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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 III : SWITCHGEAR AND PROTECTION // CHAPTER 4: UNSYMMETRICAL FAULTS IN POWER SYSTEMS // EXAMPLE : 4.11 : // Page number 520-521 clear ; clc ; close ; // Clear the work space and console // Given data n = 6.0 // Number of alternator kV_A = 6.6 // Alternator rating(kV) X_1 = 0.9 // Positive sequence reactance(ohm) X_2 = 0.72 // Negative sequence reactance(ohm) X_0 = 0.3 // Zero sequence reactance(ohm) Z_n = 0.2 // Resistance of grounding resistor(ohm) // Calculations E_a = kV_A*1000/3**0.5 // Phase voltage(V) // Case(a) Z_1_a = %i*X_1/n // Positive sequence impedance when alternators are in parallel(ohm) Z_2_a = %i*X_2/n // Negative sequence impedance when alternators are in parallel(ohm) Z_0_a = %i*X_0/n // Zero sequence impedance when alternators are in parallel(ohm) I_a_a = 3*E_a/(Z_1_a+Z_2_a+Z_0_a) // Fault current assuming 'a' phase to be fault(A) // Case(b) Z_0_b = 3*Z_n+%i*X_0 // Zero sequence impedance(ohm) I_a_b = 3*E_a/(Z_1_a+Z_2_a+Z_0_b) // Fault current(A) // Case(c) Z_0_c = %i*X_0 // Zero sequence impedance(ohm) I_a_c = 3*E_a/(Z_1_a+Z_2_a+Z_0_c) // Fault current(A) // Results disp("PART III - EXAMPLE : 4.11 : SOLUTION :-") printf("\nCase(a): Fault current if all alternator neutrals are solidly grounded, I_a = %.f A", imag(I_a_a)) printf("\nCase(b): Fault current if one alternator neutral is grounded & others isolated, I_a = %.1f∠%.1f° A", abs(I_a_b),phasemag(I_a_b)) printf("\nCase(c): Fault current if one alternator neutral is solidly grounded & others isolated, I_a = %.2fj A\n", imag(I_a_c)) printf("\nNOTE: ERROR: Calculation mistakes in the textbook solution")
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 4 //Second Law of Thermodynamics //Example 17 clear; clc; //Given: To = 275; //temperature of quenching oil (K) //To calculate loss in capacity of doing work //Referrring example 4.9 (Page no. 95) S_steel = -26.25; //change in entropy os casting (kJ/K) S_oil = 43.90; //change in entropy of oil (kJ/K) S_tot = S_steel+S_oil; //total entropy change //Let W be loss in capacity for doing work W = To*S_tot; //(kJ) mprintf('The loss in capacity for doing work is %f kJ',W); //end
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clear clc dp=2.4*(10^-3);L=dp/6; //Effective mass conductivity(m3/hr.mcat) De=5*10^-5; //Effective thermal conductivity(KJ/hr.mcat.K) Keff=1.6; //For the gas film surrounding the pellet h=160;//heat transfer coefficient(KJ/hr.m2cat.K) kg=300;//mass transfer coefficient(m3/hr.m2cat) //For the reaction Hr=-160;//KJ/molA CAg=20;//mol/m3 rA_obs=10^5;//mol/hr.m3cat kobs=rA_obs/CAg; Vp=3.14*(dp^3)/6; S=3.14*(dp^2); //Observed rate/rate if film resistance controls ratio=kobs*Vp/(kg*S); printf("\n Part a") if ratio<0.01 printf("\n Resistance to mass transport to film should not influence rate of reaction") else printf("\n Resistance to mass transport to film should influence rate of reaction") end printf("\n Part b") Mw=rA_obs*(L^2)/(De*CAg); printf("\n Mw= %f",Mw) if Mw>4 printf("\n Pore diffusion is influencing and hence strong pore diffusion") else printf("\n Pore diffusion is not influencing and hence weak pore diffuusion") end //Temp variation within pellet dt_max_pellet=De*(CAg-0)*(-Hr)/Keff; //Temp variation Across the gas film dt_max_film=L*rA_obs*(-Hr)/h; printf("\n Part c") printf("\n dTmax,pellet is %f",dt_max_pellet) printf(" degree C \n dTmax,film is %f",dt_max_film) printf(" degree C") if dt_max_pellet<1 printf("\n Pellet is close to uniform in temperature") else printf("\n There is a variation in temp within pellet") end if dt_max_film<1 printf("\n Film is close to uniform in temperature") else printf("\n There is a variation in temp within Film") end
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clc // Given that d1 = 15.2 // Initial diameter in mm d2 = 12.2 // Final diameter in mm printf("Example 10.2\n") per_CW = (d1^2 - d2^2)*100/d1^2 printf("\n The tensile strength is read directly from the curve for copper(figure 10.9b) \n as 340 MPa From figure 10.19c, the Ductility at %0.1f CW is about 7%% EL.",per_CW) // Some values are deduced from figures
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clear; clc; close; x=poly(0,'x'); p1=(x+2); p2=(x^2-x+1); p3=p1*p2;//on collecting like terms disp(p3,"product=")
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clc //initialisation of variables a=552.6 //kPa m^6/kmol^2 b= 0.03402 //m^3/kmol p= 100 //kPa R= 8.314 //J/mol K //CALCULATIONS x= poly('0',x) vector= roots('p*x^3-a*x+2*a*b') T= 2*a*(x-b)^2/(R*x^3) //RESULTS printf (' isotherm= %.1f K',T)
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// Macro for PCA of dataset -- Scilab function pcaDataset = PCA(x) width = length(x(1, :)); n = length(x(:, 1)); means = []; for i = 1:width means = [means, mean(x(:, i))]; end for i = 1:n x(i, :) = x(i, :) - means; end covarianceMat = cov(x); [eigenVector, eigenVals] = spec(covarianceMat); eigenVector = flipdim(eigenVector, 2); eigenVals = diag(eigenVals) pcaDataset = eigenVector'*x'; pcaDataset = pcaDataset'; endfunction
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total_physical_store_count = 0 total_physical_fetch_count = 65536 <cache info> items cached: 1000/1000 regions in use: 1/1 items per region: 65536 counter: 32768 temp_read_num: 65535 temp_write_num: -1 item_req_count = 65536 item_hit_count = 0 item hit ratio = 0% <cache region: 0-65535> item_cost: 1000 flush_cost: 0 counter_total: 32518000 most_recently_used: 65535 least_recently_used: 64536 most_recently_changed: -1 least_recently_changed: -1 most_recently_unchanged: 65535 least_recently_unchanged: 64536
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// Scilab code Exa6.10.2 : To calculate the radius of proton orbit in synchrotron of given energy Page 275(2011) c= 3e+08; // Speed of light in vacuum, m/s q = 1.602e-019; // Charge on proton, coulomb amu = 931; // Energy equivalent of 1 amu, MeV m = 938; // Rest mass of a proton, MeV KE = 12e+03; // Kinetic energy of proton, MeV B = 1.9; // Magnetic field, T E = m + KE; // Total energy of proton, MeV // As E = m*amu, solving for m, the mass of proton m = E/amu*1.672e-027; // Proton mass in motion, kg v = 0.9973*c; // Velocity of the proton, m/s r = m*v/(B*q); // Radius of the proton, m printf("\nRadius of the proton orbit : %4.2f m", r) // Result // Radius of the proton orbit: 22.84 m
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<?xml version="1.0" encoding="utf-8"?> <test> <description> Process npart to tecplot file </description> <executable>FieldConvert</executable> <parameters> -f -e --nparts 2 -m wss:bnd=2 Tet_channel_m3_xml:xml Tet_channel_m3.fld wss.fld </parameters> <files> <file description="Session File Directory">Tet_channel_m3_xml</file> <file description="Field File">Tet_channel_m3.fld</file> </files> <metrics> <metric type="L2" id="1"> <value variable="x" tolerance="1e-6">0.5</value> <value variable="y" tolerance="1e-6">0</value> <value variable="z" tolerance="1e-6">0.288675</value> <value variable="Shear_x" tolerance="1e-6">2.06477e-16</value> </metric> </metrics> </test>
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clc clear //Input data v1=0.056;//Initial volume of gas in m^3 v2=0.007;//Final volume of gas in m^3 p1=100;//Initial perssure compressed Isothermally in kN/m^2 //Calculations p2=(p1*v1)/v2;//Final pressure in kN/m^2 W=p1*v1*(log(v2/v1));//Work done in kJ //Output printf('(a)Final pressure p2= %3.2f kN/m^2 \n (b)The work done on gas W= %3.2f kJ',p2,W)
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// Grob's Basic Electronics 11e // Chapter No. I // Example No. I_4 clc; clear; // Express the following numbers in engineering notation: (a) 27,000 (b) 0.00047. disp ('To express the number 27,000 in engineering notation, it must be written as a number between 1 and 1000 times a power of 10 which is a multiple of 3.') disp ('Therefore 27000 = 27*10^3 in engineering') disp ('To express the number 0.00047 in engineering notation, it must be written as a number between 1 and 1000 times a power of 10 which is a multiple of 3.') disp ('Therefore 0.00047 = 470*10^-6 in engineering')
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clear //Given M=8.0*10**22 //Am**2 R=64*10**5 //m //Calculation // I=(3*M)/(4.0*%pi*R**3) //Result printf("\n earths magnetisation is %0.1f A/m",I)
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//water chemistry// //example 7.8// N=150;//amount of NaCl in solution in g/l// V=8;//volume of NaCl solution// M=N*V; printf("The amount of NaCl in 8 lit of solution is %fgms",M); V=10000;//volume of hard water// W=58.5;//molecular weight of NaCl// K=(M*100/(W*2))/V; printf("\nfor 1 litre hardness is %fg/l",K); J=K*1000; printf("\nHardness of water is %fmg/l or ppm",J);
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function inserted = insert_waveform_in_scene(waveform, scene) [nrows, ncols] = size(scene); wavesize = size(waveform, '*'); inserted = zeros(nrows, ncols); for i = [1:wavesize] inserted(nrows/2-wavesize/2+i, ncols/2) = waveform(1, i); end endfunction
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// Scilab Code Ex4.7 Page-4.16 (2004) clc;clear; a = 0.38; // Lattice constant of copper, nm h =1, k = 1, l = 0; // Miller Indices (hkl)= (110) d = a/sqrt(h^2 + k^2 + l^2); // Interplanar spacing for (110) plane, unit printf("\nInterplanar spacing for (110) plane = %4.2f nm", d); // Result // Interplanar spacing for (110) plane = 0.27 nm
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// The code was developed under Horizon2020 Framework Programme // Project: 748767 — SIMFREE function Iout=SSSoDetector(In,Ideal?) // Photodetector // // Calling Sequence // Iout=SSSoDetector(In,Ideal?) // // Parameters // In : Optical Input // Ideal? : if False (default) then ideal photodetector // Iout : Electrical Output // // Description // Converts Optical Input power to an Electrical Output waveform by taking the modulus-squared of the Optical Input field of both polarisations. // An Optical Input of 1 mW produces an Electrical Output of 1. // The photodetector does not add any electrical noise. Receiver noise, if desired, can be added by setting Ideal? = True. // The photodetector has a flat bandwidth. An alternative frequency response can be obtained by simply following the photodetector with the appropriate electrical filter. global MNT MSR MLA; [lhs,rhs]=argn(0); select rhs case 0 then error("Expect at least one argument"); case 1 then Ideal?=%T; end Xin=fft(In(:,1),1); Yin=fft(In(:,2),1); x=real(Xin.*conj(Xin)); y=real(Yin.*conj(Yin)); Iout=(x+y)*MNT^2; if ~Ideal? then q=1.6E-19; c=299792458; h=6.62607e-34; kB=1.3806488E-23; T=273; RT=20; df=1e9*MSR*MNT/MNS; R=q/(h*c/(1E-9*MLA)); Iout=R*Iout; sh=sqrt(q*df); ShotNoise=sqrt(1E-3*Iout).*grand(MNT,1,'nor',0,sh); sT=sqrt(2*kB*T*df)/RT; ThermalNoise=grand(MNT,1,'nor',0,sT); Iout=Iout+1E3*(ShotNoise+ThermalNoise); end endfunction
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function [tfr,a,f,wt] = pseudoAW(x,K,wave,tsmooth,fmin,fmax,N); // This Software is ( Copyright INRIA . 1998 1 ) // // INRIA holds all the ownership rights on the Software. // The scientific community is asked to use the SOFTWARE // in order to test and evaluate it. // // INRIA freely grants the right to use modify the Software, // integrate it in another Software. // Any use or reproduction of this Software to obtain profit or // for commercial ends being subject to obtaining the prior express // authorization of INRIA. // // INRIA authorizes any reproduction of this Software. // // - in limits defined in clauses 9 and 10 of the Berne // agreement for the protection of literary and artistic works // respectively specify in their paragraphs 2 and 3 authorizing // only the reproduction and quoting of works on the condition // that : // // - "this reproduction does not adversely affect the normal // exploitation of the work or cause any unjustified prejudice // to the legitimate interests of the author". // // - that the quotations given by way of illustration and/or // tuition conform to the proper uses and that it mentions // the source and name of the author if this name features // in the source", // // - under the condition that this file is included with // any reproduction. // // Any commercial use made without obtaining the prior express // agreement of INRIA would therefore constitute a fraudulent // imitation. // // The Software beeing currently developed, INRIA is assuming no // liability, and should not be responsible, in any manner or any // case, for any direct or indirect dammages sustained by the user. // // Any user of the software shall notify at INRIA any comments // concerning the use of the Sofware (e-mail : FracLab@inria.fr) // // This file is part of FracLab, a Fractal Analysis Software // TFRSPAW [tfr,t,f,wt] = TFRSPAW(X,K,WAVE,TSMOOTH,FMIN,FMAX,N) : // Smoothed Pseudo Afine Wigner distributions. // CHECK INPUT FORMATS [nargout,nargin] = argn(0) ; select nargin case 1 error('at least two input parameters required') ; case 2 wave = 0 ; tsmooth = 0 ; nargfixed = 4 ; case 3 tsmooth = 0 ; nargfixed = 4 ; case {4,5,6} nargfixed = 4 ; case 7 nargfixed = 7 ; end [xr,xc] = size(x) ; if xr ~= 1 & xc ~= 1 error('1-D signals only') elseif xc == 1 x = conj(x)' ; end nt = size(x,2) ; if nargfixed == 4 XTF = fft(mtlb_fftshift(x),-1) ; sp = (abs(XTF(1:nt/2))).^2 ; f = linspace(0,0.5,nt/2+1) ; f = f(1:nt/2) ; plot2d(f,sp) ; xtitle('Analyzed Signal Spectrum','frequency') ; fmin = input('lower frequency bound = ') ; fmax = input('upper frequency bound = ') ; N = input('Frequency samples = ') ; end if length(wave) > 1, error('Morlet or Mexican hat wavelet only') ; elseif wave == 0 , tsupport = (length(mexhat(fmax))-1)/2 ; elseif abs(wave) > 0 , tsupport = abs(wave) ; end Qte = fmax/fmin; umax = log(Qte) ; Teq = tsupport/(fmax*umax); if Teq<2*tsupport M0 = round((2*tsupport^2)/Teq-tsupport) ; MU = tsupport+M0 ; elseif Teq>=2*tsupport MU = tsupport ; end k = 1:N ; q = (fmax/fmin)^(1/(N-1)); a = (exp((k-1).*log(q))) ; // a is an increasing scale vector. geo_f(k) = fmin*a ; // geo_f is a geometrical increasing // frequency vector. // Wavelet computation [wt] = contwt(x,geo_f(1),geo_f(N),N,wave) ; wtnonorm = zeros(wt) ; for ptr = 1:N, wtnonorm(ptr,:) = wt(ptr,:).*sqrt(a(ptr)) ; end ; // Pseudo Affine Wigner distribution computation tfr=zeros(wt); umin = -umax ; u=linspace(umin,umax,2*MU+1) ; u(MU+1) = 0; k = 1:2*N; beta(k) = -1/(2*log(q))+(k-1)./(2*N*log(q)); for m = 1:2*MU+1 l1(m,1:2*N) = exp(-(2*%i*%pi*beta).*log(lambdak(u(m),K))); end // NEW DETERMINATION OF G(T) a_t = 3 ; // (attenuation of 10^(-a_t) at t = tmax) sigma_t = tsmooth*fmax/sqrt(2*a_t*log(10)) ; a_u = 2 * %pi^2 * sigma_t^2 * umax^2 / log(10) ; if sigma_t sigma_u = 1/(2 * %pi * sigma_t) ; else sigma_u = %inf ; end G = gauss(2*MU+1,a_u) ; G = G(1:2*MU) ; if sigma_u < umax/MU fenh = findobj('Tag','Fig_gui_fl_pseudoaw') if isempty(fenh) disp('maximum time-smoothing corresponding to the scalogram reached ') ; end end G = G(ones(1,N),:)' ; for ti = 1:nt S = zeros(1,2*N) ; S(N:-1:1) = conj(wtnonorm(:,ti)') ; S(N+1:2*N) = zeros(1,N) ; Mellin = mtlb_fftshift(fft(S,1)) ; waf = zeros(2*MU,N) ; MX1 = l1.*Mellin(ones(1,2*MU+1),:) ; X1 = fft1d(conj(MX1'),-1) ; X1 = conj(X1(1:N,:)') ; waf = real(X1(1:2*MU,:).*conj(X1(2*MU+1:-1:2,:)).*G) ; tfr(N:-1:1,ti) = conj(sum(waf,'c').*geo_f)'; end; f = sort(geo_f(:)) ;
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//CHAPTER 1- D.C. CIRCUIT ANALYSIS AND NETWORK THEOREMS //Example 54 clc; disp("CHAPTER 1"); disp("EXAMPLE 54"); //VARIABLE INITIALIZATION v=5; //voltage source in Volts r1=1; //LHS resistance in Ohms r2=5; //in Ohms r3=1; //in Ohms r4=1; //RHS resistance in Ohms I=10; //current source in Amperes //SOLUTION req1=r1+r3+r4; //on deactivating the current source, current I1 flows in the circuit I1=v/req1; vab1=v-(I1*r1); //(I1*r1) is voltage drop across 1Ω resistance I2=I/req1; vab2=vab1+(I2*r1); //(I2*r1) is voltage drop across 1Ω resistance req=r1+((r3*r4)/(r3+r4)); //'req' is the same as 'Rth' mentioned in the book I=vab2/(req+r2); RTh=(6/5)+(3/4); req2=10+2; I3=9/12; disp(sprintf("The value of the current is %f A",I3)); //END
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// ----------------------------------------------------------------------- /// \brief Calcule un terme de contrainte a partir d'une homographie. /// /// \param H: matrice 3*3 définissant l'homographie. /// \param i: premiere colonne. /// \param j: deuxieme colonne. /// \return vecteur definissant le terme de contrainte. // ----------------------------------------------------------------------- function v = ZhangConstraintTerm(H, i, j) v = rand(1, 6); v=[H(1,i) * H(1,j); H(1,i) * H(2,j) + H(2,i) * H(1,j); H(2,i) * H(2,j); H(3,i)*H(1,j) + H(1,i)*H(3,j) ; H(3,i)*H(2,j) + H(2,i) * H(3,j) ; H(3,i) * H(3,j)]'; endfunction // ----------------------------------------------------------------------- /// \brief Calcule deux equations de contrainte a partir d'une homographie /// /// \param H: matrice 3*3 définissant l'homographie. /// \return matrice 2*6 definissant les deux contraintes. // ----------------------------------------------------------------------- function v = ZhangConstraints(H) v = [ZhangConstraintTerm(H, 1, 2); ... ZhangConstraintTerm(H, 1, 1) - ZhangConstraintTerm(H, 2, 2)]; endfunction // ----------------------------------------------------------------------- /// \brief Calcule la matrice des parametres intrinseques. /// /// \param b: vecteur resultant de l'optimisation de Zhang. /// \return matrice 3*3 des parametres intrinseques. // ----------------------------------------------------------------------- function A = IntrinsicMatrix(b) A = rand(3, 3); v0 = (b(2) * b(4) - b(1) * b(5))/(b(1) * b(3) - b(2)^2); lambda = b(6) - (b(4)^2 + v0*(b(2) * b(4) - b(1) * b(5))/b(1)); alpha = sqrt(lambda/b(1)); beta = sqrt(lambda * b(1)/(b(1)*b(3)-b(2)^2)); gamma = -b(2)*(alpha^2)*beta/lambda; u0 = gamma * v0 / beta - b(4) * alpha^2/lambda; A = [alpha,gamma,u0;0,beta,v0;0,0,1] endfunction // ----------------------------------------------------------------------- /// \brief Calcule la matrice des parametres extrinseques. /// /// \param iA: inverse de la matrice intrinseque. /// \param H: matrice 3*3 definissant l'homographie. /// \return matrice 3*4 des parametres extrinseques. // ----------------------------------------------------------------------- function E = ExtrinsicMatrix(iA, H) E = rand(3, 4); lambda = 1/norm(iA * H(:,1)); r1 = lambda*iA*H(:,1); r2 = lambda*iA*H(:,2); r3 = r1 .* r2; t = lambda*iA*H(:,3); E = [r1,r2,r3,t]; endfunction
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errcatch(-1,"stop");mode(2);//Caption: Amount of Information //Example 9.3 //page no 395 //Find Amount og Information ; ; px1=1/2; px2=1/2; Ix1=log2(1/px1);//entropy Ix2=log2(1/px2); printf(" \n The amount of Information \n \n\t I(X1) = %.2d bit\n",Ix1); printf(" \n The amount of Information \n \n\t I(X2) = %.2d bit",Ix2); exit();
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Ch02Ex1.sci
// Scilab Code Ex2.1 : Page-46 (2010) function V = f(t) V = 0.2*sin(120*%pi*t); endfunction t = 0; // Time when peak value of current occurs C = 10e-012; // Capacitance of the capacitor, farad I = C*derivative(f,t); printf("\nThe peak value of displacement current = %6.4e A", I); // Result // The peak value of displacement current = 7.5398e-010 A
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//Kunii D., Levenspiel O., 1991. Fluidization Engineering(II Edition). Butterworth-Heinemann, MA, pp 491 //Chapter-11, Example 2, Page 267 //Title: The Effect of m on Bubble-Emulsion Interchange //========================================================================================================== clear clc //INPUT umf=0.12;//Velocity at minimum fluidization condition in cm/s uo=40;//Superficial gas velocity in cm/s ub=120;//Velocity of the bubble in cm/s D=0.7;//Diffusion coefficient of gas in cm^2/s abkbe1=1;//Bubble-emuslion interchange coefficient for non absorbing particles(m=0) abkbe2=18;//Bubble-emuslion interchange coefficient for highly absorbing particles(m=infinity) g=980;//Acceleration due to gravity in square cm/s^2 //CALCULATION //For non absorbing particles m=0,etad=0 Kbc=(ub/uo)*(abkbe1); dbguess=2;//Guess value of db function[fn]=solver_func(db)//Function defined for solving the system fn=abkbe1-(uo/ub)*(4.5*(umf/db)+5.85*(D^0.5*g^0.25)/(db^(5/4)));//Eqn.(10.27) endfunction [d]=fsolve(dbguess,solver_func,1E-6);//Using inbuilt function fsolve for solving Eqn.(10.27) for db //For highly absorbing particles m=infinity, etad=1 M=abkbe2-(uo/ub)*Kbc; //For intermediate condition alpha=100; m=10; etad=1/(1+(alpha/m));//Fitted adsorption efficiency factor from Eqn.(23) abkbe3=M*etad+(uo/ub)*Kbc; //OUTPUT mprintf('\nFor non absorbing particles:\n\tDiameter of bubble=%fcm\n\tBubble-cloud interchange coefficient=%fs^-1',d,Kbc); mprintf('\nFor highly absorbing partilces:\n\tM=%f',M); mprintf('\nFor intermediate condition:\n\tFitted adsorption efficiency factor:%f\n\tBubble-emuslion interchange coefficient:%fs^-1',etad,abkbe3); //====================================END OF PROGRAM ======================================================
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//Caption:Determine actual PRF and duty cycle //Ex8.3 clc; clear; close; C=0.082//Capacitance(in micro farad) Ra=3.3//Resistance(in kilo ohm) Rb=2.7//Resistance(in kilo ohm) t1=0.693*C*(Ra+Rb)*1000 t2=0.693*C*Rb*1000 T=t1+t2 P=1000/T d=t1*100/T disp(P,d,'Duty cycle(in %) and PRF(in Khz)=')
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//Example No.2.1. // Page No.59. clc;clear; p = 5*10^(-3);// output power -[W]. w = 632.8*10^(-9);//wavelength -[m]. h = 6.626*10^(-34);//Planck's constant. c = (3*10^(8));//Velocity of light. hv = ((h*c)/(w));// Energy of one photon printf("\nThe energy of one photon in joules is %3.3e J", hv); hv = hv/(1.6*10^(-19)); printf("\nThe energy of one photon in eV is %.2f eV",hv); Np = (p/(3.14*10^(-19)));//Number of photons emitted printf("\nThe number of photons emitted per second by He-Ne laser are %3.3e photons per second",Np);
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//error no output //ques11 disp('To find the inverse laplace transform of the function'); syms s t a f=s/(s^4+4*a^4); il=ilaplace(f,s,t); disp(il);
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clc; clear; delta_x=5*10^-14 //diameter of nucleus in m h=6.63*10^-34 //plancks constant m=1.675*10^-27 //mass in kg //calculation p_min=h/(4*%pi*delta_x) //minimum momentum in kg-m/s E_min=((p_min)^2/(2*m)) mprintf("The minimum kinetic energy of the nucleon is = %0.2e J",E_min)
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Ex1_4.sce
clear // // // //Variable declaration I1=10 //intensity(Wm**-2) I2=25 //intensity(Wm**-2) //Calculation a1bya2=sqrt(I1/I2) ImaxbyImin=(a1bya2+1)**2/(a1bya2-1)**2 //ratio of maximum intensity to minimum intensity //Result printf("\n ratio of maximum intensity to minimum intensity is %0.3f ",ImaxbyImin)
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errcatch(-1,"stop");mode(2);; ; Vcc = 6; Vbe = 0.7; Rx = 1.3*(10^(3)); Ix = (Vcc-Vbe)/Rx; disp(Ix,"Current through each transistor :"); exit();
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24_07.sce
//Problem 24.07: For the circuit shown in Figure 24.11, determine the value of impedance Z2. //initializing the variables: rv = 70; // in volts thetav = 30; // in degrees ri = 3.5; // in amperes thetai = -20; // in degrees //z1 consist of two resistance R1 = 4.36; // in ohms R2 = -2.1*%i; // in ohms //calculation: V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180) I = ri*cos(thetai*%pi/180) + %i*ri*sin(thetai*%pi/180) //impedance, Z Z = V/I //Total impedance Z = z1 + z2 Z1 = R1 + R2 Z2 = Z - Z1 x = real(Z2) y = imag(Z2) printf("\n\n Result \n\n") printf("\n impedance Z2 is %.2f + (%.2f) ohm\n",x,y)
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Ex16_2.sce
// Example 16_2 clc;funcprot(0); // Given data T=20+273.15;// K V=25.0;// m/s k=1.40;// The specific heat ratio p=0.101;// MPa g_c=1;// The gravitational constant c_p=1.004;// kJ/kg.K R=0.286;// kJ/kg.K // Solution p_os=p*(1+((V^2/1000)/(2*g_c*c_p*T)))^(k/(k-1));// The isentropic stagnation pressure in MPa rho=(p*10^3)/(R*T);// kg/m^3 rho_os=rho*(1+((V^2/1000)/(2*g_c*c_p*T)))^(1/(k-1));// The isentropic stagnation density in kg/m^3 printf("\nThe isentropic stagnation pressure,p_os=%0.4f MPa \nThe isentropic stagnation density,rho_os=%1.4f kg/m^3",p_os,rho_os);
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Ex3_26.sce
clc clear //Inputs //The Values in the program are as follows: //Temperature in Celcius converted to Kelvin(by adding 273) //Pressure in bar converted to kPa (by multiplying 100) //Volume in m^3 //Value of R,Cp and Cv in kJ/kg K V1=0.5; P1=0.3; V2=0.1; P2=(P1*V1)/(V2); printf('Final Pressure: %1.2f bar',P2); printf('\n'); W=(P1*100*V1)*(log(V2/V1)); printf('Work Done: %1.2f kJ',W); printf('\n'); printf('Change in Internal Energy: 0 kJ as it is Isothermal Process'); printf('\n'); printf('Change in Heat Energy: %1.2f kJ',W); printf('\n');
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10_2.sce
clc //Initialization of variables nu=0.00015 //ft^2/s L=35 //ft U=88 //fps g=32.2 //ft/s^2 b=10 //ft w=8 //ft rho=0.0725 //calculations R=L*U/nu Cf=0.455 /(log10(R))^2.58 B=2*b + w Ff=Cf*rho/g *U^2 /2 *L*B Rx=R/10^5 delta=L*0.377 /(b* Rx^(0.2)) T0=0.0587 *rho/g *U^2 /2 *(nu/(L*U))^(0.2) //results printf("Frictional drag = %.1f lb",Ff) printf("\n Thickness of boundary layer = %.3f ft",delta) printf("\n Shear stress = %.4f lb/ft^2",T0)
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sswFinish.sce
// // This function is called once at the end of the simulation. // function[] = sswFinish() endfunction
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exa_2_8.sce
//Exa 2.8 clc; clear; close; //given data k=0.175;// in W/mK h_infinite=9.3;// in W/m^2K T_infinite=30;// in degree C T_s=70;// in degree C d=10*10^-3;// in m r=d/2; L=1;// in m rc=k/h_infinite;// in m CriticalThickness = rc-r;// in meter CriticalThickness=CriticalThickness*10^3; disp(CriticalThickness,"Critical thickness in mm"); q1=2*%pi*r*L*h_infinite*(T_s-T_infinite);// in W/m q2= (T_s-T_infinite)/(log(rc/r)/(2*%pi*k*L)+1/(2*%pi*rc*h_infinite));// in W/m PerIncHeatDiss= (q2-q1)*100/q1; disp(PerIncHeatDiss,"Percentage increase in heat dissipation rate in %") //Also q1=I1^2*R with bare cable // q2=I2^2*R with insulated cable I2_by_I1 = sqrt(q2/q1); // ( I2-I1 ) / I1 = (I2_by_I1 -1) / 1 // Percentage increase in current carrying capacity PerIncCurrent = (I2_by_I1 -1) / 1 *100; disp(floor(PerIncCurrent),"Increase in current carrying capacity in %")
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5_33.sce
//Example 5.33 //Newton Raphson and Mullers Method //Page no. 202 clc;clear;close; deff('x=f(x)','x=x^4-8*x^3+18*x^2+0.12*x-24.24') deff('x=f1(x)','x=4*x^3-24*x^2+36*x+0.12') //newton raphson x9=[1.5,2.5,2.7,3.1;4,5,14,10] for h=1:4 x0=x9(1,h);e=0.00001 for i=1:x9(2,h) x1=x0-f(x0)/f1(x0) e1=abs(x0-x1) x0=x1; if abs(x0)<e then break; end end printf('\nThe solution of this equation by newton raphshon after %i Iterations is %.5f\n',i,x1) end //mullers method zx=[1,2,2.7,3.1;2,3,3.7,4.1;3,4,4.7,5.1] zi=[1;2;3]; s=["i","z0","z1","z2","f0","f1","f2","li","di","gi","li+1","hi","hi+1","zi+1","D+","D_"] li(1)=(zi(3,1)-zi(2,1))/(zi(2,1)-zi(1,1)) hi(1)=zi(3,1)-zi(2,1); for i=2:4 for j=1:3 fz(j,i-1)=f(zi(j,i-1)) end di(i-1)=1+li(i-1) gi(i-1)=fz(1,i-1)*li(i-1)^2-fz(2,i-1)*di(i-1)^2+fz(3,i-1)*(li(i-1)+di(i-1)) D1(i-1)=gi(i-1)+sqrt(gi(i-1)^2-4*fz(3,i-1)*di(i-1)*li(i-1)*(fz(1,i-1)*li(i-1)-fz(2,i-1)*di(i-1)+fz(3,i-1))) D2(i-1)=gi(i-1)-sqrt(gi(i-1)^2-4*fz(3,i-1)*di(i-1)*li(i-1)*(fz(1,i-1)*li(i-1)-fz(2,i-1)*di(i-1)+fz(3,i-1))) if abs(D1(i-1))>abs(D2(i-1)) then li(i)=-2*fz(3,i-1)*di(i-1)/D1(i-1) else li(i)=-2*fz(3,i-1)*di(i-1)/D2(i-1) end hi(i)=li(i)*hi(i-1); z(i-1)=zi(3,i-1)+hi(i) for j=1:2 zi(j,i)=zi(j+1,i-1) end zi(3,i)=z(i-1) end printf('\n\nAt the end of the %ith iteration by mullers method, the root of the equation is %.10f',j+2,z(j))
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bowtie.sce
Name=bowtie PlayerCharacters=1 BotCharacters=2.bot IsChallenge=true Timelimit=90.0 PlayerProfile=1 AddedBots=2.bot PlayerMaxLives=0 BotMaxLives=10 PlayerTeam=1 BotTeams=2 MapName=vertex.map MapScale=4.0 BlockProjectilePredictors=true BlockCheats=true InvinciblePlayer=false InvincibleBots=false Timescale=1.0 BlockHealthbars=false TimeRefilledByKill=0.0 ScoreToWin=1000.0 ScorePerDamage=1.0 ScorePerKill=0.0 ScorePerMidairDirect=0.0 ScorePerAnyDirect=0.0 ScorePerTime=1.0 ScoreLossPerDamageTaken=0.0 ScoreLossPerDeath=0.0 ScoreLossPerMidairDirected=0.0 ScoreLossPerAnyDirected=0.0 ScoreMultAccuracy=false ScoreMultDamageEfficiency=false ScoreMultKillEfficiency=false GameTag=Tracking WeaponHeroTag=Tracking Beam DifficultyTag=2 AuthorsTag=faiNt` BlockHitMarkers=false BlockHitSounds=false BlockMissSounds=true BlockFCT=false Description=HEADSHOTS ONLY // ADS GameVersion=1.0.7.2 ScorePerDistance=0.0 [Aim Profile] Name=At Feet MinReactionTime=0.3 MaxReactionTime=0.4 MinSelfMovementCorrectionTime=0.001 MaxSelfMovementCorrectionTime=0.05 FlickFOV=30.0 FlickSpeed=1.5 FlickError=15.0 TrackSpeed=3.5 TrackError=3.5 MaxTurnAngleFromPadCenter=75.0 MinRecenterTime=0.3 MaxRecenterTime=0.5 OptimalAimFOV=30.0 OuterAimPenalty=1.0 MaxError=40.0 ShootFOV=15.0 VerticalAimOffset=-200.0 MaxTolerableSpread=5.0 MinTolerableSpread=1.0 TolerableSpreadDist=2000.0 MaxSpreadDistFactor=2.0 [Aim Profile] Name=Low Skill At Feet MinReactionTime=0.35 MaxReactionTime=0.45 MinSelfMovementCorrectionTime=0.001 MaxSelfMovementCorrectionTime=0.05 FlickFOV=30.0 FlickSpeed=1.5 FlickError=20.0 TrackSpeed=3.0 TrackError=5.0 MaxTurnAngleFromPadCenter=75.0 MinRecenterTime=0.3 MaxRecenterTime=0.5 OptimalAimFOV=30.0 OuterAimPenalty=1.0 MaxError=60.0 ShootFOV=25.0 VerticalAimOffset=-200.0 MaxTolerableSpread=5.0 MinTolerableSpread=1.0 TolerableSpreadDist=2000.0 MaxSpreadDistFactor=2.0 [Aim Profile] Name=Low Skill MinReactionTime=0.35 MaxReactionTime=0.45 MinSelfMovementCorrectionTime=0.001 MaxSelfMovementCorrectionTime=0.05 FlickFOV=30.0 FlickSpeed=1.5 FlickError=20.0 TrackSpeed=3.0 TrackError=5.0 MaxTurnAngleFromPadCenter=75.0 MinRecenterTime=0.3 MaxRecenterTime=0.5 OptimalAimFOV=30.0 OuterAimPenalty=1.0 MaxError=60.0 ShootFOV=25.0 VerticalAimOffset=0.0 MaxTolerableSpread=5.0 MinTolerableSpread=1.0 TolerableSpreadDist=2000.0 MaxSpreadDistFactor=2.0 [Aim Profile] Name=Default MinReactionTime=0.3 MaxReactionTime=0.4 MinSelfMovementCorrectionTime=0.001 MaxSelfMovementCorrectionTime=0.05 FlickFOV=30.0 FlickSpeed=1.5 FlickError=15.0 TrackSpeed=3.5 TrackError=3.5 MaxTurnAngleFromPadCenter=75.0 MinRecenterTime=0.3 MaxRecenterTime=0.5 OptimalAimFOV=30.0 OuterAimPenalty=1.0 MaxError=40.0 ShootFOV=15.0 VerticalAimOffset=0.0 MaxTolerableSpread=5.0 MinTolerableSpread=1.0 TolerableSpreadDist=2000.0 MaxSpreadDistFactor=2.0 [Bot Profile] Name=2 DodgeProfileNames=Long Strafe FB DodgeProfileWeights=1.0 DodgeProfileMaxChangeTime=2.0 DodgeProfileMinChangeTime=1.5 WeaponProfileWeights=1.0;1.0;2.0;1.0;0.0;0.0;0.0;0.0 AimingProfileNames=At Feet;Low Skill At Feet;Low Skill;Default;Default;Default;Default;Default WeaponSwitchTime=3.0 UseWeapons=false CharacterProfile=Watcher SeeThroughWalls=false NoDodging=false NoAiming=false [Character Profile] Name=1 MaxHealth=300.0 WeaponProfileNames=gun;;;;;;; MinRespawnDelay=1.0 MaxRespawnDelay=5.0 StepUpHeight=75.0 CrouchHeightModifier=0.5 CrouchAnimationSpeed=2.0 CameraOffset=X=0.000 Y=0.000 Z=80.000 HeadshotOnly=false DamageKnockbackFactor=4.0 MovementType=Base MaxSpeed=700.0 MaxCrouchSpeed=500.0 Acceleration=5000.0 AirAcceleration=16000.0 Friction=4.0 BrakingFrictionFactor=2.0 JumpVelocity=800.0 Gravity=3.0 AirControl=0.25 CanCrouch=true CanPogoJump=false CanCrouchInAir=true CanJumpFromCrouch=false EnemyBodyColor=X=0.771 Y=0.000 Z=0.000 EnemyHeadColor=X=1.000 Y=1.000 Z=1.000 TeamBodyColor=X=1.000 Y=0.888 Z=0.000 TeamHeadColor=X=1.000 Y=1.000 Z=1.000 BlockSelfDamage=false InvinciblePlayer=false InvincibleBots=false BlockTeamDamage=false AirJumpCount=0 AirJumpVelocity=0.0 MainBBType=Cylindrical MainBBHeight=320.0 MainBBRadius=58.0 MainBBHasHead=false MainBBHeadRadius=45.0 MainBBHeadOffset=0.0 MainBBHide=false ProjBBType=Cylindrical ProjBBHeight=320.0 ProjBBRadius=55.0 ProjBBHasHead=false ProjBBHeadRadius=45.0 ProjBBHeadOffset=0.0 ProjBBHide=true HasJetpack=false JetpackActivationDelay=0.2 JetpackFullFuelTime=4.0 JetpackFuelIncPerSec=1.0 JetpackFuelRegensInAir=false JetpackThrust=6000.0 JetpackMaxZVelocity=400.0 JetpackAirControlWithThrust=0.25 AbilityProfileNames=;;; HideWeapon=true AerialFriction=0.0 StrafeSpeedMult=1.0 BackSpeedMult=1.0 RespawnInvulnTime=0.0 BlockedSpawnRadius=0.0 BlockSpawnFOV=0.0 BlockSpawnDistance=0.0 RespawnAnimationDuration=0.5 AllowBufferedJumps=true BounceOffWalls=false LeanAngle=0.0 LeanDisplacement=0.0 AirJumpExtraControl=0.0 ForwardSpeedBias=1.0 HealthRegainedonkill=0.0 HealthRegenPerSec=0.0 HealthRegenDelay=0.0 JumpSpeedPenaltyDuration=0.0 JumpSpeedPenaltyPercent=0.0 ThirdPersonCamera=false TPSArmLength=300.0 TPSOffset=X=0.000 Y=150.000 Z=150.000 BrakingDeceleration=2048.0 VerticalSpawnOffset=0.0 [Character Profile] Name=Watcher MaxHealth=100.0 WeaponProfileNames=;;;;;;; MinRespawnDelay=0.1 MaxRespawnDelay=5.0 StepUpHeight=45.0 CrouchHeightModifier=0.69 CrouchAnimationSpeed=2.0 CameraOffset=X=0.000 Y=0.000 Z=20.000 HeadshotOnly=true DamageKnockbackFactor=3.0 MovementType=Base MaxSpeed=200.0 MaxCrouchSpeed=270.0 Acceleration=10000.0 AirAcceleration=16000.0 Friction=100.0 BrakingFrictionFactor=0.0 JumpVelocity=300.0 Gravity=1.0 AirControl=0.16 CanCrouch=true CanPogoJump=false CanCrouchInAir=true CanJumpFromCrouch=false EnemyBodyColor=X=0.000 Y=0.000 Z=0.000 EnemyHeadColor=X=1.000 Y=0.888 Z=0.000 TeamBodyColor=X=1.000 Y=0.888 Z=0.000 TeamHeadColor=X=1.000 Y=1.000 Z=1.000 BlockSelfDamage=false InvinciblePlayer=false InvincibleBots=false BlockTeamDamage=false AirJumpCount=0 AirJumpVelocity=0.0 MainBBType=Cylindrical MainBBHeight=130.0 MainBBRadius=18.0 MainBBHasHead=true MainBBHeadRadius=10.0 MainBBHeadOffset=40.0 MainBBHide=false ProjBBType=Cylindrical ProjBBHeight=160.0 ProjBBRadius=26.0 ProjBBHasHead=false ProjBBHeadRadius=20.0 ProjBBHeadOffset=15.0 ProjBBHide=true HasJetpack=false JetpackActivationDelay=0.2 JetpackFullFuelTime=4.0 JetpackFuelIncPerSec=1.0 JetpackFuelRegensInAir=false JetpackThrust=6000.0 JetpackMaxZVelocity=400.0 JetpackAirControlWithThrust=0.25 AbilityProfileNames=;;; HideWeapon=true AerialFriction=0.0 StrafeSpeedMult=1.0 BackSpeedMult=1.0 RespawnInvulnTime=0.0 BlockedSpawnRadius=0.0 BlockSpawnFOV=0.0 BlockSpawnDistance=0.0 RespawnAnimationDuration=0.1 AllowBufferedJumps=true BounceOffWalls=false LeanAngle=0.0 LeanDisplacement=0.0 AirJumpExtraControl=0.0 ForwardSpeedBias=1.0 HealthRegainedonkill=0.0 HealthRegenPerSec=0.0 HealthRegenDelay=0.0 JumpSpeedPenaltyDuration=0.0 JumpSpeedPenaltyPercent=0.0 ThirdPersonCamera=false TPSArmLength=300.0 TPSOffset=X=0.000 Y=150.000 Z=150.000 BrakingDeceleration=2048.0 VerticalSpawnOffset=0.0 [Dodge Profile] Name=Long Strafe FB MaxTargetDistance=100000.0 MinTargetDistance=0.0 ToggleLeftRight=true ToggleForwardBack=true MinLRTimeChange=0.5 MaxLRTimeChange=2.0 MinFBTimeChange=1.0 MaxFBTimeChange=2.0 DamageReactionChangesDirection=false DamageReactionChanceToIgnore=0.5 DamageReactionMinimumDelay=0.15 DamageReactionMaximumDelay=0.18 DamageReactionCooldown=1.0 DamageReactionThreshold=0.0 DamageReactionResetTimer=0.1 JumpFrequency=0.0 CrouchInAirFrequency=0.0 CrouchOnGroundFrequency=0.0 TargetStrafeOverride=Ignore TargetStrafeMinDelay=0.125 TargetStrafeMaxDelay=0.25 MinProfileChangeTime=0.0 MaxProfileChangeTime=0.0 MinCrouchTime=0.3 MaxCrouchTime=0.6 MinJumpTime=0.3 MaxJumpTime=0.6 LeftStrafeTimeMult=1.0 RightStrafeTimeMult=1.0 StrafeSwapMinPause=0.0 StrafeSwapMaxPause=0.0 BlockedMovementPercent=0.8 BlockedMovementReactionMin=0.05 BlockedMovementReactionMax=0.1 [Weapon Profile] Name=gun Type=Hitscan ShotsPerClick=1 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//FIND THE DIRECTION OF POWER FLOW OF MICROWAVE //given clc function w=cross_prod(E,F)//function to determine the cross product of two vectors D=[E(:),F(:)] w(1)=det([[1;0;0],D]) w(2)=det([[0;1;0],D]) w(3)=det([[0;0;1],D]) endfunction E=[0 1 0] F=[1 0 0] q=cross_prod(E,F) disp(q','the cross product of the given fields')//towards az //ERROR in book as cross product of two perpendicular vector gives the third
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//Condições iniciais R = 0.001; ro = 7850; volume = 4*%pi*(R^3)/3; m = ro*volume; g = 9.8; r = 1; s0 = -4*r; v0 = 0; t = linspace(0,10,1000); s= zeros(1,1000); v = zeros(1,1000); funcprot(0); function [ds]=f(t, y) ds1=y(2); ds2=(-g/(4*r))*(y(1)); ds=[ds1;ds2]; endfunction function [z]=h(t, y) z1=16+f(1)^2; z2=f(2); z=[z1;z2] endfunction x = ode("root",[s0;v0],0,t,f,1,h); s = x(1,:); v = x(2,:); t1 = linspace(0,10,999) dv = diff(v) dt = diff(t) for i = 1:(length(dv)) a(i) = dv(i)/dt(i) end T = m*(v^2)/2; V = m*g*(s^2)/(8*r); E = T + V; //Gráficos f1=scf(1); plot(t, s); xtitle("posição em função do tempo", "t","s"); f2=scf(2); plot(t, v); xtitle("velocidade em função do tempo", "t","v"); f3=scf(3); plot(t1, a'); xtitle("aceleração em função do tempo", "t", "a"); f4=scf(4); plot(t, T); xtitle("energia cinética em função do tempo", "t", "T") f5=scf(5); plot(t, V); xtitle("energia potencial em função do tempo", "t", "V") f6=scf(6); plot(t, E); xtitle("energia mecânica em função do tempo", "t", "E") f7=scf(7); plot(s, v); xtitle("velocidade em função do espaço", "espaço", "velocidade")
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clear; clc; // Stoichiometry // Chapter 5 // Energy Balances // Example 5.19 // Page 245 printf("Example 5.19, Page 245 \n \n"); // solution H1 = 482.9 // kJ/kg H2 = 273.4 fi1 = 100*(H1-H2) // kJ/h T1 = 313.15 T2 = 403.15 fi11 = 21.3655*(T2-T1)+64.2841*10^-3*(T2^2-T1^2)/2-41.0506*10^-6*(T2^3-T1^3)/3+9.7999*10^-9*(T2^4-T1^4)/4 // kJ/h // at 20 MPa h1 = 211.1 Ts = 277.6 H11 = 427.8 x = poly(0, 'x') p = x*h1+(100-x)*H11-100*H2 a = roots(p) fi2 = (100-a)*(H11-h1) // kJ/h h2 = -148.39 H3 = 422.61 y = poly(0, 'y') p1 = 100*176.18-(100-y)*H3+h2*y b = roots(p1) fi3 = 100*(h1-176.8) H = fi3+24021 H4 = H/(100-43.16) // from ref 23 T = 262.15 printf(" (a) \n \n Yield of dry ice = "+string(b)+" kg. \n \n \n (b) \n \n Percent liquifaction = "+string(a)+". \n \n \n (c) \n \n Temp of vented gas = "+string(T)+" K.")
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clc mu_n=600 disp("mu_n = "+string(mu_n)+"cm^2(Vs)^-1") //initializing value of mobility of electron mu_p = 200 disp("mu_p = "+string(mu_p)+"cm^2(Vs)^-1") //initializing value of mobility of holes kbT = 0.026 disp("kbT = "+string(kbT)+"eV") //initializing value of kbT at 300K apsilen = 11.9*8.85*10^-14 disp("apsilen = "+string(apsilen)+"F/cm") //initializing value of relative permitivity e = 1.6*10^-19 disp("e= "+string(e)+"C")//initializing value of charge of electron Na=5*10^16 disp("Na = "+string(Na)+"cm^-3") //initializing value of doped carrier concentration ni = 1.5*10^10 disp("ni= "+string(ni)+"cm^-3")//initializing value of intrinsic carrier concentration Vfb = -0.5 disp("Vfb= "+string(Vfb)+"eV")//initializing value of flat band voltage Eox = 1.583*8.85*10^-14 disp("Eox= "+string(Eox))//initializing value of relative permitivity of oxide dox = 200*10^-8 disp("dox= "+string(dox)+"cm")//initializing value of thickness of oxide sigma_1= Na*e*mu_p disp("The channel conductivity under flat band sigma_1= Na*e*mu_p= "+string(sigma_1)+" ohm^-1cm^-1")//calculation sigma_2= Na*e*mu_n disp("The channel conductivity at inversion sigma_1= Na*e*mu_n= "+string(sigma_2)+" ohm^-1cm^-1")//calculation phi_F= (-kbT*log(Na/ni)) disp("The potential phi_F= (-kbT*log(Na/ni))= "+string(phi_F)+" V")//calculation Qs = sqrt((4*apsilen*(-phi_F))*(e*Na)) disp("The maximum depletion width is ,Qs = sqrt((4*apsilen*(-phi_F))*(e*Na))= "+string(Qs)+" C cm^-2")//calculation Vs = -(2*phi_F) disp("The surface potential is ,Vs = -(2*phi_F)= "+string(Vs)+" V")//calculation VT = Vfb+Vs+((Qs*dox)/Eox) disp("In the absence of any oxide charge, the threshold voltage is ,VT = Vfb+Vs+((Qs*dox)/Eox) = "+string(VT)+" V")//calculation // Note : due to different precisions taken by me and the author ... my answer differ
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close(); clear; clc; Beta = 100; Ibq = 20*10^(-6); Vcc = 15;//V Rc = 3000;//ohm Icbo = 0; alpha = Beta/(Beta+1); Icq = Beta*Ibq; Ieq = Icq/alpha; Vceq = Vcc-Icq*Rc; //Part (c) Rc = 6000; Vceq_c = Vcc-Icq*Rc; mprintf('(a):Ieq = %0.2f mA\n(b):Vceq = %0.0f V\n(c):Vceq = %0.0f V',Ieq*1000,Vceq,Vceq_c);
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//Function to round-up a value such that it is divisible by 5 function[v] = round_five(w) v = ceil(w) rem = pmodulo(v,5) if (rem ~= 0) then v = v + (5 - rem) end endfunction //Obtain path of solution file path = get_absolute_file_path('solution4_21.sce') //Obtain path of data file datapath = path + filesep() + 'data4_21.sci' //Clear all clc //Execute the data file exec(datapath) //Calculate permissible tensile stress sigma (N/mm2) sigma = Sut/fs //Assume value of t to be 1mm t = 1 //All dimensions in mm bi = 3 * t h = 3 * t Ri = 2 * t Ro = 5 * t ti = t to = 0.75 * t Rn = ((ti * (bi - to)) + (to * h))/((bi - to)*log((Ri + ti)/Ri)+(to * log(Ro/Ri))) R = Ri + ((((1/2)*to*(h^2)) + ((1/2)*(ti^2)*(bi - to)))/((to*h) + (ti*(bi - to)))) //Calculate eccentricity e (mm) e = R - Rn hi = Rn - Ri A = (bi * ti) + (to * Ri) //Calculate the direct tensile stress T (N/mm2) T = (P * 1000)/A //The polynomial obtained is as follows C = (((R * P * 1000 * hi)/(A * e * Ri)) + T) D = (P * 1000 * l * hi)/(A * e * Ri) p = [(-1 * sigma) 0 C D] r = roots(p) //Calculate the true value of t (mm) real_part = real(r) for i = 1:1:3 if(real_part(i)>0) t = real_part(i) break end end t = round_five(t) //Print results printf('\nThe value of t = %f mm\n',t)
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//Variable declaration m=9.1*10**-31; //mass(kg) nx=1; ny=1 nz=1 n=6; a=1; //edge(m) h=6.63*10**-34; //planck's constant k=1.38 //Calculation E1=h**2*(nx**2+ny**2+nz**2)/(8*m*a**2); E2=h**2*n/(8*m*a**2); E=E2-E1; //energy difference(J) T=(2*E2*10**37)/(3*k*10**-23) //Result printf('energy difference is%0.3f *10**-37 J \n ',(E*10**37)) printf('3/2*k*T = E2 =%0.3f *10**-37 J \n ',(E2*10**37)) printf('T =%0.3f *10**-14 K \n ',(T/10**23))
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// Copyright INRIA 2008 // This file is released into the public domain help_dir = get_absolute_file_path('builder_help.sce'); tbx_builder_help_lang("en_US", help_dir); clear help_dir;
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clear; // sampling num samplenum=500 // number of repetition repeat=1000; R=grand(repeat,samplenum,'uin',0,1); R=1-R*2; // avarage of repetition X=sum(R,'c'); //plot2d(X); xmin=-samplenum/10; xmax=samplenum/10; param=[xmin:1:xmax]; histplot(param,X) n=repeat; p=1/2; avg=0; s=sqrt(n*p*(1-p)) x=[-50:0.05:50]; plot(x,(1/(s*sqrt(2*%pi)))*exp(-((avg-x).^2)/(2*s^2)));
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clear; clc; // Illustration 1.21 // Page: 60 printf('Illustration 1.21 - Page:60 \n\n'); // Solution //*****Data*****// // a-oxygen b-nitrogen T = 293; // [K] P = 0.1; // [atm] d = 0.1*10^-6; // [m] e = 0.305; // [porosity] t = 4.39; // [tortuosity] k = 1.3806*10^-23; // [J/K] l = 2*10^-3; // [m] R = 8.314; // [cubic m.Pa/mole.K] x_a1 = 0.8; x_a2 = 0.2; M_a = 32; // [gram/mole] M_b = 28; // [gram/mole] //*****// // Using data from Appendix B for oxygen and nitrogen, and equation (1.45) sigma_a = 3.467; // [Angstrom] sigma_b = 3.798; // [Angstrom] sigma_AB = ((sigma_a+sigma_b)/2)*10^-10; // [m] lambda = k*T/(sqrt(2)*3.14*(sigma_AB^2)*P*1.01325*10^5); // [m] // From equation 1.101 K_n = lambda/d; printf("The value of a dimensionless ratio, Knudsen number is %f\n\n",K_n); // If K_n is greater than 0.05 then transport inside the pores is mainly by Knudsen diffusion // Using equation 1.103 D_Ka = (d/3)*(sqrt(8*R*T)/sqrt(3.14*M_a*10^-3)); // [square m/s] // Using equation 1.107 D_Kaeff = D_Ka*e/t; // [square m/s] p_a1 = (x_a1*P)*1.01325*10^5; // [Pa] p_a2 = (x_a2*P)*1.01325*10^5; // [Pa] // Using equation 1.108 N_a = D_Kaeff*(p_a1-p_a2)/(R*T*l); // [mole/square m.s] // Now using the Graham’s law of effusion for Knudsen diffusion // N_b/N_a = -sqrt(M_a/M_b) ,therefore N_b = -N_a*sqrt(M_a/M_b); // [mole/square m.s] printf("The diffusion fluxes of both components oxygen and nitrogen are %e mole/square m.s and %e mole/square m.s respectively\n",N_a,N_b);
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clc; clear; close; //a Reff=12.5;//in kiloohm r=2/4; Cload=4.2;//in fermifarad T=4*Reff*r*Cload; disp(T,'total delay(in picoseconds)='); //b Reff1=30;//in kiloohm Cfanout=2;//in fermifarad Cself=1;//in fermifarad r1=2/6; Cload1=Cself+Cfanout; PLH=Reff1*Cload1*r1; PHL=Reff*r*Cload1; T1=2*(PHL+PLH); disp(PLH,'rise delay(in picoseconds)='); disp(PHL,'fall delay(in picoseconds)='); disp(T1,'total delay(in picoseconds)='); //answers vary due to roundoff error
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// Demo for Ensemble classifier -- Scilab getd('../macros') // Data preparation for gaussian M = csvRead('Datasets/standard.csv') x = M(:, [1,2,3,4,5,6,7,8]); y = M(:, 9); y(or(isnan(x),'c'),:) = [] x(or(isnan(x),'c'),:) = [] probMat = naiveBayes(x, y) testx = x; pred1 = naiveBayesGaussian(x, y, probMat, testx) // Data preparation for multinomial x = x(:, 1); probMat = naiveMultBayes(x, y) testx = x; pred2 = naiveMultBayesGaussian(x, y, probMat, testx) increment = 0.01 pred = ensemble(pred1, pred2, y, increment) disp('Error through classifier 1 = ' + string(0.5*sqrt(norm(pred1 - y)))) disp('Error through classifier 2 = ' + string(0.5*sqrt(norm(pred2 - y)))) disp('Error through ensemble = ' + string(0.5*sqrt(norm(pred - y))))
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////Chapter 13 Steam Engines ////Example 13.8 Page No 288 ////Find Diagram factor //Input data clc; clear; P1=10; //Inlet pressure Pb=1; //Back pressure rc=3; //Expansion ratio a=12.1; //Area of indicator diagram b=7.5; //Length of indicator diagram S=3; //Pressure scale //Calculation Pm=round((P1/rc)*(1+log(rc))-Pb ); //Therotical mean effective pressure Pm Pma=a/b*S; //Actual mean effective pressure Pma K=Pma/Pm; //Diagram factor ///Output printf('Therotical mean effective pressure= %f bar \n',Pm); printf('Actual mean effective pressure= %f bar \n',Pma); printf('Diagram factor= %f \n',K);
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clc //initialisation of variables P= 800*10^3 sigmamax= 400 //CALCULATIONS Amin= P/sigmamax B= sqrt(Amin/2) //RESULTS printf ('Amin= %.2f mm^2',Amin) printf (' \n B=%.2f mm(Therefore the min dimensions of the column cross section are 31.6mm*63.2mm)',B)
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//Ideal Gas Equation clear; clc; printf("\t Example 5.6\n"); P1=1.2;// pressure initial, atm T1=18+273;//temperature initial, K T2=85+273;//temperature final, K //volume is constant P2=P1*T2/T1;// pressure final,atm printf("\t the final pressure is : %4.2f atm\n",P2); //End
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//Variable declaration n=1; //order of diffraction theta1=8+(35/60); //angle(degrees) d=0.282; //spacing(nm) theta2=90; //Calculation theta1=theta1*%pi/180; //angle(radian) lamda=2*d*sin(theta1)/n; //wavelength(nm) theta2=theta2*%pi/180; //angle(radian) nmax=2*d/lamda; //maximum order of diffraction //Result printf('wavelength is %0.3f nm \n',(lamda)) printf('maximum order of diffraction is %0.3f \n',(nmax))
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//To Find the Total Load and Total steady state voltage drop //Page 336 clc; clear; //Length in feet Lsd=100; //Service Drop Line Lsl=200; //Secondary Line SB=75; //Transformer Capacity in kVA pf=0.9; //Load Power Factor //From the Tables ISL=129.17; //Secondary Line Current ISD=41.67; //Service Drop Current KSD=0.01683; //Service Drop Cable Constant KSL=0.0136; //Secondary Cable Constant //for Transformer R=0.0101; //Resistance in per unit X=0.0143; //Reatance in per unit //From the Diagram ST=(3+2+8+6)+(5+6+7+4)+(6+7+8+10); //Total Load on transformer STpu=ST/SB; //In Per unit; //The Above value also corresponds to the Current as Well I=STpu; //Current in Per Unit VDT=I*((R*pf)+(X*sind(acosd(pf)))); //Voltage Drop in the Transformer VDSL=KSL*(Lsl*ISL/(10^4)); //Secondary Line VDSD=KSD*(Lsd*ISD/(10^4)); //Service Drop Line VD=VDT+VDSD+VDSL; //Total Voltage Drop printf('\na)The Load in transformer is %g kVA or %g pu\n',ST,STpu) printf('b) The Total Voltage Drop is %g pu V\n',VD)
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clc // Given that t = 15 // half-life of radioactive element in years r = 0.025 // ratio of mass of element present in specimen to the intial mass of element // Sample Problem 5 on page no. 12.33 printf("\n # PROBLEM 5 # \n") printf("Standard formula used \n") printf(" lambda = 0.693 / t_1/2 (Decay constant) \n N =N_0*e^(-lambda*t) \n") lambda = 0.693 / t T = (1 / lambda) * log(1 / r) printf("\n Period in which 2.5 percent of the initial quantity left over is %f years.",T)
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// Example 11.2 Design of two-stage CMOS op-amp uC_n=50*10^-6; // u_n*C_ox (A/V^2) uC_p=20*10^-6; // u_p*C_ox (A/V^2) V_tn0=1; // (V) V_tp0=-1; // (V) fie_f=0.6/2; // (V) y=0.5; // (V^1/2) V_DD=5; // (V) W_n=4*10^-6; // (m) L_n=2*10^-6; // (m) W_p=10*10^-6; // (m) L_p=2*10^-6; // (m) W=10*10^-6; // (m) L=10*10^-6; // (m) C_B=1*10^-12; // bit line capacitance (F) deltaV=0.2; // 0.2 V decrement WbyL_eq=1/(L_p/W_p+L_n/W_n); // WbyL_eq=(W/L)_eq // Equivalent transistor will operate in saturation I=(uC_n*WbyL_eq*(V_DD-V_tn0)^2)/2 r_DS=1/(uC_n*(W_n/L_n)*(V_DD-V_tn0)); v_Q=r_DS*I; // v_Q=r_DS*I I_5=0.5*10^-3; // (A) deltat=C_B*deltaV/I_5; disp(deltat, "The time (s) required to develop an output voltage of 0.2V")
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//Find the Amount of Heat required to raise the temperature to 400 Kelvin //Example 27.2 clc; clear; V=0.2;//Volume of tank in m^3 p=1*10^5;//Pressure of Helium Gas in N/M^2 T1=300;//Initial Temperature of Helium Gas in Kelvin T2=400;//Final Temperature of Helium Gas in Kelvin R=8.31;//Universal Gas Constant in J/mol-K n=int((p*V)/(R*T1));//Amount of moles of Helium Gas Cv=3;//Molar Heat Capacity at Constant Volume Q=n*Cv*(T2-T1);//Amount of Heat Required in calories printf("The amount of Heat required=%d cal",Q);
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//ANALOG AND DIGITAL COMMUNICATION //BY Dr.SANJAY SHARMA //CHAPTER 5 //ANGLE MODULATION clear all; clc; printf("EXAMPLE 5.19(PAGENO 222)"); //given f_m1 = 1*10^3//modulating frequency for first case f_m2 = 500//modulating frequency for second case V_m1 = 2//modulating voltage for first case V_m2 = 8//modulating volatge for second case delta_f1 = 4*10^3//frequency deviation for first case //calculations k = delta_f1/V_m1//constant delta_f2 = k*V_m2//frequency deviation for second case m_f1 = delta_f1/f_m1//modulation index for first case m_f2 = delta_f2/f_m2//modulation index for second case BW1 = 2*(delta_f1 + f_m1)//bandwidth for first case BW2 = 2*(delta_f2 + f_m2)//bandwidth for second case //results printf("\n\ni.a.Modulation index for first case = %.2f",m_f1); printf("\n\n b.Bandwidth for first case = %.2f Hz",BW1); printf("\n\nii.a.Modulation index for second case = %.2f",m_f2); printf("\n\n b .Bandwidth for second case = %.2f Hz",BW2); printf("\n\nNote: Their is error in textbook in the calculation of second case bandwidth ");
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//=========================================================================== //chapter 5 example 11 clc;clear all; //variable declaration V = 300; //voltage in V R = 12000; //coil resistance in Ω B = 6*10^-2; //flux density in Wb/m**2 l = 0.04; //length in m r = 0.03; //width in m N = 100; Tc = 25*10^-7; //torque in Nm per degree //calculations i = V/(R); //current in A Td = N*B*i*l*r; //deflecting Torque in Nm //Tc=Td; //Tc =(25*10^-7)*theta theta = Td/(Tc); //defelction in ° //result mprintf('defelction = %3.0f °",theta);
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// Exa 17.1 //To calculate- //(a) average busy-hour traffic per subscriber, //(b) traffic capacity per cell, //(c) required number of base stations per zone, and //(d) the hexagonal cell radius for the zone. clc; clear all; Susage=150;//subscriber usage per month in mins days=24;//days per month busyhrs=6;//in a day BW=4.8*10^3; //in kHz Freqreuse=4/12;//Frequency reuse plan chwidth=200; //in kHz subscriber=50000;//Present subscriber count Sgrowth=0.05;//Growth rate per year Area=500; //in km BTScapacity=30; //in Erlangs N=4; //Initial installation design years //solution Erlangspersub=Susage/(days*busyhrs*60); printf('Average busy-hour traffic per subscriber is %.4f Erlangs \n ',Erlangspersub); RFcarriers=BW/chwidth; RFcarrier_percell=RFcarriers/((Freqreuse^-1)*4); //freq reuse factor of 4 //Assuming 2 control channels per cell CC=2;//control channels TC_percell=2*RFcarriers/3-CC; //Referring Erlang-B table in Appendix A disp("Referring Erlang-B table in Appendix A,Traffic capacity of a GSM cell at 2% GoS for 14 channels = 8.2 Erlangs "); Tcapacity=8.2;// in Erlangs disp("There are 3 cells per BTS"); BTS=3; Traffic_perBTS=Tcapacity*BTS; printf(' Traffic capacity per BTS is %.1f Erlangs ',Traffic_perBTS); disp("Therefore, Traffic per BTS is less than BTS capacity(30 Erlangs)") maxsubscriber=Traffic_perBTS/Erlangspersub; initialsub=subscriber*(1+Sgrowth)^N; BTS_perZone=initialsub/maxsubscriber; printf(' The required number of base stations per zone are %d \n ',round(BTS_perZone)); cellRadius=(Area/(BTS_perZone*2.6))^0.5; printf('The hexagonal cell radius is %.1f km \n ',cellRadius);
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clc; clear all; //chapter 4 //page no 125 //example 4.7 //misprinted example number pulse_width=2*10^-6; //second rise_time=10*10^-9; //second B=.5/pulse_width; //in Hz mprintf('(a) The aproximate bandwidth for coarse reproduction is B=%i KHz\n',B*10^-3) B=.5/rise_time; mprintf(' (b) The aproximate bandwidth for fine reproduction is B=%i MHz\n',B*10^-6)
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//Chapter-4,Example4_15_19,pg 4-36 t1=4*10^-3 //thickness of 1st crystal n1=400*10^3 //frequency of 1st crystal n2=500*10^3 //frequency of 2nd crystal t2=n1*t1/n2 //since frquency is inversly proportional to thickness printf("thickness of 2nd crystal =") disp(t2) printf("meter")
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clear clc M = csvRead('winequality-white_normalizado.csv') P=M(2:700,2:12)'; //DADOS DE TREINAMENTO T=M(2:700,13)'; //SAIDAS DESEJADAS Q=M(701:1000,2:12)'; // Exemplos usados para teste TQ=M(701:1000,13)'; // Valores desejados na saída para os exemplos de teste N=[11 8 4 1]; //Tipos de funções de ativação: //ann_logsig_activ //ann_purelin_activ //ann_tansig_activ //ann_hardlim_activ //Define a função de ativação da camada intermediária e da saída //Se tiverem 2 intermediárias, tem que definir uma função para cada uma af = ['ann_tansig_activ','ann_tansig_activ','ann_purelin_activ']; lr = 0.03; //taxa de aprendizado itermax = 1000; // número máximo de iterações mse_min = 1e-5; //mínimo erro quadrático médio desejado gd_min = 1e-5; // variação mínima do gradiente W = ann_FFBP_gd(P,T,N,af,lr,itermax,mse_min,gd_min) [y] = ann_FFBP_run(Q,W, af) disp(y); xpause(30000) scf(1) plot (y) plot (TQ,'r-')
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function [x,y,typ]=CURV_f(job,arg1,arg2) x=[];y=[];typ=[]; select job case 'plot' then standard_draw(arg1) case 'getinputs' then [x,y,typ]=standard_inputs(arg1) case 'getoutputs' then [x,y,typ]=standard_outputs(arg1) case 'getorigin' then [x,y]=standard_origin(arg1) case 'set' then x=arg1 model=arg1(3) graphics=arg1(2) [rpar,ipar]=model(8:9) n=ipar(1) xx=rpar(1:n);yy=rpar(n+1:2*n) curwin=xget('window') win=maxi(windows(:,2))+1 xset('window',win);xsetech([0 0 1 1]) gc=list(rpar(2*n+1:2*n+4),ipar(2:5)) while %t do [xx,yy,ok,gc]=edit_curv(xx,yy,'axy',[' ',' ',' '],gc) if ~ok then break,end n=size(xx,'*') if or(xx(2:n)-xx(1:n-1)<0) then message('You have not defined a function') ok=%f end if ok then model(1)='intplt' model(11)=[] //compatibility rect=gc(1) model(8)=[xx(:);yy(:);rect(:)] axisdata=gc(2) model(9)=[size(xx,'*');axisdata(:)] x(2)=graphics;x(3)=model break end end xdel(win) xset('window',curwin) case 'define' then xx=[0;1;2];yy=[-5;5;0] rect=[0,-5,2,5]; axisdata=[2;10;2;10] ipar=[size(xx,1);axisdata(:)] rpar=[xx;yy;rect(:)] model=list('intplt',[],1,[],[],[],[],rpar,ipar,'c',[],[%f %t],' ',list()) gr_i=['model=arg1(3);rpar=model(8);ipar=model(9);n=ipar(1);'; 'thick=xget(''thickness'');xset(''thickness'',2);'; 'xx=rpar(1:n);yy=rpar(n+1:2*n);'; 'rect=rpar(2*n+1:2*n+4);'; 'mxx=rect(3)-rect(1);'; 'mxy=rect(4)-rect(2);'; 'xx=orig(1)+sz(1)*(1/10+(4/5)*((xx-rect(1))/mxx));'; 'yy=orig(2)+sz(2)*(1/10+(4/5)*((yy-rect(2))/mxy));'; 'xpoly(xx,yy,''lines'');'; 'xset(''thickness'',thick);'] x=standard_define([2 2],model,[],gr_i) end
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function r=%r_o_p(l1,l2) // l1==l2 with l1 rational and l2 polynomial //! // Copyright INRIA r=degree(l1('den'))==0 if r then r=l1('num')./coeff(l1('den'))==l2,end
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function afficherLogs(something) if AFFICHER_LOGS then disp(something); end endfunction
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function y = pgm5(x) y = exp(-2 * x) + 3 * sqrt(x); endfunction
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sce
ex_7_50.sce
clc; A=[0 1;2 1]; b=[1;-1]; c=[1 -1]; d=[0]; e=spec(A); if real(e(1))>0 |real(e(2))>0 then disp("the sytem is not asymptotically stable") else disp("the system is asymptotically stable") end Hs=ss2tf(syslin('c',A,b,c,d)); disp(Hs,"H(s)=") disp("there is only one pole and it is located at -1 which is in the LHP hence the system is BIBO stable")
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/6º PERÍODO/Otimizacao/UNIDADE 2/main.sce
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no_license
kaikecc/UFRN
647d09e6beeadfeabfb59f479cd77a5c2ff7bebc
821cce2e36808e890a75714d71b7cc1a24e52c27
refs/heads/master
2022-08-09T19:23:52.552924
2022-06-23T21:27:05
2022-06-23T21:27:05
177,469,993
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main.sce
clc clear all disp("MAXIMIZAÇÃO - [1]"); disp("MINIMIZAÇÃO - [2]"); disp("DETERMINAÇÃO PARA UMA SOLUÇÃO ÓTIMA - [3]"); opc = input("ESCOLHA UMA OPÇÃO: "); arq = uigetfile("*.txt","C:\Users\kaike\Desktop\UFRN\Otimizacao\UNIDADE 1\Exercicios\","Escolha um arquivo .txt: "); // Funcao para pegar um arquivo matriz_aumentada = fscanfMat(arq);// pega o arquivo de formato especificado function metodo = main(matriz_aumentada) select opc case 1 then // [Zo, A, b] = maximizacao(matriz_aumentada); disp("ENTROU EM MAX"); case 2 then // [Zo, A, b] = minimizacao(matriz_aumentada); disp("ENTROU EM MIN"); case 3 then else break; end metodo = 0; endfunction main(matriz_aumentada)
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/scilab_prof/ballon_a_sculpter.sce
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netixx/projet_meca
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a1d5794e2eeef61efadd7cbb8d8c477676610eec
refs/heads/master
2020-12-30T14:56:13.306855
2012-12-12T13:51:34
2012-12-12T13:51:34
null
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ballon_a_sculpter.sce
// ///////////////////// // Enveloppe cylindrique // ///////////////////// // ////////// // Invariants // Question 5 // ////////// function [I1,I2]=Invariants(lambda, mu) I1 = lambda^2 + mu^2 + 1/(lambda*mu)^2; I2 = 1/lambda^2 + 1/mu^2 + (lambda*mu)^2; endfunction // /////////// // Contraintes // Question 6 // /////////// function y = sigtt(Loi, AddI2,lambda, mu) [I1,I2] = Invariants(lambda, mu); [dWdI1,dWdI2]=dWdI(Loi, AddI2, I1, I2); y = (dWdI1 + dWdI2* mu^2)*(lambda^2 - 1./(lambda*mu)^2); endfunction function y = sigzz(Loi, AddI2, lambda, mu) [I1,I2] = Invariants(lambda, mu); [dWdI1,dWdI2]=dWdI(Loi, AddI2, I1, I2); y = (dWdI1 + dWdI2* lambda^2)*(mu^2 - 1./(lambda * mu)^2); endfunction // ////////////// // Delta pression // Question 7 // ////////////// function y = DeltaPression(Loi, AddI2, Epais, A, lambda, mu) y = Epais / (mu * lambda^2 ) * sigtt (Loi, AddI2, lambda, mu); endfunction // /////////////////////////// // charge axiale additionnelle // Question 7 // /////////////////////////// function y = AxialAdd(Loi, AddI2, Epais, A, lambda, mu) y = pi * A^2 * (2. * Epais / mu * sigzz (Loi, AddI2, lambda, mu) - lambda^2 * DeltaPression(Loi, AddI2, Epais, A, lambda, mu) ); endfunction // /////////////////////////////////// // pression normalisee par E/A = Epais // /////////////////////////////////// function y = DeltaPressionNorm(Loi, AddI2, Epais, A, lambda, mu) y = 1. / (mu * lambda^2 ) * sigtt (Loi, AddI2, lambda, mu); endfunction // //////////////////////////////////////////////////////////////// // charge axiale additionnelle normalisee par pi E A = Pi Epais A^2 // //////////////////////////////////////////////////////////////// function y = AxialAddNorm(Loi, AddI2, Epais, A, lambda, mu) y = 2. * sigzz (Loi, AddI2, lambda, mu) / mu - lambda^2 * DeltaPressionNorm(Loi, AddI2, Epais, A, lambda, mu) ; endfunction // // ////////////////////////////////////////////////////////////////// // Solve Ballon : Calcul de DP, MU, I1, I2 et Vol=V/V_ini=mu*lambda^2 // ////////////////////////////////////////////////////////////////// function [DP, Mu, Vol, I1, I2]=SolveBallon(lambda, Ta, Epais, A, AddI2) NomLoi = ['NH', 'Gent', 'Langevin']; NbLoi = size(NomLoi,"c"); DP = zeros(size(lambda,"c"), size(Ta,"c"), NbLoi); Mu = zeros(size(lambda,"c"), size(Ta,"c"), NbLoi); Vol = zeros(size(lambda,"c"), size(Ta,"c"), NbLoi); I1 = zeros(size(lambda,"c"), size(Ta,"c"), NbLoi); I2 = zeros(size(lambda,"c"), size(Ta,"c"), NbLoi); for kk=1:NbLoi, loi =NomLoi(kk); j = 0; for Tai=Ta, j = j+1; i = 0; for lambdai=lambda, i = i+1; deff('y=fsol(x)','y=AxialAddNorm(loi, AddI2, Epais, A, lambdai, x)-Tai'); mui = fsolve (1,fsol); DP(i,j,kk) = DeltaPression(loi, AddI2, Epais, A, lambdai, mui); Mu(i,j,kk) = mui; Vol(i,j,kk) = mui*lambdai^2; [I1(i,j,kk),I2(i,j,kk)]=Invariants(lambdai,mui); end; end; end; endfunction;
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/Resultados/results_LocGlo_16/results/16/l20-2/result2s0.tst
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Tiburtzio/TFG
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refs/heads/master
2023-01-03T12:44:56.269655
2020-10-24T18:37:02
2020-10-24T18:37:02
275,638,403
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result2s0.tst
@relation unknow @attribute sp11 integer[0,255] @attribute sp12 integer[0,255] @attribute sp13 integer[0,255] @attribute sp14 integer[0,255] @attribute sp15 integer[0,255] @attribute sp16 integer[0,255] @attribute sp17 integer[0,255] @attribute sp18 integer[0,255] @attribute sp19 integer[0,255] @attribute sp21 integer[0,255] @attribute sp22 integer[0,255] @attribute sp23 integer[0,255] @attribute sp24 integer[0,255] @attribute sp25 integer[0,255] @attribute sp26 integer[0,255] @attribute sp27 integer[0,255] @attribute sp28 integer[0,255] @attribute sp29 integer[0,255] @attribute sp31 integer[0,255] @attribute sp32 integer[0,255] @attribute sp33 integer[0,255] @attribute sp34 integer[0,255] @attribute sp35 integer[0,255] @attribute sp36 integer[0,255] @attribute sp37 integer[0,255] @attribute sp38 integer[0,255] @attribute sp39 integer[0,255] @attribute sp41 integer[0,255] @attribute sp42 integer[0,255] @attribute sp43 integer[0,255] @attribute sp44 integer[0,255] @attribute sp45 integer[0,255] @attribute sp46 integer[0,255] @attribute sp47 integer[0,255] @attribute sp48 integer[0,255] @attribute sp49 integer[0,255] @attribute class{1,2,3,4,5,6,7} @inputs sp11,sp12,sp13,sp14,sp15,sp16,sp17,sp18,sp19,sp21,sp22,sp23,sp24,sp25,sp26,sp27,sp28,sp29,sp31,sp32,sp33,sp34,sp35,sp36,sp37,sp38,sp39,sp41,sp42,sp43,sp44,sp45,sp46,sp47,sp48,sp49 @outputs class @data 3 3 3 3 3 3 3 3 4 3 3 3 4 7 2 2 3 3 3 3 3 3 3 3 7 7 7 7 2 5 3 3 3 3 7 7 3 3 7 7 7 7 2 2 7 7 7 2 2 2 5 7 3 3 7 3 7 7 7 7 5 7 3 3 2 2 7 7 2 2 2 2 4 7 4 3 2 2 7 7 7 7 7 7 7 7 4 7 4 3 4 7 5 5 4 7 1 1 7 7 5 5 5 5 5 5 7 7 1 1 1 1 1 1 1 1 7 7 7 7 1 1 7 7 3 3 3 3 1 1 1 1 1 1 7 7 1 1 7 7 7 5 1 1 7 3 1 1 1 1 3 3 3 3 7 3 1 1 2 2 1 1 1 7 1 1 1 1 5 1 2 2 1 1 1 1 3 3 3 3 1 1 1 1 1 7 1 1 5 1 3 3 3 3 3 3 2 5 3 3 4 1 2 2 7 7 2 3 2 2 3 3 7 7 5 5 7 7 5 5 4 3 3 3 7 7 7 7 4 7 1 1 4 7 5 5 3 3 5 7 1 1 1 1 1 1 1 1 1 1 1 5 5 5 5 5 5 5
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/2021/CH4/EX4.5/EX4_5.sce
5660bbd6752272bd453ab6507e5b617c0ea605aa
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FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
7bc77cb1ed33745c720952c92b3b2747c5cbf2df
refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
3
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EX4_5.sce
//Finding of Density //Given h1=0.4; h2=0.6; rho=1000; rho1=13600; g=9.81; wd=rho*0.6; md=rho1*0.4; rho2=wd+md; disp("Density is = "+string(rho2)+" Kg/m^3");
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/source/2.4.1/macros/sci2for/ins2for.sci
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permissive
clg55/Scilab-Workbench
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refs/heads/master
2023-05-31T04:06:22.931111
2022-09-13T14:41:51
2022-09-13T14:41:51
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ins2for.sci
function [txt,vnms,vtps,nwrk]=ins2for(lst,ilst,vnms,vtps,nwrk) // traduit un ensemble d'instructions debutant a l'adresse ilst de la // liste courante lst //! // Copyright INRIA nlst=size(lst) txt=[] while ilst<=nlst then if type(lst(ilst))==15 then [t1,vnms,vtps,nwrk]=cla2for(lst(ilst),vnms,vtps,nwrk) ilst=ilst+1 else [t1,ilst,vnms,vtps,nwrk]=cod2for(lst,ilst,vnms,vtps,nwrk) end txt=[txt;t1] end
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/lpf using wfir.sce
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manasdas17/Scilab-for-Signal-Processing-
6efc5adb507243c7302f7b4f3f12d12060112038
5f6e6ce941c0a11212a83674b5d35d97a2cf4396
refs/heads/master
2021-01-10T07:49:58.006357
2016-04-07T07:45:26
2016-04-07T07:45:26
55,673,271
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sce
lpf using wfir.sce
//By Manas,FOSSEE,IITB [wft,wfm,fr]=wfir('lp',33,[.23 0],'kr',[5.6 0]);//Linear phase FIR filters plot2d(fr,20*log10(wfm));//Plots filter response against grid fr xgrid(); xtitle("LPF using wfir");
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/1931/CH11/EX11.5/5.sce
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FOSSEE/Scilab-TBC-Uploads
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refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
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5.sce
clc clear //INPUT DATA ni=1.5*10^16//intrinsic charge carriers in m^-3 r1=10*10^-2//resistivity of p-type silicon in ohm m r2=10*10^-2//resistivity of n-type silicon in ohm m me=1350*10^-4//The mobility of the charge carrier in m^2 V^-1 s^-1 mh=480*10^-4//The hole charge carrier in m^2 V^-1 s^-1 e=1.6*10^-19//charge of electron in coulombs //CALCULATION Na=(1/(r1*e*mh))/10^21//The density of the intrinsic crystal for p-type in m^-3*10^21 ne=((ni^2)/(Na*10^21))/10^11//The minor carrier concentration for p-type in electrons/m^3*10^11 Nd=(1/(r2*e*me))/10^20//The density of the intrinsic crystal for n-type in m^-3*10^20 nh=((ni^2)/(Nd*10^20))/10^11//The minor carrier concentration for n-type in electrons/m^3*10^11 //OUTPUT printf('The density of the intrinsic crystal for p-type is %3.4f*10^21 m^-3 \n The minor carrier concentration for p-type is %3.3f*10^11 electrons/m^3 \n The density of the intrinsic crystal for n-type is %3.3f*10^20 m^-3 \n The minor carrier concentration for n-type is %3.4f*10^11 holes/m^3',Na,ne,Nd,nh)
8524ecb1322e1888332518ac1905565c4fc4ec96
b12941be3faf1fd1024c2c0437aa3a4ddcbbfd67
/normal2/pretest.tst
4ea72f3c3e8e5a4afd59874f9d8cc61dd3ca0b1b
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no_license
JanWielemaker/optica
950bd860825ab753236ce1daa399ee7a0b31b3ee
3a378df314b5a60926b325089edac89c00cc8c6d
refs/heads/master
2020-06-12T14:22:46.567191
2019-06-21T11:24:41
2019-06-21T11:24:41
194,328,239
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pretest.tst
/* Questionaire created by optica toolkit Date: Tue Mar 24 16:23:20 1998 */ question(1, 'pre1.11', 'Wanneer komt het snijpunt van de uittredende lichtstraal met de hoofdas verder van de lens te liggen?', [ 'Als je de hoek die de invallende straal met de hoofdas maakt vergroot.', 'Als je de lamp een stukje naar rechts beweegt.', 'Als je de lens een stukje naar rechts beweegt.' ], state(state, '', [ m6 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(10), sfere_right(10), breaking_index(1.51), pos_x(7), show_gauge(false), instrument_name(lens)), l2 = lamp1(switch(true), angle(10), pos_y(0), pos_x(0), instrument_name(lamp1)) ])). question(2, 'pre1.12', 'Wanneer komt het snijpunt van de uittredende lichtstraal met de hoofdas dichter bij de lens te liggen?', [ 'Als je de hoek die de invallende lichtstraal met de hoofdas maakt verkleint.', 'Als je de lamp een stukje naar links beweegt.', 'Als je de lens vervangt door een platbolle lens.' ], state(state, '', [ m6 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(10), sfere_right(10), breaking_index(1.51), pos_x(10), show_gauge(false), instrument_name(lens)), l2 = lamp1(switch(true), angle(10), pos_y(0), pos_x(2), instrument_name(lamp1)) ])). question(3, 'pre1.13', 'Wanneer zal de uittredende lichtstraal meer naar boven schijnen?', [ 'Als je de hoek die de invallende lichtstraal met de hoofdas maakt verkleint.', 'Als je de lens meer naar links beweegt.', 'Als je de lens vervangt door een dubbelbolle lens.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(6), sfere_left(0), sfere_right(10), breaking_index(1.51), pos_x(5), show_gauge(false), instrument_name(lens)), l1 = lamp1(switch(true), angle(10), pos_y(0), pos_x(2), instrument_name(lamp1)) ])). question(4, 'pre1.14', 'De uittredende lichtstraal loopt evenwijdig met de hoofdas. Wanneer zal de uittredende lichtstraal meer naar boven schijnen?', [ 'Als je de hoek die de invallende lichtstraal met de hoofdas maakt vergroot.', 'Als je de lens naar rechts beweegt.', 'Als je de lens vervangt door een platbolle lens.' ], state(state, '', [ m6 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(10), sfere_right(10), breaking_index(1.51), pos_x(5), show_gauge(false), instrument_name(lens)), l2 = lamp1(switch(true), angle(10), pos_y(0), pos_x(2), instrument_name(lamp1)) ])). question(5, 'pre1.15', 'De uittredende lichtstraal loopt evenwijdig met de hoofdas. Wanneer zal de uittredende lichtstraal meer naar beneden schijnen?', [ 'Als je de hoek die de invallende lichtstraal met de hoofdas maakt verkleint', 'Als je de lens een stukje naar rechts beweegt.', 'Als je de lens vervangt door een holle lens.' ], state(state, '', [ m6 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(10), sfere_right(10), breaking_index(1.51), pos_x(5), show_gauge(false), instrument_name(lens)), l2 = lamp1(switch(true), angle(10), pos_y(0), pos_x(2), instrument_name(lamp1)) ])). question(6, 'pre2.11', 'Wanneer komt het snijpunt van de uittredende lichtstralen dichter bij de lens te liggen?', [ 'Als je de invallende lichtstralen meer naar beneden richt.', 'Als je de lamp een stukje naar rechts beweegt.', 'Als je de lens vervangt door een lens met een kleinere brandpuntsafstand.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(6), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(7), show_gauge(false), instrument_name(lens)), l2 = parlamp(switch(true), angle(0), separation(1), pos_x(2), pos_y(0), instrument_name(parlamp)) ])). question(7, 'pre2.12', 'Wanneer komt het snijpunt van de uittredende lichtstralen verder van de lens af te liggen?', [ 'Als je de invallende lichtstralen iets naar boven richt.', 'Als je de lamp een stukje naar links beweegt.', 'Als je de lens vervangt door een lens met een grotere brandpuntsafstand.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(10), show_gauge(false), instrument_name(lens)), l2 = parlamp(switch(true), angle(0), separation(1), pos_x(2), pos_y(0), instrument_name(parlamp)) ])). question(8, 'pre2.13', 'Wanneer komt het snijpunt van de uittredende lichtstralen boven de hoofdas te liggen?', [ 'Als je de invallende lichtstralen meer naar boven richt.', 'Als je de lamp een stukje naar beneden beweegt.', 'Als je de lens een stukje naar rechts beweegt.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(10), show_gauge(false), instrument_name(lens)), l2 = parlamp(switch(true), angle(0), separation(1), pos_x(2), pos_y(0), instrument_name(parlamp)) ])). question(9, 'pre2.14', 'Wanneer komt het snijpunt van de uittredende lichtstralen onder de hoofdas te liggen?', [ 'Als je de stralen van de invallende lichtstralen naar beneden richt.', 'Als je de lamp een stukje naar boven beweegt.', 'Als je de lens een stukje naar links beweegt.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(6), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(7), show_gauge(false), instrument_name(lens)), l2 = parlamp(switch(true), angle(0), separation(1), pos_x(2), pos_y(0), instrument_name(parlamp)) ])). question(10, 'pre3.21', 'Wanneer komt het snijpunt van de uittredende lichtstralen boven de hoofdas te liggen?', [ 'Als je de invallende lichtstralen naar boven richt.', 'Als je de lamp een stukje naar beneden beweegt.', 'Als je de lens een stukje naar rechts beweegt.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(20), show_gauge(true), instrument_name(lens)), l1 = lamp3(switch(true), angle(0), divergence(5), pos_x(3), pos_y(0), instrument_name(lamp3)) ])). question(11, 'pre3.22', 'Wanneer komt het snijpunt van de uittredende lichtstralen onder de hoofdas te liggen?', [ 'Als je de invallende lichtstralen naar boven richt.', 'Als je de lamp een stukje naar boven beweegt.', 'Als je de lens een stukje naar links beweegt.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(6), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(16), show_gauge(true), instrument_name(lens)), l4 = lamp3(switch(true), angle(0), divergence(5), pos_x(2), pos_y(0), instrument_name(lamp3)) ])). question(12, 'pre3.23', 'Wanneer komt het virtueel beeldpunt onder de hoofdas te liggen?', [ 'Als je de invallende lichtstralen meer naar beneden richt.', 'Als je lamp een stukje naar boven beweegt.', 'Als je de lamp een stukje naar beneden beweegt.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(-6), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(16), show_gauge(true), instrument_name(lens)), l5 = lamp3(switch(true), angle(0), divergence(5), pos_x(4), pos_y(0), instrument_name(lamp3)) ])). question(13, 'pre3.24', 'De lamp wordt 2 cm boven de hoofdas geplaatst. Hoever komt het beeldpunt van de hoofdas te liggen?', [ '2 cm', '4 cm', '8 cm' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(9), show_gauge(true), instrument_name(lens)), l1 = lamp3(switch(true), angle(0), divergence(5), pos_x(3), pos_y(0), instrument_name(lamp3)), d2 = ruler(from(centerline), to(m1), offset(-2)) ])). question(14, 'pre3.25', 'De lamp wordt 2 cm boven de hoofdas geplaatst. Hoever komt het beeldpunt van de hoofdas te liggen?', [ '2 cm', '3 cm', '4 cm' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(7.5), show_gauge(true), instrument_name(lens)), l1 = lamp3(switch(true), angle(0), divergence(5), pos_x(3), pos_y(0), instrument_name(lamp3)), d1 = ruler(from(centerline), to(m1), offset(-2.4)) ])). question(15, 'pre3.26', 'De lamp wordt 4 cm boven de hoofdas geplaatst. Hoever komt het virtueel beeldpunt van de hoofdas te liggen?', [ '1 cm.', '2 cm.', '4 cm.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(-6), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(10), show_gauge(true), instrument_name(lens)), l5 = lamp3(switch(true), angle(0), divergence(5), pos_x(4), pos_y(0), instrument_name(lamp3)), d8 = ruler(from(centerline), to(m1), offset(-2.1)) ])). question(16, 'pre3.27', 'Een lamp met een divergerende lichtbundel wordt geplaatst op de afgebeelde hulplijn. Wat wordt dan de beeldsafstand?', [ '4,5 cm.', '6 cm.', '9 cm.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(6), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(18.65), show_gauge(true), instrument_name(lens)), c3 = construction_line(orientation(vertical), value(0.65), instrument_name(consline)), d6 = ruler(from(c3), to(m1), offset(3.6)) ])). question(17, 'pre3.28', 'Een lamp met een divergerende lichtbundel wordt geplaatst op de afgebeelde hulplijn. Wat wordt dan de beeldsafstand?', [ '2,25 cm.', '4,5 cm.', '12 cm.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(3), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(10.85), show_gauge(true), instrument_name(lens)), c2 = construction_line(orientation(vertical), value(1.85), instrument_name(consline)), d2 = ruler(from(c2), to(m1), offset(3)) ])). question(18, 'pre3.29', 'Een lamp met een divergerende bundel wordt geplaatst op de afgebeelde hulplijn. Wat wordt dan de beeldsafstand?', [ '9 cm.', '-3 cm.', '-6 cm.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(6), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(13.75), show_gauge(true), instrument_name(lens)), c4 = construction_line(orientation(vertical), value(10.75), instrument_name(consline)), d7 = ruler(from(c4), to(m1), offset(3.2)) ])). question(19, 'pre3.30', 'Een lamp met een divergerende lichtbundel wordt geplaatst op de afgebeelde hulplijn. Wat wordt dan de beeldsafstand?', [ '-4,5 cm.', '-6 cm.', '12 cm.' ], state(state, '', [ m1 = lens(label(''), radius(5), thickness(0.1), focal_distance(-6), sfere_left(0), sfere_right(0), breaking_index(1.51), pos_x(22.05), show_gauge(true), instrument_name(lens)), c5 = construction_line(orientation(vertical), value(4.05), instrument_name(consline)), d1 = ruler(from(c5), to(m1), offset(2.85)) ])).
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Using TensorFlow backend. Obteniendo datos... [AVISO] Usuarios: 33537 [AVISO] Restaurantes: 5881 [AVISO] Cargando datos generados previamente... Creando modelo... ################################################## MODELV4 ################################################## modelv4d2 ################################################## 0/50 1/50 2/50 3/50 4/50 5/50 6/50 7/50 8/50 9/50 10/50 11/50 12/50 13/50 14/50 15/50 16/50 17/50 18/50 19/50 20/50 21/50 22/50 23/50 24/50 25/50 26/50 27/50 28/50 29/50 30/50 31/50 32/50 33/50 34/50 35/50 36/50 37/50 38/50 39/50 40/50 41/50 42/50 43/50 44/50 45/50 46/50 47/50 48/50 49/50 /usr/lib/python3/dist-packages/xlsxwriter/worksheet.py:833: DeprecationWarning: invalid escape sequence \w if re.match('\w:', url) or re.match(r'\\', url): /usr/lib/python3/dist-packages/xlsxwriter/worksheet.py:962: DeprecationWarning: invalid escape sequence \s if re.search('^\s', token) or re.search('\s$', token): /usr/lib/python3/dist-packages/xlsxwriter/worksheet.py:962: DeprecationWarning: invalid escape sequence \s if re.search('^\s', token) or re.search('\s$', token): /usr/lib/python3/dist-packages/xlsxwriter/worksheet.py:3610: DeprecationWarning: invalid escape sequence \| elif re.match('(or|\|\|)', conditional): /usr/lib/python3/dist-packages/xlsxwriter/worksheet.py:3840: DeprecationWarning: invalid escape sequence \. name = re.sub('\..*$', '', name) /usr/lib/python3/dist-packages/xlsxwriter/worksheet.py:5136: DeprecationWarning: invalid escape sequence \s if re.search('^\s', string) or re.search('\s$', string): /usr/lib/python3/dist-packages/xlsxwriter/worksheet.py:5136: DeprecationWarning: invalid escape sequence \s if re.search('^\s', string) or re.search('\s$', string): /usr/lib/python3/dist-packages/xlsxwriter/sharedstrings.py:103: DeprecationWarning: invalid escape sequence \s if re.search('^\s', string) or re.search('\s$', string): /usr/lib/python3/dist-packages/xlsxwriter/sharedstrings.py:103: DeprecationWarning: invalid escape sequence \s if re.search('^\s', string) or re.search('\s$', string): /usr/lib/python3/dist-packages/xlsxwriter/comments.py:160: DeprecationWarning: invalid escape sequence \s if re.search('^\s', text) or re.search('\s$', text): /usr/lib/python3/dist-packages/xlsxwriter/comments.py:160: DeprecationWarning: invalid escape sequence \s if re.search('^\s', text) or re.search('\s$', text): ---------------------------------------------------------------------------------------------------- N_FOTOS_TRAIN (>=) N_ITEMS %ITEMS RND-MOD AC CNT-MOD AC MODELO 9 1987 0.22846958721398183 0.10230984578418019 0.2385226560771158 0.16153024985961978 5 3522 0.4049672300793377 0.09442727273830982 0.21646069359661552 0.18546804169907952 4 4881 0.561228009658503 0.0890243310348925 0.20124543661136085 0.19463313742005212 2 6982 0.8028055651374038 0.08792472026371356 0.19823361684304652 0.21051694367396673 1 8697 1.0 0.08328745296547176 0.19270345670093156 0.22721703661762813 ---------------------------------------------------------------------------------------------------- 50 0.0308 14.2589 0.2965 0.2272
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...given ~~~~~~~~~~~~~~~~~~~~~~~~~~ ...expected
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ExampleA7.sce
clc clear //Page number 472 //Input data d=9000;//The density of copper in kg/m^3 w=63.5;//The atomic weight of copper in kg N=6.023*10^26;//Avogadros number pi=3.14;//Mathematical constant of pi h=6.624*10^-34;//Planks constant in Js //Calculations V=w/d;//The volume of copper in m^3 Ef=((h^2/(8*9*10^-31))*((3/pi)*(N/V))^(2/3))/(1.6*10^-19);//The fermi energy in eV P=(2/3)*(N/V)*Ef;//The pressure at absolute zero for copper in N/m^2 //Output printf('(a)The Fermi energy is E = %3.3f eV \n (b)The pressure at absolute zero for copper is P = %3.6g N/m^2 ',Ef,P)
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clc //Intitalisation of variables clear p= 27.17 //mm T= 99.5 //C T1= 100.5 //C T2= 100 //C sv1= 1674 //cc per gram sv2= 1.04 //cc per gram g= 980.7 //cm/sec^2 d= 13.595 //kg/m^3 //CALCULATIONS r= (p/10)*d*g lv= (273.2+T2)*(sv1-sv2)*(p/10)*d*g/(4.184*10^7) //RESULTS printf ('Heat of vapourisation of water = %.1f cal g^-1',lv)
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Chapter12_Example11.sce
//Chapter-12, Example 12.11, Page 518 //============================================================================= clc clear //INPUT DATA mc=4;//Mass flow rate of cold water in kg/s Tci=38;//Inlet Temperature of cold water in degree C Tco=55;//Outlet Temperature of cold water in degree C D=0.02;//Diameter of the tube in m v=0.35;//Velocity of water in m/s Thi=95;//Inlet Temperature of hot water in degree C mh=2;//Mass flow rate of hot water in kg/s L=2;//Length of the tube in m U=1500;//Overall heat transfer coefficient in W/m^2.K c=4186;//Specific heat of water in J/kg.K d=1000;//Density of water in kg/m^3 //CALCULATIONS Q=(mc*c*(Tco-Tci));//Heat transfer rate for cold fluid in W Tho=(Thi-(Q/(mh*c)));//Outlet temperature of hot fluid in degree C T=Thi-Tco;//Difference of temperature between hot water inlet and cold water outlet in degree C t=Tho-Tci;//Difference of temperature between hot water outlet and cold water inlet in degree C Tlm=((T-t)/log(T/t));//LMTD for counterflow heat exchanger A=(Q/(U*Tlm));//Area of heat exchanger in m^2 A1=(mc/(d*v));//Total flow area in m^2 n=((A1*4)/(3.14*D^2));//Number of tubes L=(A/(36*3.14*D));//Length of each tube taking n=36 in m N=2;//Since this length is greater than the permitted length of 2m, we must use more than one tube pass. Let us try 2 tube passes P=((Tco-Tci)/(Thi-Tci));//P value for calculation of correction factor R=((Thi-Tho)/(Tco-Tci));//R value for calculation of correction factor F=0.9;//Corrction Factor from Fig.12.9 on page no. 514 A2=(Q/(U*F*Tlm));//Total area required for one shall pass,2 tube pass exchanger in m^2 L1=(A2/(2*36*3.14*D));//Length of tube per pass taking n=36 in m //OUTPUT mprintf('Number of tubes per pass is %i \n Number of passes is %i \n Length of tube per pass is %3.3f m',n,N,L1) //=================================END OF PROGRAM==============================
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// Example 8.8, Page No-380 clear clc Vref=7.15 Vo=5 k=Vref/Vo printf('(R1b+R2)/R2= %.2f', k) k1=k-1 printf('\nR1 = %.2f * R2', k1) // For min voltage of 2V Vom=2 km=Vref/Vom printf('\n(R1a+R1b+R2)/R2= %.3f', km) km1=km-1.43 printf('\nR1a = %.3f * R2', km1) R1a=10000 R1b=2000 R2=R1a/2.145 R2n=R2/1000 printf('\nR2= %.2f kohm', R2n) R1=6000 R3=(R1*R2)/(R1+R2) R3n=R3/1000 printf('\nR3= %.2f kohm', R3n)
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//example 5.1 //design an open wellin fine sand clc; //given Q=0.003; //required discharge H=2.5; //depression head A=Q*3600/(0.5*H); d=(4*A/%pi)^0.5; d=round(d*100)/100 mprintf("Well diameter=%f m.",d); //Alternative solution C=7.5D-5; //permeability constant from table 5.2 A=Q/(C*H); d=(16*3/%pi)^0.5; d=round(d*10)/10; mprintf("\nBy alternative solution:") mprintf("\nWell diameter=%f m",d);
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function [stk,txt,top]=sci_diag() // Copyright INRIA txt=[] if rhs==2 then stk=list('diag('+stk(top-1)(1)+','+stk(top)(1)'+')','0','?','?','1') top=top-1 else if stk(top)(3)=='1' then stk=list('diag('+stk(top)(1)+')','0',stk(top)(4),stk(top)(4),'1') elseif stk(top)(4)=='1' then stk=list('diag('+stk(top)(1)+')','0',stk(top)(3),stk(top)(3),'1') else stk=list('diag('+stk(top)(1)+')','0','?','?','1') end end
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//Example 12.2 r1=0.9*10^-2;//Radius of garden hose (m) A1=%pi*r1^2;//Cross-sectional area of hose (m^2) Q=0.5;//Flow rate (L/s) Q=Q/10^3;//Flow rate (m^3/s) v1=Q/A1;//Speed of water in the hose (m/s) printf('a.Speed of water in the hose = %0.2f m/s',v1) r2=0.25*10^-2;//Radius of nozzle (m) A2=%pi*r2^2;//Cross-sectional area of nozzle (m^2) v2=A1*v1/A2;//Speed of water in the nozzle from continuity equation (m/s) printf('\nb.Speed of water in the nozzle = %0.1f m/s',v2) //Openstax - College Physics //Download for free at http://cnx.org/content/col11406/latest
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; @Harness: disassembler ; @Result: PASS section .text size=0x00000068 vma=0x00000000 lma=0x00000000 offset=0x00000034 ;2**0 section .data size=0x00000000 vma=0x00000000 lma=0x00000000 offset=0x0000009c ;2**0 start .text: label 0x00000000 ".text": 0x0: 0x0f 0x8c ldd r0, Y+31 ; 0x1f 0x2: 0x1f 0x8c ldd r1, Y+31 ; 0x1f 0x4: 0x2f 0x8c ldd r2, Y+31 ; 0x1f 0x6: 0x3f 0x8c ldd r3, Y+31 ; 0x1f 0x8: 0x4f 0x8c ldd r4, Y+31 ; 0x1f 0xa: 0x5f 0x8c ldd r5, Y+31 ; 0x1f 0xc: 0x6f 0x8c ldd r6, Y+31 ; 0x1f 0xe: 0x7f 0x8c ldd r7, Y+31 ; 0x1f 0x10: 0x8f 0x8c ldd r8, Y+31 ; 0x1f 0x12: 0x9f 0x8c ldd r9, Y+31 ; 0x1f 0x14: 0xaf 0x8c ldd r10, Y+31 ; 0x1f 0x16: 0xbf 0x8c ldd r11, Y+31 ; 0x1f 0x18: 0xcf 0x8c ldd r12, Y+31 ; 0x1f 0x1a: 0xdf 0x8c ldd r13, Y+31 ; 0x1f 0x1c: 0xef 0x8c ldd r14, Y+31 ; 0x1f 0x1e: 0xff 0x8c ldd r15, Y+31 ; 0x1f 0x20: 0x0f 0x8d ldd r16, Y+31 ; 0x1f 0x22: 0x1f 0x8d ldd r17, Y+31 ; 0x1f 0x24: 0x2f 0x8d ldd r18, Y+31 ; 0x1f 0x26: 0x3f 0x8d ldd r19, Y+31 ; 0x1f 0x28: 0x4f 0x8d ldd r20, Y+31 ; 0x1f 0x2a: 0x5f 0x8d ldd r21, Y+31 ; 0x1f 0x2c: 0x6f 0x8d ldd r22, Y+31 ; 0x1f 0x2e: 0x7f 0x8d ldd r23, Y+31 ; 0x1f 0x30: 0x8f 0x8d ldd r24, Y+31 ; 0x1f 0x32: 0x9f 0x8d ldd r25, Y+31 ; 0x1f 0x34: 0xaf 0x8d ldd r26, Y+31 ; 0x1f 0x36: 0xbf 0x8d ldd r27, Y+31 ; 0x1f 0x38: 0xcf 0x8d ldd r28, Y+31 ; 0x1f 0x3a: 0xdf 0x8d ldd r29, Y+31 ; 0x1f 0x3c: 0xef 0x8d ldd r30, Y+31 ; 0x1f 0x3e: 0xff 0x8d ldd r31, Y+31 ; 0x1f 0x40: 0x0f 0x8c ldd r0, Y+31 ; 0x1f 0x42: 0x07 0x8c ldd r0, Z+31 ; 0x1f 0x44: 0x0f 0xac ldd r0, Y+63 ; 0x3f 0x46: 0x08 0x80 ld r0, Y 0x48: 0x0f 0x8c ldd r0, Y+31 ; 0x1f 0x4a: 0x0f 0x84 ldd r0, Y+15 ; 0x0f 0x4c: 0x0f 0x80 ldd r0, Y+7 ; 0x07 0x4e: 0x0b 0x80 ldd r0, Y+3 ; 0x03 0x50: 0x09 0x80 ldd r0, Y+1 ; 0x01 0x52: 0x0c 0xac ldd r0, Y+60 ; 0x3c 0x54: 0x0e 0x8c ldd r0, Y+30 ; 0x1e 0x56: 0x0b 0xa8 ldd r0, Y+51 ; 0x33 0x58: 0x09 0x8c ldd r0, Y+25 ; 0x19 0x5a: 0x0c 0x84 ldd r0, Y+12 ; 0x0c 0x5c: 0x0e 0x80 ldd r0, Y+6 ; 0x06 0x5e: 0x0a 0xa4 ldd r0, Y+42 ; 0x2a 0x60: 0x0d 0x88 ldd r0, Y+21 ; 0x15 0x62: 0x0a 0x84 ldd r0, Y+10 ; 0x0a 0x64: 0x0d 0x80 ldd r0, Y+5 ; 0x05 0x66: 0x0a 0x80 ldd r0, Y+2 ; 0x02 start .data:
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//Example 2.2.8 page 2.15 clc; clear; Tf = 1400; //fictive temperature BETA = 7*10^-11; n= 1.46; //RI p= 0.286; //photo elastic constant Kb = 1.381*10^-23; //Boltzmann's constant lamda = 850*10^-9; //wavelength alpha_scat = 8*%pi^3*n^8*p^2*Kb*Tf*BETA/(3*lamda^4); l= 1000; //fibre length TL = exp(-alpha_scat*l); //transmission loss attenuation = 10*log10(1/TL); printf("The attenuation is %.3f dB/Km",attenuation);
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clc //initialization of new variables clear u=20 //km/h Cd=0.15 S=4 //m^2 r=1025 //kg/m^3 //calculations D=Cd*1/2*r*(u/3.6)^2*(S) P=D*u/3.6 //results printf('Drag force = %d N',D) printf('\n Power = %.2f KW',P/1000)
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clc; clear; R=50 //resistance in ohm C=25 //capacitance in micro-F L=0.15 //inductance in H V=230 //voltage in Volts f=50 //frequency in Hz //calculation XL=2*%pi*f*L //in ohm XC=(10^6)/(2*%pi*f*C) //in ohm X=XL-XC //in ohm Z=sqrt(R^2+X^2) I=V/Z pf=R/Z power_consumed=V*I*pf mprintf("(i)Impedance = %2.1f ohm\n",Z) //The answers vary due to round off error mprintf("(ii)Current = %1.2f A\n",I) //The answers vary due to round off error mprintf("(iii)Power Factor = %1.2f (Lead)\n",pf) mprintf("(iv)Power Consumed = %d W",power_consumed) //The answers vary due to round off error
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//CHAPTER 6- MAGNETIC CIRCUITS //Example 4 clc; disp("CHAPTER 6"); disp("EXAMPLE 4"); //VARIABLE INITIALIZATION di=10; //diameter of iron ring in cm dr=1.5; //diameter of iron rod in cm mui=900; //relative permeability of rod mu0=4*%pi*10^(-7); //absolute permeability in Henry/m lg=5/10; //length of air-gap in cm N=400; //number of turns I=3.4; //current through the winding in Amperes //SOLUTION li=(di*%pi)-lg; //length of iron path area=((dr^2)*%pi)/4; //area of iron cross-section //solution (a) mmf=(4*%pi*N*I)/10; //in gilberts, since 1 AT=(4*pi)/10 mmf=round(mmf); //to round off the value disp(sprintf("(a) MMF is %d Gilberts",mmf)); //solution (b) //tot reluctance = iron reluctance + air gap reluctance(mur=1 for air) totR=(li/(area*mu0*mui))+(lg/(area*mu0*1)); disp(sprintf("(b) The total reluctance is %E Gilberts/Maxwell",totR)); //solution (c) phi=mmf/totR; disp(sprintf("(c) The flux in the circuit is %f Maxwell",phi)); //solution (d) b=phi/area; disp(sprintf("(d) The flux density in the circuit is %f Gauss",b)); //Answers of (b), (c) & (d) are different because absolute permeability is not included in (b) //END
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2021-01-01T11:10:20.053388
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Koza_3.tst
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// Data Reconciliation Benchmark Problems From Lietrature Review // Author: Edson Cordeiro do Valle // Contact - edsoncv@{gmail.com}{vrtech.com.br} // Skype: edson.cv // Authors //Dovì, V G, and C Solisio. 2001. Reconciliation of censored measurements in chemical processes: an alternative approach. Chemical Engineering Journal 84, no. 3 (December): 309-314. http://www.sciencedirect.com/science/article/B6TFJ-45KNHB1-F/2/199f358469628f600f10b394d2b55a8b. //Bibtex Citation //@article{Dovi2001, //annote = { The importance of considering the censoring of measured data in the reconciliation of process flow rates has been shown in a previous paper [Chem. Eng. Sci. 52 (17) (1997) 3047]. The purpose of the present paper is to introduce a new technique for carrying out the actual reconciliation procedure and compare its significance and performance with those of previous methods. A numerical example shows how nontrivial differences are to be expected.}, //author = {Dov\`{\i}, V G and Solisio, C}, //isbn = {1385-8947}, //journal = {Chemical Engineering Journal}, //keywords = {Censored data,Data reconciliation,Detection limits}, //month = dec, //number = {3}, //pages = {309--314}, //title = {{Reconciliation of censored measurements in chemical processes: an alternative approach}}, //url = {http://www.sciencedirect.com/science/article/B6TFJ-45KNHB1-F/2/199f358469628f600f10b394d2b55a8b}, //volume = {84}, //year = {2001} //} // 6 Streams // 3 Equipments getd('../functions/'); getd('../jacobians/'); clear xr sd sds x_sol xfinal jac jac_col jac_col rj sigma sigam_inv res V V_inv diag_diag_V Wbar gama zr_nt adj zadj Wbar_alt adjustability detect resi Qglr betaglr xchiglr ge_glr op_glr; clear avti_gt_mt op_gt_mt op_gt_nt_tmp avt1_mt1 avt1_mt2 op_mt1 op_mt2 avti_glr op_glr_mt aee_mt aee_nt_tmp op_glr_nt_tmp avti_glr_nt_tmp avti_gt_mt_tmp op_gt_mt_tmp op_gt_nt avt1_nt1 avt1_nt2 op_nt1 op_nt2 avti_glr_tmp op_glr_mt_tmp aee_mt_tmp aee_nt op_glr_nt avti_glr_nt; //stacksize('max'); tic; xr =[11;10;21;11;5.5;5.5]; szx = size(xr,1); runsize = 1000; //the variance proposed by this work sd =[0.032 0.026 0.120 0.033 0.052 0.015].^(0.5); sds = sd; var=sd.^2; jac=jacP2(); rj=rank(jac); jac_col = size(jac,2); jac_row = size(jac,1); sigma=diag(sds.^2); [adj, detect, V, V_inv, sigma_inv, diag_diag_V, Wbar] = adjust(sigma, jac); bias_min = 2; bias_max = 7; leak_min = 0.02; leak_max = 0.07; bias_multi_error = [0 0 1 1 0 0 ]; leak_multi_error = [ 0 1 1]; sum_bias_and_leak = 0; [xfinal, resRand, resGrossErrorNodalRand]=generate_data(xr, sd, jac, runsize, bias_min, bias_max, leak_min, leak_max); [xfinal2, resRand2, resGrossErrorNodalRand2]=generate_data_multiple(xfinal(1:runsize,:), sd, jac, 2, 10, 0.5, 0.9,bias_multi_error,leak_multi_error, sum_bias_and_leak); xfinal = xfinal2; resGrossErrorNodalRandFi2 = [ resRand2;resGrossErrorNodalRand2]; //observability/redundancy tests //user can set unmeasured streams here, if this vector is empty, all streams are measured umeas_P2 = []; [red_P2, just_measured_P2, observ_P2, non_obs_P2, spec_cand_P2] = qrlinclass(jac,umeas_P2); measured_P2 = setdiff([1:length(xr)], umeas_P2); red = measured_P2;// // to run robust reconciliation,, one must choose between the folowing objective functions to set up the functions path and function parameters: //WLS analytical = -1 WLS numerical = 0 ; Absolute sum of squares = 1 ; Cauchy = 2 ;Contamined Normal = 3 ; Fair = 4 //Hampel = 5 Logistic = 6 ; Lorenztian = 7 ; Quasi Weighted = 8 // run the configuration functions with the desired objective function type obj_function_type = -1; [x_sol] = calc_results_DR(xfinal2, jac, sigma, resGrossErrorNodalRandFi2, obj_function_type); [res, gamaMeasuremts,gamaNodal,zr_nt_nodal, zr_nt_nodal_rand, zadj ] = calc_results_index(x_sol, jac, sigma, resGrossErrorNodalRandFi); // user must implemment the multiple gross error routines bellow with (or without the indexes above calculated) // global test for multi GE [avti_gt_mt, op_gt_mt, op_gt_nt] = global_test_multi(0.1, 0.1, gamaMeasuremts, runsize, rj, jac_col, jac_row); runtime=toc(); //streamNames =generateStreamName(szx); //prettyprinttable([tokens(streamNames), string([xr, rrn(4,sd), rrn(3,adj), rrn(3,detect), rrn(3,op_mt1), rrn(3,op_mt2), rrn(3,op_glr_mt), rrn(7,aee_mt)])],"latex") //eqpNames = generateEqpName('', jac_row); //prettyprinttable([tokens(eqpNames), string([rrn(3,op_nt1), rrn(3,op_nt2), rrn(3,op_glr_nt), rrn(7,aee_nt)])],"latex") //[ op_gt_mt avti_gt_mt avt1_mt1 avt1_mt2 avti_glr avt1_nt1 avt1_nt2 avti_glr_nt runtime ] //prettyprinttable(string([rrn(3,avt1_mt1), rrn(3,avt1_mt2), rrn(3,avti_glr), rrn(3,avt1_nt1), rrn(3,avt1_nt2), rrn(7,avti_glr_nt)])) //[rrn(3,op_mt1), rrn(3,op_mt2), rrn(3,op_glr_mt), rrn(7,aee_mt)] //[rrn(3,op_nt1), rrn(3,op_nt2), rrn(3,op_glr_nt), rrn(7,aee_nt)] // ////saving results //aa = clock(); //nowtime = '_' + string(aa(4)) + '-'+ string(aa(5)); //save ('P_resumed_' + date() + nowtime +'.sav', runtime, adj, detect, op_nt1, op_nt2, avt1_nt1, avt1_nt2, op_mt1, op_mt2, avt1_mt1, avt1_mt2, op_gt_mt, op_gt_nt, avti_gt_mt, op_glr_mt, op_glr_nt, avti_glr, avti_glr_nt, aee_nt, aee_mt); //
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clc // Given that lambda = 6000 // wavelength of light in angstrom e = 0.3 // Width of slit in mm m = 1 // Order for first dark band n = 3/2 // Order for first bright band // Sample Problem 8 on page no. 143 printf("\n # PROBLEM 8 # \n") printf(" Standard formula used \n") printf(" lambda = e*sin(theta) \n") theta_d = m*asin(lambda*1e-10/(e*1e-3)) // Calculation of angle in radian theta_b = n*asin(lambda*1e-10/(e*1e-3)) // Calculation of angle in radian printf("\n First dark band is formed at angle %e rad. \n First bright band is formed at angle %e rad.",theta_d,theta_b)
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clc //Given that m = 2e-3 // mass of linear harmonic oscillator in kg k = 100 // spring constant in N/m h = 6.6e-34 // Plank's constant //Sample Problem 15 page No. 142 printf("\n\n\n # Problem 15 # \n") printf("\n Standard formula Used \n f = sqrt(k / m ) \n U = 1/2* h * nu ") nu = sqrt(k / m ) / (2 * %pi) //calculation of frequency of linear harmonic oscillator U = 1/2* h * nu //calculation of Zero point energy of a linear harmonic oscillator printf ("\n Zero point energy of a linear harmonic oscillator is %e J.", U )
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// Example 1.11 page no-33 clear clc l=2 //cm D=18 //cm s=0.5 //cm //(a) va1=500 //volts ds1=l*D/(2*s*va1)//Deflection Sensitivity //(b) va2=1000 //Volts ds2=l*D/(2*s*va2) //(c) va3=1500 //Volts ds3=l*D/(2*s*va3) printf("\n(a)Va=%dV\nDeflection Sensitivity S_E=%.3f cm/V \n\n(b)Va=%dV\nDeflection Sensitivity S_E=%.3f cm/V\n\n(c)Va=%dV\nDeflection Sensitivity S_E=%.3f cm/V",va1,ds1,va2,ds2,va3,ds3)
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scenario = "One-Back"; # This name is recorded in the log file scenario_type = trials; response_matching = simple_matching; no_logfile = true; active_buttons = 3; button_codes = 1, 2, 3; # These values will be used to code participant responses default_font_size = 56; default_font = "Arial"; default_background_color = 0,0,0; #Black# default_text_color = 255,255,255; #White# begin; #Thank you screen at end of experiment trial{ trial_duration=forever; trial_type=specific_response; terminator_button = 3; picture{ text{font_size = 20; caption = "Ende des Experiments Vielen Dank!"; }; x = 0; y = 0; }; time = 0; }thankyou; begin_pcl; thankyou.present(); output_file rating = new output_file;
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/1682/CH8/EX8.10/Exa8_10.sce
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2018-02-03T05:31:52
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Exa8_10.sce
//Exa 8.10 clc; clear; close; //Given data : p1=(100-96)/100;//unitless p2=(96-89)/100;//unitless p3=(89-68)/100;//unitless p4=(68-37)/100;//unitless p5=(37-13)/100;//unitless p6=(13-0)/100;//unitless N0=1000;//no. of resistors N1=N0*p1;//no. of resistors N2=N0*p2+N1*p1;//no. of resistors N3=N0*p3+N1*p2+N2*p1;//no. of resistors N4=N0*p4+N1*p3+N2*p2+N3*p1;//no. of resistors N5=N0*p5+N1*p4+N2*p3+N3*p2+N4*p1;//no. of resistors N6=N0*p6+N1*p5+N2*p4+N3*p3+N4*p2+N5*p1 ;//no. of resistors //Calculation of individual replacement cost Life=0;//in months p=[p1 p2 p3 p4 p5 p6];//Unitless for i=1:6 Life=Life+i*p(i); end disp(Life,"Expected life of each transistor in months : ") disp(round(1000/Life),"Average No. of failures/month : "); disp(round(1000/Life)*10,"Therefore, cost of individual replacement in Rs. : ") //Calculation of group replacement cost disp("Cost/transistor when replaced simultaneously = Rs. 4"); disp("Cost/transistor when replaced individually = Rs. 10"); disp("The cost of group replacement policy for several replacement periods are summarized in Table 8.7. This table can be seen from the book."); disp("From table it is clear that the avg cost/month is minimum for the 3rd month. Hence, the group replacement period is 3 months."); disp("Individual replacement cost/month = Rs. 2480"); disp("Minimum group replacement ost/month = Rs. 2426.67"); disp("Since the min group replacement cost/month is less than the individual replacement cost/month, the group replacement policy is the best, and hence all the transistors should be replaced in 3 months.")