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// Exa 6.7 clc; clear; close; format('v',9) // Given data Im = 20;// in mA Im = Im * 10^-3;// in A Vm = 50;// in mV Vm = Vm * 10^-3;// in V V = 500;// in V Rm = Vm/Im;// in ohm Rs = (V/Im)-Rm;// in ohm disp(Rs,"The series resistance in ohm is");
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clear clc E_RHE=(0.5335-(-2.363));//reduction reaction at RHE in V RT_F=0.05915;// E_LHE=((RT_F/2)*log10(0.1*0.2^2));//reduction reaction at LHE in V Ecell=E_RHE-E_LHE;//cell reaction in V printf('Ecell=%.4f V',Ecell) E_RHE=(0.0-0.0713);//reduction reaction at RHE in V RT_F=0.05915;// E_LHE=((RT_F)*log10((0.5^(1/2))/(0.02*0.02)));//reduction reaction at LHE in V Ecell=E_RHE-E_LHE;//cell reaction in V printf('\nEcell=%.4f V',Ecell) E_RHE=(0.337-(-0.441));//reduction reaction at RHE in V RT_F=0.05915;// E_LHE=((RT_F/2)*log10(0.05/0.01));//reduction reaction at LHE in V Ecell=E_RHE-E_LHE;//cell reaction in V printf('\nEcell=%.4f V',Ecell) E_RHE=(0.0-0.0);//reduction reaction at RHE in V RT_F=0.05915;// E_LHE=((RT_F/2)*log10(6.43/0.127));//reduction reaction at LHE in V Ecell=E_RHE-E_LHE;//cell reaction in V printf('\nEcell=%.4f V',Ecell) E_RHE=(-0.763-0.337);//reduction reaction at RHE in V RT_F=0.05915;// E_LHE=((RT_F/2)*log10((0.1^2)*0.732));//reduction reaction at LHE in V Ecell=E_RHE+E_LHE;//cell reaction in V printf('\nEcell=%.3f V',Ecell) //There are some errors in the solution given in textbook //page 455
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clc; p=3;//bar v=0.18;//m^2/kg p2=0.6;//bar c=p*v^2; v2=(c/p2)^0.5; W=-c*(10^5)*[(1/v)-(1/v2)]; disp("Work done by the fluid is:"); disp("N m/kg",-W); //Answers vary more than than +/-5 : //Answers in the textbook is wrong
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clc //initialisation of variables h= 10 //ft l= 50 //ft d= 1 //in lm= 5 //in f= 0.01 sm= 13.6 g=32.2 //CALCULATIONS ps= sm*lm/12 v= sqrt((ps+h)*2*g*(d/12)/(4*f*l)) Q= v*%pi*(d/12)^2/4 //RESULTS printf ('Discharge through the pipe= %.3f cuses',Q)
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function [simulator_filepath, input_filepath,output_filepath]=simulator_properties() simulator_filepath='/Users/mcfly/Desktop/INP-SCILAB/' input_filepath='/Users/mcfly/Desktop/INP-SCILAB/input/' output_filepath='/Users/mcfly/Desktop/INP-SCILAB/output/' endfunction
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//for a matrix - increasing the neighbourhood mat = [0 0 0 0 0 0;0 0 0 1 0 0;0 0 7 0 0 0 ;0 0 0 0 0 0]; loc=localMaximaFinder(mat,[3 3],4,1); loc
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// Example 4.17.1 clc; clear; L=10; //length of optical link n1=1.49 //refractive index c=3d8; //speed of light delta=1/100; //relative refractive index delTS=L*n1*delta/c; //computing delay difference delTS=delTS*10^12; sigmaS=L*n1*delta/(2*sqrt(3)*c); //computing rms pulse broadning sigmaS=sigmaS*10^12; B=1/(2*delTS); //computing maximum bit rate B=B*10^3; B_acc=0.2/(sigmaS); //computing accurate bit rate B_acc=B_acc*10^3; BLP=B_acc*L; //computing Bandwidth length product printf("\nDelay difference is %d ns.\nRMS pulse broadning is %.1f ns.\nBit rate is %.1f Mbit/s.\nAccurate bit rate is %.3f Mbits/s.\nBandwidth length product is %.1f MHz.km",delTS,sigmaS,B,B_acc,BLP); //answer for maximum bit rate is given as 1.008 Mb/s, deviation of 0.008 Mb/s.
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t=linspace (0,10,1000); p=poly(0,'p'); a=95/13 S=%pi*(2.5^2) Ts=100 //tps de simulation en min Qs=40 G=1 Tau=S/(a*G); G=syslin('c',10/(1+Tau*p)) h=csim('step',t,G); plot2d(t,h);
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 12.6 //writing the equation giving angular displacement as a function of time //given data theta0=%pi/10//amplitude(in rad) of motion theta=%pi/10//displacement(in rad) at t=0 s T=.05//time period(in s) //calculation //required equation is ......theta = theta0*sind((w*t) + delta) w=(2*%pi)/T//value of w in above equation delta=asind(theta/theta0)//value of delta in above equation...i.e at t=0 printf('equation giving angular displacement as a function of time is \n theta = (%3.2f rad)*sin[(%3.2f s^-1)t + %d]',theta0,w,delta)
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clc(); clear; // To calculatevthe amount of water evaporated per hour per square feet from the water surface u = 10; // Flow of air stream in fps r = 33.3; // Relative humidity T = 519; // Temperature in Rankine p = 0.1130; // Partial pressure of water vapour x = 4/12; // Water surface in the wind direction n = 15.99*10^-5; // Kinematic viscosity k = 0.0149; // Thermal conductivity in Btu/hr-ft-F Re = u*x/n; // reynolds number D = 1.127; // Diffusion coefficient in ft^2/sec R = 85.74; // Gas constant in Imperial in Imperial units hd =0.664*Re^0.5*(n*3600/D)^(1/3)*D/x; // Heat transfer coefficient Pr = 0.710; // Prandtls number Nu = 0.664*sqrt(Re)*Pr^(1/3); // Nusselt number h = Nu*k/x; // Heat transfer coefficient ps = 0.2473; // Saturation pressure of water vapour m = hd*(ps-p)*144/(R*T); // Water vapour formation rate in lb/hr-ft^2 printf("The rate of amount of water evaporated per sq. foot is %.3f lb/hr-ft^2",m);
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//Faults and Protection// //Example 16.3// P=100;//input power in KVA// Xt=0.04;//limiting ac reactance value// Fov=2;//current ovarload factor// Pc=Xt*P*Fov;//choke power of the converter in KVA// printf('choke power of the converter=Pc=%fKVA',Pc);
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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 : 7.11 : // Page number 216-217 clear ; clc ; close ; // Clear the work space and console // Given data V = 85.0 // Line Voltage(kV) g_max = 55.0 // Maximum stress(kV/cm) // Calculations V_1 = 0.632*V // Intersheath potential(kV) d = 0.736*V/g_max // Core diameter(cm) d_1 = 2*V/g_max // Intersheath diameter(cm) D = 3.76*V/g_max // Overall diameter(cm) d_un = 2*V/g_max // Core diameter of ungraded cable(cm) D_un = 2.718*d_1 // Overall diameter of ungraded cable(cm) // Results disp("PART II - EXAMPLE : 7.11 : SOLUTION :-") printf("\nDiameter of intersheath, d_1 = %.2f cm", d_1) printf("\nVoltage of intersheath, V_1 = %.2f kV, to neutral", V_1) printf("\nConductor diameter of graded cable, d = %.2f cm", d) printf("\nOutside diameter of graded cable, D = %.2f cm", D) printf("\nConductor diameter of ungraded cable, d = %.2f cm", d_un) printf("\nOutside diameter of ungraded cable, D = %.2f cm", D_un)
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clc //given that r = 0.53 // radius of hydrogen atom in angstrom m_e = 9.1e-31 // mass of electron in kg h = 6.63e-34 // Plank constant printf("Example 2.10") h_bar = h / (2*%pi) // constant del_x = 2*r // calculation of uncertainty in position del_p = h_bar/(2*del_x*1e-10) // calculation of uncertainty in momentum p = del_p E = p^2/(2*m_e*1.6e-19)// Calculation of energy in eV printf("\n Kinetic energy needed by an electron to be \n confined in electron is %f eV.\n\n\n",E) // When problem is solved by del_x*del_p = h_bar, then minimum value of kinetic energy will become 13.6eV
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x(1)=1 // c(1)fn +1 x(2)=-0 // c(2)fn +0 x(3)=-1 // c(3)fn -1 b(1)=1 b(2)=0 b(3)=1/3 for i=1:3 M(1,i)=1 M(2,i)=x(i) M(3,i)=x(i)^2 end c=inv(M)*b S=c(1)^2+c(2)^2+c(3)^2 //solução //disp(S) disp('Coeficientes:') disp(c) //resp.: C = coeficientes /* Resposta está em C, são os coeficientes Ex (1): Un+1 = Un + h (C1fn<+0> + C2fn<-1> + C3fn<-2>) //valores em <> 1: x(1)=0 x(2)=-1 x(3)=-2 b(1)=1 b(2)=1/2 b(3)=1/3 //se tiver b(4) é 1/4 2: x(1)=1 x(2)=-1 x(3)=-2 b(1)=1 b(2)=1/2 b(3)=1/3 3: igual à 2 4: igual à 2/3, mas pega a norma 5: acho que essa merda tá errada, tomar no cu x(1)=1 x(2)=0 x(3)=-1 b(1)=1 b(2)=0 b(3)=1/3 6: igual à 5, mas usa max Teste: 57 (questão em print) x(1)=1 x(2)=0 x(3)=-1 b(1)=1 b(2)=0 b(3)=1/3
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//function [] = sunPosition() clc clear exec('NumberDays.sce'); exec('ConvertTime.sce'); exec('getDepthDiffuse.sce'); //latitudeSign is +1 if N of equator, -1 if S of equator latitudeSign = 1; degreesLat = 29; minutesLat = 39; secondsLat = 7.19; longitudeSign = 1; degreesLong = 82; minutesLong = 19; secondsLong = 29.97; standardTimeMeridian = 75; month = 2; // number e.g. 01 for January, 02 for February day = 1; //solarHourAngle = -10.7; // in degrees solarHourAngle = 79; panelTiltAngle = 30; panelAzimuthAngle = 10; groundReflectance = 0.2; daysPassed = numberDays(month,day); latitude = latitudeSign*(degreesLat + minutesLat/60 + secondsLat/3600); longitude = longitudeSign*(degreesLong + minutesLong/60 + secondsLong/3600); solarDec = 23.45*sind(360*(284+daysPassed)/365); solarAltitudeAngle = asind(sind(latitude)*sind(solarDec)+cosd(latitude)*cosd(solarDec)*cosd(solarHourAngle)); solarAzimuthAngle = asind(cosd(solarDec)*sind(solarHourAngle)/cosd(solarAltitudeAngle)); i = acosd(cosd(solarAltitudeAngle)*cosd(solarAzimuthAngle-panelAzimuthAngle)*sind(panelTiltAngle)+sind(solarAltitudeAngle)*cosd(panelTiltAngle)) [opticalDepth, diffuseFactor] = getDepthDiffuse(month); insolationExtra =1353*(1+.034*cosd(360*daysPassed/365.25)) instantRadiation = insolationExtra*exp(-opticalDepth/sind(solarAltitudeAngle))// Cn omitted (assume Cn = 1) beamRadiation = instantRadiation*cosd(i); diffuseRadiation = diffuseFactor*instantRadiation*cosd(panelTiltAngle/2)^2; groundRadiation = groundReflectance*instantRadiation*(sind(solarAltitudeAngle)+diffuseFactor)*sind(panelTiltAngle/2)^2; insolationTotal = beamRadiation + diffuseRadiation +groundRadiation; hourAngleSunriseDif = acosd(-tand(latitude)*tand(solarDec)); //sunset is with +coef. timeDifNoon = hourAngleSunriseDif*4/60; // unit in hours (float) solarRiseTime = 12-timeDifNoon; solarSetTime = 12+timeDifNoon; solarRiseTimeFormatted = convertTime(solarRiseTime); solarSetTimeFormatted = convertTime(solarSetTime); n=daysPassed; b=360/364*(n-81); et =9.87*sind(2*b)-7.53*cosd(b)-1.5*sind(b); localSunriseTime = solarRiseTime+(-et-4*(standardTimeMeridian-longitude))/60; //divid ed by 60 to match units in hours localSunsetTime = solarSetTime+(-et-4*(standardTimeMeridian-longitude))/60; localRiseTimeFormatted = convertTime(localSunriseTime); localSetTimeFormatted = convertTime(localSunsetTime); //endfunction
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refs/heads/master
2020-05-20T13:36:05.842840
2013-07-31T06:53:59
2013-07-31T06:53:59
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//=========================================================================== //chapter 10 example 15 clc;clear all; //variable declaration P = 200; //resistance in arm in Ω Q = 200; //resistance in arm in Ω S = 200; //resistance in arm in Ω R = 200; //resistance in arm in Ω p = 0.5; //power in W r = 2; //r is internal resistance of battery in Ω E = 24; //voltage in V //calculations //P = (I^2)*R; power disiipation in W I = sqrt(p/R); V = I*2*R; //the maximum voltage ,that can be appliedto the bridge in V I1 = 2*I; //current through series resistor in A //E = V+(2*I*(r+R) battery emf E R1 = ((E-V)/I1)-r; //series resistance in Ω //result mprintf("current = %3.2f A",I); mprintf("\nseries resistance = %3.2f Ω",R1);
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Example9_7_b.sce
//Example 9.7 (b) //Program To Determine Record Length of Bartlett, //Welch(50% Overlap) and Blackmann-Tukey Methods clear; clc; close; //Data Q=10;//Quality Factor N=1000;//Samples //RECORD LENGTH CALCULATION lb=N/Q; lw=16*N/(9*Q); lbt=3*N/(2*Q); //Display the result in command window disp(lb,"Record Length of Bartlett Method"); disp(lw,"Record Length of Welch(50% overlap) Method"); disp(lbt,"Record Length of Blackmann-Tukey Method");
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Vtol_Parameters.sce
//Parameters of the Plane m=4; //mass of the plane Kg J=0.05; //Inertia Kgm2 r=0.3; //distance at which the plane operates m g= 9.81; //Gravity m/s2 c=0.07; //Damping constant Ns/m
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Ex4_5.sce
clc T=300 //K Nn=2.8*10^19 //cm^-3 Np=1.04*10^19 //cm^-3 //a=Ef-Ev an=0.25 //eV ap=0.87 //eV k=8.617*10^-5 //eV/K n0=Nn*exp(-an/(k*T)) disp(n0,"n0 in cm^-3 is=") p0=Np*exp(-ap/(k*T)) disp(p0,"p0 in cm^-3 is=")
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13_13.sce
clear clc Pi=1 Pma=1.75 Pmb=.4 Pmc=1.25 d0=asin(Pi/Pma) dm=%pi - asin(Pi/Pmc) dcc=acosd(((Pi*(dm-d0))- (Pmb*cos(d0))+ (Pmc*cos(dm)))/(Pmc-Pmb)) mprintf("Critical Clearing angle = %.1f deg", dcc)
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Example_9_13.sce
//A Textbook of Chemical Engineering Thermodynamics //Chapter 9 //Chemical Reaction Equilibria //Example 13 clear; clc; //Given: //Reaction: N2 + 3H2 --> 2NH3 K = 2*10^-4; //equilibrium constant of reaction //To determine the percent conversion: //Basis: //1 mole nitrogen and 3 moles of hydrogen in the reactant mixture //Let e be the extent of reaction //Using eq. 9.3 (Page no. 400) //mol fraction of nitrogen is (1-e)/(4-2e) //mol fraction of hydrogen is (3-3e)/(4-2e) //mol fraction of ammonia is 2e/(4-2e) //so, ([2e/(4-2e)]^2)/[(1-e)/(4-2e)][3(1-e)/(4-2e)]^3 = K*P^2 //(a) P = 20; //(bar) //e(4-2e)/(1-e)^2 = 0.73485 e = poly(0,'e'); f = 2.73845*e^2 - 5.4697*e + 0.73485; x = roots(f); mprintf('(a) Percentage conversion is %f percent',x(2)*100); //(b) P = 200; //(bar) //e(4-2e)/(1-e)^2 = 7.3485 e = poly(0,'e'); f = 9.3485*e^2 - 18.697*e + 7.3485; x = roots(f); mprintf('\n\n (b) Percentage conversion is %f percent',x(2)*100); //end
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clc; s1=5.615;//kJ/kg K t1=311;//C t2=300;//C t3=350;//C s2=7.124+(t1-t2)/(t3-t2)*(7.301-7.124); T=t1+273;//K Q=T*(s2-s1); disp("heat supplied is:"); disp("kJ/kg",Q) u1=2545;//kJ/kg u2=2794+(t1-t2)/(t3-t2)*(2875-2794); W=(u2-u1)-Q disp("work done by the steam is:"); disp("kJ/kg",-W)
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function ok=check_mac(txt) //ok=%t,return //errcatch doesnt work poperly ok=%t errcatch(-1,'continue') comp(mac) errcatch(-1) if iserror(-1)==1 then errclear(-1) message('Incorrect syntax: see message in Scilab window') ok=%f end
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// Coding gain plot for n-1 redundant receivers // // (c)2011 L. Rayzman // Created : 10/18/2011 // Last Modified: 10/18/2011 // // TODO: // clear; getd("inc"); // Include Q-function definition //////////////////////////////////////SPECIFY////////////////////////////////////// n=[2:1:10]; // Vector representing number of receivers x=[5, 7, 10, 12, 15]; // Vector representing SNR(dB) /////////////////////////////////////////////////////////////////////////////////// Pe=Qfunc(sqrt(2)*sqrt(10^(x/10))); leg_string=emptystr(); for i=1:length(x), plot2d(n-1, Pe(i).^(1-n), style=i+1, logflag='nl'); //, rect=[n(1)-1, 1e-100 , n($)-1, 1] a=gca(); a.grid=[4 4 -1]; // Prettify a.children(1).children.line_mode="on"; a.children(1).children.mark_mode="on"; a.children(1).children.mark_size=1; a.children(1).children.mark_foreground=(i+1); a.box='on' a.tight_limits='off' leg_string(i)=strcat(["SNR=" sci2exp(x(i)) " dB"]); end legend(leg_string, 2, %t); xtitle("Probability of Error Gain for n-1 Redundant Receivers", "Number of redundant receivers", "Probability of Error Gain "); a.title.font_size=4; // Prettify
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[[i= partials/header ]] [[i= partials/navbar ]] <div class="container" style="width: 100%;flex-flow:row;"> [[i= static/sidebar ]] <div class="container center" style="width:78%;"> <h1>Contact Info</h1> <p>You can contact us at the details below:</p> <ul> <li>Mobile - 0412 356 789</li> <li>Email - <a href="mailto:admin@quickmark.net">admin@quickmark.net</a></li> <li>Author - Jarod Brennfleck (z5217759)</li> </ul> </div> </div> [[i= partials/footer ]]
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clear //Given R0=5 //ohm R100=5.23 //ohm Rt=5.795 //ohm //Calculation t=((Rt-R0)/(R100-R0))*100 //Result printf("\n The temperature of the bath is %0.2f degree C",t)
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clear all; clc; disp("Ex 2_12") disp(" Vector r = (-3i + 2j +6k) m") r=sqrt((-3)^2+2^2+6^2) printf('\n\nthe magnitude of r is: r = %.0f m',r) disp(" ") disp("u = (-3/7)i + (2/7)j + (6/7)k") a1=acos((-3)/7) a=a1*180/%pi printf('\n\nalpha = %.0f degrees',a) b1=acos(2/7) b=b1*180/%pi printf('\n\nbeta = %.1f degrees',b) c1=acos(6/7) c=c1*180/%pi printf('\n\ngamma = %.0f degrees',c)
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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_V2-en.txt // Author:Rohit Suri // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in function [imgSet]=imageSet(imageFolder,varargin) // This function is used to create a collection of images. // // Calling Sequence // imgSet = imageSet(location) // imgSet = imageSet(location, 'recursive') // // Parameters // imgSet: Structure containing collection of images // location: Address of the folder from which collection is to be created // // Description // This function creates an imageSet structure with the following attributes- Description, ImageLocation, and Count. // // Examples // imgSet = imageSet(directory); // // imgSet = imageSet(directory,'recursive'); // // Authors // Rohit Suri [lhs rhs]=argn(0); if lhs>1 error(msprintf(" Too many output arguments")); elseif rhs<1 error(msprintf(" Not enough input arguments")); elseif rhs>2 error(msprintf(" Too many input arguments")); end if rhs==1 then imgSetList=raw_imageSet(imageFolder); for i=1:imgSetList(3) imgLocations(i)=imgSetList(4)(1)(1,i); end imgSet=struct('Description',imgSetList(2),'ImageLocation',imgLocations,'Count',double(imgSetList(3))); else imgSetList=raw_imageSet(imageFolder,varargin(1)); for i=1:length(imgSetList(3)) for j=1:imgSetList(3)(i) imgLocations(j)=imgSetList(4)(i)(1,j); end imgSet(1,i)=struct('Description',imgSetList(2)(i),'ImageLocation',imgLocations,'Count',int32(imgSetList(3)(i))); imgLocations=[]; end end endfunction
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## Test of strip command set echo read <simple.fi strip blobs write -
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PL/SQL Developer Test script 3.0 5 begin -- Call the procedure personas_por_educacion(peducacion => :peducacion, p_recordset => :p_recordset); end; 2 peducacion 1 Tercer ciclo completo 5 p_recordset 1 <Cursor> 116 0
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clear all; clc; disp("Scilab Code Ex 8.2 : ") //Given: P = 15000; //N a = 40; //mm b = 100; //mm //Stress Components: //Normal Force: A = a*b; sigma = P/A; //Bending Moment: I = (a*b^3)/12; //I = (1/12)*bh^3 M = P*(b/2);(b/2); c = b/2; sigma_max =(M*c)/I; //Superposition: x = ((sigma_max-sigma)*b)/((sigma_max+sigma)+(sigma_max-sigma)); sigma_b = (sigma_max-sigma); sigma_c = (sigma_max + sigma); //Display: printf("\n\nThe state of stress at B = %1.1f MPa (tensile)',sigma_b); printf('\nThe state of stress at C = %1.1f MPa (compressive)',sigma_c); //----------------------------------------------------------------------END--------------------------------------------------------------------------------
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//clear// clc clear //exec("9.8data.sci"); t = 0:.01:.5; function w=f(t,Y) w =zeros(2,1); d(X)/d(z}=-ra/U/Ca0 Ka=0.05; Kb=.15; Pao=12; eps=1; A=7.6; R=0.082; T=400+273; Kc=.1; rho=80; kprime=0.0014; D=1.5; Uo=2.5 U:Uo*(l+eps*X) Pa=PBo*(l-X)/(ltepstX) Pb=Pao*X/(l+eps*X) vo=Uo*3.1416*D*D/4 Ca0=PBo/R/T Kca=Ka*R*T Pc=Pb a=l/(l+At(z/U)**O.S) raprime=at (-kprirne*Pa/(l t Kat Pa+Kb:l:Pb+Kct Pc» ra:rhotraprime endfunction x=ode([1;.8],t0,t,f); Ca0=.8; Ct0=1 ya0=Ca0/Ct0; for i=1:length(t) X(i)=1-(1+ya0)/(1+x(2,i)/Ct0)*x(2,i)/Ca0; end plot2d(t,x(1,:)); plot2d(t,x(2,:)); plot2d(t,X);
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//page 70 clc;funcprot(0);//EXAMPLE 3.11 // Initialisation of Variables E=12;......//No. of Edges in the octahedral sites of the unit cell S=1/4;.......//so only 1/4 of each site belongs uniquelyto each unit cell N=E*S+1;.....//No.of site belongs uniquely to each unit cell disp(N,"No.of octahedral site belongs uniquely to each unit cell:")
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// Exa 2.23 clc; clear; close; // Given data e = 1.6*10^-19;// in C R_H = 0.0145;// in m^3/coulomb Miu_e = 0.36;// m^2/v-s E = 100;// V/m n = 1/(e*R_H);// in /m^3 J= n*e*Miu_e*E;// in A/m^2 disp(J,"The current density in A/m^2 is");
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// // plot_linear2dmap.sci // // Plots n iterations of the linear map with matrix A. // function x = plot_linear2dmap(x0,A,n) x = x0; for i = 1:n, plot(x(1),x(2),'diamondred') x = A*x; end endfunction
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// chapter 3 // example 3.5 // Design free-running UJT relaxation oscillator // page-105-106 clear; clc; // given fmin=5; // in Hz (minimum frequency) fmax=50; // in Hz (maximum frequency) E_dc=12; // in V (DC supply) I_P=80; // in mA (peak current) T=8; // in us (trigger time) V_drop=1; // in V (voltage drop across PUT) Rs=39; // in ohm (Assumption as done in the book) Ig=1; // in mA (assumption as done in the book // calculate // since T=Rs*C C=T/Rs; // calculation of capacitance (in uF) printf("\nThe value of capacitance is \t\t C=%.2f uF",C); I_P=I_P*1E-3; // changing unit from mA to A V_P=(I_P*Rs)+V_drop; // calculation of peak point voltage // since V_P=neta*E_dc+V_D, neglecting V_D, we get neta=V_P/E_dc; // calculation of intrinsic stand-off ratio Tmax=1/fmin; // calculation of maximum time period Tmin=1/fmax; // calculation of maximum time period C=C*1E-6; // changing unit from uF to F Rmax=Tmax/(C*log(E_dc/(E_dc-V_P))); // calculation of maximum value of R Rmin=Tmin/(C*log(E_dc/(E_dc-V_P))); // calculation of minimum value of R I_V_max=E_dc/Rmin; // calculation of maximum anode current I_V_min=E_dc/Rmax; // calculation of minimum anode current // since Ig=2*neta*E_dc/Rg , therefore Rg=2*neta*E_dc/Ig; // calculation of gate resistance in k-ohm R1=Rg/neta; // calculation of R1 resistance R2=Rg/(1-neta); // calculation of R1 resistance Rmax=Rmax*1E-6; // changing unit from ohm to M-ohm Rmin=Rmin*1E-6; // changing unit from ohm to M-ohm I_V_max=I_V_max*1E6; // changing unit from A to uA I_V_min=I_V_min*1E6; // changing unit from A to uA printf("\nThe peak point voltage is \t\t V_P=%.2f V",V_P); printf("\nThe intrinsic stand-off ratio is \t neta=%.2f",neta); printf("\nThe maximum value of R is \t\t Rmax=%.2f M-ohm",Rmax); printf("\nThe minimum value of R is \t\t Rmin=%.2f M-ohm",Rmin); printf("\nThe maximum value of anode current is \t I_V_max=%.2f uA",I_V_max); printf("\nThe minimum value of anode current is \t I_V_min=%.f uA",I_V_min); printf("\nThe value of gate resistance is \t Rg=%.2f k-ohm",Rg); printf("\nThe value of R1 is \t\t\t R1=%.f k-ohm",R1); printf("\nThe value of R2 is \t\t\t Rg=%.2f k-ohm",R2); // Note : the answee of Rmax, I_V_max, Rg and R2 varies slightly due to exact calculation
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//EXAMPLE 27.2 //8-POLE GENERATOR clc; funcprot(0); //Variable Initialisation P=8;......//Total number of poles Z=722;.....//Total number of conductors V=500;.......//Termiinal voltage in Volts Ia=200;........//Armature current in Amperes Z=1280;......//Total number of conductors as=160;........//Total number of armature segments ba=4;..........//Advancement in brushes from no-load neutral axis Al=P;...........//Number of parallel paths in a I=Ia/P;.....//Current per path in Amperes b=ba*360/as;......//Brush lead in degrees ATdpole=Z*I*b/360;.....//Armatuue demagnetizing ampere-turns per pole disp(ATdpole,"Armatuue demagnetizing ampere-turns per pole:"); ATepole=Z*I*((1/(2*P))-(b/360));.......//Armature cross-magnetizing ampere-turns per pole disp(ATepole,"Armature cross-magnetizing ampere-turns per pole:");
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//A program to read formatted data from files. fd=mopen("sine.dat","r"); s=mfscanf (fd, "%s %s"); // Reads two words. The %s reads up to the first white space. [n, x1, y1] =mfscanf (4, fd, "%d,%f"); // Reads the next four data set. mclose(fd) disp (y1, x1, n, s); //Reading embedded data fd=mopen("test.dat","r"); A=mfscanf(-1,fd,"Name:%3s,Age:%d,Weight:%f\n"); mclose(fd) disp("The data from the test.dat file") disp(A);
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//Graphical// //Example 4.4.4 //Frequency Response of First Order Difference Equation //a = 0.9 and b = 1-a //Impulse Response h(n) = b.(a^n).u(n) clear; clc; close; a = input('Enter the constant value of Ist order Difference Equation'); b= 1-a; //Calculation of Impulse Response n =0:50; h =b*(a.^n) ; //Discrete-time Fourier transform K = 500; k = 0:1:K; w = %pi*k/K; H = h * exp(-sqrt(-1)*n'*w); //phasemag used to calculate phase and magnitude in dB [Phase_H,m] = phasemag(H); H = real(H); subplot(2,1,1) plot2d(w/%pi,H) xlabel('Frequency in Radians') ylabel('abs(H)') title('Magnitude Response') subplot(2,1,2) plot2d(w/%pi,Phase_H) xlabel('Frequency in Radians') ylabel('<(H)') title('Phase Response')
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Ex8_11.sce
//=========================================================================== //chapter 8 example 11 clc;clear all; //variable declaration V = 230; //voltage in volts I = 4; //current in A I1 = 5; //current in A cosphi = 1; //power factor h = 6; //hours R = 2208; //revolutios made by meter R1 = 1472; //revolutios made by meter E1 = 400; //energy consumption h1 =4; //calculations E = (V*I*cosphi*h)/(1000); //energy consumption in kWh M = R/(E); //meter constant in rev/kWh cosphi2 = (R1/(E1)*(1000/(V*I1*h1))); //power factor of the load is cosphi2 for second measuremnet //result mprintf("meter constant = %3.2f revolutions/kWhr",M); mprintf("\npower factor of the load is cosphi2 for second measuremnet = %3.2f",cosphi2);
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//Exa 8.5 clc; clear; close; //Given data disp("Put alfa=sqrt(6) to find the gain"); alfa=sqrt(6);//unitless Beta=1/(1-5*alfa^2); //Barkhausen critera : A*|Beta|>=1 Beta=-Beta;// A=1/Beta;//unitless disp(A,"Minimum Gain of Amplifier must be : ");
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function traceRoad() clf; isoview(-3,-3,3,3) plot2d(0,0,rect=[-3,-3,3,3]); n=0; x=0; y=0; bt=0; for n=1:2 [bt,x(n),y(n)]=xclick() plot(x(n),y(n),"ro"); if n>1 //plot([x(n-1),x(n)],[y(n-1),y(n)],3,'LineWidth', 2) end end plot(x,y) z=calcZ(x,y) fXYz(x,y,z) endfunction function roadInter() clf; isoview(-3,-3,3,3) plot2d(0,0,rect=[-3,-3,3,3]); n=0; x1=0; y1=0; bt=0; for n=1:2 [bt,x1(n),y1(n)]=xclick() plot(x1(n),y1(n),"ro"); end z1=calcZ(x1,y1) [fX1,fY1]=fXYz(x1,y1,z1) tracefXYzw(fX1,fY1,z1,0.5,"b") x2=0; y2=0; bt=0; for n=1:2 [bt,x2(n),y2(n)]=xclick() plot(x2(n),y2(n),"go"); end z2=calcZ(x1,y1) [fX2,fY2]=fXYz(x2,y2,z2) tracefXYzw(fX2,fY2,z2,0,"m") [p1,p2]=findInter(fX1,fY1,fX2,fY2,0.5,0) tracefXYzwPoint(fX1,fY1,p1,0.5) tracefXYzwPoint(fX2,fY2,p2,0) endfunction function findSide() clf; isoview(-3,-3,3,3) plot2d(0,0,rect=[-3,-3,3,3]); n=0; x1=0; y1=0; bt=0; for n=1:2 [bt,x1(n),y1(n)]=xclick() plot(x1(n),y1(n),"ro"); end z1=calcZ(x1,y1) [fX1,fY1]=fXYz(x1,y1,z1) tracefXYzw(fX1,fY1,z1,0,"b") x2=0; y2=0; bt=0; for n=1:2 [bt,x2(n),y2(n)]=xclick() plot(x2(n),y2(n),"go"); end z2=calcZ(x1,y1) [fX2,fY2]=fXYz(x2,y2,z2) tracefXYzw(fX2,fY2,z2,0,"m") disp(deter(fX1,fY1,fX2,fY2)) endfunction function v=deter(fX1,fY1,fX2,fY2) v=(fX1(2)*fY2(2))-(fX2(2)*fY1(1)) endfunction function [z]=calcZ(x,y) z(1)=0; z(2)=sqrt((x(1)-x(2))^2+(y(1)-y(2))^2); endfunction function [fXz,fYz]=fXYz(x,y,z) fXz(1)=x(1) fXz(2)=(x(2)-x(1))/z(2) fYz(1)=y(1) fYz(2)=(y(2)-y(1))/z(2) fXz(3)=fYz(2) fYz(3)=-fXz(2) endfunction function tracefXYzw(fX,fY,z,w,c) k=1 for n=0:0.01:z(2) Xz(k)=fX(1)+n*fX(2)+w*fX(3) Yz(k)=fY(1)+n*fY(2)+w*fY(3) k=k+1 end plot(Xz,Yz,c); endfunction function tracefXYzwPoint(fX,fY,z,w) Xz=fX(1)+z*fX(2)+w*fX(3) Yz=fY(1)+z*fY(2)+w*fY(3) plot(Xz,Yz,"Ro"); endfunction function [p1,p2]=findInter(fX1,fY1,fX2,fY2,w1,w2) M1=[fX1(2),-fX2(2);fY1(2),-fY2(2)] R1=[fX2(1)-fX1(1)+fX2(3)*w2-fX1(3)*w1; fY2(1)-fY1(1)+fY2(3)*w2-fY1(3)*w1] disp(M1) disp(R1) k=M1(1,1)/M1(2,1) M1(1,:)=M1(1,:)-M1(2,:)*k disp(M1) R1(1)=R1(1)-R1(2)*k disp(R1) p2=R1(1)/M1(1,2); p1=(R1(2)-p2*M1(2,2))/M1(2,1) endfunction //traceRoad(); //roadInter(); findSide()
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//Strength Of Material By G.H.Ryder //Chapter 1 //Example 4 // To Calulate Stress & Extension g=9.8; //Acceleration due to Gravity, Unit in m/sec^2 m=100; //Falling Mass , Unit in Kg W=m*g; //Falling weight , Unit in N D1=1; // diameter of first part of bar, Unit in cm l1=1.5; //Lenght fo first part of bar, Unit in m D2=2; // diameter of second part of bar, Unit in cm l2=1.5; //Lenght fo second part of bar, Unit in m A1=%pi*(D1^2)/4*100; //Area of first part of bar, Unit in mm^2 A2=%pi*(D2^2)/4*100; //A;rea of Second part of bar, Unit in mm^2 E=205,000; //Young's Modulus of the bar, Unit in N/mm^2 h=4; //height from which weight is falling, Unit in cm P=W*(1+(1+2*h*10*E/((l1*1000/A1)+(l2*1000/A2)))^(1/2)); //Formula for Equivalent load, from energy equation, Unit in N x=P*l1/A1*E+P*l2/(A2*E); //Extension in rod, unit in mm //The maximum stress will occur in smallest section. so, maxstress=P/A1; printf("maximum stress=%f N/mm^2",maxstress) printf("Extension =%f mm",x)
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clc // Given that n = 4 // order of bright fringe x_n = 10 // Separation of 4th bright fringe from center in mm D = 1// Separation between source and screen in meter d = 0.2 // Separation between coherent sources in mm // Sample Problem 10 on page no. 97 printf("\n # PROBLEM 10 # \n") printf("\n Standard formula used \n x = D*n*lambda/d \n") lambda = x_n*1e-3*d*1e-3/(n*D) // Calculation of wavelength of sauce in meter printf("\n Wavelength of sauce is %d Angstrom.",lambda*1e10)
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function [x,iter] = jacobiL(A,b,x0,e,it) [l,c] = size(A); erro = 1; x = x0, iter = 0; while erro > e & iter < it xa = x; iter = iter + 1; for i = 1:l soma= 0; for j = 1:l if j ~= i then soma = soma + A(i,j)*xa(j); end end x(i) = (b(i)-soma)/A(i,i); end erro = max(abs(x-xa))/max(abs(x)); end endfunction
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//Given that R = 200 //in ohm C = 15*10^-6 //in F L = 230*10^-3 //in H Em = 36.0 //in volts fd = 60.0 //in Hz //Sample Problem 33-7a printf("**Sample Problem 33-7a**\n") w = 2*%pi*fd Xl = w*L Xc = 1/(w*C) Z = sqrt(R^2 + (Xl - Xc)^2) Imax = Em/Z printf("The amplitude of current in the circuit is %1.2fA, Imax\n", Imax) //Sample Problem 33-7b printf("\n**Sample Problem 33-7a**\n") phi = atan((Xl-Xc)/R) printf("The phase constant is equal to %fdegrees", phi)
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printf("\t example 12.2 \n"); printf("\t approximate values are mentioned in the book \n"); T1=244; // inlet hot fluid,F T2=244; // outlet hot fluid,F t1=85; // inlet cold fluid,F t2=120; // outlet cold fluid,F W=60000; // lb/hr w=488000; // lb/hr printf("\t 1.for heat balance \n"); printf("\t for propanol \n"); l=285; // Btu/(lb) Q=((W)*(l)); // Btu/hr printf("\t total heat required for propanol is : %.2e Btu/hr \n",Q); printf("\t for water \n"); c=1; // Btu/(lb)*(F) Q=((w)*(c)*(t2-t1)); // Btu/hr printf("\t total heat required for water is : %.2e Btu/hr \n",Q); delt1=T2-t1; //F delt2=T1-t2; // F printf("\t delt1 is : %.0f F \n",delt1); printf("\t delt2 is : %.0f F \n",delt2); LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1)))); printf("\t LMTD is :%.0f F \n",LMTD); Tc=((T2)+(T1))/(2); // caloric temperature of hot fluid,F printf("\t caloric temperature of hot fluid is : %.0f F \n",Tc); tc=((t1)+(t2))/(2); // caloric temperature of cold fluid,F printf("\t caloric temperature of cold fluid is : %.1f F \n",tc); UD1=70; // assume, from table 8 A1=((Q)/((UD1)*(LMTD))); printf("\t A1 is : %.2e ft^2 \n",A1); N2=766; // assuming 4 tube passes, from table 9 a1=0.1963; // ft^2/lin ft L=(A1/(N2*a1)); printf("\t L is : %.1f ft \n",L); A2=(N2*12*a1); // ft^2 printf("\t total surface area is : %.0f ft^2 \n",A2); UD=((Q)/((A2)*(LMTD))); printf("\t correct design overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",UD); printf("\t hot fluid:shell side,propanol \n"); Do=0.0625; // ft G1=(W/(3.14*N2*Do)); // from eq.12.36 printf("\t G1 is : %.0f lb/(hr)*(lin ft) \n",G1); printf("\t cold fluid:inner tube side,water \n"); Nt=766; n=4; // number of passes L=12; //ft at1=0.302; // flow area, in^2 at=((Nt*at1)/(144*n)); // total area,ft^2,from eq.7.48 printf("\t flow area is : %.3f ft^2 \n",at); Gt=(w/(at)); // mass velocity,lb/(hr)*(ft^2) printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gt); V=(Gt/(3600*62.5)); printf("\t V is : %.2f fps \n",V); mu2=1.74; // at 102.5F,lb/(ft)*(hr) D=0.0517; // ft Ret=((D)*(Gt)/mu2); // reynolds number printf("\t reynolds number is : %.2e \n",Ret); hi=1300; //Btu/(hr)*(ft^2)*(F) printf("\t hi is : %.0f Btu/(hr)*(ft^2)*(F) \n",hi); ID=0.62; // ft OD=0.75; //ft hio=((hi)*(ID/OD)); // using eq.6.5 printf("\t Correct hi0 to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n",hio); ho=100; // assumption tw=(tc)+(((ho)/(hio+ho))*(Tc-tc)); // from eq.5.31 printf("\t tw is : %.1f F \n",tw); tf=(Tc+tw)/(2); // from eq 12.19 printf("\t tf is : %.0f F \n",tf); kf=0.0945; // Btu/(hr)*(ft^2)*(F/ft), from table 4 sf=0.76; // from table 6 muf=0.65; // cp, from fig 14 ho=102; // Btu/(hr)*(ft^2)*(F), from fig 12.9 printf("\t Correct ho to the surface at the OD is : %.0f Btu/(hr)*(ft^2)*(F) \n",ho); printf("\t pressure drop for annulus \n"); ID=31; // in C=0.1875; // clearance B=29; // baffle spacing,in PT=0.937; as=((ID*C*B)/(144*PT)); // flow area,from eq 7.1,ft^2 printf("\t flow area is : %.2f ft^2 \n",as); Gs=(W/as); // mass velocity,from eq 7.2,lb/(hr)*(ft^2) printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gs); mu1=0.0242; // lb/(ft)*(hr), fig 15 De=0.0458; // fig 28 Res=((De)*(Gs)/mu1); // reynolds number printf("\t reynolds number is : %.1e \n",Res); f=0.0014; // friction factor for reynolds number 91000, using fig.29 s=0.00381; // for reynolds number 91000,using fig.6 Ds=31/12; // ft phys=1; N=(5); // number of crosses,using eq.7.43 printf("\t number of crosses are : %.0f \n",N); delPs=((f*(Gs^2)*(Ds)*(N))/(5.22*(10^10)*(De)*(s)*(phys)))/(2); // using eq.12.47,psi printf("\t delPs is : %.1f psi \n",delPs); printf("\t allowable delPa is 2 psi \n"); printf("\t pressure drop for inner pipe \n"); f=0.00019; // friction factor for reynolds number 36200, using fig.26 s=1; phyt=1; delPt=((f*(Gt^2)*(L)*(n))/(5.22*(10^10)*(D)*(s)*(phyt))); // using eq.7.45,psi printf("\t delPt is : %.1f psi \n",delPt); X1=0.2; // X1=((V^2)/(2*g)),using fig.27 delPr=((4*n*X1)/(s)); // using eq.7.46,psi printf("\t delPr is : %.1f psi \n",delPr); delPT=delPt+delPr; // using eq.7.47,psi printf("\t delPT is : %.1f psi \n",delPT); printf("\t allowable delPT is 10 psi \n"); Uc=((hio)*(ho)/(hio+ho)); // clean overall coefficient,eq 6.38,Btu/(hr)*(ft^2)*(F) printf("\t clean overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",Uc); Rd=((Uc-UD)/((UD)*(Uc))); // eq 6.13,(hr)*(ft^2)*(F)/Btu printf("\t actual Rd is : %.5f (hr)*(ft^2)*(F)/Btu \n",Rd); // end
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//Exa 1.23 clc; clear; close; format('v',8); //Given Data : Rdegree=8314.3;//Universal Gas Constant r=12;//meter Patm=75;//cm of Hg Patm=Patm/76*1.01325*10^5;//N/m^2 V=4/3*%pi*r^3;//m^3 M_air=28.97; M_H2=2 Tair=18+273;//K g=9.81;//gravity constant Rair=Rdegree/M_air;//Nm/KgK RH2=Rdegree/M_H2;//Nm/KgK //p*V=m*R*T m_air=Patm*V/Rair/Tair;//Kg disp(m_air,"Mass of air in kg : "); n_air=m_air/M_air;//moles disp(n_air,"No. of moles : "); m_H2=n_air*M_H2;//Kg disp(m_H2,"Mass of H2 in kg : "); Load=g*(m_air-m_H2);//N disp(Load,"Load balloon can lift in N ; ");
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//pagenumber 113 example 18 clear resacu=0.1*10^-12;//ampere u=20+273;//kelvin voltaf=0.55;//volt w=1.38*10^-23; q=1.6*10^-19; for z=1:2 if z==2 then u=100+273; disp("current at 100celsius rise"); end voltag=w*u/q; i1=(10^-13)*(exp((voltaf/voltag))-1); if z==2 then i1=(256*10^-13)*((exp(voltaf/voltag)-1)); end disp("current = "+string((i1))+"ampere"); end
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(set-strategy depth) (unwatch all) ; tceplace.bat test (clear) (open "Results//tceplace.rsl" tceplace "w") (load "compline.clp") (dribble-on "Actual//tceplace.out") (batch "tceplace.bat") (dribble-off) (printout tceplace "tceplace.bat differences are as follows:" crlf) (compare-files "Expected//tceplace.out" "Actual//tceplace.out" tceplace) ; close result file (close tceplace)
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// Example 3.1 // Computation for mobility of the free electrons in aluminium// // Page no.61 clc; clear; close; //Given data ; d=2.70*10^3//2.70*10^3 kg/m3 v=3;//3 electrons/atom A=26.98; M=1.660*10^-27;//1.660*10^-27 kg/atom e=1.60*10^-19; R=3.44*10^-8;//R=resistivity //...................................(B)....................................// //Calculation for concentration of the free electrons in aluminium// n=(d*v)/(A*M); //Calculation for mobility of the free electrons in aluminium// mu=10^4/(n*e*R);//mu=mobility of the free electrons //Displaying the result in command window printf('\n Concentration of the free electrons in aluminium = %0.3f x 10^29 electron/m3',n*10^-29); printf('\n \n Mobility of the free electrons in aluminium = %0.2f cm2/V sec',mu); //Answers are varying due to round off error//
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clc clear //Initialization of variables Rj=1.985 N=1 T1=540+460 //R T2=3540+460 //R //calculations Q=N*(14.215*(T2-T1)-6.53*10^3 *log(T2/T1) -1.41*10^6 *(1/T2-1/T1)) Tm=(T1+T2)/2 Cv=14.215-6.53*10^3 /Tm +1.41*10^6 /Tm^2 Q2=N*Cv*(T2-T1) //results printf("Heat added in case 1 = %.1f Btu",Q) printf("\n Heat added in case 2 = %.1f Btu",Q2)
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//Effect of variable specific heat on efficiency clc,clear //Given: r=7 //Compression ratio g=1.4 //Specific heat ratio(gamma) cv=0.718 //(Assume)Specific heat at constant volume in kJ/kgK dcv=1*cv/100 //Change in specific heat in kJ/kgK //Solution: R=cv*(g-1) //Specific gas constant in kJ/kgK eta=round(100*(1-1/r^(g-1)))/100 //Efficiency when there is no change in specific heat function [eta]=Otto(cv) //Defining efficiency as a function of specific heat eta=1-1/r^(R/cv) endfunction funcprot(0) detaBydcv=derivative(Otto,cv) //Derivative of efficiency wrt to specific heat at initial value of specific heat detaByeta=detaBydcv*dcv/eta //Change in efficiency wrt to initial value of efficiency //Results: printf("\n The percentage change in the efficiency of Otto cycle = %.3f percent",detaByeta*100) if (detaByeta < 0) then disp("decrease") end
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 7.3\n\n\n"); // Chapter 7 : Mixtures Of Ideal Gases // Problem 7.3 (page no. 323) // Solution //Ten pounds of air,1 lb of carbon dioxide,and 5 lb of nitrogen are mixed at constant temperature until the mixture pressure is constant nair=10/29; //no of moles of air=ratio of mass and molecular weight //10 lb of nitrogen per pound //molecular weight of air=29 printf("The moles of air is %f mole/lbm of mixture\n",nair); nCO2=1/44; //no of moles of carbon dioxide=ratio of mass and molecular weight //1 lb of per pound //molecular weight of CO2=44 printf("The moles of carbon dioxide is %f mole/lbm of mixture\n",nCO2); nN2=5/28; //no of moles of nitrogen=ratio of mass and molecular weight //5 lb of nitrogen per pound //molecular weight of N2=28 printf("The moles of nitrogen is %f mole/lbm of mixture\n",nN2); nm=nair+nCO2+nN2; //Unit:Mole/lbm //number of moles of gas mixture is sum of the moles of its constituent gases printf("The total number of moles is %f mole/lbm\n\n",nm); xair=nair/nm //mole fraction of air=ratio of no of moles of air and total moles in mixture xCO2=nCO2/nm; //mole fraction of carbon dioxide=ratio of no of moles of carbon dioxide and total moles in mixture xN2=nN2/nm; //mole fraction of nitrogen=ratio of no of moles of oxygen and total moles in mixture printf("The mole fraction of air is %f \n",xair); printf("The mole fraction of carbon dioxide is %f\n",xCO2) printf("The mole fraction of nitrogen is %f\n\n",xN2); //final pressure of is 100 psia pair=xair*100; //the partial pressure of air= final pressure * the mole fraction of air //psia printf("The partial pressure of air is %f psia\n",pair); pCO2=xCO2*100; //the partial pressure of carbon dioxide= final pressure * the mole fraction of CO2 //psia printf("The partial pressure of carbon dioxide is %f psia\n",pCO2); pN2=xN2*100; //the partial pressure of nitrogen=final pressure * the mole fraction of nitrogen //psia printf("The partial pressure of nitrogen is %f psia\n\n",pN2); //the molecular weight of mixture=sum of products of mole fraction of each gas component MWm=(xair*29) + (xCO2*44) + (xN2*28); //The molecular weight of air printf("The molecular weight of air is %f\n\n",MWm); Rm=1545/MWm; //the gas constant of air printf("The gas constant of air is %f\n\n",Rm);
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//clear// clear; clc; //Example 9.6 //Given Dt = 2; //[m] Da = 0.667; //[m] n = 180/60; //[rps] T = 20; //[C] qg = 100; //[m^3/h] rho = 1000; //[kg/m^3] mu = 10^-3; //[kg/m-s] ut = 0.2; //[m/s] //(a) //The power input is calculated and followed by correction of gas effect Nre = n*Da^2*rho/mu; //For a flat blade turbine, from Table 9.3 KT = 5.75; //Using Eq.(9.24) Po = KT*n^3*Da^5*rho/1000; //[kW] At = %pi/4*Dt^2; //[m^2] //Superficial gas velocity Vs_bar = At*qg/3600/10 //[m/s] //From Fig. 9.20 Pg/Po = 0.60 Pg = Po*0.6; //[kW] //From Fig.9.7, depth of liquid is equal to diameter of the tank //Hence, liquid volume V = %pi/4*Dt^2*Dt; //[m^3] //The input power per unit volume PgbyV = Pg/V ; //[kW/m^3] //(b) sigma = 72.75; //[g/s^2] rho_L = 10^-3; //[g/mm] PgbyV = PgbyV*10^3 ; //[g/mm-s^2] //Using Eq.(9.46) //Let x = shi^(0.5) //solving the equation as quadratic equation a = 1; b = -(Vs_bar/ut)^0.5; c = -0.216*((PgbyV)^0.4)*(rho_L^0.2)/(sigma^0.6)*(Vs_bar/ut)^(0.5); x = (-b+sqrt(b^2-4*a*c))/(2*a); shi = x^2; //(c) //To find out mean bubble diameter //Using Eq.(9.44) Ds_bar = 4.15*sigma^0.6/(PgbyV^0.4*rho_L^0.2)*shi^0.5+0.9 // [mm] //(d) //From Eq.(9.40) aprime = 6*shi/Ds_bar //[mm^-1]
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//Example 2_14 clc; clear;close; //Given data: Vs=230;//V f=50;//Hz alfa=30*%pi/180;//radian I=4;//A //Solution : disp("part (a)"); Vm=Vs*sqrt(2);//V Vdc=2*Vm/%pi*cos(alfa)//V RL=Vdc/I;//ohm IL=I*2*sqrt(2)/%pi;//A Pin_active=Vs*IL*cos(alfa);//W Pin_reactive=Vs*IL*sin(alfa);//vars Pin_appearent=Vs*IL;//VA disp(Vdc,"dc output voltage(V)"); disp(Pin_active,"Active power input(W)"); disp(Pin_reactive,"Reactive power input(vars)"); disp(Pin_appearent,"Appearent power input(VA)"); disp("part (b)"); Vdc=Vm/%pi*(1+cos(alfa))//V IL=Vdc/RL;//A I_fund=2*sqrt(2)/%pi*IL*cos(alfa/2);//A Pin_active=Vs*I_fund*cos(alfa/2);//W Pin_reactive=Vs*I_fund*sin(alfa/2);//vars Pin_appearent=Vs*I_fund;//VA disp(Vdc,"dc output voltage(V)"); disp(Pin_active,"Active power input(W)"); disp(Pin_reactive,"Reactive power input(vars)"); disp(Pin_appearent,"Appearent power input(VA)"); disp("part (c)"); Vdc=Vs/sqrt(2)/%pi*(1+cos(alfa))//V Idc=Vdc/RL;//A disp(Vdc,"dc output voltage(V)"); disp(Idc,"dc output current(A)");
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function [y]=f(t) y = 4*sin(t) - 3*t endfunction function [y]=fp(t) y = 4*cos(t) - 3 endfunction function [raiz, x, iter, ea]=newtonraphson(x0,f,fp,tol,imax) iter = 0; // inicializa numero de iteracoes xr = x0; // inicializa raiz aproximada com a inicial x(iter+1)=x0; // insere raiz inicial no vetor de raizes while (1) xrold = xr; xr = xrold - f(xrold)/fp(xrold); // aplica formula de Newton iter = iter+1; // incrementa numero de iteracoes x(iter+1) = xr; // insere raiz aproximada no respectivo vetor if(xr ~= 0) then // calcula erro relativo ea(iter)=abs((xr-xrold)/xr); end; if(ea(iter) <= tol) then // se erro relativo menor que tol, FIM raiz = xr; return; end; if(iter >= imax) then // se excedeu num. maximo de iteracoes, FIM error('Número Máximo de Iterações Alcançado'); end; end end function [h]=altura(t) r = 4/t; x = r*cos(t); h = r - x; endfunction x0 = 4; tol = 0.0001; imax = 100; mprintf("i Gráfico plotado\n"); interv = [-2*%pi:%pi/8:2*%pi]; plot(interv,f); xgrid; [raiz, x, iter, ea] = newtonraphson(x0,f,fp,tol,imax); mprintf("ii Raiz utilizando método de Newton: %f\n",raiz); h = altura(raiz); mprintf("iii Altura máxima da peça: %f\n",h);
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clear; printf('************** mta_1.sci Start! ****************'); printf('\n'); printf('Enter a File Name of UNIT SPACE Material'); UnitSpaceFile = input('File Name(.xls)?: ',"string"); //scanf('%s',UnitSpaceFile); printf('./' +UnitSpaceFile+'.xls\n'); //f=findfiles(SCI,UnitSpaceFile+'.xls'); //MT_Mat_Sheets = readxls('./mt_mat.xls'); // EXELファイルの読み出し MTA_Mat_Sheets = readxls('./' + UnitSpaceFile + '.xls'); // EXELファイルの読み出し Sheet = MTA_Mat_Sheets(1); // Sheetの抜き出し MTAMate = Sheet.value; // 数値の取り出し SampleCount = size( MTAMate, 1); ItemCount = size( MTAMate, 2); // printf('SampleCount ='); disp(string(SampleCount)); printf('ItemCount ='); disp(string(ItemCount)); printf('\n'); // 予備計算 for j = 1: ItemCount, x1( 1, j) = 0, //行列の初期化 x2( 1, j) = 0, //行列の初期化 for i = 1: SampleCount, x1( 1, j) = x1( 1, j) + MTAMate( i, j), // 1乗の総和を求める x2( 1, j) = x2( 1, j) + MTAMate( i, j)^2; // 2乗の総和を求める end, end // 算術平均 for j = 1: ItemCount, Ave( 1, j) = x1( 1, j) / SampleCount; end // 標準偏差(MT法) for j = 1: ItemCount, StDevM( 1, j) = (( x2( 1, j) - (x1( 1, j)^2)/SampleCount) /SampleCount)^0.5; end // 正規化 for j = 1: ItemCount, for i = 1: SampleCount, u( i, j) = MTAMate( i, j) - Ave( 1, j); end; end //分散共分散行列 for i = 1: ItemCount, ui = (u(:,i))', for j = 1: ItemCount, uj = u(:,j), S( i, j) = ui*uj/SampleCount, end; end //disp(S); //printf('\n'); //余因子行列 m=1; n=1; for i = 1: ItemCount, for j = 1: ItemCount, for k = 1: ItemCount, for l = 1: ItemCount, if (k~=i && l~=j) then A(m,n) = S(k,l), n=n+1, if(n>ItemCount-1) then n=1, m=m+1, if(m>ItemCount-1) then m=1, AJMt(i,j) = det(A)*((-1)^(i+j)), //disp(A); //printf('\n'); end; end; end; end; end; end; end AJM = AJMt'; printf('AJM ='); disp(AJM); printf('\n'); for i=1:SampleCount, Ut = u( i, :), U = Ut', D2(i,1) = Ut * AJM * U / ItemCount; end /* 信号空間の検証 */ printf('Enter a File Name of MT Signal Material'); MTSigFile = input('File Name(.xls)?: ',"string"); printf('./' +MTSigFile+'.xls\n'); MTSig_Sheets = readxls('./' + MTSigFile + '.xls'); // EXELファイルの読み出し SigSheet = MTSig_Sheets(1); // Sheetの抜き出し MTSig = SigSheet.value; // 数値の取り出し SampleCount = size( MTSig, 1); ItemCount = size( MTSig, 2); printf('SampleCount ='); disp(string(SampleCount)); printf('ItemCount ='); disp(string(ItemCount)); printf('\n'); // 正規化 for j = 1: ItemCount, for i = 1: SampleCount, v( i, j) = MTSig( i, j) - Ave( 1, j); end; end for i=1:SampleCount, Vt = v( i, :), V = Vt', SD2(i,1) = Vt * AJM * V / ItemCount; end //clf; // clear scf; // add subplot(2,2,1); plot2d(D2); subplot(2,2,2); histplot( 30, D2(:,1)'); subplot(2,2,3); plot2d(SD2); subplot(2,2,4); histplot( 30, SD2(:,1)');
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//Example 4.28//number ,spacing,mounting height and total wattafe clc; clear; close; format('v',6) h=5;// in meters el=120;//in lux ef=40;//efficiency in lumens/watt tw=80;//in watts df=1.4;//depreciation factor uf=0.5;//utiliazation factor l=30;// in meters b=15;// in meters a=l*b;//arean in m^2 glr=(a*el*df)/(uf);//gross lumens required twr=glr/ef;//total wattage required nt=twr/tw;//no. of tubes required disp(twr,"total wattage required in watts") disp(" number of tubes required is "+string(nt)+" equivalent to 48 tubes")
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function [ fieldE, rvector ,rnorm ] = Efield(q0,P0,P); Efield calcula o campo elétrico estático no ponto P, devido a uma carga q0 localizada no ponto P0 Uso: [ fieldE, rvector ,rnorm ] = Efield(q0,P0,P); eps0 = 8.854187817e-12; rvector = P-P0; rnorm = norm(rvector); fieldE = 1/(4*pi*eps0)*q0*(P-P0)./(norm(P-P0).^3); end
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clear; clc; //Example 14.10 //Caption : Program to illustrate the Concepts of Pue Gas Adsorption subplot(2,1,1) m=4.7087; b=2.1941; t=0.3984; P=linspace(0,40,10); N=(m.*P)./((b+(P.^t)).^(1/t)); plot(P,N) m=0.6206; b=1.5454; t=1; n=(m.*P)./((b+(P.^t)).^(1/t)); plot(P,n,'b--') legend('Toth Equation','Langmuir Equation') xtitle('Adsorption Isotherm(n vs P)','P(kPa)','n(mol/kg)') subplot(2,1,2) C0=0.4016; C1=-0.6471; C2=0.4567; C3=-0.12; n=linspace(0,1.6,20); K=C0+(C1*n)+(C2*(n^2))+(C3*(n^3)); plot(n,K) n=linspace(0,0.5,20); K=C0+(C1*n); plot(n,K,'b--') legend('Cubic Polynomial fit','Langmuir Equation') xtitle('n/P vs n for Ethylene','n(mol/kg)','n/P(mol/kg/kPa)') //End
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clear // lal=7.5 lcu=6 rcu=0.017*(10**-6) ral=0.028*(10**-6) d=(10**-6) a=((3.14*d))/(4) Ral=(lal*ral)/(a) printf("\n R= %0.1f ohm",Ral) ial=3 pv=Ral*ial Rcu=pv/(2) printf("\n Rcu") a=(rcu*lcu)/(Rcu) dcu=(((a*4)/3.14)**0.5) printf("\n dcu= %e nm",dcu)
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//Example 9.3:change in refractive index ,net phase shiftand Vpi clc; clear; close; format('v',6) v=5;//kV l=1;//cm ez=(v*10^3)/(l*10^-2);//in V/m no=1.51;// r63=10.5*10^-12;//m/V dn=((1/2)*no^3*r63*ez);// h=550;//nm dfi=((2*%pi*dn*l*10^-2)/(h*10^-9));// fi=2*dfi;// vpi=((h*10^-9)/(2*no^3*r63))*10^-3;//kV disp(dfi,"change in refrative index is") disp(fi,"net phase shift is") format('v',4) disp(vpi,"Vpi in kV is") //refractive index and phase shift is in the form of pi in the textbook
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// Ex 43 Page 387 clc;clear;close; // Given R2=0.03;//ohm X2=0.18;//ohm Ns=100;//rpm s1=3;//% Nfl=(100-s1);//rpm (full load speed) N2=Nfl/2;//rpm s2=(Ns-N2)/Ns*100;//% V1BYV2=sqrt(s2/s1*(R2**2+(s1/100*X2)**2)/(R2**2+(s2/100*X2)**2));//from torque equation //let V1=V12BYV1 V2=1 V1=V1BYV2;//V V2=1;//V V12BYV1=(V1-1)/V1*100;// % reduction in the stator (V12=V1-V2) printf("Percentage reduction in stator voltage = %.f percent",V12BYV1) fi=atan(s2/100*X2/R2);//radian pf=cos(fi);//power factor printf("\n power factor = %.1f",pf)
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function [stk,txt,top]=sci_erf() // Copyright INRIA txt=[] stk=list('erf('+stk(top)(1)+')','0',stk(top)(3),stk(top)(4),stk(top)(5))
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function [K,Y,err]=leqe(P21,Qx) [A,B1,C2,D21,xo,dom]=P21(2:7) [KT,Y,err]=leqr(syslin(dom,A',C2',B1',D21'),Qx); K=KT';
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//Example 9.7 (a) //Program To Determine Frequency Resolution of Bartlett, //Welch(50% Overlap) and Blackmann-Tukey Methods clear; clc; close; //Data Q=10;//Quality Factor N=1000;//Samples //FREQUENCY RESOLUTION CALCULATION K=Q; rb=0.89*(2*%pi*K/N); rw=1.28*(2*%pi*9*Q)/(16*N); rbt=0.64*(2*%pi*2*Q)/(3*N); //Display the result in command window disp(rb,"Resolution of Bartlett Method"); disp(rw,"Resolution of Welch(50% overlap) Method"); disp(rbt,"Resolution of Blackmann-Tukey Method");
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// Example 18.1, page no-460 clear clc atom=4 kci=0.629*10^-9//m alfk=1.264*10^-40//m^2 alfCl=3.408*10^-40//m^2 eps0=8.854*10^-12 pol=alfk+alfCl N=atom/kci^3 epsr=(N*pol/eps0)+1 printf("\nThe electronic polarisability for KCL = %.3f *10^-40 F m^2\n",pol*10^40) printf("\nThe no of Dipoles per m^3 = %.3f * 10^28 atoms m^-3\n",N/10^28) printf("\nThe dielectric constant of KCL is %.3f",epsr)
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// Example 2.7, page no-119 clear clc printf("1 kg/cm^2 = 10 mWG\n") //(a) press=10+2 printf("\n(a)Bourdon Gauge is mounted 20 meters below water line:\nPressure read by the Gauge = %d kg/cm^2",press) //(b) press2=10-3 printf("\n\n(b)Bourdon Gauge is located 30 meters above the water line:\nPressure read by the Gauge = %d kg/cm^2",press2)
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clc // Given that d = 4 // distance of star from the earth in light years v = 3e8 * sqrt(0.9999) // speed of rocket in meter/sec // Sample Problem 33 on page no. 11.31 printf("\n # PROBLEM 33 # \n") printf(" Standard formula used \n") printf(" t = t_0/((1-v^2/c^2)^1/2) \n") t = (2 * d * 3e8) / v T_ = t * sqrt(1 - (v / 3e8)^2) printf("\n Time taken by the rocket is %f year.",T_)
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//Example 1_40 clc; clear; close; format('v',5); //given data : V=24;//V R1=7;//ohm R2=7;//ohm R3=7;//ohm R4=7;//ohm R5=8;//ohm R6=10;//ohm RAB=(R5*R6/(R5+R6)+R4)*(R2+R3)/(R5*R6/(R5+R6)+R4+R2+R3)+R1;//ohm I=V/RAB;//A I2=I*(R2+R3)/(R2+R3+R5*R6/(R5+R6)+R4);//A VPQ=I2*(R5*R6/(R5+R6));//V disp(VPQ,"Voltage drop across the 10 ohm resistor(V)"); //Answer in the book is not accurate.
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clc; clear; p=60;//psia T=1000;//degree R px=12;//psia k=1.4; R=53.3;//ft*lb/(lbm*degree R) pratio=p/px; //for this value of pratio, Max is calculated as Max=1.9; //using this value of Max, Tx/T0,x is found as Tratio=0.59; //T=T0,x=T0,y Tx=Tratio*T;//degree R cx=(R*Tx*k)^0.5;//ft/sec Vx=1.87*cx*(32.2^0.5);//ft/sec disp(Max,"The Mach number for the flow=") disp("ft/sec",Vx,"The velocity of the flow=")
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clc clear //Input data t=5;//Time taken for a body to cool from 60 to 40 degree centigrade in minutes t11=60;//The initial temperature of the body in degree centigrade t12=40;//The final temperature of the body in degree centigrade ts=10;//The temperature of the surrounding in degree centigrade //Calculations K=log((t12-ts)/(t11-ts));//The constant value for the first case at ts x=((exp(K))*(t12-ts))+ts;//The temperature after the next 5 minutes in degree centigrade //Output printf('The temperature after the next 5 minutes is x = %3.0f degree centigrade ',x)
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//Problem 2.09: Calculate the power dissipated when a current of 4 mA flows through a resistance of 5 k //initializing the variables: I = 0.004; // in ampere R = 5000; // in ohms //calculation: P = I*I*R printf("\n\nResult\n\n") printf("\nPower(P): %.2f Watt(W)\n",P)
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// Copyright (C) 2021 - UGA - JIANG Yilun // // Date of creation: 2021-9-15 // deff("y = f3(x)", "y = log(x + sqrt((x) .^ 2 - 1))") deff("y = f4(x)", "y = log(x + sqrt((x) .^ 2 + 1))") deff("y = f5(x)", "y = 1/2 * log((1 + x) ./ (1 - x))") t = 0:0.2:5 u = f3(cosh(t)) v = f4(sinh(t)) w = f5(tanh(t)) M = [t;u;v;w]' disp(M)
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example3_15.sce
clear; clc; //Example3.15[Cost of Heat Loss through walls in winter] //Given:- R_va_insu=2.3;//thickness to thermal conductivity ratio[m^2.degreeCelcius/W] L1=12;//length of first wall of house[m] L2=12;//length of second wall of house[m] L3=9;//length of third wall of house[m] L4=9;//length of fourth wall of house[m] H=3;//height of all the walls[m] T_in=25;//Temperature inside house[degree Celcius] T_out=7;;//average temperature of outdoors on a certain day[degree Celcius] ucost=0.075;//Unit Cost of elctricity[$/kWh] h_in=8.29,h_out=34.0;//Heat transfer coefficients for inner and outer surface of the walls respectively[W/m^2.degree Celcius] v=24*(3600/1000);//velocity of wind[m/s] //Solution:- //Heat transfer Area of walls=(Perimeter*Height) A=(L1+L2+L3+L4)*H;//[m^2] //Individual Resistances R_conv_in=1/(h_in*A);//Convection Resistance on inner surface of wall[degree Celcius/W] R_conv_out=1/(h_out*A);//Convection Resistance on outer surface of wall[degree Celcius/W] R_wall=R_va_insu/A;//Conduction resistance to wall[degree Celcius/W] //All resistances are in series R_total=R_conv_in+R_wall+R_conv_out;//[degree Celcius/W] Q_=(T_in-T_out)/R_total;//[W] disp("W",Q_,"The steady rate of heat transfer through the walls of the house is") delta_t=24;//Time period[h] Q=(Q_/1000)*delta_t;//[kWh/day] disp("kWh/day",Q,"The total amount of heat lost through the walss during a 24 hour period ") cost=Q*ucost;//[$/day] disp("per day",cost,"Cost of heat consumption is $")
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//NETWORK PROJECT //IMPLEMENT NETWOK TOPOLOGY WITH DATA TRANSFER INCLUDING SECURITY STANDARDS // BY SHREYA SOMKUWAR name='TOPOLOGY';// graph name n=7;//graph parameters tail=[1 1 1 2 2 3 3 4 4 4 5 6]; head=[2 3 4 4 5 4 6 5 6 7 7 7]; node_x=[100 275 275 500 750 750 900]; node_y=[500 200 800 500 200 800 500]; [g]=NL_G_MakeGraph(name,n,tail,head,node_x,node_y)//application of NL_G_MakeGraph windowIndex=1; f=NL_G_ShowGraphN(g,windowIndex); [rt]=NL_R_DijkstraRT(g)//application of NL_R_DijkstraRT //application of presence table [l c]=size(rt); [pp]=NL_R_RTPathPresence(rt,l); //implementation of data transfer n=7; bs=15;//constant buffer size [nd,nf]=NL_F_RandIntNiNj(n);//generation of connection extreme nodes L=1000;//network square area side t=1;//current time [probroute]=NL_I_RouteManagerInit(rt,rt,rt,rt,rt,pp,n,L);//initialization of the route manager //Generation network matrix network=NL_I_NetworkMatrixInit(n,bs);//initialization of the reception network matrix networks=NL_I_NetworkMatrixInit(n,bs);//initialization of the emission network matrix tpmax=n*bs;//maximal quantity of packets simultaneously supported by each network matrix rp=NL_I_PacketManagerInit(tpmax);//initialization of the packet manager cpmax=10;//maximal quantity of packets per connection ct=1;//connection type selection index: creation of Tcp connections swmin=1; rtmin=1; rtmax=50; pr=0.90;//probability threshold [swi,rti]=NL_I_TCPNetworkInit(n,swmin,rtmin);//initialization of the TCP parameters for each node [networks,rp]=NL_I_ConnectionManager(nd,n,bs,cpmax,networks,rp,ct,pr);//generation of connections [swi,rti,network,networks,rp]=NL_I_Emission2Reception(swi,rti,rtmax,network,networks,n,bs,rp,t,probroute);//emission of packets on the reception network //selection of source nodes ind=find(network(:,$) <> 0);//TCP packets present on the network disp(network,"Network matrix before any transmission:"); disp(ind,"Nodes in which packets are available:"); [ind_r,ind_size]=size(ind); w=2; //select destination nodes for b=1:ind_size i=ind(b);//selection of the first one p=network(i,1);//first TCP packet //j=7;//Destination node disp(i,"Source node:"); disp(network(i,:),':',i,'Network buffer of source node,');//buffer of the node i mod=33;//modulus pu=3;//public exponent pr=7;//private exponent disp(p,'Original packet:'); p1=p+15; //encryption of data [en]=NL_S_RSAEncryption(mod,pu,p1);//application of NL_S_RSAEncryption disp(en,'After encryption:'); for j=1:7 if(j~=i) then disp(j,"Destination node:"); [path]=NL_R_DijkstraNiNj(g,i,j);//application of NL_R_DijkstraNiNj cack=0.8; closs=0.8; [n1,n2]=size(path); disp(path,'Path to be followed by the packet:'); i1=i; for k=1:(n2-1) [j,ack]=NL_I_PathNextNode(i1,path); if(j==path(n2)) then //decryption of data [de]=NL_S_RSADecryption(mod,pr,en)//application of NL_S_RSADecryption de=de-15; disp(network(j,:),'before transmission:',j,'Network buffer of '); [network,rp,swi,rti]=NL_I_PacketTCPIntraNet(j,de,network,rp,t,swi,rti,rtmax); disp(network(j,:),'after transmission:',j,'Network buffer of '); disp(en,'Recieved Packet:') disp(de,'After decryption:'); else disp(network(j,:),'before transmission:',j,'Network buffer of '); [network,rp,swi,rti]=NL_I_PacketTCPIntraNet(j,en,network,rp,t,swi,rti,rtmax); disp(network(j,:),'after transmission:',j,'Network buffer of '); end i1=j; end [g] = NL_G_HighlightPath(path,g,2,3,3,10,w); w=w+1; end end end
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vmm_offc.sce
style.fontSize=12; style.displayedLabel="vmm_offc"; pal1_1=xcosPalAddBlock(pal1_1,"vmm_offc",[],style);
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Ex10_2.sce
//Example 10.2, Page Number 498 //Sensitivity Calculation clc; a=5*(10**-7) //Thermal expansion Coefficient per Kelvin b=6.8*(10**-6) //Thermal Expansion Coefficient per Kelvin l=1.55*(10**-6) //Wavelength in meter p11=0.126 //Constant Coeffiecient p12=0.274 //Constant Coeffiecient u=0.17 n=1.46//cladding refractive index dl=l*(a+b); // dl is the wavelength sensitivity to temp. changes disp(dl,"The Wavelength Sensitivity to temperature changes of the filter structure in nm/K is:"); pe=((n**2)/2)*(((1-u)*p12)-(u*p11)); //pe is the effective photoelastic coefficient disp(pe," The Effective Photoelastic Coefficient is:"); dl=l*(1-pe) disp(dl," As far as Strain is concerned the Sensitivity in m/ε is:");
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hohiroki/Scilab_TBC
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eg24_1.sce
errcatch(-1,"stop");mode(2);; //; f=50; cap=1.2*10^(-6); xl=1/(3*2*(%pi)*cap*f); printf("The inductive reactance to neutralize 100 percent of the capacitance is:%.2f Ohm\n",xl); xl1=xl/0.9; printf("The inductive reactance to neutralize 90 percent of the capacitance is:%.2f Ohm\n",xl1); xl2=xl/0.8; printf("The inductive reactance to neutralize 80 percent of the capacitance is:%.2f Ohm",xl2) exit();
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ex6_31.sce
// Three-Phase Circuits :example 6.31 :(pg 6.32) VL=220; Po=11.2*10^3; N=0.88;//efficiency IL=38; Pi=(Po/N); x=(Pi/(sqrt(3)*VL*IL)); phi=acosd(x); W1=(VL*IL*cosd(30-phi)); W2=(VL*IL*cosd(30+phi)); printf("\nVL=220 V \nPo=11.2kW \nN=0.88 \nIL=38A \N=(Po/Pi)= %.2f W",Pi); printf("\nPi=sqrt(3)*VL*IL*cos(phi) \ncos(phi)=%.2f lagging",x); printf("\nphi=%.2f degrees",phi); printf("\nW1 =VL*IL*cos(30-phi) =%.2f W",W1); printf("\nW2 =VL*IL*cos(30+phi) =%.2f W",W2);
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1_8.sce
clc //initialisation of variables clear r= 96 T= 10.5 //C K1= 288 //C K2= 0.0015 //C^-1 h= 3000 //ft P1= 14.69 //CALCULATIONS P2= P1*10^(((1/(r*K2))*log10((K1-K2*h)/K1))) w= P2*144/(r*(273+T)) //RESULTS printf ('Density = %.4f lb/ft^3 ',w)
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plot-3.sce
// Graphics of some histograms // Nonlinear initial regions // Z=1, 3, 4, 10 model scf(0);clf(); a = get("current_axes"); a.x_label.font_size=4; a.x_label.text="$\log_{10}t$"; a.y_label.font_size=4; a.y_label.text="$\log_{10}P(t)$"; a.title.foreground=9; a.title.font_size=4; a.title.text="$\textrm{Time distribution probability}$"; DIR='th/z01/'; t = read(DIR+'1000',-1,3); nt=max(size(t)); plot2d(log10(t(2:nt,1)), log10(t(2:nt,3)),[1]); DIR='th/z03/'; t = read(DIR+'1000n1',-1,3); nt=max(size(t)); plot2d(log10(t(2:nt,1)), log10(t(2:nt,3)),[3]); DIR='th/z04/'; t = read(DIR+'1000n1',-1,3); nt=max(size(t)); plot2d(log10(t(2:nt,1)), log10(t(2:nt,3)),[5]); DIR='th/z10/'; t = read(DIR+'1000n1',-1,3); nt=max(size(t)); plot2d(log10(t(2:nt,1)), log10(t(2:nt,3)),[6]); a.data_bounds=[0,-4;2.5,0]; ht = legend(['$Z=1$','$Z=3$', '$Z=4$', '$Z=10$']); ht.font_size=3; ht.visible='on'; tname = "graphics/histograms/tt.eps"; unix('rm '+tname); xs2eps(gcf(), tname); scf(1);clf(); a = get("current_axes"); a.x_label.font_size=4; a.x_label.text="$\log_{10}s$"; a.y_label.font_size=4; a.y_label.text="$\log_{10}P(s)$"; a.title.foreground=9; a.title.font_size=4; a.title.text="$\textrm{Size distribution probability}$"; DIR='sh/z01/'; t = read(DIR+'1000',-1,3); nt=max(size(t)); plot2d(log10(t(2:nt,1)), log10(t(2:nt,3)),[1]); DIR='sh/z03/'; t = read(DIR+'1000n1',-1,3); nt=max(size(t)); plot2d(log10(t(2:nt,1)), log10(t(2:nt,3)),[3]); DIR='sh/z04/'; t = read(DIR+'1000n1',-1,3); nt=max(size(t)); plot2d(log10(t(2:nt,1)), log10(t(2:nt,3)),[5]); DIR='sh/z10/'; t = read(DIR+'1000n1',-1,3); nt=max(size(t)); plot2d(log10(t(2:nt,1)), log10(t(2:nt,3)),[6]); a.data_bounds=[0,-4;2.5,0]; hs = legend(['$Z=1$','$Z=3$', '$Z=4$', '$Z=10$']); hs.font_size=3; hs.visible='on'; sname = "graphics/histograms/ss.eps"; unix('rm '+sname); xs2eps(gcf(), sname);
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ex_27_4.sce
//find clc //solution //given Wr=2500//N Wa=1500//N //Wa/Wr=0.6 //refer table 27.4 X=1 V=1 Y=0 W=X*V*Wr + Y*Wa//N //from table 27.5,Ks=1.5... Ks=1.5 W1=W*Ks//N //ref table 27.6 C=53000//N L=(C/W)^(3)*10^6 printf("rating life is,%f rev\n",L)
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gauss.sce
clc;clear;close; function[x,a]=gaussElimination(A,b) A_aug=[A b] a=A_aug n=3; for i=2:n for j=2:n+1 a(i,j)=a(i,j)-a(1,j)*a(i,1)/a(1,1); end a(i,1)=0; end for i=3:n for j=3:n+1 a(i,j)=a(i,j)-a(2,j)*a(i,2)/a(2,2); end a(i,2)=0; end x(n)=a(n,n+1)/a(n,n); for i=n-1:-1:1 sumk=0; for k=i+1:n sumk=sumk+a(i,k)*x(k); end x(i)=(a(i,n+1)-sumk)/a(i,i); end endfunction function main() A=[0,0,0;0,0,0;0,0,0] A(1,1)=input("enter a11: ") A(1,2)=input("enter a12: ") A(1,3)=input("enter a13: ") A(2,1)=input("enter a21: ") A(2,2)=input("enter a22: ") A(2,3)=input("enter a23: ") A(3,1)=input("enter a31: ") A(3,2)=input("enter a32: ") A(3,3)=input("enter a33: ") disp('1.Gaussian Elimination\n') b=[0;0;0] b(1,1)=input("enter b1: ") b(2,1)=input("enter b2: ") b(3,1)=input("enter b3: ") [x,a]=gaussElimination(A,b) disp(x(3),x(2),x(1),'The values of x,y,z are '); disp(a(1,1),a(2,2),a(3,3),'The pivots are'); endfunction main();
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Ex12_22.sce
//EX12_22 Pg-41 clc clear R=5;//frequency deviation constant in KHz/V fm=10;//modulation frequency in kHz V=15;//amplitude of the modulating signal fd=R*V;//frequency deviation printf("\n maximum frequency deviation fd=%.0f KHz/V \n",fd) mf=fd/fm; printf(" \n modulation index mf=%.1f",mf)
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<?xml version="1.0" encoding="UTF-8" standalone="yes"?> <ProtoParser> <scan finState="FINISH"><sym id="30" cat="24">/*------------------------------------------------------*/</sym></scan> <scan finState="FINISH"><sym id="31" cat="22">\n</sym></scan> <scan finState="FINISH"><sym id="32" cat="26">EOF</sym></scan> <scan finState="FINISH"><sym id="33" cat="23"> </sym></scan> <scan finState="FINISH"><sym id="34" cat="26">IDENTIFIER</sym></scan> <scan finState="FINISH"><sym id="33" cat="23"> </sym></scan> <scan finState="FINISH"><sym id="35" cat="26">NUMBER</sym></scan> <scan finState="FINISH"><sym id="33" cat="23"> </sym></scan> <scan finState="FINISH"><sym id="36" cat="26">STRING</sym></scan> <scan finState="FINISH"><sym id="33" cat="23"> </sym></scan> <scan finState="FINISH"><sym id="14" cat="14">;</sym></scan> <scan finState="FINISH"><sym id="31" cat="22">\n</sym></scan> <scan finState="FINISH"><sym id="37" cat="24">/* Meta Grammar for Parsing of Transformation Grammars \n Georg Fischer 1980-08-01 */</sym></scan> <scan finState="FINISH"><sym id="31" cat="22">\n</sym></scan> <scan finState="FINISH"><sym id="38" cat="24">/*------------------------------------------------------*/</sym></scan> <scan finState="FINISH"><sym id="31" cat="22">\n</sym></scan> <store>AXIOM = EXTRA_INPUT</store> <store>EXTRA_INPUT = '[' GRAMMAR ']'</store> <store>GRAMMAR = RULES</store> <store>RULES = RULE</store> <store>RULES = RULES '.' RULE</store> <store>RULE = LEFT_SIDE '=' RIGHT_SIDES</store> <store>LEFT_SIDE = IDENTIFIER</store> <store>RIGHT_SIDES = RIGHT_SIDE</store> <store>RIGHT_SIDES = RIGHT_SIDES '|' RIGHT_SIDE</store> <store>RIGHT_SIDE = SYNTAX_PART SEMANTIC_PART</store> <store>SYNTAX_PART = MEMBERETIES</store> <store>MEMBERETIES =</store> <store>MEMBERETIES = MEMBERETIES MEMBER</store> <store>MEMBER = PRIMARY</store> <store>PRIMARY = IDENTIFIER</store> <store>PRIMARY = STRING</store> <store>PRIMARY = NUMBER</store> <store>SEMANTIC_PART = TRANSFORMATIONS</store> <store>TRANSFORMATIONS =</store> <store>TRANSFORMATIONS = '=&gt;' TRANSFORMATION</store> <store>TRANSFORMATIONS = TRANSFORMATIONS '-' TRANSFORMATION</store> <store>TRANSFORMATION = DESTINATION</store> <store>TRANSFORMATION = TRANSFORMATION ELEMENT</store> <store>DESTINATION = '='</store> <store>DESTINATION = ELEMENT</store> <store>DESTINATION = SYMBOL '='</store> <store>ELEMENT = SYMBOL</store> <store>ELEMENT = '#' NUMBER</store> <store>ELEMENT = NUMBER</store> <store>ELEMENT = STRING</store> <store>ELEMENT = '@'</store> <store>ELEMENT = SYMBOL '(' COMBINED_LIST ')'</store> <store>SYMBOL = INCARNATION</store> <store>SYMBOL = INCARNATION '$' IDENTIFIER</store> <store>INCARNATION = IDENTIFIER</store> <store>INCARNATION = IDENTIFIER ':' NUMBER</store> <store>COMBINED_LIST =</store> <store>COMBINED_LIST = COMBINED_LIST SYMBOL</store> <store>COMBINED_LIST = COMBINED_LIST NUMBER</store> <store>COMBINED_LIST = COMBINED_LIST STRING</store> <store>COMBINED_LIST = COMBINED_LIST '#' NUMBER</store> <grammar axiom="AXIOM"> <symbolList> <sym id="0" cat="0" type="EOP">EOP</sym> <sym id="1" cat="1">=&gt;</sym> <sym id="2" cat="2">|</sym> <sym id="3" cat="3">=</sym> <sym id="4" cat="4">-</sym> <sym id="5" cat="5">.</sym> <sym id="6" cat="6">[</sym> <sym id="7" cat="7">]</sym> <sym id="8" cat="8">#</sym> <sym id="9" cat="9">(</sym> <sym id="10" cat="10">)</sym> <sym id="11" cat="11">+</sym> <sym id="12" cat="12">*</sym> <sym id="13" cat="13">/</sym> <sym id="14" cat="14">;</sym> <sym id="15" cat="15">@</sym> <sym id="16" cat="16">:</sym> <sym id="17" cat="17">$</sym> <sym id="18" cat="18">/*</sym> <sym id="19" cat="19">*/</sym> <sym id="20" cat="20">//</sym> <sym id="21" cat="21" type="EOF">EOF</sym> <sym id="22" cat="22" type="EOL">EOL</sym> <sym id="23" cat="23" type="SPACE">SPACE</sym> <sym id="24" cat="24" type="NESTCOM">NESTCOM</sym> <sym id="25" cat="25" type="EOLCOM">EOLCOM</sym> <sym id="26" cat="26" type="IDENTIFIER">IDENTIFIER</sym> <sym id="27" cat="27" type="NUMBER">NUMBER</sym> <sym id="28" cat="28" type="STRING">STRING</sym> <sym id="29" cat="29" type="axiom">axiom</sym> <sym id="30" cat="24" type="NESTCOM">/*------------------------------------------------------*/</sym> <sym id="31" cat="22" type="EOL">\n</sym> <sym id="32" cat="26" type="IDENTIFIER">EOF</sym> <sym id="33" cat="23" type="SPACE"> </sym> <sym id="34" cat="26" type="IDENTIFIER">IDENTIFIER</sym> <sym id="35" cat="26" type="IDENTIFIER">NUMBER</sym> <sym id="36" cat="26" type="IDENTIFIER">STRING</sym> <sym id="37" cat="24" type="NESTCOM">/* Meta Grammar for Parsing of Transformation Grammars \n Georg Fischer 1980-08-01 */</sym> <sym id="38" cat="24" type="NESTCOM">/*------------------------------------------------------*/</sym> <sym id="39" cat="26" type="IDENTIFIER">AXIOM</sym> <sym id="40" cat="26" type="IDENTIFIER">EXTRA_INPUT</sym> <sym id="41" cat="23" type="SPACE"> </sym> <sym id="42" cat="28" type="STRING">[</sym> <sym id="43" cat="26" type="IDENTIFIER">GRAMMAR</sym> <sym id="44" cat="28" type="STRING">]</sym> <sym id="45" cat="23" type="SPACE"> </sym> <sym id="46" cat="26" type="IDENTIFIER">RULES</sym> <sym id="47" cat="23" type="SPACE"> </sym> <sym id="48" cat="27" type="NUMBER">2</sym> <sym id="49" cat="23" type="SPACE"> </sym> <sym id="50" cat="26" type="IDENTIFIER">RULE</sym> <sym id="51" cat="23" type="SPACE"> </sym> <sym id="52" cat="28" type="STRING">.</sym> <sym id="53" cat="23" type="SPACE"> </sym> <sym id="54" cat="26" type="IDENTIFIER">LEFT_SIDE</sym> <sym id="55" cat="28" type="STRING">=</sym> <sym id="56" cat="26" type="IDENTIFIER">RIGHT_SIDES</sym> <sym id="57" cat="23" type="SPACE"> </sym> <sym id="58" cat="23" type="SPACE"> </sym> <sym id="59" cat="27" type="NUMBER">3</sym> <sym id="60" cat="26" type="IDENTIFIER">RIGHT_SIDE</sym> <sym id="61" cat="28" type="STRING">|</sym> <sym id="62" cat="23" type="SPACE"> </sym> <sym id="63" cat="26" type="IDENTIFIER">SYNTAX_PART</sym> <sym id="64" cat="26" type="IDENTIFIER">SEMANTIC_PART</sym> <sym id="65" cat="26" type="IDENTIFIER">MEMBERETIES</sym> <sym id="66" cat="23" type="SPACE"> </sym> <sym id="67" cat="27" type="NUMBER">6</sym> <sym id="68" cat="23" type="SPACE"> </sym> <sym id="69" cat="27" type="NUMBER">7</sym> <sym id="70" cat="26" type="IDENTIFIER">MEMBER</sym> <sym id="71" cat="23" type="SPACE"> </sym> <sym id="72" cat="26" type="IDENTIFIER">PRIMARY</sym> <sym id="73" cat="27" type="NUMBER">8</sym> <sym id="74" cat="23" type="SPACE"> </sym> <sym id="75" cat="27" type="NUMBER">9</sym> <sym id="76" cat="23" type="SPACE"> </sym> <sym id="77" cat="26" type="IDENTIFIER">TRANSFORMATIONS</sym> <sym id="78" cat="23" type="SPACE"> </sym> <sym id="79" cat="27" type="NUMBER">11</sym> <sym id="80" cat="27" type="NUMBER">12</sym> <sym id="81" cat="28" type="STRING">=&gt;</sym> <sym id="82" cat="26" type="IDENTIFIER">TRANSFORMATION</sym> <sym id="83" cat="27" type="NUMBER">13</sym> <sym id="84" cat="28" type="STRING">-&gt;</sym> <sym id="85" cat="23" type="SPACE"> </sym> <sym id="86" cat="27" type="NUMBER">14</sym> <sym id="87" cat="26" type="IDENTIFIER">DESTINATION</sym> <sym id="88" cat="26" type="IDENTIFIER">ELEMENT</sym> <sym id="89" cat="27" type="NUMBER">16</sym> <sym id="90" cat="23" type="SPACE"> </sym> <sym id="91" cat="27" type="NUMBER">17</sym> <sym id="92" cat="23" type="SPACE"> </sym> <sym id="93" cat="27" type="NUMBER">18</sym> <sym id="94" cat="26" type="IDENTIFIER">SYMBOL</sym> <sym id="95" cat="27" type="NUMBER">19</sym> <sym id="96" cat="27" type="NUMBER">20</sym> <sym id="97" cat="28" type="STRING">#</sym> <sym id="98" cat="27" type="NUMBER">21</sym> <sym id="99" cat="27" type="NUMBER">22</sym> <sym id="100" cat="27" type="NUMBER">23</sym> <sym id="101" cat="28" type="STRING">@</sym> <sym id="102" cat="27" type="NUMBER">24</sym> <sym id="103" cat="28" type="STRING">(</sym> <sym id="104" cat="26" type="IDENTIFIER">COMBINED_LIST</sym> <sym id="105" cat="28" type="STRING">)</sym> <sym id="106" cat="27" type="NUMBER">25</sym> <sym id="107" cat="26" type="IDENTIFIER">INCARNATION</sym> <sym id="108" cat="28" type="STRING">$</sym> <sym id="109" cat="27" type="NUMBER">27</sym> <sym id="110" cat="27" type="NUMBER">28</sym> <sym id="111" cat="28" type="STRING">:</sym> <sym id="112" cat="27" type="NUMBER">29</sym> <sym id="113" cat="27" type="NUMBER">30</sym> <sym id="114" cat="27" type="NUMBER">32</sym> <sym id="115" cat="27" type="NUMBER">33</sym> <sym id="116" 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<sema act="20" inf="0" /> </prod> <prod left="TRANSFORMATION" size="2"> <sym id="82" cat="26">TRANSFORMATION</sym> <sym id="88" cat="26">ELEMENT</sym> <sym id="0" cat="0">EOP</sym> <sema act="16" inf="16" /> <sema act="20" inf="0" /> </prod> </rule> <rule left="TRANSFORMATIONS"> <prod left="TRANSFORMATIONS" size="0"> <sym id="0" cat="0">EOP</sym> <sema act="16" inf="12" /> <sema act="20" inf="0" /> </prod> <prod left="TRANSFORMATIONS" size="2"> <sym id="1" cat="1">=&gt;</sym> <sym id="82" cat="26">TRANSFORMATION</sym> <sym id="0" cat="0">EOP</sym> <sema act="16" inf="13" /> <sema act="20" inf="0" /> </prod> <prod left="TRANSFORMATIONS" size="3"> <sym id="77" cat="26">TRANSFORMATIONS</sym> <sym id="4" cat="4">-</sym> <sym id="82" cat="26">TRANSFORMATION</sym> <sym id="0" cat="0">EOP</sym> <sema act="16" inf="14" /> <sema act="20" inf="0" /> </prod> </rule> </rules> </grammar> <legibleGrammar> [AXIOM = EXTRA_INPUT .EXTRA_INPUT = '[' GRAMMAR ']' .GRAMMAR = RULES .RULES = RULE | RULES '.' RULE .RULE = LEFT_SIDE '=' RIGHT_SIDES .LEFT_SIDE = IDENTIFIER .RIGHT_SIDES = RIGHT_SIDE | RIGHT_SIDES '|' RIGHT_SIDE .RIGHT_SIDE = SYNTAX_PART SEMANTIC_PART .SYNTAX_PART = MEMBERETIES .SEMANTIC_PART = TRANSFORMATIONS .MEMBERETIES = | MEMBERETIES MEMBER .MEMBER = PRIMARY .PRIMARY = IDENTIFIER | STRING | NUMBER .TRANSFORMATIONS = | '=&gt;' TRANSFORMATION | TRANSFORMATIONS '-' TRANSFORMATION .TRANSFORMATION = DESTINATION | TRANSFORMATION ELEMENT .DESTINATION = '=' | ELEMENT | SYMBOL '=' .ELEMENT = SYMBOL | '#' NUMBER | NUMBER | STRING | '@' | SYMBOL '(' COMBINED_LIST ')' .SYMBOL = INCARNATION | INCARNATION '$' IDENTIFIER .COMBINED_LIST = | COMBINED_LIST SYMBOL | COMBINED_LIST NUMBER | COMBINED_LIST STRING | COMBINED_LIST '#' NUMBER .INCARNATION = IDENTIFIER | IDENTIFIER ':' NUMBER ] </legibleGrammar> <legibleTable> </legibleTable> </ProtoParser>
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function f=%rs(f) // %rs(f) -f, f rational //! f(2)=-f(2)
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// O ponto do quadrado correspondente à extremidade de um vetor w do plano // será desenhado como a extremidade do vetor u0+M*w function [] = desenhaQuad (M, u0) a = gca() //obtem posição dos eixos a.isoview = "on" //usa a mesma escala nos dois eixos // a.axes_visible = "on" //defina o vetor w0 da translação que centra os eixos w0 = [1,1]' //defina a matriz V dos vértices V = [ [0,0]',[1,0]',[0,1]',[1,1]' ] //defina a matriz U0 = [u0-w0, u0-w0, u0-w0, u0-w0] U0 = [u0-w0, u0-w0, u0-w0, u0-w0] //defina o escalar b b = 2 //calcule C=(M*V+U0)/b C = (M*V + U0)/b // 0 1 0 1 // 0 0 1 1 // xsegs([V(1,1), V(1,2)], [V(2,1), V(2,2)]) // x0 = 0, x1 = 1, y0 = 0, y1 = 0 // xsegs([V(1,1), V(1,3)], [V(2,1), V(2,3)]) // x0 = 0, x1 = 0, y0 = 0, y1 = 1 // // xsegs([V(1,2), V(1,4)], [V(2,2), V(2,4)]) // x0 = 1, x1 = 1, y0 = 0, y1 = 1 // xsegs([V(1,3), V(1,4)], [V(2,3), V(2,4)]) // x0 = 0, x1 = 1, y0 = 1, y1 = 1 // // xsegs([V(1,4), V(1,2)], [V(2,4), V(2,2)]) // xsegs([V(1,4), V(1,3)], [V(2,4), V(2,3)]) for i=1:2 //plote dois lados do quadrilátero xsegs([C(1,i^2), C(1,2)], [C(2,i^2), C(2,2)]) //plote os outros dois lados do quadrilátero xsegs([C(1,i^2), C(1,3)], [C(2,i^2), C(2,3)]) end endfunction function [] = I () // segmento vertical M = [0, 0; 1, 0] // zera o tamanho dos segmentos horizontais U = [.25,0]' // desloca o segmento 0.5 unidade para a direita desenhaQuad(M, U) // segmento horizontal inferior M = [.5, 0; // deixa o segmento horizontal inferior com 0.5 unidade de comprimento 0, 0] // zera o tamanho dos segmentos verticais U = [0,0]' // não desloca o segmento desenhaQuad(M, U) // segmento horizontal superior M = M U = [0,1]' // desloca o segmento 1 unidade para cima desenhaQuad(M, U) endfunction function [] = I_italico () // segmento vertical M = [0, sin(%pi/10); // zera o tamanho dos segmentos horizontais 0, 1] // e inclina o segmento vertical em 18 graus no sentido horario U = [0,0]' // não desloca o segmento desenhaQuad(M, U) // segmento horizontal inferior M = [.6, 0; // deixa o segmento horizontal inferior com 0.5 unidade de comprimento 0, 0] // zera o tamanho dos segmentos verticais U = [-.3,0]' // desloca o segmento 0.3 unidade para a esquerda desenhaQuad(M, U) // segmento horizontal superior M = M U = [0,1]' // desloca o segmento 1 unidade para cima desenhaQuad(M, U) endfunction function [] = L () // segmento vertical M = [0, 0; 1, 0] // zera o tamanho dos segmentos horizontais U = [,0]' // desloca o segmento 0.5 unidade para a direita desenhaQuad(M, U) // segmento horizontal inferior M = [.6, 0; // deixa o segmento horizontal inferior com 0.5 unidade de comprimento 0, 0] // zera o tamanho dos segmentos verticais U = [0,0]' // não desloca o segmento desenhaQuad(M, U) endfunction function [] = F () // segmento vertical M = [0, 0; 1, 0] // zera o tamanho dos segmentos horizontais U = [0,0]' // desloca o segmento 0.5 unidade para a direita desenhaQuad(M, U) // segmento horizontal maior M = [.6, 0; // deixa o segmento horizontal maior com 0.5 unidade de comprimento 0, 0] // zera o tamanho dos segmentos verticais U = [0,1]' // desloca o segmento 1 unidade para cima desenhaQuad(M, U) // segmento horizontal menor M = [.4, 0; // deixa o segmento horizontal menor com 0.25 unidade de comprimento 0, 0] // zera o tamanho dos segmentos verticais U = [0,.5]' //não desloca o segmento 0.5 unidade para cima desenhaQuad(M, U) endfunction function [] = H () M = [0, 0; 1, 0] U = [0,0]' desenhaQuad(M, U) M = [.5, 0; 0, 0] U = [0,.5]' desenhaQuad(M, U) M = [0, 0; 1, 0] U = [.5,0]' desenhaQuad(M, U) endfunction function [] = V () M = [0, sin(%pi/10); 0, 1] U = [0,0]' desenhaQuad(M, U) M = [0, -sin(%pi/10); 0, 1] U = [0,0]' desenhaQuad(M, U) endfunction function [] = Y () M=[0,0; 0,.5] U=[.5,0]' desenhaQuad(M,U) M=[0,-sin(%pi/10); 0,.5] U=[.5,.5]' desenhaQuad(M,U) M=[0,sin(%pi/10); 0,.5] U=[.5,.5]' desenhaQuad(M,U) endfunction function [] = Y_italico () ita = sin(%pi/20) mat_ita = [0,ita; 0,0] M=[0,0; 0,.5] U=[.5-ita,0]' desenhaQuad(M+mat_ita,U) M=[0,-sin(%pi/10); 0,.5] U=[.5,.5]' desenhaQuad(M+mat_ita,U) M=[0,sin(%pi/10); 0,.5] U=[.5,.5]' desenhaQuad(M+mat_ita,U) endfunction function [] = W () M = [0,-sin(%pi/12); 0,1] U = [0,0]' desenhaQuad(M,U) M = [0,sin(%pi/12); 0,1] U = [.5,0]' desenhaQuad(M,U) M = [0,sin(%pi/12); 0,.5] U = [0,0]' desenhaQuad(M,U) M = [0,-sin(%pi/12); 0,.5] U = [.5,0]' desenhaQuad(M,U) endfunction function [] = S () M=[0,sin(%pi/6); 0,.5] U=[0,.5]' desenhaQuad(M,U) M=[0,-sin(%pi/6); 0,.5] U=[0,.5]' desenhaQuad(M,U) M=[0,-2*sin(%pi/6); 0,1] U=[.5,1]' desenhaQuad(M,U) M=[0,sin(%pi/6); 0,.5] U=[-.5,2]' desenhaQuad(M,U) M=[0,-sin(%pi/6); 0,.5] U=[.5,2]' desenhaQuad(M,U) endfunction function letras() titlepage(["Letras";"Alexandre Pierre"]) sleep(1000) clf() I() sleep(500) clf() I_italico() sleep(500) clf() L() sleep(500) clf() F() sleep(500) clf() H() sleep(500) clf() V() sleep(500) clf() Y() sleep(500) clf() Y_italico() sleep(500) clf() W() sleep(500) clf() S() sleep(500) clf() // exit(0) endfunction
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Ex2_1.sce
//Example 2.1, Page Number 51 //Conductivity Calculation clc; dc=8.93*(10**3) //Density of Copper in Kg/meter cube N=63.54 //Atomic Mass Number of Copper in amu t=2.6*(10**-14)//Mean free time between collision (in seconds) m=9.1*(10**-31) //Mass of electron in kilogram em=0.135 //Electron Mobility in meter square per volt second hm=0.048 //Hole Mobility in meter square per volt second n=1.6*(10**16) //Concentration per meter cube an=6*(10**26) //Avogadro's number per mole e=1.6*(10**-19) //Charge of an electron in Coulombs n1=(an*dc)/N //Free electron concentration/No. of atoms per unit volume rhoc=(n1*e*em)/3 //Conductivity of Copper in per ohm m //From equation 2.24 rhos=n*e*(em+hm) //Conductivity of Copperintrinsic silicon in per ohm m mprintf("Free Electron Concentration is: %.2e per meter cube\n",n1); mprintf(" Conductivity of copper is:%.2e per ohm meter\n",rhoc)//The answer provided for rhoc in the textbook is wrong mprintf(" Conductivity of intrinsic silicon is:%.2e per ohm meter\n",rhos)
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clear; clc; // Illustration 12.9 // Page: 709 printf('Illustration 12.9 - Page: 709\n\n'); // Solution //***Data***// x1 = 0.46;// [fraction moisture] x2 = 0.085;// [fraction moisture] Y1 = 0.08;// [kg water/kg dry solid] Y2 = 0.03;// [kg water/kg dry solid] G = 1.36;// [kg/square m.s] //**********// X1 = x1/(1-x1);// [kg water/kg dry solid] X2 = x2/(1-x2);// [kg water/kg dry solid] // By water balance: SsByGs = (Y1-Y2)/(X1-X2);// [kg dry solid/kg air] // Since the initial moisture content of the rayon is less than the critical, drying takes place entirely within zone III. // Comparing with Eqn. 12.22: // (kY*A/(Ss(Xc-X*)))=0.0137*G^1.47 // thetha=integrate('(1/(0.0137*G^1.47))*(1/((X-X_star)*(Yw-Y)))','X',X2,X1) // [s] X = [X1 0.80 0.60 0.40 0.20 X2];// [kg water/kg dry solid] Y = zeros(6); for i = 1:6 // From Eqn. 12.54: Y(i) = Y2+((X(i)-X2)*SsByGs);// [kg water/kg dry gas] end // From Fig. 7.5 (Pg 232): Yw = [0.0950 0.0920 0.0790 0.0680 0.0550 0.0490];// [kg water/kg dry gas] X_star = zeros(6); Val = zeros(6); P = 51780;// [vapour pressure, kN/square m] for i = 1:6 // From Eqn 7.8: deff('[y]=f(p)','y=Y(i)-((p/(101330-p))*(18/29))'); p = fsolve(7,f);// [kN/square m] RH(i) = (p/P)*100; X_star(i) = (RH(i)/4)/(100-(RH(i)/4));// [kg water/kg dry solid] Val(i) = 1/((X(i)-X_star(i))*(Yw(i)-Y(i))); end scf(41); plot(X,Val); xgrid(); xlabel("X kg water/kg dry solid"); ylabel("1/((X-X*)*(Yw-Y))"); title("Graphical Integration"); // Area Under the curve: Area = 151.6; // From Eqn. 12.59: thetha = Area/(0.0137*G^1.47); printf("Time required for drying: %f h\n",thetha/3600);
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clc clear //Input data To=27+273 //Stagnation temperature in K Po=8 //Stagnation Pressure in bar P=5.6 //Static pressure in bar, taken from diagram given m=2 //Mass flow rate in kg/s k=1.4 //Adiabaatic constant Cp=1005 //Specific heat capacity at constant pressure in J/kg-K R=287 //Specific gas constant in J/kg-k //Calculation T=To*(P/Po)^((k-1)/k) //Static temperature in K a=sqrt(k*R*T) //Sound velocity in m/s C=sqrt(2*Cp*(To-T)) //Velocity in m/s M=C/a //Mach number A=((m*R*T)/(P*10^5*C))*10^4 //Area at a point in the channal in cm^2 //Output printf('(A)Mach number is %3.4f\n (B)Velocity is %3.1f m/s\n (C)Area at a point in the channal is %3.3f cm^2',M,C,A)
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Ex7_7.sce
printf("\t example 7.7 \n"); printf("\t approximate values are mentioned in the book \n"); U=50; // Btu/(hr)*(ft^2)*(F) TP=328; // F TE=228; // F CP=(0.30/(888.8*1000)); CE=(0.05/(960*1000)); CF=1.20; theta=8000; // annual hours X=((CF*(TP-TE))/((CP-CE)*U*theta)); // from eq 7.53 printf("\t X is : %.9f \n",X); a=(1); // coefficient of t^2 b=(-556); // coefficient of t c=(74784-X); // constant printf("\t coefficient of t^2 is : %.2f \n",a); printf("\t coefficient of t is : %.2f \n",b); printf("\t constant term is : %.9f \n",c); P=poly([c b a], 't','c'); t=roots(P); printf("\t t is :%.0f \n",t); printf("\t t cannot be greater than 328F \n \t t is 218F \n"); //end
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Ex3_12.sce
// Calculating the specific iron loss clc; disp('Example 3.12, Page No. = 3.35') // Given Data Bm = 1.0;// Maximum flux density in Wb per meter square f = 100;// Frequency in Hz t = 0.3*10^(-3);// Thickness of sheet in mm p = .5*10^(-6);// Resistivity of alloy steel in ohm*meter D = 7650;// Density in kg per meter cube pi_quoted = 1.2;// Quoted iron loss in W per Kg // Calculation of total iron loss S1 = 2*12;// Sides of hysteresis loop in A/m S2 = 2*1;// Sides of hysteresis loop in Wb per meter square A = S1*S2;// Area of hysteresis loop in W-s per meter cube ph_each = A;// Hysteresis loss in each cycle in Joule per meter cube ph = ph_each*f/D;// Hysterseis loss in W per Kg pe = %pi*%pi*f*f*Bm*Bm*t*t/(6*p*D);// Eddy current loss in W per Kg pi = pe+ph;// Total iron loss in W per Kg disp(pi,'Specific iron loss(W per Kg)='); disp('The calculated iron loss is smaller than the quoted.') //in book answer is 1.014 W per Kg. The answers vary due to round off error
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ex_2_36.sce
clear; clc; n=-5:5; for i=1:length(n) if(n(i)>=-1) h(i)=2^-(n(i)+1); else h(i)=0; end end causal=%t; for i=1:length(n) if n(i)<0 & h(i)~=0 then causal=%f; end end disp(causal,"the statement that the system is causal is");
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Ex9_9.sce
//Example 9.9// xl=12.6;//wt % //liquid solution composition xa=1.6;//wt %// composition of two phases x1=10;//wt % //x1 is the overall composition xb=100;//wt %//composition of two phases a=1;//kg ma=((xl-x1)/(xl-xa))*a mprintf("ma = %f kg ",ma) b=10^3;//g //As 1kg = 10^3grams ma2=ma*b mprintf("\nma2= %i g",ma2) //At 576degree C, the overall microstructure is alpha+beta, the amount of each are ma1=((xb-x1)/(xb-xa))*a mprintf("\nma1 = %f kg ",ma1) ma3=ma1*b mprintf("\nma3= %i g",ma3) mb=((x1-xa)/(xb-xa))*a mprintf("\nmb = %f kg ",mb) mb1=mb*b mprintf("\nmb1= %i g",mb1) ae= ma3-ma2 mprintf("\nae = %i g",ae) a1=0.016;//wieght fraction a2=1.000;//wieght fraction si1=(a1)*(ma2) mprintf("\nsi1 = %f g",si1) si2=(a1)*(ae) mprintf("\nsi2 = %f g",si2) si3=(a2)*(mb1) mprintf("\nsi3 = %i g",si3)
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function [x0,f0,iter]=naive(x0,maxiter,krok) exec('cel.sci'); clf exec('mapa.sci',0); [f0,df0]=cel(x0); iter=0; kryt=norm(df0); g=df0; while(kryt>0.0001)&(iter<maxiter) do x1=x0-krok*g; [f1,df1]=cel(x1); iter=iter+1; xsegs([x0(1);x1(1)]',[x0(2);x1(2)]'); x0=x1; f0=f1; kryt=norm(df1); g=df1; end endfunction
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Ex14_1.sce
clc D=2*10^-14//cm^2/sec t=3600//K Cx=10^19 A=sqrt(D*t) disp(A,"A in cm is= ") Qt=1.13*Cx*A disp(Qt,"Q(t) in atoms/cm^3") //dC/dx=b b=-(Cx/sqrt(%pi*D*t)) disp(b,"dC/dx in cm^-4 is= ") xj=2*sqrt(D*t)*2.75 disp(xj,"xj in meter is= ") b=-(Cx/sqrt(%pi*D*t))*exp(-xj^2/(4*D*t)) disp(b,"dC/dx in cm^-4 is= ")