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clc //initialisation of variables d= 2 //in F= 1000 //lb t= 10 //sec L= 48 //in S= 24 //in //CALCULATIONS ohp= F*L/(t*6600) Ac= %pi*d^2/4 P= ohp*t*6600/(S*Ac) //RESULTS printf ('Pressure within the system = %.f psi',P)
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//clear// clear; clc; //Example 28.1 //Given rho_p = 0.002650; //[g/mm^3] a = 2; phi_s = 0.571; //Solution //(a) //For the 4/6-mesh increment, from Table 28.2 x = [0,2.51,12.5,32.07,25.7,15.9,5.38,2.10,1.02,0.77,0.58,0.41,0.31,0.75]'*10^-2; //[mass fraction] Dp = [4.699,3.327,2.362,1.651,1.168,0.833,0.589...
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errcatch(-1,"stop");mode(2); //example 3.3 //page 120 ; funcprot(0); // Initialization of Variable pi=3.14; Q=integrate('10*r-1000*r^3','r',0,0.1); V=2*Q/0.1^2; disp(V,"velocity(m/s)="); exit();
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clear clc //Example 14.11 FRANCIS TURBINE r1=0.6; //[m] beta1=110; //degrees w=600*(2*%pi)/60 //angular speed[rad/s] Q=4; //discharge[m^3/s] B=0.1; //blade height[m] //Radial velocity at inlet Vr1=Q/(2*%pi*r1*B) //[m/s] //Inlet guide vane angle alpha1=acotd((r1*w/Vr1)+cotd(beta1)) printf("\nThe guide vane ...
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clc; printf("\n Example 4.2\n"); l=30;//Length of the tube d=150e-3;//Diameter of the tube P1=0.4e3;//Initial Pressure P2=0.13e3;//final Pressure //X=e/d, Relative roughness //Y=R/(rho*u^2) = 0.004 X=0.003; Y=0.005; v1=21.15e1; G_A=poly([0],'G_A'); f=(G_A^2*log(P1/P2))+((P2^2-P1^2)/(2*P1*v1))+(4*(Y*l/...
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clc; clear; vis=0.4;//Ns/(m^2) d=900;//kg/(m^3) D=0.02;//m Q=2.0*(10^-5);//(m^3)/s x1=0; x2=10;//m p1=200;//kPa x3=5;//m V=Q/(%pi*(D^2)/4);//m/s Re=d*V*D/vis; disp("Hence the flow is laminar.",Re,"a) Reynolds number =") pdiff=128*vis*(x2-x1)*Q/(%pi*(D^4)*1000); //for part b0 p1=p2; Q=%pi*(pdiff-(sw*l*sin(...
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//Chapter 3 //Example 3.2 //page 88 //To claculate the capacitance to neutral and charging current of a three phase transmission line clear;clc; d=350; //distance between adjacent lines r=1.05/2; //radius of the conductor v=110e3; //line voltage; f=50; Deq=(d*d*2*d)^(1/3); //GMD or equivalent Cn=(0.0242/l...
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clear clc x0 = [39.8858212875252,0.1,0,0]; u0 = [-0.15625,1112.82]; //C=[1 0;0 1] A = [-0.05321913896678525,-7.309325638062395,-0,-0,-9.809999999999766,-0,-0,-0.1176174797773325,-0,-0,-0,-0;-0.01219893117742086,-1.718678333264892,0,0,-5.377762026288438e-08,0,0,0.972369652215591,0,0,0,0;0,0,-0.1173471007801458,0,0,0...
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clc //Example 4.1 //calculate change in pot. energy per unit mass and total change in pot. energy g=9.81;//m/s^2 acc. due to gravity dh=23;//m change in height dpe=g*dh//m^2/s^2 change in pot energy per unit mass printf("change in pot. energy per unit mass is %f m^2/s^2\n",dpe); m=10;//kg dPE=m*dpe//kgm^2/s^2 o...
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errcatch(-1,"stop");mode(2);//Example 2_23 ; ; //To calculate the possible order of spectra N=5.095*10^3 //units in lines per inch lemda=6000*10^-8 //units in cm m=(1/N)/lemda printf("The possible order of the spectra is %.0f",m) exit();
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// find frequency and peak-to-peak voltage // Electronic Principles // By Albert Malvino , David Bates // Seventh Edition // The McGraw-Hill Companies // Example 22-12, page 873 clear;clc; close; // Given data Vsat=13;// in volts R1=1*10^3;// resistance in ohms R2=100*10^3;// resistance in ohms R3=10*1...
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function [t,x] = alpha_mixed_euler(f, x0, ti, tf, h, alpha) t=[ti:h:tf]; x=zeros(length(x0),length(t)); x(:,1)=x0; k=0 for tk=ti:h:tf-h k = k+1; // forward euler xf = step_feuler(f, x(:,k), tk, h); // backward euler xb = step_beuler(f, x(:,k), tk, h); ...
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//Problem 1.02: A mass of 5000 g is accelerated at 2 m/s2 by a force. Determine the force needed. //initializing the variables: M = 5; // in Kg a = 2; // in m/s2 //calculation: F = M*a printf("\n\nResult\n\n") printf("\nForce: %.0f Newton(N)\n",F)
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// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART II : TRANSMISSION AND DISTRIBUTION // CHAPTER 7: UNDERGROUND CABLES // EXAMPLE : 7.13 : // Page number 219 clear ; clc ; close ; // Clear the work space and console ...
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// (PG 607) A = [1 2 3;2 3 4;3 4 5] lam = spec(A)' lam1 = lam(1,3) lam2 = lam(1,1) lam3 = lam(1,2) // Theoretical ratio of convergence lam2/lam1 b = 0.5*(lam2+lam3) B = A-b*eye(3,3) // Eigen values of A-bI are: lamb = spec(B)' lamb1 = lamb(1,3) lamb2 = lamb(1,2) lamb3 = lamb(1,1) // ...
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// Copyright (C) 2016 - Corporation - Author // // About your license if you have any // // Date of creation: 16 juin 2016 // //SCI2C: DEFAULT_PRECISION= DOUBLE // // ReadWave : FFT_Mat.sce // function Sac = GetSac() Type = ['D' ;'W5' ;'H3' ;'5' ;'6' ;'7' ;'8' ;'9' ;'P...
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function L = cholesky(A) // Autores: Hector E. Gomez Morales // Funcion que realiza la factorizacion de cholesky ie A = LL^t (L triangular inferior) // Regresa L una matriz triangular inferior // A debe ser una matriz cuadrada, simetrica y definida positiva para que se pueda realizar esta factorizacion //**************...
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clc; H0=282990; HRo=(1*1018)+(0.5*1036); HRr=(1*86115)+(0.5*90144); HPo=1*1368; HPr=1*140440; HT=H0+(HRr-HRo)-(HPr-HPo); disp(HT,"Δh at 2800 K is:")
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clear; clc; printf("\n Example 3.5"); //rate of dissolution of salt function[x]=dissolution(d) x = (3*10^(-6))-(2*10^(-4)*3.406*10^(5)*d^2); funcprot(0) endfunction //rate of falling of the particle in stokes law region function[y]=rate_h(d) y = 3.406*d^(2)/(-3*10^(-6)-68.1*d^2);//y is in m/sec ...
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function [] = kiks_gui_eventhandler(win,x,y,ibut) printf("Event handler %d\n", ibut); if ibut == -1000 then kiks_quit; end; endfunction;
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.VECTORS 125 .PATTERNS 010010111011001001110111001010001lhllhlhhhlhhllhllhhhlhhhllhlhlllhlhhhlhlhhlllhlllhhhhlllhhhlhhhllhhlhhhlhlhhlllhlllhhhhhhlhlhhhllhllllhhlllhhlllhllhhhhhhhhhhllhhlhllhhhhllllhhllhhhlllllhhhhhhhhlhllhhhhhhhhhhhlhhhhhhlhhhhhllhhlhlhllllhhhhlhlhhhhhhhlhhhhhlhhhhhlllhhhhhhhhlllllhllllhhllllllhhll...
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 5.15\n\n\n"); // Chapter 5 : Properties Of Liquids And Gases // Problem 5.15 (page no. 202) // Solution //The values of temperature and pressure are listed in Table 4(Figure 5.10) and can be read directly. printf("Specific volume of subcooled water at ...
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// Example 3_5 clc;funcprot(0); // Given data m=100;// kg d=3;// m V=0.002;// m^3 P_gage=100;// The gage pressure in kPa g=9.81;// m/s^2 // Calculation F=m*g;// N W=-(F)*(d);// J P_abs=200;// The absolute pressure in kPa W_p=P_abs*10^3*V;// The work done on the system in J W_net=W+W_p;// The net work don...
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clc; //A circuit is midpoint bised when the Q-point value of Vce is one half of Vcc. //from example and 7.3 Vcc=8; //volt Vbe=0.7; //volt Rb=360000; //ohm Ib=(Vcc-Vbe)/Rb; //Ampere Hfe=100; Ic=Hfe*Ib;//Ampere Rc=2000; //ohm Vce=Vcc-Ic*Rc; //volt disp('V',Vce,"Vce=");//The answers vary due to round off error...
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clc //initialisation of variables a=12//in h=121//in p=11//in s=220//in //CALCULATIONS B={a/[p*(h-1)]}*s//per unit //RESULTS printf('the interval of time a noted before=% f per unit',B)
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//ques2 //The Ideal Otto Cycle clear clc //the temperature and pressure of air at the end of the isentropic compression process (state 2), using data from Table A–17 T1=290;//initial temp in K u1=206.9;//initial internal energy in kJ/kg vr1=676.1;//initial reduced volume //Process 1-2 (isentropic compression of...
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//CHAPTER 10- THREE-PHASE INDUCTION MACHINES //Example 6 disp("CHAPTER 10"); disp("EXAMPLE 6"); //VARIABLE INITIALIZATION P=4; //number of poles f_r=2; //rotor frequency in Hertz f_s=50; //stator frequency in Hertz v=400; //in Vo...
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clear; clc; N1=2; N2=3; //n=1; //while(n<=N1) // i(n)=input("enter the elements of first set"); // n=n+1; //end; i=['p' 'q']; //set A //n=1; //while(n<=N2) // j(n)=input("enter the elements of second set"); // n=n+1; //end; j=['r' 's' 't']; //set B c=1;d=1; for a=1:1:N1 //realtion ...
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//clear// clear; clc; //Example 9.8 //Given D1 = 1; //[ft] D6 = 6 Nre_i = 10^4; Da = 4; //[in.] t1 = 15; //[s] P = 2; //[hp/gal] //(a) //Using Fig. 9.15 //the mixing factor ntT is constant and time tT is asumed constant, //speed n will be the same in both vessels. //Using Eq.(9.24) with consant dens...
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clear; fc = 12800; // freq de coupure filtre passe-bas fs = 44100; // freq echantillonnage // filtre passe bas analogique //hs=analpf(20,'ellip',[1E-4 1E-4],fc); //fr=20:100:20000; //hf=freq(hs(2),hs(3),%i*fr); //hm=abs(hf); //plot(fr,hm) // filtre numerique RII reponse impulsionnelle infinie (passe-bande) //hz=iir(...
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## Regression test authors write format # Expected format: USER = Name <USER@DOMAIN> read <<EOF blob mark :1 data 3 Hi reset refs/heads/master commit refs/heads/master mark :2 author Kevin O. Grover <kevin@kevingrover.net> 1472305515 -0700 committer Kevin O. Grover <kevin@kevingrover.net> 1472305515 -0700 data 13 Just...
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<?php include ('knights.php'); $array=newBoard(); printBoard($array); ?>
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//exapple 1.10 clc; funcprot(0); // Initialization of Variable pi=3.14159; theta=20+30/60; H=42+6/60;//hour angle delta=50; //in triangle ZPM(figure in book) PZ=(90-delta)*pi/180; H=H*pi/180; PM=(90-theta)*pi/180; ZM=acos((cos(PZ)*cos(PM)+sin(PM)*sin(PZ)*cos(H))); alpha=pi/2-ZM; alpha=alpha*180/pi; d...
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clc pathname=get_absolute_file_path('9_1_2.sce') filename=pathname+filesep()+'912.sci' exec(filename) printf(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook") disp("From reaction, only gaseous are counted") left=1+2 right=1+1 deltaUr=deltaHr-R*T*(rig...
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//----------------------------------Gauss-------------------------------------// function [res]=usolve(U,b) n=size(b,1) x=zeros(n,1) x(n)=b(n)/U(n,n) for i= n-1:-1:1 x(i)=(b(i)-U(i,(i+1):n) * x((i+1):n))/U(i,i) end res=x endfunction function [res]= gauss(A,b) t=[] n=size(...
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clear; clc; R=55;L=0.6*(10^-3);G=1*(10^-6);C=0.04*(10^-6);f=800;r=8;l=0.1;d=2.5; //value of l(loading coil inductance) as taken in solution w=2*%pi*f; Z=round(R+(%i*w*L)); Y=G+(%i*w*C); Zo=sqrt(Z/Y); P=round(sqrt(Z*Y)*10^4)/10^4; Zc=r+(%i*w*l); A=fix(((cosh(P*d))+(Zc*(sinh(P*d))/(2*Zo)))*10^3)/10^3; Pl=(acosh...
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// Example 6_3 clc;funcprot(0); // Given data p_1=2.00;// MPa p_2=0.100;// MPa T_2=150;// °C h_1=2776.4;// kJ/kg h_2=2776.4;// kJ/kg // Calculation h_f1=908.8;// kJ/kg h_fg1=1890.7;// kJ/kg h_g1=2799.5;// kJ/kg x_1=(h_1-h_f1)/h_fg1;// The quality of steam x_1=x_1*100;// The quality of steam in % T_1=212...
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clear; clc; //To find Approx Value function[A]=approx(V,n) A=round(V*10^n)/10^n;//V-Value n-To what place funcprot(0) endfunction //Example 11.8 //Caption : Program to Find the Fugacity of 1-butene vapor T=473.15;//[K] P=70;//[bar] Tc=420;//[K] Pc=40.43;//[bar] omega=0.191; //By interpolati...
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//Example 17_6 clc(); clear; //To calculate the cost needed to operate power=0.7 //Units in KW time=0.5 //Units in h heat=power*time //Units in K Wh cost=0.10 //Units in Dollars tcost=cost*heat //Units in Dollars printf("Cost needed to operate is=%.4f Dollars",tcost)
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//pathname=get_absolute_file_path('18.07.sce') //filename=pathname+filesep()+'18.07-data.sci' //exec(filename) p2 = 4; Cp = 1.005; //Pressure(in bar): p1=1.2 p6=p1 p3=4 p3=p2 p4=1 p7=0.9 //Temperatures(in K): T1=288 T6=T1 T5=25+273 T3=323 T8=30+273 n=1.45 n1=1.3 //Temperature at state 2(in K): T2=T1*(p2/p1)^((n-1)/n) T...
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//example 14 clear alpha1=0.99; ib=25*10^-6;//ampere icb=200*10^-9;//ampere beta1=alpha1/(1-alpha1); ic=beta1*ib+(beta1+1)*icb; disp("collector current = "+string((ic))+"ampere"); ie1=(ic-icb)/alpha1; disp("emitter current = "+string((ie1))+"ampere"); ic=beta1*ib; disp("collector current with ib = ...
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function s = from_idx_to_name(path_idx,filename_idx,path_names,filename_names) //retrieve indexes idx = csvRead(path_idx+filename_idx,",",".","double"); idx($,:)=[]; //s=idx; //retrieve names all_data = csvRead(path_names+filename_names,",",".","string"); all_data(find(all_data(:,1)=="[...
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clear // // // //Variable declaration L=3*10**-10 //length(m) m=9.1*10**-31 //mass(kg) e=1.6*10**-19 //charge(c) h=6.63*10**-34 //plank constant //Calculation E1=h**2/(8*m*e*L**2) //minimum energy(eV) //Result printf("\n minimum energy is %0.1f eV",E1)
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clc(); clear; // To compute the temprature distribution h=1; // Heat transfer coefficient in Btu/hr-ft^2-degF x=1; // Assumed thickness in ft k=1; // Thermal conductivity in Btu/hr-ft-degF N=h*x/...
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//Chapter 1 //Example 2.1 //Page 16 clear; clc; n_overall = 20; W = 0.6; printf("Let x kcal/kg be the calorific value of fuel.\n") printf("Heat produced by 0.6 kg of coal = 0.6 x kcal\n") printf("Heat equivalent of 1 kWh = 860 k cal\n") //Calculation of calorific value of coal printf("Now, n_overall = Electrical ou...
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//Caption:Find (a)the rotor copper loss per phase if motor is running at slip of 4% (b)Mechanical power developed //Exa:11.9 clc; clear; close; P_i=100000//Input power(in watt) P_sc=2000//Stator copper loss(in watt) s=4//slip(in %) P_r=P_i-P_sc P_rc=((s/100)*P_r)/3 disp(P_rc,'(a)Rotor copper lossper phase(in ...
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x=[30 40 30 50 20 60] y=[1 2 3 4 5 6] subplot(3,1,1) bar(x,y,'c') xlabel('x-axis') ylabel('y-axis') xtitle('Bargrap 1') xgrid() subplot(3,1,2) bar(x,y,'c') xlabel('x-axis') ylabel('y-axis') xtitle('bargraph 2') xgrid() subplot(3,1,3) bar(x,y,'c') xlabel('x-axis') ylabel('y-axis') xtitle('bargraph 3') xgrid()
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============= Simple datums ============= #lang test 32 ; this is a line comment 'akaibara "a little string" `define #t #F #T --- (source_file (lang_line (symbol)) (datum (number)) (datum (quoted_datum (datum (symbol)))) (datum (string)) (datum (quoted_datum (datum (symbol)))) (datum (bool...
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clc clear //Input data M=2.5 //Mach number h=10 //Height in km //Calculation alp=asind(1/M) //Mach cone angle in degree d=10/tand(alp) //Distance the jet would cover before a sonic boom is heard on ground in km //Output printf('Distance the jet would cover before a sonic boom is heard on ground is %3.2f...
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// Example 4.5 // Computation of (a) Shaft speed (b) Mechanical power developed // (c) Developed torque // Page No. 152 clc; clear; close; // Given data Prcl=263; // Rotor copper loss Pgap=14580; // Power input to the rotor fs=60; // Frequency p=4; ...
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//to find voltage available b/w slip rings and its freq clc; disp('(a)'); f=50; p=6; n_s=120*f/p; n=-1000; s=(n_s-n)/n_s; f_s=f*s; disp(f_s,'slip freq(Hz)'); v2=100; V2=s*v2; disp(V2,'slip ring voltage(V)'); disp('(b)'); n=1500; s=(n_s-n)/n_s; f_s=abs(f*s); disp(f_s,'slip freq(Hz)'); v2=100; V2=s...
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b = 6 ; // Outer dimension of the pole in inch t = 0.5 ; // thickness of the pole P1 = 20*(6.75*24); // Load acting at the midpoint of the platform d = 9 ; // Distance between longitudinal axis of the post and midpoint of platform P2 = 800; // Load in lb h = 52 ; // Distance between base and point of action of P2 ...
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 6 //Thermodynamic Properties of Pure Fluids //Example 8 clear; clc; //Given: //Equation of state: P(V-B) = RT + (A*P^2)/T Cp = 33.6; //mean specific heat at atmosheric pressure (J/mol K) A = 1*10^-3; //m^3 K/(bar)mol B = 8.0*10^-5; //m^3/mo...
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A=[0 10 30 40;10 0 20 30;30 20 0 -50;40 30 50 0]; b=[-50;-40;120;50]; A1=A;A1(:,1)=b; A2=A;A2(:,2)=b; A3=A;A3(:,3)=b; A4=A;A4(:,4)=b; D=det(A); d(1)=det(A1); d(2)=det(A2); d(3)=det(A3); d(4)=det(A4); x=d/D P=A*x-b
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clear; clc; //Example 3.4 Vbb=1.5;//(V) Rb=580;//(KOhm) Veb=0.6;//(V) Vcc=5;//(V) b=100; //writing Kirchhoff voltage law equation around E-B loop Ib=(Vcc-Veb-Vbb)/Rb; printf('\nbase current=%.3f mA\n',Ib) Ic=b*Ib; printf('\ncollector current=%.2f mA\n',Ic) Ie=(1+b)*Ib; printf('\nemitter current=%.3f mA\n'...
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function [m,b] = gerade(x, y) m1 = (y(2)-y(1))/(x(2)-x(1)); m2 = (y(3)-y(2))/(x(3)-x(2)); m = (m1 + m2); b = y(1) - m*x(1); endfunction
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//pagenumber 581 example 5 clear voltag=2000;//volt d=2*10^-2;//metre //(1) frequency vx=sqrt(2*1.6*10^-19*(voltag)/(9.11*10^-31)); durati=d/vx; freque=1/(2*durati); disp("max frequency "+string((freque))+"hertz"); //(2) durati=60*durati; disp("duration electron between the plates = "+string((durati))+...
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*Testcase dfp-080-to-packed.tst: CPDT, CPXT sysclear archmode esame # # Following suppresses logging of program checks. This test program, as part # of its normal operation, generates 16 expected program specification checks from # packed lengths greater than allowed. This is part value of the validation process. ...
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//Caption:Determine the internal losses,torque & efficiency of the motor //Exam:2.48 clc; clear; close; V=240;//supply voltage(in V) I=80;//motor taking current(in Amp) H.p=20;//motor giving H.P. I_p=V*I;//input (in watts) O_p=H.p*735.5;//output (in watts) L_i=I_p-O_p;//internal losses(in watts) disp(L_i,'in...
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errcatch(-1,"stop");mode(2);//Caption:Find (a)Starting current (b)Starting torque //Exa:12.2 ; ; P=3000//Power of motor(in watt) V=415//Voltage supplied(in volts) f=50//Frequency(in hertz) p=6//Number of poles pf=0.8//Power factor pfs=0.64//Power factor on short circuit pfn=0.1//No load power factor I_n=3....
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function [z]= g1(X) z = X.^0 endfunction function [z] = g2(X) z = X endfunction function [z] = g3(X) z = X.^2 endfunction exec('quadrados_minimos.sci'); // definindo os pontos tabelados da função X = [0, 1, 2, 3]; F = [3, 5, 8, 13] ; GLista = list(g1,g2,g3) [a] = quadrados_mi...
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clc //initialisation of variables clear v= 1 m= 0.5 //CALCULATIONS m1 = 2*m m2 = 1*m v1 = 2*v v2 = 1*v M = (m1^2*m2)^(1/(v1+v2)) //RESULTS printf ('mean ionic molality = %.1f ',m2) printf ('\n mean ionic molality = %.3f ',M)
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//pathname=get_absolute_file_path('3.17.sce') //filename=pathname+filesep()+'3.17-data.sci' //exec(filename) //Initial pressure(in MPa): p1=0.5; CpH2 = 14.307; CpN2 = 1.039; RN2 = 0.2968; RH2 = 4.124; T1 = 300; v1=0.5; vN1 = 0.5; vN2 = 0.75; v2 = 0.25; //Initial volume(in m^3): vi=0.5 //Final pressure(in MPa): pf...
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clc //Given that alpha_300 = 2.5e-39 // total polarisability in C^2m/N at 300 K alpha_600 = 1.75e-39 // total polarisability in C^2m/N at 600 K T1 = 300 // Initial temperature in Kelvin T2 = 600 // Final Temperature in Kelvin printf("Example 4.12\n") b = (alpha_300-alpha_600)*T2 al_def_300 = alpha_300 - b/30...
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//example 1.4 //best approximation //page 10 clc;clear;close; A_X=1/3;//the actual number X1=0.30; X2=0.33; X3=0.34; A_E1=abs(A_X-X1); A_E2=abs(A_X-X2); A_E3=abs(A_X-X3); if(A_E1<A_E2) if(A_E1<A_E3) B_A=X1; end end if(A_E2<A_E1) if(A_E2<A_E3) B_A=X2; end end if(A_E3<A_E2) if(A_E3<A_E1) ...
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Turbomachinery Design and Theory,Rama S. R. Gorla and Aijaz A. Khan, Chapter 3, Example 16") disp("If D1 is the diameter of pipe, then discharge is Q = pi*D1^2*C2/4") Q = 0.245 D1 = 0.28 C2 = 4*0.245/(%pi*0.28^...
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clc // Given that d = 0.0001 // distance between two slits in meter Beta = 0.005 // width of the fringes formed in meter D = 1 // distance between slit and screen in meter // Sample Problem 10 on page no. 1.43 printf("\n # PROBLEM 10 # \n") lambda = (Beta * d) / D // calculation for wavelength of light = %e met...
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l=1//span, in m t=0.27//tread in m sigma_cbc=5//in MPa sigma_st=140//in MPa MF=1.6 a=MF*7 D=l*10^3/a//in mm D=100//assume, in mm W1=D/10^3*t*25//in kN/m M1=W1*l/2//in kN-m M2=t*3*l/2//in kN-m M3=1.3*l//in kN-m M=M1+max(M2,M3)//in kN-m d=sqrt(M*10^6/0.87/t/10^3)//in mm d=83//in mm //assume 8 mm dia bars ...
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Sy=4.8*10^3;//shear load,given,in N A=300;//Boom area,given,in mm^2 L=200;//leangth,given,in mm
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'successful test OPEN "B", #1, "mouse2.dll" FOR i = 1 TO LOF(1) SEEK #1, i k$ = INPUT$(1, #1) mouse$ = mouse$ + k$ NEXT CLOSE #1 SCREEN 13 DEF SEG = VARSEG(mouse$) a% = SADD(mouse$) CALL absolute(a%) DO k% = INP(&H60) LOOP UNTIL k% = 129
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clc,clear printf('Example 6.27\n\n') kW=[800,500,1000,600] cosphi=[1,0.9,0.8,0.9] tanphi=tan(acos(cosphi)) kVAR=kW.*tanphi kW_total=kW(1)+kW(2)+kW(3)+kW(4) kVAR_total=kVAR(1)+kVAR(2)+kVAR(3)+-1*kVAR(4) //4th case is leading phi_c=atan(kVAR_total/kW_total) //total power factor angle phi_1=acos(0.95)//pf...
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//ex_9 to show that e^iwt is periodic with T=2*pi/W0 clear; clc; close; t=0:1/100:10; w0=1; T=2*%pi/w0; x=exp(%i*w0*t); y=exp(%i*w0*(t+T)); if ceil(x)==ceil(y) then disp('e^iwt is periodic with T=2*pi/W0') else disp('nonperiodic') end
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clc pathname=get_absolute_file_path('8_3_4.sce') filename=pathname+filesep()+'834.sci' exec(filename) printf(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook \n ") function[Cp1]=fun1(T) Cp1=0.04937 + T*13.92*10^(-5) - T^2 *5.816*10^(-8) + T^3 *7.280...
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clc clear printf("Example 2.6 | Page number 41 \n\n"); //(a)Find the work of compression of air. //(b)What would be the work done on air. //Given Data p1 = 1.0 //bar //initial pressure V1 = 0.1 //m^3 //initial volume p2 = 6 //bar //final pressure //and p1*(V1^1.4) = p2*(V2^1.4) printf("Initial ...
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clc clear Eith=0.29; Em=0.77; BP=5.5; SG=0.87; CV=43000; Ebth=Em*Eith; Mf=(BP*3600)/(Ebth*CV); D=SG*1000; Mff=(Mf*1000)/D printf('Mf= %3.2f litre/hr',Mff); printf('\n');
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//Example 20.2 delta_Q_electrons=-0.300*10^-3;//Charge (C) delta_t=1;//Time (s) I_electrons=delta_Q_electrons/delta_t;//Current due to flow of electrons (C/s) q=-1.60*10^-19;//Charge per electron (C) printf('Number of electrons passing per second = %0.2e electrons/s',I_electrons/q) //Answer varies due to round of...
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// Scilab code Ex4.31: : Pg:181 (2008) clc;clear; Lambda = 6000e-08; // Wavelength of light, cm mu = 1.35; // Refractive index of thin wedge shaped film omega = 0.20; // Fringe width, cm // As omega = Lambda/(2*mu*theta), solving for theta theta = Lambda/(2*mu*omega)*180/%pi; // Angle of the wedge, de...
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// Electric Machinery and Transformers // Irving L kosow // Prentice Hall of India // 2nd editiom // Chapter 4: DC Dynamo Torque Relations-DC Motors // Example 4-1 clear; clc; close; // Clear the work space and console. // Given data d = 0.5; // diameter of the coil in m l = 0.6; // axial length of the ...
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errcatch(-1,"stop");mode(2);//Example 9.2 : the fracture strength and compare ; ; format('v',10) //given data : E=70*10^9; // in N/m^2 C=(4.2*10^-6)/2;// in m gama=1.1; // in J/m^2 sigma=sqrt((2*E*gama)/(%pi*C)); disp(sigma,"fracture strength,sigma(N/m^2) = ") exit();
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//Example 5.33 //Newton Raphson and Mullers Method //Page no. 202 clc;clear;close; deff('x=f(x)','x=x^4-8*x^3+18*x^2+0.12*x-24.24') deff('x=f1(x)','x=4*x^3-24*x^2+36*x+0.12') //newton raphson x9=[1.5,2.5,2.7,3.1;4,5,14,10] for h=1:4 x0=x9(1,h);e=0.00001 for i=1:x9(2,h) x1=x0-f(x0)/f1(x0) e1=ab...
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// This file is part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. // File name: projects/02/ALU.tst load ALU.hdl, output-file czALU.out, compare-to czALU.cmp, output-list x%B1.16.1 y%B1.16.1 zx%B1.1.1 nx%B1.1.1 zy%B1.1.1 ny%B1.1.1 f%B1.1.1...
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//Caption: Weighted Arithmetic mean of ungrouped data //Example3.4 //page43 clear; clc; X = input('Enter the demand in units ='); w = input('Corresponding Weights='); n = length(X); num =0; den =0; for i = 1:n num = num+w(i)*X(i); den = den+w(i); end Xw = num/den; disp(Xw,'The Estimated demand for the Year 2003...
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scenario = "One-Back-Arm"; # This name is recorded in the log file scenario_type = trials; response_matching = simple_matching; response_logging = log_all; active_buttons = 3; #button_codes = 1, 2, 3; # These values will be used to code participant responses #target_button_codes = 0, 0, 2; button_codes = 0, 0, 2; tar...
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//Example 17.1 //Forces in rolling //Page No. 596 clc;clear;close; mu=0.08; //no unit R=12; //in inches alpha=atand(mu); dh=mu^2*R; printf('\nMaximum possible reduction when mu is 0.08 = %g in\n',dh); mu=0.5; //no unit dh=mu^2*R; printf('Maximum ...
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// PG (405) deff('[y]=f(x,y)','y=lamda*y+(1-lamda)*cos(x)-(1+lamda)*sin(x)') lamda = -1; [x,y]=Euler1(1,1,5,0.5,f) lamda = -10; [x,y]=Euler1(1,1,5,0.1,f) lamda = -50; [x,y]=Euler1(1,1,5,0.01,f)
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//EX10_8 PG-10.40 clc A=61;//gain required for the non inverting amplifier R1=1e3; printf("Refer to the figure-10.36 shown\n") printf(" \n The gain of the non inverting amplifier is A=1+Rf/R1") //the gain of the non inverting amplifier is A=1+Rf/R1 x=A-1;//x=Rf/R1 Rf=x*R1; printf("\n\n Therefore feedback resis...
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Ex9_6.sce
//Chapter 9 Ionic Equilibria and Buffer Action clc; clear; //Initialisation of Variables C= 0.050 //M K= 2.4*10**-17 c= 0.1 //M //CALCULATIONS c1= K*C/c**2 //RESULTS mprintf("Concentration of carbonate-ion = %.1e mole per litre",c1)
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[fd, SST, Sheetnames, Sheetpos] = xls_open("book1.xls"); s= ['The number of sheets are:', string(size(Sheetnames,2)), ' and their names are:', Sheetnames]; disp(s); [Value, TextInd] = xls_read (fd, Sheetpos (1)); mclose(fd); a=Value(TextInd(:)==0); // Extracting the numeric values s= ['The numeric values in the E...
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//Example 27_6 clc(); clear; //To find how many radium atoms in the vial undergo decay t1=5.1*10^10 //Units in sec lamda=0.693/t1 //Units in sec^-1 n1=6.02*10^26 //Units in atoms/Kmol n2=226 //Units in Kg/Kmol m1=0.001 //Units in Kg N=n1*m1/n2 //Units in number of atoms deltat=1 ...
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\name ARDUINO_SETUP \palette Arduino \smalldescription Permet de configuration le port de communication série entre l'arduino et scilab. \description Ce bloc doit \bold{obligatoirement} être placé sur le schéma lors de l'utilisation d'autres blocs de la toolbox. Il permet de définir le port de communication entre l...
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//Example 4.14 //illumination under each lamp and midway between lamps clc; clear; close; format('v',5 ) CP=100;// h=6;//in meter d=16;// meter x=sqrt(h^2+d^2);// em=2*((CP/h^2)*(h/(d-h))^3);//illumination in the middle in lux ee=((CP/h^2)*(1+(h/x)^3));//illumination iunder each lamp in lux disp(ee,"illumination under ...
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l2r.sci
// function [Rnum,Rnumdeg,Rden,Rdendeg] = l2r(N,degN,D,degD) // given Numerator and Denominator polynomial matrices in left form, // not necessarily coprime, finds right coprime factorisation. function [Rnum,Rnumdeg,Rden,Rdendeg] = l2r(N,degN,D,degD) [N,degN] = transp(N,degN); [D,degD] = transp(D,degD); [Rn...
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function []=interpolation(t0,vt0,t1,vt1,t2,vt2,t) b0=vt0; b1=(vt1-vt0)/(t1-t0); b2=((vt2-vt1)/(t2-t1)-(vt1-vt0)/(t1-t0))/t2-t0; pol=b0+b1*(t-t0)+b2((t-t0)*(t-t1)); disp(horner(pol,t)) endfunction
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clc clear //input e=235;//e.m.f generated by an armature of a d.c. machine in volts v=100;//velocity of armature of a d.c. machine in rad/s i=16;//current in amperes //calculations p=e*i;//power of armature in watts t=p/v;//required torque in newton meter //output mprintf('required torque is %3.1f Nm',t...
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clear clc A=[0 1 2;1 2 3;2 3 4] B=[1 -2;-1 0;2 -1] disp("AB= ") A*B disp("BA= ") B*A
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//example 11.7// clc //clears the screen// clear //clears already existing variables// //given A1A0=11, I1I0=01// disp('the association operation is performed with keyword 01. The memory locations 1, 5 and 7 match the keyword giving out logic 0 at the corresponding Y outputs. Therefore,') disp('Y=01011101')
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//Chapter 13_Voltage Regulators //Caption : Inductor and Capacitor //Example13.14: A switching voltage regulator operates at a switching frequency of 30kHz and is to supply a load current Io of 1 A at a dc output voltage Vo of +10V.The dc input voltage is Vin=20V and the output(peak-peak) ripple factor is not to exce...
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//Determine following parameters N = 7; C = 395; Nc = C/N; Se1 = 39.8/63.1; Se2 = 5.8648/1.384; disp(Nc, 'No. of voice channels/cell site') disp(Se1, 'Spectral Efficiency in analog system') disp(Se2, 'Spectral Efficiency in digital system')
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errcatch(-1,"stop");mode(2);//ex8.1 g_m=4*10^-3; R_d=1.5*10^3; A_v=g_m*R_d; disp(A_v,'Voltage gain') exit();