum height reached (km):");
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clear; clc; //Example 17.4 Vx=-0.7; Vy=-0.7; iCxy=3.22;//(mA) iCR=0; i5=1.40; i1=1.40; Vor=-0.7; R4=1.500; Vnor=-1.4; V2=-5.2; R3=1.500; i3=(Vor-V2)/R3; printf('\ncurrent i3=%.2f mA\n',i3) i4=(Vnor-V2)/R4; printf('\ncurrent i4 =%.2fmA\',i4) P=(iCxy+iCR+i5+i1+i3+i4)*(0-V2); printf('\npower dissipation=%.2f mW\n',P)
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clc clear //input v1=7.5//initial voltag v2=11.5//final voltage ic=18*10^-6//collector current //calculation r=(v2-v1)/ic//output resistance //output printf("the output resistance is %2.2e ohm ",r)
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Example10_12.sce
clear ; clc; printf("\t Example 10.12\n"); T1=291; //temp.of sky,K T2=308; //temp of air,K e1=0.9; //emissivity 0f black paint h=8; //heat transfer coefficient,W/(m^2*K) P=600 ; //heat flux,W/m^2 //heat loss from the roof to the inside of the barn will lower the roof temp., since we dont have enough information to evaluate the loss, we can make an upper bound on roof temp. by assuming that no heat is transferred to the interior. x=poly([0],'x'); x=roots(8*(e1*5.67*10^-8*(x^4-T1^4)+(x-T2)-e1*P)); //for white acrylic paint, by using table, e=0.9 and absorptivity is 0.26,Troof T=poly([0],'T'); T=roots(8*(e1*5.67*10^-8*(T^4-T1^4)+(T-T2)-0.26*P)); Tn=T(2)+0.6 printf("\t temp. of the root is :%.1f C or 312 K ,the white painted roof is only a few degrees warmer than the air.\n",Tn); //end
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2021-07-10T08:39:13.955398
2017-01-19T01:24:51
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ADC_onchip.sce
function [Out,out_volt] = ADC(samples) //samples is actually number of shifts [path,fname,extension] = fileparts(file_name); hid_dir = path + '.' + fname; select board_num case 2 then brdtype = ''; case 3 then brdtype = '_30a'; case 4 then brdtype = '_30n'; case 5 then brdtype = '_30h'; else messagebox('Please select the FPAA board that you are using.', "No Selected FPAA Board", "error"); abort; end //exec("~/rasp30/prog_assembly/libs/scilab_code/characterization/char_miteADC.sce",-1); [a1,b1]=unix_g("sudo ~/rasp30/prog_assembly/libs/sh/asm2ihex.sh ./sftreg_miteADC ./Ramp_ADC_DAC.s43 16384 16384 16384") if b1==1 disp('There is an error in your assembly file') abort; end [a1,b1]=unix_g("sudo tclsh ~/rasp30/prog_assembly/libs/tcl/write_mem2_NoRelease.tcl -start_address 0x4300 -input_file_name "+hid_dir+"/input_vector"); [a2,b2]=unix_g("sudo tclsh ~/rasp30/prog_assembly/libs/tcl/write_mem2_NoRelease.tcl -start_address 0x4200 -input_file_name "+hid_dir+"/output_info"); err=1; while err==1 [y,err]=unix_g('sudo tclsh ~/rasp30/prog_assembly/libs/tcl/run.tcl '+path+'sftreg_miteADC.elf'); end //************************* unix_w("sudo tclsh ~/rasp30/prog_assembly/libs/tcl/read_mem2_NoRelease.tcl -start_address 0x5000 -length 1 -output_file_name "+path+"output_vector.txt"); y = mopen(path+'output_vector.txt','rt') Output=mgetl(y); New_output(1,1) = part(Output(1,1),3:7); Output_dec(1,:) = msscanf(New_output(1,:),'%x'); //scan with hexadecimal format length_out= Output_dec(1) - 24576; length_out=length_out/2; disp(length_out) unix_w("sudo tclsh ~/rasp30/prog_assembly/libs/tcl/read_mem2_NoRelease.tcl -start_address 0x6000 -length "+string(length_out)+" -output_file_name "+path+"output_vector.txt"); y = mopen(path+"output_vector.txt','rt') Output=mgetl(y); m=1 i=3 while m<length_out+1 New_output(m,1) = part(Output(1,1),i:i+3); m=m+1; i=i+7; end clear Output_dec m=1 while m<length_out+1 Output_dec(m,:) = msscanf(New_output(m,:),'%x'); //scan with hexadecimal format m=m+1; end m=1 Out=Output_dec; out_volt=Out endfunction
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EX12_2.sce
//Examle12.2 // Determine the following parameter of 8-bit A/D converter a) Normalized step size b) Actual step size c) Normalized maximum quantization level d) Actual maximum quantization e) Normalized peak quantization error f) Actual peak quantization error g) Percentage of quantization error clc; clear; close; N = 8 ; Vin = 12 ; //a) Normalized step size of A/D converter Ns = 2^-N ; disp('Normalized step size of A/D converter is = '+string(Ns)+ ' '); // b) Actual step size of A/D converter As = Vin*Ns ; disp('Actual step size of A/D converter is = '+string(As)+ ' '); // c) Normalized maximum quantization level of A/D converter Qmax = 1-2^-N ; disp('Normalized maximum quantization level of A/D converter is = '+string(Qmax)+ ' '); // d) Actual maximum quantization level of A/D converter QAmax = Qmax*Vin ; disp('Actual maximum quantization level of A/D converter is = '+string(QAmax)+ ' '); // e) Normalized peak quantization error of A/D converter Qp = 2^-(N+1); disp('Normalized peak quantization error of A/D converter is = '+string(Qp)+ ' ' ); // f) Actual peak quantization error of A/D converter Qe = Qp*Vin ; disp('Actual peak quantization error of A/D converter is = '+string(Qe)+ ' V '); // g) Percentage of quantization error of A/D converter %Qp = 2^-(N+1)*100 ; disp('Percentage of quantization error of A/D converter is = '+string(%Qp)+ ' ') ;
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aproximacion_derivada.sce
// deff('y=g(x)', 'y=1/x') // Derivada: // d = numderivative(g, numero) // p = poly([coef...], 'x', 'c') // Derivada: // d = derivat(p)
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champ.sce
// Program to plot using champ function for plotting a vector field x = linspace(-1,1,10); y = linspace(-1,1,10); [X,Y] = meshgrid(x,y); fy = 3.*Y; fx = 0.5.*X; champ(x,y,fx',fy') xtitle('Using champ function to plot vector field') xlabel('x') ylabel('y')
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8_2.sce
clc; h1=3442.6; s1=7.066; s2=s1; sf2=0.391; sfg2=8.13; x2=(s2-sf2)/sfg2 hf2=112; hfg2=2438; h2=hf2+x2*hfg2; h3=112; W12_=h1-h2; Q=h1-h3; Ceff=(h1-h2)/(h1-h3); disp(Ceff,"cycle efficiency is:"); ssc=1/(h1-h2); disp("kg/kW h",ssc,"specific steam consumption is:"); disp("cycle efficiency has increased due to superheating and the improvement in specific steam consumption is even more marked:")
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akylKerimbekov/oracle_pl_sql_programming
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/* Compare performance of repeated querying of data to caching in the PGA (packaged collection) and the new Oracle 11g Result Cache. To compile and run this test script, you will first need to run the following script. Note that to compile the my_session package and display PGA usage statistics, you will need SELECT authority on: sys.v_$session sys.v_$sesstat sys.v_$statname Author: Steven Feuerstein */ @@plvtmr.pkg @@mysess.pkg @@11g_emplu.pkg @@11g_emplu_compare.sp SET SERVEROUTPUT ON BEGIN test_emplu (100000); /* With 100000 iterations: PGA before tests are run: session PGA: 2057168 Execute query each time Elapsed: 5.65 seconds. Factored: .00006 seconds. session PGA: 1139664 Oracle 11g result cache Elapsed: .3 seconds. Factored: 0 seconds. session PGA: 1139664 Cache table in PGA memory Elapsed: .12 seconds. Factored: 0 seconds. session PGA: 1336272 */ END; / /*====================================================================== | Supplement to the fifth edition of Oracle PL/SQL Programming by Steven | Feuerstein with Bill Pribyl, Copyright (c) 1997-2009 O'Reilly Media, Inc. | To submit corrections or find more code samples visit | http://oreilly.com/catalog/9780596514464/ */
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// Impedance profile of a single capacitor (simplified LRC model) // Output being a csv file that can be used to drive SPICE simulations stacksize(64*1024*1024); clear; //Clear user variables //////////////////////////////////////////////////SPECIFY////////////////////////////////////////////////////// C1 = 100e-9; //Series capacitance L1 = 1e-9; //Series inductance R1 = 0.01; //Series resistance NumOfCaps=2; //Number of parallel caps FreqRange=1e6:1e6:10e9; //Frequency range ////////////////////////////////////////////////////////////////////////////////////////////////////////////// C1=C1*NumOfCaps; L1=L1/NumOfCaps; R1=R1/NumOfCaps //Plot single cap Z_freq=sqrt(R1^2+FreqRange^2*L1^2-2*L1/C1+(FreqRange^2*C1^2)^(-1)); plot2d(FreqRange, Z_freq , logflag="ll", style=5);
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clear clc I=200 r1=.05 r2=.06 r3=.02 r4=.04 r5=.03 r6=.01 ra=.02 rb=.03 I1=100 I3=30 I4=50 I5=20 //(a) dv=0 A=[ (ra) (-rb) (r6+r5 +r4+r3) 1 1 0 (ra+r1+r2) (-rb) -(r1+r2) ] B=[ dv+ (r5*(I5) +r4*(I5+I4)+r3*(I5+I4+I3)) I dv+(r2*I1) ] i=inv(A)*B mprintf("\n(a) Ia = %.0f A, Ib= %.0f A", i(1), i(2)) //(b) dv=-5 A=[ (ra) (-rb) (r6+r5 +r4+r3) 1 1 0 (ra+r1+r2) (-rb) -(r1+r2) ] B=[ dv+(r5*(I5) +r4*(I5+I4)+r3*(I5+I4+I3)) I dv+(r2*I1) ] i=inv(A)*B mprintf("\n(b) Ia = %d A, Ib= %d A", i(1), i(2))
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clc; close(); //page no 210 //prob no. 6.17 //All power in Watts //All voltage in volts //All current in ampere R=50; m=0.5; P=1125; //for m=0.5 Vrms=sqrt(R*P); Irms=sqrt(P/R); disp('A',Irms,'V',Vrms,'(a) For m=0.5, Vrms and Irms are:'); m=1; P=1500; //For m=1 Vrms=sqrt(R*P); Irms=sqrt(P/R); disp('A',Irms,'V',Vrms,'(b) For m=1, Vrms and Irms are:');
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stochastic_SIR.sce
N=100 s=floor(rand()*100) i=N-s r=0 mu=0.01 b=0.05 g=0.09 steps=5000 S=zeros(1,steps) I=zeros(1,steps) R=zeros(1,steps) for j=1:steps do if rand() < mu then s=s+1, N=N+1; end, if rand() <mu then a=rand() if a<s/N then s=s-1; elseif s/N<a & a<(s+i)/N then i=i-1; else r=r-1; end, N=N-1 end if rand() < b*s*i/(N**2) then s=s-1, i=i+1; end if rand() < g*i/N then i=i-1, r=r+1; end S(j)=s; I(j)=i; R(j)=r; end clf() plot(1:steps,S,"blue") plot(1:steps,I,"green") plot(1:steps,R,"red")
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Pressure vessel design.sce
// Reference: E. Sandgren,Nonlinear Integer and Discrete Programming in Mechanical I Design Optimization,Journal of Mechanical Design,JUNE 1990, Vol. 112/223 //A cylindrical pressure vessel is capped at both ends by hemispherical heads.The total cost, //including the cost of material, cost of forming and welding,is to be minimized. The design variables are Ts and Th are the thicknesses of the shell and head, and R and L, the inner radius and length of the cylindrical section.These variables are denoted by X1 x2 , x3, and x4, respectively, and units for each are inches. The variables are such that R and L are continuous while Ts and Th are integer multiples of 0.0625 inch, the available thicknesses of rolled steel plates. // Min f = 0.6224*X(1)*X(3)*X(4) + 1.7781*X(2)*X(3)^2 + 3.1661*X(1)^2*X(4) + 19.84*X(1)^2*X(3) // subject to // g1(X) - X1 + 0.0193 X3 < = 0 // g2(X) - x2 + 0.00954X3 <= 0 // g3(X) -%pi*X3^2*X4 - (4/3)*%pi*X3^3 + 1296000 <= 0 // g4(X) X4 - 240 <= 0 //====================================================================== // Copyright (C) 2018 - IIT Bombay - FOSSEE // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt // Author: Remya Kommadath // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in //====================================================================== clc; // Onjective fucntion function f = ObjectiveFunction (X) X(1:2) = X(1:2)*0.0625; f = 0.6224*X(1)*X(3)*X(4) + 1.7781*X(2)*X(3)^2 + 3.1661*X(1)^2*X(4) + 19.84*X(1)^2*X(3); endfunction // Non linear equality and inequality constraints function [C,Ceq] = NLconstraints(X) X(1:2) = (X(1:2))*0.0625; C = -%pi*X(3)^2*X(4) - (4/3)*%pi*X(3)^3 + 1296000; Ceq = []; endfunction // Linear inequality constraints A = [-0.0625 0 0.0193 0;0 -0.0625 0.00954 0;0 0 0 1]; b = [0 0 240]'; // Bounds of the variables lb = [1 1 10 10]; ub = [99 99 200 200]; nVar = length(lb); // Initial guess given to the solver x0 = [20 10 58.291 43.69]; // indices of the integer decision variables int = [1 2]; // Calling the solver [xopt,fopt,exitflag,output,lambda] = intfmincon(ObjectiveFunction,x0,int,A,b,[],[],lb,ub,NLconstraints) // Result representation // Converting the integer variables to the discrete variable x0(1:2) = x0(1:2)*0.0625; clc; disp(x0,"Initial guess given to the solver") select exitflag case 0 disp("Optimal Solution Found") disp(xopt',"The optimum solution obtained") disp(fopt,"The optimum value of the objective function") case 1 disp("Converged to a point of local infeasibility") case 2 disp("Objective Function is Continuous Unbounded") case 3 disp("Limit Exceeded") case 4 disp("User Interupt") case 5 disp("MINLP Errors") end
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clc //Variable Initialisation Ea=230//Input Voltage of motor in volts d1=0.8//Duty Ratio d2=0.75//Duty Ratio Ia1=80//Armature Current in Ampere Ra=0.25//Armature resistance in ohm N2=750//Rated Speed of Motor in rpm N3=600//Rated Speed of Motor in rpm Ia2=70 Eb2=210//Average Value of Back EMF //Solution E01=d1*Ea Eb1=E01-(Ia1*Ra) N1=(Eb1/Eb2)*N2 Ia2=86 E02=d2*Ea Eb3=E02-(Ia2*Ra) Wm=2*%pi*N3/60 T=Eb3*Ia2/Wm printf('\n\n Motor Speed=%0.1f rpm\n\n',N1) printf('\n\n Torque produced=%0.1f N-m\n\n',T)
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function [X_n]=F_alog_remez(Cprime,X_n) l=0 if X_n(1)<=Cprime & Cprime<X_n(2) if sign(fmoinsp(X_n(2))) == sign(fmoinsp(Cprime)) X_n(2)=Cprime disp('display',10) else [X_n]=F_permut(X_n,Cprime,1) disp('display',20) end l=1 end if l==0 if X_n(n)< Cprime & Cprime <=X_n(n+1) if sign(fmoinsp(X_n(n))) == sign(fmoinsp(Cprime)) X_n(n)=Cprime disp('display',30) else [X_n]=F_permut(X_n,Cprime,-1) disp('display',40) end l=1 end end if l==0 k=3 while (X_n(k)<Cprime) k=k+1 end if sign(fmoinsp(X_n(k)))== -sign(fmoinsp(Cprime)) X_n(k-1)=Cprime disp('display',50) l=1 else X_n(k)=Cprime disp('display',60) l=1 end end endfunction
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-- CE3107A.TST -- Grant of Unlimited Rights -- -- Under contracts F33600-87-D-0337, F33600-84-D-0280, MDA903-79-C-0687, -- F08630-91-C-0015, and DCA100-97-D-0025, the U.S. Government obtained -- unlimited rights in the software and documentation contained herein. -- Unlimited rights are defined in DFAR 252.227-7013(a)(19). By making -- this public release, the Government intends to confer upon all -- recipients unlimited rights equal to those held by the Government. -- These rights include rights to use, duplicate, release or disclose the -- released technical data and computer software in whole or in part, in -- any manner and for any purpose whatsoever, and to have or permit others -- to do so. -- -- DISCLAIMER -- -- ALL MATERIALS OR INFORMATION HEREIN RELEASED, MADE AVAILABLE OR -- DISCLOSED ARE AS IS. THE GOVERNMENT MAKES NO EXPRESS OR IMPLIED -- WARRANTY AS TO ANY MATTER WHATSOEVER, INCLUDING THE CONDITIONS OF THE -- SOFTWARE, DOCUMENTATION OR OTHER INFORMATION RELEASED, MADE AVAILABLE -- OR DISCLOSED, OR THE OWNERSHIP, MERCHANTABILITY, OR FITNESS FOR A -- PARTICULAR PURPOSE OF SAID MATERIAL. --* -- OBJECTIVE: -- CHECK THAT IS_OPEN RETURNS THE PROPER VALUES FOR FILES OF -- TYPE TEXT_IO. -- HISTORY: -- DLD 08/10/82 -- SPS 11/09/82 -- JBG 03/24/83 -- EG 05/29/85 -- DWC 08/17/87 SPLIT OUT CASES WHICH DEPEND ON A TEXT FILE -- BEING CREATED OR SUCCESSFULLY OPENED. PLACED -- CASES INTO CE3107B.ADA. -- PWB 03/07/97 ADDED CHECK FOR FILE SUPPORT. WITH REPORT; USE REPORT; WITH TEXT_IO; USE TEXT_IO; PROCEDURE CE3107A IS TEST_FILE_ZERO : FILE_TYPE; TEST_FILE_ONE : FILE_TYPE; TEST_FILE_TWO : FILE_TYPE; TEST_FILE_THREE : FILE_TYPE; VAL : BOOLEAN; INCOMPLETE : EXCEPTION; BEGIN TEST("CE3107A", "CHECK THAT IS_OPEN RETURNS THE PROPER " & "VALUES FOR UNOPENED FILES OF TYPE TEXT_IO"); -- FIRST TEST WHETHER IMPLEMENTATION SUPPORTS TEXT FILES AT ALL BEGIN TEXT_IO.CREATE ( TEST_FILE_ZERO, TEXT_IO.OUT_FILE, REPORT.LEGAL_FILE_NAME ); EXCEPTION WHEN TEXT_IO.USE_ERROR | TEXT_IO.NAME_ERROR => REPORT.NOT_APPLICABLE ( "TEXT FILES NOT SUPPORTED -- CREATE OUT-FILE" ); RAISE INCOMPLETE; END; TEXT_IO.DELETE ( TEST_FILE_ZERO ); -- WHEN FILE IS DECLARED BUT NOT OPEN VAL := TRUE; VAL := IS_OPEN(TEST_FILE_ONE); IF VAL = TRUE THEN FAILED("FILE NOT OPEN BUT IS_OPEN RETURNS TRUE"); END IF; -- FOLLOWING UNSUCCESSFUL CREATE BEGIN VAL := TRUE; CREATE(TEST_FILE_TWO, OUT_FILE, "$ILLEGAL_EXTERNAL_FILE_NAME1"); FAILED("NAME_ERROR NOT RAISED - UNSUCCESSFUL CREATE"); EXCEPTION WHEN NAME_ERROR => VAL := IS_OPEN(TEST_FILE_TWO); IF VAL = TRUE THEN FAILED("IS_OPEN GIVES TRUE AFTER AN " & "UNSUCCESSFUL CREATE"); END IF; END; -- FOLLOWING UNSUCCESSFUL OPEN BEGIN VAL := FALSE; OPEN(TEST_FILE_TWO, IN_FILE, LEGAL_FILE_NAME); FAILED("NAME_ERROR NOT RAISED - " & "UNSUCCESSFUL OPEN"); EXCEPTION WHEN NAME_ERROR => VAL := IS_OPEN(TEST_FILE_TWO); IF VAL = TRUE THEN FAILED("IS_OPEN GIVES TRUE - " & "UNSUCCESSFUL OPEN"); END IF; END; -- CLOSE FILE WHILE NOT OPEN BEGIN VAL := TRUE; CLOSE(TEST_FILE_THREE); -- STATUS ERROR FAILED("STATUS_ERROR NOT RAISED - UNSUCCESSFUL CLOSE"); EXCEPTION WHEN OTHERS => VAL := IS_OPEN(TEST_FILE_THREE); IF VAL = TRUE THEN FAILED("IS_OPEN GIVES TRUE - UNSUCCESSFUL " & "CLOSE"); END IF; END; RESULT; EXCEPTION WHEN INCOMPLETE => NULL; REPORT.RESULT; WHEN OTHERS => REPORT.FAILED ( "UNEXPECTED EXCEPTION" ); REPORT.RESULT; END CE3107A;
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//clear// clear; clc; //Example 4.1 // (a) // density of the fluid rho = 0.887*62.37;// [lb/ft^3] // total volumetric flow rate q = 30*60/7.48; //[ft^3/hr] // mass flow rate in pipe A and pipe B is same mdot = rho*q //[lb/hr] // mass flow rate in each pipe of C is half of the total flow mdot_C = mdot/2 //[lb/hr] disp('lb/hr',mdot,'mass flow rate pipe A = ') disp('lb/hr',mdot,'mass flow rate pipe B = ') disp('lb/hr',mdot_C,'mass flow rate pipe C = ') // (b) // Using Eq.(4.4), // velocity through pipe A V_Abar = 240.7/(3600*0.0233) //[ft/s] // velocity through pipe B V_Bbar = 240.7/(3600*0.0513) //[ft/s] // velocity through each pipe of C V_Cbar = 240.7/(2*3600*0.01414) //[ft/s] disp('ft/s',V_Abar,'velocity through pipe A = ') disp('ft/s',V_Bbar,'velocity through pipe B = ') disp('ft/s',V_Cbar,'velocity through pipe C = ') // (c) // Using Eq.(4.8), // mass velocity through pipe A GA = mdot/0.0233 // [kg/m^2-s] // mass velocity through pipe B GB = mdot/0.0513 //[kg/m^2-s] // mass velocity through each pipe of C GC = mdot/(2*0.01414) //[kg/m^2-s] disp('kg/m^2-s',GA,'mass velocity through pipe A = ') disp('kg/m^2-s',GB,'mass velocity through pipe B = ') disp('kg/m^2-s',GC,'mass velocity through pipe C = ')
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clear //compute brake mean effective pressure //given T=350. D=4**0.25 L=5 M=4 //bmep for 4-cycle engine=192*t bmep=192*(T/(D**2)*L*M) //bmep for 2-cycle engine bmep2=bmep/2 printf("\n \n bmep for 4-cycle %.2f psi",bmep) printf("\n \n bmep for 2-cycle %.2f psi",bmep)
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clc; clear; //Example 3.11 mu=1.906*10^-5 //[kg/(m.s)] k=0.02723 //W/m.K Cp=1.007 //[kJ/(kg.K)] rho=1.129 //[kg/cubic m] Npr=0.70 Mavg=29 u_inf=35 //[m/s] L=0.75 //[m] Tm=313 //[K] P=101.325 //[kPa] Nre_l=rho*u_inf*L/mu //Reynold's number >5*10^5 Nnu=0.0366*Nre_l^(0.8)*Npr^(1.0/3.0); h=Nnu*k/L //[W/s m.K] A=1*L //[sq m] Tw=333 //[K] T_inf=293 //[K] Q=h*A*(Tw-T_inf); //[W] printf("Heat transfer from the plate is %f W",Q);
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clc; clear; exec ("C:\Program Files\scilab-5.3.0\bin\TCP\1.2data.sci"); W=m*9.81; disp("W=") disp(W) //F=W+m*acc //1 ft= 0.3048 m F=W+(m*acc*0.3048); disp("N",F,"F=")
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//exapple 13.1 clc; funcprot(0); // Initialization of Variable rho=998; g=9.81; pi=3.1428; omega=2*pi*1055/60;//angular rotation r=2.55/100//radius outer ld=1.55/100;//liq. depth l=10.25/100; //calculation //part1 a=r*omega^2/g; disp(a,"ratio of cetrifugal force & gravitational force is:"); //part2 ri=r-ld;//radius internal V=pi*(r^2-ri^2)*l; sigma=(omega^2*V)/(g*log(r/ri)); disp(sigma,"equivalent to gravity settling tank of crossectional area of in (m^2):")
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// potencia (%), giro esquerdo (graus/%s), giro direito (graus/%s) datalog = [ 0 0 0; 1 0 0; 2 0 0; 3 8 0; 4 0 0; 5 7 13; 6 27 30; // 7 0 10; 8 53 43; 9 59 58; // 10 15 0; 11 72 86; 12 92 91; 13 95 96; 14 99 97; 15 112 111; 16 116 118; 17 116 115; 18 132 131; 19 134 136; 20 142 139; 21 152 154; 22 147 147; 23 167 167; 24 163 164; 25 183 184; 26 179 180; 27 195 197; 28 194 193; 29 208 211; 30 217 214; 31 222 224; 32 262 269; 33 270 271; 34 269 269; 35 294 273; 36 285 280; 37 287 288; 38 291 293; 39 298 300; 40 305 304; 41 307 307; 42 315 314; 43 319 319; 44 327 327; 45 362 364; 46 359 358; 47 361 362; 48 395 396; 49 365 391; 50 412 412; 51 404 405; 52 428 411; 53 429 426; 54 419 420; 55 424 425; 56 429 428; 57 432 434; 58 463 472; 59 456 458; 60 461 464; 61 465 468; 62 502 504; 63 495 497; 64 498 496; 65 498 501; 66 537 534; 67 521 526; 68 529 533; 69 527 531; 70 571 572; 71 556 556; 72 560 563; 73 595 595; 74 580 581; 75 583 585; 76 594 598; 77 608 610; 78 629 644; // 79 27 461; 80 627 620; 81 649 652; // 82 30 483; 83 660 661; 84 662 664; 85 670 674; 86 689 688; 87 680 684; 88 688 690; 89 696 700; 90 707 707; 91 717 720; 92 719 723; 93 728 731; 94 729 730; //95 565 35; //96 28 617; 97 769 764; 98 763 764; 99 764 769; 100 766 769; ];
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function X = inverse(y,k) X = []; for(i=1:k-1) X = [X y.^(i-1)] end endfunction
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//example:-5.5,page no.-226. //program to determine the reciprccity and lossless of two port network and find return loss. syms S Rl tao; S=[0.1 0.8*%i;0.8*%i 0.2]; // s-parameter matrix. if (S(1,2)==S(2,1)) disp("the network is reciprocal.") else disp("the network is not reciprocal.") end if (S(1,1)^2+S(1,2)^2==1) disp("the network is lossless.") else disp("the network is lossy.") end tao=S(1,1)-(S(1,2)*S(2,1))/(1+S(2,2)); //input reflection coefficient. Rl=-20*log10(abs(tao)); // return loss in dB. //result disp(Rl,'return loss at port 1 in dB=')
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clc; clear all; disp("heat transfer rate") La=0.22;//m Lb=0.22;//m kA=3.5;// W/(m*C) kB=0.65;// W/(m*C) thf=1300;// degree C tcf=40;// degree C hconvi=16.4;//W/(m^2*C) hconvo=11.5;//W/(m^2*C) hradi=17.5;//W/(m^2*C) hrado=7.2;//W/(m^2*C) //q= qconvi+qradi=qA+qB=qconvo+qrado //q=delT/Rtotal=(thf-tcf)/Rtotal hi=hconvi+hradi; ho=hconvo+hrado; Rtotal=1/hi+1/ho+La/kA+Lb/kB; q=(thf-tcf)/Rtotal;// W disp("W/m^2",q,"rate of heat transfer through wall are unit area =") //q=hi*(thf-t1)=(t1-t2)/(La/kA) t1=thf-q/hi;// degree C t2=t1-q*(La/kA);// degree C disp ("degree C",t2,"maximum temperature to which common brick is subjected t2 =")
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//Chapter-5,Example 4,Page 123 clc(); close(); t= 30 //time in minutes a=100 x= 25 k=(2.303/t)*log10(a/(a-x)) t_half=0.693/k printf('the time of 50 percent completion of reaction is %.2f mins',t_half)
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// Example 7_6 clc;funcprot(0); // Given data T_1=1000;// K P_1=0.5;// The inlet pressure in MPa P_2=0.1;// The exit pressure in MPa T_0=298;// K R=0.286;// kJ/kg.K // Calculation // From the air tables phi_1=2.968;// kJ/kg.K phi_2=phi_1-(R*log(P_1/P_2));// kJ/kg.K // Thus T_2=657.5;// K h_2=667.8;// kJ/kg h_1=1046.1;// kJ/kg h_0=298.2;// kJ/kg V_2=sqrt(2)*((h_1-h_2)*10^3)^(0.5);// m/s P_0=P_2;// MPa phi_0=1.695;// kJ/kg.K X_2=(h_2-h_0)+((V_2)^2/(2*1000))-(T_0*(phi_2-phi_0-(R*log(P_2/P_0))));// kJ/kg X_1=h_1-h_0-(T_0*(phi_1-phi_0-(R*log(P_1/P_0))));// The availability supplied in kJ/kg e_II=X_2/X_1;// The second law effectiveness for an ideal isentropic nozzle printf("\nThe second law effectiveness for an ideal isentropic nozzle,e_II=%1.2f",e_II); // The answer provided in the textbook is wrong
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exec('myfunc.sci'); //y=ode("rk",y0,t0,t,f) //initial condition at r=0 //chidashi //chii //final condition at tub wal r=R (tube radius) //chidashf //chif n=120; tchi(1:n)=zeros(n); tpi(1:n)=zeros(n); rw(1:4)=zeros(4); dchi(1:4)=zeros(4); dpi(1:4)=zeros(4); delta=0.01; rtube=4.2e6; //4.2Mm 4.2e6 params.m=1; params.k=0.8e-6; params.om=5e-4; //s^-1rad params.p=2.4e4; params.rpi=1e-6; //Uuse initial bc params.rchi=1e-6; //use initial bc params.b0=1.0e4; params.a=rtube; params.b=1.1*rtube; params.beta=2*params.p/(params.b0^2); params.mu1hat=mu1(params); tchi(1)=params.rchi; tpi(1)=params.rpi; maxits=3200; //oma=.00115; //omb=.00118; oma=0.00001; omb=0.2; ka=1.0e-10; kb=5.0e-6; nkval=60; params.om=oma; deltaom=(omb-oma)/10; chiopia= -%inf ; chiopib= %inf; h=(rtube)/n; for ki=1:nkval params.k=ka+ki*(kb-ka)/nkval; kval(ki)=params.k; soln=pionchi(params,params.m,params.k); fx=10*soln; nits(ki)=0; tol=0.000001; while abs(soln-fx)>=tol //while abs(omb-oma)>(%eps*omb) //for iom=1:10 //params.om=(oma+omb)/2; oldom=params.om; params.om=params.om+deltaom; for i=2:n rt=(i-1)*h; for j=1:4 rrt=rt+rw(j).*h; dchi(j)=h*myfuncchi(params,tchi(i-1),tpi(i-1),rrt); dpi(j)=h*myfuncpi(params,tchi(i-1),tpi(i-1),rrt); end tpi(i)=tpi(i-1)+(dpi(1)+dpi(4)+2*(dpi(2)+dpi(3)))/6.0; tchi(i)=tchi(i-1)+(dchi(1)+dchi(4)+2*(dchi(2)+dchi(3)))/6.0; params.rpi=tpi(i); params.rchi=tchi(i); end nits(ki)=nits(ki)+1; fx=params.rpi/params.rchi; if fx>soln then params.om=oldom; deltaom=deltaom/2; else deltaom=deltaom*2; end omval(ki)=params.om; // disp(params.om); // disp( params.rpi/params.rchi); // disp(soln); // disp('next'); if nits(ki)>maxits then tol=%inf; end solnval(ki)=soln; fxval(ki)=fx; end //while loop //disp(params.om); //disp(params.k); //disp(nits(ki)); //disp(soln); //disp(fx); //disp('next'); save('m_1_120_3200its.mat',nits,omval,kval,solnval,fxval); end // loop over kvalues
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clc clear //DATA GIVEN m=1; //mass of wet steam in kg p=6; //pressure of steam in bar x=0.8; //dryness fraction //At 6 bar, from steam tables Ts=158.8+273; //in K hfg=2085; //kJ/kg swet=4.18*log(Ts/273)+x*hfg/Ts; //entropy of wet steam in kJ/kgK printf('The Entropy of wet steam is: %1.4f kJ/kgK.',swet); //NOTE; //the exact ans is 5.7794, while in TB it is given as 5.7865 kJ/kgK
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clc; clear; P0=2/3; //P(X=0) P1=2/3; //P(Y=0) H_x=0.919; H_y=0.919; H_b=0.919; //Hb(2/3) //since X,Y pair is uniformly distributed on three values H_xy=log2(3); // H(X,Y) H_xdivy=H_xy-H_y; //H(X/Y)=H(X,Y)-H(Y) I_xdivy=H_x-H_xdivy; //I(X,Y)=H(X)-H(X/Y) disp(I_xdivy,"I(X,Y)=");
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//Example 2 Page 59 clc clear //creating C(x) function function y=C(x) y=100000+160*x-0.2*x^2 endfunction //the variable cost is the part of the cost function that depends on x VariableCost=poly([0 160 -0.2],'x','c')//variablecost polynomial disp(VariableCost,'VariableCost in dollars=') FixedCost=100000 disp(FixedCost,'FixedCost in dollars=') x=poly(0,'x')//x polynomial creation Rx=800*x//annual revence of x members //for the profit we use the formula Px=Rx-C(x)//displaying the P(x) value by subtracting C(x) from Rx disp('P(x)=R(x)-C(x)') disp(Px) r=roots(Px)//finding the roots of the quadratic equation obtained in P(x) disp(r) disp('since members cannot be in negative value we consider the positive value') x=[0 250]//taking random variables of x for graph y=C(x)//function calling plot(400,350000,x,y,'red')//plotting graph //creating the P(x) function function y=P(x) y=-100000+640*x+0.2*x^2 endfunction x=([0 150 350])//taking values of x for graph y=P(x) plot(x,y,'blue')//plotting graph xtitle(' ','x','y');
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clear; clc; csv = csvRead("observacoes.csv",' ') pontos = [3.6118 2.9033 1.7140 1.7016 2.6102 2.3943 3.5217 1.1030 2.1144 2.8743 3.1330 3.4240 5.3695 2.6703 2.8835 2.4096 4.6215 2.9022 5.0193 2.8179 3.1008 3.2167 2.3034 1.4748 3.6472 2.6745 4.6394 5.2640 2.8566 4.1702 3.8175 2.6147 2.2934 4.2012 2.5310 3.7031 3.7229 1.8605 3.5397 3.8148 2.6236 1.3864 3.1297 4.0862 5.3783 2.5705 2.6732 2.0909 2.3348 2.4896 2.7638 2.3800 0.6342 3.4446 3.3705 3.2042 3.2886 3.5406 3.3370 4.8267 3.4636 3.6535 3.4020 2.5565 4.4431 3.9920 1.6055 3.6936 2.4774 3.3368 0.7867 0.8749 2.6352 4.1560 3.2858 1.5949 3.7392 2.8120 2.3117 1.1348 4.1271 2.7336 2.8331 3.8487 3.9682 2.8423 3.1162 3.6531 4.0666 1.4985 2.8107 3.5702 4.6343 2.2260 3.8566 2.9034 -0.5429 2.3434 3.7257 2.1756 7.3158 6.8513 4.4364 4.9454 7.2912 6.0339 7.2389 6.2518 6.4749 5.5465 6.2102 6.4830 7.6684 5.1612 3.9020 3.6581 6.3501 3.3559 6.5372 6.3260 6.9188 5.6233 4.2438 7.6421 5.0076 5.1970 7.5322 6.7750 4.8007 5.6805 7.0233 5.7450 6.1705 5.9890 7.1068 6.6353 5.8883 5.3207 5.9186 5.9272 4.8481 5.9914 5.2151 4.5259 5.0553 6.1791 5.9917 8.0680 5.7569 6.0470 5.7304 6.3432 5.6364 6.1037 4.9605 4.8067 6.2229 5.5165 6.5714 6.2241 7.6794 6.6357 6.6964 5.5870 7.9882 8.0092 6.1868 5.3676 6.2800 4.2854 4.7845 7.8715 5.6212 6.2646 6.7731 5.7555 5.6720 6.0163 5.2168 7.6996 7.3287 6.4373 5.1401 7.2426 6.6416 4.6958 5.8749 5.7205 6.4459 6.1071 6.3143 6.4245 7.0774 6.2737 6.2029 5.5610 4.8296 7.3827 3.7052 6.7945 1.7929 6.7778 4.2065 5.2426 3.5766 6.1978 4.2324 6.6047 1.4872 4.6725 4.4523 4.9796 1.5541 5.5445 3.0522 6.1081 3.4205 6.9626 3.8244 7.1447 3.7479 6.7558 3.3164 4.3651 2.7089 6.2450 2.2006 4.6953 2.7406 5.2457 1.6868 4.9766 1.7925 8.1089 2.2501 8.2444 2.2835 5.9420 1.4198 7.3276 4.2358 4.2813 1.9965 9.3107 4.0336 5.5681 4.3984 5.8426 2.3934 6.7423 3.4478 7.4134 2.7141 8.3087 2.3565 5.3853 2.7511 6.8783 4.7119 6.0405 3.7889 5.5342 2.1840 6.0321 3.3701 5.3050 3.0671 5.7734 2.9267 7.3887 3.5408 6.8923 2.3032 5.2765 2.5115 7.5541 3.0926 6.4991 4.0144 5.2921 1.8681 6.1830 3.3216 6.5363 3.1158 6.0099 2.8416 5.6921 2.9440 8.9717 4.1624 5.5926 3.5246 6.8471 2.2795 5.1525 2.5550 6.2115 3.9852 6.3222 5.7991 3.7095 5.8397 2.6598 4.6287 5.1226 5.5106 2.9931 7.5461 2.5471 4.8117 4.0446 6.3646 2.9572 5.8535 1.5515 7.8161 1.9857 7.4761 2.9690 7.9240 2.1706 4.2917 3.2975 5.8123 2.0199 5.9155 3.2743 5.3924 4.1475 4.0223 3.0908 7.1925 3.8504 6.6647 2.5534 4.9922 3.6519 5.9101 3.7208 5.8788 2.4348 6.9995 3.0777 6.3702 3.2633 7.2753 3.1483 6.4220 2.5780 3.9440 2.8523 9.4274 5.2018 4.3413 4.8142 7.5378 3.4213 6.8213 3.3570 6.1389 2.9317 6.1057 2.6351 5.7291 4.3498 6.5174 3.4393 6.4982 3.7073 6.0232 3.0007 5.7020 3.1841 5.9035 3.0020 5.2193 4.0240 7.1192 2.0947 6.8281 2.3275 5.3324 1.2517 6.2047 3.7529 6.4696 2.8792 3.9511 1.1018 6.1152 1.4202 7.8509 3.2019 6.0825 3.3396 6.4342 3.5129 6.4441 2.6522 ] clf(); xlabel('x'); ylabel('y'); plot(pontos(:,1),pontos(:,2),'r*'); k = input("k-means: ") b=zeros(200,1); bnovo=ones(200,1); numcentroides=2 for j=1:k cluster(j,:)=pontos(round(200*rand(1,1,'uniform')),:); end xlabel('Ciclos'); ylabel('Disp.'); cor = input("INPUT COR: ") iteracao=0; while ~isequal(b, bnovo) && (iteracao<1000) b=bnovo; iteracao=iteracao+1; disp(iteracao); for entrada=1:200 maisproximo=1000; for j=1:k distancia=sqrt((cluster(j,1)-pontos(entrada,1))^2 + (cluster(j,2)-pontos(entrada,2))^2); disp(distancia); if distancia<maisproximo maisproximo=distancia; centroide_maisproximo=j; end bnovo(entrada)=centroide_maisproximo; end end for j=1:k soma=[0 0]; elementos=0; for entrada=1:200 if b(entrada)==j soma=soma+pontos(entrada,:) elementos=elementos+1; end end if elementos ~=0 cluster(j,:)=soma/elementos; end end plot(iteracao,distancia,cor'); end timestep = input("Precione qualquer tecla."); for entrada=1:200 if bnovo(entrada)==1 plot(pontos(entrada, 1), pontos(entrada,2), 'k*'); elseif bnovo(entrada)==2 plot(pontos(entrada, 1), pontos(entrada,2), 'r*'); elseif bnovo(entrada)==3 plot(pontos(entrada, 1), pontos(entrada,2), 'b*'); else bnovo(entrada)==4 plot(pontos(entrada, 1), pontos(entrada,2), 'm*'); end end xlabel('x'); ylabel('y'); ts = input("Precione qualquer tecla.") for j=1:k plot(cluster(j,1),cluster(j,2), 'g+'); end
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clear ; clc; // Example 2.7 printf('Example 2.7\n\n'); //Page no.58 // Solution // Basis 500 L solution containing 35g/L // (NH4)2SO4 is the only nitrogen source cn = 35 ;//[g/L] wt = 9 ;//[wt % N] m_wt1 = 132 ;//[g] m_wt2 = 14 ;//[g] amt = (500*(35)*.09*1*1*m_wt1)/(1*m_wt2*1*1); printf('Total amount of (NH4)2SO4 consumed is %.1f g.',amt);
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<?xml version="1.0" encoding="utf-8"?> <test> <description>Kovasznay Flow 3D homogeneous 1D, adaptive P, 16 Fourier modes, using explicit mapping</description> <executable>IncNavierStokesSolver</executable> <parameters>KovaFlow_3DH1D_adaptive_16modes_FFTW_Mapping.xml</parameters> <files> <file description="Session File">KovaFlow_3DH1D_adaptive_16modes_FFTW_Mapping.xml</file> </files> <metrics> <metric type="L2" id="1"> <value variable="u" tolerance="1e-9">4.33062e-05</value> <value variable="v" tolerance="1e-9">1.16121e-05</value> <value variable="w" tolerance="1e-9">1.29819e-05</value> <value variable="p" tolerance="1e-9">0.000260747</value> </metric> <metric type="Linf" id="2"> <value variable="u" tolerance="1e-9">8.96493e-05</value> <value variable="v" tolerance="1e-9">2.47788e-05</value> <value variable="w" tolerance="1e-9">2.85904e-05</value> <value variable="p" tolerance="1e-9">0.000805358</value> </metric> </metrics> </test>
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//Section-5,Example-1,Page no.-D.11 //To show that the given reaction is a 2nd order reaction and calculate the fraction of ester decomposed in 30 minutes. clc; k_5=(1/5)*((1/10.2)-(1/16)) kbar_5=k_5*10^2 k_25=(1/25)*((1/4.3)-(1/16)) kbar_25=k_25*10^2 k_55=(1/55)*((1/2.3)-(1/16)) kbar_55=k_55*10^2 k_120=(1/120)*((1/1.1)-(1/16)) kbar_120=k_120*10^2 //Constant value of kbar shows that the given reaction is a 2nd order reaction
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function [new_generation, new_eval] = elitism(parent_pop, descend_pop, parent_eval, descend_eval, percentage) [prows, pcols] = size(parent_pop); new_pop = zeros(prows, pcols); new_evaluated_pop = zeros(prows, 1); [psorted_eval, pindex] = gsort(parent_eval); [dsorted_eval, dindex] = gsort(descend_eval); new_prows = ceil(percentage*prows) new_pop(1:new_prows,:) = parent_pop(1:new_prows,:); new_evaluated_pop(1:new_prows,:) = parent_eval(1:new_prows,:); new_pop((new_prows + 1):$,:) = descend_pop(1:(prows - new_prows),:); new_evaluated_pop((new_prows + 1):$,:) = descend_eval(1:(prows - new_prows),:) new_generation = new_pop; new_eval = new_evaluated_pop; endfunction
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type newType = short; var b : int; void sym1() { type newType = int; var a : newType; const c = 11; b = c; a = b + c; } main() { var a : newType; const c = 11; b = c; a = b + 1; }
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// Exa 17.15 // To calculate No of RNC required. clc; clear all; BTS=800;//No of BTS sites Sec=3;//No of sectors per BTS freq_sec=2;//No of frequencies used per sector cellsRNC=1152;//Maximum capacity of cellRNC btsRNC=384;//One RNC can support btsRNC(BTSs) VE=25;//Voice service(mErl/subscriber) BRV=16;// bitrate Voice CS1=10;//CS data service 1(mErl/subscriber) BRC1=32;//bit rate for CS1 in kbps CS2=5;//CS data service 2(mErl/subscriber) BRC2=64;////bit rate for CS2 in kbps PSdata=0.2;//PS data service(kbps per subscriber) PSoverhead=0.15; SHO=0.4;//40% Totalsub=350000;//Total subsribers Maxcap=196;//Maximum Iub capacity of tpRNC (in Mbps) FR1=0.9;FR2=0.9;FR3=0.9;//Filler rates //solution RNCreqd=(BTS*Sec*freq_sec)/(cellsRNC*FR1);//from eqn 17.28 printf('using equation 17.28,Number of RNC required are %d \n ',round(RNCreqd)); RNC_reqd=BTS/(btsRNC*FR2);//from eqn 17.29 printf('using equation 17.29,Number of RNC required are %d \n ',round(RNC_reqd)); RNCrequired=((VE/1000*BRV+CS1/1000*BRC1+CS2/1000*BRC2+(PSdata/(1-PSoverhead)))*(1+SHO)*Totalsub)/(Maxcap*1000*FR3);//from eqn 17.30 printf('using equation 17.30,Number of RNC required are %d \n ',round(RNCrequired)); printf(' We select first value which is %d RNCs \n ',round(RNCreqd));
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clc clear //Initialization of variables m1=121 m2=18 p1=0.0042 p2=0.0858 //calculations massfrac= (p1*m1)/(p1*m1+p2*m2) //results printf("mass fractions of DMA and water are %.3f and %.3f respectively",massfrac,1-massfrac)
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path_rna_ifr = get_absolute_file_path('RNA_IFR.sce'); N_IFR = getN( path_rna_ifr + "\N_IFR.txt" ); W_IFR = getW( path_rna_ifr, "IFR" ); function saida_da_rna = rna_ifr( ifr ) saida_da_rna = ann_FF_run( [ifr], N_IFR, W_IFR ); endfunction
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clear //Given l=1.4*10**-10 //m h=6.63*10**-34 l1=2.0*10**-10 c=3*10**8 //m/s //Calculation E=h*c*(1/l-1/l1) //Result printf("\n Energy of the scattered electron is %0.2f *10**-16 J",E*10**16)
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// Example 10.7: (a) Midband gain, Upper half-power frequency // (b) Zi clc, clear ICQ=1e-3; // in amperes RS=300; // in ohms RC=1.2e3; // in ohms bta=125; fT=300e6; // in hertz C_mu=0.5e-12; // in farads VT=25e-3; // Voltage equivalent to temperatue at room temperature in volts disp("Part (a)"); gm=ICQ/VT; // in mho r_pi=bta/gm; // in ohms // To find C_pi C_pi=gm/(2*%pi*fT)-C_mu; // in farads AVo=-bta*RC/(RS+r_pi); // Midband gain disp(AVo,"Midband gain ="); R_pi0=RS*r_pi/(RS+r_pi); a1=R_pi0*C_pi+(R_pi0+RC*(1+gm*R_pi0))*C_mu; // in seconds a2=R_pi0*RC*C_pi*C_mu; // in seconds p1=1/a1; // in rad/sec p2=a1/a2; // in rad/sec disp(p2/p1,"p2/p1 ="); disp("Since p2/p1 >> 8, therefore dominant-pole approximation holds good."); wH=p1*1e-6; // in M rad/sec disp(wH,"Upper half-power frequency (M rad/sec) ="); disp("Part (b)"); CM=C_pi+C_mu*(1+gm*RC); // in farads Zi=r_pi/(1+%i*wH*1e6*CM*r_pi); // in ohms disp(Zi,"Zi (Ω) =");
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// DC motor with PID control, tuned through pole placement technique, as in Example 9.18. // 9.21 exec('desired.sci',-1); exec('pp_pid.sci',-1); exec('cosfil_ip.sci',-1); exec('pd.sci',-1); exec('polyno.sci',-1); exec('myc2d.sci',-1); exec('zpowk.sci',-1); exec('polmul.sci',-1); exec('polsize.sci',-1); exec('xdync.sci',-1); exec('rowjoin.sci',-1); exec('left_prm.sci',-1); exec('t1calc.sci',-1); exec('indep.sci',-1); exec('seshft.sci',-1); exec('makezero.sci',-1); exec('move_sci.sci',-1); exec('colsplit.sci',-1); exec('clcoef.sci',-1); exec('cindep.sci',-1); // Motor control problem // Transfer function a = [-1 0; 1 0]; b = [1; 0]; c = [0 1]; d = 0; G = syslin('c',a,b,c,d); Ts = 0.25; [B,A,k] = myc2d(G,Ts); [Ds,num,den] = ss2tf(G); // Transient specifications rise = 3; epsilon = 0.05; phi = desired(Ts,rise,epsilon); // Controller design Delta = 1; //No internal model of step used [Rc,Sc] = pp_pid(B,A,k,phi,Delta); // continuous time controller [K,taud,N] = pd(Rc,Sc,Ts); numb = K*[1 taud*(1+1/N)]; denb = [1 taud/N]; numf = 1; denf = 1; // simulation parameters st = 1; // desired change in position t_init = 0; // simulation start time t_final = 20; // simulation end time st1 = 0; // continuous controller simulation: g_s_cl3.xcos num1 = 0; den1 = 1; // discrete controller simulation: g_s_cl2.xcos // u1: -0.1 to 0.8 // y1: 0 to 1.4 C = 0; D = 1; N = 1; gamm = 1; Tc = Sc; [Tcp1,Tcp2] = cosfil_ip(Tc,1); // Tc/1 [Np,Rcp] = cosfil_ip(N,Rc); // N/Rc [Scp1,Scp2] = cosfil_ip(Sc,1); // Sc/1 [Cp,Dp] = cosfil_ip(C,D); // C/D Numb = polyno(numb,'s'); Denb = polyno(denb,'s'); Numf = polyno(numf,'s'); Denf = polyno(denf,'s'); Num1 = polyno(num1,'s'); Den1 = polyno(den1,'s');
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clc() n=input("Enter the no. of terms:"); for a=1:n M(a)=input("Enter value of x"+string(a)+" :"); end for b=1:n N(b)=input("Enter value of y"+string(b)+" :"); end X=input("Enter the value of X "); for o=1:n for p=1:n DD(o,p)=0; end end p=1; for k=1:n DD(1,k)=N(k); end for j=2:n for i = 1:n+1-j DD(j,i)=(DD(j-1,i+1)-DD(j-1,i))/(M(i+j-1)-M(i)); end end disp("The Divided differences Table is"); disp(DD); Ans=DD(1,1); p=X-M(1); for q=2:n Ans=Ans+(p*DD(q,1)); p=p*(X-M(q)); end disp("The value of corresponding Y for the given X is :"); disp(Ans);
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//Ex:5.1 clc; clear; close; ht=100;// transmitter height in m hr=9;// receiver height in m D=3550*(sqrt(ht)+sqrt(hr));// distance to horizon in m f=60;// frequency in MHz y=300/f;// wavelength in m p=10*1000;// power in watt d=10*1000;// distance in m h=5; Et=(88*sqrt(p)*hr*ht)/(h*d^2);// the field strength in V/m et=10^-3;// field strength in V/m d2=(88*sqrt(p)*hr*ht)/(h*et); d1=sqrt(d2);// distance at which the field strength reuces to 1 mV/meter printf("The field strength = %f mV/m", Et*1000); printf("\n The distance at which the field strength reuces to 1 mV/meter = %f*10^3 meter", d1/1000);
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function f = flutuante(v,b,e,m,M) f = 0 cont=0 decimal=0 if b ==2 while (v~=0) if (v>1)&(decimal==0) rd=modulo(v,2) f = f+(rd*(10^cont)) v=int(v/2) cont=cont +1 else decimal=1 if v>0 cont=cont-1 v=v*2 if (int(v)==1) f=f+(1*10^(cont)) v=v-1 end end if abs(cont)==e v=0 end end end end endfunction
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//Chapter 12 : Solutions to the Exercises //Scilab 6.0.1 //Windows 10 clear; clc; //Solution for 1.14 //(a) mat_prod=[1;2;3;4;]*[1 2 3 4] disp(mat_prod,'(a)Matrix product=') //(b) mat_prod=[1 2 3 4]*[1;2;3;4;] disp(mat_prod,'(b)Matrix product=')
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//Example 21.2 clc; C=8*10^-6 X_c=1/(377*C) disp(X_c,"Resistance in ohm=") I_rms=150/X_c disp(I_rms,"Current in Amps=")
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clc; clear all; disp("temperature and time") R=40/2000;//m ti=20;// degree C tau=4*60;//s ta=100;// degree C k=10;// W/m.C rho=1200;// kg/m^3 c=2000;// J/kg.C h=100;//W/m^2.C Lc=R/3;// for sphere Bi=h*Lc/k t=ta+(ti-ta)*exp(-h*3*tau/(R*rho*c)) disp("degree C") ti=5;// degree C X=(t-ta)/(ti-ta); Y=-h*3/(R*rho*c) tau=(log (X))/Y;// sec disp("min",tau/60,"time taken tau =")
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clc //initialisation of variables d= 3 //in P= 2000 //psi s= 20 //strokes //CALCULATIONS Cl= s*d/2 F= P*%pi*d^2/4 stl= (Cl-40)/10 //RESULTS printf ('length of the stop tube= %.f in',Cl) printf (' \n thrust on the rod= %.f lb',F+3) printf (' \n Stop Tube length= %.f stl',stl)
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//Example 3_14 clc; clear; close; format('v',5); //given data : V=230;//V f=50;//Hz V1=120*expm(%i*30*%pi/180);//V Z1=15*expm(%i*40*%pi/180);//ohm V2=V-V1;//V I=V1/Z1;//A Z2=V2/I;//ohm R=real(Z2);//ohm XC=imag(Z2);//ohm C=-1/2/%pi/f/XC*10^6;//micro F disp(Z2,"Value of Z2(ohm) : "); disp(R,"Resistance(ohm)"); format('v',7); disp(C,"Capacitance(micro F)"); //Answer is not accurate in the book.
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// Exa 2.18 clc; clear; close; // Given data miu_n= 0.13;// in m^2/Vs miu_p= 0.05;// in m^2/Vs q=1.6*10^-19;// in C ni=1.5*10^16;// per m^3 sigma_intrinsic= q*ni*(miu_n+miu_p);// in (Ωm)^-1 disp(sigma_intrinsic,"The conductivity of silicon in Intrinsic condition in (Ωm)^-1 is : ") // Part (b) n= 5*10^28/10^9; sigma= q*n*miu_n;// in (Ωm)^-1 disp(sigma,"The conductivity with donar impurity in (Ωm)^-1 is : ") // Part (c) p= 5*10^28/10^8; sigma= q*p*miu_p;// in (Ωm)^-1 disp(sigma,"The conductivity with acceptor impurity in (Ωm)^-1 is : ") // Part (d) p_desh= p-n;// in /m^3 sigma= q*p_desh*miu_p;// in (Ωm)^-1 disp(sigma,"The conductivity with donar and acceptor impurity in (Ωm)^-1 is : ") // Note : Answer in the book of part (a) may be miss printed or wrong
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clc //variable initialisation Vm=400 //Input Voltage in volt F=50 //supply frequency in Hz P1=4 //number of poles R1=0.15 //resistance of stator in ohm R2=0.12 //resistance of rotor in ohm X1=0.45 //reactance of Motor in ohm X2=0.45 //reactance of Motor in ohm Xm=28.5 //reactance of Motor in ohm S=0.04 //Slip Of Motor //Solution Rl=R2*((1/S)-1) Vph=Vm/sqrt(3) I2=Vph/((R1+R2+Rl)+%i*(X1+X2)) I0=Vph/(%i*Xm) I1=I0+I2 y=imag(I1) x=real(I1) phi=atand(y/x) pf=cosd(phi) printf('\n\n The Stator Current=%0.1f Amp\n\n',I1) printf('\n\n The Power Factor=%0.1f lag\n\n',pf)
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# GeoProver test file for Maple, created on Jan 18 2003 read("GeoProver.mpl"): read("supp.mpl"): with(geoprover): interface(prettyprint=0): # Example Arnon # # The problem: # Let $ABCD$ be a square and $P$ a point on the line parallel to $BD$ # through $C$ such that $l(BD)=l(BP)$, where $l(BD)$ denotes the # distance between $B$ and $D$. Let $Q$ be the intersection point of # $BF$ and $CD$. Show that $l(DP)=l(DQ)$. # # The solution: vars_:=List(x1, x2, x3); # Points A_:=Point(0,0); B_:=Point(1,0); P_:=Point(x1,x2); # coordinates D_:=rotate(A_,B_,1/2); C_:=par_point(D_,A_,B_); Q_:=varpoint(D_,C_,x3); # polynomials polys_:=List(on_line(P_,par_line(C_,pp_line(B_,D_))), eq_dist(B_,D_,B_,P_), on_line(Q_,pp_line(B_,P_))); # conclusion con_:=eq_dist(D_,P_,D_,Q_); # solution gb_:=geo_gbasis(polys_,vars_); result_:=geo_normalf(con_,gb_,vars_); # Example CircumCenter_1 # # The problem: # The intersection point of the midpoint perpendiculars is the # center of the circumscribed circle. # # The solution: parameters_:=List(a1, a2, b1, b2, c1, c2); # Points A_:=Point(a1,a2); B_:=Point(b1,b2); C_:=Point(c1,c2); # coordinates M_:=intersection_point(p_bisector(A_,B_), p_bisector(B_,C_)); # conclusion result_:=List( eq_dist(M_,A_,M_,B_), eq_dist(M_,A_,M_,C_) ); # Example EulerLine_1 # # The problem: # Euler's line: The center $M$ of the circumscribed circle, # the orthocenter $H$ and the barycenter $S$ are collinear and $S$ # divides $MH$ with ratio 1:2. # # The solution: parameters_:=List(a1, a2, b1, b2, c1, c2); # Points A_:=Point(a1,a2); B_:=Point(b1,b2); C_:=Point(c1,c2); # coordinates S_:=intersection_point(median(A_,B_,C_),median(B_,C_,A_)); M_:=intersection_point(p_bisector(A_,B_), p_bisector(B_,C_)); H_:=intersection_point(altitude(A_,B_,C_),altitude(B_,C_,A_)); # conclusion result_:=List(is_collinear(M_,H_,S_), sqrdist(S_,fixedpoint(M_,H_,1/3))); # Example Brocard_3 # # The problem: # Theorem about the Brocard points: # Let $\Delta\,ABC$ be a triangle. The circles $c_1$ through $A,B$ and # tangent to $g(AC)$, $c_2$ through $B,C$ and tangent to $g(AB)$, and # $c_3$ through $A,C$ and tangent to $g(BC)$ pass through a common # point. # # The solution: parameters_:=List(u1, u2); # Points A_:=Point(0,0); B_:=Point(1,0); C_:=Point(u1,u2); # coordinates M1_:=intersection_point(altitude(A_,A_,C_),p_bisector(A_,B_)); M2_:=intersection_point(altitude(B_,B_,A_),p_bisector(B_,C_)); M3_:=intersection_point(altitude(C_,C_,B_),p_bisector(A_,C_)); c1_:=pc_circle(M1_,A_); c2_:=pc_circle(M2_,B_); c3_:=pc_circle(M3_,C_); P_:=other_cc_point(B_,c1_,c2_); # conclusion result_:= on_circle(P_,c3_); # Example Feuerbach_1 # # The problem: # Feuerbach's circle or nine-point circle: The midpoint $N$ of $MH$ is # the center of a circle that passes through nine special points, the # three pedal points of the altitudes, the midpoints of the sides of the # triangle and the midpoints of the upper parts of the three altitudes. # # The solution: parameters_:=List(u1, u2, u3); # Points A_:=Point(0,0); B_:=Point(u1,0); C_:=Point(u2,u3); # coordinates H_:=intersection_point(altitude(A_,B_,C_),altitude(B_,C_,A_)); D_:=intersection_point(pp_line(A_,B_),pp_line(H_,C_)); M_:=intersection_point(p_bisector(A_,B_), p_bisector(B_,C_)); N_:=midpoint(M_,H_); # conclusion result_:=List( eq_dist(N_,midpoint(A_,B_),N_,midpoint(B_,C_)), eq_dist(N_,midpoint(A_,B_),N_,midpoint(H_,C_)), eq_dist(N_,midpoint(A_,B_),N_,D_) ); # Example FeuerbachTangency_1 # # The problem: # For an arbitrary triangle $\Delta\,ABC$ Feuerbach's circle (nine-point # circle) is tangent to its 4 tangent circles. # # The solution: vars_:=List(x1, x2); parameters_:=List(u1, u2); # Points A_:=Point(0,0); B_:=Point(2,0); C_:=Point(u1,u2); P_:=Point(x1,x2); # coordinates M_:=intersection_point(p_bisector(A_,B_), p_bisector(B_,C_)); H_:=intersection_point(altitude(A_,B_,C_),altitude(B_,C_,A_)); N_:=midpoint(M_,H_); c1_:=pc_circle(N_,midpoint(A_,B_)); Q_:=pedalpoint(P_,pp_line(A_,B_)); # polynomials polys_:=List(on_bisector(P_,A_,B_,C_), on_bisector(P_,B_,C_,A_)); # conclusion con_:=is_cc_tangent(pc_circle(P_,Q_),c1_); # solution gb_:=geo_gbasis(polys_,vars_); result_:=geo_normalf(con_,gb_,vars_); # Example GeneralizedFermatPoint_1 # # The problem: # A generalized theorem about Napoleon triangles: # Let $\Delta\,ABC$ be an arbitrary triangle and $P,Q$ and $R$ the third # vertex of isosceles triangles with equal base angles erected # externally on the sides $BC, AC$ and $AB$ of the triangle. Then the # lines $g(AP), g(BQ)$ and $g(CR)$ pass through a common point. # # The solution: vars_:=List(x1, x2, x3, x4, x5); parameters_:=List(u1, u2, u3); # Points A_:=Point(0,0); B_:=Point(2,0); C_:=Point(u1,u2); P_:=Point(x1,x2); Q_:=Point(x3,x4); R_:=Point(x5,u3); # polynomials polys_:=List(eq_dist(P_,B_,P_,C_), eq_dist(Q_,A_,Q_,C_), eq_dist(R_,A_,R_,B_), eq_angle(R_,A_,B_,P_,B_,C_), eq_angle(Q_,C_,A_,P_,B_,C_)); # conclusion con_:=is_concurrent(pp_line(A_,P_), pp_line(B_,Q_), pp_line(C_,R_)); # solution sol_:=geo_solve(polys_,vars_); result_:=geo_eval(con_,sol_); # Example TaylorCircle_1 # # The problem: # Let $\Delta\,ABC$ be an arbitrary triangle. Consider the three # altitude pedal points and the pedal points of the perpendiculars from # these points onto the the opposite sides of the triangle. Show that # these 6 points are on a common circle, the {\em Taylor circle}. # # The solution: parameters_:=List(u1, u2, u3); # Points A_:=Point(u1,0); B_:=Point(u2,0); C_:=Point(0,u3); # coordinates P_:=pedalpoint(A_,pp_line(B_,C_)); Q_:=pedalpoint(B_,pp_line(A_,C_)); R_:=pedalpoint(C_,pp_line(A_,B_)); P1_:=pedalpoint(P_,pp_line(A_,B_)); P2_:=pedalpoint(P_,pp_line(A_,C_)); Q1_:=pedalpoint(Q_,pp_line(A_,B_)); Q2_:=pedalpoint(Q_,pp_line(B_,C_)); R1_:=pedalpoint(R_,pp_line(A_,C_)); R2_:=pedalpoint(R_,pp_line(B_,C_)); # conclusion result_:=List( is_concyclic(P1_,P2_,Q1_,Q2_), is_concyclic(P1_,P2_,Q1_,R1_), is_concyclic(P1_,P2_,Q1_,R2_)); # Example Miquel_1 # # The problem: # Miquels theorem: Let $\Delta\,ABC$ be a triangle. Fix arbitrary points # $P,Q,R$ on the sides $AB, BC, AC$. Then the three circles through each # vertex and the chosen points on adjacent sides pass through a common # point. # # The solution: parameters_:=List(c1, c2, u1, u2, u3); # Points A_:=Point(0,0); B_:=Point(1,0); C_:=Point(c1,c2); # coordinates P_:=varpoint(A_,B_,u1); Q_:=varpoint(B_,C_,u2); R_:=varpoint(A_,C_,u3); X_:=other_cc_point(P_,p3_circle(A_,P_,R_),p3_circle(B_,P_,Q_)); # conclusion result_:=on_circle(X_,p3_circle(C_,Q_,R_)); # Example PappusPoint_1 # # The problem: # Let $A,B,C$ and $P,Q,R$ be two triples of collinear points. Then by # the Theorem of Pappus the intersection points $g(AQ)\wedge g(BP), # g(AR)\wedge g(CP)$ and $g(BR)\wedge g(CQ)$ are collinear. # # Permuting $P,Q,R$ we get six such {\em Pappus lines}. Those # corresponding to even resp. odd permutations are concurrent. # # The solution: parameters_:=List(u1, u2, u3, u4, u5, u6, u7, u8); # Points A_:=Point(u1,0); B_:=Point(u2,0); P_:=Point(u4,u5); Q_:=Point(u6,u7); # coordinates C_:=varpoint(A_,B_,u3); R_:=varpoint(P_,Q_,u8); # conclusion result_:=is_concurrent(pappus_line(A_,B_,C_,P_,Q_,R_), pappus_line(A_,B_,C_,Q_,R_,P_), pappus_line(A_,B_,C_,R_,P_,Q_)); # Example IMO/36_1 # # The problem: # Let $A,B,C,D$ be four distinct points on a line, in that order. The # circles with diameters $AC$ and $BD$ intersect at the points $X$ and # $Y$. The line $XY$ meets $BC$ at the point $Z$. Let $P$ be a point on # the line $XY$ different from $Z$. The line $CP$ intersects the circle # with diameter $AC$ at the points $C$ and $M$, and the line $BP$ # intersects the circle with diameter $BD$ at the points $B$ and # $N$. Prove that the lines $AM, DN$ and $XY$ are concurrent. # # The solution: vars_:=List(x1, x2, x3, x4, x5, x6); parameters_:=List(u1, u2, u3); # Points X_:=Point(0,1); Y_:=Point(0,-1); M_:=Point(x1,x2); N_:=Point(x3,x4); # coordinates P_:=varpoint(X_,Y_,u3); Z_:=midpoint(X_,Y_); l_:=p_bisector(X_,Y_); B_:=line_slider(l_,u1); C_:=line_slider(l_,u2); A_:=line_slider(l_,x5); D_:=line_slider(l_,x6); # polynomials polys_:=List(is_concyclic(X_,Y_,B_,N_), is_concyclic(X_,Y_,C_,M_), is_concyclic(X_,Y_,B_,D_), is_concyclic(X_,Y_,C_,A_), is_collinear(B_,P_,N_), is_collinear(C_,P_,M_)); # constraints nondeg_:=List(x5-u2,x1-u2,x6-u1,x3-u1); # conclusion con_:=is_concurrent(pp_line(A_,M_),pp_line(D_,N_),pp_line(X_,Y_)); # solution sol_:=geo_solveconstrained(polys_,vars_,nondeg_); result_:=geo_eval(con_,sol_); # Example IMO/43_2 # # The problem: # # No verbal problem description available # # The solution: vars_:=List(x1, x2); parameters_:=List(u1); # Points B_:=Point(-1,0); C_:=Point(1,0); # coordinates O_:=midpoint(B_,C_); gamma_:=pc_circle(O_,B_); D_:=circle_slider(O_,B_,u1); E_:=circle_slider(O_,B_,x1); F_:=circle_slider(O_,B_,x2); A_:=sym_point(B_,pp_line(O_,D_)); J_:=intersection_point(pp_line(A_,C_), par_line(O_, pp_line(A_,D_))); m_:=p_bisector(O_,A_); P1_:=pedalpoint(J_,m_); P2_:=pedalpoint(J_,pp_line(C_,E_)); P3_:=pedalpoint(J_,pp_line(C_,F_)); # polynomials polys_:=List(on_line(E_,m_), on_line(F_,m_)); # constraints nondegs_:=List(x1-x2); # conclusion con_:=List(eq_dist(J_,P1_,J_,P2_), eq_dist(J_,P1_,J_,P3_)); # solution sol_:=geo_solveconstrained(polys_,vars_,nondegs_); result_:=geo_simplify(geo_eval(con_,sol_)); quit;
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P=5000 Vt=215 n=1000 Ra=0.4 Ia=P/Vt Eag=Vt+Ra*Ia Eam=Vt-Ra*Ia newn=Eam/Eag*n/1.1 disp(newn)
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I=20 w=2000 R=200 L=0.25 Xl=w*L*%i Ir=I*Xl/(Xl+R) Il=I-Ir Vl=Xl*Il t=1E-3 ir=sqrt(2)*real(Ir*exp(%i*w*t)) il=sqrt(2)*real(Il*exp(%i*w*t)) vl=sqrt(2)*real(Vl*exp(%i*w*t)) is=sqrt(2)*real(I*exp(%i*w*t)) vs=vl Pr=ir*ir*R Pl=vl*il Ps=is*ir*R Pr=ir*vl disp(Ps,Pl,Pr)
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# A simple recycle test # set up thermo - the name can be anything, I just use # 'thermo' for convenience. Essentially the rhs causes # a thermo package to be created and assigned to the unit op # owning the name thermo - in the case the base flowsheet # Also note that for now spaces are needed around the operators (= + etc) # A further also is that case is always significant $thermo = VirtualMaterials.IdealLiquid/Ideal/HC / -> $thermo thermo + METHANOL ETHANOL units SI # Add a stream # for now creating a unit op requires module.class(), but this # will be stream lined in the future stream = Stream.Stream_Material() # Make the stream In port current to save typing # You can use cd (named because it is similar to change directory in # Unix and DOS) to sub objects in this case first to the unit op stream # and then to its port In. This is just a typing convenience as everything # could be done from the top level with full names i.e. stream.In.T = 360.15 cd stream.In # Mole fractions can be enter indivually (Fraction.METHANOL = .25) or all # together as below. Fraction = .5 .5 VapFrac = 0.4 T = 300 K MoleFlow = 3000 # Now create a recycle stream cd / # return to top level - only place a slash is used recycle = Stream.Stream_Material() cd recycle.In # Estimate the values in the stream # Estimates use the ~= operator in place of the normal = which # fixes values T ~= 460.15 K P ~= 715 MoleFlow ~= 300 Fraction # any object without an operator displays itself - here to get order Fraction ~= 0 .5 . # a dot represents the current obj for display purposes # add a mixer to combine the first stream with the recycle cd / mixer = Mixer.Mixer() # ports are connected with the -> operator. They would be disconnected # by having an empty rhs. Similarly "stream.In.T =" would remove any value # for the stream In port Temperature stream.Out -> mixer.In0 recycle.Out -> mixer.In1 mixer.Out # add a separator flash = Flash.SimpleFlash() mixer.Out -> flash.In # split the liquid from the flash splitter = Split.Splitter() flash.Liq0 -> splitter.In # set the flow in one of the splitter outlets splitter.Out1.MoleFlow = 200 # close the recycle splitter.Out1 -> recycle.In # All done - check some streams recycle.Out splitter.Liq0 #splitter.Liq0.Out splitter.Out0 flash.In
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///Chapter No 7 Fluid Mechanics ///Example 7.17 Page No:128 /// Find bernoulli's equation for discharge //input data //refer figure 12 clc; clear; Q1=0.04; //Water flows at rate in m**2/s DA=0.22; //Pipe diameter at section A in m DB=0.12; //Pipe diameter at section B in m PA=400*10^3; //Intensity of pressure at setion A in kPa PB=150*10^3; //Intensity of pressure at setion B in kPa pi1=3.14; //Pi constant g1=9.81; //Gravity constant rho=1000; ///Calculation VA=Q1/(pi1/4*(DA)^2); //contuity equation for discharge VB=Q1/(pi1/4*(DB)^2); //bernoulli's equation for discharge ///Z=ZB-ZA Z=(PA/(rho*g1))+(VA^2/(2*g1))-(PB/(rho*g1))-(VB^2/(2*g1)); ///Output printf('Contuity equation for discharge= %f m63 \n ',VA); printf('Contuity equation for discharge= %f m^3 \n ',VB); printf('Bernoullis equation for discharge=%f m \n',Z);
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clc; close(); clear(); //page no 506 //prob no. 15.7 Pt=2000; //W Irms=5; Rrad=Pt/Irms^2; mprintf('The radiation resistance is ,Rrad= %i ohm',Rrad);
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// Example 2.3 // Rp=(4+4)||(8+4) Rp=(8*12)/(8+12); // By Voltage divider rule disp(' voltage Across Foue resisrance = '+string(Rp)+' Ohm'); // p 20 2.3
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function [r] = kProximity(ref) // Ouput variables initialisation (not found in input variables) r=[]; // Display mode mode(0); // Display warning for floating point exception ieee(1); //KPROXIMITY Get the value of the proximity sensors of Khepera // //value = kProximity(ref) // Return a vector of 8 values corresponding to the 8 // proximity sensors of Khepera. // Use the reference obtained with kopen. // Written by Skye Legon, 2/99 retries = 3; retry = 1; errmsg = "none yet"; while asciimat(errmsg)&bool2s(retry<=retries) reply = kcmd(ref,"N"); // !! L.16: Matlab function sscanf not yet converted, original calling sequence used [r,count,errmsg] = sscanf(reply,"n,%d,%d,%d,%d,%d,%d,%d,%d"); if errmsg then disp("Error reading sensors. Retrying...") retry = retry+1; end; end; if errmsg then disp("Sensors failed.") r = -1; end; endfunction
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//example 8.7 clc; clear; close; printf("For input J and K = 0 otput Qn+1 = Qn i.e output does not change its state And for J = K = 1, The Output Qn+1 = Qn` i.e output toggles " );
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exec('C:\Users\Julien Guégan\Desktop\PFE\fonctions test.sce',-1) function [xm,n] = evolutionstrategie(f,population,lambda,mu,itermax,tol,C) // μ < λ rand("normal") xm0 = sum(population)./length(population)//barycentre sm0 = sqrt(1/length(population)*sum((population-xm0)^2))//xm0 - max(population) //ecart type moyen // ou utiliser la fonction nanstdev(x) de scilab tau = 1 //facteur d'apprentissage n = 1 //nbr iteration xmp1 = xm0 xm = xm0 smp1 = sm0 while (((n<itermax)&(norm(xm-xmp1)>tol))| (n == 1)) then xm = xmp1 sm = smp1 /////* AFFICHAGE *///// clf() domaine=-20:0.1:20 for i=-20:0.1:20 ind_i=i*10+201 z(ind_i) = f(i) end plot(domaine,z') /////////////////////////////// for i = 1:lambda //creation s(i) = sm * exp(tau*rand()) x(i) = xm + s(i)*rand() if (-20<x(i)&x(i)<20) plot(x(i),f(x(i)),'k.','markersize',4) end end plot([2.9 2.9], [0 300],'g-') //disp('xm0 = '+string(xm)+' ; sm0 = '+string(sm)) Z = 0 for i = 1:length(x) //Une matrice Z = [x f(x) σ] Z(i,1) = x(i) Z(i,2) = f(x(i)) Z(i,3) = s(i) end xp = 0 //vecteur des x parents sp = 0 // les ecarts types associés slct = find(0 < C(Z(:,1)))// ceux qui respectent la contraintes W = 0 for i = 1:length(slct) W(i,1) = Z(slct(i),1) W(i,2) = Z(slct(i),2) W(i,3) = Z(slct(i),3) end mu2 = mu if length(slct) < mu mu2 = length(slct) end for j = 1:mu2 //selection mini = find(W(:,2) == min(W(:,2)))// indice de minf(x) xp(j) = W(mini,1)//vecteur des mu minimum sp(j) = W(mini,3) W(mini,:) = []//on l'enleve pour la prochaine iteration end xmp1 = sum(xp)/mu2 //recombinaison smp1 = sum(sp)/mu2 //autoadaptation plot(xmp1,f(xmp1),'r*') xs2png(gcf(),"image"+string(n)) n = n+1 end endfunction function C = contraintes(x) C = x-3 endfunction //population = -10:10//;-25:0.1:-10]' lambda = 15 mu = 5 fonction = rastr itermax = 200 tol = 0.001 //population initiale rand("uniform") for i = 1:10 population(i) = 28*rand()-14 end plot(population,fonction(population),'k.','markersize',5) [sol n] = evolutionstrategie(fonction,population,lambda,mu,itermax,tol,contraintes) disp('le minimum est x = ') disp(sol) disp(' au bout de '+string(n)+' iterations') /* x=-15:0.1:15 for i=-15:0.1:15 ind_i = i*10+151 z(ind_i) = fonction(i) end plot(x',z,'b-')
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//Area of 1 (in ft^2): A1=0.2; //Area of 2 (in ft^2): A2=0.5; //Area of 3 (in ft^2): A3=0.4; //Area of 4 (in ft^2): A4=0.4; //Density of water (in slug/ft^3): d=1.94; //Mass flow rate out of section 3(in slug/sec): m3=3.88; //Volme flow rate in section 4 (in ft^3/sec): Q4=1; //Velocity at 1(in ft/sec): V1=10;
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//ques-5.31 //Calculating dissociation constant of HCN clc c=0.02//molarity of KCN h=4.9;//percentage of hydrolysis Kw=10^-14; Kh=(h/100)^2*c;//hydrolysis constant Ka=Kw/Kh;//dissociation constant printf("Dissociation constant for HCN is %.2f*10^-10.",Ka*10000000000);
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function [features]=binaryFeatures(featureVectors) // Object for storing Feature Vectors // // Calling Sequence // features = binaryFeatures(featureVector); // // Parameters // featureVector: M-by-N matrix consisting of M features stored in N uint8 containers // features: Binary Feature object for the extracted Features // // Description // This object provides the ability to pass data between extractFeatures and matchFeatures function // // Examples // image_1 = imread('sample1.jpg'); // points_1 = detectFASTFeatures(image_1); // [ f1 vpts_1 ] = extractFeatures(image_1, points_1); // features1 = binaryFeatures(f1); // // Authors // Umang Agrawal // Sridhar Reddy [lhs rhs]=argn(0); if lhs>1 error(msprintf(" Too many output arguments")); elseif rhs>1 error(msprintf(" Too many input arguments")); elseif inttype(featureVectors) <> 11 then error(msprintf("wrong argument #%d: FeatureVectors must be uint8",1)); end [rows cols]=size(featureVectors); features=struct('Type','binaryFeatures','Features',featureVectors,'NumBits',cols*8,'NumFeatures',rows); endfunction
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data=[...]; //ici les donnees [data_tri]=gsort(data,'lr','i'); //permet de trier les donnees t=data_tri(:,1); y=data_tri(:,2); clf; plot(t,y,'ko'); //Levenberg-Marquardt function r=res(theta,n) r=exp(theta(1)+t*theta(2)+t.^2*theta(3))-y; endfunction function J=jac(theta,n) J=ones(n,3); for i=1:n for k=1:3 J(i,k)=t(i)^(k-1)*exp(theta(1)+t(i)*theta(2)+t(i)^2*theta(3)); end end endfunction //Logtrick function return=constrA(t,y,d) n=length(t); A=ones(n,d+1); for i=1:d A(:,i+1)=t.^i; end return=A; endfunction function [theta,reg]=reglin(t,y,d) A=constrA(t,y,d); theta=A\y; reg=A*theta; endfunction //Levenberg-Marquardt n=length(t); theta=[1;-1;-1]; //theta_0 proche de la solution lambda=1; for i=1:50 Jr=jac(theta,n); newtheta=theta-(Jr'*Jr+lambda*eye(3,3))\(Jr'*res(theta,n)) theta=newtheta; end plot(t,exp(theta(1)+t*theta(2)+t.^2*theta(3)),"b"); //Logtrick [theta_bis,reg]=reglin(t,log(y),2); plot(t,exp(reg),"r"); //A la place de Levenberg-Marquardt, on aurait pu utiliser lsqrsolve //theta_0=[1;-1;-1]; //theta_0 proche de la solution //theta_second=lsqrsolve(theta_0,res,length(t),jac); //plot(t,exp(theta_second(1)+t*theta_second(2)+t.^2*theta_second(3)),"m"); title("Approximations par le log-trick et par Levenberg-Marquardt", 'fontsize',3); legend(["Donnees";"Levenberg-Marquardt";"Log-trick"],opt=1);
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function [matrix] = berawgn(EbNo,scheme) funcprot(0); if (string(scheme) == "bpsk") then [nr, nc]= size (EbNo) matrix = zeros(1, nc); for i = 1 : nc var = sqrt(EbNo(1,i)); errorfunc = erfc(var); matrix(1,i) = 1/2*errorfunc; end end endfunction