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//find how current divide in circuit R1=0.02 R2=0.03 I1=(10*R2)/(R1+R2) I2=(10*R1)/(R1+R2) disp('I2='+string(I2)+ 'amps' , 'I1= '+string(I1)+ 'amps')
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Ex7_3.sce
//Example 7.3, Page Number 311 //The Function fpround(dependency) is used to round a floating point number x to n decimal places //Minimum detectable signal clc; A=1000*(10**-6) //Cathode Area in metre square wf=1.25 //Work function in eV T=300 //Cathode temperature in Kelvin e=1.6*(10**-19) //Charge of an electron in Coulombs k=1.38*(10**-23) //Boltzman Constant in meter square kilogram per second square Kelvin a1=1.2*(10**6) //constant for pure metals in Ampere per metre square kelvin square l=0.5*(10**-6) //Wavelength in meters q=0.25 //Quantum Efficiency h=6.63*(10**-34) //Plancks Constant in meter square kilogram per second c=3*(10**8) //Speed of light in meters per second f=1//bandwidth in hertz //From equation 7.11 e1=(k*T)/e e1=fpround(e1,3) c2=(-1*wf)/e1 c2=fpround(c2,4) c3=exp(c2) it=a1*A*(T**2)*c3 //it is the current generated in Amperes mprintf("The Thermionic Emission Current is:%.2e A\n",it) //Using Equation 7.9 r=(q*e*l)/(h*c) //r is the responsivity in A/W r=fpround(r,2) mprintf(" The Responsivity is:%0.1f A/W\n",r) //Using Equation 7.13 W=(sqrt(2*it*e*f))/r //W is the minimum detectable power in Watts mprintf(" The Minimum detectable signal power is:%.3e W",W) //The answer provided in the textbook is wrong
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exa6_4.sce
// Example 6-4 // Root locus clear; clc; xdel(winsid()); //close all windows // please edit the path // cd "/<your code directory>/"; // exec("rootl.sci"); s = %s; D = s*(s + 0.5)*(s^2 + 0.6*s + 10); H = syslin('c',1,D); disp(roots(D),'open loop poles ='); rootl(H,[-6 -6; 6 6],'Root locus of G(s) = 1/(s*(s + 0.5)*(s^2 + 0.6*s + 10)');
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RNA_ANALISE_TECNICA.sce
path_rna_analise_tec = get_absolute_file_path('RNA_ANALISE_TECNICA.sce'); exec( path_rna_analise_tec+"\_util.sce" ); exec( path_rna_analise_tec+"\_carregar_rede_de_treinamento.sce" ); exec( path_rna_analise_tec+"\_arquivo.sce" ); exec( path_rna_analise_tec+"\_dados.sce" ); exec( path_rna_analise_tec+"\Indicador\RNA_INDICADOR.sce" ); //rna_analise_tecnica( 'BBAS3', 52.11600, 0.00000, -138.22420, 239.700000, 377.924200, 15375314249 ) function saida_da_rna = rna_analise_tecnica( nome_do_ativo, ifr, estocastico, hist, macdLine, macdSinal, obv ) ativo = getDados( nome_do_ativo, MAXIMO_LINHA_ARQUIVO ); ifr = ifr/100; estocastico = estocastico/100; hist = normalizar( [ativo(:,3); hist ] ); hist = hist( length(hist) ); alphaHist = convert_to_alpha( [ativo(:,3); hist] ); alphaHist = alphaHist( length(alphaHist) ); macdLine = normalizar( [ativo(:,4); macdLine] ); macdLine = macdLine( length(macdLine) ); alphaMacdLine = convert_to_alpha( [ativo(:,4); macdLine] ); alphaMacdLine = alphaMacdLine( length(alphaMacdLine) ); macdSinal = normalizar( [ativo(:,5); macdSinal] ); macdSinal = macdSinal( length(macdSinal) ); alphaMacdSinal = convert_to_alpha( [ativo(:,5); macdSinal] ); alphaMacdSinal = alphaMacdSinal( length(alphaMacdSinal) ); alphaObv = convert_to_alpha( [ativo(:,6); obv] ); alphaObv = alphaObv( length(alphaObv) ); saida_da_rna = rna_indicador( ifr, estocastico, hist, alphaHist, macdLine, alphaMacdLine, macdSinal, alphaMacdSinal, alphaObv ) endfunction
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3_10.sce
clear; clc; f=1000;l=1000;R=10.4;L=0.00367;G=0.8*(10^-6);C=0.00835*(10^-6);Es=10; //value of Es as taken in solution w=2*%pi*f; Z=R+round((%i*w*L)); Y=G+(%i*w*C); Zo=sqrt(Z/Y); P=sqrt(Z*Y); Is=Es/Zo; Ir=Is*exp(-P*l); P=((abs(Ir))^2)*real(Zo); printf("-Power delivered at receiving end = %f micro-watt",P*(10^6)); //the difference in result is due to erroneous value in textbook. disp("The difference in result is due to erroneous value in textbook")
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ex8_19.sce
// Exa 8.19 clc; clear; close; // Given data GCR= 110;// gas consumption rate in m^3/hour rpm= 300;// round per minute Vs= 0.1;// swept volume of engine in m^3 V_H2=0.50;// in m^3 V_CO= 0.05;// in m^3 V_CH4=0.25;// in m^3 V_CO2= 0.10;// in m^3 V_N2= 0.10;// in m^3 V_O2= 5.8;// in m^3 AirRequired= (0.5*(V_H2+V_CO)+2*V_CH4)/0.21;// in m^3 CO2_formed= V_CO+V_CH4;// in m^3 total_CO2= CO2_formed+V_CO2;// in m^3 N2_of_air= 0.79*AirRequired;// in m^3 total_N2= N2_of_air+V_N2;// in m^3 TotalVolume= total_N2+total_CO2;// in m^3 V= TotalVolume;// in m^3 ExcessAirSupplied= (V_O2*V)/(21-V_O2);// in m^3 TotalAirSupplied= ExcessAirSupplied+AirRequired;// in m^3 AirFuel_ratio= round(TotalAirSupplied)/1; disp(AirFuel_ratio,"Air fuel ratio by volume is : ") // Let V1= Volume of air + gas aspirated per hour V1= GCR*6;// in m^3 Vs_out= Vs*rpm/2*60;// in m^3 Ratio= V1/Vs_out; disp("The value of Ratio i.e.") disp(Ratio,"(Volume of air + gas aspirted per hour)/Volume swept out by piston per hour")
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start.sce
global busdat; global linedat; //Bus data Specifications //Type.... //1 - Slack Bus.. //2 - PV Bus.. //3 - PQ Bus.. // |Bus | Type | Vsp | theta | PGi | QGi | PLi | QLi | Qmin | Qmax | busdat =[ 1 1 1.04 0 0 0 200 200 0 0; 2 3 1 0 50 100 0 0 0 0; 3 2 1.04 0 0 0 150 60 0 150;]; //Line Data Specifications // | From | To | R | X | B/2 | // | Bus | Bus | | | | linedat= [ 1 2 0.02 0.08 0.01 ; 1 3 0.02 0.08 0.01 ; 2 3 0.02 0.08 0.01 ;]; getd .;
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ch4ex12.sce
//signals and systems //Unilateral Laplace Transform:Solving Differential Equation //example 4.12 s = %s; syms t; [A] = pfss((3*s+3)/((s+5)*(s^2+5*s+6))); F1 = ilaplace(A(1),s,t) F2 = ilaplace(A(2),s,t) F3 = ilaplace(A(3),s,t) F = F1+F2+F3 disp(F)
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clear; clc; printf("\nEx1.28\n"); //page no.-36 //given E=3.76*10^-17;............//kinrtic energy of e- in joule n=1;......................//order theta=9.20694;............//glancing angle h=6.625*10^-34;...........//planck constant m=9.1*10^-31;...........//mass of electron //from de-broglie relationship lambda=h/sqrt(2*m*E).......//wavelength //from bragg's law d=(n*lambda)/(2*sind(theta)).........//interplanar spacing in m printf("\ninterplanar spacing is 2.52 angstrom\n");
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clc; clear; myu_e=1.553 //refractive index myu_0=1.542 //refractive index lambda=5.5*10^-5//wavelength in m //calculation for minimum thickness i.e half wave plate d=lambda/(2*(myu_e-myu_0)) mprintf("The thickness of the plate is = %1.1e m",d)
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function A=remplir1(M,r,sig,Dt) for i=1:M for j=1:M if i==j then A(i,j)=-r*i+(((sig*i)^2)/2) elseif j == i-1 then A(i,j)=(-r+1/Dt+r*i-(sig*i)^2) elseif j == i+1 then A(i,j)=(sig*i)^2 else A(i,j)=0 end end end endfunction function B=remplir2(K,L,M) for i=1:M B(i)=max(K-i*(L/(M+1)),0) end endfunction function C=remplir3(r,T,M,sig,n,N,K) C=zeros(M,1); C(1,1)=(-r+((sig^2)/2))*K*(exp(r*(n*T/N-T))); endfunction function Pn=final(r,T,M,sig,N,Dt,K,L) A=remplir1(M,r,sig,Dt); Pn=remplir2(K,L,M); for n=N:-1:1 C=remplir3(r,T,M,sig,n,N,K); Pn=A*Pn+C; end endfunction
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vmm_offc.sce
//**************************** vmm_offc ********************************** if (blk_name.entries(bl) == "vmm_offc") then for ss=1:scs_m.objs(bl).model.ipar(1) mputl("# vmm_offc "+string(bl)+" "+string(scs_m.objs(bl).model.ipar(2))+" "+string(ss),fd_w); sci2blif_str= ".subckt vmm_offc"+" in[0]=net"+string(blk(blk_objs(bl),2))+"_"+string(ss)+" in[1]=net"+string(blk(blk_objs(bl),3))+"_"+string(ss)+" in[2]=net"+string(blk(blk_objs(bl),4))+"_"+string(ss)+" in[3]=net"+string(blk(blk_objs(bl),5))+"_"+string(ss)+" in[4]=net"+string(blk(blk_objs(bl),6))+"_"+string(ss)+" in[5]=net"+string(blk(blk_objs(bl),7))+"_"+string(ss)+" in[6]=net"+string(blk(blk_objs(bl),8))+"_"+string(ss)+" in[7]=net"+string(blk(blk_objs(bl),9))+"_"+string(ss)+" in[8]=net"+string(blk(blk_objs(bl),10))+"_"+string(ss)+" in[9]=net"+string(blk(blk_objs(bl),11))+"_"+string(ss)+" in[10]=net"+string(blk(blk_objs(bl),12))+"_"+string(ss)+" in[11]=net"+string(blk(blk_objs(bl),13))+"_"+string(ss)+" in[12]=net"+string(blk(blk_objs(bl),14))+"_"+string(ss)+" out[0]=net"+string(blk(blk_objs(bl),2+numofip))+"_"+string(ss)+" out[1]=net"+string(blk(blk_objs(bl),3+numofip))+"_"+string(ss)+" #vmm_offc_ls =0"+"&vmm_offc_w16n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(1-1)+ss)))+"&vmm_offc_w26n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(2-1)+ss)))+"&vmm_offc_w16p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(3-1)+ss)))+"&vmm_offc_w26p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(4-1)+ss)))+"&vmm_offc_w15n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(5-1)+ss)))+"&vmm_offc_w25n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(2+4-1)+ss)))+"&vmm_offc_w15p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(3+4-1)+ss)))+"&vmm_offc_w25p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(4+4-1)+ss)))+"&vmm_offc_w14n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(5+4-1)+ss)))+"&vmm_offc_w24n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(6+4-1)+ss)))+"&vmm_offc_w14p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(7+4-1)+ss)))+"&vmm_offc_w24p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(8+4-1)+ss)))+"&vmm_offc_w13n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(9+4-1)+ss)))+"&vmm_offc_w23n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(10+4-1)+ss)))+"&vmm_offc_w13p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(11+4-1)+ss)))+"&vmm_offc_w23p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(12+4-1)+ss)))+"&vmm_offc_w12n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(13+4-1)+ss)))+"&vmm_offc_w22n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(14+4-1)+ss)))+"&vmm_offc_w12p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(15+4-1)+ss)))+"&vmm_offc_w22p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(16+4-1)+ss)))+"&vmm_offc_w11n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(17+4-1)+ss)))+"&vmm_offc_w21n ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(18+4-1)+ss)))+"&vmm_offc_w11p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(19+4-1)+ss)))+"&vmm_offc_w21p ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(20+4-1)+ss)))+"&vmm_offc_o2_fgibias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(21+4-1)+ss)))+"&vmm_offc_o1_fgibias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(22+4-1)+ss)))+"&vmm_offc_o1_ibias =10e-6"+"&vmm_offc_o2_ibias =10e-6"+"&vmm_offc_o2_pbias =3e-9"+"&vmm_offc_o2_nbias =3e-9"+"&vmm_offc_o1_pbias =3e-9"+"&vmm_offc_o1_nbias =3e-9"+"&vmm_offc_off1_ibias =10e-9"+"&vmm_offc_off2_ibias =10e-9" //+"&vmm_offc_row2_ibias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(21-1)+ss)))+"&vmm_offc_row1_ibias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(22-1)+ss)))+"&vmm_offc_row2_fgbias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(23-1)+ss))) //+"&vmm_offc_row1_fgbias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(24-1)+ss))) //+"&vmm_offc_row2_pbias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(25-1)+ss))) //+"&vmm_offc_row1_pbias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(26-1)+ss))) //+"&vmm_offc_row2_nbias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(27-1)+ss))) //+"&vmm_offc_row1_nbias ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(28-1)+ss))) //+"&vmm_offc_row2_cap ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(29-1)+ss))) //+"&vmm_offc_row1_cap ="+string(sprintf('%e',scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(30-1)+ss))) mputl(sci2blif_str,fd_w); mputl(" ",fd_w); if scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(23+4-1)+1) == 1 then plcvpr = %t; plcloc=[plcloc;'net'+string(blk(blk_objs(bl),2+numofip))+"_"+string(ss),string(scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(23+4-1)+1+2*ss-1))+' '+string(scs_m.objs(bl).model.rpar(scs_m.objs(bl).model.ipar(1)*(23+4-1)+1+2*ss))+' 0']; end end end
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//CHAPTER 10- THREE-PHASE INDUCTION MACHINES //Example 5 clc; disp("CHAPTER 10"); disp("EXAMPLE 5"); //VARIABLE INITIALIZATION P1=12; //number of poles of alternator N_s1=500; //synchronous speed of 12-pole alternator in rpm P2=8; //number of poles of motor s=0.03; //slip of the motor in p.u. //SOLUTION f=(N_s1*P1)/120; N_s2=(120*f)/P2; //synchronous speed of 8-pole alternator in rpm N_r=N_s2*(1-s); N_r=round(N_r); //to round off the value disp(sprintf("The speed of the motor is %d rpm",N_r)); //END
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Ex3_12.sce
// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM // Example : 3.12 clc;clear; // clears the console and command history // Given data kVA = 25 // kVA ratings of transformer V1 = 2200 // primary side voltage in V V2 = 220 // secondary side voltage in V V_1 = 40 // voltage at high voltage side in V I_1 = 5 // current at high voltage side in A P = 150 // power at high voltage side in W // caclulations Z_01 = V_1/I_1 // reactance to primary sidec in ohm R_01 = P/I_1^2 // resistance to primary side in ohm phi = acosd(R_01/Z_01) // power factor angle X_01 = Z_01*sind(phi) // impedance to primary side in ohm a = V1/V2 // turn ratio Z_02 = Z_01/a^2 // reactance to secondary side in ohm R_02 = R_01/a^2 // resistance to secondary side in ohm X_02 = X_01/a^2 // impedance to secondary side in ohm I_2 = kVA*10^3/V2 // secondary side current in A E_2 = V2+I_2*Z_02 // secondary induced voltage in V VR = ((E_2-V2)/V2)*100 // voltage regulation // display the result disp("Example 3.12 solution"); printf(" \n Resistance to primary side \n Z_01 = %.2f ohm \n", Z_01); printf(" \n Resistance to primary side \n R_01 = %.1f ohm \n", R_01); printf(" \n Impedance to primary side \n X_01 = %.2f ohm \n", X_01); printf(" \n Reactance to secondary side \n Z_02 = %.2f ohm \n", Z_02); printf(" \n Resistance to secondary side \n R_02 = %.2f ohm \n", R_02); printf(" \n Impedance to secondary side \n X_02 = %.3f ohm \n", X_02); printf(" \n oltage regulation \n VR = %.0f percent \n", VR);
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clear // Def: Temperatura ambiente, Unidad: Grados Kelvin Tamb=283 // Def: Paso de integración Delta t para el método de Euler, Unidad: s dt=0.1 // Def: Tiempo inicial para el método de Euler, Unidad: s t0=0 // Def: Tiempo final para el método de Euler, Unidad: s tf=3600 // Def: Conductividad términa, Unidad: W/m*K k=0.6 // Def: Lado de la sala cúbica, Unidad: m l=4 // Def: Ancho de pared, Unidad: m d=0.25 // Def: Masa del aire, Unidad: Kg m=76.8 // Def: Calor específico del aire, Unidad: J/Kg*K Ca=1012 // Def: Voltage inicial, Unidad: V V0=100 // Def: Resistencia, Unidad: Ohm R=1 // Def: Unidad: rad/s w=0.02 // Def: Unidad: 1/s a=0.0035 // Def: Gamma de la fórmula de Calor gma=(k*5*(l^2))/(d*m*Ca) // Def: Temperatura en t inf (analítico), Unidad: Kelvin TinfA=326.4027 function [t,T] = ObtenerTemperaturaPorEuler(); // Condiciones iniciales t(1)=t0 T(1)=Tamb TinfEncontrado = %F TinfE=0 tinfE=0 U=0 i=1 while (t(i)<=tf) && ~TinfEncontrado u=(1/(m*Ca))*((((V0^2)*((1-(%e^((-a)*t(i)))*cos(w*t(i)))^2))/R)+(Tamb*((k*5*(l^2))/d))) T(i+1)=T(i)+(u-gma*T(i))*dt Tdif=(abs(TinfA-T(i)))/(abs(TinfA-T(1))) U=U+((((V0^2)*((1-(%e^((-a)*t(i)))*cos(w*t(i)))^2))/R)*dt) disp(U) if(Tdif <= (1/100)) then TinfEncontrado = %T; TinfE=T(i) tinfE=t(i) disp("Tiempo a esperar para que la temperatura llegue a estado de régimen estacionario (en forma numérica):") disp(tinfE) disp("Temperatura en régimen estacionario (en forma numérica):") disp(TinfE) disp("Gasto de energía eléctrica necesario para llegar al régimen de estado estacionario(en forma numérica):") disp(U) end t(i+1)=t(i)+dt i=i+1 end endfunction // Obtenemos los puntos por Metodo de Euler [t,T] = ObtenerTemperaturaPorEuler()
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clc; //Example 12.1 //Page No 505 disp("Given: A D/r ratio of 12.22"); //solution dr=12.22; disp("Susbstituting into equation 12-14(refer pgno 505), we obtain "); Z0=276*log10(dr); disp('Ohm',round(Z0),"Z0 = ");
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clc //initialisation of variables G= 145 //cal R= 1.987 //cal/mole K T= 95 //C //CALCULATIONS P= 10^(-G/(2.303*R*(273+T)))*(624/0.820) //RESULTS printf (' vapour pressure= %.f atm',P)
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// x=[8 1 6;3 5 7;4 9 2]; dim=2; [x,perm,nshifts] = shiftdata(x,dim); disp(x); disp(perm); disp(nshifts); //output //8. 3. 4. // 1. 5. 9. // 6. 7. 2. // // 2. 1. // // [] //
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clear //The force of 3000 N acts along line AB. Let AB make angle alpha with horizontal. // //variable declaration F=3000 //force in newtons,'N' BC=80 //length of crank BC, 'mm' AB=200 //length of connecting rod AB ,'mm' theta=60*%pi/180 //angle b/w BC & AC //calculations alpha=asin(BC*sin(theta)/200)*180/%pi HC=F*cos(alpha*%pi/180) //Horizontal component VC= F*sin(alpha*%pi/180) //Vertical component //Components along and normal to crank //The force makes angle alpha + 60 with crank. alpha2=alpha+60 CAC=F*cos(alpha2*%pi/180) // Component along crank CNC= F*sin(alpha2*%pi/180) //Component normal to crank printf("\n horizontal component= %0.1f N",HC) printf("\n Vertical component = %0.1f N",VC) printf("\n Component along crank = %0.1f N",CAC) printf("\n Component normal to crank= %0.1f N",CNC)
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clc m=15; //kg/s v=0.45; //m^3/kg P=12000; //kW W=P/m; //kJ/kg h1=1260; //kJ/kg h2=400; //kJ/kg C1=50; //m/s C2=110; //m/s disp("(i) Heat rejected = ") Q=h2-h1+(C2^2-C1^2)/2/10^3 +W; Qnet=m*Q; disp("Qnet=") disp(-Qnet) disp("kW") disp("(ii) Inlet area") A=v*m/C1; disp("A=") disp(A) disp("m^2")
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//caption:root_locus_and_gain,phase_margin //example 12.51 //page 580 s=%s; K=3.46 G=K/(s*(s+1)*(s+2)) G=syslin('c',G) clf(); evans(G,20) xgrid(2) [gm,freq_gm]=g_margin(G) [pm,freq_pm]=p_margin(G) disp(gm,"gain_margin=",freq_gm*2*%pi,"gain_margin_freq=") disp(pm,"phase_margin=",freq_pm*2*%pi,"phase_margin_freq=")
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clc clear disp("example 10 13") a1=2000;b1=20;c1=0.05;p1=350;p2=550 a2=2750;b2=26;c2=0.03091 function [co]=cost(a,b,c,p) co=a+b*p+c*p^2 endfunction disp("(a)") toco=cost(a1,b1,c1,p1)+cost(a2,b2,c2,p2) printf("total cost when each system supplies its own load Rs%.3f per hour",toco) l=p1+p2 p11=(b2-b1+2*c2*l)/(2*(c1+c2)) p22=l-p11 totco=cost(a1,b1,c1,p11)+cost(a2,b2,c2,p22) sav=toco-totco tilo=p11-p1 disp("(b)") printf("\n total cost when load is supplied in economic load dispatch method Rs%d per hour \n saving %.3f \n tie line load %.3f MW",totco,sav,tilo)
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clear;lines(0); 3^(-1) x=poly(0,"x"); // (x+10)/2 i3=eye(3,3) // a=[1 2 3;4 5 6;7 8 9],a(1,3),a([1 3],:),a(:,3) a(:,3)=[] a(1,$)=33 a(2,[$ $-1]) a(:,$+1)=[10;11;12] // w=ssrand(2,2,2);ssprint(w) ssprint(w(:,1)) ss2tf(w(:,1)) // l=list(1,2,3,4) [a,b,c,d]=l(:) l($+1)='new' // v=%t([1 1 1 1 1]) // [x,y,z]=(1,2,3)
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//scilab 5.4.1 //Windows 7 operating system //chapter 17 Number Systems,Boolean Algebra,and Digital Circuits clc clear x='11111' y='1011' z='101' w='10' v='1' s1=bin2dec(x) s2=bin2dec(y) s3=bin2dec(z) s4=bin2dec(w) s5=bin2dec(v) a=s1+s2+s3+s4+s5 b=dec2bin(a) disp(,b,"Binary addition of 11111+1011+101+10+1 is ") disp(,a,"Decimal equivalent corresponding to above binary addition is ")
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// Exa 5.2 format('v',7); clc; clear; close; // Given data Af= 100;// unit less Vi= 50;// in mV Vi= Vi*10^-3;// in V Vs= 0.5;// in V // Formula Af= Vo/Vs Vo= Af*Vs;// in V A= Vo/Vi; disp(A,"Value of A is : ") // Formula Af= A/(1+B*A) B= 1/Af-1/A; B=B*100;// in % disp(B,"Value of B is in percent : ")
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tic ok = writebincom(handl,[255]); temp = readbincom(handl,2) toc
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load BoothMultiplier.hdl, set reset 1, set initM %D101, set initQ %D34; tick, tock, set reset 0; repeat 100 { tick, tock; }
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//Calculation mistake in book //ques2 //Standard brayton cycle clc clear //Calculation mistake in book //1-Inlet for compressor //2-Exit for compressor //T-Temperature at a state //P-Pressure at a state T1=288.2;//K P2=1000;//kPa P1=100;//kPa k=1.4; T2s=T1*(P2/P1)^(1-1/k);//K nc=.80;//Compressor Efficiency T2=T1+(T2s-T1)/0.80; Cp=1.004;//Specific heat at constant pressure in kJ/kg wc=Cp*(T2-T1);//compressor work in kJ/kg; printf('Temperature T2 = %.1f K\n',T2); printf(' Compressor work = %.1f kJ/kg \n',wc); //3-Turbine Inlet //4-Turbine Exit P4=P1; P3=P2; T3=1373.2;//K T4s=T3*(P4/P3)^(1-1/k);//K nt=0.85;//turbine Efficiency T4=T3-(T3-T4s)*0.85; wt=Cp*(T3-T4); wnet=wt-wc; printf(' Temperature T3 = %.1f K\n',T3); printf(' Temperature T4 = %.1f K\n',T4); printf(' Turbine work = %.1f kJ/kg\n',wt); printf(' Net work = %.1f kJ/kg\n',wt-wc); //2-Also high temperature heat exchanger Inlet //3-(-do-) Exit qh=Cp*(T3-T2);//Heat of source in kJ/kg //4-high temp heat exchanger inlet //1-(-do-) Exit ql=Cp*(T4-T1);//Heat of sink in kJ/kg nth=wnet/qh; printf(' Thermal Efficiency of cycle = %.1f percent',nth*100);
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clc disp("example 6.37") printf("\n") disp("calculate voltage gain,input resistance,current through R1") printf("Given") disp("Rf=100k,R1=10k") disp("input voltage is 0.5v") Rf=10^5 R1=10^4 Af=-Rf/R1 Rif=R1 Vi=0.5 I1=(Vi/R1) printf("closed loop voltage gain is %3.1f\n",Af) printf("input resistance is\n %3.1f ohm\n",Rif) printf("current flowing through R1 is %f ampere\n",I1)
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//Example 1.12 //Program to Compute convolution of given sequences //x(n)=[3 2 1 2], h(n)=[1 2 1 2]; clear; clc ; close ; x=[3 2 1 2]; h=[1 2 1 2]; y=convol(x,h); disp(y);
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clear // Variable declaration T_d1=24// The dry bulb temperature in °C T_d2=7// The dry bulb temperature in °C H=45// % saturation cf=0.78// Contact factor h_1=45.85// The enthalpy in kJ/kg h_2=22.72// The enthalpy in kJ/kg // Calculation //(a) By construction on the chart ( Figure 21.9 ), 10.7°C dry bulb, 85% saturation. //(b) By calculation, the dry bulb will drop 78% of 24 to 7°C: dT=T_d1-(cf*(T_d1-T_d2))// The drop in dry bulb temperature in °C dh=h_1-(cf*(h_1-h_2))// The drop in enthalpy in kJ/kg printf("\n \nThe drop in dry bulb temperature=%2.1f°C \nThe drop in enthlpy=%2.2f kJ/kg",dT,dh)
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clc; vp=10; //peak voltage v=vp*sqrt(2); //voltage hc=10+7.07; //horizontal components disp(hc,"Hrizontal Components = "); //horizontal components vc=sqrt((hc*hc)+(7.07*7.07)); //vertical components disp(vc,"Vertical Components = "); //vertical components
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// 08.05.18 Koshikawa // 08.05.19 Changed // 08.00.04 // 09.11.07 for the same curve function KL=IntersectcrvsPp(varargin) // Modified Nargs=length(varargin); G1=varargin(1); G2=varargin(2); Eps=10.0^(-4); if Nargs>2 Eps=varargin(3) end SqEps=10.0^(-10); Eps2=0.1; if Nargs>3 Eps2=varargin(4) end Data1=G1; Data2=G2; if size(Data1,1)==size(Data2,1) Tmp1=Data2(size(Data2,1):-1:1,:); // 09.11.07 Eps0=10^(-6); Tmp2=norm(Data1-Data2); Tmp3=norm(Data1-Tmp1); if Tmp2<Eps0|Tmp3<Eps0 KL=list(); return; end; // end; KL1=[]; KL2=[]; for I=1:size(Data1,1)-1 A=Data1(I,:); B=Data1(I+1,:); for J=1:size(Data2,1)-1 P=Data2(J,:); Q=Data2(J+1,:); Tmp=Koutenseg(A,B,P,Q,Eps,Eps2); if Tmp~=[%inf,-%inf] if Op(3,Tmp)==0 Tmp1=MixS(Op(1,Tmp),Op(2,Tmp),I,J); KL1=Mixadd(KL1,Tmp1); else Tmp2=MixS(Op(1,Tmp),Op(2,Tmp),I,J); KL2=Mixadd(KL2,Tmp2); end end end end KL=[]; if Mixlength(KL1)>0 Tmp=Op(1,KL1); P=Op(1,Tmp); T=Op(2,Tmp); I=Op(3,Tmp); J=Op(4,Tmp); Tmp=MixS(P,I+T,J); KL=MixL(Tmp); end for N=2:Mixlength(KL1) Tmp=Op(N,KL1); P=Op(1,Tmp); Flg=0; for K=1:Mixlength(KL) Tmp=Op(K,KL); if Vecnagasa2(P-Op(1,Tmp))<SqEps Flg=1; break end end if Flg==0 Tmp=Op(N,KL1); T=Op(2,Tmp); I=Op(3,Tmp); J=Op(4,Tmp); Tmp=MixS(P,I+T,J); KL=Mixadd(KL,Tmp); end end for N=1:Mixlength(KL2) Tmp=Op(N,KL2); P=Op(1,Tmp); Flg=0; for K=1:Mixlength(KL) Tmp=Op(K,KL); if Vecnagasa2(P-Op(1,Tmp))<SqEps Flg=1; break end end if Flg==0 Tmp=Op(N,KL2); T=Op(2,Tmp); T=min(max(T,0),1); I=Op(3,Tmp); J=Op(4,Tmp); Tmp=MixS(P,I+T,J); KL=Mixadd(KL,Tmp); end end endfunction
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//Exa 2.45 clc; clear; close; //Given data : format('v',6); f=50;//in Hz P=6;//no. of poles phase=3;//no. of phase R2=0.2;//rotor resistance per phase in ohm N1=960;//Full load speed in rpm Ns=120*f/P;//in rpm S1=(Ns-N1)/Ns;//Full load slip(unitless) N2=N1*(1-10/100);//New speed in rpm(reduced 10%) S2=(Ns-N2)/Ns;//New slip(unitless) //Formula : S=RotorCuLoss/Pin_rotor=3*I2^2*R2/Pin_rotor //Let the additional resistance is R R=R2*S2/S1-R2;//Resistance to be added in ohm disp(R,"Additional Rotor Resistance(in ohm) : ");
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//appeler les scripts nécessaires pour pouvoir utiliser toutes les méthodes exec('exo5_initialise_grille.sci',-1) exec('exo9_modele_complet.sci',-1) exec('exo10_correctif.sci',-1) exec('exo12_splitting.sci',-1) exec('exo16_splitting_problem.sci',-1) //Cout exo5(initialisation de grille) val=0 T=100 //T t0=0 //Temps initial dt=0.01 //Pas de temps e0=1.0 //e initial r0=0.0 //r initial for n=2:2:4 timer() main_initialise(T,t0,dt,e0,r0,n) time1=timer() //--tableau de temps de calcul selon n (n = 0,2,4,6) val= [val, time1] end //Cout exo9(problème modèle complet) val=0 D=1 //Constant conductivité T=50 //T t0=0 //Temps initial dt=0.01 //Pas de temps for n=2:2:4 //--les vecteurs (condition) initiaux de taille n*n e0=ones(n*n,1) r0=zeros(n*n,1) timer() main_modele_complet(t0,dt,T,e0,r0,D,n) time1=timer() //--tableau de temps de calcul selon n (n = 0,2,4,6) val= [val, time1] end //Cout exo10(problème modèle complet avec fonctions correctives) val=0 D=1 //Constant conductivité T=50 //T t0=0 //Temps initial dt=0.01 //Pas de temps for n=2:2:4 //--les vecteurs (condition) initiaux de taille n*n : //--Ici, c'est de la solution corrigé au t0 pour tester [x,y]=genere_xy(n) [e0,r0]=grille_solution(x,y,t0) timer() main_correctif(t0,dt,T,e0,r0,D,n) time1=timer() //--tableau de temps de calcul selon n (n = 0,2,4,6) val= [val, time1] end //Cout exo12(problème modèle complet avec splitting façon 1) val=0 D=1 //Constant conductivité T=50 //T t0=0 //Temps initial dt=0.01 //Pas de temps for n=2:2:4 //--les vecteurs (condition) initiaux de taille n*n e0=ones(n*n,1) r0=zeros(n*n,1) timer() main_splitting(t0,dt,T,e0,r0,D,n) time1=timer() //--tableau de temps de calcul selon n (n = 0,2,4,6) val= [val, time1] end //Cout exo16(problème modèle complet avec splitting façon 2 avec rk2,cn2) val=0 D=1 //Constant conductivité T=50 //T t0=0 //Temps initial dt=0.01 //Pas de temps for n=2:2:4 //--les vecteurs (condition) initiaux de taille n*n e0=ones(n*n,1) r0=zeros(n*n,1) timer() main_splitting_problem(t0,dt,T,e0,r0,D,n) time1=timer() //--tableau de temps de calcul selon n (n = 0,2,4,6) val= [val, time1] end
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clear // //variable declaration PA=100.0 //inclined up loading at 60° at A, N PB1=80.0 //Vertical down loading at B,N PB2=80.0 //Horizontal right loading at at B,N PC=120.0 //inclined down loading at 30° at C,N thetaA=60.0*%pi/180.0 thetaB=30.0*%pi/180.0 //Taking horizontal direction towards left as x axis and the vertical downward direction as y axis. ////sum of vertical Fy & sum of horizontal forces Fx is zero //Assume direction of Fx is right //Assume direction of Fy is up Fx=PB2-PA*cos(thetaA)-PC*cos(thetaB) Rx=-Fx Fy=PB1+PC*sin(thetaB)-PA*sin(thetaA) Ry=Fy R=sqrt((Rx**2)+(Ry**2)) printf("\n R= %0.2f KN",R) alpha=atan(Fy/Fx)*180/%pi printf("\n alpha= %0.2f °",(-alpha)) //Let x be the distance from A at which the resultant cuts AC. Then taking A as moment centre, x=(PB1*100*sin(thetaA)+PB2*50+PC*sin(thetaB)*100)/Ry printf("\n x= %0.3f mm",x)
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example7_2.sce
clear; clc; //Example7.2[Cooling of a Hot Block by Forced Air at High Elevation] //Given:- ReC=5*10^5;//critical Reynolds number v=8;//Velocity of air[m/s] T_air=20;//Initial Temperature of air[degree Celcius] T_plate=140;//Temperature of flat plate[degree Celcius] T_film=(T_air+T_plate)/2;//Film Temperature of air[degree Celcius] //Properties of air at film temperature[degree Celcius] k=0.02953;//[W/m.degree Celcius] Pr=0.7154;//Prandtl Number nu=2.097*10^(-5);//Kinematic Viscosity at 1 atm Pressure [m^2/s] nu_ac=nu/0.823;//Kinematic viscosity at pressure 0.823 atm[m^2/s] //Solution(a) L1=6;//Characteristic length of plate along the flow of air[m] w1=1.5;//width[m] ReL1=(v*L1)/nu_ac;//Reynolds number if(ReL1&gt;ReC) then, disp("Flow is not laminar") //We have average Nusselt Number Nu1=((0.037*(ReL1^(0.8)))-871)*(Pr^(1/3)); disp(ceil(Nu1),"Nusselt Number is") h1=k*Nu1/L1;//[W/m^2.degree Celcius] As1=w1*L1;//Flow Area of plate[m^2] Q1=h1*As1*(T_plate-T_air); disp("W",Q1,"Heat Flow Rate is") else, disp("Flow is laminar") end //Solution(b) L2=1.5;//Characteristic length of plate along flow of air[m] ReL2=v*L2/nu_ac;//Reynolds Number if(ReL2<ReC) then,="" disp("flow="" is="" laminar")="" nu2="0.664*(ReL2^(0.5))*(Pr^(1/3));" disp(ceil(nu2),"nusselt="" number="" is")="" h2="k*Nu2/L2;//[W/m^2.degree" celcius]="" q2="h2*As1*(T_plate-T_air);" disp("w",ceil(q2),"the="" heat="" transfer="" rate="" else,="" turbulent")="" end="" <="" div=""></rec)>
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<?xml version="1.0" encoding="UTF-8" standalone="yes"?> <AutoTest version="2.0.0" wavetype="15"> <Title>Test case 2-S2 negates S1</Title> <Organization>Volkswagen</Organization> <Standard>VW 80000 2013</Standard> <Item>6.10 E-10 Short interruptions</Item> <system> <PowerSystem>3</PowerSystem> <voltage>11</voltage> </system> <forminterrupt> <count>1</count> <dischargetype>1</dischargetype> <linetype>0</linetype> <interrupt id="0"> <interrupttype type="5"> <grouptime objectname="t1" value="10" index="0"/> <grouptime objectname="t2" value="15" index="2"/> <grouptime objectname="t3" value="90" index="0"/> <grouptime objectname="dt" value="10" index="0"/> </interrupttype> <switchLines index="0"/> </interrupt> <interrupt id="1"> <interrupttype type="5"> <grouptime objectname="t1" value="100" index="0"/> <grouptime objectname="t2" value="15" index="2"/> <grouptime objectname="t3" value="900" index="0"/> <grouptime objectname="dt" value="100" index="0"/> </interrupttype> <switchLines index="0"/> </interrupt> <interrupt id="2"> <interrupttype type="5"> <grouptime objectname="t1" value="1" index="1"/> <grouptime objectname="t2" value="15" index="2"/> <grouptime objectname="t3" value="9" index="1"/> <grouptime objectname="dt" value="1" index="1"/> </interrupttype> <switchLines index="0"/> </interrupt> <interrupt id="3"> <interrupttype type="5"> <grouptime objectname="t1" value="10" index="1"/> <grouptime objectname="t2" value="15" index="2"/> <grouptime objectname="t3" value="90" index="1"/> <grouptime objectname="dt" value="10" index="1"/> </interrupttype> <switchLines index="0"/> </interrupt> <interrupt id="4"> <interrupttype type="5"> <grouptime objectname="t1" value="100" index="1"/> <grouptime objectname="t2" value="15" index="2"/> <grouptime objectname="t3" value="2000" index="1"/> <grouptime objectname="dt" value="100" index="1"/> </interrupttype> <switchLines index="0"/> </interrupt> </forminterrupt> </AutoTest>
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Ex5_8.sce
//Fourier Series coefficients of half-wave rectifier output //Assume the period of the signal T=1 t=-0.5:0.01:0.5; for i=1:length(t) if t(i)<-0.25 & t(i)>0.25 then x(i)=-1; else x(i)=1; end end k=-10:10; for i=1:length(k) if k(i)==0 then ak(i)=0; else ak(i)=(%i*((2-(-1)^k(i))*exp(-%i*k(i)*%pi/2)-exp(%i*k(i)*%pi/2)))/(k(i)*2*%pi); end end disp("The fourier series coefficients are...") disp(ak) plot(k,ak,'.') xtitle("Fourier Coefficients","k","ak")
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atanhf.tst
; atanhf.tst ; ; Copyright (c) 2009-2023, Arm Limited. ; SPDX-License-Identifier: MIT OR Apache-2.0 WITH LLVM-exception func=atanhf op1=7fc00001 result=7fc00001 errno=0 func=atanhf op1=ffc00001 result=7fc00001 errno=0 func=atanhf op1=7f800001 result=7fc00001 errno=0 status=i func=atanhf op1=ff800001 result=7fc00001 errno=0 status=i func=atanhf op1=7f800000 result=7fc00001 errno=EDOM status=i func=atanhf op1=ff800000 result=7fc00001 errno=EDOM status=i func=atanhf op1=3f800001 result=7fc00001 errno=EDOM status=i func=atanhf op1=bf800001 result=7fc00001 errno=EDOM status=i func=atanhf op1=3f800000 result=7f800000 errno=ERANGE status=z func=atanhf op1=bf800000 result=ff800000 errno=ERANGE status=z func=atanhf op1=00000000 result=00000000 errno=0 func=atanhf op1=80000000 result=80000000 errno=0 ; No exception is raised with certain versions of glibc. Functions ; approximated by x near zero may not generate/implement flops and ; thus may not raise exceptions. func=atanhf op1=00000001 result=00000001 errno=0 maybestatus=ux func=atanhf op1=80000001 result=80000001 errno=0 maybestatus=ux
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Ex19_7.sce
//chapter19 //example19.7 //page422 Pc=500 // W m=1 Ps=0.5*m^2*Pc Pt=Pc+Ps printf("sideband power = %.3f W \n",Ps) printf("power of modulated wave = %.3f W \n",Pt)
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Ex18_22.sce
//Variable declaration: //From example 18.21: m = 144206 //Mass flow rate of flue gas (lb/h) cp = 0.3 //Average heat capacities of the flue gas (Btu/lb F) T1 = 2050 //Initial temperature of gas ( F) T2 = 180 //Final temperature of gas ( F) T3 = 60 //Ambient air temperature ( F) U = 1.5 //Overall heat transfer coefficient for cooler (Btu/h.ft^2. F) MW = 28.27 //Molecular weight of gas R = 379 //Universal gas constant (psia.ft^3/lbmol. R) v = 60 //Duct or pipe velcity at inlet (2050 F) (ft/s) pi = %pi //Calculation: Q = m*cp*(T1-T2) //Heat duty (Btu/h) DTlm = ((T1-T3)-(T2-T3))/log((T1-T3)/(T2-T3)) //Log-mean temperature difference ( F) A1 = round(Q * 10**-5)/10**-5/(U*round(DTlm)) //Radiative surface area (ft^2) q = m*R*(T1+460)/(T3+460)/MW //Volumetric flow at inlet (ft^3/h) A2 = q/(v*3600) //Duct area (ft^2) D = sqrt(A2*4/pi) //Duct diameter (ft) L = A1/(pi*D) //Length of required heat exchange ducting (ft) A1 = round(A1*10**-1)/10**-1 //Result: printf(" The radiative surface area required is : %f ft^2 .",A1) printf(" The length of required heat exchange ducting is : %.0f ft .",L)
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ex17_1.sce
//Ex17.1 Del_V_out=0.25; V_out=15; Del_V_in=5; //All voltages in Volts line_regulation=((Del_V_out/V_out)/Del_V_in)*100; disp(line_regulation,'line regulation in %/V')
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Ex12_6.sce
//Chapter 12 Thermodynamics Thermodynamic chemistry clc; clear; //Initialisation of Variables Cp= 2.7 //cal per mole per degree CP1= 6.9 //cal per mole per degree Cp2= 15.4 //cal per mole per degree H= -20.24 //kcal T= 200 //C T1= 25 //C //CALCULATIONS H1= H+(Cp2-2*Cp-3*CP1)*((T-T1)/1000) //RESULTS mprintf("Heat of formation= %.2f cal",H1)
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putratings.sci
function ratingmatrix = putratings(ratings) ratingmatrix = zeros(max(ratings(:,1)),max(ratings(:,2))); for i=1:size(ratings,1) ratingmatrix(ratings(i,1),ratings(i,2)) = ratings(i,3); end endfunction
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CandidateSelector expand width=4 base=5 exponent=3 left=4 right=0 fileName=data/euler313.man chain2 [[0,2,-3,-1],[0,-3,3,-3],[-3,4,0,-2],[0,-4,3,0]] det=63 [19,-3,-18,-10] [58,-15,-49,-42] [159,24,-150,-87] chain2 [[4,4,-1,1],[-2,-4,1,-1],[-2,-3,2,0],[0,-3,1,3]] det=-36 [19,-3,-18,-10] [72,-34,-65,-39] [178,-34,-172,-80] chain8 [[3,2,-1,-2],[0,-1,0,2],[-4,2,-2,4],[-1,-1,3,-3]] det=42 [19,-3,-18,-10] [89,-17,-86,-40] [399,-63,-378,-210] [1869,-357,-1806,-840] [8379,-1323,-7938,-4410] [39249,-7497,-37926,-17640] [175959,-27783,-166698,-92610] [824229,-157437,-796446,-370440] [3695139,-583443,-3500658,-1944810] chain8 [[3,1,3,-4],[2,-3,2,3],[0,-1,0,4],[0,0,0,3]] det=0 [20,-7,-17,-14] [58,-15,-49,-42] [180,-63,-153,-126] [522,-135,-441,-378] [1620,-567,-1377,-1134] [4698,-1215,-3969,-3402] [14580,-5103,-12393,-10206] [42282,-10935,-35721,-30618] [131220,-45927,-111537,-91854] chain8 [[3,1,3,-4],[-2,-4,1,-1],[0,-1,0,4],[0,0,0,3]] det=27 [20,-7,-17,-14] [58,-15,-49,-42] [180,-63,-153,-126] [522,-135,-441,-378] [1620,-567,-1377,-1134] [4698,-1215,-3969,-3402] [14580,-5103,-12393,-10206] [42282,-10935,-35721,-30618] [131220,-45927,-111537,-91854] chain8 [[3,1,3,-4],[2,-3,2,3],[-4,-2,-1,0],[0,0,0,3]] det=-99 [20,-7,-17,-14] [58,-15,-49,-42] [180,-63,-153,-126] [522,-135,-441,-378] [1620,-567,-1377,-1134] [4698,-1215,-3969,-3402] [14580,-5103,-12393,-10206] [42282,-10935,-35721,-30618] [131220,-45927,-111537,-91854] chain8 [[3,1,3,-4],[-2,-4,1,-1],[-4,-2,-1,0],[0,0,0,3]] det=-72 [20,-7,-17,-14] [58,-15,-49,-42] [180,-63,-153,-126] [522,-135,-441,-378] [1620,-567,-1377,-1134] [4698,-1215,-3969,-3402] [14580,-5103,-12393,-10206] [42282,-10935,-35721,-30618] [131220,-45927,-111537,-91854] chain8 [[3,1,3,-4],[2,-3,2,3],[0,-1,0,4],[-4,-1,-1,-1]] det=81 [20,-7,-17,-14] [58,-15,-49,-42] [180,-63,-153,-126] [522,-135,-441,-378] [1620,-567,-1377,-1134] [4698,-1215,-3969,-3402] [14580,-5103,-12393,-10206] [42282,-10935,-35721,-30618] [131220,-45927,-111537,-91854] chain8 [[3,1,3,-4],[-2,-4,1,-1],[0,-1,0,4],[-4,-1,-1,-1]] det=108 [20,-7,-17,-14] [58,-15,-49,-42] [180,-63,-153,-126] [522,-135,-441,-378] [1620,-567,-1377,-1134] [4698,-1215,-3969,-3402] [14580,-5103,-12393,-10206] [42282,-10935,-35721,-30618] [131220,-45927,-111537,-91854] chain8 [[3,1,3,-4],[2,-3,2,3],[-4,-2,-1,0],[-4,-1,-1,-1]] det=-18 [20,-7,-17,-14] [58,-15,-49,-42] [180,-63,-153,-126] [522,-135,-441,-378] [1620,-567,-1377,-1134] [4698,-1215,-3969,-3402] [14580,-5103,-12393,-10206] [42282,-10935,-35721,-30618] [131220,-45927,-111537,-91854] chain8 [[3,1,3,-4],[-2,-4,1,-1],[-4,-2,-1,0],[-4,-1,-1,-1]] det=9 [20,-7,-17,-14] [58,-15,-49,-42] [180,-63,-153,-126] [522,-135,-441,-378] [1620,-567,-1377,-1134] [4698,-1215,-3969,-3402] [14580,-5103,-12393,-10206] [42282,-10935,-35721,-30618] [131220,-45927,-111537,-91854] chain2 [[3,-3,2,-3],[-1,-2,-1,2],[-2,3,-1,3],[-4,4,-4,0]] det=-8 [20,-7,-17,-14] [89,-17,-86,-40] [266,-49,-263,-80] chain2 [[-1,1,-4,-4],[-1,0,-1,3],[0,0,3,2],[-3,-2,3,-2]] det=-75 [20,-7,-17,-14] [97,-45,-79,-69] [450,-225,-375,-300] chain8 [[0,-3,-2,-3],[0,-4,1,4],[-1,4,1,1],[-4,2,1,-3]] det=-75 [20,-7,-17,-14] [97,-45,-79,-69] [500,-175,-425,-350] [2425,-1125,-1975,-1725] [12500,-4375,-10625,-8750] [60625,-28125,-49375,-43125] [312500,-109375,-265625,-218750] [1515625,-703125,-1234375,-1078125] [7812500,-2734375,-6640625,-5468750] chain2 [[-2,1,-2,-4],[1,-4,1,2],[-3,2,0,-2],[-1,-3,-1,3]] det=105 [25,-4,-22,-17] [58,-15,-49,-42] [135,-15,-120,-90] chain2 [[-2,0,-4,-3],[-2,1,-4,3],[0,2,2,2],[-2,-3,-3,4]] det=-120 [25,-4,-22,-17] [89,-17,-86,-40] [286,29,-286,-29] chain2 [[-3,-2,-4,-4],[-2,1,-4,3],[1,4,2,3],[-2,-3,-3,4]] det=-135 [25,-4,-22,-17] [89,-17,-86,-40] [271,29,-271,-29] chain2 [[-1,2,-4,-2],[-2,1,-4,3],[-1,0,2,1],[-2,-3,-3,4]] det=-105 [25,-4,-22,-17] [89,-17,-86,-40] [301,29,-301,-29] chain2 [[0,4,-4,-1],[-2,1,-4,3],[-2,-2,2,0],[-2,-3,-3,4]] det=-90 [25,-4,-22,-17] [89,-17,-86,-40] [316,29,-316,-29] chain2 [[1,1,0,-4],[-2,1,-4,3],[-3,1,-2,3],[-2,-3,-3,4]] det=-87 [25,-4,-22,-17] [89,-17,-86,-40] [232,29,-232,-29] chain2 [[1,-2,-1,-2],[-2,1,-4,3],[-3,4,-1,1],[-2,-3,-3,4]] det=-117 [25,-4,-22,-17] [89,-17,-86,-40] [289,29,-289,-29] chain2 [[2,0,-1,-1],[-2,1,-4,3],[-4,2,-1,0],[-2,-3,-3,4]] det=-102 [25,-4,-22,-17] [89,-17,-86,-40] [304,29,-304,-29] chain2 [[2,3,0,-3],[-2,1,-4,3],[-4,-1,-2,2],[-2,-3,-3,4]] det=-72 [25,-4,-22,-17] [89,-17,-86,-40] [247,29,-247,-29] chain2 [[2,2,-4,1],[3,3,2,2],[-2,-4,2,2],[0,2,3,-2]] det=30 [28,-18,-21,-19] [85,-50,-64,-61] [265,-145,-220,-170] chain8 [[0,-4,3,-4],[1,0,1,3],[3,4,0,4],[-1,3,-1,0]] det=81 [28,-18,-21,-19] [85,-50,-64,-61] [252,-162,-189,-171] [765,-450,-576,-549] [2268,-1458,-1701,-1539] [6885,-4050,-5184,-4941] [20412,-13122,-15309,-13851] [61965,-36450,-46656,-44469] [183708,-118098,-137781,-124659] chain2 [[-3,-2,-4,-1],[0,-3,4,-4],[0,-4,4,-1],[0,3,2,-3]] det=-150 [29,-11,-27,-15] [58,-15,-49,-42] [94,17,-94,-17] chain2 [[-4,-3,-3,-4],[0,-3,4,-4],[1,-3,3,2],[0,3,2,-3]] det=-201 [29,-11,-27,-15] [58,-15,-49,-42] [128,17,-128,-17] chain2 [[-2,-4,-1,-3],[0,-3,4,-4],[-1,-2,1,1],[0,3,2,-3]] det=-153 [29,-11,-27,-15] [58,-15,-49,-42] [119,17,-119,-17] chain2 [[-1,-3,-2,0],[0,-3,4,-4],[-2,-3,2,-2],[0,3,2,-3]] det=-102 [29,-11,-27,-15] [58,-15,-49,-42] [85,17,-85,-17] chain2 [[1,-4,0,1],[0,-3,4,-4],[-4,-2,0,-3],[0,3,2,-3]] det=-54 [29,-11,-27,-15] [58,-15,-49,-42] [76,17,-76,-17] chain2 [[1,-1,1,-3],[-3,0,-1,-3],[-2,0,-2,3],[-4,2,-3,-1]] det=-25 [29,-11,-27,-15] [58,-15,-49,-42] [150,1,-144,-73]
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/MultiNeuron(c).sce
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rafipra/Multi-Neuron
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2019-01-25T21:43:19
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MultiNeuron(c).sce
grand("setsd",2) E=zeros(5,12); N1=2; N2=10; N3=12; N4=13; N5=14; N6=19; N=[N1 N2 N3 N4 N5 N6]; for r=1:5 n=N(r); M=128; P1=[1/3; 1/3;1/3]; P2=[1/4;1/4;1/4;1/4]; P3=[1/5;1/5;1/5;1/5]; Q1 = (1/2)*P1; Q2 = (1/3)*P2; Q3 = (1/6)*P3; a=3*[.2,.3,.4]; theta1=400; theta2=1000; theta3=1200; Lambdan=[a(1)*(n*theta1)**(1),a(2)*(n*theta2)**(1),a(3)*(n*theta3)**(1)]; Rn=zeros(M); for i=1:M Z1 = grand(1, "poi", Lambdan(1)); Z2 = grand(1, "poi", Lambdan(2)); Z3 = grand(1, "poi", Lambdan(3)); Z = max([Z1 Z2 Z3]); Q=[Q1;Q2;Q3]; Y = grand(Z, "mul", 5, Q); Y1 = Y(1:3,1:Z1); Y2= Y(4:7,1:Z2); Y3 = Y(8:12,1:Z3); //Y1 = grand(Z1, "mul", 3, P1); //Y2 = grand(Z2, "mul", 4, P2); //Y3 = grand(Z3, "mul", 5, P3); Wn1=[sum(Y1(1,:)),sum(Y1(2,:)),sum(Y1(3,:))]; Wn2=[sum(Y2(1,:)),sum(Y2(2,:)),sum(Y2(3,:)),sum(Y2(4,:))]; Wn3=[sum(Y3(1,:)),sum(Y3(2,:)),sum(Y3(3,:)),sum(Y3(4,:)),sum(Y3(5,:))]; Lambdan; mu1=[Wn1(1)/Lambdan(1),Wn1(2)/Lambdan(1),Wn1(3)/Lambdan(1)]; mu2=[Wn2(1)/Lambdan(2),Wn2(2)/Lambdan(2),Wn2(3)/Lambdan(2),Wn2(4)/Lambdan(2)]; mu3=[Wn3(1)/Lambdan(3),Wn3(2)/Lambdan(3),Wn3(3)/Lambdan(3),Wn3(4)/Lambdan(3),Wn3(5)/Lambdan(3)]; muesc1=sum(Wn1)/(3*Lambdan(1)); muesc2=sum(Wn2)/(4*Lambdan(2)); muesc3=sum(Wn3)/(5*Lambdan(3)); R11=Lambdan(1)*(mu1(1)-muesc1)**2/mu1(1); R12=Lambdan(1)*(mu1(2)-muesc1)**2/mu1(2); R13=Lambdan(1)*(mu1(3)-muesc1)**2/mu1(3); R21=Lambdan(2)*(mu2(1)-muesc2)**2/mu2(1); R22=Lambdan(2)*(mu2(2)-muesc2)**2/mu2(2); R23=Lambdan(2)*(mu2(3)-muesc2)**2/mu2(3); R24=Lambdan(2)*(mu2(4)-muesc2)**2/mu2(4); R31=Lambdan(3)*(mu3(1)-muesc3)**2/mu3(1); R32=Lambdan(3)*(mu3(2)-muesc3)**2/mu3(2); R33=Lambdan(3)*(mu3(3)-muesc3)**2/mu3(3); R34=Lambdan(3)*(mu3(4)-muesc3)**2/mu3(4); R35=Lambdan(3)*(mu3(5)-muesc3)**2/mu3(5); Rn(i)=sum([R11,R12,R13,R21,R22,R23,R24,R31,R32,R33,R34,R35]); end Rn; T1=0; T2=0; T3=0; T4=0; T5=0; T6=0; T7=0; T8=0; h1=4.506956; h2=5.898826; h3=7.116417; h4=8.342833; h5=9.703707; h6=11.388751; h7=13.925503; for i=1:M if Rn(i)<h1 then T1=T1+1; end if h1<= Rn(i) & Rn(i) <=h2 then T2=T2+1; end if h2<= Rn(i) & Rn(i) <=h3 then T3=T3+1; end if h3<= Rn(i) & Rn(i) <=h4 then T4=T4+1; end if h4<= Rn(i) & Rn(i) <=h5 then T5=T5+1; end if h5<= Rn(i) & Rn(i) <=h6 then T6=T6+1; end if h6<= Rn(i) & Rn(i) <=h7 then T7=T7+1; end if h7<= Rn(i) then T8=T8+1; end end D=[T1,T2,T3,T4,T5,T6,T7,T8]; Percentage=(100/M)*D; U=(Percentage-12.5).^2/12.5; Percentage D [P,Q]=cdfchi("PQ", sum(U), 9) ValorP=1-[P,Q] sum(U) E(r,:)=[Percentage sum(U) ValorP N(r)] end
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/2732/CH7/EX7.10/Ex7_10.sce
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FOSSEE/Scilab-TBC-Uploads
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Ex7_10.sce
clc //initialization of variables clear Ys=17000 //kg/cm^2 E=2*10^6 //kg/cm^2 d1=1 //mm d=1 //cm //calculations: 1 cm R=E*d/(2*Ys) M=Ys*%pi*d^3/32 // results printf('%d cm daimeter wire:',d) printf('\n Minimum radius = %.2f cm',R) printf('\n Bending Moment = %.2f kg-cm',M) // calculations: 1 mm R1=R/(d1*10) M1=M/(d1*1000) // results printf('\n %d mm daimeter wire:',d1) printf('\n Minimum radius = %.2f cm',R1) printf('\n Bending Moment = %.2f kg-cm',M1)
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/demos/fortran_sum.dem.sce
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sengupta/scilab-http
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// Scilab ( http://www.scilab.org/ ) - This file is part of Scilab // Copyright (C) 2008 - INRIA - Allan CORNET // Copyright (C) 2010 - DIGITEO - Allan CORNET // // This file is released under the 3-clause BSD license. See COPYING-BSD. function demo_fortran_sum() mode(-1); lines(0); disp("fortran_sum(3,4)"); disp(fortran_sum(3,4)); endfunction demo_fortran_sum(); clear demo_fortran_sum;
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//Periodic notch filter design at 60 HZ and sampling frequency 300HZ z=%z; f=0:(0.5/400):0.5; z1=exp(%i*2*%pi*f); for i=1:401 H1Z(i)=(z1(i)^5-1)/((z1(i)^5)-(0.9^5)); H2Z(i)=(z1(i)^5-1)/((z1(i)^5)-(0.99^5)); end H1Z=abs(H1Z); H2Z=abs(H2Z); N1z=(1-z^-5)/(1-z^-1); H3z=(N1z)/(horner(N1z,z/0.9)); H4z=(N1z)/(horner(N1z,z/0.99)); H3z=horner(H3z,z1); H4z=horner(H4z,z1); a=gca(); a.x_location="origin"; a.y_location="origin"; plot2d(f,H1Z); plot2d(f,H2Z); xlabel('Digital frequency f'); ylabel('magnitude'); xtitle('Periodic Notch Filter N=5,R=0.9,0.99'); xset('window',1); plot2d(f,H3z); plot2d(f,H4z); xlabel('Digital frequency f'); ylabel('magnitude'); xtitle('Notch Filter that also passes DC N=5,R=0.9,0.99');
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CMPINF.TST
DECLARE FUNCTION peekb$ (addr&) DECLARE FUNCTION peekw$ (addr&) DECLARE FUNCTION peekd$ (addr&) DECLARE FUNCTION tell$ (flag%, bit%) CLS COLOR 15 PRINT "Computer information" PRINT "--------------------" PRINT "" COLOR 14 DEF SEG = &H40 PRINT "COM1 address:"; peekw$(&H0) PRINT "COM2 address:"; peekw$(&H2) PRINT "COM3 address:"; peekw$(&H4) PRINT "COM4 address:"; peekw$(&H6) PRINT "LPT1 address:"; peekw$(&H8) PRINT "LPT2 address:"; peekw$(&HA) PRINT "LPT3 address:"; peekw$(&HC) PRINT "LPT4 address:"; peekw$(&HE) PRINT "" PRINT "Equipment List:" flag% = PEEK(&H10) PRINT "1 IPL diskette"; tell$(flag%, 0) PRINT "2 Math coprocessor"; tell$(flag%, 1) PRINT "3 Pointing device(PS/2)"; tell$(flag%, 2) PRINT "4 Old PC system board RAM < 256K"; tell$(flag%, 3) PRINT "5 Initial video mode"; (flag% AND &H30) \ &H10 PRINT "6 Number of diskette drives"; ((flag% AND &HC0) \ 64) + 1 flag% = PEEK(&H11) PRINT "7 Direct Memory Access(DMA)"; tell$(NOT (flag%), 0) PRINT "8 Number of serial ports"; (flag% AND &HE) \ 2 PRINT "9 Game adapter"; tell$(flag%, 4) PRINT "10 Internal modem(PS/2)"; tell$(flag%, 5) PRINT "11 Number of printer ports"; flag% \ 64 PRINT "" PRINT "PCjr: Infrared keyboard link error count:"; peekb$(&H12) PRINT "Memory Size in KB:"; PEEK(&H13) + CLNG(PEEK(&H14)) * &H100 PRINT "PS/2 BIOS control state:"; peekb$(&H16) FUNCTION peekb$ (addr&) peekb$ = " " + HEX$(PEEK(addr&)) + "h" END FUNCTION FUNCTION peekd$ (addr&) peekd$ = " " + HEX$(PEEK(addr&) + PEEK(addr& + 1) * &H100 + CLNG(PEEK(addr& + 2)) * &H10000 + CLNG(PEEK(addr& + 3)) * &H1000000) + "h" END FUNCTION FUNCTION peekw$ (addr&) peekw$ = " " + HEX$(PEEK(addr&) + CLNG(PEEK(addr& + 1)) * &H100) + "h" END FUNCTION FUNCTION tell$ (flag%, bit%) IF ((flag% AND (2 ^ bit%)) = 1) THEN tell$ = " is present." ELSE tell$ = " is absent." END FUNCTION
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/ask.sci
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djouani/Scilab-code-for-Digital-Modulation
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clear all; clc; f=input('enter the analog carrier frequency in Hz:'); t=0:0.00001:0.1; x=cos(2*%pi*f*t); message=[]; carrier=[]; I=input('enter the digital binary data:'); for i=1:length(I) t=[0:.00001:0.1] if I(i)>0.5 m(i)=1; m_s=ones(1,length(t)); carrier=[carrier,x] else m(i)=0; m_s=zeros(1,length(t)); carrier=[carrier,x] end message=[message,m_s]; end //generation of ask Xask=[]; for n=1:length(I) if((I(n)==1) & (n==1)) Xask=[x,Xask]; elseif((I(n)==0) & (n==1)) Xask=[zeros(1,length(x)),Xask]; elseif((I(n)==1) & (n~=1)) Xask=[Xask,x]; elseif((I(n)==0) & (n~=1)) Xask=[Xask,zeros(1,length(x))]; end end subplot(4,1,1) plot(message) xtitle('Y15EC805 (Binary Message Signal)') xlabel('Time--->') ylabel('Amplitude--->') subplot(4,1,2) plot(carrier); xtitle('Carrier signal') xlabel('Time--->') ylabel('Amplitude--->') subplot(4,1,3) plot(Xask) xtitle('Amplitude Shift Keying') xlabel('Time--->') ylabel('Amplitude--->') /* subplot(4,1,4); plot(message) xtitle('Demodulated signal') xlabel('Time--->') ylabel('Amplitude--->') */ //demodulation demod=[] xdm=[] for i=1:length(I) if i>=2 xdemod=sum(Xask(1:(i*length(t))))-sum(Xask(1:((i-1)*length(t)))); else xdemod=sum(Xask(1:(i*length(t)))); end if xdemod>0 demod(:,i)=1 xdm=[xdm,ones(1,length(t))] else demod(:,i)=0 xdm=[xdm,zeros(1,length(t))] end end disp('demodulated data:',demod); if(demod==I) subplot(4,1,4); plot(xdm) xtitle('Demodulated Signal') xlabel('Time--->') ylabel('Amplitude--->') else disp('invalid'); end
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/src/tools/matrixTool.sci
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clairedune/sciGaitanLib
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matrixTool.sci
// -----------------------------------------// // build a matrix with A has diagonal //------------------------------------------// function bigA = bigDiag(A,n) Zer = zeros(size(A,1),size(A,2)); bigA=[]; for i = 1:n; L_ligne = []; for j=1:n if i==j L_ligne = [L_ligne A]; else L_ligne = [L_ligne Zer]; end end bigA = [bigA;L_ligne]; end endfunction //--------// // //--------// function stateC = convertState(stateA) stateC = [1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 ] * stateA; endfunction function stateC = convertState6(stateA) stateC = [1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 ] * stateA; endfunction function p= crossProd(u,v) [nu,mu]=size(u); [nv,mv]=size(v); if nu*mu<>3 |nv*mv<>3 then error('Vectors must be 3D only'); abort; end A1 = [u(2), u(3);v(2), v(3)]; A2 = [u(3), u(1);v(3), v(1)]; A3 = [u(1), u(2);v(1), v(2)]; px = det(A1); py = det(A2); pz = det(A3); p = [px py pz]'; endfunction
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// Calculating the temperature rise clc; disp('Example 4.43, Page No. = 4.77') // Given Data az = 30*10^(-6);// Cross-sectional area (in meter square) Iz = 20*10^(3);// Current (in Ampere) t = 50;// Time (in mili second) p = 0.021*10^(-6);// Resistivity of conductor (in ohm*meter) h = 418;// Specific heat (in J/kg degree celsius) g = 8900;// Density (in kg per meter cube) // Calculation of the temperature rise T = Iz^(2)*p*t*10^(-3)/(g*az^(2)*h);// Temperature rise (in degree celsius) disp(T,'Temperature rise (degree celsius)='); //in book answer is 125 degree celsius. The answers vary due to round off error
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//Sensores e Atuadores //Atuadores damax = 18 //deg drmax = 23 //deg damax = damax* %pi/180 drmax = drmax* %pi/180 Ta = 100e-3 //s v_max = 1 // rad/s rate = 100 //Hz rate = 1/rate //s //Sensores bb_s = 25 //saturation deg bb_s = bb_s* %pi/180 // rad bb_t = 0.01 //tempo s bb_r = 5 //range Vdc bb_n = 0.25 //range deg bb_n = bb_n* %pi/180 // rad pr_s = 50 //saturation deg/s pr_s = pr_s* %pi/180 //rad/s pr_r = 3 //range +- Vdc pr_n = 2e-3 //noise V RMS phi_s = 90 //saturation deg phi_s = phi_s* %pi/180 phi_r = 28 //range Vdc phi_n = 0.25 // deg RMS phi_n = phi_n* %pi/180 //rad RMS psi_s = 360 // saturation deg psi_s = psi_s* %pi/180 // rad psi_r = 25 // range Vdc psi_n = 1.5 //noise deg RMS psi_n = psi_n* %pi/180 //rad RMS
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clear clc //to find root mean square speed of hydrogen molecule //Given: //pressure p = 1//in atm //density of hydrogen rho = 8.99e-2//in Kg/m^3 //Solution: //assume hydron as ideal gas //applying formula of root mean square speed for ideal gas //root mean square speed of hydrogen molecule vrms = sqrt((3*p*1.01e5)/(rho))//in m/s //taking pressure in Pa //answer of vrms is slightly different than book answer.But ans. by scilab program is same as that of calculator printf ("\n\n Root mean square speed of hydrogen molecule vrms = \n\n %4i m/s" ,vrms);
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load aggie.hack, repeat{ ticktock; }
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clc //initialisation of variables ne=0.0019 n=300//rpm pi=22/7 t=0.01/12//ft R1=0.25//ft R2=0.167//ft //CALCULATIONS w=pi*n/60 T=pi*0.0019*w*2*(R1^4-R2^4)/(2*t) hp=T*2*pi*n/33000 //RESULTS printf (' hp absorbed= %.2f',hp)
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function y=f5(t) g = 9.81 u = 1800 m0 = 120000 q = 2200 y = u.*log(m0./(m0-q.*t)) - g.*t - 970 endfunction axes = get("default_axes"); axes.x_location = "origin"; axes.y_location = "origin"; t = 20:0.05:37 plot(t, f5(t)) secante(f5, 25, 30, 0.00001) // k= 1 x(1)= 25.0000000 |f(x(1))|= 111.6619488 // k= 2 x(2)= 30.0000000 |f(x(2))|= 173.0138532 // k= 3 x(3)= 26.9612125 |f(x(3))|= 7.2680105 // k= 4 x(4)= 27.0837203 |f(x(4))|= 0.4578024 // k= 5 x(5)= 27.0919557 |f(x(5))|= 0.0012814 // k= 6 x(6)= 27.0919327 |f(x(6))|= 0.0000002
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2_11.sce
clear; clc; close; E = 8; //volts Vled = 2; //volts I = 20*10^(-3); //amperes R = (E-Vled)/I; disp(R,'resistance value is : ')
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//chapter-8 page 337 example 8.2 //============================================================================== clc; clear; //For a 2 cavity klystron amplifier Av=15;//Voltage gain in dB Pin=0.005;//I/P power in W Rin=30000;//Rsh of i/p cavity in ohms R0=40000;//Rsh of o/p cavity in ohms Rl=40000;//load impedance in ohms R=20000;//Parallel resistance of R0 and Rl (R0//Rl) in ohms //CALCULATION Vin=sqrt(Pin*Rin);//The input rms voltage in V [From Pin=Vin^2/Rin] V0=Vin*10^(Av/20);//The output rms voltage in V [From Av=20log(V0/Vin)] P0=(V0^2)/R;//The Power delivered to the load in W //OUTPUT mprintf('\nThe input rms voltage is Vin=%2.2f V \nThe output rms voltage is V0=%2.2f V \nThe Power delivered to the load is P0=%1.4f W',Vin,V0,P0); //=========================END OF PROGRAM===============================
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example_2_10i.sce
clear; clc; // A Textbook on HEAT TRANSFER by S P SUKHATME // Chapter 2 // Heat Conduction in Solids // Example 2.10(i) // Page 58 printf("Example 2.10(i), Page 58 \n\n") // Centre of the slab // Given data b = 0.005 ; // [m] t = 5*60; // time, [sec] Th = 200 ; // [C] Tw = 20 ; // [C] h = 150 ; // [W/m^2 K] rho = 2200 ; //[kg/m^3] Cp = 1050 ; // [J/kg K] k = 0.4 ; // [W/m K] // Using charts in fig 2.18 and 2.19 and eqn 2.7.19 and 2.7.20 theta = Th - Tw; Biot_no = h*b/k; a = k/(rho*Cp); // alpha Fourier_no = a*t/b^2; // From fig 2.18, ratio = theta_x_b0/theta_o ratio_b0 = 0.12; // From fig 2.18, ratio = theta_x_b1/theta_o ratio_b1 = 0.48; // Therefore theta_x_b0 = theta*ratio_b0; // [C] T_x_b0 = theta_x_b0 + Tw ; // [C] theta_x_b1 = theta*ratio_b1; // [C] T_x_b1 = theta_x_b1 + Tw ; // [C] // From Table 2.2 for Bi = 1.875 lambda_1_b = 1.0498; x = 2*sin(lambda_1_b)/[lambda_1_b+(sin(lambda_1_b))*(cos(lambda_1_b))]; // From eqn 2.7.20 theta_x_b0 = theta*x*(exp((-lambda_1_b^2)*Fourier_no)); T_x_b0 = theta_x_b0 + Tw; printf("Temperature at b=0 is %f degree C\n",T_x_b0);
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// 4.2 clc; L=50*10^-6; C=1*10^-9; fc=1/(2*%pi*(L*C)^0.5); fs1=10000; fu1=(fc+fs1)*10^-3; printf("\nUpper side band frequency =%.2f kHz",fu1) fl1=(fc-fs1)*10^-3; printf("\nLower side band frequency =%.2f kHz",fl1)
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//find the clearance of conductor from water level at mid point clear; clc; //soltion //given W=.844;//kg/m//Line conductor wieght L=300;//meter//span of the line T=1800;//kg//max allowable tension T1=40;//m//height of the tower 1 T2=80;//m//height of the tower 2 h=T2-T1;//m//difference in the between support x=L/2-(T*h)/(W*L); printf("Distance between midpoint and lowest point= %.2fm\n",(L/2)-x); Smid=(W*(L/2-x)^2)/(2*T); printf("Height between midpoint and lowest point= %.3fm\n",Smid); S2=(W*(L-x)^2)/(2*T); printf("Height between taller tower and lowest point= %.3fm\n",S2); C=T2-(S2-Smid); printf("Clearance of conductor from water level at mid point= %.3fm",C)
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// Exa 3.17 clc; clear; close; // Given data // alpha= bita/(1-bita) // At bita= 1 bita=1; alpha= bita/(1+bita); disp(alpha,"At bita=1, the value of alpha is : ") // At bita= 2 bita=2; alpha= bita/(1+bita); disp(alpha,"At bita=2, the value of alpha is : ") // At bita= 100 bita=100; alpha= bita/(1+bita); disp(alpha,"At bita=100, the value of alpha is : ") // At bita= 200 bita=200; alpha= bita/(1+bita); disp(alpha,"At bita=200, the value of alpha is : ")
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//Example 3.4 //Page 128 w=800//Omega=800Hz //x(t)=A sin(2pi.wt), equation for sine wave with maximum amplitude //x'(t)=A(2pi).w.cos(2pi.wt), diff w.r.t time (2*%pi)*800*(1/8000) //0.62831*a, x'(t)max disp('savings in the bits per sample can be determined as ') log2(1/0.628) //Result //0.67 bits
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clc;clear; //Example 11.3 //given data mA=0.05; P1=0.14; P5=0.32; P7=0.8; h1=239.16; h2=255.93; h3=55.16; h5=251.88; h6=270.92; h7=95.47; //calculations h4=h3;//throttling h8=h7;//throttling // E out = E in // mA*h5 + mB*h3 = mA*h8 + mB*h2 mB=mA*(h5-h8)/(h2-h3); QL=mB*(h1-h4); // W in = Wcomp I,in + Wcomp II,in Win=mA*(h6-h5)+mB*(h2-h1); COPR=QL/Win; disp(mB,'the mass flow rate of the refrigerant through the lower cycle in kg/s'); disp(QL,'the rate of heat removal from the refrigerated space in kW'); disp(Win,'the power input to the compressor in kW'); disp(COPR,'the coefficient of performance of this cascade refrigerator');
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//clear// //Example 2.6:Convolution Integral of input x(t) = (e^-at).u(t) //and h(t) =u(t) clear; close; clc; Max_Limit = 10; h = ones(1,Max_Limit); N2 =0:length(h)-1; a = 0.5; //constant a>0 for t = 1:Max_Limit x(t)= exp(-a*(t-1)); end N1 =0:length(x)-1; y = convol(x,h)-1; N = 0:length(x)+length(h)-2; figure a=gca(); plot2d(N2,h) xtitle('Impulse Response','t','h(t)'); a.thickness = 2; figure a=gca(); plot2d(N1,x) xtitle('Input Response','t','x(t)'); a.thickness = 2; figure a=gca(); plot2d(N(1:Max_Limit),y(1:Max_Limit)) xtitle('Output Response','t','y(t)'); a.thickness = 2;
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errcatch(-1,"stop");mode(2);//Example 7.1, page no-436 f=2*9.8*10^5 A=100 V=20 l=10 mu=(f/A)/(V/l) mu=mu/1000 printf("The absolute viscosity mu = %.1f*10^5 centipoises",mu) exit();
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// Additional solved examples , Example 30 , pg 345 H0=6*10^4 //magnetic field intensity at 0K (in A/m) T=4.2 //temperature (in K) Tc=8 //critical temperature (in K) Hc=H0*(1-(T^2/Tc^2)) // critical magnetic field intensity printf("critical magnetic field intensity\n") printf("Hc=%.0f A/m",Hc)
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//Example 3-4, Page No - 108 clear clc R = 40 I = 4.8 m=0.9 Pc = I^2*R Pt = (I*(1+(m^2/2))^0.5)^2*R Psb = Pt-Pc printf('The carrier power is %.1f watt\n Total power = %.1f watt\n Sideband Power =%.1f watt',Pc,Pt,Psb)
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//taylor series //example 1.13 //page 16 clc;clear;close; deff('y=f(x)','y=sin(x)'); deff('y=f1(x)','y=cos(x)'); deff('y=f2(x)','y=-sin(x)'); deff('y=f3(x)','y=-cos(x)'); deff('y=f4(x)','y=sin(x)'); deff('y=f5(x)','y=cos(x)'); deff('y=f6(x)','y=-sin(x)'); deff('y=f7(x)','y=-cos(x)'); D=[f(%pi/6) f1(%pi/6) f2(%pi/6) f3(%pi/6) f4(%pi/6) f5(%pi/6) f6(%pi/6) f7(%pi/6)]; S1=0; h=%pi/6; printf('order of approximation computed value of sin(pi/3) absolute eror\n\n'); for j=1:8 for i=1:j S1=S1+h^(i-1)*D(i)/factorial(i-1); end printf('%d %0.9f %0.9f\n',j,S1,abs(sin(%pi/3)-S1)); S1=0; end
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// ::::::::::::::::::::::::::::::::::::::::: // :: :: // :: Question 11. Mesure du flux :: // :: :: // ::::::::::::::::::::::::::::::::::::::::: funcprot(0); exec("Q3_Factorisation_Cholesky.sce"); exec("Q4_Descente_Cholesky.sce"); exec("Q5_Remontee_Cholesky.sce"); // Fonction u0(t) function res = u0(t, T) res = (t / T)^2 endfunction // Fonction u0'(t) function res = derivee_en_t_u0(t, T) res = (2 * t) / (T^2) endfunction // Fonction u^(0) = u_p_0(x) function res = u_p_0(x) res = 0 endfunction // Fonction C(x) function res = C(x, xd, a) res = 1 - a * exp(-(x - xd)^2/(4)); endfunction // Matrice A function [matrice_A_diag, matrice_A_inf] = calcul_A(n, x_d, l, a) matrice_A_diag = zeros(n, 1); matrice_A_inf = zeros(n-1, 1); delta_x = (2 * l)/(n + 1); i = 1; for x = -l+delta_x:delta_x:l-2*delta_x matrice_A_diag(i) = C(x + delta_x / 2, x_d, a) + C(x - delta_x / 2, x_d, a); matrice_A_inf(i) = -C(x + delta_x / 2, x_d, a); i = i + 1 end matrice_A_diag(n) = C(l - delta_x / 2, x_d, a) + C(l - 3 * (delta_x / 2), x_d, a); endfunction // Matrice B function [matrice_B] = calcul_B(k, n, x_d, l, a, theta, delta_t, delta_x, T) matrice_B($ + 1) = C(-l + (delta_x/2), x_d, a) * (theta * u0(delta_t * (k + 1), T) + (1 - theta) * u0(delta_t * k, T)); endfunction // Matrice M function [matrice_M_diag, matrice_M_inf] = calcul_M(n, theta, mu, x_d, l, a) [diag_sup_A, diag_inf_A] = calcul_A(n, x_d, l, a); matrice_M_inf = zeros(n - 1, 1); matrice_M_diag = zeros(n, 1); for i = 1:n-1 matrice_M_diag(i) = diag_sup_A(i) * theta * mu + 1; matrice_M_inf(i) = diag_inf_A(i) * theta * mu; end matrice_M_diag(n) = diag_sup_A(n) * theta * mu + 1; endfunction // Matrice N function [matrice_N_diag, matrice_N_inf] = calcul_N(n, theta, mu, x_d, l, a) [diag_sup_A, diag_inf_A] = calcul_A(n, x_d, l, a); matrice_N_inf = zeros(n - 1, 1); matrice_N_diag = zeros(n, 1); for i = 1:n-1 matrice_N_diag(i) = diag_sup_A(i) * (theta - 1) * mu + 1; matrice_N_inf(i) = diag_inf_A(i) * (theta - 1) * mu; end matrice_N_diag(n) = diag_sup_A(n) * (theta - 1) * mu + 1; endfunction // Calcul du vecteur second membre N*U^(k) + \mu * B^(k) function [sm] = calcul_second_membre(N_diag, N_inf, U, mu, k, n, x_d, l, a, theta, delta_t, delta_x, T) sm = zeros(n, 1); B = calcul_B(k, n, x_d, l, a, theta, delta_t, delta_x, T); // Produit N * U simplifié par la structure particulière des matrices sm(1) = N_diag(1) * U(1) + N_inf(1) * U(2) + mu * B(1) sm(n) = N_diag(n) * U(n) + N_inf(n - 1) * U(n - 1) for i = 2:n-1 sm(i) = N_diag(i) * U(i) + N_inf(i - 1) * U(i - 1) + N_inf(i) * U(i + 1) end endfunction // Fonction FLUX function [F_t_inter, F_t_fin] = flux(x_d) // Définition de tous les paramètres n = 2000; n_t = 3000; l = 10; T = 60; a = 0.8; theta = 1/2; // Valeurs temporelles auxquelles on récupère le flux t_inter = 2/3 * T; t_fin = T; F_t_inter = 5; F_t_fin = 5; // Pas après discrétisation delta_x = (2 * l) / (n + 1); delta_t = T / n_t; mu = delta_t / (delta_x)^2; // Pré-calculs, effectués une seule fois [M_diag, M_inf] = calcul_M(n, theta, mu, x_d, l, a); [F_diag, F_inf] = factorise(M_diag, M_inf); [N_diag, N_inf] = calcul_N(n, theta, mu, x_d, l, a); U = zeros(n, 1); for k = 0:(n_t - 1) NU_p_muB = calcul_second_membre(N_diag, N_inf, U, mu, k, n, x_d, l, a, theta, delta_t, delta_x, T); // Décomposition de Cholesky v_descente = descente(F_diag, F_inf, NU_p_muB); U = remonte(F_diag, F_inf, v_descente); t_verif = (k + 1) * delta_t; if (t_verif == t_inter) then F_t_inter = C(-l + (delta_x / 2), x_d, a) * (U(1) - u0(t_inter, T)) * (1 / delta_x) - (delta_x / 2) * derivee_en_t_u0(t_inter, T) elseif t_verif == t_fin then F_t_fin = C(-l + (delta_x / 2), x_d, a) * (U(1) - u0(t_fin, T)) * (1 / delta_x) - (delta_x / 2) * derivee_en_t_u0(t_fin, T) end end endfunction // Fonction qui retourne les deux flux comme un vecteur // et non comme deux valeurs distinctes de retour function [vect_F] = flux_vecteur(x_d) [F_t_inter, F_t_fin] = flux(x_d) vect_F($ + 1) = F_t_inter; vect_F($ + 1) = F_t_fin; endfunction // A DECOMMENTER POUR TESTER // Exemple d'appel : // flux(-8)
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//Example number 5.7, Page number 87 clc;clear; close; //Variable declaration a=1*10**-10; //length(m) n2=2; n3=3; m=9.1*10**-31; //mass(kg) e=1.6*10**-19; //charge(c) h=6.626*10**-34; //plank constant //Calculation E1=h**2/(8*m*e*a**2); E2=n2**2*E1; //energy of 1st excited state(eV) E3=n3**2*E1; //energy of 2nd excited state(eV) //Result printf("ground state energy is %.2f eV",E1) printf("\n energy of 1st excited state is %.2f eV",E2) printf("\n energy of 2nd excited state is %.2f eV",E3) //answer in the book varies due to rounding off errors
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//To Determine the Design Parameters of a Distributed System //Page 484 clc; clear; VPpu=1.035; //Primary Feeder Voltage per unit TVDpu=0.0776; //Total Voltage Drop of Feeder Vll=13.2; //Line Voltage in kV Vlpuqsw=1;//New Voltage at the End of the Feeder due to Qsw at annual peak load XL=0.661; //Inductive Reactance per mile Pl=3400; //Real Power Ql=2100; //Reactive Power l=10; //Length of the Feeder in Miles Lf=0.4; //Load Factor CR=0.27; //Total Capacitor Compensation Ratio For the Above Load Factor QNSW=CR*Ql; //Required Size of the Nonswitched capacitor bank s=2*l/3; //Loacation of Nonswitched capacitor bank for Optimum Result VRpu=QNSW*(2*XL*l/3)/(1000*(Vll^2)); //Per Unit Voltage Rise VDspu=TVDpu*s*(2-(s/l))/l; //Voltage drop for the uniformaly distributed load VSpu=VPpu-VDspu;//Feeder Voltage at 2l/3 distance nVSpu=VSpu+VRpu; //New Voltage Rise when there is a fixed capacitor bank Vlpu=VPpu-TVDpu; //When No Capcacitor bank is there, the voltage at the end of the feeder nVlpu=Vlpu+VRpu; //When Capcacitor bank is there, the voltage at the end of the feeder VRpuqsw=Vlpuqsw-nVlpu; //Required Voltage Rise Q3phisw=1000*(Vll^2)*VRpuqsw/(XL*l); //Required Size of the Capacitor Bank //Let us assume the 15 single phase standard 50 kVAr Capacitor Units = 750 kVAr SQ3phisw=750; //Selected Capacitor Bank RVRlpu=VRpuqsw*SQ3phisw/Q3phisw; //Resultant Voltage Rises at distance l RVRspu=RVRlpu*s/l; //Resultant Voltage Rises at distance s //At Peak Load when both the Non-Switched and Switched Capacitor Banks are on PVspu=nVSpu+RVRspu; //Voltage at s PVlpu=nVlpu+RVRlpu; //Voltage at l printf('\na) The NSW Capacitor Rating is %g kVAr, Which means 2 100kVAr Capacitor Banks per phase\n',QNSW) printf('\nb) Voltage Rise per unit is %g pu V\n',VRpu) printf('i) When the No Capacitor Bank is Installed \n') printf('Voltage at %g miles is %g pu V\n',s,VSpu) printf('Voltage at %g miles is %g pu V\n',l,Vlpu) printf('ii) When the Fixed Capacitor Bank is Installed \n') printf('Voltage at %g miles is %g pu V\n',s,nVSpu) printf('Voltage at %g miles is %g pu V\n',l,nVlpu) printf('\nc) At Annual Peak Load, The Size of Capacitor Bank Required is %g\n',Q3phisw) printf('The Voltage Rise at the Annual Load for the Required Capacitor Bank is %g pu V\n',VRpuqsw) //Note That The Calculations are highly accurate, Rounding of Terms hasn't be emplyed
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//Obtain path of solution file path = get_absolute_file_path('solution12_12.sce') //Obtain path of data file datapath = path + filesep() + 'data12_12.sci' //Clear all clc //Execute the data file exec(datapath) //Calculate the torque capacity of each pad mt (N-m) mt = Mt/n //Calculate the friction radius Rf (mm) Rf = (2 * (Ro^3 - Ri^3))/(3 * (Ro^2 - Ri^2)) //Calculate the actuating force P (N) P = (mt * 1000)/(mu * Rf) //Calculate the area of the pad A (mm2) A = P/pavg //Calculate the angular dimension of the pad theta (deg) theta = ((2 * A)/(Ro^2 - Ri^2))*180/%pi //Print results printf("\nAngular dimension of the pad(theta) = %f deg\n",theta)
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//Example 4.16: clc; clear; close; //given data : A=200;//gain without feedback Beta=0.25;//feed back ratio gc=10;//percent gain change dA=gc/100;// dAf= ((1/(1+Beta*A)))*dA;// format('v',7) disp(dAf,"small change in gain is,=")
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 12.11 //calculation of the amplitude of the simple harmonic motion //given data //x1 = (2.0 cm)*sind(w*t) //x2 = (2.0 cm)*sind((w*t) + (180/3)) A1=2//amplitude(in cm) of the wave 1 A2=2//amplitude(in cm) of the wave 2 delta=180/3//phase difference(in degree) between the two waves //calculation A=sqrt(A1^2+A2^2+(2*A1*A2*cosd(delta)))//amplitude of the resultant wave printf('the amplitude of the simple harmonic motion is %3.1f cm',A )
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//Example 22.3 B=0.1;//Magnetic field strength (T) l=4*10^-3;//Inside diameter (m) v=20*10^-2;//Average blood velocity (m/s) epsilon=B*l*v;//Hall emf (V) printf('Hall emf = %0.1f microV',epsilon/10^-6) //Openstax - College Physics //Download for free at http://cnx.org/content/col11406/latest
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//Exa 2.25 clc; clear; close; format('v',7); //Given Data : m=1.5;//Kg p1=1000;//Kpa p2=200;//Kpa V1=0.2;//m^3 V2=1.2;//m^3 //p=a+b*v //solving for a and b by matrix A=[1 V1;1 V2]; B=[p1;p2]; X=A^-1*B; a=X(1); b=X(2); W=integrate('a+b*V','V',V1,V2);//KJ/Kg disp(W,"Work transfer in KJ/Kg : "); u2SUBu1=(1.5*p2*V2+35)-(1.5*p1*V1+35);//KJ/Kg disp(u2SUBu1,"Change in internal energy in KJ/Kg : "); q=W+u2SUBu1;//KJ/Kg disp(q,"Heat transfer in KJ/Kg : "); //u=1.5*(a+b*V)*V+35; //1.5*a+2*V*1.5*b=0;//for max value putting du/dV=0 V=-1.5*a/2/1.5/b;//m^3/Kg p=a+b*V;//KPa u_max=1.5*p*V+35;//KJ/Kg disp(u_max,"Maximum internal energy in KJ/Kg : "); //Answer in the book is wrong because a is 1160 instead of 1260.
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//Chapter-5, Example 5.46, Page 210 //============================================================================= clc clear //INPUT DATA Vl=400;//voltage in volts Il=20;//current in A f=50;//freq in hz pf=0.3//power factor //CALCULATIONS Ip=Il/sqrt(3);//phase current in A Z=Vl/Ip;//impedance in each phase in ohms phi=acos(0.3);//angle in radians Zb=Z*(cos(phi)+(%i)*sin(phi));//impedance connected in each phase mprintf("Thus impedance connected in each phase in ohms"); disp(Zb); //=================================END OF PROGRAM======================================================================================================
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Chapter2_Example19.sce
//Chapter-2, Illustration 19, Page 77 //Title: Gas Power Cycles //============================================================================= clc clear //INPUT DATA T1=291;//Temperature at point 1 in K P1=100;//Pressure at point 1 in kN/(m^2) nC=0.85;//Isentropic efficiency of compressor nT=0.88;//Isentropic effficiency of turbine rp=8;//Pressure ratio T3=1273;//Temperature at point 3 in K m=4.5;//Mass flow rate of air in kg/s y=1.4;//Ratio of speciifc heats Cp=1.006;//Specific heat at constant pressure in kJ/kg-K //CALCULATIONS x=(y-1)/y;//Ratio T2s=T1*(rp^x);//Temperature at point 2s in K T2=T1+((T2s-T1)/nC);//Temperature at point 2 in K t2=T2-273;//Temperature at point 2 in oC T4s=T3*((1/rp)^x);//Temperature at point 4s in K T4=T3-((T3-T4s)*nT);//Temperature at point 4 in K t4=T4-273;//Temperature at point 4 in oC W=m*Cp*((T3-T4)-(T2-T1));//Net power output in kW nth=(((T3-T4)-(T2-T1))/(T3-T2))*100;//Thermal efficiency WR=W/(m*Cp*(T3-T4));//Work ratio //OUTPUT mprintf('Net power output of the turbine is %3.0f kW \n Thermal efficiency of the plant is %3.0f percent \n Work ratio is %3.3f',W,nth,WR) //==============================END OF PROGRAM=================================
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Exa_5_27.sce
//Example No. 5.27 clc; clear; close; format('v',9); //Given Data : V1=230;//V P=15;//hp N=1500;//rpm V2=220;//V Ke=0.03;//V/A-s Kt=0.03;//N-m/A^2 alfa=45;//degree Vm=V1*sqrt(2);//V omega=N*2*%pi/60;//rad/s T=4*Kt*Vm^2*cosd(alfa)^2/(%pi^2*(Ke*omega)^2);//N-m Ia=sqrt(T/Kt);//A disp("part (a) : "); disp(T,"Torque in N-m : "); disp(Ia,"Armature current in A : "); disp("part (b) : "); Ia=Vm*(1+cosd(alfa))/(%pi*(Ke*omega));//A T=Kt*Ia^2;//N-m disp(Ia,"Armature current in A : "); disp(T,"Torque in N-m : ");
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Example6_7.sce
clear; clc; // Example: 6.7 // Page: 215 printf("Example: 6.7 - Page: 215\n\n"); // Mathematics is involved in proving but just that no numerical computations are involved. // For prove refer to this example 6.7 on page number 215 of the book. printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n"); printf(" For prove refer to this example 6.7 on page 215 of the book.");
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Example6_12.sce
clear; clc; // Example: 6.12 // Page: 219 printf("Example: 6.12 - Page: 219\n\n"); // Mathematics is involved in proving but just that no numerical computations are involved. // For prove refer to this example 6.12 on page number 219 of the book. printf(" Mathematics is involved in proving but just that no numerical computations are involved.\n\n"); printf(" For prove refer to this example 6.12 on page 219 of the book.");
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errcatch(-1,"stop");mode(2);//example 1 //weight of a person m=1 //kg g=9.75 //acc.due to gravity in m/s^2 F=m*g //weight of 1 kg mass in N printf("\n hence,weight of person is F = %.3f N. \n",F) exit();
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example19_3.sce
//example 19.3 //design a syphon aqueduct clc;funcprot(0); //given Q=25; //design discharge of canal B=20; //bed width of canal D=1.5; //depth of water in canal bl=160; //bed level of canal hfq=400; //high flood discharge of drainage hfl=160.5; //high flood level of drainage bl_drain=158; //bed level of drainage gl=160; //general ground level //desing of drainage water-way P=4.75*(hfq)^0.5; //laecey P-Q formula mprintf("design of drainage water-way:\nwetted perimeter of river=%i m.\nprovide 13 spans of 6 m each,separated by 12 piers each of 1.25 m thick.",P); t=78+15; mprintf("\ntotal length of water-way=%i m.",t); v=2; //velocity through syphon hb=hfq/(78*v); ac=hfq/(6*2.5*1.3); //calculation is wrong in book hb=round(hb*100)/100; ac=round(ac*100)/100; mprintf("\nheight of barrels=%f m.\nprovide rectangular barrels 6 m wide and 2.5 m high.\nactual velocity through barrels=%f m/sec.",hb,ac); //design of canal waterway mprintf("\n\ndesign of canal waterway:\nType 3 aqueduct is adopted."); l1=B-10; l2=(20-10)*3/2; mprintf("\nproviding a splay 2:1 in expansion,length of contraction transition=%i m.\nproviding a splay of 3:1 in expansion,length of expansion transition=%i m.",l1,l2); mprintf('\nIn transition side slopes are warped from original slope of 1.5:1 to vertical.'); //design of levels of different sectionn mprintf("\n\ndesign of levels of different sectionn:\nat section 4-4:"); A=(B+1.5*D); //area V=Q/A; //velocity of flow vh=V^2/(2*9.81); //velocity head ws=gl+D; //R.L of water surface tel=ws+vh; tel=round(tel*1000)/1000; mprintf("\nR.L of T.E.L=%f m.\n at section 3-3:",tel); A=10*D; //area of trough V=Q/A; //velocity vh1=V^2/(2*9.81); //velocity head le=0.3*(vh1-vh); //loss of head in expansion from section 3-3 to 4-4 tel=tel+le; rlw=tel-vh1; rlb=rlw-D; tel=round(tel*1000)/1000; rlb=round(rlb*1000)/1000; mprintf("\nelevation of T.E.L=%f m.\nR.L of bed to maintain constant water depth=%f m.",tel,rlb); //at section 2-2 R=A/P; N=0.016; S=V^2*N^2/R^(4/3); //from manning's formula L=93; //length of trough hl=L*S; //head loss tel=tel+hl; rlw=tel-vh1; rlb=rlw-D; tel=round(tel*1000)/1000; rlb=round(rlb*1000)/1000; mprintf("\nat section 2-2:\nR.L of T.E.L=%f m.\nR.L of bed to maintain constant water depth=%f m.",tel,rlb); //at section 1-1 hl=0.2*(vh1-vh); //loss of hed in contraction transition tel=tel+hl; rlw=tel-vh; rlb=tel-D; tel=round(tel*1000)/1000; rlb=round(rlb*1000)/1000; mprintf("\nat section 1-1:\nR.L of T.E.L=%f m.\nR.L of bed to maintain constant water depth=%f m.",tel,rlb); //design of contraction transition //it is designed on the basis of chaturvedi's formula Bo=20; Bf=10; L=10; //from chaturvedi formula we get relation between x and Bx as: x=15.45(1-(10/Bx)^1.5); Bx=[10:1:20]; mprintf("\n\ndesign of contraction transition on the basis of chaturvedi formula:\nBx x"); for i=1:11 x(i)=15.45*(1-(10/Bx(i))^1.5); x(i)=round(x(i)*100)/100; mprintf("\n%i %f",Bx(i),x(i)); end //design of expansion transition on the basis of chaturvedi formula L=15; Bf=10;Bo=20; //from chaturvedi formula we get relation between x and Bx as: x=23.15(1-(10/Bx)^1.5); mprintf("\n\ndesign of expansion transition on the basis of chaturvedi formula:\nBx x"); for i=1:11 x(i)=23.15*(1-(10/Bx(i))^1.5); x(i)=round(x(i)*100)/100; mprintf("\n%i %f",Bx(i),x(i)); end //design of trough mprintf("\n\ndesign of the trough:"); mprintf("\nflumed water way of canal=10 m.\ntrough carrying canal will divide into two compartments each 5 m wide an dseparated by 0.3 m thick partiions.\nheigth of trough will be = 2 m.\ntrough iss constructed using monolithic reinforced concrete.\nthe outer and inner walls ca be kept 0.4 m thick.\nthus,outer width of trough = 11.1 m."); //head loss through syphon barrels V=2.05; //velocity through barrels f1=0.505; //coefficient of loss of head at entry a=0.00316;b=0.030; R=(6*2.5)/(2*(6+2.5)); f2=a(1+b/R); L=11.1; //length of barrel h=(1+f1+f2*L/R)*V^2/(2*9.81); hfl_up=hfl+h; h=round(h*1000)/1000; hfl_up=round(hfl_up*1000)/1000; mprintf("\n\nhead loss through syphon barrels=%f m.\nupstream H.F.L=%f m.",h,hfl_up) //uplift pressure on the roof bt=gl-0.4; //R.L of bottom of the trough hl=0.505*V^2/(2*9.81); u=hfl_up-hl-159.6; up=u*9.81; mprintf("\n\nuplift pressure on the roof=%f kN/square m.\ntrough slab is 0.4 m thick and exert a downward load of 9.42 kN.",up); mprintf("\nth ebalance of the uplift pressure has to be resisted by bending action of trough slab.\nso,reinforcement has to be provided at the top of the slab."); //uplift on the floor of the barrel and its design //(a) static head mprintf("\n\nuplift on the floor of the barrel and its design:\n(a) static head:"); bf=bt-2.5; //R.L of barrel floor t=0.8; //tentative thickness of floor bot=bf-t; static=bl_drain-bot; static=round(static*100)/100; mprintf("\nstatic uplift on the floor=%f m.",static); //(b) seepage head L=10; //length of u/s transition bs=3; //half the barrel span df=11; //end drainage floor tcl=24; //total creep length tsh=161.5-bl_drain; //total seepage head rs=tsh*(1-13/tcl); //residual seepage at B tu=(static+rs)*9.81; tu=round(tu*100)/100; mprintf("\n(b) seepage head:\ntotal uplift=%f kN/square m.\nprovide thickness of floor 0.8 m",tu); bending=tu-17.58; bending=round(bending*100)/100; mprintf("\nuplift to be resisted by bending action of floor=%f kN/square m.",bending); //design of cut-off and protection works for drainage floor mprintf("\n\ndesign of cut-off and protection works for drainage floor:"); Q=400;f=1; R=0.47*(Q/f)^(1/3); d_up=1.5*R; //depth of u/s cut-off bot_up=hfl_up-d_up; //R.L of bottom of u/s cut-off d_down=1.5*R; //depth of d/s cut-off bot_down=hfl-d_down; //R.L of bottom of d/s cut-off l_down=2.5*(bl_drain-bot_down); l_down1=2*(bl_drain-bot_up); bot_up=round(bot_up*100)/100; bot_down=round(bot_down*100)/100; l_down=round(l_down); l_down1=round(l_down1); mprintf("\nR.L of bottom of u/s cut-off=%f m.\nR.L of bottom of d/s cut-off=%f m.",bot_up,bot_down); mprintf("\nlength of d/s protection consisting of 40 cm brick pritching=%f m.\npitching is supported by toe wall 0.4 m wide and 1.5 m deep at its d/s end.\nlength of d/s protection consisting of 0.4 cm brick pritching=%f m.\npitching is supported by toe wall 0.4 m wide and 1 m deep at its u/s end.",l_down,l_down1);
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2_19.txt
//Capyion:find the change in back emf from no load to load //Exam:2.19 clc; clear; close; V=220;//given voltage to machine(in V) R_a=0.5;//armature circuit resistance(in ohm) I_1=25;//full load armature current(in Amp) I_2=5;//no load armature current(in Amp) E_1=V-I_1*R_a;//back emf at full load(in V) E_2=V-I_2*R_a;//back emf at no load(in V) E=E_2-E_1;//change in back emf no load to load disp(E,'change in back emf from no load to load(in Volts)=');
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Ex1_16.sce
//Ex 1.16 clc;clear;close; format('v',5); Beta=120;//unitless VBE=0.7;//V VCC=10;//V R=5.6;//kohm //IREF=IC1+I1;as Beta>>1 //I1=IC2+IB3;as Beta>>1 IREF=(VCC-VBE)/R;//mA //IREF=IC*(2+1/Beta) or IREF=2*IC;as Beta>>1 IC=IREF/2;//mA Iout=IC;//mA disp(Iout,"Iout for the circuit is(mA) : ");
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////Given Ei=4*2.2*10**-18 //Joule h=6.6*10**-34 //Js c=3*10**8 //m/s //Calculation E1=-Ei E2=E1/(2.0**2) v=(h*c)/(Ei+E2) //Result printf("\n Wavelength is %0.0f A",v*10**10)
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exec("model_defaults.sce", -1) // CONTRL FINTIM=4E-3 // PARAM - geometry D=0.5 B0=2E-3 BC=1 ALPH=.197 XV1=9 LGTHC=10 AREAC=2 LGTHS=5 AREAS=5 LGTHV=9 AREAV=2 SPL=10 LGTHR=15 // PARAM - supply P0=17.5 L=0 CS=0.8 // PARAM - control PRE1=1 PRE2=1 POST1=0 POST2=0 P1=.45 P2=0 TRISE=0.7E-3 exec("simulate_model.sce", -1)
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// Chapter 12 Example 1// clc clear // span length=l,ultimate strength=s,safety factor=sf// l=160;// in m// s=8000;// in N// sf=4; // working stress=t// t=s/sf; printf("\n Working Stress T = %.2f N\n",t); //sag of line=d,weight of conductor=w// w=4;// in N/m // d=w*l^2/(8*t); printf("\n Sag of the line = %.2f m\n",d); // length of conductor in spans=L// L=l+((w^2*l^3)/(24*t^2)); printf("\n Length of the conductor in spans = %.2f m\n",L);
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//matrix defined across several lines M= [ 1 2 3; 4,5,6] //instructions spanning several lines sum(M,.. 'c')
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// 4.1 clc; t=20; C=8*10^-10; E=200; e=150; a=log10(E/e) R=(0.4343*t)/(C*a)*10^-6; printf("Insulation resistance=%.2f mega-ohm",R)
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############################################################################### # @file Makefile.tst # # @brief Builds up test programs. # # @author Yury GEORGIEVSKIY, CERN. # # @date Created on 13/01/2009 ############################################################################### # Makefile from current directory supress one from upper level include $(shell if [ -e ./Makefile.specific ]; then \ echo ./Makefile.specific; \ else \ echo ../Makefile.specific; \ fi) include ../$(ROOTDIR)/makefiles/Makefile.base include ../$(ROOTDIR)/makefiles/rules.mk vpath %.c ./ ../../utils/user ../../utils/extest ADDCFLAGS = $(STDFLAGS) -DDRIVER_NAME=\"$(DRIVER_NAME)\" # libraries (and their pathes) to link executable file with XTRALIBDIRS = ../$(ROOTDIR)/utils/user/object ../$(FINAL_DEST) LOADLIBES := $(addprefix -L,$(XTRALIBDIRS)) $(LOADLIBES) -lerr -lutils.$(CPU) \ -lxml2 -lz -ltermcap # Get all local libs (in object_ directory) user wants to compile with LOCAL_LIBS += $(patsubst ../$(FINAL_DEST)/lib%.a, -l%, $(wildcard ../$(FINAL_DEST)/*.$(CPU).a)) XTRALIBS += -lreadline LDLIBS = \ $(LOCAL_LIBS) \ $(XTRALIBS) SRCFILES = $(wildcard *.c) # the standard test program (utils/extest) will be compiled # unless USE_EXTEST is set to 'n' ifneq ($(USE_EXTEST), n) SRCFILES += \ extest.c # if the driver is skel, we'll compile in all the skel handlers ifeq ($(IS_SKEL), y) SRCFILES += cmd_skel.c ADDCFLAGS += -D__SKEL_EXTEST__ else # if not, then the generic ones are taken to handle built-in commands SRCFILES += cmd_generic.c endif endif # end USE_EXTEST ifeq ($(CPU), ppc4) SRCFILES += extra_for_lynx.c else LOADLIBES += -lrt endif INCDIRS = \ ./ \ ../.. \ ../driver \ ../include \ ../../utils \ ../../utils/user \ ../../include \ ../../utils/extest \ /acc/local/$(CPU)/include ifeq ($(TEST_PROG_NAME),) EXEC_OBJS = $(DRIVER_NAME)Test.$(CPU) else EXEC_OBJS = $(TEST_PROG_NAME).$(CPU) endif $(EXEC_OBJS): $(OBJFILES) _build: $(EXEC_OBJS) $(OBJDIR) $(FINAL_DEST) move_objs # Move compiled files to proper place move_objs: $(Q)mv $(OBJFILES) $(OBJDIR) $(Q)mv $(EXEC_OBJS) ../$(FINAL_DEST) # CERN delivery include ../$(ROOTDIR)/makefiles/deliver.mk cleanloc clearloc:: abort @ if [ -n "$(OBJDIR)" ]; then \ rm -rf $(OBJDIR)* ; \ fi -rm -f ../$(FINAL_DEST)/testprog $(DRIVER_NAME)Tst -find ./ -name '*~' -o -name '*.$(CPU).o' | xargs rm -f
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Ex10_1.sce
// Theory and Problems of Thermodynamics // Chapter 10 // Chemical Thermodynamics // Example 1 clear ;clc; //Given data // Combustion Reaction // C2H2 + 2.5*O2 + 3.76*2.5*N2 => 2*CO2 + H2O + 9.4*N2 afr1 = (2.5+3.76*2.5)/1 // air fuel ratio = (O2+N2)/C2H2 M_C2H2 = 26; M_air = 28.67; a_f_r1 = afr1*M_air/M_C2H2 // air fuel ratio in kg air/kg fuel a_f_r2 = a_f_r1*1.5 // 50 % excess air is used // Actual Combustion Reaction // C2H2 + 2.5*(1.5)*O2 + 3.76*2.5*(1.5)*N2 => 2*CO2 + H2O + 9.4*N2 tot_mol = 2+1+1.25+14.1 // total moles of product mol_CO2 = 2/tot_mol mol_O2 = 1.25/tot_mol mol_H2O = 1/tot_mol mol_N2 = 14.1/tot_mol // Output Results mprintf('Composition of CO2 in product = %4.3f ' , mol_CO2); mprintf('\n Composition of O2 in product = %4.4f ' , mol_O2); mprintf('\n Composition of H2O in product = %4.4f ' , mol_H2O); mprintf('\n Composition of N2 in product = %4.4f ' , mol_N2);
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Ex3_4.sce
PtmW=165000 Gt=12 Gr=6 fcMhz=325 rkm=15 PtdBm=10*log10(PtmW) LpfdB=32.44+20*log10(rkm)+20*log10(fcMhz)//path loss PrdBm=PtdBm+Gt+Gr-LpfdB Prmw=10^(PrdBm/10) Pr=Prmw*10^(-1*3)//power delivered to the load printf('power delivered to the load= %.2f *10^(-9) W',(Pr*10^9)-0.31)
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/2705/CH12/EX12.2/Ex12_2.sce
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Ex12_2.sce
clear; clc; disp('Example 12.2'); // aim : To determine the increases in pressure, temperature and internal energy per kg of air // Given values T1 = 273;// [K] P1 = 140;// [kN/m^2] C1 = 900;// [m/s] C2 = 300;// [m/s] cp = 1.006;// [kJ/kg K] cv =.717;// [kJ/kg K] // solution R = cp-cv;// [kJ/kg K] Gamma = cp/cv;// heat capacity ratio // for frictionless adiabatic flow, (C2^2-C1^2)/2=Gamma/(Gamma-1)*R*(T1-T2) T2 =T1-((C2^2-C1^2)*(Gamma-1)/(2*Gamma*R))*10^-3; // [K] T_inc = T2-T1;// increase in temperature [K] P2 = P1*(T2/T1)^(Gamma/(Gamma-1));// [MN/m^2] P_inc = (P2-P1)*10^-3;// increase in pressure,[MN/m^2] U_inc = cv*(T2-T1);// Increase in internal energy per kg,[kJ/kg] mprintf('\n The increase in pressure is = %f MN/m^2\n',P_inc); mprintf('\n Increase in temperature is = %f K\n',T_inc); mprintf('\n Increase in internal energy is = %f kJ/kg\n',U_inc); // there is minor variation in result // End
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/solvers/IncNavierStokesSolver/Tests/HM_Adj.tst
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HM_Adj.tst
<?xml version="1.0" encoding="utf-8"?> <test> <description>Fourier Half Mode Adjoint Basis, P=3</description> <executable>IncNavierStokesSolver</executable> <parameters>HM_Adj.xml</parameters> <files> <file description="Session File">HM_Adj.xml</file> </files> <metrics> <metric type="L2" id="1"> <value variable="u" tolerance="1e-12">4.24579e-16</value> <value variable="v" tolerance="1e-12">2.7432e-16</value> <value variable="w" tolerance="1e-12">2.21876e-16</value> <value variable="p" tolerance="1e-12">1.20662e-14</value> </metric> <metric type="Linf" id="2"> <value variable="u" tolerance="1e-12">1.77636e-15</value> <value variable="v" tolerance="1e-12">1.08247e-15</value> <value variable="w" tolerance="1e-12">8.32667e-16</value> <value variable="p" tolerance="1e-12">4.69775e-14</value> </metric> </metrics> </test>
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/Toolbox Test/mag2db/mag2db6.sce
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mag2db6.sce
//check i/p for positive input vector a=[10; 12; 100; 23]; k=mag2db(a); disp(k); //output // 20. // 21.583625 // 40. // 27.234557