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//Example 3.15 Cruise Control Transfer Function. //Coefficients of numerator and denominator of the transfer function clear; clc; //------------------------------------------------------------------ // Transfer function coefficients num=[0.001 0]; den=[0 0.05 1]; // Transfer function Ns=poly(num,'s','coeff...
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// **** Purpose **** // This function generates ham file for postprocess calculations in PiLab // **** Variables **** //==== << PiLab inputs >> ==== //[ham.builder]: n x 1, string //<= tasks that used to build Hr. Currently, PiLab support the follwing //tasks to build Hr. // ['lat','hop','scc']: conventional Slaster...
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clc //Chapter 8:Linear model of phase locked loop //example 8.6 page no 349 //given //The VCO is designed to oscillate at 100kHz R1=10.8*10^3 R2=10.8*10^3 C=1*10^-9 N=2//order of filter fmin=(R2*(C+32*10^-12))^-1//minimum frequency fmax=fmin+(R1*(C+32*10^-12))^-1//maximum frequency VDD=9//regulated power sup...
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//Analysis of fuel from exhaust gas analysis clc,clear //Given: pCO2=12/100,pCO=2/100,pCH4=4/100,pH2=1/100,pO2=4.5/100 //Composition of Carbon di oxide(CO2), Carbon mono oxide(CO), Methane(CH4), Hydrogen(H2), Oxygen(O2) in dry exhaust gas pN2=1-(pCO2+pCO+pCH4+pH2+pO2) //Composition of Nitrogen(N2) in dry exhaust gas //...
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//chapter 10 //example 10.4 //page 387 printf("\n") printf("given") Idss=8*10^-3;Vpmax=6;Vgs=2.3;Vgsmax=6; Id=Idss*(1-(Vgs/Vgsmax))^2 Idss=4*10^-3;Vp=3; Idmin=Idss*(1-(Vgs/Vp))^2
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Initial=23 //lb.ft/min^2
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clc clear //Input data d1=6000//Diffraction grating have number of lines per cm w=6000//Wavelength in Angstrom //Calculations n=(1/(d1*w*10^-8))//Maxmum order of diffraction //Output printf('Maximum order of diffraction that can be observed is %i',n)
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// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 2 : BESICS OF MAGNETIC CIRCUITS // Example : 2.3 clc;clear; // clears the console and command history // Given data myu_r = 625 // relative permiability of rectangular core N = 25 // numbe...
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clc clear //Input data ng=0.97//Efficiency of electric generator nt=0.95//Efficiency of turbine nb=0.92//Efficiency of boiler nc=0.42//Efficiency of cycle no=0.33//Efficiency of overall plant //Calculations na=(no/(ng*nt*nb*nc))//Efficiency of auxiliaries n=(1-na)*100//Percentage of total electricity gene...
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f=50 N1=500 Pin=60 Io=0.4 Vin=220 r=0.8 Pci=Io*Io*r Pi=Pin-Pci disp(Pi) theta=acos(Pin/Vin/Io) Im=Io*sin(theta) Xm=Vin/Im disp(Xm) Iio=Io*cos(theta) Ri=Vin/Iio disp(Ri)
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clc clear //Input data r=8.5;//Compression ratio e=5.5;//Expansion ratio P1=1;//Pressure at the beginning of compression stroke in bar T1=313;//Temperature at the beginning of compression stroke in K n=1.3;//polytropic constant Cp=1.004;//Specific heat of air at constant pressure in kJ/kg K Cv=0.717;//Specific...
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disp("Name: Pradyumna YM") disp("SRN: PES1201700986") disp("Enter elements of A :\n"); a=input("Enter the Matrix A") [m,n]=size(a) //take inputs for the matrix b b=input("Enter the matrix B(semicolon seperated ") disp('The Solution is:'); disp((inv((a'*a))*a')*b); //we know that x=(a'*a)-1 *a'b
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clc T=2773; //K lambda=1.2*10^(-6); //m e=0.9; a=5.67*10^(-8); disp("(i) Monochromatic emissive power at 1.2 μm length") C1=0.3742*10^(-15); //W.m^4/m^2 C2=1.4388*10^(-4); //mK E_lambda_b=C1*lambda^(-5)/(exp(C2/lambda/T)-1); disp("E_lambda_b =") disp(E_lambda_b) disp("W/m^2") disp("(ii) Wavelength at ...
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//Controle de Posição de um sistema de levitação pneumática clc; clear; s = %s; //Configurações básicas para a simulação m = 0.1; C = 0.5; rho = 1; r = 0.1; A = %pi*r^2; alfa = C*rho*A/2*m; //Calculando a velocidade do ar para o equilibrio g = 9.81; vo = sqrt(g/alfa) //FT do movimento da bolinha Gb = syslin('c...
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//Problem 17.01: A 15 μF uncharged capacitor is connected in series with a 47 kohm resistor across a 120 V, d.c. supply. Use the tangential graphical method to draw the capacitor voltage/time characteristic of the circuit. From the characteristic, determine the capacitor voltage at a time equal to one time constant af...
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clc clear //Input data R0=5;//The resistance of the platinum wire of a platinum resistance thermometer at the ice point in ohms R100=5.93;//The resistance of the platinum wire of a platinum resistance thermometer at the steam point in ohms Rt=5.795;//The resistance of the platinum wire when both the thermometers a...
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//Example Non 2.5 Determine the pressure at the end of the artery when the head is// clc Bh=1.8 //in m Ah=2.4 // in m rhoHg=13.6 gHg=1000 Hhg=0.212 rhoBlood=1 gBlood=1000 gama=1000*9.8 z1=0 z2=2.4 //Calculation hBlood=(rhoHg*gHg*Hhg)/(rhoBlood*gBlood) P2=(hBlood+(z1-z2))*gama //when the head is 1.8m below the hear...
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clc; clear; mprintf('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-13.3 Page No.286\n'); Np=24; Ng=36; Pd=8; Yp=33.7*%pi/180; Yg=56.3*%pi/180; theta=14.5*%pi/180; //Pitch diameter Dp=Np/Pd; mprintf('\n Pitch diameter = %f in.',Dp); //Transmitted force n=2200; P=8; T=63000*P/n; Ft=2*T/Dp; mprintf('\n Tran...
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// exa 3.11 Pg 72 clc;clear;close; // Given Data N=200;// rpm P=25;// kW tau_d=42;// MPa W=900;// N L=3;// m Syt=56;// MPa Syc=56;// MPa sigma_d=56;// MPa T=P*60*10**3/(2*%pi*N);// N.m M=W*L/4;// N.m Te=sqrt(M**2+T**2);// N.m // Te=(%pi/16)*d**3*tau_d d=(Te*10**3/((%pi/16)*tau_d))**(1/3);// mm printf('\n shaft diame...
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// Сохранение графиков function[]=save_graphics(b,z) b_ = 0 for i=2:length(b) b_ = b_ + (i-1)*b(i) end b_ = 1/b_ a = 0.001:0.001:b_-0.001 l = length(a) for i=1:l N1(i) = get_data(a(i),b,z) end properties = gda() // x labels d...
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function [resultado] = erro_relativo(x, xl) resultado = abs((x-xl)/x) endfunction function [resultado] = digitos_significativos(x, xl) erro_rel = erro_relativo(x,xl) resultado = 0 while erro_rel <= 5*10^(resultado * (-1)) resultado = resultado + 1 end resultado = resultado - 1 endf...
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// Exa 4.2 format('v',5) clc; clear; close; // Given data V_BB = 15;// in V V_CC = 15;// in V I_CO = 0.1;// in µA I_CO = I_CO * 10^-6;// in A Beta = 60; I_B = 50;// in µA I_B = I_B * 10^-6;// in A V_CE = 8;// in V I_C = (Beta*I_B)+((1+Beta)*I_CO);// in A I_C = round(I_C * 10^3);// in mA disp("Part (i) ...
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//Chapter 9 //Example 9.3 //page 340 //To calculate subtransient current in Generator,Motor and fault clear;clc; mvab=25; kvb=11; Vo=10.6/kvb; //PU Prefault voltage printf('\nPrefault Voltage = %0.4fpu\n',Vo); Load=15/mvab; //load PU with 0.8pf leading Io=(Load/(Vo*0.8))*(cosd(36.9)+%i*sind(36.9)); //Prefault current...
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clc; clear all; X=8; //volt t =10*1e-6; //second thita=-90; //degree //part-a: find x for f= 90 Mhz f=90*1e+6; //Hz x=X*sin((3.14/180)*(6.28*f*t)+(thita)); //volt disp(+'volt',x,"for part-a x="); //part-b: find x for f= 102 Mhz f=102*1e+6; //Hz x=X*sin((180/3.14)*(6.28*f*t)+(thita)); //volt disp(+...
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// Example 4.3: Plot the min and max transconductance curve clc, clear //given data //VGS_off=-2 to -6 volt //IDSS=8 to 20 mA //Formula : ID=IDSS*[1-VGS/VGS_off]^2 //Let take some values for plotting IDSS=8;//mA VGS=0:-0.5:-2;//in Volt VGS_off=-2 ;// in Volt ID=zeros(1,5) for i=1:5 ID(i)=IDSS*[1-VGS(i...
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clear; outdir = 'cachestudy/algo/ufget'; rmdir(outdir, 's'); mkdir('cachestudy'); mkdir('cachestudy', 'algo'); mkdir('cachestudy/algo', 'ufget'); lfiles = []; outfiles = []; curdir = 'mat'; dirs = ls(curdir); l = size(dirs); l = l(1); count = 0; for i = 1:l d = dirs(i) curdir2 = strcat([curdir '/' d]); files = ls(c...
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//control systems by Nagoor Kani A //Edition 3 //Year of publication 2015 //Scilab version 6.0.0 //operating systems windows 10 // Example 6.4 clc; clear; s=poly(0,'s') K=240//the value of K h=syslin('c',240/(s*(s+10)^2)) evans(h) xtitle("uncompensated system") kvd=20//given desired velocity error cons...
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//Variable declaration lamda=5460*10**-10; //wavelength(m) d=1*10**-4; //seperation(m) D=2; //distance(m) n=10; //position //Calculation Xmax10=n*lamda*D/d; tan_phi=Xmax10/D; phi_max10=atan(tan_phi); phi_max10=phi_max10*180/%pi; //angular position of 10th maximum(degrees) phim=60*(p...
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//Chemical Engineering Thermodynamics //Chapter 13 //Thermodynamics in Phase Equilibria //Example 13.2 clear; clc; //Given P1 = 50*1.03*10^4;//Initial pressure in Kgf/sq m T = 373;//Temperature in K P2 = 1.03*10^4;//Final pressure in Kgf/sq m V = 0.001*18;//Volume in cubic meter R = 848;//gas constant in...
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//exapple 2.4 clc; funcprot(0); // Initialization of Variable //part1 pi=3.14259; H=1200;//altitude h=80;//elevation of hill f=15/100; R80=f/(H-h); disp(R80,"representative fraction of hill is (times):"); //part2 h=300;//elevation of hill R300=f/(H-h); disp(R300,"representative fraction of hill is (ti...
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ScreenName String 'Error Matching Server' ImplName String 'Dialog Screen' ElementChunkArray Int 4 ScreenElementType Int 0 ImplName String 'Front End Dialog Screen Backdrop' TabIndex Int 1 Selectable Bool False Enabled Bool True ReferenceArea Rect( 173, 170, 484, 449 ) # left,top,right,bottom ScreenElementType Int 1 Im...
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// Example 4.8 clear; clc; close; format('v',6); // Given data P=4;//no.of poles phase=3;//no. of phase slots=36;//no. of stator slotes turns=20;//turns per coil conductors=10;//per slot fi_m=1.8;//in m wb N=3000//in rpm //Calculations f=P*N/120;//in Hz Tph=turns*phase*P;//no. of turns per phase m=s...
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//Example 20_5 clc(); clear; //To find the current in circuit and voltmeters reading across R C and L f=600 //Units in Hz l=4*10^-3 //Units in H xl=2*%pi*f*l //Units in Ohms c=10*10^-6 //Units in F xc=1/(2*%pi*f*c) //Units in Ohms r=20 //Units in Ohms z=sqrt(r^2+(xl-...
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//Example 3_11 clc(); clear; //To find out how fast a ball must be thrown a=9.8 //units in meters/sec^2 t=3 //units in sec v=(0.5*a*t^2)/t printf("The speed by which the ball has to be thrown is v=%.1f meters/sec",v)
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// C-Exercise 18 // Jurian Kahl // Nattawut Phanrattinon funcprot(0); function X = GARCH11(k, m, theta, sigma1) //simulation for X X = zeros(m,k); var = zeros(m,k); // starting sigma var(:,1) = sigma1^2 // simulation Y from N~(0,1) Y = grand(m,k,"nor",0,1); // Given starting...
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clc(); clear; //To calculate the power radiated by the filament r=0.05; //radius of the wire in mm l=4; //length of the wire in cm A=2*%pi*r*l*10^-5; //in m^2 //According to Stephen-Boltzmann law R=e*s*(T^4) //P=R*A e=1; T=3000;...
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//exec('Setup_Parameters_Finish.sce'); exec('Setup_Parameters.sce'); stacksize('max') //Determine Boundary Profile (profile of last roughing cut) listFinalRoughingPass = list(OffsetPolyline(vPoints, vNormalOffsetRegime, mNormalOffsetComponents)); //Boundary plBoundWithRad = AddInsideRad(listFinalRoughingPass(1), vORB...
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//chapter 5 Ex 8 clc; clear; close; //let the value to be found out be x and y y=32.4/2^2; //after squaring both sides x=sqrt((12.3-sqrt(86.49))^2-5); //from the given equation mprintf("(i)y=%.1f",y); mprintf("\n(ii)x=%.0f",x);
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//Initilizatin of variables W=6 //lb l=8 //ft v=10 //ft/s g=32.2 //ft/s^2 theta1=60 //degrees theta2=30 //degrees //Calculations Fe=(W*v^2)/(g*l*0.5) //lb //Using equations of motion //Solving for C and T A=[cosd(theta1),-cosd(theta2);cosd(theta2),cosd(theta1)] B=[-Fe;W] P=inv(A)*B //lb C=P(1) //lb T=P(2...
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@relation haberman @attribute Age integer[30,83] @attribute Year integer[58,69] @attribute Positive integer[0,52] @attribute Survival{positive,negative} @inputs Age,Year,Positive @outputs Survival @data negative negative negative negative negative negative positive negative positive negative negative negative negative...
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module option. data option (A : ★) : ★ = | Some : A ➔ option | None : option.
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//ques-16.1 //Calculating rate constant of a reaction clc //t = Time in minutes //c = KMnO4 in mL t1=0; c1=37; t2=5; c2=29.8; t3=15; c3=19.6; t4=25; c4=12.3; t5=45; c5=5; a=c1; k2=(2.303/t2)*log10(a/c2); k3=(2.303/t3)*log10(a/c3); k4=(2.303/t4)*log10(a/c4); k5=(2.303/t5)*log10(a/c5); k=(k2+k3+k4+k5)/4; ...
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//Variaveis do Problema D1 = 0.2//Diâmetro na extremidade fixa, quando x = 1, D = ao mesmo valor D2 = 0.15//Diâmetro na outra extremidade, vamos alterar para os outros valores especificados no trabalho L = 2.5//Comprimento da Haste E = 208*(10^9) //Módulo da elasticidade //Solução //Criando uma função com os valore...
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clc; p5=6000; // Pressure of superheated steam in kPa T5=723.15; // Temperature of superheated steam in kelvin p1=0.6113; // Pressure at reference state in kPa T1=273.16; // Temperature at reference state in kelvin hfg1=2501.3; // Latent heat of vapourization of water at reference state in kJ/kg R_1=8.3143; // Un...
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//exercicio 2 clear a = input (" Entre com o n° para calcular a raiz "); x_old = input ("Entre com a extimativa da raiz"); n_sig = input ("Entre com o n° de algarismos significativos"); es = 0.005*10^(2-n_sig); //Condição de parada. x_new = 0; while 1 do x_new = (x_old + (a/x_old))/2 erro = abs((x_new - x_old...
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//Chapter-1,Example1_15_7,pg 1-71 n=1 //first order maximum l=1.54*10^-10 //wavelength of rock salt crystal q=21.7 //glancing angle in degree //using Bragg's law n*l=2*d*sin(q) d=n*l/(2*sind(q)) printf("latt...
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clear; clc; l = 20;//feet W = 500;// lb per foot run c = 750;// lb/in^2 t = 18000;// lb/in^2 m = 15; BM_max = W*l*l*12/8 ;// lb-inches //by making the effective deapth d twice the width b d = (BM_max/(126*0.5))^(1/3);//inches b = 0.5*d;//inches //necessary reinforcement is 0.8% of concrete section A_t = 0.0...
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clc(); clear; // To calculate the critical current T=4.2; //temp in kelvin Tc=7.18; //critical temp in kelvin Hc_0=6.5*10^4; //in amp/m d=1; //diameter in mm d=d*10^-3; //diameter in m r=d/2; Hc_T=Hc_0*(1-((T/Tc)^2)); Ic=2*%pi*r*Hc_T; printf("the critical current is %f Amp",Ic); ...
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clc //Intitalisation of variables clear ci= 0.1896 //mole per liter cKI= 0.02832 //mole per liter r= 625 //CALCULATIONS CI2= ci/r dc= cKI-CI2 //RESULTS printf ('Conc of I2 in KI layer = %.6f mole per litre',CI2) printf ('\n Conc of I3- ions = %.5f mole per litre',dc)
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clc; clear; //Example 5.20 Cpo=1.9 //Heat capacity for oil[kJ/kg.K] Cps=1.86 //Heat capacity for steam [kJ/kg.K] ms_dot=5.2 //Mass flow rate in [kg/s] T1=403 //[K] T2=383 //[K] Q=ms_dot*Cps*(T1-T2) //[kJ/s] Q=Q*1000 //[W] t1=288; ...
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//Exa 10.10 clc; clear; close; //given data : n=0.6;//refractive index N=4.23*10^4;//in m^-3 //Formula : n=sqrt(1-81*N/f^2) f=sqrt(81*N/(1-n^2));//in Hz disp(f/1000,"Frequency of wave propagation in KHz : ");
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// Example 11.7: Low frequency gain, Upper 3 dB frequency clc, clear // Without feedback AM=1e4; // Low frequency values of A wH=100; // Upper 3 dB frequency in hertz // With feedback R1=1; // in kilo-ohms R2=9; // in kilo-ohms bta=R1/(R1+R2); // Feedback factor AfM=AM/(1+bta*AM); // Low frequency gain wHf=wH...
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//The ac Motor Control// //Example 15.12// PFi=0.6;//input powerfactor// DF=0.7;//distortion factor// IDF=PFi/DF;//input displacement factor// printf('input displacement factor=%f',IDF);
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//Example 4 // Frequency clc; clear; close; //given data : T=625;// in N T1=100;// in N l=1/2; n=240;// in Hz n1=1/l*(sqrt(T1/T))*n; disp(n1,"The frequency,n1(Hz) = ")
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// Theory and Problems of Thermodynamics // Chapter 4 // Energy Analysis of Process // Example 9 clear ;clc; //Given data T_cal = 110 // Calorimeter Temperature in C P_cal = 0.1 // Calorimeter Pressure in MPa P = 0.5 // Pressure wet steam in MPa // from super...
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$thermo = VirtualMaterials.NRTL/Ideal/HC / -> $thermo $thermo + HYDROGEN METHANE PROPANE N-PENTANE WATER units SI V101 = Flash.SimpleFlash() V101.LiquidPhases = 2 V101.In.T = 45 V101.In.P = 165 psia V101.In.MassFlow = 852143 V101.In.Fraction = 2 2 1 1 1 V101.Vap V101.Liq0 V101.Liq1 V101.Vertical...
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//pathname=get_absolute_file_path('3.11.sce') //filename=pathname+filesep()+'3.11-data.sci' //exec(filename) //Enthalpy of steam entering the injector(in kcal/kg): h1=720; g = 9.81; //Enthalpy of water entering(in kcal/kg): h2=24.6 //Enthalpy of water and steam mixture leaving the injector(in kcal/kg): h3=100 //Depth o...
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java -ea trip.Main -m trip-tests/travel01 <<EOF C2, Santa_Cruz, Berkeley EOF
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//Calculate the work done if the expansion is carried out Adiabatically and Reversibly //Example 4.7 clc; clear; n=0.850; //Number of moles of momnoatomic ideal gas in mol R=0.08206; //Gas constant in L atm K^-1 T1=300; //Initial temperature in K P1=15; //Initial pressure in atm V1=(n*R*T1)...
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//pathname=get_absolute_file_path('9.02.sce') //filename=pathname+filesep()+'9.02-data.sci' //exec(filename) //Pressure at A(in kPa): pa=138 //Pressure at B(in kPa): pb=1380 //Thermal efficiency: nt=0.5 //Mechanical efficiency: nm=0.8 //Calorific value of fuel(in kJ/kg): c=41800 //Adiabatic compressive ind...
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Veo = 1.777; // Ve/Vo e = 1-Veo; // Degree of dissociation P = 0.124; // in atm K = (4*e^2*P)/(1-e^2); disp("atm",K,"The value of equillibrium constant is")
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clear all; clc; disp("Scilab Code Ex 8.3 : ") //Given: ri = 600/1000; //m t = 12/1000; //m ro = ri+t; sp_wt_water = 10; //kN/m^3 sp_wt_steel = 78; //kN/m^3 l_a = 1; //m depth of point A from the top //Internal Loadings: v = (%pi*l_a)*(ro^2 - ri^2); W_st = sp_wt_steel*v; p = sp_wt_water*l_a; //Pascal...
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//EX2_8 To find complement of (AB' + C)D' + E //clears the screen clc //clears already existing variables clear //By using 2.12 i.e. (X + Y)' = X'Y' disp('((AB'' + C)D'' + E)'' = [(AB'' + C)D'']''E''') //By using 2.13 i.e. (XY)' = X' + Y' disp('= [(AB'' + C) + D]E''') //Using 2.12 disp('= [(AB'')'' C'' ...
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clear clc disp('Ex-14.7'); mpc2=938; //rest energy of proton Q=mpc2+mpc2-(4*mpc2); //Q value of reaction Kth=(-Q)*(6*mpc2/(2*mpc2)); // thershold kinetic energy printf('The threshold kinetic energy is %.2f MeV',Kth);
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// ControlSynthesisBodeCT.sce function VerifySynthesisBodeCT(Grat, Gdel, R, wmin, ndec, wcmin,wcmax, pmmin, wa1, wa2, AadB, wr1, wr2, ArdB, nfig) dmin = floor(log10(wmin)) w = logspace(dmin,dmin+ndec,500); f = w/2/%pi; hGrat = repfreq(Grat,f); ...
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//Engineering and Chemical Thermodynamics //Example 6.5 //Page no :271 clear ; clc ; disp(" Example 6.5 Page no:271") disp(" There is no numerical part involved in this problem . Users can refer Figure 6.5.")
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clc clear //DATA GIVEN Mw=2; //mass of water separated out in kg Ms=20.5; //amount of steam (condensate) discharged from throttling calorimeter in kg Tsup3=110+273; //temp. of steam afetr throttling in K p1=12; //initial pr...
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//Ex14_16 Pg-701 clc n1=1.46 //core refracrive index r=4.5*10^(-6) //radius of the core del=0.25/100 //fractional difference of refractive indices Vc=2.405 //normalzed frequency disp("We have, cut-off wavelength expression") lamda=(2*%pi*r*n1*sqrt(2*del))/Vc printf(" = %.3f um",lamda*1e6)
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//ANALOG AND DIGITAL COMMUNICATION //BY Dr.SANJAY SHARMA //CHAPTER 11 //Information Theory clear all; clc; printf("EXAMPLE 11.55(PAGENO 538)"); //given P_x1 = 1/2//probability of first signal P_x2 = 1/4//probability of second signal P_x3 = 1/8//probability of third signal P_x4 = 1/16//probability of fourth...
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clc; //Example 23.2 page no 323 printf("Example 23.2 page no 323\n\n"); //calculation of aerodynamic diameter of irregular saped sphere d_e=1.3//eq. diameter,micron sg=2.35 d_pa=d_e*sqrt(sg)//aerodynamic diameter printf("\n aerodynamic diameter d_pa=%f micron",d_pa);
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function generate_schema() MAX_DISTANCE = 5; MIN_DISTANCE = 1; AREA=MAX_DISTANCE^2; //10% ~ 80% //AVERAGE_DISTANCE=20; //STD_DISTANCE=AVERAGE_DISTANCE*0.3; //10%, 30%, 60%, 80% NODES_DENSITY=0.1; //40,120,240,320 NUM_NODES = round(AREA*NODES_DENSITY); //Initia...
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clear all; clc; disp("Ex 6_13") disp("The free body diagrams are shown in figures 6-25c, 6-25d and 6-25e")
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//Shubham Sharma //Msc PhD OR //Roll no: 18i190002 clc clear function i=JobsDoneInOneArrival(lembda,mu,x) summ=0 y=grand(1,1,"exp",1/lembda) for i=0:x p= grand(1,1, "exp",1/mu) summ=summ+p if summ>=y then break end end endfunction x0=10 c1=JobsDoneInOneArriv...
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clc clear //Initialization of variables m=1 //lbm p=50.9 //atm t=212+460 //R R=0.73 //calculations pc=72.9 //atm tc=87.9 +460 //R pr=p/pc Tr=t/tc z=0.88 v=z*R*t/p //results printf("volume = %.3f ft^3/mole",v)
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clc clear //Input data Pa=1;//Atmospheric pressure in bar g=9.81;//Gravity in m/sec^2 do=0.8*1000;//Density of oil in kg/m^3 Zo=0.8;//Level of oil in m dw=1000;//Density of water in kg/m^3 Zw=0.65;//Level of water in m d=13.6*10^3;//Density of Hg in kg/m^3 Z=0.45;//Level of Hg in m //Calculations Po=(d...
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clc; k=5; // Effective turns ratio E1=400; // supply voltage for primary il=10; // load current E2=E1/k; // magnitude of maximum secondary induced E1o=E1+E2; // maximum possibe value of output voltage P=(il*E2)/1000; // Rating of secondary winding ip=(il*E2)/E1; // neglecting noload current, primary winding curr...
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// Grob's Basic Electronics 11e // Chapter No. 29 // Example No. 29_1 clc; clear; // For the diode circuit, calculate the ac resistance, rac, for the following values of R: (a) 10 kOhms, (b) 5 kOhms, and (c) 1 kOhms. Use the second approximation of a diode. // Given data R1 = 10*10^3; // Resistance 1=10 k...
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clc;funcprot(0);//Example 8.8 //Initilisation of Variables mh=10;....//Mass flow rate of water in kg/s L=6.7;.....//Lemgth of the tubes in m Do=0.026;....//Diameter of tube in m hi=470;....//Inside heat transfer coefficient in W/m^2 K ho=210;.....//Outside heat transfer coefficient in W/m^2 K Tci=15;......//Inle...
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//All the quantities are expressed in SI units rho = 1.225; //freestream density of air along the streamline delta_p = 335.16; //pressure difference between inlet and throat ratio = 0.8; //throat-to-inlet area ratio //The velocity at the inlet can be given as v_1 = sqrt(2*delta_p/rho/(((1/ratio)^...
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clc;clear; //Example 16.6 //given data E=0.51;//kinetic energy in MeV R=0.15;//radius in m e=1.6*10^-19;//the charge on electron in C mo=9.12*10^-31;//mass of electron in kg c=3*10^8;//speed of light in m/s //calculation Eo=E; m=mo*(1+(E/Eo)); b=sqrt(1-(mo/m)^2); v=b*c; B=mo*v/(e*R); disp(B,'magnetic ...
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<?xml version="1.0" encoding="utf-8"?> <test> <description>Kovasznay Flow P=8 Weak Pressure VSS</description> <executable>IncNavierStokesSolver</executable> <parameters>KovaFlow_m8_short_ConOBC_VCSWeakPress.xml</parameters> <files> <file description="Session File">KovaFlow_m8_short_ConOBC_VCSWea...
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//Optoelectronics - An Introduction, 2nd Edition by J. Wilson and J.F.B. Hawkes //Example 8.12 //OS=Windows XP sp3 //Scilab version 5.5.2 clc; clear; //given n1=1.48;//Dimensionless refractive index of fiber core n0=1;//Dimensionless refractive index of air Rf=((n1-n0)/(n1+n0))^2;//Fraction of light reflec...
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//check o/p for more than one i/p arg r = [5.0000]; z=[5]; [a,efinal] = ac2poly(r,z); disp(a); disp(efinal); // Output
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//ques3 //Gas Flow through a Converging–Diverging Duct clear clc Cp=0.846;//specific heat at constant pressure in kJ/kg/K R=0.1889;//gas constant for substance T0=473;//temp at state 0 in K T1=T0;//temp at state 1 in K P0=1400;//pressure at state 0 in kPa P1=P0;//pressure at state 1 in kPa //from Eqn 17-5 P=...
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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[de...
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//ques-6.8 //Calculating standard emf and emf generated clc e1=1.7; e2=-0.31;//emfs of two half-cell reactions (in V) C1=0.1;//concentration of hydrogen ion (in M) C2=2;//concentration of sulphate ion (in M) n=2;//number of electrons Es=e1-e2;//standard emf (in V) E=Es-(0.0592/n)*log10(1/(C1^4*C2^2)); printf("...
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clc // Given that N = 4000 // Grating lines per cm theta = 90 // for maximum order lambda_min = 5000 // minimum wavelength of light in angestrom lambda_max = 7500 // maximum wavelength of light in angestrom // Sample Problem 24 on page no. 165 printf("\n # PROBLEM 24 # \n") printf(" Standard formula used \n") ...
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clear //Given V=sqrt(3) //calculation // a=atan(V)*180/3.14 //Result printf("\n Angle of dip is %0.0f Degree",a)
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clear; clc; //Example - 10.23 //Page number - 366 printf("Example - 10.23 and Page number - 366\n\n"); //Given T = 320 + 273.15;//[K] R = 8.314;//[J/mol*K] - Universal gas constant // For water Tc = 647.1;//[K] Pc = 220.55;//[bar] Pc = Pc*10^(5);//[Pa] // The cubic form of Redlich Kwong equation of ...
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clc; clf; clear all; l=10; a=1.5; n=-l:l; x=a^n; plot2d3(x); title('Exponential Signal'); xlabel('n'); ylabel('x');
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x=[0 %pi/3 2*%pi/3 %pi/4 3*%pi/4 %pi/6 5*%pi/6 %pi]; sec(x) x=linspace(-%pi,%pi,100) plot(x,sec(x))
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clc // given data w=0.6 // in km h2=2.5 // in km p=5/100.0 // porosity rhor=3000.0 // density of sediment in kg/m^3 cr=750.0 // specific heat of sediment in J/kg-K rhow=1000.0 // density of water in kg/m^3 cw=4200.0 // specific heat of water in J/kg-K G=35.0 // temperature gradient in degree C/km T1=45.0 // ...
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// Example 7.12;//OPTICAL POWER EMITTED & EXTERNAL POWER EFFICIENCY clc; clear; close; e=1.6*10^-19;//Electronic charge ht=6.62*10^-34;//Constt C=3*10^8;//sPPED OF LIGHT IN M/S h=0.85*10^-5;//wavelength in meter V=1;//VOLTAGE F=0.68;//transmiison factor nx=3.6;//refractive index n=1;;//refrative index of air...
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// Example 7-28-1 // Design of Lag - lead compensation with Bode plots clear; clc; xdel(winsid()); //close all windows mode(0); // please edit the path // cd "/<your code directory>/"; // exec("shmargins.sci"); s = %s /2 /%pi ; G = 1 / (s * (s + 1) * (s + 2)); Kv = 10; K = Kv / horner(s * G,0) GK = syslin('c',K * G...