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//Example 10.15 //Fourth Order Runge Kutta Method //Page no. 324 clc;clear;close; deff('y=f(x,y)','y=x^2+y^2') y=1;h=0.1; for i=1:2 x=(i-1)*h K1=h*f(x,y); K2=h*f(x+h/2,y+K1/2); K3=h*f(x+h/2,y+K2/2); K4=h*f(x+h,y+K3); disp(K4,'K4 =',K3,'K3 =',K2,'K2 =',K1,'K1 =') y=y+(K1+2*K2+2*K3+K4)/6 printf('\ny(%g) = %.13f\n\n\n\n',x+h,y) end
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* * This file was put into the public domain 2016-11-29 * by John P. Hartmann. You can use it for anything you like, * as long as this notice remains. * *Testcase privopisk sysclear archmode z loadcore "$(testpath)/privop.core" *Program 1 runtest .1 *Done *Testcase privopiske *Program 2 runtest program .1 *Done *Testcase privopgo r 151=00 runtest program .1 *Compare gpr *Gpr 0 0000000000000f06 r 88.4 *Want 00020000 *Done
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clear// //Variables W = 64000 //Heat produced (in Joules) t = 40 //time (in seconds) //Calculation P = W/t //Rate at which electrical energy is converted into heat energy (in watt) //Result printf("\n The rate at which electrical energy is converted into heat energy is : %0.3f W.",P)
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//Example 3.2// rCu=0.128;//nm //atomic radius copper (From appendix 2) a=(4/sqrt(2))*rCu mprintf("a = %f nm",a) //The density of the unit cells is a1=4;// atoms b1=63.55;//gram //atomic mass of copper c1=0.6023*10^24;//atoms// Avogardo's number d=10^7;//nm/cm p=(a1/a^3)*(b1/c1)*d^3 mprintf("\n p = %f g/cm^3",p)
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clear; clc; p1=acos(-1); format(16) a=0; b=%pi / 2; function [f]=my(x) f = sqrt(x+1)*sin(x)*cos(x); endfunction X=linspace(a,b,100); for i=1:100 [Y(i)]=my(X(i)); end figure(1); plot(X,Y,'k'); xlabel('x'); ylabel('y'); n=43; xT=linspace(a,b,n); hT=xT(2)-xT(1); for i=1:n [yT(i)]=my(xT(i)); end IT(1)=yT(1); IT(n)=yT(n); S=0; for i=2:n-1 IT(i)=2*yT(i); S=S+IT(i); end TRAP=hT*(IT(1)+S+IT(n))/2; printf('TRAP=%f \n ', TRAP)
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clc(); clear; //To determine the Q-value mLi=7.016004; //mass of Lithium(A=7) mH=1.007825; //mass of Hydrogen(A=1) mHe=4.002603; //mass of helium(A=4) Q=[mLi+mH-2*(mHe)]*931.5 //Q is the energy balance of the reaction p=0.5; //energy of proton in MeV //The energy of 2 alpha particles is equal to the Q-value + energy of proton. Ealpha=(Q+p)/2 //energy of each alpha particle printf("The Q-value of the reaction is %f MeV and energy of each alpha particle is %f MeV",Q,Ealpha)
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:: XCOM Version 4.03 :: :: (1) Call Request Packet : Replace calling address by 440821000010 :: :: (2) Call Accept Packet : Replace called address by 440821000010 :: :: (3) Calledmaping function : :: a) If called address of call request packet have sub address, :: 4406200008XX convention is used. :: b) If sub address is not given, :: only 4406 is submit and enable TKSUP option. :: :: PATCH(881215,1300,NIS.TSUJI,PA0PTR,,0C) WS 0 :Word boundary NISADR XC 0C440821000010 :Length & Address WS 0 : NIS4406 XC 044406 :Length & DNIC of NIS WS 0 : CONPATCH(CRQ300-42,,6) J PA1PTR,, CONPATCH(PA1PTR,,14) JAL R9,VENCLG,, :Put address LB R0,PFXCLL, :Get calling address length LB R8,PFXCLD, :Get called address length J CRQ300-3A,, :return CONPATCH(ESP912-48,,6) J PA1PTR,, CONPATCH(PA1PTR,,14) JAL R9,VENCLG,, :Put address LB R0,PFXCLL, :Get calling address length LB R8,PFXCLD, :Get called address length J ESP912-40,, :return CONPATCH(MCA020-1A,,6) J PA1PTR,, CONPATCH(PA1PTR,,14) JAL R9,VENCLD,, :Put address LB R5,PFXCLD,R6, :Get called address length LR R7,R5 : From source J MCA020-14,, :return CONPATCH(PA1PTR,,1A) VENCLG L R0,NISADR :Get NIS's address on Venus-p ST R0,PFXCLL, :Store into calling address LH R0,NISADR+4 : STH R0,PFXCLL+4, : LB R0,NISADR+6 : STB R0,PFXCLL+6, : JR R9 : CONPATCH(PA1PTR,,1A) VENCLD L R5,NISADR :Get NIS's address on Venus-p ST R5,PFXCLD, :Store into called address LH R5,NISADR+4 : STH R5,PFXCLD+4, : LB R5,NISADR+6 : STB R5,PFXCLD+6, : JR R9 : CONPATCH(FND2ND+1A,,6) J PA1PTR,, CONPATCH(PA1PTR,,28) LB R4,REMBUF :Get length of remaining digits buffer JEFS NOSUBA :Skip if no digits remains AR R4,R6 :Add two length up J FND2ND+20,, :Return back NOSUBA LH R0,NIS4406 :Get NIS's DNIC number STH R0,DTESAX,R7, :Store it LB R0,NIS4406+2 : STB R0,DTESAX+2,R7, : J CLDEXI,, : ENDPATCH(Replace calling/called address by NIS's address on Venus-P)
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//Ex:13.7 clc; clear; close; r=12;//in ohms i=0.5;//in amps P_r=i*i*r;//in W printf("Power radiated = %d W",P_r);
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clc w1=1 //Assigning values to parameters w2=2*w1 t=atan(sqrt(3)*(w2-w1)/(w1+w2)) pf=cos(t) disp(pf,"Power factor is")
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//Hougen O.A., Watson K.M., Ragatz R.A., 2004. Chemical process principles Part-1: Material and Energy Balances(II Edition). CBS Publishers & Distributors, New Delhi, pp 504 //Chapter-2, illustration 4, Page 36 //Title: Expressing weight percent into mole percent //============================================================================= clear clc //INPUT W = 100; //Weight of solution in grams(Basis of calculation) w1 = 40; //Weight of sodium carbonate present in solution in grams MW = [106,18.02]; //Molecular weight of sodium carbonate and water respectively in g/g-mole //CALCULATION n1 = w1/MW(1); //To find the no of moles of sodium carbonate in g mole n2 = (W-w1)/MW(2); //To find the no of moles of water in g mole N = n1+n2; //Calculation of total no of moles in g mole x1 = n1*100/N; //Mole % of sodium carbonate x2 = n2*100/N; //Mole % of water //OUTPUT mprintf('\n mole percent of Na2CO3 = %4.2f',x1); mprintf('\n mole percent of H2O = %3.1f',x2); //================================END OF PROGRAM===============================
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// Prob 1.17 clc; clear; close; format('v',6); // Given data : Poles=4;//no. of poles m=3;//no. of phase f=50;//in Hz V=7000;//in volt/phase I=1400;//in A/phase Xs=1.2;//in ohm/phase E=sqrt(V^2+(I*Xs)^2);//in volt disp(E,"Induced emf in volt : "); cosfi=1;//for resistive load P=3*V*I*cosfi;//in watts P=P/10^6;//in MWatts N=120*f/Poles;//in rpm w=2*%pi*N/60;//in radian per sec T=P*10^6/w;//in Nw-m T=T/9.81;//in Kg-m disp(T,"Torque in Kg-m : "); //Note : Answers in the book is not as much accurate as calculated by Scilab. //Note : Figure given in this question is not a plot. It is just drawn to represent data and can't be plotted.
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//Exa 6.25 clc; clear; close; format('v',7); //Given Data : p1=10;//bar x1=0.9;//dryness p2=1;//bar hf1=762.6;//KJ/Kg(at 10bar) hfg1=2013.6;//KJ/Kg(at 10bar) h1=hf1+x1*hfg1;//KJ/Kg h2=h1;//KJ/Kg hg2=h2;//KJ/Kg p2=0.075;//bar(from steam table) disp(p2,"Pressure at exit in bar : "); //Steam table is used to get some data.
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clear;lines(0); s=poly(0,'s'); derivat(1/s) // -1/s^2;
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A=[1 2 3; 4 5 6; 7 8 9] // 3x3 matrix (3 rows, 3 columns) A(2,3) // value stored at row 2, column 3 A(2,3)=-1 // modify the value at row 2 column 3 A(4,5)=10 // this assignment increases the size of A // entry that doesn't exist in A A(10,10) // this call returns error 21
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clear// //Variable Declaration d=150 //Depth of the web in mm wf=100 //Width of the flange in mm df=20 //Depth of the flange in mm t=20 //Thickness of the web in mm //Calculations y_bar=10**-3*(((wf*df*(d+df*0.5))+(d*t*d*0.5))/(wf*df+d*t)) //Distance of Neutral Axis in m //Simplfying the computation a=wf*df**3*12**-1 b=wf*df*((d+df*0.5)-y_bar*10**3)**2 c=t*d**3*12**-1 f=t*d*((d*0.5)-y_bar*10**3)**2 I=(a+b+c+f)*10**-12 //Moment of inertia in mm^3 //Limit Moment yp=(wf*df+d*t)/(2*t) //Plastic Neutral Axis in mm Myp=I/y_bar //Yielding will start at moment without the stress term to ease computation mom=10**-9*((t*yp**2*0.5)+(wf*df*(d-yp+10))+(t*25**2*0.5)) //Sum of 1st moments Ml_Myp=mom*Myp**-1 //Ratio //Result printf("\n The ratio ML/Myp= %0.3f ",Ml_Myp)
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//bad recursion function y=Fibonacci1(n) if n<=1 then y=1 else y=Fibonacci1(n-1)+Fibonacci1(n-2) end endfunction tic() Fibonacci1(25) time=toc() //matrix structure use function y=Fibonacci2(n) F=[0 1; 1 1] u=(F^n)*[1;1] y=u(1) endfunction tic() Fibonacci2(25) time=toc()
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clc; Vcc=10; //volt R1=18000; //ohm R2=4700; //Ohm Vb=(R2/(R1+R2))*Vcc; //volt //voltage divider rule Ve=Vb-0.7; //volt Re=1100; //ohm Icq=Ve/Re; //Ampere//assumption Icq=Ie Rc=3000; //Ohm Re=1100; //Ohm Vceq=Vcc-Icq*(Rc+Re); //Volt disp('A',Icq,"Icq=");//The answers vary due to round off error disp('V',Vceq,"Vceq=");//The answers vary due to round off error
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 13.8w //calculation of the angle that the plank makes with the vertical in equilibrium //given data l=1//length(in m) of the planck h=0.5//height(in m) of the water level in the tank s=0.5//specific gravity of the planck //calculation //A = OC/2 = l/(2*cosd(theta) // mg = 2*l*rho*g //buoyant force Fb=(2*l*rho*g)/cosd(theta) //m*g*(OB)*sind(theta) = F(OA)*sind(theta) theta=acosd(sqrt(1/2)) printf('the angle that the plank makes with the vertical in equilibrium is %d degree',theta)
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clc; // page no 200 // prob no 6.3.2 // RC phase shift oscillator // all resistors are in Kohm f=800;R0=18; // R>>Ro should be chosen to minimize the effect of Ro on frequency. A number of values for R can be tried, and it will be found that R=100Kohm is reasonable. R=100; c=1/(2*%pi*f*R*sqrt(6+(4*R0/R)))*10^9;// C in pF disp('pF',c,+'The value of capacitor is ');
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// Group_13: Bhargava B // Surya K // S K Savant // Question: //Compute and plot the cross-correlation rxy (m), 0<= m<= 99. //Use the plot to estimate the value of the delay D. exec("../2prob/2_delay_noise_seq.sce",-1) exec("../1prob/sinusoidal_vector.sci",-1) exec("../1prob/normalnoisevec.sci",-1) function[delay,corrvec]=cross_correlation(xseq,yseq,M) newseq=[] for i=1:length(yseq) if (i<=0 | i>length(xseq)) then newseq=[newseq,0] else newseq=[newseq,xseq(i)] end end N=length(newseq) corrvec=[] for m=-99:99 corrval=0 for n=1:200 if (n-m>0 & n-m<=200) then corrval=corrval+newseq(n)*yseq(n-m) //disp(n,n-m) end end corrvec=[corrvec,corrval] end maxf=0 index=i nvec=linspace(-99,99,199) for i=1:length(corrvec) if(corrvec(i)>maxf) then maxf=corrvec(i) index=i end end //disp(nvec(index)) //disp(corrvec) delay=-nvec(index) disp('Delay is') disp(delay) //disp(length(corrvec)) //disp(length(nvec)) //disp(length(corrvec)) xset('window',5) plot(nvec,corrvec) endfunction //xvec=sinusoidalvec(0.1,200) xvec=barker yvec=delay_noise_seq(xvec,0.001,0.9,20,-1) cross_correlation(xvec,yvec,100)
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// f = ab ^ x // ln(f) = ln(a) + xln(b) // ln(f) = a0 + x * a1 // a = e^a0 // b = e^a1 function xr = mmqPolinomial(x, y, grau) for i = 1 : (grau + 1) for j = 1 : (grau + 1) A(i, j) = sum( x .^ (j + i - 2)) end end for i = 1 : (grau + 1) b(i) = sum(y .* (x .^ (i - 1))) end xr = A\b endfunction function xr = vandermondMatrix(x, y, grau) n = length(x) for i = 1 : n for j = 1 : grau A(i,j) = x(i) ^ (j - 1) end b(i) = y(i) end xr = (A'*A)\(A'*b) endfunction
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// Lecture de l'image img = readpbm('A1.pbm'); //Attribution de la taille de l'image [colonne, ligne]=size(img) // Déclaration & initialisation d'une variable pour déterminer // la valeur maximale de gris (le plus haut niveau de gris) max_gray=max(img); // Déclaration & initialisation d'un compteur pour déterminer // le nombre de coordonnées de zones d'atterrissage possibles. coordonnées=0; // Déclaration & initialisation de deux tableaux pour consigner // les coordonnées de la d'atterrissage possibles. posX= [0,0] // Déclaration & initialisation d'un deuxième compteur. i=1 // Affichage de texte. disp("Les coordonnées optimales pour atterrissage robot ou fusée sont : ") // Boucle pour parcourir l'image. for c=1:colonne for l=1:ligne // Condition pour déterminer si le pixel choisi possède la valeur // attendue. if img(c,l)==max_gray // Affichage de texte. disp("(" + string(c) + "," + string(l) +") ;") // Incrémentation du compteur. coordonnées=coordonnées+1 posX(i)=(c,l) i=i+1 end end end // Affichage de texte. disp("Il y a " + string(coordonnées) + " position possible.")
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clc //Initialization of variables K2=1.0008*10^-14 //m^2 K1=1.754*10^-5 //m c=0.1 //calculations disp("Neglecting x w.r.t c,") x2=c*K2/K1 x=sqrt(x2) //results printf("Concentration of OH minus ions = %.1e m",x)
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3 1.Яке з твердження правильне? +а)Фігура F1 називається подібною до фігури F (), якщо існує відображення фігури F на фігуру F1, при якому для будь-яких двох точок А і В фігури F та їх образів А1 і В1 фігури F1 відношення відстаней АВ і А1В1 є величиною сталою. б)Фігура F1 називається подібною до фігури F (), якщо існує відображення фігури F на фігуру F1, при якому для будь-яких двох точок А і В фігури F та їх образів А1 і В1 фігури F1 відношення відстаней АВ і А1В1 є величиною не сталою. в)Фігура F1 називається подібною до фігури F (), якщо існує відображення фігури F1 на фігуру F, при якому для будь-яких двох точок А і В фігури F та їх образів А1 і В1 фігури F1 відношення відстаней АВ і А1В1 є величиною сталою. 2.У подібних фігур відповідні кути рівні, а відповідні відрізки пропорційні, це ? +а)правда б)брехня в)спірне питання 3.Який з прикладів найкраще описує цю тему? +а)явище тіні в реальному житті б)закон гепарда: якщо не наздогнав за 100м, то вже не цікаво в)українська народна пісня "Ой у лузі червона калина"
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function y = fitness(hydrozoan) y = 0; for i=1:size(hydrozoan,2)-1 y = y + (hydrozoan(i)-1)^2 + 100*(hydrozoan(i+1) - hydrozoan(i)^2)^2; end endfunction D = 2; N = 2; low = 0; high = 1; Iteration = 5; H = zeros(N,D); for i=1:N for j=1:D H(i, j) = rand()*(high-low) + low; end end growth_low = 0.01; growth_high = 0.1; while(Iteration > 0) Fit = []; for i = 1:N Fit = [Fit; fitness(H(i,:))]; end G = []; for i=1:N G(i) = rand()*(growth_high-growth_low) + growth_low; end Med = median(G); Swarm = zeros(N); Swarm = G - Med; Split = zeros(N); Min = min(Swarm); Max = max(Swarm); for i = 1:N if(Swarm(i) < 0) then Split(i) = 0; elseif (Swarm(i) == 0) then Split(i) = 1; elseif (Swarm(i) > 0 & Swarm(i) == Min) then Split(i) = 1; elseif (Swarm(i) > 0 & Swarm(i) == Max) then Split(i) = 3; else Split(i) = 2; end end Clone = []; for i = 1:N for j = 1:Split(i) Clone = [Clone; H(i,:)]; end end Min = 1e-9; Max = 1e-7; for i = 1:size(Clone,1) for j = 1:size(Clone,2) RP = (Max - Min)*rand() + Min; Clone(i,j) = Clone(i,j)*(1 + RP); end end Fit = []; for i = 1:size(Clone, 1) Fit = [Fit; fitness(Clone(i,:))]; end sum_of_fitness = 0; for i = 1:size(Clone ,1) sum_of_fitness = sum_of_fitness + Fit(i); end Probability = []; for i = 1:size(Clone ,1) Probability = [Probability; (Fit(i)/sum_of_fitness)]; end max_probability = max(Probability); counter = zeros(size(Clone,1),1); n_select = 1000; index = -1; for i=1:n_select while(1) index = ceil(size(Clone,1)*rand()); if (rand() < Probability(index)) break; end end counter(index) = counter(index)+1; end par1 = -1; par2 = -1; first_max = 0; second_max = 0; for i=1:size(Clone,1) if(first_max < counter(i)) second_max = first_max; par2 = par1; par1 = i; first_max = counter(i); elseif(second_max < counter(i)) second_max = counter(i); par2 = i; end end swap_probability = 0.5; for i=1:D if(swap_probability < rand()) swap_val = Clone(par1,i); Clone(par1,i) = Clone(par2,i); Clone(par2,i) = swap_val; end end for i=1:size(Clone, 1) Clone(i,1+ceil(rand()*D)) = low + rand()*(high-low); end Fit = []; for i=1:size(Clone, 1) Fit = [Fit; fitness(Clone(i,:))]; end gsort(Fit); bestcount = 2; Fit_threshold = Fit(bestcount); for i=1:size(Clone, 1) if(fitness(Clone(i,:)) == Fit_threshold) Ibest = Clone(i,:); break; end end for i=1:size(Clone, 1) if(fitness(Clone(i,:)) >= Fit_threshold) Clone(i, :) = Ibest; end end H = Clone; Iteration = Iteration-1; end //disp(Ibest); disp(fitness(Ibest));
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clear; clc; close; Vm = 20; //volts Vdc = -0.318*(Vm-0.7); //volts disp(Vdc,'Dc voltage for silicon diode :'); t = 0:0.1:4*%pi; x = (20-0.7)*sin(t); for i=1:length(t) if(x(i)<=0) y(i) = x(i); else y(i)=0 end end plot(t,y); xtitle('output for silicon diode','t','Vo');
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EXAMPLE6_10.SCE
//ANALOG AND DIGITAL COMMUNICATION //BY Dr.SANJAY SHARMA //CHAPTER 6 //NOISE clear all; clc; printf("EXAMPLE 6.10(PAGENO 305)"); //given F_1 = 2//noise figure of first stage in dB A_1 = 12//gain in first stage in dB F_2 = 6//noise figure of second stage in dB A_2 = 10//gain in first second in dB //calculations F_1ratio = exp((F_1/10)*log(10));//noise figure of first stage in ratio F_2ratio = exp((F_2/10)*log(10));//noise figure of second stage in ratio A_1ratio = exp((A_1/10)*log(10));//gain of first stage in ratio A_2ratio = exp((A_2/10)*log(10));//gain of second stage in ratio F = F_1ratio + ((F_2ratio - 1)/(A_1ratio));//Overall noise figure F_dB = 10*log10(F);//Overall noise figure in dB //results printf("\n\nOverall noise figure = %.2f dB",F_dB );
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//inverse systems z=%z; H=(1+2*(z^(-1)))/(1+3*(z^(-1))); //inverse of H is H1=1/H H=1+2*(z^(-1))+3*(z^(-2)); H1=1/H
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//Exa 7.10 clc; clear; close; //given data : ID_on=5;//in mA VGS=6;//in Volt VGS_on=8;//in Volt VGST=4;//in Volt K=ID_on/(VGS_on-VGST)^2;//in mA/V^2 ID=K*(VGS-VGST)^2;//in mA disp(ID,"When VGS=6V the drain current in mA : ");
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showMatchedFeatures.sce
//************************************************************************** //output-> shows the error on console //-> means that output is correct and no modification needs to be done //************************************************************************** I1 = imread("left.jpg"); I2 = imread("right.jpg"); //I3 = showMatchedFeatures(I1,I2); //-> output is correct //I3 = showMatchedFeatures(I2,I); //I variable hasn't been defined //output->Undefined variable: I //I3 = showMatchedFeatures(I2); //output->showMatchedFeatures: Wrong number of input argument(s): 2 expected. //I3 = showMatchedFeatures(I2,3); //output-> API Error: // in getListItemAddress: Unable to get address of item #2 in argument #2 // in getListItemNumber: Invalid argument type, list expected //I3 = showMatchedFeatures(); //output-> showMatchedFeatures: Wrong number of input argument(s): 2 expected.
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Ch08Ex10.sce
// Scilab code Ex8.10 Page:250 (2006) clc; clear; a0 = 5.3; // Bohr radius, nm rs_a0_ratio = 3.93; // Ratio of solid radius to the lattice parameter chi_Pauli = 2.59/rs_a0_ratio; // Pauli's spin susceptibility, cgs units printf("\nThe Pauli spin susceptibility for Na in terms of free electron gas parameter = %4.2f", chi_Pauli); // Result // The Pauli spin susceptibility for Na in terms of free electron gas parameter = 0.66
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3_7.sce
// example:-3.7,page no.-101. // NOTE:-this example is a method for calculating unknown load impedence of slotted line section.all data are given and preassumed. // program to determine unknown load impedence. clc clear exec("DEPENDENCIES/smith_chart_tao.sci") Zl=0;Zo=50; // for short circuitting the load. SWR=%inf; // short circuit is removed and replace with unknown load. SWR=1.5;lamda=0.04; lmin=4.2-2.72; tao=(1.5-1)/(1.5+1); theta=(%pi+((4*%pi)/4)*1.48); tao=abs(tao)*exp(%i*theta); Zl=50*((1+tao)/(1-tao)); // result disp(Zl,'load impedence = ') smith_chart(tao) // when analyse with the help of smith chart.see the angle from x=0 axis i.e Tao_real axis.if it is above this axis take angle anticlockwise and if it is below this axis.take angle clockwise from Tao_real axis below.
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rc2poly11.sce
//check o/p when negative i/p vector is passed to the function as i/p X = [-7 -6 -5 -8 -3 -6]; r=[1]; [a, efinal] = rc2poly(X,r); disp(a) disp(efinal) //output //1. 147. 4780. 21630. - 17375. - 777. - 6. // // 7.112D+08
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ex8_3.sce
//Solutions to Problems In applied mechanics //A N Gobby clear all; clc //initialisation of variables b=3*6^3/12//in^4 d=b+3*6*6^2//in^4 b2=%pi*2^4/64//in^4 h=b2+%pi*1^2*6^2//in^4 //CALCULATIONS P=d-h//in^4 //RESULTS printf('the rectangular plate with circular hole=% f in^4',P)
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Chapter4_Example7.sce
clc clear //Input data N=5000//Number of lines drawn on the grating per m w=(5890*10^-10)//Wavelength of the light used in m //Calculations n=(1/(w*N*100))//Order of spectrum x=ceil(n)//Rounding off to next integer //Output printf('Since n < %i, it is not possible to observe the fourth or higher order of diffraction',x)
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Ex21_6.sce
errcatch(-1,"stop");mode(2);//Ex21_6 Pg-1070 C=0.01*10^(-6) //capacitance in farad f0=2000 //frequency in Hz Req=1.45/(f0*C) //equivalent resistance or R1+R2 disp(" Because a square wavw has duty cycle of 50% each resistor must be the same") R1=Req/2 R2=R1 printf(" R1 = R2 = %.2f kohm",R2*1e-3) exit();
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Decision Tree_Demo1.sce
// Demo for decision tree -- Scilab getd('../macros') // Data preparation M = csvRead('Datasets/forestfires.csv') x = M(:,[5,6,7,8,9]); y = M(:, 13); y(or(isnan(x),'c'),:) = [] x(or(isnan(x),'c'),:) = [] n = length(y(:, 1)) for i = 1:n if(y(i)>0) y(i) = 1 end end [questions,flag] = decisionTreeFit(x, y); pred = decisionTreePredict(x, questions, flag); disp(0.5*norm(pred' - y))
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Ex18_4.sce
//Example 18.4// n=8;//numbers Ni2+/ unit cell n1=2; //moment of Ni2+ m=n*n1 mprintf("m = %i ",m) a=18.4;// measured value of nickel ferrite e=((a-m)/a)*100 mprintf("\ne = %i percent",e)
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// Algoritmo Genético (configurado para maximização) // Autor: Denis Ricardo da Silva Medeiros clear; // Parâmetros do AG TAM_POP = 40; NUM_GER = 30; TAXA_CROSS = 0.8; TAXA_MUT = 0.04; L_MIN = -500; L_MAX = 500; QNT_BITS = 16; ELITISMO = 0.02; DIMENSOES = 2; //rand('seed', 2330099) //RESPOSTA = X perto de 440 e Y perto de -500. // Função de avaliação. function saida = fa(xn) x = xn(1); y = xn(2); z=-x.*sin(sqrt(abs(x)))-y.*sin(sqrt(abs(y))); x = x/250; y = y/250; r = 100*(y-x.^2).^2+(1-x).^2; r1 = (y-x.^2).^2+(1-x).^2; w = r .* z; w2 = z - r1; w6 = w + w2; saida = -w6; endfunction // ################################################################# // // Calcula o ganho normalizado. GANHO_NORM = (L_MAX - L_MIN)/(2^QNT_BITS - 1); // Cria a população inicial com strings. pop_bin = []; for d=1:DIMENSOES pop_bin_t = []; for i = 1:TAM_POP pop_bin_t = [pop_bin_t; strcat(string(round(rand(1, QNT_BITS))))]; end pop_bin = [pop_bin pop_bin_t]; end pop_dec = bin2dec(pop_bin); nova_pop_bin = pop_bin; aptidao = zeros(TAM_POP, 1); melhores = zeros(NUM_GER, 1); media = zeros(NUM_GER, 1); // Para o relatório primeira = L_MIN + GANHO_NORM * pop_dec;; intermediaria1 = primeira; intermediaria2 = primeira; ultima = primeira; // Inicia o processamento das gerações. for i = 1:NUM_GER // Normaliza os indivíduos. pop_norm = L_MIN + GANHO_NORM * pop_dec; if i == 3 then intermediaria1 = pop_norm; elseif i == 10 then intermediaria2 = pop_norm; end // Avalia os indivíduos. for j=1:TAM_POP aptidao(j) = fa(pop_norm(j,:)); end // Garante que todas as aptidões sejam positivas. aptidao = aptidao - min(aptidao); aptidao = aptidao + 0.01*max(aptidao); aptidao_acc = cumsum(aptidao); // Armazena o melhor da geração atual. [aptidao_ord, indices_ord] = gsort(aptidao); melhores(i) = aptidao_ord(1); medias(i) = mean(aptidao); // Faz a seleção dos indivíduos através do método da roleta. for j = 1:2:TAM_POP // Faz o primeiro giro da roleta. valor = aptidao_acc(TAM_POP) * rand(1); for k1 = 1:TAM_POP if valor < aptidao_acc(k1) then break; end end // Faz o segundo giro da roleta. valor = aptidao_acc(TAM_POP) * rand(1); for k2 = 1:TAM_POP if valor < aptidao_acc(k2) then break; end end // //Faz a seleção por torneio, com 2 indivíduos. // inds = 1 + floor(rand(2, 1)*TAM_POP); // if aptidao(inds(1)) > aptidao(inds(2)) then // k1 = inds(1); // else // k1 = inds(2); // end // // inds = 1 + floor(rand(2, 1)*TAM_POP); // if aptidao(inds(1)) > aptidao(inds(2)) then // k2 = inds(1); // else // k2 = inds(2); // end // Realiza o cruzamento dos dois indivíduos selecionados // com base na taxa de cruzamento. filho1 = pop_bin(1, :); filho2 = pop_bin(1, :); for d=1:DIMENSOES // Testa se passa da taxa de cruzamento. if rand(1) < TAXA_CROSS then // Define o ponto de corte. pos = 1 + floor((QNT_BITS-1)*rand(1)); // Faz o cruzamento (quebra as strings e depois as une). partes1 = strsplit(pop_bin(k1, d), pos); partes2 = strsplit(pop_bin(k2, d), pos); filho1(1, d) = strcat([partes1(1), partes2(2)]); filho2(1, d) = strcat([partes2(1), partes1(2)]); else // Se não passou na taxa de cruzamento, passa os indivíduos // diretamente para a próxima população. filho1(1, d) = pop_bin(k1, d); filho2(1, d) = pop_bin(k2, d); end end // Adiciona os novos filhos na nova população. nova_pop_bin(j,:) = filho1; nova_pop_bin(j+1,:) = filho2; end // Operação de mutação. for j=1:TAM_POP for d=1:DIMENSOES if rand(1) < TAXA_MUT then // Encontra o bit a ser mutado. pos = 1 + floor(QNT_BITS*rand(1)); // Verifica se este bit é 1 ou 0 para alterar seu valor. bit = part(nova_pop_bin(j,d), pos); if bit == '1' then nova_pop_bin(j, d) = strcat([part(nova_pop_bin(j, d), 1:pos-1), '0', part(nova_pop_bin(j, d), pos+1:QNT_BITS)]) else nova_pop_bin(j, d) = strcat([part(nova_pop_bin(j, d), 1:pos-1), '1', part(nova_pop_bin(j, d), pos+1:QNT_BITS)]) end end end end // Aplica o elitismo. inds_elite = round(ELITISMO*TAM_POP); nova_pop_bin(1:inds_elite) = pop_bin(indices_ord(1:inds_elite)); // Substitui a população antiga. pop_bin = nova_pop_bin // Gera a população de decimais. pop_dec = bin2dec(pop_bin) end // Obtém o melhor indivíduo. pop_norm = L_MIN + GANHO_NORM * pop_dec; ultima = pop_norm; for j=1:TAM_POP aptidao(j) = fa(pop_norm(j,:)); end [_, indice] = max(aptidao); resposta = pop_norm(indice, :); //disp(pop_norm); disp(['Resposta: ', string(resposta)]); //erros = abs(fa(melhores_norm) - RESULTADO); // Gráfico das aptidões. function _ = plotar_aptidoes() clf(); plot([1:NUM_GER]', melhores, '*-'); plot([1:NUM_GER]', medias, 'go-'); legend(['Melhor aptidão'; 'Média das aptidoẽs']); xlabel("Gerações"); ylabel("Aptidão"); title("Convergência da resposta"); //grafico = gca() ; //grafico.box="on"; //grafico.data_bounds=[0, X_MIN; NUM_GER, X_MAX]; //define the bounds endfunction // Gráficos das populações. function _ = plotar_populacoes() scf(); plot(primeira(:,1), primeira(:,2), 'ob'); legend(['Indivíduos']); xlabel("X"); ylabel("Y"); title("População inicial"); grafico = gca() ; grafico.box = "on"; grafico.data_bounds=[-500, -500; 500, 500]; scf(); plot(intermediaria1(:,1), intermediaria1(:,2), 'or'); legend(['Indivíduos']); xlabel("X"); ylabel("Y"); title("População intermediária (após 3 gerações)"); grafico = gca() ; grafico.box = "on"; grafico.data_bounds=[-500, -500; 500, 500]; scf(); plot(intermediaria2(:,1), intermediaria2(:,2), 'om'); legend(['Indivíduos']); xlabel("X"); ylabel("Y"); title("População intermediária (após 10 gerações)"); grafico = gca() ; grafico.box = "on"; grafico.data_bounds=[-500, -500; 500, 500]; scf(); plot(ultima(:,1), ultima(:,2), 'og'); legend(['Indivíduos']); xlabel("X"); ylabel("Y"); title("População final"); grafico = gca() ; grafico.box = "on"; grafico.data_bounds=[-500, -500; 500, 500]; endfunction // Gráfico da função de avaliação. function plotar_fa() [x, y] = meshgrid(-500:5:500,-500:5:500); z = -x.*sin(sqrt(abs(x)))-y.*sin(sqrt(abs(y))); x = x/250; y = y/250; // r: Rosenbrock's function r = 100*(y-x.^2).^2+(1-x).^2; r1 = (y-x.^2).^2+(1-x).^2; w = r .* z; w2 = z - r1; w6 = w + w2; x = x * 250; y = y * 250; surf(x, y, w6); endfunction
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P2_Derivation_of_torque_uniform_wear.sce
clc //Example 14.2 //Derivation of torque capacity for one pair of surfaces subjected to uniform wear //------------------------------------------------------------------------------ //This example is derivation based, hence the code will comprise only of statements printed to text file //Printing result file to .txt res2=mopen(TMPDIR+'2_derivation_of_torque_uniform_wear.txt','wt') mfprintf(res2,"When a clutch is new, the pressure may be rather uniform.\nIf the surfaces are relatively rigid, the outer portion, where velocity is high,\nwill wear more than inner portion.\n") mfprintf(res2,"After the initial wearing-in,it is reasonable to assume that the curve of the profile will maintain its shape;\nor, the wear thereafter may be considered to be uniform.\n") mfprintf(res2,"Uniform wear can be expressed in a different way by saying that\nat any time interval, the work done per unit area is constant:\n") mfprintf(res2,"[(frictional force)*(velocity)]/area = \n\t[(f*p*2*pi*r dr)*(rw)]/(2*pi*r dr) = constant C\n") mfprintf(res2,"or\n\tp=C''/f*r*w\t Since f and w are constant,") mfprintf(res2,"\n\tp=C/r, where C is constant.\n") mfprintf(res2,"An alternate method of showing that pressure varies inversely as the radius is\nto consider that wear (delta) is proportional to pressure p and velocity V.\n") mfprintf(res2,"Thus \n\t(delta)=K*p*V=K*p*(r*w),\nor\n\tp=C/r\t\tsince (delta) and K are constants and w is fixed for a given clutch.\n") mfprintf(res2,"The differential frictional torque=dT= r(f*p(2*pi*r dR));\n\n") mfprintf(res2,"Intergrating with respect to r over r=Ri to r=Ro,\nwe get the total torque as\n") mfprintf(res2,"\t\tT=2*pi*f*C[((Ro^2)-(Ri^2))/2]\n\n") mfprintf(res2,"To find C, we can integrate p(2*pi*r dr) with respect to r over r=Ri to r=Ro\n") mfprintf(res2,"We get\n\t\tC=F/(2*pi*(Ro-Ri))\n\n") mfprintf(res2,"Substituting this value of C into T\n") mfprintf(res2,"We obtain\n\n") mfprintf(res2,"\t\tT=F*f*[(1/2)*(Ro+Ri)] = F*f*Rf") mclose(res2) editor(TMPDIR+'2_derivation_of_torque_uniform_wear.txt') //------------------------------------------------------------------------------ //-------------------------------End of program---------------------------------
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function [SQNR,xq,en_code] = uniform_pcm(x,L) //x = input sequence //L = number of qunatization levels xmax = max(abs(x)); xq = x/xmax; en_code = xq; d = 2/L; q = d*[0:L-1]; q = q-((L-1)/2)*d; for i = 1:L xq(find(((q(i)-d/2)<= xq)&(xq<=(q(i)+d/2))))=... q(i).*ones(1,length(find(((q(i)-d/2)<=xq)&(xq<=(q(i)+d/2))))); en_code(find(xq == q(i)))= (i-1).*ones(1,length(find(xq == q(i)))); end xq = xq*xmax; SQNR = 20*log10(norm(x)/norm(x-xq)); endfunction
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//Example 21.3 //Trapezoidal Rule and Simpsons Rule in Parallel Computing //Page no. 726 clc;close;clear; n=8;a=0;b=8; h=(b-a)/n deff('y=f(x)','y=1/(1+x)') for i=0:8 x(i+1)=i; y(i+1)=f(x(i+1)) end printf('xi\t ') for i=1:9 printf('%i\t ',x(i)) end printf('\n yi\t') for i=1:9 printf('1/%i\t',i) end //trapezoidal rule S=0; for i=1:9 if(i==1 | i==9) S=S+y(i) else S=S+2*y(i) end end S=S*h/2 printf('\n\nTrapezoidal Rule Sum = %g',S) //Simpsons 1/3rd Rule S=0; for i=1:9 if(i==1 | i==9) S=S+y(i) elseif(((i)/2)-fix((i)/2)==0) S=S+4*y(i) else S=S+2*y(i) end end S=S*h/3 printf('\n\nSimpsons 1/3rd Rule Sum = %g',S)
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//Example 10_33 //Find the inverse Z-transform clc; clear; z=poly(0,'z'); x=ldiv((z+1),(z-1/3),4); disp(x,'x[n]=');
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// Example 9_9 clc;funcprot(0); // Given data gamma=0.500;// The specific heat ratio for air T_in=70.0;// °F p_in_psig=[0.000,20.00,40.00,60.00,80.00,100.00,120.00,140.00];// psig p_in=[14.7,34.7,54.7,74.7,94.7,114.7,134.7,154.7];// psia T_hot=[70.0,119.0,141.0,150.0,156.0,161.0,164.0,166.0];// °F T_cold=[70.0,19.5,-3.00,-14.0,-22.0,-29.0,-34.0,-39.0];// °F T_r=[1.000,1.209,1.315,1.368,1.406,1.441,1.465,1.487];// Note:T_r=(T_hot+460)/(T_cold+460) p_e=14.7;// The exit pressure in psia R=0.0685;// Btu/(lbm.R) c_p=0.240;// Btu/(lbm.R) // Calculation Sdot_pbymdot_3_1=((c_p*log(((T_r(1)^gamma)/(1+(gamma*(T_r(1)-1))))))+(R*log(p_in(1)/p_e)));// Btu/(lbm.R) Sdot_pbymdot_3_2=((c_p*log(((T_r(2)^gamma)/(1+(gamma*(T_r(2)-1))))))+(R*log(p_in(2)/p_e)));// Btu/(lbm.R) Sdot_pbymdot_3_3=((c_p*log(((T_r(3)^gamma)/(1+(gamma*(T_r(3)-1))))))+(R*log(p_in(3)/p_e)));// Btu/(lbm.R) Sdot_pbymdot_3_4=((c_p*log(((T_r(4)^gamma)/(1+(gamma*(T_r(4)-1))))))+(R*log(p_in(4)/p_e)));// Btu/(lbm.R) Sdot_pbymdot_3_5=((c_p*log(((T_r(5)^gamma)/(1+(gamma*(T_r(5)-1))))))+(R*log(p_in(5)/p_e)));// Btu/(lbm.R) Sdot_pbymdot_3_6=((c_p*log(((T_r(6)^gamma)/(1+(gamma*(T_r(6)-1))))))+(R*log(p_in(6)/p_e)));// Btu/(lbm.R) Sdot_pbymdot_3_7=((c_p*log(((T_r(7)^gamma)/(1+(gamma*(T_r(7)-1))))))+(R*log(p_in(7)/p_e)));// Btu/(lbm.R) Sdot_pbymdot_3_8=((c_p*log(((T_r(8)^gamma)/(1+(gamma*(T_r(8)-1))))))+(R*log(p_in(8)/p_e)));// Btu/(lbm.R) Sdot_pbymdot_3=[Sdot_pbymdot_3_1,Sdot_pbymdot_3_2,Sdot_pbymdot_3_3,Sdot_pbymdot_3_4,Sdot_pbymdot_3_5,Sdot_pbymdot_3_6,Sdot_pbymdot_3_7,Sdot_pbymdot_3_8];// Btu/(lbm.R) plot(p_in_psig,Sdot_pbymdot_3); xlabel('Inlet pressure(psig)'); ylabel('Sdot_p/mdot_3(Btu/lbm.R)'); xtitle('Sdot_p/mdot_3 vs. inlet pressure for a vortex tube'); disp('Remaining Results for Example 9.9'); disp('The entropy production rate per unit mass flow rate for each pressure shown'); disp('Inlet pressure psig'); disp(p_in_psig); disp('T_1/T_2'); disp(T_r); disp('Sdot_P/mdot_3 Btu/(lbm⋅R)'); disp(Sdot_pbymdot_3);
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clc(); clear; //Given : RBE = 0.7 ; //RBE factor for cobalt 60 gamma rays dose = 1000 ; // dose in rad e = RBE*dose; // equivalent dose in rem printf("Equivalent dose is %d rem",e);
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G32KOFS1.TST
[program] revize=32KOFS136601Z21 dat_akt=2012-01-19 [files] @(GINADR)@\KOF\FRM 00000478.ALV s=5015 c=49411 00000AUR.XME s=2903 c=31131 000002K8.ALF s=5349 c=50419 000001NV.ALV s=-2 00000A56.XME s=-2 00000060.ALF s=-2 00000060.ZIP s=-2 000000DZ.ALV s=-2 000002BT.ALV s=-2 00000A9T.XME s=-2 000000FQ.ALF s=-2 000002R6.ALV s=-2 000002R8.ALV s=-2 000001PJ.ALV s=-2 00000A5E.XME s=-2 0000006A.ALF s=-2 0000006A.ZIP s=-2 000000DY.ALV s=-2 000000E0.ALV s=-2 0000008H.ALV s=-2 0000008D.ALV s=-2 00000211.ALV s=-2 000000AF.ALF s=-2 000002QA.alf s=-2 00000A6R.XME s=-2 000000XJ.ALV s=-2 0000000X.alf s=-2 00000013.alf s=-2 00000018.alf s=-2 0000013I.alf s=-2 0000013J.alf s=-2 000002A3.alf s=-2 000002A4.alf s=-2 000002A5.alf s=-2 000002A6.alf s=-2 000002A7.alf s=-2 00000A0U.xme s=-2 @(GINADR)@\KOF G32KOFS1.TST
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test_cache_2_b.tst
total_physical_store_count = 0 total_physical_fetch_count = 40 <cache info> items cached: 20/20 regions in use: 4/4 items per region: 10 counter: 40 temp_read_num: 39 temp_write_num: -1 item_req_count = 40 item_hit_count = 0 item hit ratio = 0% <cache region: 0-9> item_cost: 0 flush_cost: 0 counter_total: 0 most_recently_used: -1 least_recently_used: -1 most_recently_changed: -1 least_recently_changed: -1 most_recently_unchanged: -1 least_recently_unchanged: -1 <cache region: 10-19> item_cost: 0 flush_cost: 0 counter_total: 0 most_recently_used: -1 least_recently_used: -1 most_recently_changed: -1 least_recently_changed: -1 most_recently_unchanged: -1 least_recently_unchanged: -1 <cache region: 20-29> item_cost: 10 flush_cost: 0 counter_total: 255 most_recently_used: 9 least_recently_used: 0 most_recently_changed: -1 least_recently_changed: -1 most_recently_unchanged: 9 least_recently_unchanged: 0 <cache region items> item #20, chg: 0, counter: 21, used (-1, 1), chg (-1, -1), unchg (-1, 1) item #21, chg: 0, counter: 22, used (0, 2), chg (-1, -1), unchg (0, 2) item #22, chg: 0, counter: 23, used (1, 3), chg (-1, -1), unchg (1, 3) item #23, chg: 0, counter: 24, used (2, 4), chg (-1, -1), unchg (2, 4) item #24, chg: 0, counter: 25, used (3, 5), chg (-1, -1), unchg (3, 5) item #25, chg: 0, counter: 26, used (4, 6), chg (-1, -1), unchg (4, 6) item #26, chg: 0, counter: 27, used (5, 7), chg (-1, -1), unchg (5, 7) item #27, chg: 0, counter: 28, used (6, 8), chg (-1, -1), unchg (6, 8) item #28, chg: 0, counter: 29, used (7, 9), chg (-1, -1), unchg (7, 9) item #29, chg: 0, counter: 30, used (8, -1), chg (-1, -1), unchg (8, -1) <cache region: 30-39> item_cost: 10 flush_cost: 0 counter_total: 355 most_recently_used: 9 least_recently_used: 0 most_recently_changed: -1 least_recently_changed: -1 most_recently_unchanged: 9 least_recently_unchanged: 0 <cache region items> item #30, chg: 0, counter: 31, used (-1, 1), chg (-1, -1), unchg (-1, 1) item #31, chg: 0, counter: 32, used (0, 2), chg (-1, -1), unchg (0, 2) item #32, chg: 0, counter: 33, used (1, 3), chg (-1, -1), unchg (1, 3) item #33, chg: 0, counter: 34, used (2, 4), chg (-1, -1), unchg (2, 4) item #34, chg: 0, counter: 35, used (3, 5), chg (-1, -1), unchg (3, 5) item #35, chg: 0, counter: 36, used (4, 6), chg (-1, -1), unchg (4, 6) item #36, chg: 0, counter: 37, used (5, 7), chg (-1, -1), unchg (5, 7) item #37, chg: 0, counter: 38, used (6, 8), chg (-1, -1), unchg (6, 8) item #38, chg: 0, counter: 39, used (7, 9), chg (-1, -1), unchg (7, 9) item #39, chg: 0, counter: 40, used (8, -1), chg (-1, -1), unchg (8, -1)
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Example_6_1.sce
//Example 6.1 clear; clc; //To calculate the number of ways of distributing distinguishable molecules a,b,c between 3 energy levels w=(3*2*1)/(1*1*1);//ways of distributing distinguishable molecules a,b,c between 3 energy levels mprintf('ways of distributing distinguishable molecules a,b,c between 3 energy levels = %i',w); //end
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//Ex:63 clc; clear; close; h=35800;//height in km r=6364;//earth's radius in km r_o=r+h;//orbital radius in km i=2;//angle of inclination in degree w_m=0.0175; y_m=i;//max latitude deviation d_m=r_o*i*(3.14/180);//max displacement due to latitude deviation in km D_m=d_m*(w_m/y_m);//max displacement due to longitude deviation in km printf("max displacement due to latitude deviation=%d km",d_m); printf("\n max displacement due to longitude deviation=%f km",D_m);
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18_2_1.sce
clc; // page no 671 // prob no 18_2_1 //A drum of facsimile machine with diameter=70.4mm & scanning pitch=0.2mm/scan D=70.4;P=0.2; //Determination of index of co-operation IOC_CCITT=D/P; IOC_IEEE=IOC_CCITT*(%pi); disp(IOC_IEEE,'The index of co-operation is');
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Ex14_4.sce
// Exa 14.4 // Refer circuit 14.25 given on page no. 484 clc; clear all; // Given data E=10;// Volts R=50;// Unstrained gauge resistance(Ohms) Gain=100;// Amplifier gain Vo=1.5;// Output Voltage // Solution // Using the formula: Vo=E*(Delta_R/R)*gain Delta_R=Vo*R/(E*Gain);// Change in resistance printf('The change in resistance =%.2f Ohms\n This means that Rt1 and Rt3 decrease by 0.07 ohms \n and Rt2 and Rt4 increase by 0.07 ohms when a certain weight is placed on the scale platform\n',Delta_R); // The answer mentioned in the textbook is incorrect as R=50 Ohms and not 100 Ohms.
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q13.sce
clear stacksize(268435454) exec("q3a5.sci") exec("ressources_q11aq14.sci") //Question 13 res_dich = dichotomie(J, 1e-5, -l, l) disp(res_dich)
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clc(); clear; // To calculate the lag of thermometer used in initial example while the oven is heating r = 0.01; // Radius of cylindrical tube in ft a = 0.178; // Thermal diffusivity in ft^2/hr k = 5; // Thermal conductivity in Btu/hr-ft-F h = 2; // Heat transfer coefficient in Btu/hr-ft^2-F s = 400; // Rate of temperature change tlag = r*k*s/(2*a*h); printf("The lag of thermometer while the oven is heating at the rate of 400F/hr is %.1f F",tlag);
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Ex3_15.sce
clc //to calculate order when visible light of wavelength in the range 4000 to 7500 angstrom //let E=(e+d) E=1/4000 //in cm lambda1=4*10^-5 //wavelength in cm lambda2=7.5*10^-5 n1=E*sin(%pi/2)/lambda1 n2=E*sin(%pi/2)/lambda2 disp("order when wavelength of 4000 angstrom is n1="+string(n1)+"unitless") disp("order when wavelength of 7500 angstrom is n2="+string(n2)+"unitless")
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vibrotact_test.sce
################################################################################################################################### # Test script for the fMRI compatible piezoelectric vibratory stimulation system build by Mag Design & Engineering # # Tests all stimulators individually with all available frequencies # # # # Script written by Jasper Wijnen # # Requested by Yael Salzer # # TOP, University of Amsterdam, February 2016 # # Using NBS Presentation 18.2 # ################################################################################################################################### default_font = "calibri"; default_font_size = 26; begin; text{caption="stimulator test";}test1_txt; text{caption="testing";}test2_txt; picture{ text test1_txt;x=0;y=300; text test2_txt;x=0;y=0; }test_pic; begin_pcl; output_port outport = output_port_manager.get_port(1); #LPT data I/O register, switches individual stims on/off array <int> portcodes [8] = {1,2,4,8,16,32,64,128}; dio_device freq_ctrl = new dio_device(memory_dio_device,890,4); #LPT control I/O register, controls vibratory frequency #value 890 is dec conversion from hexadecimal adress 0378, this is the I/O range for LPT port on adress D050 + 2 #if LPT adress is different for your PC you should change this value array <int> freqcodes [16]; freqcodes.fill(1,0,0,1); array <int> freqs [16];freqs.fill(1,0,30,30); loop int f=1; until f>16 begin freq_ctrl.write(1,freqcodes[f]); loop int x=1 until x>portcodes.count() begin test2_txt.set_caption("testing stim: " + string(x) + "\nport code: " + string(portcodes[x]) + "\n f value: " + string(freqcodes[f]) + "\nsupposed frequency: " + string(freqs[f]),true); test_pic.present(); outport.send_code(portcodes[x],150); wait_interval(200); x=x+1; end; f=f+1; end;
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18_18.sce
//ques-18.18 //Calculating entropy change and free energy change of the reaction clc T1=300; T2=330;//temperature (in K) G1=-16;//free energy change (in kcal) H=-10;//enthalpy change (in kcal) S=(H-G1)/T1;//entropy change (in kcal/K) G2=H-T2*S;//free energy change (in kcal) printf("The entropy change is %d cal/K and free energy change at 330K is %.1f kcal.",S*1000,G2);
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Bc1=30*10^3;cimin1=18 Bc2=25*10^3;cimin2=14 Bc3=12.5*10^3;cimin3=12 Bc4=6.25*10^3;cimin4=9 Y=4//path propogation constant BcI=6.25*10^3 cieq1=cimin1+20*log10(Bc1/BcI) cieq2=cimin2+20*log10(Bc2/BcI) cieq3=cimin3+20*log10(Bc3/BcI) cieq4=cimin4+20*log10(Bc4/BcI) disp(cieq1,'(C/I)eq in dB for system I') disp(cieq2,'(C/I)eq in dB for system II') disp(cieq3,'(C/I)eq in dB for system III') disp(cieq4,'(C/I)eq in dB for system IV') if cieq1<cieq2 then if cieq1<cieq3 then if cieq1<cieq4 then disp(,'System I offers the best capacity') end end elseif cieq2<cieq3 then if cieq2<cieq4 then if cieq2<cieq1 then disp(,'System II offers the best capacity') end end elseif cieq3<cieq4 then if cieq3<cieq1 then if cieq3<cieq2 then disp(,'System II offers the best capacity') end end elseif cieq4<cieq3 then if cieq4<cieq1 then if cieq4<cieq2 then disp(,'System IV offers the best capacity') end end end
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~BivLCM-SR-bfas_co_hrz_col_d-PLin-VLin.tst
THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.285986D+00 2 -0.365888D-02 0.232624D-02 3 0.295377D-01 0.192935D-04 0.283314D+00 4 0.435274D-03 0.292983D-03 -0.468214D-02 0.236014D-02 5 0.127699D-02 -0.152202D-03 -0.158820D-02 0.280635D-03 0.315068D-02 6 -0.103394D-02 0.104635D-03 0.305555D-03 -0.987353D-04 0.215378D-03 7 0.788336D-03 -0.386811D-04 0.163429D-02 -0.660310D-04 0.934584D-03 8 0.192105D-03 0.152746D-04 0.455220D-03 0.100884D-03 -0.318100D-03 9 -0.156285D+00 0.102618D-01 -0.107058D+00 0.285500D-01 0.739509D-01 10 -0.107776D-01 -0.107176D-01 -0.320989D-01 0.792110D-02 0.143253D+00 11 -0.132683D+00 0.158078D-01 0.380396D-01 0.884639D-02 -0.383237D-01 12 0.145225D+00 0.647701D-03 -0.856904D+00 0.471203D-01 0.794258D-01 13 0.767782D-01 -0.445828D-02 0.529905D-01 -0.634074D-02 0.429555D-01 14 -0.205047D+00 0.740541D-02 -0.489296D+00 0.166168D-01 -0.133521D-01 15 -0.149684D+01 -0.365183D-01 -0.219210D+00 0.891186D-02 -0.118137D+00 16 -0.333545D-01 -0.773625D-03 0.114798D-01 -0.192731D-03 -0.115389D-02 17 0.895769D-02 0.219728D-03 0.266597D-04 -0.824447D-04 -0.314929D-03 18 -0.242470D+00 0.220470D-01 -0.479064D+00 -0.534321D-01 -0.327232D-01 19 0.467804D-02 -0.460576D-02 0.107663D+00 0.618935D-03 -0.758263D-03 20 -0.183476D+00 -0.427393D-02 -0.247596D+01 0.716522D-01 -0.467392D-02 21 0.133603D-01 0.454826D-02 -0.141286D+00 -0.135689D-02 -0.542522D-03 22 0.989281D-04 -0.134322D-03 0.347860D-02 0.342238D-03 0.275361D-05 23 -0.138828D-01 -0.134131D-02 0.416991D-01 -0.117551D-01 -0.153054D-02 24 0.163436D-02 0.587741D-04 -0.152437D-02 -0.246977D-03 -0.134277D-04 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 0.887967D-03 7 0.686539D-03 0.325451D-02 8 0.204869D-03 -0.445758D-04 0.278103D-02 9 -0.157835D-01 0.277894D-01 0.484880D-02 0.298609D+02 10 0.126769D-01 0.483679D-01 -0.281945D-02 0.202572D+01 0.164424D+02 11 0.403962D-01 0.132346D-01 -0.961864D-02 0.334986D+01 -0.220657D+01 12 -0.515301D-01 -0.416781D-01 0.689086D-02 0.112765D+02 0.605599D+01 13 0.462892D-01 0.102873D+00 0.180242D-02 -0.518013D+00 0.384223D+01 14 0.176911D-01 0.170013D-01 0.185845D+00 0.926509D+00 0.249660D+01 15 0.535234D-02 -0.974207D-01 0.485787D-01 -0.160079D+01 -0.810455D+01 16 -0.712048D-03 0.347687D-04 -0.205982D-02 0.500170D+00 -0.628702D-01 17 0.162487D-04 0.154432D-03 0.184561D-03 -0.102238D+00 -0.367963D-01 18 -0.470927D-01 -0.117076D+00 -0.114208D-01 -0.146841D+01 -0.270169D+01 19 -0.105979D-01 -0.388110D-02 -0.177920D-01 0.355743D+00 -0.279382D+00 20 -0.131258D-03 -0.753031D-01 -0.215994D+00 -0.681554D+01 -0.549305D+01 21 0.938658D-02 0.527775D-02 0.176550D-01 -0.359854D+00 0.162665D+00 22 -0.224294D-04 -0.989355D-04 0.852359D-04 -0.411671D-02 0.325023D-02 23 0.659517D-03 -0.555594D-03 -0.569456D-02 0.181720D+00 -0.272134D+00 24 -0.210326D-04 0.430328D-03 0.889246D-03 -0.446550D-02 0.397081D-01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 0.355006D+02 12 -0.558279D+01 0.129781D+03 13 -0.102582D+01 -0.344138D+01 0.106811D+02 14 -0.368404D+01 0.128889D+01 0.365503D+01 0.466388D+02 15 -0.367128D+00 -0.679789D+01 -0.290183D+01 0.588825D+01 0.205527D+03 16 0.367330D+00 0.108796D+00 -0.960420D-01 -0.151663D+00 0.145217D+01 17 -0.349316D-01 0.544468D-02 0.137488D-01 -0.216208D-02 -0.107083D+01 18 -0.628799D+01 -0.447822D+00 -0.524485D+01 -0.163312D+01 0.114613D+01 19 0.124225D+01 -0.136480D+01 -0.854841D+00 -0.251439D+01 -0.172054D+01 20 0.221079D+01 -0.242743D+02 -0.178761D+01 -0.184483D+02 0.307184D+02 21 -0.605486D+00 0.271137D+01 0.822130D+00 0.231801D+01 0.724803D+00 22 -0.459689D-01 -0.577615D-01 -0.379428D-02 0.224763D-01 0.966223D-01 23 -0.281607D-01 0.935897D+00 0.652617D-01 -0.471439D+00 0.347326D+00 24 -0.535107D-01 -0.110498D+00 0.133537D-01 0.949201D-01 -0.174375D+00 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 0.348220D+00 17 -0.267494D-01 0.131686D-01 18 -0.346620D+00 0.654902D-01 0.154817D+03 19 0.546208D-01 -0.429547D-02 -0.969455D+00 0.354951D+01 20 -0.144544D+00 -0.105851D+00 0.231705D+02 -0.140096D+01 0.305610D+03 21 0.566927D-01 0.535040D-03 0.299018D+01 -0.317485D+01 0.125608D+01 22 -0.344289D-02 0.102331D-04 -0.716771D+00 -0.887067D-02 -0.540586D-01 23 0.411233D-01 -0.458184D-02 0.428320D+00 0.132739D+00 0.199731D+01 24 -0.236594D-02 0.135211D-02 -0.668529D-01 -0.235771D-02 -0.143949D+01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 0.382530D+01 22 -0.384212D-01 0.803639D-02 23 0.378008D-01 -0.661655D-02 0.613070D+00 24 -0.597186D-02 0.149346D-02 -0.571120D-01 0.177597D-01 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.142 1.000 3 0.104 0.001 1.000 4 0.017 0.125 -0.181 1.000 5 0.043 -0.056 -0.053 0.103 1.000 6 -0.065 0.073 0.019 -0.068 0.129 7 0.026 -0.014 0.054 -0.024 0.292 8 0.007 0.006 0.016 0.039 -0.107 9 -0.053 0.039 -0.037 0.108 0.241 10 -0.005 -0.055 -0.015 0.040 0.629 11 -0.042 0.055 0.012 0.031 -0.115 12 0.024 0.001 -0.141 0.085 0.124 13 0.044 -0.028 0.030 -0.040 0.234 14 -0.056 0.022 -0.135 0.050 -0.035 15 -0.195 -0.053 -0.029 0.013 -0.147 16 -0.106 -0.027 0.037 -0.007 -0.035 17 0.146 0.040 0.000 -0.015 -0.049 18 -0.036 0.037 -0.072 -0.088 -0.047 19 0.005 -0.051 0.107 0.007 -0.007 20 -0.020 -0.005 -0.266 0.084 -0.005 21 0.013 0.048 -0.136 -0.014 -0.005 22 0.002 -0.031 0.073 0.079 0.001 23 -0.033 -0.036 0.100 -0.309 -0.035 24 0.023 0.009 -0.021 -0.038 -0.002 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 1.000 7 0.404 1.000 8 0.130 -0.015 1.000 9 -0.097 0.089 0.017 1.000 10 0.105 0.209 -0.013 0.091 1.000 11 0.228 0.039 -0.031 0.103 -0.091 12 -0.152 -0.064 0.011 0.181 0.131 13 0.475 0.552 0.010 -0.029 0.290 14 0.087 0.044 0.516 0.025 0.090 15 0.013 -0.119 0.064 -0.020 -0.139 16 -0.040 0.001 -0.066 0.155 -0.026 17 0.005 0.024 0.030 -0.163 -0.079 18 -0.127 -0.165 -0.017 -0.022 -0.054 19 -0.189 -0.036 -0.179 0.035 -0.037 20 0.000 -0.076 -0.234 -0.071 -0.077 21 0.161 0.047 0.171 -0.034 0.021 22 -0.008 -0.019 0.018 -0.008 0.009 23 0.028 -0.012 -0.138 0.042 -0.086 24 -0.005 0.057 0.127 -0.006 0.073 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 1.000 12 -0.082 1.000 13 -0.053 -0.092 1.000 14 -0.091 0.017 0.164 1.000 15 -0.004 -0.042 -0.062 0.060 1.000 16 0.104 0.016 -0.050 -0.038 0.172 17 -0.051 0.004 0.037 -0.003 -0.651 18 -0.085 -0.003 -0.129 -0.019 0.006 19 0.111 -0.064 -0.139 -0.195 -0.064 20 0.021 -0.122 -0.031 -0.155 0.123 21 -0.052 0.122 0.129 0.174 0.026 22 -0.086 -0.057 -0.013 0.037 0.075 23 -0.006 0.105 0.026 -0.088 0.031 24 -0.067 -0.073 0.031 0.104 -0.091 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 1.000 17 -0.395 1.000 18 -0.047 0.046 1.000 19 0.049 -0.020 -0.041 1.000 20 -0.014 -0.053 0.107 -0.043 1.000 21 0.049 0.002 0.123 -0.862 0.037 22 -0.065 0.001 -0.643 -0.053 -0.034 23 0.089 -0.051 0.044 0.090 0.146 24 -0.030 0.088 -0.040 -0.009 -0.618 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 1.000 22 -0.219 1.000 23 0.025 -0.094 1.000 24 -0.023 0.125 -0.547 1.000
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gain_in_decibel_power.sci
function A_p_dB=gain_in_decibel_power(A_p) A_p_dB=10*log10(A_p) endfunction
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//Ex4_12 Pg-245 clc V=20 //source voltage Vz=12 //zener voltage Vr=V-Vz //voltage across resistor Rs=330 //series resistance disp("Voltage across resistor ") printf(" = %.0f V \n ",Vr) disp("Current through series resistor") Iser=Vr/Rs //Current through series resistor printf(" = %.1f mA \n ",Iser*10^3) disp("Since Zener diode is in series with resistor, current through it is equal to current flowing through resistor,i.e 24.2mA ")
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//Caption: Probability //Example 2.9 //page no 46 //find the probability clc; clear; PA=1/8; PB=1/12; probability_makingerror=1/10001; probability=(PA*PB)/((PA*PB)+((1-PA)*(1-PB)*probability_makingerror)); disp(probability,"Probability of program is correct");
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//Chapter-1,Example 1_2,Page 1-16 clc() //Given Data: theta=40/3600*%pi/180 //angle of wedge in radians B=0.12*10^-2 //fringe spacing //Calculations: //We know, B=lam/(2*u*theta). Here u=1 lam=2*B*theta //wavelength of light used printf('Wavelength of light used is =%.10f m',lam)
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function [ar]=armac(a,b,d,ny,nu,sig) // just build a tlist for storing armacx coefficients // A(z^-1)y= B(z^-1)u + D(z^-1)sig*e(t) // a=<Id,a1,..,a_r>; matrix (ny,r*ny) // b=<b0,.....,b_s>; matrix (ny,(s+1)*nu) // d=<Id,d1,..,d_p>; matrix (ny,p*ny); // ny : dim of observation y // nu : dim of control u // sig : standard deviation (ny,ny); // //! // Copyright INRIA [na,la]=size(a); if na<>ny then write(%io(2),"armac: a(:,1) must be of dimension "+string(ny)); return; end [nb,lb]=size(b); if nb<>0 & nb<>ny then write(%io(2),"armac: b(:,1) must be of dimension "+string(ny)); return; end; if lb<>0 & nu<>0 then if modulo(lb,nu)<>0 then write(%io(2),"armac: number of columns of b are incompatible with nu'); return; end; end [nd,ld]=size(d); if nd<>ny then write(%io(2),"armac: d(:,1) must be of dimension "+string(ny)); return; end ar=tlist(['ar','a','b','d','ny','nu','sig'],a,b,d,ny,nu,sig);
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clc clear //Input data sa1=10;//Cross section of nucleus in barns N=2200;//Neutrons in m/s En1=0.1;//Kinetic energy of neutrons increases in eV En2=0.02525;//Kinetic energy of neutron in eV //Calculations sa2=sa1/[(En1/En2)^0.5];//The cross section of neutrons in barns //Output printf('The cross section of neutrons = %3.2f barns ',sa2)
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//Exa 2.9 clc; clear; close; format('v',7); //Given Data deltaU=-4000;//KJ W=-1.2;//KWh W=W*3600;//KJ Q=W+deltaU;//KJ/hr disp(Q,"Net heat transfer in KJ/hr : ");
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clc //to find net force on passenger ang scale reading while descending and ascending // GIVEN:: //refer to figure 3-19(a) and3-19(b) from page no. 56 //mass of passenger m = 72.2 // in Kg //acceleration of elevator while descending a0y = 0// in m/s^2 // acceleration of elevator while ascending ay = 3.20//in m/s^2 //acceleration due to gravity g = 9.81//in m/s^2 // SOLUTION: //passenger while descending //applying newton's second law Fps_d = m*(g+a0y)//in m/s^2 Fps_d1 = Fps_d/(g*.4535)//in lb //passenger while ascending //applying newton's second law Fps_a = m*(g+ay)//in m/s^2 Fps_a1 = Fps_a/(g*.4535)//in lb printf ("\n\n Net force on passenger while descending Fps_d = \n\n %3i N" ,Fps_d); printf ("\n\n Net force on passenger while descending Fps_d1 = \n\n %3i lb" ,Fps_d1); printf ("\n\n Net force on passenger while ascending Fps_a = \n\n %3i N" ,Fps_a); printf ("\n\n Net force on passenger while ascending Fps_a1 = \n\n %3i lb" ,Fps_a1); printf ("\n\n Scale raeding will not change while descending due to constant acceleration whilescale reading will increase while ascending due to increase in acceleration");
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clc; disp("Example A.19") n=100 // in rpm omega=2*%pi*n/60 r=0.05 // radius in m u=r*omega // velocity in m/s gap=0.001 // in m mew=0.5 // in kg/ms tau=mew*u/gap disp(tau,"Shear stress is ")
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//error no output //ques18 disp('To find the inverse laplace transform of the function'); syms s t a f=s/(s^2+a^2)^2; il=ilaplace(f,s,t); disp(il);
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//Calculations on two stroke engine clc,clear //Given: N=450 //Engine speed in rpm P=450 //Net load on brake in N imep=2.9 //Indicated mean effective pressure in bar m_f=5.4 //Fuel consumption in kg/h deltaT_w=36.1 //Cooling water temperature rise in degreeC m_w=440 //Mass of cooling water used in kg/h A_F=31 //Air-fuel ratio T1_g=20+273,T2_g=355+273 //Inlet and outlet temperature of exhaust gases blown in K P1=76 //Atmospheric pressure in cm of Hg d=22,l=27 //Bore and stroke in cm D_b=1.5 //Effective diameter of the brake wheel in m CV=44000 //Calorific value in kJ/kg p=15 //Percentage of hydrogen by mass contained by the fuel R=0.287 //Specific gas constant in kJ/kgK cp_g=1.005,cp_s=2.05 //Specific heat for dry exhaust gases and superheated steam in kJ/kgK //Solution: ip=imep*10^2*l*%pi/4*d^2*N/(60)*10^-6 //Indicated power in kW eta_it=ip*3600/(m_f*CV) //Indicated thermal efficiency bp=2*%pi*N/60*(P*D_b/2)*10^-3 //Brake power in kW bp=round(10*bp)/10 bsfc=m_f/bp*1000 //Brake specific fuel consumption in gm/kWh V_s=(%pi/4)*d^2*l*10^-6*N //Swept volume in m^3/min m_a=m_f*A_F/60 //Mass of air inhaled in kg/min P1=1.0132 //Atmospheric pressure equivalent to 76 cm of Hg in bar T1=293 //Atmospheric temperature in K V_a=m_a*R*T1/(P1*100) //Volume of air inhaled in m^3/min V_a=round(100*V_a)/100 eta_vol=V_a/V_s //Volumetric efficiency //Heat balance sheet Q1=m_f/60*CV //Heat input in kJ/min Q_bp=bp*60 //Heat equivalent to brake power in kJ/min cp_w=4.1868 //Specfic heat of water in kJ/kgK Q_w=m_w/60*cp_w*deltaT_w //Heat in cooling water in kJ/min m_e=m_a+m_f/60 //Mass of exhaust gases in kg/min //Since, 2 mole of hydrogen gives 1 mole of water on combine with 1 mole of oxygen //Thus, 1 mole of hydrogen gives 1/2 mole or 9 unit mass of water m_h=m_f/60*p/100 //Mass of hydrogen in kg/min m_s=9*m_h //Mass of steam in exhaust gases in kg/min m_d=m_e-m_s //Mass of dry exhaust gases in kg/min Q_d=m_d*cp_g*(T2_g-T1_g) //Heat in dry exhaust gases kJ/min lv=2256.9 //Latent heat of vapourisation of water in kJ/kg Q_s=m_s*((373-T1_g)+lv+cp_s*(T2_g-373)) //Heat in steam in exhaust gases in kJ/min Q_r=Q1-Q_bp-Q_w-Q_d-Q_s //Heat in radiation in kJ/min //Results: printf("\n (a)The indicated thermal efficiency, eta_it = %.1f percent",eta_it*100) printf("\n (b)Brake specific fuel consumption = %.1f gm/kWh",bsfc) printf("\n (c)The volumetric efficiency, eta_vol = %.1f percent",eta_vol*100) printf("\n\n Heat balance sheet\n\t Heat input = %.1f kJ/min, %d percent",Q1,Q1/Q1*100) printf("\n\t Heat equivalent to b.p. = %.1f kJ/min, %.1f percent",Q_bp,Q_bp/Q1*100) printf("\n\t Heat in cooling water = %.1f kJ/min, %.1f percent",Q_w,Q_w/Q1*100) printf("\n\t Heat in dry exhaust gases = %.1f kJ/min, %.1f percent",Q_d,Q_d/Q1*100) printf("\n\t Heat in steam in exhaust gases = %.1f kJ/min, %.1f percent",Q_s,Q_s/Q1*100) printf("\n\t Heat in radiation = %.1f kJ/min, %.1f percent",Q_r,Q_r/Q1*100)
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Ex10_10.sce
// Theory and Problems of Thermodynamics // Chapter 10 // Chemical Thermodynamics // Example 10 clear ;clc; //Given data // CO2(g) + H2O(g) => CO2(g) + H2(g) T1 = 298 // intial temperature in K T2 = 1000 // final temperature in K del_H_1 = -110.53 // heat of formation of CO in kJ at 298 K del_G_1 = -137.17 // Gibbs free energy of CO in kJ at 298 K del_H_2 = -393.51 // heat of formation of CO2 in kJ at 298 K del_G_2 = -394.36 // Gibbs free energy of CO2 in kJ at 298 K del_H_3 = -241.82 // heat of formation of H2O in kJ at 298 K del_G_3 = -228.57 // Gibbs free energy of H2O in kJ at 298 K R = 8.314 // gas constant // Calculations del_G = del_G_2 - del_G_1 - del_G_3 // Gibbs free energy K1 = exp(-del_G*1e3/(R*T1)) // equilibrium constant del_H = del_H_2 - del_H_1 - del_H_3 // heat of reaction //log(K2/K1) = (-del_H/R)*(1/T2-1/T1) deff('y=cons(K2)', 'y = log(K2/K1) - (-del_H*1e3/R)*(1/T2-1/T1)') K2 = fsolve(1,cons) // equilibrium constant // Output Results mprintf('(a) Gibbs free energy at 298 K and 0.1 MPa = %4.2f kJ' , del_G); mprintf('\n Equilibrium constant at 298 K and 0.1 MPa = %4.4f E+05' , K1*1e-5); mprintf('\n (b) Standard heat of reaction at 298 K and 0.1 MPa = %4.2f kJ' , del_H); mprintf('\n Equilibrium constant at 1000 K and 0.1 MPa = %4.4f ' , K2);
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11_4_1.sce
//chapter_no.-11, page_no.-508 //Example_no.11-4-1 clc; //(a)Calculate_the_K_factor er=2.56//relative_dielectric_constant w=25;//strip_width t=14;//strip_thickness d=70;//shield_depth K=1/(1-(t/d)); disp(K,'the_K_factor is ='); //(b)Calculate_the_fringe_capacitance Cf=((8.854*er)*((2*K*log(K+1))-((K-1)*log((K^2)-1))))/%pi; disp(Cf,'the_fringe_capacitance(in pF/m)is ='); //(c) Calculate_the_characteristic_impedance_of_the_line Z0=94.15/((((w/d)*K)+(Cf/(8.854*er)))*(sqrt(er))); disp(Z0,'the_characteristic_impedance_of_the_line(in ohms)is =');
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Ex12_1.sce
errcatch(-1,"stop");mode(2);//Caption:Calculate (a)No load power factor (b)Core and friction loss (c)r_m (d)power factor on short circuit (e)Equivalent impedance in series circuit (f)Rotor resistance referred to stator (g)Stator leakage reactance (h)Rotor leakage reactance referred to stator //Exa:12.1 ; ; P=3000//Power of motor(in watt) V=415//Voltage supplied(in volts) f=50//Frequency(in hertz) p=6//Number of poles pf=0.8//Power factor I_n=3.5//No load current(in A) P_n=250//Power input on no load test(in watt) r_s=1.5//Stator resistance per phase(in ohm) V_r=115//Reduced voltage applied at short circuit test(in volts) I_s=13//Current supplied on short circuit test(in A) P_s=1660//Voltage applied at short circuit test(in watt) pfn=P_n/(sqrt(3)*V*I_n) disp(pfn,'(a)Noload power factor=') P_wf=P_n-(3*(I_n^2)*r_s) disp(P_wf,'(b)Core and friction loss(in watt)=') r_m=(V/sqrt(3))/(I_n*pfn) disp(r_m,'(c)Resistance(in ohms)=') pfs=P_s/(sqrt(3)*V_r*I_s) disp(pfs,'(d)Power factor on short circuit=') Ze=(V/sqrt(3))/((I_s*V)/V_r) disp(Ze,'(e)Equivalent impedance in series circuit(in ohms)=') R=(Ze*pfs)-r_s disp(R,'(f)Rotor resistance referred to stator(in ohm)=') X=(sqrt((Ze^2)-((Ze*pfs)^2))) disp(X,'(g)Stator leakage reactance(in ohms)=') x=X/2 disp(x,'(h)Rotor leakage reactance referred to stator(in ohms)=') exit();
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Example4_23.sce
//Example 4.23. clc f=(1/(2*%pi*sqrt(0.33*0.065*10^-12)))*10^-6 // in MHz format(6) disp(f,"(i) f(in MHz) = 1 / 2*pi*sqrt(L*C) =") ceq=0.065/1.065 // in pF disp(ceq,"(ii) C_eq(in pF) = C*C_M / C+C_M =") fp=(1/(2*%pi*sqrt(0.33*0.061*10^-12)))*10^-6 // in MHz disp(fp,"(i) f_p(in MHz) = 1 / 2*pi*sqrt(L*C_eq) =") pi=((1.121-1.087)/1.087)*100 // in percentage disp(pi,"(iii) % increase =") q=(2*%pi*1.087*0.33*10^6)/(5.5*10^3) format(8) disp(q,"(iv) Q = omega_x*L / R = 2*pi*f_s*L / R =")
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Ex1_1.sce
//Mohr's circle clc; sigma=((40+80)/2) disp(sigma,"center of the circle in MPa = ") //solution a x=((80-40)^2); y=30^2; sigma1=60+sqrt((.25*x)+y) disp(sigma1,"maxi pricipal stress in MPa = ");// displaying result sigma2=60-sqrt((.25*x)+y) disp(sigma2,"mini pricipal stress in MPa = ");// displaying result theta1=((atand(30/20))/2) disp(theta1,"pricipal stresses in degree");// displaying result theta2=(((atand(30/20))+180)/2) disp(theta2,"pricipal stresses in degree");// displaying result //solution b tau=sqrt((.25*x)+y) disp(tau,"maxi shearing stress in MPa = ");// displaying result theta3=theta1+45 disp(theta3,"stress in MPa = ");// displaying result theta4=theta2+45 disp(theta4,"stress in MPa = ");// displaying result //final solution in matrix form p=[80 30 ;30 40] disp(p) q=[sigma1 0 ; 0 sigma2] disp(q) r=[sigma -tau ; -tau sigma] disp(r)
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example2_17.sce
clc // Given that lambda = 5e-7 // wavelength of light in meter theta = %pi / 6 // half angular width of central maximum in first case in radian theta_ = %pi / 2 // half angular width of central maximum in second case in radian // Sample Problem 17 on page no. 2.44 printf("\n # PROBLEM 17 # \n") m = 1 // for first minima b1 = (lambda * m) / sin(theta) // calculation for slit width in first case b2 = (lambda * m) / sin(theta_) // calculation for slit width in second case printf("\n Standard formula used \n b = (lambda * m) / sin(theta). \n") printf("\n Slit width in first case = %e meter. \n Slit width in second case = %e meter",b1,b2)
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7_2.sce
clc //initialisation of variables m= 1 //gms M= 63.54 //gms e= 2 //farady F= 96493 n= 3 //CALCULATIONS t= (m/M)*(e*F/n) //RESULTS printf (' Time = %.f sec',t)
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EX4_9_2.sce
//Example No. 4.9.2 clc; clear; close; format('v',6); le=61.4;//m Irms=50;//A lambda=625;//m P=160*%pi^2*(le/lambda)^2*Irms^2;//kW Rr=160*%pi^2*(le/lambda)^2;//Ω disp(P*10^-3,"Power radiated in kW : "); disp(Rr,"Radiation resistance in Ω : ");
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Exa11_2.sce
//Exa 11.2 clc; clear; close; disp("The method of finding the economic life of the machine witha discounting factor of 20% at zero inflation rate is summarized in table below. From the table it is clear that total annual cost is minimum if the machine is used for 14 years. Hence the economic life of the machine is 14 years."); disp("End of year Op_cost Main_cost Op+Main P/F,i,n PW Cummulative Salvage PW_S TPW A/P,i,n AEM"); i=20;//in per year Cum=0;//initialising Op_cost=40000;//in RS. Main_cost=60000;//in Rs. OpMain=Op_cost+Main_cost;//in Rs. S=400000;//in Rs. for n=1:15 PF=1/((1+i/100)^n); PW=OpMain*PF;//in Rs. Cum=Cum+PW PW_S=PF*S;//in RS. TPW=500000+Cum-PW_S;//in Rs. AP=((i/100)*(1+i/100)^n)/(((1+i/100)^n)-1); AEM=TPW*AP;//in RS disp(" "+string(n)+" "+string(Op_cost)+" "+string(Main_cost)+" "+string(OpMain)+" "+string(PF)+" "+string(PW)+" "+string(Cum)+" "+string(S)+" "+string(PW_S)+" "+string(TPW)+" "+string(AP)+" "+string(AEM)); Op_cost=Op_cost+5000;//in Rs. Main_cost=Main_cost+6000;//in Rs. S=S-50000;//in Rs. end
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example6_5_a.sce
clear; clc; printf("\t Example 6_5_a\n"); //table 6.5.1 //S.NO. Time (Hr) weight of wet material(kg) // 0.0 5.314 // 0.4 5.238 // 0.8 5.162 // 1.0 5.124 // 1.4 5.048 // 1.8 4.972 // 2.2 4.895 // 2.6 4.819 // 3.0 4.743 // 3.4 4.667 // 4.2 4.524 // 4.6 4.468 // 5.0 4.426 // 6.0 4.340 // infinite 4.120 w=[5.314 5.238 5.162 5.124 5.048 4.972 4.895 4.819 4.743 4.667 4.524 4.468 4.426 4.340 4.120] t=[0.0 0.4 0.8 1.0 1.4 1.8 2.2 2.6 3.0 3.4 4.2 4.6 5.0 6.0] //part(i) x=4.120; //weight of the dried material printf("\n moisture content (dry basis) "); i=1; //looping starts while(i<16) //calculation of moisture content p(i)=(w(i)-x)/x; printf("\n :%f",p(i)); i=i+1; end printf("\n \n Drying rate kg/hr*m^2"); i=2; while(i<15) a(i)=(p(i-1)-p(i))*4.12/(t(i)-t(i-1)); printf("\n :%f ",a(i)); i=i+1; end a(1)=.19; a(15)=0; printf("\n\n from the above data it is clear that critical moisture content Xcr=0.11"); plot(p,a,"o-"); title("Fig.6.19(a) Example3 Drying Rate curve"); xlabel("X-- Moisture content, X(kg/kg) ---->"); ylabel("Y-- Drying Rate, N(kg/hr.m^2 ---->"); //end
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x = [-3 -2 -1; 0 1 2 ;4 1 2]; y = [-1 -1 -1; 0 1 1;2 3 1]; t = -3:.01:3; p = pchip(x,y,t); disp(p); ////output //!--error 9999 //Inconsistent element-wise operationat line 40 of function pchip called by : //p = pchip(x,y,t); //
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clc y1=0.75 //H2 y2=0.25 //N2 CP1=28.6455 CP2=29.1783 CP=(y1*CP1)+(y2*CP2) mprintf("CP=%fkJ/kmol K\n",CP)//ans vary due to roundoff error Cv1=20.3311 Cv2=20.8641 Cv=(y1*Cv1)+(y2*Cv2) mprintf("Cv=%fkJ/kmol K\n",Cv)//ans vary due to roundoff error gama=CP/Cv mprintf("gamma=%f\n",gama)//ans vary due to roundoff error P1=100 //pressure in kPa P2=500 //pressure in kPa T1=300 T2=T1*((P2/P1)^((gama-1)/gama)) mprintf("T2=%fK\n",T2)//ans vary due to roundoff error ws=-CP*(T2-T1) mprintf("-ws=%fkJ/kmol\n",-ws)//ans vary due to roundoff error M1=2.016 M2=28.013 M=(y1*M1)+(y2*M2) mprintf("Molar mass=%fkg/kmol\n",M)//ans vary due to roundoff error Ws=-(-ws/M) mprintf("-Ws=%fkJ/kg of mixture\n",-Ws)//ans vary due to roundoff error R=8.314 deltas1=(CP1*log(T2/T1))-(R*log(P2/P1)) mprintf("s2-s1=%fkJ/kmol K\n",deltas1)//ans vary due to roundoff error deltas2=(CP2*log(T2/T1))-(R*log(P2/P1)) mprintf("s2-s1=%fkJ/kmol K\n",deltas2)//ans vary due to roundoff error deltas=(y1*deltas1)+(y2*deltas2) mprintf("s2-s1=%fkJ/kmol K",deltas)//ans vary due to roundoff error
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// Grob's Basic Electronics 11e // Chapter No. I // Example No. I_7 clc; clear; // Express the power value of 250-W using the appropriate metric prefix from Table I–2. disp ('In this case, it is not necessary to use any of the metric prefixes listed in Table I–2. The reason is that 250-W cannot be expressed as a number between 1 and 1000 times a power of 10 which is a multiple of 3.') disp ('250 W cannot be expressed in engineering notation. The closest we can come is 0.25*10^3-W, which is not representative of engineering notation. Although 10^3 can be replaced with the metric prefix kilo (k)') disp ('It is usually preferable to express the power as 250-W and not as 0.25-kW.')
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border.left: 7 border.right: 7 border.bottom: 7 border.top: 7 source: imgbackground.png
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//calculating the Resolution clc; Fs=200; D=100; SD=Fs/D; R=SD/10; disp(R,'resolution (V)=')
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function s = lagrange(x,y,p) tam=length(x) s=0 for i=1:tam l=1 for j=1:tam if j~=i l=l*(p-x(j))/(x(i)-x(j)) end end s=s+y(i)*l; end endfunction
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function C=borda2cadeia(I,x,y) //I é a imagem binária e x,y são as coordenadas //colocação de bordas adicionais [M,N]=size(I); Ii=zeros(M+2,N+2); Ii(2:M+1,2:N+1)=I(1:M,1:N); I=Ii; //posição inicial de busca x=x+1; y=y+1; X=x; Y=y; //preparação C=zeros(1,2); i=0; B=4; FLAG=0 //inicio das buscas while(FLAG==0) i=i+1; if((I(X,Y+1)==1)&(B~=0)) //direção 0 C(i)=0; Y=Y+1; B=4; //direção contrária (não consid. na prox.iteração) elseif((I(X+1,Y+1)==1)&(B~=7)) //direção 7 C(i)=7; B=3; X=X+1; Y=Y+1; elseif((I(X-1,Y+1)==1)&(B~=1)) //direção 1 C(i)=1; B=5; X=X-1; Y=Y+1; elseif((I(X+1,Y)==1)&(B~=6)) //direção 6 C(i)=6; B=2; X=X+1; elseif((I(X-1,Y)==1)&(B~=2)) //direção 2 C(i)=2; B=6; X=X-1; elseif((I(X+1,Y-1)==1)&(B~=5)) //direção 5 C(i)=5; B=1; X=X+1; Y=Y-1; elseif((I(X-1,Y-1)==1)&(B~=3)) //direção 3 C(i)=3; B=7; X=X-1; Y=Y-1; elseif((I(X,Y-1)==1)&(B~=4)) //direção 4 C(i)=4; B=0; Y=Y-1; end if((X==x)&(Y==y)) FLAG=1; end end endfunction
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//Problem 29.03: A two-branch parallel network is shown in Figure 29.8. Determine the resonant frequency of the network. //initializing the variables: RL = 5; // in ohms L = 0.002; // IN Henry C = 25e-6; // IN fARADS Rc = 3; // in ohms //calculation: //Resonant frequency, for parallel fr = (1/(2*%pi*((L*C)^0.5)))*((RL^2 - (L/C))/(Rc^2 - (L/C)))^0.5 printf("\n\n Result \n\n") printf("\n resonant frequency, fr is %.2f Hz",fr)
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10_10.sce
clear clc //Example 10.10 disp('Example 10.10') s=%s; Gp=1/(5*s+1); Gm=1/(s+1); Gv=1/(2*s+1); Ys=Gv*Gp*Gm Routh=routh_t(Ys,poly(0,"Kc")); // produces routh table for polynomial 1+Kc*Ys disp(Routh) K1=roots(numer(Routh(3,1))); K2=roots(numer(Routh(4,1))); mprintf('K lies between %f and %f for system to be stable', K2,K1)
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Ex13_15.sce
//Initilization of variables g=9.8 //m/s^2 //Calculations //Simplfying the equations we can solve for T2 and aA first to obtain the solution //Solving by matrix method A=[-1.5,-4;-3.5,24] B=[-4*g;-24*g] C=inv(A)*B T2=C(1) //N T1=T2/2 //N T3=T2/2 //N //Acceleration calculations a1=1*g-T1 //m/s^2 a2=(2*g-T1)/2 //m/s^2 a3=(3*g-T3)/3 //m/s^2 a4=(4*g-T3)/4 //m/s^2 //Tension in fixed cord T_f=2*T2 //N //Result clc printf('The acceleration values are a1=%f,a2=%f,a3=%f and a4=%f m/s^2\n The tension in the fixed cord is %fN',a1,a2,a3,a4,T_f)
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Example9_2.sce
///Chapter 9 Law Of Thermodynamics ///Example 9.2 Page No:166 ///Find Quantity of heat transferred ///Input data clc; clear; //During compression W1=-9200; //Stroke work done by the piston in Nm Nm1=-9.2; //Nm of work done Q1=-50; //Heat rejected during copression in KJ //During expansion W2=8400; //Stroke work done by the piston in Nm Nm2=8.4; //Nm of work done ///Calculation; //Quantity of heat transferred Q2=-((Nm1+Nm2)+Q1); //-sign for indicate heat is transferred ///Output printf('Quantity of heat transferred= %f KJ \n',Q2);
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example14_5.sce
//Chapter 14 //Example 14_5 //Page 363 clear;clc; i1=5; i2=14.08; pf1=0.8; pf2=0.85; l1=600; l2=400; hp=10; n=0.90; vb=400; r=1; x=0.5; z=r+%i*x; Zac=z*l1/1000; Zcb=z*l2/1000; printf("Impedance of distributor/km = %.2f+j(%.2f) ohm \n\n", real(z), imag(z)); printf("Impedance of section AC = Zac = %.2f+j(%.2f) ohm \n", real(Zac), imag(Zac)); printf("Impedance of section CB = Zcb = %.2f+j(%.2f) ohm \n\n\n", real(Zcb), imag(Zcb)); Vb=vb/sqrt(3)+%i*0; printf("Voltage at point B taken as the reference vector = %.0f+j%.0f \n", real(Vb), imag(Vb)); Ib=hp*746/sqrt(3)/vb/n/pf2; I2=i2*(pf2-%i*sin(acos(pf2))); I1=i1*(pf1-%i*sin(acos(pf1))); Iac=I2+I1; Icb=I2; Vcb=Icb*Zcb; Vac=Iac*Zac; Va=Vb+Vcb+Vac; printf("Line current at B = %.2f A \n\n", Ib); printf("Load current at point B = %.2f+j(%.2f) A \n", real(I2), imag(I2)); printf("Load current at point C = %.2f+j(%.2f) A \n\n", real(I1), imag(I1)); printf("Current in section CB = %.2f+j(%.2f) A \n", real(Icb), imag(Icb)); printf("Current in section AC = %.2f+j(%.2f) A \n\n", real(Iac), imag(Iac)); printf("Voltage drop in section CB = %.2f+j(%.2f) A \n", real(Vcb), imag(Vcb)); printf("Voltage drop in section AC = %.2f+j(%.2f) A \n\n", real(Vac), imag(Vac)); printf("Voltage at A/phase = %.2f+j(%.2f) A \n\n", real(Va), imag(Va)); printf("Magnitude of Va/phase = %.2f V \n\n", abs(Va)); printf("Line voltage at A = %.2f V \n\n", abs(Va)*sqrt(3));
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clc //Intitalisation of variables clear Kc= 1.08*10^-5 n= 2 //moles v= 0.45 //lit n1= 0.5 //mole //CALCULATIONS y= (-Kc*v+sqrt(Kc^2*v^2+4*Kc*v*n1*n^2))/(2*n^2) c= 2*y/v //RESULTS printf ('y = %.2e mole',y) printf ('\n concentration of NO2 = %.2e mole per liter',c)
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clc; clear; close; x1=input('x(n)='); x2=input('h(n)='); L1=length(x1); L2=length(x2); N=max(L1,L2); x1=[x1,zeros(1,N-L1)]; x2=[x2,zeros(1,N-L2)]; //circular convolution in freq domain X1=fft(x1); X2=fft(x2); Y=X1.*X2; y=ifft(Y); disp(y,'Circular Convolution y=') subplot(3,1,1);plot2d3(x1);xtitle('input signal x1','n','x1[n]'); subplot(3,1,2);plot2d3(x2);xtitle('input signal x2','n','x2[n]'); subplot(3,1,3);plot2d3(Y);xtitle('output signal Y','n','Y[n]');
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ex_5_45a.sce
clear; clc; close; disp("dy(t)/dt+2y(t)=x(t)"); w=0:0.1:10; t=w; dw=.1; Xw=ones(1,length(w))./(1+%i*w); Hw=ones(1,length(w))./(2+%i*w); Yw=Xw.*Hw; y=Yw*exp(%i*t'*w)*dw*.31; d=gca() plot(t,y); poly1=d.children.children; poly1.thickness=3; poly1.foreground=2; xtitle('y(t)','t') yy=exp(-t)-exp(-2*t); disp("y(t)=exp(-t)-exp(-2*t)") figure d=gca() plot(t,yy); poly1=d.children.children; poly1.thickness=3; poly1.foreground=2; xtitle('y(t)','t')
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Ex34_12.sce
// Example 34_12 clc;funcprot(0); //Given data C_1=5000;//Cost of first unit in Rupees MD_1=100;// Maximum demand in kW C_2=14000;//Cost of second unit in Rupees MD_2=60;// Maximum demand in kW n=40000;// Useful life in hours C_e=80;//Energy charge per kW in Rupees/year C_kwh=5/100;//Energy charge per kW-hr in Rupees //Calculation //(a)First unit Cc=C_1/n;// Capital cost of unit per hour in Rupees C_MD=((MD_1*C_e)/8760);// Charge for maximum demand per hour in Rupees C_eh=MD_1*1*C_kwh;// Energy charge per hour in Rupees TC_1=Cc+C_MD+C_eh;// Total charges per hour for the operation of first unit in Rupees //(b)Second unit Cc=C_2/n;// Capital cost of unit per hour in Rupees C_MD=((MD_2*C_e)/8760);// Charge for maximum demand per hour in Rupees C_eh=MD_2*1*C_kwh;// Energy charge per hour in Rupee TC_2=Cc+C_MD+C_eh;// Total charges per hour for the operation of second unit in Rupees printf('\n(a)Total charges per hour for the operation of first unit=Rs.%0.3f\n(b)Total charges per hour for the operation of second unit=Rs.%0.3f',TC_1,TC_2); if(TC_1>TC_2) printf('\n The second unit is more economical than first unit in this case.'); else printf('\n The first unit is more economical than second unit in this case.'); end
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dHfDiam=-94.50;//heat of formation value of Diamond in kcal// dHfGrap=-94.05;//Heat of formation value of Graphite in kcal// dHf=dHfGrap-dHfDiam;//Enthalpy change when graphite converted to diamond in Kcal// printf('Enthalpy change when graphite converted to diamond=%fKcal',dHf);