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// CRIAR FUNÇÕES DE OBJETO // Função que cria um objeto do tipo "FuncName", i.e., figure, uicontrol, etc... // com campos de propriedade definidos na estrutura "PropStruc", cujo nomes // dos campos devem ser os mesmos nomes das propriedades do objeto function obj = CriarObjeto(FuncName,PropStruc) PropNam = "...
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clc; alpha_a=2800; lambda=10^3*50; x=10; alpha_b=11; U=1/[1/alpha_a+x/lambda+1/alpha_b]; tA=90; tB=15; q=(tA-tB)*U; disp("rate of heat lost per sq m of surface") disp("kW",q) //part b t2=q/alpha_b+tB; disp("temperature of outsede surface:"); disp("C",t2)
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//Chapter-4,Example 15,Page 96 clc; close; P1= 528 // pressure in mm of Hg P2= 760 // pressure in mm of Hg T2=100+273 //teperature in Kelvin delta_Hv= 545.5 *18 // latent heat of vapourisation of water in J/mol R= 1.987 //gas constant //from the integrated form of Clausius-Clapeyron equati...
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//chapter-14,Example14_7,pg 510 d1=4*10^-2//diameter of inner cylinder d2=4.4*10^-2//diameter of outer cylinder h=2.2//level of water H=4//height of tank eps1=((80.37*10^11)/((4*%pi*10^8)^2))//dielectric const. in free space(SI) epsv=0.013*10^-5//dielectric const. of medium(SI) C=(((H*epsv)+(h*(eps...
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//Problem 14.06: //initializing the variables: xCO2 = 0.0314; xO2 = 0.0584; P = 1; // in atm T = 2050; // in deg F //calculation: //from example 13.10, at 2050 deg F K = 9.156E-7 yCO = xCO2*K/xO2^0.5 printf("\n\nResult\n\n") printf("\n the mole fraction of CO is %.2E",yCO)
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// Copyright (C) 2015 - IIT Bombay - FOSSEE // // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_...
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//Example 14.5: time difference clc; clear; close; format('v',5) n=1.33;// x=2;// l=50;//m c=3*10^8;//m/s dt=((n*x*l)/c);//s disp(dt*10^6,"time difference is,(micro-seconds)=")
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// Example 17_5 clc;funcprot(0); // Given data m_h=1.00;// kg E_me=33.1;// MJ E_na=10.5;// MJ m_fat=10.0;// kg // Calculation // (a) mdot_fat=E_na/E_me;// The mass of body fat consumed per day in kg of body/d // (b) t=m_fat/mdot_fat;// d printf("\n(a)The mass of body fat consumed per day,mdot_fat=%0.3f kg...
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n=10; m=50; x=[1 2 3 4 5 6 7 8 9 10]; y=[9 8 5 8 4 6 2 3 5 1]; k=2; for i=1:n if(i>1) p1(i)=(y(i)-y(i-1))/(x(i)-x(i-1)) else p1(i)=(y(i+1)-y(i))/(x(i+1)-x(i)) end if(i<n) p2(i)=(y(i+1)-y(i))/(x(i+1)-x(i)) else p2(i)=(y(i)-y(i-1))/(x(i)-x(i-1)) ...
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//Example 16.3// n=1.59;// Average refractive index Polystyrene R=((n-1)/(n+1))^2;//Fresnel's formula disp(R)
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//Example 4.7 //Jacobi Method //Page no. 99 clc;close;clear; A=[2,3,1;3,2,2;1,2,1]; n=3; for k=1:10 max1=0 for i=1:n for j=1:n if A(i,j)>max1 & i~=j then max1=A(i,j) i1=i;j1=j; end end end fi=(atan((2*A(i1,j1))/(A(i1,i1)-A(j1,j1)+10^-20)))/2 disp(fi...
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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 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES // EXAMPLE : 3.17 : // Page number 147-148 clear ;...
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//Section-1,Example-3,Page no.-AC.205 //To calculate the percentage of excess air used for combustion. clc; C=0.54 H=0.065 O=0.03 N=0.018 M_W=(((32/12)*C)+((16/2)*H)-O)*(100/23)//Minimum weight of air required for combustion W_CO2=(C*(44/12)) W_N2=N+(M_W*(77/100)) T_W=(W_CO2+W_N2) //Total weight of dry products of ...
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//Example 8,Chapter 3 clc; f=50 Irms=10 //Current in amperes //(i) Im=Irms*sqrt(2) disp('14.14sin(18000t)') //(ii) t=0.0025 t=(1/(4*f)) + t printf("\n t=%.1f ms \n",t*10^3) i=14.14*sin(18000*7.5*10^-3) printf("\n i=%.0f A \n",i) //(ii) t=0.0075 t=(1/(2*f))+t printf("\n t=%.1f ms \n",t*10^3) i=14.14*sin...
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//Chemical Engineering Thermodynamics //Chapter 14 //Thermodynamics of Chemical Reactions //Example 14.13 clear; clc; //Given T1 = 273+110;//Temperature in K T = 298;//Room temperature in K P = 1;//Pressure in atm R = 1.98;//gas constant in Kcal/Kgmole //Ag2CO3(s) (A) = Ag2O(s) (B) + CO2(g) (C) a = 1;//...
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Ftrue2 = [ 1 1 1 1 1 1 1 1 2 2 1 1 1 2 3 6 2 1 1 2 5 12 4 1 1 2 4 9 3 1 1 1 1 1 1 1]'; Ftrue=matrix(Ftrue2,36,1); W = [ 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0...
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//example 5.41 clear; clc; disp("Zn(s)|ZnCl2(aq)||CdSO4(aq)|Cd(s)"); //For Zn(s)|ZnCl2(aq)||CdSO4(aq)|Cd(s) //Given: T=298;//Temperature[K] R=8.314;//Universal gas constant[J/K/mol] E1=-0.7618;//Standard electrode potential for Zn2+/Zn [volts] E2=-0.403;//Standard electrode potential for Cd2+/Cd [volts] ...
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// Example 6_5 clc;funcprot(0); // Given data P_1=100;// The initial pressure in psia T_1=600;// The initial temperature in °F P_2=10;// The final pressure in psia // Calculation // From steam tables v_2=6.216;// ft^3/lbm v_1=v_2;// ft^3/lbm v_f2=0.0166;// ft^3/lbm v_g2=38.42;// ft^3/lbm x=(v_2-v_f2)/(v_g...
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//1)The answer of Fx in Part(a) is 1532N and not !530N as stated. //2)The answer of Fy in Part(a) is 1286N and not 129N as stated. //3)The answer of Fx! in pPart(b) is 1769.10N and not 1770N as stated clear all; clc; disp("Ex 2_2") disp("Part (a)") //refer figure 2-11b printf('\nVector addition is F=Fx+Fy') f=2000//ma...
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clc //Example 10.7 //Calculate the estimated pressure rise in the first stage of mutisatge centrifugal compressor rho=0.075//lbm/ft^3 omega=1047//rad/sec d=2//ft dP=(1/2)*(rho)*(omega*d/2)^2/32.2/144//psia //1 lbf.s^2 = 32.2 lbm into feed //1 ft = 144 in^2 printf("the estimated pressure rise in the first stage...
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// Exa 6.10 clc; clear; close; // Given data I_DSS= 10;// in mA I_DSS= I_DSS*10^-3;// in A gm= 10;// in ms gm=gm*10^-3;// in s // V_GSoff = V_GS = Vp so , gm = gmo = -2*I_DSS/V_GSSoff V_GSoff= -2*I_DSS/gm;// in volt disp(V_GSoff,"The value of V_GS(off) in volts is : ")
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before x, phead=0, pbody=1, ptail= - 476*y - 448*y^2 - 112, mlead= + x^2, flead=4, root2=2, widev=4 after x, phead=15*x^2, pbody=1, ptail= - 476*y - 448*y^2 - 127, vmapt={x=>1+2*x,y=>y} before y, phead=15360*x^2, pbody=68, ptail= - 130048, mlead= + 16*y^2, flead=16384, root2=128, widev=1024 after y, phead=15360*...
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clc // In the (001) surface the top atoms are either Ga or As //A square of area a^2 has 4 atoms on the edges of square shared by 4 other square and 1 atom in centre N=2 disp("N = "+string(N)) //initializing value of total number of atoms per square a = 5.65*10^-8 disp("a = "+string(a)+"cm^-1") //initializing val...
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//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS //Example 7 disp("CHAPTER 1"); disp("EXAMPLE 7"); //VARIABLE INITIALIZATION I1=1; //current source in Amperes v1=4; //voltage source in Volts v2=3; //voltage source in Volts v3=6; ...
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clc D=0.25; //m r=9; L=0.3; //m cv=0.71; //kJ/kg K cp=1; //kJ/kg K p1=1; //bar T1=303; //K p3=60; //bar p4=p3; n=3; //number of working cycles/ sec y=1.4; R=287; disp("(i) Air standard efficiency") Vs=%pi/4*D^2*L; Vc=Vs/(r-1); V1=Vs+Vc; p2=p1*(r)^y; T2=T1*r^(y-1); T3=T2*p3/p2; rho=4/100*(r-1)+1...
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// Test #5 : Input Argument #1 range test exec('./allpasslp2bsc.sci',-1); [n,d]=allpasslp2bsc(-32,[0.5,0.89]); //!--error 10000 //Wo must lie between 0 and 1 //at line 39 of function allpasslp2bsc called by : //[n,d]=allpasslp2bsc(-32,[0.5,0.89]);
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// Example 2.12.3 clc; clear; core_diameter=8d-6; //core diameter delta=0.92/100; //relative index difference lamda=1550d-9; //operating wavelength n1=1.45; //core refractive index a=core_diameter/2; //computing core radius v= 2*%pi*a*n1*sqrt(2*delta)/lamda; //computing normaliz...
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clear; clc; //Example5.5[Transient Heat Conduction in a Large Uranium Plate] //Given:- k=28;//[W/m.degree Celcius] a=12.5*10^(-6);//Thermal diffusivity[m^2/s] T1_0=200,T2_0=200;//Initial Temperature[degree Celcius] e_gen=5*10^6;//Heat generated per unit volume[W/m^3] h=45;//heat transfer coefficient[W/m^2.deg...
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////Ex 4.13 clc; clear; close; format('v',5); Beta=0.04;//feedback factor AOL=5000;//unitless(at dc) Rio=40;//kohm Ro=1;//kohm SF=1+AOL*Beta;//sacrifice factor at dc Rif=Rio/SF*1000;//ohm disp(Rif,"Input impedence(ohm)"); Rof=Ro*1000/SF;//ohm disp(Rof,"Output impedence(ohm)");
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// Example 9.5 //Write a program to calculate standard deviation of an array values. //Array elements are read from terminal.Use functions to calulate- //standard deviation and mean funcprot(0); //passing array named 'value' to function std_dev at 'a' function[std]=std_dev(a,n) sum1=0; x=...
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clc clear //Input data a=14;//Air fuel ratio T1=288;//The ambient temperature of air in K T2=(288-23);//The evaporation of fuel cause 23 degree C drop in mixture temperature in K p=1.3;//Pressure ratio nc=75;//The isentropic efficiency of the compressor in percent Cpm=1.05;//The specific heat of the mixtu...
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clear;lines(0); x=[1,%i,-1,-%i] tanh(x) sinh(x)./cosh(x)
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//water// //page 1.9 example 4// clc H=210.5;//hardness in ppm// M1=100;//molecular weight of CaCO3// M2=136;//molecular weight of FeSO4// M=M1/M2;//multiplication factor of FeSO4// W=H/M;//weight of FeSO4 required// printf("\nFeSO4 required is %.1f ppm",W);
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errcatch(-1,"stop");mode(2);//Example 20.1 ; B=.5//in T A=3.24*10^-4//in m^2 Flux=B*A N=25 delta_t=.8 disp(Flux,"Magnetic flux in T.m^2=") e=(N*Flux)/(delta_t) disp(e,"Induced emf in volt=") exit();
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a= 2*(10^(-12)); //de Broglie wavelength, mts h= 4.136*(10^(-15)); //Planck's constant, eV.s c= 3*(10^8); //velocity of light, m/s pc= (h*c)/a; //p is momentum, pc is electron's energy, eV pc= pc/1000; //convert to keV Eo= 511; //rest energy, keV E= sqrt((Eo^2)+(pc^2)); //Total Energy, keV KE= E-Eo; //Kinetic ...
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// Exa 5.1 clc; clear; // Given E1 = 1/100; // exposure set for grid line impression(sec) E2 = 10; // second exposure duration(sec) R = 10^-4; // persistence of CRO screen(sec) I1 = 1; // Trace intensity for exposure 1(candle power) I2_normal = 4 ; // trace intensity for normal settings(candle power) /...
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clc //Initialization of variables ken=0.5 kex=0.2 f=0.0018 l=10 //ft dia=3 //in z1=8 z2=5 //calculations x1=ken+kex+f*l*12/dia t=35.5*2/3 *(z1^(3/2) - z2^(3/2)) //results printf("Time reqired = %d s",t)
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clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 4.4 Page 230 \n'); //Example 4.4 // Temperature Field and Rate of Heat Transfer //Operating Conditions ho = 1000; //[W/m^2.K] Heat Convection coefficient hi = 200; //[W/m^2.K] Hea...
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//maximum impedance of the line //given clc Zo=75//ohm VSWR=3//voltage standing wave ratio Zmax=VSWR*Zo//ohm disp(Zmax,'the maximum impedance of the line for the given VSWR IN ohm')//ohm
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// To calculate primary and scondary side impedences,current and their pf and real power // and calculate terminal voltage clc; N_1=150; N_2=75; a=N_1/N_2; Z_2=[5,30]; //polar(magnitude,phase diff) disp(Z_2,'secondary impedence(ohm)'); Z_1=[a^2*Z_2(1),Z_2(2)]; disp(Z_1,'primary impedence(o...
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//exapple 2.5 clc; funcprot(0); // Initialization of Variable T=273+15; rho=999; rhom=13559;//density of mercury g=9.81; P2=764.3/1000*rhom*g; R=8.314; M=16.04/1000; d=4.5/1000; A=pi*d^2/4; G=0.75/1000;//mass flow rate delP=(1-exp(R*T*G^2/2/P2^2/M/A^2))*P2; h=-delP/rho/g; disp(h*100,"height of manom...
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//Chapter 12 //Example 12.3 //page 450 //To calculate maximum power transferred clear;clc; Vt=1.0; //generator terminal voltage V=1.0 ; //infinite bus voltage Pe=1.0 ; //power delivered Xd=0.25*%i ; //generator's transient reactance Xl=0.5*%i ; //transmission line's reactance Xt=0.1*%i; //transformer's reactance //to...
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clear // // // //Variable declaration Vm=20 //voltage(V) RL=500 //load resistance(ohm) rf=10 //forward resistance(ohm) VB=0.7 //bias voltage(V) //Calculation Im=(Vm-VB)*10**3/(rf+RL) //peak current(mA) Vo=Im*RL/10**3 //peak output voltage(V) //Result printf("\n peak curre...
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//variable initialization v1=214330 //fundamental band for CO molecule (m-1) v2=425970 //first overtone for CO molecule (m-1) A=[1 -2;2 -6]; ...
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function [] = fprintnum(str,arr) // Display mode mode(0); // Display warning for floating point exception ieee(1); // Fprint array of integers with a string in front of it, no ending newline if isempty(arr) then tmp = "[]"; else // !! L.8: Matlab function sprintf not yet converted, original calling sequence us...
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// 08.09.10 // 08.09.13 function Out5=Sfbdrawparadata(varargin) global IMPLICITDATA CUSPDATA CUSPPT CUSPSPLITPT; Nargs=length(varargin); Fd=varargin(1); FdL=Fullformfunc(Fd); Np=[50,50]; if Nargs>=2 Np=varargin(2); if type(Np)==1 & length(Np)==1 Np=[Np,Np]; end; end; Eps=0.05; if Na...
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// Exa 4.3 clc; clear; close; // Given data lamda= 670;// in nm h_int= 1/100; EpIn_eV= 1248/lamda;// in eV I=50;// in mA P= h_int*EpIn_eV*I;// in mW disp(P,"Power radiated by an LED in mW is : ") // Note : There is a calculation error in evaluating the value of P so the answer in the book is wrong
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// Example 5.8:3-db frequency and bandwidth clc; clear; close; Cp=1;//PARALLEL capacitance IN PICO FARAD Cs=2;//series capacitance IN micro FARAD rs=1;//series resistance in killo ohms rp=10;//PARALLEL resistance in killo ohms ts= ((rs+rp)*10^3*Cp*10^-12);//time constant tp= ((rs*rp)/(rs+rp)*10^3*Cp*10^-12);//...
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clc // Given that p1 = 8 // Pressure of entrance in bar t1 = 1125 // Temperature of entrance in K p2 = 1.5 // Pressure of exit in bar n = 11 // No of stages Vf = 110 // Axial velocity of flow in m/s n_p = 0.85 // Polytropic efficiency Vb = 140 // Mean velocity in m/s gama = 1.33 // Heat capacity ratio for gas...
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clear; clc; //Example 11.8 CMRRdB=90;//dB CMRR=3.16*10^4; b=100; Vt=0.026; Iq=0.8; Ro=(2*CMRR-1)*Vt*b/((1+b)*Iq); Ro=Ro*10^-3;//Mohm disp(Ro,"output resistance (MOhm)")
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// to find base current // Electronic Principles // By Albert Malvino , David Bates // Seventh Edition // The McGraw-Hill Companies // Example 6-4, page 197 clear;clc; close; // Given data Bdc=200;// current gain Vbb=2;// base source voltage in volts Vbe=0.7;// emitter diode in volts Rb=100*10^3;// resistance in ohm...
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clc clear //Page number 491 //Input data d=2*10^-10;//The molecular diameter of an ideal gas in m t=20;//The temperature of the gas in degree centigrade p=1;//The pressure of the gas in atmosphere pi=3.142;//The mathematical constant of pi //Calculations T=t+273;//The temperature of the gas in K P=1....
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// ELECTRICAL MACHINES // R.K.Srivastava // First Impression 2011 // CENGAGE LEARNING INDIA PVT. LTD // CHAPTER : 6 : SYNCHRONOUS MACHINES // EXAMPLE : 6.22 clear ; clc ; close ; // Clear the work space and console // GIVEN DATA E1 = 1100 + (%i*0); // EMFs of two identicel sy...
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// Poisson PMF // Le Thu Huong ADEO1 clc N = 10000; lamda = 4; x0 = -2; delx = 0.05; xmax = 12; x = [x0:delx:xmax]; for k = 1:length(x) c = 0; for j = 1:N; cumul = exp(-lamda); proba = cumul; u = rand (); alpha = 0; while u > cumul then ...
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function y = trapezio(f,a,b,n) // integral de f usando método dos trapezios e n repeticoes h=(b-a)/n; Soma = f(a) + f(b); for k=1:n-1 Soma=Soma + 2*f(a+k*h); end; y= (h/2)*Soma; endfunction
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// Chapter4 // Page.No-141, Figure.No-4.28(b) // Example_4_12 // Output ripple voltage // Given clear;clc; delta_Vio=15.85*10^-6; // Change in input offset voltage delta_V=1; // Unit change in supply voltage V=10*10^-3; // Change in supply voltage R1=1*10^3;Rf=100*10^3; delta_Voo=(1+Rf/R1)*(delta_Vio/delta_V)...
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jan.immediate_drawing = "off" delete(HistCargas); clear HistCargas HistCargas = []; T1 = evstr(NewmarkBeta(4).string); if isempty(T1) then; T1=3*t1; end Tipo = [" - FX" " - FY" " - MZ"] for i=1:length(Cargas) x = Cargas(i).user_data(1); y = Cargas(i).user_data(2); CarDir = Cargas(i).user_data(3);...
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par=input('de [X,Y,l,N,dt,tmax,T]\n'); X=par(1); Y=par(2); l=par(3); N=par(4); dt=par(5); tmax=par(6); T=par(7); xset('auto clear','on') l2=l^2; S=X*Y; a=sqrt(S/N); Nx=floor(X/a); Ny=floor(Y/a); //a=min(X/Nx, Y/Ny); x=zeros(1,N); y=zeros(1,N); for n=1:N; z=(n-1)/Nx; x(n)=a*(Nx*(z-floor(z))+.5); y(n)...
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exec('Gravitation.sci',-1) exec('degree_rad.sci', -1) //Given that //masses in kg m1 = 8 m2 = 2 m3 = 2 m4 = 2 m5 = 2 a = 2*(10^-2); //in meter Theta = dtor(30) //in radians //Sample Problem 14-2 printf("**Sample Problem 14-2**\n") //The net force will be equal to the vector eum of all the forces acting...
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// Example 6.1, page no-370 clear clc //(a) p=1.5 a=4 b=20 wh=(((b-a)/2)*p)+a printf("(a)just at the bottom level of the tank\nWater head applied to the transmitter =%d mA ",wh) //(b) wh2=(((b-a)/2)*p)+2*a printf("\n\n(b)5m below the bottom of the tank\nWater head applied to the transmitter =%d mA ",wh2) //...
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clear close clc valor = 5 x = [1; 2; 3; 4; 5] y = [15; 28.4; 45.3; 58.6; 77.4] X = [size(x,1) sum(x) sum(x^2); sum(x) sum(x^2) sum(x^3); sum(x^2) sum(x^3) sum(x^4)] Y = [sum(y); sum(y.*x); sum(y.*x^2)] A = X\Y resultado = A(1,1)+A(2,1)*valor+A(3,1)*valor^2 disp (resultado, "Resultado: ") disp (A, "A: ") disp ("f(x)...
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//Rappel function [p] = d1(x,t,K,T,r,sigma) a = log(x/K)+(r+sigma**2/2)*(T-t) p = a /(sigma*sqrt(T-t)) endfunction test = d1(10,100,100,30,0.05,0.1) function [p] = d2(x,t,K,T,r,sigma) p = d1(x,t,K,T,r,sigma) - sigma*sqrt(T-t) endfunction test = d2(10,100,100,30,0.05,0.1) function [p] = Ca...
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//Ex9_3 clc disp("Vp = (a^2)*sigma/(2*apsilent*micro_p)")//piunch off voltage h = 2*10^-4 //channel height in centimeters a= h/2 //channel width in centimeters rho = 1 //resistivity in ohm_cm sigma = 1/rho //conductivity in mho/cm micro_p = 1800 //mobility in cm_sq/Vs apsilent_r = 16 ...
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//Section-1,Example-2,Page no.-AC.34 // To find number average molecular masses(Mn_bar) and weight averge molecular masses(Mw_bar) clc; WA=200 WB=200 WC=100 MA_bar=1.2*10^5 MB_bar=5.6*10^5 MC_bar=10*10^5 Mn_bar_mixture=(WA+WB+WC)/(WA/MA_bar+WB/MB_bar+WC/MC_bar) disp (Mn_bar_mixture,'number average molecular m...
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// Example 6.4, page no-373 clear clc //(a) a=5*10^-4 l=8 dens=6*1000 w=a*l*dens printf("(a)\nWeight of the displacer if weighed in air = %d kg",w) //(i) sbr1=23 wloss1=w-sbr1 L1=wloss1/(1000*a) printf("\n(i)\tL1=%dm",L1) //(ii) sbr2=22 wloss2=w-sbr2 L2=wloss2/(1000*a) printf("\n(ii)\tL2=%dm",L2) //(...
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function h = cl2bp (m, w1, w2, up, lo, gridsize) //Constrained L2 bandpass FIR filter design. //Calling Sequence //h = cl2bp (m, w1, w2, up, lo, gridsize) //h = cl2bp (m, w1, w2, up, lo) //Parameters //m: degree of cosine polynomial, i.e. the number of output coefficients will be m*2+1 //w1 and w2: bandpass filter cut...
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// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal // Chapter 2 - Pressure and its measurements // Problem 2.21 //Given Data Set in the Problem dens=1000 g=9.81 h1=0.35 h2=0.3 SG=0.8 //calculations //pC=pD //pC=pA-dens*g*h1.....adn pD=pB-dens*g*h1-dens*g*h2 pB_pA=SG*dens*g*h2 mprin...
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//ques-34.8 //Calculating absorbance and molecular absorption coefficient of sample clc ratio=1/0.16;//ratio = Io/I C=0.05;//concentration of benzene solution (in M) l1=0.1; l2=0.2;//length (in cm) EC=log10(ratio)/(C*l1); A=EC*C*l1; //On solving, log10(tran) = EC*C*l2 tran=0.025; printf("The absorbance is %.1...
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//Example 5_6 clc(); clear; //To find out how fast the car is going f=4000 //units in Newtons s=50 //units in meters theta=180 //units in degrees m=2000 //units in Kg v0=20 //units in meter/sec vf=sqrt((2*((f*s*cos(theta*%pi/180))+(0.5*m*v0^2)))/m) //units in meter/sec printf("The speed of the c...
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clc; clear; mprintf('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-15.1 Page No.332\n'); //Torque P=5; n=1750; T=63000*P/n; mprintf('\n Torque = %f in-lb.',T); //Length of key for shear Su=61000; Ss=0.5*Su; b=0.125; D=0.5; Ls1=2*T/(Ss*b*D); SF=2.5; Ls=SF*Ls1; mprintf('\n Length of key for shear = %f in.',...
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 4.10\n\n\n"); // Chapter 4 : The Second Law Of Thermodynamics // Problem 4.10 (page no. 159) // Solution hfg=1959.7; //Unit:kJ/kg //Evaporative enthalpy T=195.07+273; //Converted into Kelvin //Temperature deltaS=hfg/T; //Change in entropy //kJ/kg*K p...
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//Problem 5.15: //initializing the variables: MWDCB = 147; MWTCB = 290 //calculation: //for 1 lb of dichlorobenzene (DCB), the following mass of HCl is produced: HCLpd1 = 2/MWDCB //for 1lb of tetrachlorobiphenyl (TCB), the following mass of HCl is produced HCLpd2 = 4/MWTCB x = (HCLpd2 - HCLpd1)*100/HCLpd1 ...
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//Caption:Determine the value of resistance //Exa:2.28 clc; clear; close; V=220;//in volts R_a=0.1;//in ohms N_1=800;//in rpm N_2=520;//in rpm I_a1=20;//in ampers E_1=V-(I_a1*R_a);//in volts E_2=N_2*E_1/N_1;//in volts R_A=-(E_2-V+I_a1*R_a)/20; disp(R_A,'Additional resistance(in ohms)=');
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 15.4w //calculation of the extension of the wire over its natural length //given data m=5*10^-3//mass(in kg) of the wire L=50*10^-2//length(in cm) of the wire v=80//speed(in m/s) of the wave Y=16*10^11//Young modulu...
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//Exa 4.8 clc; clear; close; format('v',7); //Given Data : m1=5;//Kg T1=200+273;//K Cp1=0.4;//KJ/KgK m2=100;//Kg T2=30+273;//K Cp2=2.1;//KJ/KgK //m1*Cp1*(T1-T)=m2*Cp2*(T-T2) T=(m1*Cp1*T1+T2*m2*Cp2)/(m2*Cp2+m1*Cp1);//K deltaS1=integrate('m1*Cp1/T','T',T1,T);//KJ/K deltaS2=integrate('m2*Cp2/T','T',T2,T);...
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// RO05 - tp noté // Antoine Hars // Exercice 1 ///////////////////////////////////////////////////////////////////// N = 500;
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//=========================================================================== //chapter 5 example 35 clc; clear all; //variable declaration e = 8.85*10^-12; V = 10000; //voltage in V r = 40*10^-3; //radius in m //calcaulations d = (4/2)*10^-3; ...
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function [y] = RaisedCosinetxfilter(in,bet,span,sps,varargin) y=[]; // Display mode mode(0); // Display warning for floating point exception ieee(1); //RaisedCosinetxfilter Apply pulse shaping by upsampling signal using raised cosine FIR filter //Y = RaisedCosinetxfilter(in,bet...
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// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal // Chapter 4-Buoyancy and Floatation //// Problem 4.17 //Derivation asked(Theoretrical Work)
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// Finding value of f(x) given (x0,y0),(x1,y1) and (x2,y2) function [y] = quad_interpolation(x0,y0,x1,y1,x2,y2,x) b0 = y0 b1 = (y1 - y0)/(x1 - x0) b2 = ((y2 - y1)/(x2 - x1) - b1)/(x2 - x0) y = b0 + b1*(x - x0) + b2*(x - x0)*(x - x1) endfunction
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clc; clear;close; EI = 250e9; // szywnosc na zginanie, Nmm^2 P = 1000; // sila skupiona, N L = 2000; // dlugosc belki, mm xp = 0.5; y = [0;0] x = 0:10:L // dla i=1 i=1; printf('----- KROK %i -----\n',i) h = x(i+1) - x(i); //k1 = f(x(i), y(:,i)); printf("Wartosc x(i)= %i \n",x(i)); printf("Wartosc y(:,i)= "...
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//Section-9,Example-4,Page no.-E.14 //To find the potential of Daniel cell. clc; C_Zn=1.52 C_Cu=0.48 E0_cell=1.10 n=2 E_cell=E0_cell-((0.0592/n)*log10(C_Zn/C_Cu)) disp(E_cell,' potential of Daniel cell')
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clc; funcprot(0); //Example 20.5 //Initializing the variables f = 0; Atunnel = 1.227; Ashaft = 12.57; Q =2; L = 200; g = 9.81; //Calculations Zmax = (Q/Ashaft)*sqrt(Ashaft*L/(Atunnel*g)); T = 2*%pi*sqrt(Ashaft*L/(Atunnel*g)); disp(T,"Mass Oscillation Period (s) : ",Zmax,"Peak water level (m):");
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//Flywheel //alpha=12-t //omega=12*t-(t^2)/2+C //When t=4 sec omega=60 rad/sec C1=20 //When t=6 sec omega=12*6-((6*6)/(2))+20 //rad/sec //theta=6*t^2-(t^3)/6+20*t+C2 //When t=0 theta0=C2 //When t=6 sec theta6=180+C2 //Angular displacement during 6 seconds=180 rad //Number of revolution N=180/(2*%pi) pr...
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clc; a=0.8/12; b=0.12/2; x=a+b/2; s_AF=32*x/0.233; disp(s_AF,"stoichiometric A/F ratio is:"); Twp=a+b+3.76*x; C=a/Twp*100; H=b/Twp*100; N=.365/Twp*100; disp(N,H,C,"wet analysis of C,H, and N respectively is:")
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//Exa1 clc; clear; close; //given data : Production=1000//units CostOfProduction=1850;//in Rs. NormalLoss=10//in % ActualLoss=150;//in Units ScrapValue=50;//in Paise/unit NLoss=Production*NormalLoss/100;//in Units UnitsProduced=Production-NLoss;//in Units CostPerUnit=(CostOfProduction-50*10^-2*NLoss)/UnitsP...
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names = ['mb_init_tcp',.. 'mb_init_rtu',.. 'mb_read',.. 'mb_write',.. 'mb_write2']; files = ['block_common.o',.. 'mb_common.o',.. 'mb_init_tcp.o',.. 'mb_init_rtu.o',.. 'mb_read.o',.. 'mb_write.o',.. 'mb_write2.o']; ldflags="-L. ...
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clc disp("Problem 10.7") printf("\n") printf("Voltage v1=5*cos(w1*t)") printf("Voltage v2=10*cos(w2*t+60)") //The circuit is modeled as disp("Resistance is 10ohm and inductance is 5mH") R=10;L=5*10^-3; disp("a)") w1=2000;w2=2000; //Let Z be the impedance of the coil Z1=R+%i*L*w1 Z2=R+%i*L*w2 //Let V be p...
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ex6.sce
//Gerar e apresentar histogramas, normais e equalizados //leitura das imagens a = imread('C:\Users\marco\OneDrive\Documentos\GitHub\PDI\aula3\parte 2\1.bmp'); //captura de dimensões da img [rows,columns] = size(a); //cálculo do número de pixels da img t = double(rows*columns); //definição inicial do maior valor com...
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errcatch(-1,"stop");mode(2);//Caption:Determine the (a)full load speed (b)Speed regulation (c)HP rating (d)Full load efficiency //Exa:2.36 ; ; V=240;//in volts R_f=120;//in ohms R_a=0.25;//in ohms I_1=60;//in amperes I_f=V/R_f;//in amperes I_a1=I_1-I_f;//in amperes E_b1=V-I_a1*R_a;//in volts N_o=1000;//in...
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x=[2.30256624769934; 2.29071803023829; 2.26283604900314; 2.35145015316178; 2.27686291358213; 2.29805616201205; 2.32805830340568; 2.30878734371402; 2.29343801980763; 2.23019030245799]; fs=4e6; t=(1/fs); [F,LT,UT]=falltime(x,fs); disp(F); disp(LT); disp(UT); //output // 0.0000002 // // 0.0000022 // // 0.0000...
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p=2.94//g/cm^3(density) W=9.81//kN/m^3(Specific weight of water)
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6_6.sce
clc //initialisation of variables g= 32.2 //ft/sec^2 v= 4 //ft/sec K= 300000 //lb/in^2 d= 6 //in t= 0.25 //in E= 30*10^6 //lb/in^2 w= 62.4 //lb/ft^3 //CALCULATIONS P= sqrt((w*v^2/g)/((d/(E*144*t))+(1/(K*144))))/144 Sm= P*d/(2*t) //RESULTS printf ('Hoop stress = %.f lb/in^2',Sm)
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Engineering Thermodynamics by Onkar Singh Chapter 7 Example 15") disp("In question no. 15 prove for ideal gas satisfies the cyclic relation is done which cannot be solve using scilab software.")
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clc d_p=0.05; // diameter of piston in m d_c=0.0504; // diameter of cylinder in m SG=0.87; rho_w=1000; // kg/m^3 v=10^-4; // m^2/s dp=1.4*10^6; // Pa l=0.13; // m c=(d_c-d_p)/2; // clearance u=SG*rho_w*v; // Dynamice viscocity Vp=dp*c^3/(6*u*l*(d_p/2+c)); disp("Velocity of the dashpot =") disp(Vp) ...