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clear //Given A=0.9 Ie=1 //mA //Calculation Ic=A*Ie Ib=Ie-Ic //Result printf("\n Base current is %0.3f mA",Ib)
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aaa=[631200 780143.766831 2373680.0708 631196 780143.766831 2373680.0708 630915 780535.103072 2373588.16789 630934 781272.611854 2373443.69326 631143 781272.611854 2373443.69326 631177 780535.103072 2373588.16789 631181 780535.103072 2373588.16789 631219 781272.611854 2373443.69326 631257 780535.103072 2373588.16789 631272 781731.820967 2373387.22376 631276 781731.820967 2373387.22376 631295 782271.440297 2373268.56034 631310 781272.611854 2373443.69326 631447 781272.611854 2373443.69326 631466 782271.440297 2373332.34712 631485 780535.103072 2373588.16789 631519 781272.611854 2373443.69326 631542 780535.103072 2373608.16789 631580 782461.025319 2373300.35196 631599 782271.440297 2373268.56034 631618 782271.440297 2373268.56034 631637 780143.766831 2373680.0708 631656 781272.611854 2373443.69326 631675 781272.611854 2373443.69326 631694 781272.611854 2373443.69326 631728 780535.103072 2373588.16789 631732 780535.103072 2373588.16789 631751 781272.611854 2373443.69326 631808 781272.611854 2373443.69326 631846 780535.103072 2373608.16789 631865 781272.611854 2373443.69326 631884 780535.103072 2373608.16789 631903 780143.766831 2373680.0708 630896 781272.611854 2373393.69326 630892 781272.611854 2373393.69326 630911 780535.103072 2373588.16789 630930 781272.611854 2373443.69326 630953 781272.611854 2373443.69326 630949 781272.611854 2373443.69326 630972 782271.440297 2373268.56034 630968 782271.440297 2373268.56034 630991 781272.611854 2373443.69326 630987 781272.611854 2373443.69326 631010 781272.611854 2373443.69326 631006 781272.611854 2373443.69326 631029 781272.611854 2373443.69326 631025 781272.611854 2373443.69326 631048 780143.766831 2373700.0708 631044 780143.766831 2373700.0708 631063 781272.611854 2373443.69326 631086 781272.611854 2373443.69326 631082 781272.611854 2373443.69326 631101 781272.611854 2373443.69326 631124 781272.611854 2373443.69326 631120 781272.611854 2373443.69326 631139 781272.611854 2373443.69326 631158 781272.611854 2373443.69326 631215 781272.611854 2373443.69326 631238 781731.820967 2373387.22376 631234 781731.820967 2373387.22376 631253 780535.103072 2373588.16789 631291 782271.440297 2373268.56034 631314 781272.611854 2373443.69326 631333 780143.766831 2373680.0708 631329 780143.766831 2373680.0708 631352 780143.766831 2373740.0708 631348 780143.766831 2373740.0708 631371 780535.103072 2373673.16789 631367 780535.103072 2373673.16789 631390 781272.611854 2373443.69326 631386 781272.611854 2373443.69326 631409 782271.440297 2373268.56034 631405 782271.440297 2373268.56034 631428 781272.611854 2373443.69326 631424 781272.611854 2373443.69326 631443 781272.611854 2373443.69326 631462 782271.440297 2373332.34712 631481 780535.103072 2373588.16789 631504 781272.611854 2373443.69326 631500 781272.611854 2373443.69326 631523 781272.611854 2373443.69326 631538 780535.103072 2373608.16789 631561 781272.611854 2373443.69326 631557 781272.611854 2373443.69326 631576 782461.025319 2373300.35196 631595 782271.440297 2373268.56034 631614 782271.440297 2373268.56034 631633 780143.766831 2373680.0708 631652 781272.611854 2373443.69326 631671 781272.611854 2373443.69326 631690 781272.611854 2373443.69326 631713 781272.611854 2373443.69326 631709 781272.611854 2373443.69326 631747 781272.611854 2373443.69326 631770 781272.611854 2373443.69326 631766 781272.611854 2373443.69326 631789 781272.611854 2373523.69326 631785 781272.611854 2373523.69326 631804 781272.611854 2373443.69326 631827 781272.611854 2373443.69326 631823 781272.611854 2373443.69326 631842 780535.103072 2373608.16789 631861 781272.611854 2373443.69326 631880 780535.103072 2373608.16789 631899 780143.766831 2373680.0708 631067 781272.611854 2373443.69326 631105 781272.611854 2373443.69326 631162 781272.611854 2373443.69326]; bbb=aaa; naaa = max(size(aaa)); function [X,Y]=UTMtoGeog(x,y,utmz,hemi) if (x<160000 | x>840000) then disp("Outside permissible range of easting values \n Results may be unreliable \n Use with caution"); return; end; if (y<0) then disp("Negative values not allowed \n Results may be unreliable \n Use with caution"); return; end; if (y>10000000) then disp("Northing may not exceed 10,000,000 \n Results may be unreliable \n Use with caution"); return; end; DatumEqRad = [6378137.0,6378137.0,6378137.0,6378135.0,6378160.0,6378245.0,6378206.4,6378388.0,6378388.0,6378249.1,6378206.4,6377563.4,6377397.2,6377276.3]; DatumFlat = [298.2572236, 298.2572236, 298.2572215, 298.2597208, 298.2497323, 298.2997381, 294.9786982,296.9993621, 296.9993621, 293.4660167, 294.9786982, 299.3247788, 299.1527052, 300.8021499]; Item = 1;//Default a = DatumEqRad(Item);//equatorial radius, meters. f = 1/DatumFlat(Item);//polar flattening. drad = 4*atan(1.0)/180;//Convert degrees to radians) k0 = 0.9996;//scale on central meridian b = a*(1-f);//polar axis. e = sqrt(1 - (b/a)*(b/a));//eccentricity e0 = e/sqrt(1 - e*e);//Called e prime in reference esq = (1 - (b/a)*(b/a));//e squared for use in expansions e0sq = e*e/(1-e*e);// e0 squared - always even powers zcm = 3 + 6*(utmz-1) - 180;//Central meridian of zone e1 = (1 - sqrt(1 - e*e))/(1 + sqrt(1 - e*e));//Called e1 in USGS PP 1395 also M0 = 0;//In case origin other than zero lat - not needed for standard UTM M = M0 + y/k0;//Arc length along standard meridian. if hemi then M=M0+(y-10000000)/k; end; mu = M/(a*(1 - esq*(1/4 + esq*(3/64 + 5*esq/256)))); phi1 = mu + e1*(3/2 - 27*e1*e1/32)*sin(2*mu) + e1*e1*(21/16 -55*e1*e1/32)*sin(4*mu);//Footprint Latitude phi1 = phi1 + e1*e1*e1*(sin(6*mu)*151/96 + e1*sin(8*mu)*1097/512); C1 = e0sq*(cos(phi1)**2); T1 = (tan(phi1)**2); N1 = a/sqrt(1-(e*sin(phi1))**2); R1 = N1*(1-e*e)/(1-(e*sin(phi1))**2); D = (x-500000)/(N1*k0); phi = (D*D)*(1/2 - D*D*(5 + 3*T1 + 10*C1 - 4*C1*C1 - 9*e0sq)/24); phi = phi + (D**6)*(61 + 90*T1 + 298*C1 + 45*T1*T1 -252*e0sq - 3*C1*C1)/720; phi = phi1 - (N1*tan(phi1)/R1)*phi; //Output Latitude X = floor(1000000*phi/drad)/1000000; //Longitude lng = D*(1 + D*D*((-1 -2*T1 -C1)/6 + D*D*(5 - 2*C1 + 28*T1 - 3*C1*C1 +8*e0sq + 24*T1*T1)/120))/cos(phi1); lngd = zcm+lng/drad; //Output Longitude Y = floor(1000000*lngd)/1000000; endfunction for i=1:naaa, [bbb(i,2),bbb(i,3)]=UTMtoGeog(aaa(i,2),aaa(i,3),15,%f);end;
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//EXAMPLE 5.12 //Linear convolution using Circular convolution clc; clear; g = [1 2 0 1]; disp(g,'g[n] = '); h = [2 2 1 1]; disp(h,'h[n] = '); //linea convolution length = 4+4-1 = 7 //appending the two signals with zeros g = [g,zeros(1,3)] h = [h,zeros(1,3)] G = fft(g,-1); H = fft(h,-1); Y = G.*H; //element wise multiplication y = fft(Y,1);//IDFT //Plotting linear convolution n=0:6; figure(0); clf(); a = gca(); a.x_location = 'origin'; a.y_location = 'origin'; plot2d3(n,y,2); plot(n,y,'r.'); poly1 = a . children (1) . children (1) ; poly1.thickness = 2; xtitle('Linear convolution','n','y'); disp(y," linear convolution ,y = ");
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//Example 6_3 clc(); clear; //To find out how fast and the direction car moving m1=30000 //units in Kg m2=1200 //units in Kg v10=10 //units in meters/sec v20=-25 //units in meters/sec vf=((m1*v10)+(m2*v20))/(m1+m2) //unis in meters/sec printf("The car is moving at vf=%.2f Meters/sec\n",vf) printf("The positive sign of vf Indicate the car is moving in the direction the truck was moving")
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// SAMPLE PROBLEM 5/10 clc;funcprot(0); // Given data v_B=0.8;// The velocity in m/s theta=30;// degree d_co=18;// The distance in inch // Calculation v_A=v_B*cosd(theta);// ft/sec OAbar=(d_co/12)/(cosd(theta));// ft omega=v_A/(OAbar);// rad/sec CCW printf("\nThe angular velocity of the slotted arm,omega=%0.3f rad/sec CCW",omega);
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//chapter13 //example13.14 //page288 R1=45 // kilo ohm R2=15 // kilo ohm Re=7.5 // kilo ohm Vcc=30 // V Vbe=0.7 // V gain_beta=200 V2=Vcc*R2/(R1+R2) // voltage across R2 Ve=V2-Vbe // voltage across Re Ie=Ve/Re re_dash=1d-3*25/Ie // in kilo ohm Zin_base=gain_beta*re_dash Zin=Zin_base*(R1*R2/(R1+R2))/(Zin_base+R1*R2/(R1+R2)) printf("input impedence of amplifier circuit = %.3f kilo ohm \n",Zin) // the accurate answer for input impedence is 3.701 kilo ohm but in book it is given as 3.45 kilo ohm
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main var inp, outp; { let outp <- call inputnum(); let inp <- outp * outp / outp - 22; if outp == inp then call outputnum(inp); let outp <- call inputnum(); call outputnewline() fi; call outputnum(inp); call outputnewline(); call outputnum(outp); call outputnewline() }.
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clc; clear; mprintf('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-18.4 Page No.405\n'); //Torque L=0.5; F=5800/2; T=0.177*F*L; mprintf('\n Torque = %f in-lb.',T); //Power n=175*2/3; P=T*n/63000; mprintf('\n Power = %f hp.',P);
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clear; clc; printf("\t Example 3.4\n"); NA=7.5*10^-7; //mass flux in gmol/cm^2*s Dab=1.7*10^-5; //diffusivity if SO2 in water in cm^2/s c=1/18.02; //concentration is density/molecular weight in gmol/cm^2*s //SO2 is absorbed from air into water xa1=0.0025; //liquid phase mole fraction at 1 xa2=0.0003; //liquid phase mole fraction at 2 //NA=kc(Ca1-Ca2)=Dab*(Ca1-Ca2)/d k_c=NA/(c*(xa1-xa2)); //k_c=Dab/d=NA/c(xa1-xa2) printf("\nmass transfer coefficient k_c is:%f cm/s",k_c); d=Dab/k_c; printf("\nfilm thickness d is :%f cm",d); //end
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clc clear deff('y=f(x)','y=x/(x*x+5)') a=input("enter lower limit") b=input("enter upper limit") n=input("enter number of sum intervals") h=(b-a)/n x(1)=a; sum=f(a); for i=2:n x(i)=x(i-1)+h end for j=2:2:n sum=sum+4*f(x(j)); end for k=3:2:n sum=sum+2*f(x(k)); end sum=sum+f(b); val=sum*h/3; disp(val,"vlaue of integral by simpsons 1/3rd rule is :");
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PL/SQL Developer Test script 3.0 5 begin -- Call the procedure personas_por_ciudad(pciudad_id => :pciudad_id, p_recordset => :p_recordset); end; 2 pciudad_id 1 1 4 p_recordset 1 <Cursor> 116 0
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function[r]=mag(A) x=real(A) y=imag(A) r=sqrt(x^2+y^2) endfunction j=%i Zl=8+6.2*j//load impedance //for transformer A Ea=6600//secondary induced emf Za=.3+3.2*j//equivalent impedance referred to secondary //for transformer B Eb=6400//secondary induced emf Zb=.2+1.2*j//equivalent impedance referred to secondary Ia=(Ea*Zb+(Ea-Eb)*Zl)/(Za*Zb+(Za+Zb)*Zl) Ib=(Eb*Za-(Ea-Eb)*Zl)/(Za*Zb+(Za+Zb)*Zl) mprintf("Current delivered by transformer A is %f A\nCurrent delivered by transformer B is %f A",mag(Ia),mag(Ib))
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//===================================================================================== //Chapter 12 example 27 clc;clear all; //variable declaration f = 450*10^3; //resistance inHz C = 250*10^-12; //capcaitance in F Rsh = 0.75; //resistance in Ω Q = 105; //Q-factor //calculations w = 2*(%pi)*f; L = 1/(((w)^2)*(C)); //inductance in uH R = ((w*L)/(Q))-Rsh; //resistance of the coil in Ω //result mprintf("inductance = %3.2f uH",(L*10^6)); mprintf("\n resistance of the coil = %3.2f Ω",R);
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//Example 3.10 // quantity of electricity and time taken clc; clear; close; //given data : d=0.1;//in m l=.25;// in m Tc=2;// thickness of coating in mm D=8.9;//density of metal in gm/CC C_density=160;//in A/sq I_efficiency=0.9; S=%pi*d*l; m=S*Tc*10^-3*D*10^3; Z=30.43*10^-8;// in kg/C Q=(m/Z)/3600;// in A-h Q_dash=Q/I_efficiency; disp(Q_dash,"quantity of electricity,Q_dash(A-h) = ") I=C_density*S; t=Q_dash/I; disp(t,"time required,t(hours) = ")
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//Chapter 4 Example 1// clc clear //from table given in the problem we take the required values directly// //thus the values of various loads are taken as l1,l2,l3........ln// //total energy produced=te,average demand=ad,total time=t// l1=400;l2=380;l3=350;l4=300;l5=350;l6=500;l7=700;l8=750;l9=900;l10=1200;l11=1350;l12=1200;l13=1000;l14=950;l15=1250;l16=1300;l17=1400;l18=1300;l19=1500;l20=1800;l21=2000;l22=1950;l23=1000;l24=800;// in kWh// t=24;// in hrs// ad=(l1+l2+l3+l4+l5+l6+l7+l8+l9+l10+l11+l12+l13+l14+l15+l16+l17+l18+l19+l20+l21+l22+l23+l24)/t; printf("\n Average Demand = %.2f kW\n",ad); // load factod=lf,max demand=md// md=l21;//max demand is the highest of all individual demands// lf=ad/md; printf("\n Load factor = %.6f \n",lf); // loss factor=lf,peak loss at peak load=pl,average power loss=apl// lf=0.14; pl=108;// in kW// apl=lf*pl; printf("\n Average power loss = %.2f kW\n",apl); // annual power loss= average power loss*365// apl1=apl*365; printf("\n Annual Power loss = %.2f kW\n",apl1); // demand factor=df,connected demand=cd// cd1=2500;// in kW// df=md/cd1; printf("\n Demand Factor= %.2f \n",df);
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//example 1 //Balancing the Combustion Equation clear clc Mair=28.97 //Molar mass of air in kg/kmol x=8 //no. of moles of CO2 in products y=9 //no. of moles of H2O in products z=7.5 //no. of moles of O2 in products w=75.2 // no. of moles of N2 in products NMair=20*4.76*29 //mass of air in kg NMc=8*12 //mass of carbon in fuel in kg NMh2=2*9 //mass of hydrogen in fuel in kg AF=NMair/(NMc+NMh2) //air fuel ratio in kg air/kg fuel printf("\n Hence, the air fuel ratio for this combustion process is = %.1f kg air/kg fuel. \n",AF);
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clc clear //Initialization of variables M2=18 //Molar mass of water M=170 //Molar mass of octane p=0.4593 //Pressure of octane //psia disp("from steam tables,") vfg=694.9 J=778.2 m=9*18 //Mass of water u1=-2363996 //Btu //calculations hfg=1050.4 //Btu/lbm ufg= hfg- p*vfg*144/J dU=ufg*m Lhv=u1+dU //results printf("Lower heating value = %d Btu/lbm",Lhv) disp("The answers are a bit different due to rounding off error in textbook.")
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t=0:%pi/20:2*%pi; plot2d(t,sin(t),1) plot2d(t,exp(sin(t)),8) title('sin et exp(sint)') xlabel('t axis') ylabel('y axis') legend('sin(t)','exp(sin(t))')
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function [x]= soma(a,b) x = a+b; endfunction
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// Parametres du modele d'Aliev-Paniflov a=0.2; b=0.1; kk=8.0; M1= 0.07; M2=0.3; epsilon=0.01; d=5e-5; // Definition des parametres d'execution nx = 200; // Nombre de points de discretisation dans la direction x ny = 200; // Nombre de points de discretisation dans la direction y lx = 1.0; // Taille de la boite dans la direction x ly = 1.0; // Taille de la boite dans la direction y T = 1000.0; // Temps final // Initialise les inconnues du probleme E = zeros(nx,ny); R = zeros(nx,ny); // Sauvegarde du champs entre 2 iterations du schema temporel E_prec = zeros(nx,ny); // Pas d'espace dx = lx/(nx); x = dx/2.:dx:lx; X = repmat(x,nx,1); Y = repmat(x',1,nx); // Definie une condition initiale for i = 1:nx, for j = 1:nx, if((dx*i-0.5)*(dx*i-0.5)+(dx*j-0.5)*(dx*j-0.5) < 0.005) then E(i,j) = 1.0; end end end //R(1:nx, ny/2:ny) = 1.; E_prec = E; // Contrainte pour le pas de temps rp = kk*(b+1)*(b+1)/4.; dte = (dx*dx)/(d*4+(dx*dx)*(rp+kk)); dtr = 1.0/(epsilon + (M1/M2) *rp); // Pas de temps dt = 0.95*min(dte,dtr); // Coefficient multuplicateur du schema difference finie alpha = d*dt/(dx*dx); Ex=sparse([2:nx;1:nx-1]',ones(nx-1,1)',[nx,nx]); Ax=Ex+Ex'- 2*speye(nx,nx); Ax(1,1) = -1; Ax(nx,nx) = -1; // Conditions de Neumann Ey=sparse([2:ny;1:ny-1]',ones(ny-1,1)',[ny,ny]); Ay=Ey+Ey'-2*speye(ny,ny); Ay(1,1) = -1; Ay(ny,ny) = -1; // Conditions de Neumann A=speye(nx*ny,nx*ny) + alpha*((Ay .*. speye(nx,nx))+(speye(ny,ny) .*. Ax)); C = ((Ay .*. speye(nx,nx))+(speye(ny,ny) .*. Ax))/(dx*dx); // Instant initial t = 0.; R = R(:); i = 0; k = 0; // Resolution du systeme d'equation while (t<T) //while (t<2*dt) t = t + dt; i = i+1; // Applique l'equation du champ electrique E = A*E_prec(:); // Resout les equations explicites E = E - dt*(kk*E .* (E-a) .* (E-1) + E .*R); R = R - dt*(epsilon + M1 *R ./ (E + M2)) .* (R+kk*E .* (E-b-1)); // Mise à jour du champ E_prec = E; if(modulo(i,10)==0) fprintfMat("Model2D/E_" + string(k) + ".dat", matrix(E,nx,ny), "%5.2f"); fprintfMat("Model2D/R_" + string(k) + ".dat", matrix(R,nx,ny), "%5.2f"); //if(k<10) then //fprintfMat("Model2D/E_0000" + string(k) + ".dat", matrix(E,nx,ny), "%5.2f"); //fprintfMat("Model2D/R_0000" + string(k) + ".dat", matrix(R,nx,ny), "%5.2f"); //elseif (k<100) then //fprintfMat("Model2D/E_000" + string(k) + ".dat", matrix(E,nx,ny), "%5.2f"); //fprintfMat("Model2D/R_000" + string(k) + ".dat", matrix(R,nx,ny), "%5.2f"); //elseif (k<1000) then //fprintfMat("Model2D/E_00" + string(k) + ".dat", matrix(E,nx,ny), "%5.2f"); //fprintfMat("Model2D/R_00" + string(k) + ".dat", matrix(R,nx,ny), "%5.2f"); //elseif (k<10000) then //fprintfMat("Model2D/E_0" + string(k) + ".dat", matrix(E,nx,ny), "%5.2f"); //fprintfMat("Model2D/R_0" + string(k) + ".dat", matrix(R,nx,ny), "%5.2f"); //else //fprintfMat("Model2D/E_" + string(k) + ".dat", matrix(E,nx,ny), "%5.2f"); //fprintfMat("Model2D/R_" + string(k) + ".dat", matrix(R,nx,ny), "%5.2f"); //end k = k + 1; end end //E = matrix(E,nx,ny); //R = matrix(R,nx,ny); //surf(E,'interp') //f=gcf(); //f.color_map = jetcolormap(256); //a = gca(); //a.view = "2d";
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//Chapter 13 example 11 //------------------------------------------------------------------------------ clc; clear; // Given data DFM1 = 50; // dispersive fade margin FFM = 30; // flat fade margin DFM2 = 40; // dispersive fade margin // Calculations CFM1 = -10*log10(10^(-FFM/10) + 10^(-DFM1/10)); CFM2 = -10*log10(10^(-FFM/10) + 10^(-DFM2/10)); d_CFM = CFM1 -CFM2; // Output mprintf('CFM increases by %3.2f dB for a 10 dB increase in DFM',d_CFM); //------------------------------------------------------------------------------
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clc; funcprot(0); //Example 14.5 //Initializing the variables D = 0.3; Q = 0.8; rho = 1.2; f = 0.008; L_entry = 10; L_exit = 30; Lt = 20*D;//Transition may be represented by a separation loss equivalent length of 20 × the approach duct diameter K_entry = 4; K_exit = 10 l = 0.4; // length of cross section b = 0.2; // width of cross section //Calculations A = %pi*D^2/4; Dp1 = 0.5*rho*Q^2/A^2*(K_entry + 4*f*(L_entry+Lt)/D); area = l*b; perimeter =2*(l+b); m = area/perimeter; Dp2 = 0.5*rho*Q^2/area^2*(K_exit + f*L_exit/m); Dfan = Dp1+Dp2; disp(Dfan,"fan Pressure input (N/m2) :");
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//Example 14_6 clc(); clear; //To find the difference between the frequency of wave reaching the officer and the car fo=10^10 //Units in Hz vw=3*10^8 //Units in meters/sec vc=25 //Units in meters/sec f1=fo*((vw+vc)/(vw-vc)) //Units in Hz f1=f1-10^10 //Units in Hertz printf("The difference between the both frequencies is=%d Hz",f1) //In text book answer printed wrong as 1670 Hz correct answer is 1666 Hz
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%helps = [%helps; "../../examples/man-examples/helpdir1", "Title1";"../../examples/man-examples/helpdir2", "Title2";];
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//person throws a ball //refer fig. 13.23 //(a) Up the plane atheta=35 //degree aalpha=atheta+20 //degree //maximum range aRangemax=((30*30)/(9.81*(cosd(20))^2))*(sind(2*55-20)-sind(20)) //m //(b) Down the plane //refer fig. 13.24 btheta=(90+20)/2 //degree balpha=55-20 //degree //maximum range bRangemax=((30*30)/(9.81*(cosd(-20))^2))*(sind(2*35+20)-sind(-20)) //m printf("\nUp the plane\nMax Range=%.3f m",aRangemax) printf("\nDown the plane\nMax Range=%.3f m",bRangemax)
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//Problem 6.09: A parallel plate capacitor has nineteen interleaved plates each 75 mm by 75 mm separated by mica sheets 0.2 mm thick. Assuming the relative permittivity of the mica is 5, calculate the capacitance of the capacitor. //initializing the variables: n = 19; // no. of plates L = 75E-3; // in m B = 75E-3; // in m d = 0.2E-3; // in m e0 = 8.85E-12; // in F/m er = 5; //calculation: A = L*B C = e0*er*A*(n-1)/d printf("\n\nResult\n\n") printf("\n Capacitance %.2E F\n",C)
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Example11_7.sce
// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART II : TRANSMISSION AND DISTRIBUTION // CHAPTER 4: OVERHEAD LINE INSULATORS // EXAMPLE : 4.7 : // Page number 186-187 clear ; clc ; close ; // Clear the work space and console // Given data n = 3.0 // Number of insulators C_1 = 0.2 // Capacitance in terms of C C_2 = 0.1 // Capacitance in terms of C // Calculations // Without guard ring e_2_a = 13.0/13.3 // Potential across middle unit as top unit e_1_a = 8.3/6.5*e_2_a // Potential across bottom unit E_a = 1+(1/(8.3/6.5))+(1/e_1_a) // Voltage in terms of e_1 eff_a = E_a/n*100 // String efficiency(%) e1_a = 1/E_a // Voltage across bottom unit as a % of line voltage e2_a = 1/(8.3/6.5)*e1_a // Voltage across middle unit as a % of line voltage e3_a = 1/e_1_a*e1_a // Voltage across top unit as a % of line voltage // With guard ring e_2_b = 15.4/15.5 // Potential across middle unit as top unit e_1_b = 8.3/7.7*e_2_b // Potential across bottom unit E_b = 1+(1/(8.3/7.7))+(1/e_1_b) // Voltage in terms of e_1 eff_b = E_b/n*100 // String efficiency(%) e1_b = 1/E_b // Voltage across bottom unit as a % of line voltage e2_b = 1/(8.3/7.7)*e1_b // Voltage across middle unit as a % of line voltage e3_b = 1/e_1_b*e1_b // Voltage across top unit as a % of line voltage // Results disp("PART II - EXAMPLE : 4.7 : SOLUTION :-") printf("\nWithout guard ring:") printf("\n Voltage across bottom unit, e_1 = %.2f*E", e1_a) printf("\n Voltage across bottom unit, e_2 = %.2f*E", e2_a) printf("\n Voltage across bottom unit, e_3 = %.2f*E", e3_a) printf("\n String efficiency = %.1f percent \n", eff_a) printf("\nWith guard ring:") printf("\n Voltage across bottom unit, e_1 = %.2f*E", e1_b) printf("\n Voltage across bottom unit, e_2 = %.2f*E", e2_b) printf("\n Voltage across bottom unit, e_3 = %.3f*E", e3_b) printf("\n String efficiency = %.2f percent", eff_b)
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Ex4_9.sce
//Example 4 . 9 //MAXIMA SCILAB TOOLBOX REQUIRED FOR THIS PROGRAM //Program to Ca l c u l a t e Group Delay and Phase Delay // y ( n ) =0.25 x ( n )+x ( n􀀀1)+0.25 x ( n􀀀2) clc ; //w=po l y ( 0 , "w") ; syms w; theeta =-w; gd= -diff( theeta ,w); //Group Delay pd=- theeta /w; // Phase Delay disp (gd , 'GROUP DELAY =' ); disp (pd , 'PHASE DELAY =' );
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16_5.sce
clc //initialisation of variables p=14.0//psia t=140//F r=6//ratio w=1//ratio Q=1300//Btu per lb t1=1229//R c=0.24//ft w1=600//ft f=517500//ft-lb per lb v=15.86//ft v1=2.64//cu ft //CALCULATIONS T=Q/(w*c)//R T1=t1+T//R T2=T1/(r)^0.4//R Q1=w*(c)*(T2-w1)//Btu per lb Qs=Q-Q1//Btu per lb W=Qs/Q*100//percent V=v*(T2/w1)//cu ft M1=f/((V-v1)*144)//psia //RESULTS printf('the cylinder volume is=% f psia',M1)
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Ex5_1.sce
clc // Example 5.1.py // Consider the subsonic-supersonic flow through a convergent-divergent nozzle. The // reservoir pressure and temperature are 10 atm and 300 K, repectively. There are // two locations in the nozzle where A/Astar = 6, one in the convergent section and // the other in the divergent section. At each location calculate M, p, T, u. // Variable declaration po = 10.0 // reservoir pressure (in atm) To = 300.0 // reservoir temperature (in K) A_by_Astar = 6.0 // area ratio gamma1 = 1.4 // ratio of specific heat R = 287.0 // gas constant (in J/ Kg K) // Calculations // from table A1 for subsonic flow with A/Astar = 6.0 Msub = 0.097 // mach number in converging section po_by_p = 1.006 // po/p in converging section To_by_T = 1.002 // To/T in converging section psub = 1 / po_by_p * po // pressure (in atm) in converging section Tsub = 1 / To_by_T * To // temperature (in K) in converging section asub = (gamma1*R*Tsub** 0.5) // speed of sound (in m/s) in converging section usub = Msub*asub // velocity (in m/s) in converging section // from table A1 for supersonic flow with A/Astar = 6.0 Msup = 3.368 // mach number in diverging section po_by_p = 63.13 // po/p in diverging section To_by_T = 3.269 // To/T in diverging section psup = 1 / po_by_p * po // pressure (in atm) in diverging section Tsup = 1 / To_by_T * To // temperature (in K) in diverging section asup = (gamma1*R*Tsup** 0.5) // speed of sound (in m/s) in diverging section usup = Msup*asup // velocity (in m/s) in diverging section // Results printf("\n Converging section") printf("\n M = %.3f", Msub) printf("\n p = %.2f atm", psub) printf("\n T = %.1f K", Tsub) printf("\n u = %.2f m/s", usub) printf("\n Divering section") printf("\n M = %.3f", Msup) printf("\n p = %.4f atm", psup) printf("\n T = %.2f K", Tsup) printf("\n u = %.2f m/s", usup)
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ex1_2.sce
// Example 1.2, page no-53 clear clc span=1000 accuracy=1/100 err=span*accuracy printf("(a)\nAs error can be either positive or negative ,\n the probable error at any point on the scale = %d°C",err) max_scale=1200 Range_instr=max_scale+span printf("\n(b)\nRange of the Instrument = %d°C",Range_instr) meter_reading=700 per_of_err=(err/meter_reading)*100 printf("\n(c)\nPercentage of Error = ± %.2f%% ",per_of_err)
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Example_2_9.sce
clear; clc; //Example 2.9 //Caption : Program To Find Work,Heat,del U and del H //Given values //Initial P1=1;//Pressure=1bar T1=298.15;//Temp=298.15K(25`C) V1=0.02479;//Molar Volume=0.02479m^3/mol //Final P2=5;//Pressure=5bar Cv=20.78;//J/mol/K Cp=29.10;//J/mol/K //to Find del_U,del_H by two processes V2=V1*(P1/P2);//m^3(1 mol) disp('m^3',V2,'Final Volume') //Solution //(a)-Cooling at const pressure followed by heating at const Volume T2=T1*(V2/V1);//K disp('K',T2,'Final Temperature') del_H=round(Cp*(T2-T1));//J Q1=del_H;//J del_U1=round(del_H-(P1*(10^5)*(V2-V1)));//J //Second Step del_U2=round(Cv*(T1-T2));//J Q2=del_U2; Q=Q1+Q2; del_U=0; W=del_U-Q;//J del_H=0;//const Temperature disp('(a) Cooling at const Pressure Followed by Heating at const Volume') disp('J',Q,'Heat Required') disp('J',W,'Work Required') disp('J',del_H,'Change in enthalpy') disp('J',del_U,'Change in Energy') //(b)-heating at Const Volume Followed by cooling at const Pressure T2=T1*(P2/P1);//K del_U1=round(Cv*(T2-T1));//J Q1=del_U1; del_H=round(Cp*(T1-T2));//J Q2=del_H; del_U2=round(del_H-(P2*(10^5)*(V2-V1)));//J Q=Q1+Q2; del_U=0; W=del_U-Q;//J del_H=0;//const Temperature disp('(b) Heating at const Volume Followed by Cooling at const Pressure') disp('J',Q,'Heat Required') disp('J',W,'Work Required') disp('J',del_H,'Change in enthalpy') disp('J',del_U,'Change in Energy') //Note disp('Note : The Answer varies From That in the book because in Book 4956.44 has been rounded to 4958 which is absurd') //End
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example1_sce.sce
// chapter 4 //example 4.1 // page 193 ,figure 4.20 R1=120;R2=51*10^3;//given Vsat=15;Vcc=15;Vee=15;Vin=1;//given Vut=((Vsat*R1)/(R1+R2)); disp(Vut)//result threshold in ampere Vult=((-Vsat*R1)/(R1+R2)); disp(Vult)//ampere
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Ex5_3.sce
//Ex 5.3 page 185 clc; clear; close; Vs=400;//V alfa=0.25;// duty cycle delta_I=10;// A L=0.5;// H R=0;// ohm Vo=alfa*Vs;//V //Vo+L*di/dt=Vs -- putting dt=Ton & di=delta_I Ton=delta_I/((Vs-Vo)/L)*1000;// ms T=Ton/alfa;// ms f=1/T*1000;//Hz printf('\n chopping frequency = %d Hz',f)
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10_28.sce
clear; clc; funcprot(0); //Example - 10.28 //Page number - 374 printf("Example - 10.28 and Page number - 375\n\n"); //Given T = 150 + 273.15;//[K] Tc = 647.1;//[K] Pc = 220.55;//[bar] Pc = Pc*10^(5);//[Pa] w = 0.345; R = 8.314;//[J/mol-K] - Universal gas constant // Let us assume a pressure of 100 kPa. P_1 = 100*10^(3);//[Pa] // At 100 kPa and 423.15 K, from Peng-Robinson equation of stste m = 0.37464 + 1.54226*w - 0.26992*w^(2); Tr = T/Tc; alpha = (1 + m*(1 - Tr^(1/2)))^(2); a = ((0.45724*(R*Tc)^(2))/Pc)*alpha;//[Pa*m^(6)/mol^(2)] b = (0.07780*R*Tc)/Pc;//[m^(3)/mol] // Cubic form of Peng-Robinson equation of stste is given by // V^(3)+(b-(R*T)/P)*V^(2)-((3*b^(2))+((2*R*T*b)/P)-(a/P))*V+b^(3)+((R*T*(b^(2))/P)-((a*b)/P)=0; // Solving the cubic equation deff('[y]=f(V)','y=V^(3)+(b-(R*T)/P_1)*V^(2)-((3*b^(2))+((2*R*T*b)/P_1)-(a/P_1))*V+b^(3)+((R*T*(b^(2)))/P_1)-((a*b)/P_1)'); V1 = fsolve(-1,f); V2 = fsolve(0,f); V3 = fsolve(1,f); // The largest root and the smallest root is considered for liquid phase and vapour phase respectively. V_liq = V1;//[m^(3)/mol] - Molar volume in liquid phase V_vap = V3;//[m^(3)/mol] - Molar volume in vapour phase // The compressibility factor is given by Z_vap = (P_1*V_vap)/(R*T);// For liquid phase Z_liq = (P_1*V_liq)/(R*T);// For vapour phase // The expression for fugacity of Peng Robinson equation is // log(f/P) = (Z-1) - log(Z-((P*b)/(R*T))) - (a/(2*2^(1/2)*b*R*T))*log((Z+(1+2^(1/2))*((P*b)/(R*T)))/((Z+(1-2^(1/2))*((P*b)/(R*T))) // For vapour phase f_P_vap = exp((Z_vap-1) - log(Z_vap-((P_1*b)/(R*T))) - (a/(2*2^(1/2)*b*R*T))*log((Z_vap+(1+2^(1/2))*((P_1*b)/(R*T)))/(Z_vap+(1-2^(1/2))*((P_1*b)/(R*T))))); // For liquid phase f_P_liq = exp((Z_liq-1) - log(Z_liq-((P_1*b)/(R*T))) - (a/(2*2^(1/2)*b*R*T))*log((Z_liq+(1+2^(1/2))*((P_1*b)/(R*T)))/(Z_liq+(1-2^(1/2))*((P_1*b)/(R*T))))); // Therefore f_liq/f_vap can be calculated as fL_fV = (f_P_liq/f_P_vap); // The two values (f/P)_vap and (f/P)_vap are not same [ (f_P_liq/f_P_vap) >1 ]; therefore another pressure is to be assumed. The new pressure be P_new = P_1*(f_P_liq/f_P_vap);//[Pa] // At P_new and 423.15 K, from Peng-Robinson equation of stste // V^(3)+(b-(R*T)/P)*V^(2)-((3*b^(2))+((2*R*T*b)/P)-(a/P))*V+b^(3)+((R*T*(b^(2))/P)-((a*b)/P)=0; // Solving the cubic equation deff('[y]=f(V)','y=V^(3)+(b-(R*T)/P_new)*V^(2)-((3*b^(2))+((2*R*T*b)/P_new)-(a/P_new))*V+b^(3)+((R*T*(b^(2)))/P_new)-((a*b)/P_new)'); V4 = fsolve(-1,f); V5 = fsolve(0,f); V6 = fsolve(1,f); // The largest root and the smallest root is considered for liquid phase and vapour phase respectively. V_liq_2 = V4;//[m^(3)/mol] - Molar volume in liquid phase V_vap_2 = V6;//[m^(3)/mol] - Molar volume in vapour phase // The compressibility factor is given by Z_vap_2 = (P_new*V_vap_2)/(R*T);// For liquid phase Z_liq_2 = (P_new*V_liq_2)/(R*T);// For vapour phase // For vapour phase f_P_vap_2 = exp((Z_vap_2-1) - log(Z_vap_2-((P_new*b)/(R*T))) - (a/(2*2^(1/2)*b*R*T))*log((Z_vap_2+(1+2^(1/2))*((P_new*b)/(R*T)))/(Z_vap_2+(1-2^(1/2))*((P_new*b)/(R*T))))); // For liquid phase f_P_liq_2 = exp((Z_liq_2-1) - log(Z_liq_2-((P_new*b)/(R*T))) - (a/(2*2^(1/2)*b*R*T))*log((Z_liq_2+(1+2^(1/2))*((P_new*b)/(R*T)))/(Z_liq_2+(1-2^(1/2))*((P_new*b)/(R*T))))); // Therefore f_liq/f_vap can be calculated as fL_fV_2 = (f_P_liq_2/f_P_vap_2); // And new pressure is given by P_new_prime = P_new*(f_P_liq_2/f_P_vap_2);//[Pa] P_new_prime = P_new_prime*10^(-5); // Since the change in pressure is small, so we can take this to be the vapour pressure at 150 C printf(" The vapour pressure of water using Peng-Robinson equation of stste is %f bar\n",P_new_prime);
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6_12.sce
clc //initialisation of variables v= 5 //ft/sec Q= 500 //cuses w= 25 //ft g= 32.2 //ft/sec^2 //CALCULATIONS h= (Q/v)/w E= h+(v^2/(2*g)) //RESULTS printf ('Specific energy = %.2f ft ',E)
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ExampleA3.sce
clc clear //Page number 470 //Input data t=0;//The given temperature in degree centigrade E=5.64*10^-21;//The mean kinetic energy of molecules of hydrogen in J R=8.32;//Universal gas constant in J/mole-K //Calculations T=t+273;//The given temperature in K N=(3/2)*(R/E)*(T);//Avogadros number //Output printf('The Avogadro number is N = %3.4g ',N)
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Exa4_4.sce
//Exa 4.4 clc; clear; close; //Alternative 1 : disp("In 1st alternative down payment : Rs. 16,00,000"); //Alternative 2 : P0=400000;//in Rs P=200000;//in Rs i=18;//in % per annum n=10;//in years //Formula : (P/A,i,n) : (((1+i/100)^n)-1)/((i/100)*(1+i/100)^n) PW=P0+P*(((1+i/100)^n)-1)/((i/100)*(1+i/100)^n);//in RS disp(PW,"The present worth of alternative 2 in RS. : "); disp("The present worth of 2nd alternative is less than that of first one i.e., complete downpayment of Rs. 1600000. Hence, select 2nd alternative.");
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clear; clc; disp('Example 7.8'); // aim : To determine // the change of entropy // Given values m = .3;// [kg] P1 = 350;// [kN/m^2] T1 = 273+35;// [K] P2 = 700;// [kN/m^2] V3 = .2289;// [m^3] cp = 1.006;// [kJ/kg K] cv = .717;// [kJ/kg K] // solution // for constant volume process R = cp-cv;// [kJ/kg K] // using PV=mRT V1 = m*R*T1/P1;// [m^3] // for constant volume process P/T=constant,so T2 = T1*P2/P1;// [K] s21 = m*cv*log(P2/P1);// formula for entropy change for constant volume process mprintf('\n The change of entropy in constant volume process is = %f kJ/kg K\n',s21); // 'For the above part result given in the book is wrong V2 = V1; // for constant pressure process T3 = T2*V3/V2;// [K] s32 = m*cp*log(V3/V2);// [kJ/kg K] mprintf('\n The change of entropy in constant pressure process is = %f kJ/kg K\n',s32); // there is misprint in the book's result // End
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//Chapter 21, Problem 6 clc; N=10; //turns ratio v1=2.5e3; //primary voltage P=5000; //power v2=v1/N; //secondary voltage i2=P/v2; //secondary current Rl=v2/i2; //resistance in ohm i1=i2/N; //primary current printf("(a) Full-load secondary current = %d A\n\n",i2); printf("(b) Minimum value of load resistance = %.1f ohms\n\n",Rl); printf("(c) Primary current = %d A\n\n",i1);
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errcatch(-1,"stop");mode(2);//laplace// syms t s; y=laplace('13',t,s); disp(y,"ans=") exit();
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errcatch(-1,"stop");mode(2);//Example 5_9_u1 ; ; //To determine the interplanar spacing h=6.63*10^-34 //units in m^2 kg s^-1 m=9.1*10^-31 //units in Kgs e=1.6*10^-19 //units in coulombs v=844 //units in Volts lamda=h/sqrt(2*m*e*v) //units in meters n=1 theta=58 //units in degrees d=(n*lamda)/(2*sin(theta*(%pi/180))) //units in meters printf("The interplanar spacing d=") disp(d) printf("meters") exit();
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//Chapter 18 //Example 18_19 //Page 452 clear;clc; mva=20; kv=11; xn=5; x1=20; x2=10; Er=kv*1000/sqrt(3); printf("Phase emf of red phase = %d V \n\n", Er); //from the reactance diagram given in the text; r_x1=x1/2; r_x2=x2/2; r_xn=30; X1=r_x1*kv^2*10/mva/1000; X2=r_x2*kv^2*10/mva/1000; X0=r_xn*kv^2*10/mva/1000; Ir=3*Er/(X1+X2+X0)/%i; printf("X1 = %.3f ohm \n\n", X1); printf("X2 = %.3f ohm \n\n", X2); printf("X0 = %.3f ohm \n\n", X0); printf("Fault current = j(%d) A \n\n", imag(Ir));
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//To calculate the distance between two adjacent atoms MW = 23+35.5; //molecular weight of NaCl, gm N = 6.023*10^23; //avagadro number, mol-1 rho = 2.18; //density of NaCl, gm/cm^3 M = MW/N; //mass of NaCl molecule, gm n = rho/M; //number of molecules per unit volume n = 2*n; //since NaCl is diatomic, atoms/cm^3 //length of edge of unit cube is n*a //volume V = n^3*a^3 = 1 cm^3 V = 1; //volume of unit cube, cm^3 a = (V/n)^(1/3); //distance between two adjacent atoms, cm a = a*10^8; //distance between two adjacent atoms,A printf("distance between two adjacent atoms is %5.2f A",a);
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function y=f(x) y= exp(-x)-x endfunction a=-1; b=1; c=0; d=0; tol=1E-4; i=0; while(abs(a-b)>=tol) i=i+1; c=(a+b)/2; d=abs(a-b); fa=f(a); fb=f(b); fc=f(c); if(fa*fc<0)then a = a; else a = c; end if(fb*fc<0)then b = b; else b = c; end printf("%d) %.6f\n",i,c) end
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//Example 11_5 clc;clear; //Properties rho_ag=1.20;// kg/m^3 rho_ac=0.312;// kg/m^3 C_Lmax1=1.52;// The maximum lift coefficient of the wing with flaps C_Lmax2=3.48;// The maximum lift coefficient of the wing without flaps //Given values m=70000;// kg A=150;// m^2 V=558;/// km/h g=9.81;// m/s^2 // Calculation //(a) W=m*g;// N V=V/3.6;// m/s V_min1=sqrt((2*W)/(rho_ag*C_Lmax1*A));// m/s V_min2=sqrt((2*W)/(rho_ag*C_Lmax2*A));// m/s V_1s=1.2*V_min1*3.6;// 1 m/s=3.6 km/h printf('(a)Without flaps:V_min1,safe =%0.0f km/h\n',V_1s); V_2s=1.2*V_min2*3.6;// 1 m/s=3.6 km/h printf(' With flaps:V_min2,safe =%0.0f km/h\n',V_2s); //(b) F_l=W;// N C_l=F_l/(1/2*rho_ac*V^2*A);// The lift coefficient //For the case with no flaps, the angle of attack corresponding to this value of C_L is determined from Fig. 11–45 to be alpha=10;// The angle of attack in degree printf('(b)The angle of attack,alpha~=%0.0f degree\n',alpha); //(c) // From Fig.11-45,C_d~=0.03 C_d=0.03;// The drag coefficient F_d=(C_d*A*rho_ac*(V^2/2))/1000;//kN P=F_d*V;// kW printf('(c)The power that needs to be supplied to provide enough thrust to overcome wing drag,P=%0.0f kW\n',P); // The answer vary due to round off error
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function xdot = myfunc4(t,x) xdot = zeros(2,1); a=0.5; xdot(1) = -a*x(1)-sin(x(2)); xdot(2) = x(1); endfunction
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//Chemical Engineering Thermodynamics //Chapter 3 //First Law of Thermodynamics //Example 3.10 clear; clc; //Given V = 0.3;//Volume of the tank in m^3 P1 = 1;//Initial pressure of the tank in atm P2 = 0;//Final pressure of the tank in atm T = 298;//Temperature of the tank in K t = 10;//evacuation time in min //delN=(V/(R*T)*delP)..(a) change in moles as V and T are constant //delW=delN*R*T*lnP..(b)pump work required //From (a)&(b),delW=V*delP*lnP //To calculate the pump work required //On doing integration of dW we will get W = V*(P1-P2);//pump work done in J/sec W1=(W*(1.033*10^4))/(75*600); mprintf('The pump work required is %f hp',W1); //end
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clc // Given that thetaF= 1540 // Temperature of mould face in degree centigrate ThetaO = 28 // Initial temperature of mould in Degree centigrate L= 272e3 // Latent heat of iron in J/Kg Dm = 7850 // Density of iron in Kg/m^3 Cs = 0.67e+3 //Specific heat of iron in J/Kg-K C = 0.376e3 //Specific heat of copper in J/Kg-K Ks = 83 // Conductivity of iron in W/m-K K = 398 // Conductivity of copper in W/m-K D= 8960 // Density of copper in Kg/m^3 h = .1 // Height in m // Sample Problem 6 on page no. 73 printf("\n # PROBLEM 2.6 # \n") zeta1=0.98//By solving eqauation- zeta*exp(zeta^2)*erf(zeta)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.98 AlphaS = Ks /(Dm*Cs) ts1 = h^2 / (16*(zeta1^2) * AlphaS)//In sec ts1_=ts1/3600 // In hour Phi = sqrt((Ks*Dm*Cs)/(K*D*C)) zeta2=0.815//By solving eqauation- zeta*exp(zeta^2)*(erf(zeta)+Phi)=((thetaF-thetaO)*Cs)/(sqrt(pi)*L), zeta = 0.815 ts2 = h^2 / (16*(zeta2^2) * AlphaS)//In sec ts2_=ts2/3600 // In hour thetaS= (thetaF-(L*(sqrt(%pi))*zeta2*(exp(zeta2^2))*erf(zeta2))/Cs) printf("\n Solidification time for slab-shaped casting when the casting is done in a water cooled copper mould = %f hr,\n Solidification time for slab-shaped casting when the casting is done in a very thick copper mould = %f hr,\n The surface temperature of the mould = %f° C", ts1_,ts2_,thetaS)
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#yaml dsp_debug_scenario: name: IV3_cmdq_test local: model_file: dsp_iv3/iv3_dirty.nnc input: - dsp_iv3/nnc_input_0.bin golden: - dsp_iv3/nnc_output_0.bin target: model_file: /data/raw/iv3_dirty.nnc input: - /data/raw/nnc_input_0.bin golden: - /data/raw/nnc_output_0.bin max_layer: 108
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//(Threaded Joints) Example 7.23 //Refer Fig.7.47 on page 267 //Maximum pressure in the vessel Pmax (MPa) Pmax = 1 //Minimum pressure in the vessel Pmin (MPa) Pmin = 0 //Seating pressure for the gasket PSeat (MPa) PSeat = 5 //Number of bolts N N = 8 //Assume the stiffness of the bolts to be 1N/mm kb kb = 1 //Calculate the stiffness of the parts kc kc = 4 * kb //Factor of safety fs fs = 2 //Ultimate tensile strength of bolt material Sut (N/mm2) Sut = 780 //Yield tensile strength of the bolt material Syt (N/mm2) Syt = 580 //Endurance limit in bending Sdash (N/mm2) Sdash = 260 //Fatigue stress concentration factor Kf Kf = 3 //Inner diameter of the gasket Di (mm) Di = 300 //Outer diameter of the gasket Do (mm) Do = 300 + (2 * 50)
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//Tested on Windows 7 Ultimate 32-bit //Chapter 9 Frequency Response of Amplifier Pg no. 310 clear; clc; //Given VCC=15;//collector supply voltage in volts RC=2.2D3;//collector resistance in ohms RE=470;//emitter resistance in ohms R1=33D3;//divider network resistance R1 in ohms R2=10D3;//divider network resistance R2 in ohms VBE=0.7;//forward voltage drop of emitter diode in volts B=150;//DC CE current gain beta Rs=600;//source internal impedance in ohms RL=4.7D3;//load resistance in ohms C1=0.1D-6;//input coupling capacitance in farads C2=50D-6;//emitter bypass capacitance in farads C3=0.1D-6;//output coupling capacitance in farads re=4;//a.c. emitter resistance in ohms //Solution Rin=1/(1/R1+1/R2+1/(B*re));//thevenised input network resistance in ohms fc_input=1/(2*%pi*(Rs+Rin)*C1);//input cutoff frequency in hertz Rth=1/(1/R1+1/R2+1/Rs);//thevenised bypass network resistance in ohms Rin_emitter=7.7;//resistance looking into the emitter in ohms fc_bypass=1/(2*%pi*1/(1/RE+1/Rin_emitter)*C2);//bypass cutoff frequency in hertz Rout=RC+RL;//thevenised output network resistance in ohms fc_output=1/(2*%pi*Rout*C3);//output cutoff frequency in hertz s=poly(0,'s') F=syslin('c',8*%pi^3*(fc_input*fc_bypass*fc_output)/(s+2*%pi*fc_output)/(s+2*%pi*fc_bypass)/(s+2*%pi*fc_input)); clf; gainplot(F,100,10000,"Bode Plot for given amplifier in Example 9.8");
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function [x,y,typ]=IFTHEL_f(job,arg1,arg2) // Copyright INRIA 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; graphics=arg1(2);label=graphics(4) model=arg1(3); if label==[] then label=string(1);end while %t do [ok,inh,label]=getvalue('Set parameters',.. ['Inherit (1: no, 0: yes)'],list('vec',1),label) if ~ok then break,end if inh==0 then inh=[]; else inh=1;end [model,graphics,ok]=check_io(model,graphics,1,[],inh,[1;1]) if ok then graphics(4)=label; model(4)=inh; model(1)(2)=-1 x(2)=graphics;x(3)=model break end end case 'define' then model=list(list('ifthel',-1),1,[],1,[1;1],[],[],[],[],'l',[-1 -1],[%f %f],' ',list()) gr_i=['txt=[''If in>=0'';'' '';'' then else''];'; 'xstringb(orig(1),orig(2),txt,sz(1),sz(2),''fill'');'] label=string(1); x=standard_define([3 3],model,label,gr_i) end
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Example_2_7_11.sce
// Example 2.7.11 page 2.29 clc; clear all; n1=1.5; //refractive index of core n2=1.38; //refractive index of cladding lamda=1300d-9; //Wavelength a=25d-6; //core radius NA=sqrt(n1^2 - n2^2); //computing Numerical aperture theta= asind(NA); //computing acceptance angle solid_angle=%pi*(NA)^2; //computing solid angle v= 2*%pi*a*NA/lamda; //computing normalized frequency M=(v)^2/2; //computing guided modes M=round(M); printf("\nNumerical aperture is %.2f.\nNormalized frequency is %.2f.\nAcceptance angle is %.2f degrees.\nSolid angle is %.3f radians.\nTotal number of modes are %d.",NA,v,theta,solid_angle,M); printf("\n\n NOTE - Calculation error in the book.\n(2.25-1.9)^0.5=0.59; they have taken 0.35"); //Calculation error in the book.(2.25-1.9)^0.5=0.59; they have taken 0.35 //answers in the book, //Numerical aperture is 0.35.(incorrect) //Normalized frequency is 42.26.(incorrect) //Acceptance angle is 20.48 degrees.(incorrect) //Solid angle is 0.384 radians.(incorrect)
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1_4.sce
//Eg-1.4 //pg-13 clear clc a=input("enter any number") r=a-round(a/2)*2; if r==0 then disp("even number") else disp("odd number") end
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Ex6_2.sce
clc clear //DATA GIVEN Ms=5.4; //mass of steam used in kg/kWh p=50; //pressure of steam in bar Tsup=350; //temp. of steam in deg celsius eta=82; //boiler efficiency in % Tfw=150; //feed water temp. in deg cel;sius C=28100; //calorific value of coal in kJ rate=500; //cost of coal/tonne in Rs //boiler efficiency is given by, eta=Ms*(hsup-hf1)/(Mf*C) //from steam table, at 45 bar and 350deg celsius, hsup=3068.4 kJ/kg h=3068.4; //enthalpy at 45 bar and 350 deg celsius hf1=4.18*(Tfw-0); //hf1 at 150 deg celsius in kJ/kg //subs. these in eq. of boiler efficiency Mf=Ms*(h-hf1)/((eta/100)*C); //mass of coal required in kg/kWh cost=(Mf/1000)*rate*100; //cost of coal in paisa/kWh printf(' (i) The mass of coal required is: %5.3f kg/kWh. \n',Mf); printf(' (ii) The Total cost of fuel(coal) is: %2.1f paisa/kWh. \n',cost); //NOTE:in text book //in question pressure is given as =50 bar //but from steam table enthalpy is found at 45 bar
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Ex3_13.sce
// Examle 3.13 // From Diagram (3.26) Apply KVL to get 24-4I-2I+18I= 0 I=(-24/12); // Current disp(' The value of Current = '+string(I)+' Amp'); V1=4*I; // Voltage across 4 Ohm Resistor p=-(4.5*V1*I); // Power absorbed disp(' Power absorbed by dependent source = '+string(p)+' Watt'); V=24; // Independent voltage source R=V/I; // Resistence Seen from Independent source disp(' Resistence Seen from Independent source = '+string(R)+' Ohm'); // p 67 3.13
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autoAndCrossCorrelationProperties.sce
//auto and cross correlation properties clc clear close /////////////////////////////// //auto correlation properties// /////////////////////////////// x = input("sequence x(n): ") y = input("sequence y(n): ") disp("Auto correlation properties: ") //1. rxx(l) = rxx(-l) disp("1. rxx(l) = rxx(-l)") [lhs lag1] = xcorr(x,x) [rhs lag2] = xcorr(x,x) rhs = flipdim(rhs,2) disp("rxx(l) : ",lhs) disp("rxx(-l): ",rhs) //2. |rxx(l)| <= sqrt(rxx(0).rxx(0)) disp("2. |rxx(l)| <= sqrt(rxx(0).rxx(0))") [lhs1 lag3] = xcorr(x,x) rhs1 = sqrt(max(xcorr(x,x))*max(xcorr(x,x))) disp("lhs: ",lhs1) disp("no samples in lhs in greater than : ",rhs1) //plots figure(0) subplot(211) plot2d3(lag1,lhs) plot(lag1,lhs,'red.') title("rxx(l)") xlabel("---> samples n") ylabel("Amplitude") a1 = gca() a1.x_location = 'origin' a1.y_location = 'origin' subplot(212) plot2d3(lag2,rhs) plot(lag2,rhs,'red.') title("rxx(-l)") xlabel("---> samples n") ylabel("Amplitude") a2 = gca() a2.x_location = 'origin' a2.y_location = 'origin' disp("--------------------------------------") /////////////////////////////// //cross correlation properties// /////////////////////////////// disp("Cross correlation properties: ") //1. rxy(l) = rxy(-l) disp("1. rxy(l) = rxy(-l)") [lhs lag1] = xcorr(x,y) [rhs lag2] = xcorr(x,y) rhs = flipdim(rhs,2) disp("rxy(l) : ",lhs) disp("rxy(-l): ",rhs) //2. |rxy(l)| <= sqrt(rxx(0).ryy(0)) disp("2. |rxy(l)| <= sqrt(rxx(0).ryy(0))") [lhs1 lag3] = xcorr(x,y) rhs1 = sqrt(max(xcorr(x,x))*max(xcorr(y,y))) disp("lhs: ",lhs1) disp("no samples in lhs in greater than : ",rhs1) //plots figure(1) subplot(211) plot2d3(lag1,lhs) plot(lag1,lhs,'red.') title("rxy(l)") xlabel("---> samples n") ylabel("Amplitude") a1 = gca() a1.x_location = 'origin' a1.y_location = 'origin' subplot(212) plot2d3(lag2,rhs) plot(lag2,rhs,'red.') title("rxy(-l)") xlabel("---> samples n") ylabel("Amplitude") a2 = gca() a2.x_location = 'origin' a2.y_location = 'origin'
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Ex4_4.sce
// Example 4.4 clear all; clc; // Given data ratpower = 1075; // Output rated electrical power in MWe of the reactor delpower_yr = 255000; // Net output power delivered in one year in terms of MWd time_refuel = 28; // Number of days the plant was shutdown for refuelling time_repairs = 45; // Number of days the plant was shutdown for repairs time_convrepairs = 18; // Number of days the plant was shutdown for conventional repairs // 1. // 1 year = 365 days ratpower_yr = ratpower*365; // Net output rated power in one year in terms of MWd // Calculation cap_factor = delpower_yr/ratpower_yr; // Result printf(" \n Plant capacity factor = %d percent\n",ceil(cap_factor*100)); // 2. // Number of days the plant was shutdown in one year total_shutdown = time_refuel+time_repairs+time_convrepairs; // Number of days the plant was operable in one year total_operation = 365-total_shutdown; // Calculation ava_factor = total_operation/365; // Result printf(" \n Plant availability factor = %d percent\n",ava_factor*100);
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Ex18_25.sce
//Initilisation of variables d=2/12 //ft v=80 //ft/s g=32.2 //ft/s^2 //Calculations //Mass flow reate without time m=(1/4)*%pi*d^2*v*(62.4/g) //Let P=force of plate on mass m of water P=m*(0-v) //lb //Result clc printf('The force water exerts on the plate is %f lb',-P )
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Inc16.tst
load Inc16.hdl, output-file Inc16.out, compare-to Inc16.cmp, output-list x%B1.16.1 out%B1.16.1; set x %B0000000000000000, // in = 0 eval, output; set x %B1111111111111111, // in = -1 eval, output; set x %B0000000000000101, // in = 5 eval, output; set x %B1111111111111011, // in = -5 eval, output;
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chapter3_ex14.sce
clc clear //input v=240;//voltage of a d.c. shunt motor in volts ra=0.4;//armature resistance of d.c. shunt motor in ohms rf=120;//armature resistance of d.c. shunt motor in ohms is=22;//supply current in amperes w=600;//angular velocity of motor in rev/min il=30;//load current in amperes //calculations //armature reaction is neglected W=(w*(2*%pi))/60;//angular velocity in rad/s fi=v/rf;//feild current in amperes ai=is-fi;//armature current in amperes e=v-(ai*ra);//e.m.f. in volts t1=(e*ai)/W;//torque when current is 20A in newton meter aI=il-fi;//changed armature current in amperes t2=t1*(aI/is);//torque when current is 30A in newton meter //output mprintf('with a supply current of 30A the motor produces a total torque of %3.1f Nm',t2)
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Ex4_15.sce
clear // //Given //Variable declaration L=1.82*1000 //Length of rod in mm h1=30 //Height through which load falls in mm h2=47.5 //Fallen height in mm sigma=157 //Maximum stress induced in N/sq.mm E=2.1e5 //Youngs modulus in N/sq.mm //Calculation U=sigma**2/(2*E) //Strain energy stored in the rod in N-m delL=sigma*L/E //Extension of the rod in mm Tot_dist=h1+delL //Total distance in mm //case(i):Stress induced in the rod if the load is applied gradually sigma1=((U/Tot_dist)*L) //case(ii):Maximum stress if the load had fallen from a height of 47.5 mm sigma2=((sigma1)*(1+(sqrt(1+((2*E*h2)/(sigma1*L)))))) //Result printf("\n Stress induced in the rod = %.1f N/mm^2",sigma1) printf("\n NOTE:The given answer for stress(2nd case) in the book is wrong.The correct answer is,") printf("\n Maximum stress if the load has fallen = %.2f N/mm^2",sigma2)
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Ex15_9.sce
// Exa 15.9 clc; clear all; // Given data // Refering Fig. 15.2(a)- All pass filter f=2.5;// Input frequency in kHz // Solution disp(" Let C=0.01 micro farads and R= 15 k Ohms"); C=0.01;// micro farads R=15;// k Ohms Phi=2*atan(2*%pi*f*C*R); // phase angle in radians printf(' This means that the output voltage Vo has the same frequency and amplitude as the input voltage but lags it by - %d degrees\n',Phi*180/%pi);
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12_6.sce
Tmax=200 Tstart=80 p=poly([1,-2*Tmax/Tstart,1],"smaxT","coeff") smaxT=roots(p) smaxT=smaxT(2) disp(smaxT) p=poly([1,-4,1],"w","coeff") w=roots(p) sfl=smaxT/w(1) disp(sfl) ratio=sqrt(((smaxT/sfl)^2+1)/(smaxT^2+1)) disp(ratio) p=poly([1,-((smaxT/sfl)^2+1)*sfl,smaxT^2],"k","coeff") k=roots(p) k=k(2) disp(k)
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ex7_5.sce
// Exa 7.5 format('v',7);clc;clear;close; // Given data y1 = 8;// in units y2 = 10;// in units phi = asind(y1/y2);// phase difference in degree disp(phi,"The phase difference in degree is");
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7_9.sce
clc //Initialization of variables g=9.81 //m/s^2 rho=10^3 //kg/m^3 d=0.3 //m per=25/100 Q=0.1 //m^3/s k0=0.025*10^-2 //m nu=0.000001 year=10 //calculations V=Q/(%pi/4 *d^2) RN=V*d/nu e1=k0/d f1=0.019 f2=(1+per)*f1 e2=0.002 k2=e2*d rate = (k2-k0)*100/year //results printf("Rate of increase =%.4f cm/year",rate)
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fact.sci
function [f] = fact (n) //cette fonction calcule la factorielle d'un nombre entier if (n-floor(n)~= 0 | n<0) error("erreur dans fact : l""argument doit etre entier"); end if n==0 f=1; else f = prod(1:n) end endfunction
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Ex5_10.sce
clc; v=6600; // rated voltage of motor xs=20 ; // per phase synchronous reactance p=500000; // VA rating of motor il=p/(sqrt(3)*v); // rated armature current vt=v/sqrt(3); // per phase rated voltage disp('case a'); de=10; // load angle c1=1; c2=-2*vt*cosd(de); c3=vt^2-(il*xs)^2; // coefficients of quadratic equation in Ef p= [ c1 c2 c3 ]; Ef=roots(p); printf('Per phase excitation EMF at lagging pf is %f v\n',Ef(2)); printf('Excitation line EMF at lagging pf is %f v\n',sqrt(3)*Ef(2)); printf('Per phase excitation EMF at leading pf is %f v\n',Ef(1)); printf('Excitation line EMF at leading pf is %f v\n',sqrt(3)*Ef(1)); disp('case b'); disp('For lagging pf'); pd=(3*vt*Ef(2)*sind(de))/xs; pf=pd/(sqrt(3)*v*il); printf('Mechanical power developed is %f W\n',pd); printf('Power factor is %f lagging\n',pf); disp('For leading pf'); pd=(3*vt*Ef(1)*sind(de))/xs; pf=pd/(sqrt(3)*v*il); printf('Mechanical power developed is %f W\n',pd); printf('Power factor is %f leading\n',pf); disp('case c'); p=200000; // delivered power de=90; // load angle for falling out of step // motor falls out of step at de= 90 degrees Ef=(p*xs)/(3*sind(de)*vt); printf('Minimum excitation voltage per phase is %f v',Ef);
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Example3_8.sce
//clear// //Example3.8:Fourier Series Representation of Periodic Impulse Train clear; clc; close; T =4; T1 = T/4; t = [-T,0,T]; xt = [1,1,1]; //Generation of Periodic train of Impulses t1 = -T1:T1/100:T1; gt = ones(1,length(t1));//Generation of periodic square wave t2 = [-T1,0,T1]; qt = [1,0,-1];//Derivative of periodic square wave Wo = 2*%pi/T; ak = 1/T; b(1) = 0; c(1) = 2*T1/T; for k =1:5 b(k+1) = ak*(exp(sqrt(-1)*k*Wo*T1)-exp(-sqrt(-1)*k*Wo*T1)); if(abs(b(k+1)) < =0.1) b(k+1) =0; end c(k+1) = b(k+1)/(sqrt(-1)*k*Wo); if(abs(c(k+1)) < =0.1) c(k+1) =0; end end k = 0:5 disp('Fourier Series Coefficients of periodic Square Wave') disp(b) disp('Fourier Series Coefficients of derivative of periodic square wave') disp(c) //Plotting the periodic train of impulses figure subplot(3,1,1) a = gca(); a.y_location = "origin"; a.x_location = "origin"; a.data_bounds=[-6,0;6,2]; plot2d3('gnn',t,xt,5) poly1 = a.children(1).children(1); poly1.thickness = 3; title('x(t)') //Plotting the periodic square waveform subplot(3,1,2) a = gca(); a.y_location = "origin"; a.x_location = "origin"; a.data_bounds=[-6,0;6,2]; plot2d(t1,gt,5) poly1 = a.children(1).children(1); poly1.thickness = 3; plot2d(T+t1,gt,5) poly1 = a.children(1).children(1); poly1.thickness = 3; plot2d(-T+t1,gt,5) poly1 = a.children(1).children(1); poly1.thickness = 3; title('g(t)') //Plotting the periodic square waveform subplot(3,1,3) a = gca(); a.y_location = "origin"; a.x_location = "origin"; a.data_bounds=[-6,-2;6,2]; poly1.thickness = 3; plot2d3('gnn',t2,qt,5) poly1 = a.children(1).children(1); poly1.thickness = 3; plot2d3('gnn',T+t2,qt,5) poly1 = a.children(1).children(1); poly1.thickness = 3; plot2d3('gnn',-T+t2,qt,5) poly1 = a.children(1).children(1); poly1.thickness = 3; title('q(t)')
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Ex10_2.sce
printf("\t example 10.2 \n"); printf("\t approximate values are mentioned in the book \n"); T1=250; // inlet hot fluid,F T2=250; // outlet hot fluid,F t1=95; // inlet cold fluid,F t2=145; // outlet cold fluid,F W=16000; // lb/hr w=423; // lb/hr printf("\t 1.for heat balance \n"); printf("\t for kerosene \n"); c=0.5; // Btu/(lb)*(F) Q=((W)*(c)*(t2-t1)); // Btu/hr printf("\t total heat required for kerosene is : %.0f Btu/hr \n",Q); printf("\t for steam \n"); l=945.5; // Btu/(lb) Q=((w)*(l)); // Btu/hr printf("\t total heat required for steam is : %.2e Btu/hr \n",Q); delt1=T2-t1; //F delt2=T1-t2; // F printf("\t delt1 is : %.0f F \n",delt1); printf("\t delt2 is : %.0f F \n",delt2); LMTD=((delt2-delt1)/((2.3)*(log10(delt2/delt1)))); printf("\t LMTD is :%.0f F \n",LMTD); tc=((t1)+(t2))/2; // caloric temperature of cold fluid,F printf("\t caloric temperature of cold fluid is : %.0f F \n",tc); printf("\t hot fluid:shell side,steam \n"); ho=(1500); // condensation of steam Btu/(hr)*(ft^2)*(F) printf("\t individual heat transfer coefficient is : %.0f Btu/(hr)*(ft^2)*(F) \n",ho); printf("\t cold fluid:inner tube side,kerosene \n"); Nt=86; n=2; // number of passes L=12; //ft at1=0.594; // flow area, in^2,from table 10 at=((Nt*at1)/(144*n)); // total area,ft^2,from eq.7.48 printf("\t flow area is : %.3f ft^2 \n",at); Gt=(W/(.177)); // mass velocity,lb/(hr)*(ft^2) printf("\t mass velocity is : %.2e lb/(hr)*(ft^2) \n",Gt); mu2=1.36*2.42; // at 145F,lb/(ft)*(hr) D=(0.87/12); // ft Ret1=((D)*(Gt)/mu2); // reynolds number printf("\t reynolds number is : %.0f \n",Ret1); mu3=1.75*2.42; // at 120F,lb/(ft)*(hr) D=(0.87/12); // ft Ret2=((D)*(Gt)/mu3); // reynolds number printf("\t reynolds number is : %.1e \n",Ret2); Z1=331; // Z1=(L*n/D) jH=3.1; // from fig 24 mu4=1.75; // cp and 40 API Z2=0.24; // Z2=((k)*(c*mu4/k)^(1/3)), from fig 16 Hi=((jH)*(1/D)*(Z2)); // using eq.6.15a,Btu/(hr)*(ft^2)*(F) printf("\t Hi is : %.2f Btu/(hr)*(ft^2)*(F) \n",Hi); ID=0.87; // ft OD=1; //ft Hio=(Hi*(ID/OD)); //Btu/(hr)*(ft^2)*(F), from eq.6.5 printf("\t Hio is : %.2f Btu/(hr)*(ft^2)*(F) \n",Hio); tw=(tc)+(((ho)/(Hio+ho))*(T1-tc)); // from eq.5.31 printf("\t tw is : %.0f F \n",tw); muw=1.45; // lb/(ft)*(hr),at 249F from fig.14 phyt=(mu3/muw)^0.14; printf("\t phyt is : %.1f \n",phyt); // from fig.24 hio=(Hio)*(phyt); // from eq.6.37 printf("\t Correct hio to the surface at the OD is : %.1f Btu/(hr)*(ft^2)*(F) \n",hio); delt=tw-tc; //F printf("\t delt is : %.0f F \n",delt); printf("\t Since the kerosene has a viscosity of only 1.75 cp at the caloric temperature and delt=129F, free convection should be investigated. \n"); s=0.8; row=50; // lb/ft^3, from fig 6 s1=0.810; // at 95F s2=0.792; // at 145F bita=((s1^2-s2^2)/(2*(t2-t1)*s1*s2)); // /F printf("\t beta is : %.6f /F \n",bita); G=((D^3)*(row^2)*(bita)*(delt)*(4.18*10^8)/(mu3^2)); printf("\t G is : %.1e \n",G); psy=((2.25)*(1+(0.01*G^(1/3)))/(log10(Ret2))); printf("\t psy is : %.2f \n",psy); hio1=(hio*psy); printf("\t corrected hio1 is : %.1f Btu/(hr)*(ft^2)*(F) \n",hio1); Uc=((hio1)*(ho)/(hio1+ho)); // clean overall coefficient,Btu/(hr)*(ft^2)*(F) printf("\t clean overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",Uc); A2=0.2618; // actual surface supplied for each tube,ft^2,from table 10 A=(Nt*L*A2); // ft^2 printf("\t total surface area is : %.0f ft^2 \n",A); UD=((Q)/((A)*(delt))); printf("\t actual design overall coefficient is : %.1f Btu/(hr)*(ft^2)*(F) \n",UD); Rd=((Uc-UD)/((UD)*(Uc))); // (hr)*(ft^2)*(F)/Btu printf("\t actual Rd is : %.2f (hr)*(ft^2)*(F)/Btu \n",Rd); // end
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EX2_18.sce
clc;funcprot(0);//Example 2.18 //Initilisation of Variables d1=0.5;....//diameter of container in m T1=80;....//inner temparature of spherical container in degrees celcius t=0.025;....//thickness of insulating materials in m Ta=303;....//outer surface temparature of spherical container in degrees celcius K1=0.042;....//thermal conductivity of first insulating layer in W/m*K K=0.0017;....//thermal conductivity of second insulating layer in W/m*K h=20;....//heat transfer coefficient in W/m^2 hfg=2*10^2;....//latent heat of vapouraisation o f liquid nitrogen in kJ/Kg*degrees celcius //calculations r1=d1/2;....//radius of container in m r2=r1+t;....//radius of inner layer in m r3=r2+t;....//radius of outer layer in m R1=(r2-r1)/(4*%pi*K1*r1*r2);.....//resistance of first layer in degrees celcius/W R2=(r3-r2)/(4*%pi*K*r2*r3);.....//resistance of second layer in degrees celcius/W R3=1/(4*%pi*h*r3^2);.....//resistance of inner layer in degrees celcius/W Q=(T1-Ta)/(R1+R2+R3);....//heat flows from the airbient air to nitrogen in W m=Q/(hfg*10^3);....//rate of vapouraisation of liquid nitrogen per hour Kg/s disp(m*3600,"rate of vapouraisation of liquid nitrogen per hour Kg/hr:")
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errcatch(-1,"stop");mode(2);// Exa 20.3 ; all; // Given data Vmax=8;//Maximum value of voltage Vmin=2;//minimum value of voltage //Solution SWR=(Vmax+Vmin)/(Vmax-Vmin);//Standing wave ratio printf('Standing Wave Ratio = %.2f \n ',SWR); exit();
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Exa3_11.sce
//Exa 3.11 clc; clear; close; //Given Data : format('v',5); L=200;//in meter W=684/1000;//in Kg/m T=1450;//in Kg S=W*L^2/(8*T);//in meter disp(S,"Sag(in meter) : ");
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Ex9_1.sce
// scilab Code Exa 9.1 Calculation on multi stage turbine d=1; // mean diameter of the impeller blade in m T1=500; // Initial Temperature in degree C t1=T1+273; // in Kelvin p1=100; // Initial Pressure in bar N=3e3; // Speed in RPM m=100; // in kg/s alpha2=70; // exit angle of the first stage nozzle blades // part(a) single stage impulse nsti=0.78; u=%pi*d*N/60; sigma=0.5*(sind(alpha2)); // maximum utilization factor c2=u/sigma; cx=c2*(cosd(alpha2)); beta2=atand(0.5*(tand(alpha2))); // beta2=beta3 wst=2*(u^2)*1e-3; P=m*wst; disp("(a)for single stage impulse") disp("degree",beta2,"blade angles are beta2=beta3= ") disp("MW",P*1e-3,"Power developed is") sv=0.04; // specific volume of steam after expansion in m3/kg h=(m*sv)/(cx*%pi*d); // h2=h3=h disp("cm",h*1e2,"blade height is") delhs=wst/nsti; disp("final state of the steam is") p=81.5; // from enthalpy-entropy diagram T=470; disp("bar",p,"p=") disp("degree C",T,"T=") // part(b) Two-stage Curtis wheel nstc=0.65; u=%pi*d*N/60; sigma2=0.25*(sind(alpha2)); c2_2=u/sigma2; cx2=c2_2*(cosd(alpha2)); beta2_2=atand((3*u)/cx2); // beta2=beta3 alpha3=atand((2*u)/(c2_2*cosd(alpha2))); // alpha2'=alpha3 beta2_s=atand((u)/cx2); // beta2'=beta3' wI=6*(u^2)*1e-3; wII=2*(u^2)*1e-3; wst2=wI+wII; P2=m*wst2; disp("(b)for Two-stage Curtis wheel") disp("degree",alpha3,"air angles are alpha2s=alpha3= ") disp("degree",beta2_2,"for first stage blade angles are beta2=beta3= ") disp("degree",beta2_s,"for second stage blade angles are beta2s=beta3s= ") disp("MW",P2*1e-3,"Power developed is") delhs2=wst2/nstc; // from enthalpy-entropy diagram for the expansion disp("final state of the steam is") p2=27; T2=365; v2=0.105; // specific volume of steam after expansion in m3/kg disp("bar",p2,"p=") disp("degree C",T2,"T=") disp("m3/kg",v2,"v=") h2=(m*v2)/(cx2*%pi*d); disp("cm",h2*1e2,"blade height is") // part(c) Two-stage Reateau wheel nst1=0.78; wI3=2*(u^2)*1e-3; wII3=2*(u^2)*1e-3; wst3=wI3+wII3; P3=m*wst3; disp("(c)for Two-stage Reateau wheel") disp("degree",beta2,"blade angles are beta2=beta3= ") disp("MW",P3*1e-3,"Power developed is") delhs3=wst3/nst1; disp("final state of the steam is") p3=65; // from enthalpy-entropy diagram T3=445; v3=0.05; // specific volume of steam after expansion in m3/kg disp("bar",p3,"p=") disp("degree C",T3,"T=") disp("m3/kg",v3,"v=") h3=(m*v3)/(cx*%pi*d); disp("cm",h3*1e2,"blade height for the second stage is") // part(d) single stage 50% reaction nstr=0.85; sigma4=sind(alpha2); // maximum utilization factor c2_4=u/sigma4; // c2_4=w_3 cx4=c2_4*(cosd(alpha2)); // alpha2=beta3; beta2_4=0; // beta2=alpha3 wst4=(u^2)*1e-3; P4=m*wst4; disp("(d)for single stage 50% reaction") disp("degree",beta2_4,"blade angles are beta2=alpha3= ") disp("degree",alpha2,"and beta3=alpha2= ") disp("MW",P4*1e-3,"Power developed is") delhs4=wst4/nstr; // from enthalpy-entropy diagram disp("final state of the steam is") p4=90; T4=485; v4=0.035; disp("bar",p4,"p=") disp("degree C",T4,"T=") disp("m3/kg",v4,"v=") h4=(m*v4)/(cx4*%pi*d); disp("cm",h4*1e2,"the rotor blade height at exit is")
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Example1_13.sce
// Electric Machinery and Transformers // Irving L kosow // Prentice Hall of India // 2nd editiom // Chapter 1: Electromechanical Fundamentals // Example 1-13 clear; clc; close; // Clear the work space and console. // Given data R_a = 0.25; // Armature resistance V_a = 125; // dc bus voltage I_a = 60; // Armature current // Calculations E_c = V_a - I_a * R_a; // Counter EMF generated in the armature conductors of motor // Display the result disp("Example 1-13 Solution : "); printf("\n Ec = % d V ", E_c );
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3_6.sce
clear clc disp('Exa-3.6(a)'); w1=0.24;wc=0.00243;theta=60; //given values w=wavelength(lambeda) w2=w1+(wc*(1-cosd(theta))); printf('The wavelength of x-rays after scattering is %.4f nm\n',w2); disp('Exa-3.6(b)'); hc=1240; E2=hc/w2;E1=hc/w1; printf('The energy of scattered x-rays is %.0f eV\n',E2); disp('Exa-3.6(c)'); K= E1-E2; //The kinetic energy is the difference in the energy before and after the collision; printf('The kinetic energy of the x-rays is %.3f eV\n',K); disp('Exa-3.6(d)'); phi2=atand(E2*sind(theta)/(E1-E2*cosd(theta))) printf('The direction of the scattered eletron is %.1f degrees',phi2);
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Example_3_18.sce
//Caption: Measure of Skewness //Karl Pearson's coefficient of skewness //Example3.18 //Page61 clear; clc; X = [0,4;4,8;8,12;12,16;16,20;20,24;24,28;28,32]; //Absenteeism in days f = [10,76,100,150,24,36,14,2]; //No. of employees [m,n] = size(X); for i = 1:m Xi(i) = sum(X(i,:))/2; //Mid point end if (modulo(m,2)==1) then //to check even or odd mid = m/2; else mid = (m+1)/2; end A = Xi(mid); //assumed mean N = sum(f); //total frequency C = diff(X(1,:)); //class interval for i = 1:m d(i) = (Xi(i)-A)/C; fd(i)= f(i)*d(i); fd2(i) = f(i)*(d(i)^2); end Xmean = A+(sum(fd)*C/N); //Mean value [m1,n1] = max(f); //maximum frequency L = X(n1,1); //Lower limit of the modal class f1 = abs(f(n1)-f(n1-1)); //Abs difference between freq. of modal class & its //immediately preceding class f2 = abs(f(n1)-f(n1+1)); //Abs difference between freq. of modal class & its //immediately succeeding class Mode = L+((f1/(f1+f2))*C);//Mode Std = sqrt((sum(fd2)/N)-(sum(fd)/N)^2)*C;//standard deviation CS = (Xmean-Mode)/Std; //coefficient of skewness disp(Xmean,'Mean Value =') disp(Mode,'Mode value =') disp(CS,'coefficient of skewness = ') //Result if (CS<0) then disp('Since the coefficient of skewness is negative, the distribution is') disp('skewed to the left & extent of distortion is very small') end //Result //Mean Value = // // 12.679612 // // Mode value = // // 13.136364 // // coefficient of skewness = // // - 0.0832508 // // Since the coefficient of skewness is negative, the distribution is // // skewed to the left & extent of distortion is very small
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clc Msi=28.9//g/mole Dsi=2.33//g/cm^3 Msidi=60.08//g/mole Dsidi=2.21//g/cm^3 vsi=Msi/Dsi disp(vsi,"vsi in cm^3/mole is= ") vsidi=Msidi/Dsidi disp(vsidi,"vsidi in cm^3/mole is= ") T=vsi/vsidi disp(T,"T is ratio of Thickness of Si to SiO2 is= ")
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getd ../../../common_files/ exec ../../../common_files/loader.sce exec ser_init.sce exec pi_bda_tuned_dist.sci xcos pi_tuned_dist.xcos
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//Setup GaAs=1; Ge=2; Si=3; Eg = [1.42 0.66 1.12];//eV B = [2.1E14 1.66E15 5.23E15];//cm^{-3}K^{-3/2} k=86E-6;//eV/K //user selections T=300;//Kelvin material = Ge; ni = B(material) * T^1.5 * exp(-Eg(material) / (2*k*T)) // in cm^{-3}
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clc //initialization of variables clear l=20 //cm dL=1 //m dl=0.004 //cm //calculations L=l*dL/dl //m //results printf('The depth of the clay bed is %d m',L)
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function arbit_response(g_closedloop,t,u) // This program is free software; you can redistribute it and/or modify // it under the terms of the GNU General Public License as published by // the Free Software Foundation; either version 2 of the License, or // (at your option) any later version. // // This program is distributed in the hope that it will be useful, // but WITHOUT ANY WARRANTY; without even the implied warranty of // MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the // GNU General Public License for more details. // // You should have received a copy of the GNU General Public License // along with this program; if not, write to the Free Software // Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA // Authors // Holger Nahrstaedt - 2010 // Ishan Pendharkar - 2001-2007 // //RLTOOL for scilab (c) Ishan Pendharkar. //function simulate for piecewise linear inputs. global gridon tstep //check if input is NOT single-valued [x,y]=size(t); for i=1:x-1 if t(i+1)-t(i)<= 0 then messagebox(['Error entering input.';'Input is not single valued']); return; end; end; // interpolate to get values of input interp_u(1)=0; interp_t(1)=0; interp_size=ceil(max(t)/tstep); for i=1:interp_size interp_u(i)=interpln([t';u'],(i-1)*tstep) interp_t(i)=(i-1)*tstep; end; // plotting interpolated values t=interp_t; u=interp_u; if roots(denom(g_closedloop))<>[] then resp=csim(u',t',g_closedloop); //this is the response vector plot2d(t,[resp' u],[1,2],leg="Response@Input"); if gridon==1 then xgrid(4); end; xtitle('Dynamic Response','Time (sec)','Magnitude') else messagebox(['Sorry! I cannot plot the response';'Due to numerical tolerances, a pole-zero cancellation has occured.';' Please reselect point.']); end; //xselect(); show_window(); clear interp_t,interp_u,t,resp //return endfunction
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// Collector Current in C-E mode // Basic Electronics // By Debashis De // First Edition, 2010 // Dorling Kindersley Pvt. Ltd. India // Example 4-3 in page 209 clear; clc; close; // Given Data alpha=0.90; // Current Gain of BJT Ico=15*10^-6; // Reverse Saturation Current of BJT in micro-A Ib=0.5*10^-3; // Base Current in C-E mode in mA // Calculations beta_bjt=alpha/(1-alpha); Ic=(beta_bjt*Ib)+(beta_bjt+1)*Ico; printf("(a)The value of Current gain beta for BJT is %0.0f \n",beta_bjt); printf("(b)The value of the Collector Current is %0.2e A \n",Ic); // Results // (a) The value of Current Gain beta for BJT is 9 // (b) The value of the Collector Current is 4.65 mA
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clear; clc; printf("\t Example 7.5\n"); p1=.3; //percentage of the solute in the solution w1=1000; //weight of the solution taken w2=142; //molecular weight of Na2SO4. M1=(w2/(180+w2)); //solute (Na2SO4) present in the Na2CO3.10H2O solution s1=40.8; //solubility of Na2SO4 at 30 degree per 100 gm of water s2=9.0; //solubility of Na2SO4 at 10 degree per 100 gm of water //percent weight of solute in Na2SO4.10H2O= 144/322 //let x be the weight of crystal formed x=poly([0],'x'); //calc. x the weight of crystal t=roots((w1*40.8/140.8)-(.442*x+(w1-x)*(s2/(100+s2)))); printf("\n the weight of crystal formed after crystallisation :%f kg",t); //end
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function [stk,txt,top]=sci_diff() // Copyright INRIA txt=[] if stk(top-rhs+1)(5)=='4' then v='bool2s('+stk(top-rhs+1)(1)+')', else v=stk(top-rhs+1)(1), end if rhs==1 then [m,n]=checkdims(stk(top)) x=stk(top)(1) if m==-1&n==-1 then set_infos(['mtlb_diff('+x+') may be replaced by ' ' '+x+'(2:$)-'+x+'(1:$-1) if '+x+'is a vector' ' '+x+'(2:$,:)-'+x+'(1:$-1,:) if '+x+'is a matrix'],1) stk=list('mtlb_diff('+x+')','0','?','?','1') elseif m==1|n==1 then if ~isname(x) then x=gettempvar() txt=[txt;x+'='+stk(top)(1)] end stk=list(x+'(2:$)-'+x+'(1:$-1)','1','1','1','1') else if ~isname(x) then x=gettempvar() txt=[txt;x+'='+stk(top)(1)] end stk=list(x+'(2:$,:)-'+x+'(1:$-1,:)','1','1','1','1') end else //diff(x,N) N=stk(top)(1) top=top-1 [m,n]=checkdims(stk(top)) x=stk(top)(1) if m==-1&n==-1 then set_infos([ 'mtlb_diff('+x+','+N+') may be replaced by ' ' '+x+'('+addf('1',N)+':$)-'+x+'(1:$-'+N+') if '+x+' is a vector' ' '+x+'('+addf('1',N)+':$,:)-'+x+'(1:$-'+N+',:) if '+x+' is a matrix'],1) stk=list('mtlb_diff('+x+','+N+')','0','?','?','1') elseif m==1|n==1 then if ~isname(x) then x=gettempvar() txt=[txt;x+'='+stk(top)(1)] end stk=list(x+'('+addf('1',N)+':$)-'+x+'(1:$-'+N+')','1','1','1','1') else if ~isname(x) then x=gettempvar() txt=[txt;x+'='+stk(top)(1)] end stk=list(x+'('+addf('1',N)+':$,:)-'+x+'(1:$-'+N+',:)','1','1','1','1') end end
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//chapter 4 //may june 2013 n=50; d=0.5; lamda=1;//assume L=n*d; D=2*(L/lamda); printf("the directivity is %g",D);
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clc; s = %s; num = 2; den = s*(s+1)*(s+2); g = syslin('c', num/den); evans(g, 10) sgrid() // kp < 3 | tanteando kp = 1.2 ku = 2.998; wu = 1.414; tu = (2*%pi)/wu; kp = 0.6*ku ki = 1.2*(ku/tu) kd = (0.6/8)*(ku*tu) // Manual // kp = 1.7988 // ki = 0.045 // kd = 1.2
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// Example 11.10 //Bode Plot of a Narrowband Filter s=%s; num=20*s; den=(s^2+20*s+10^4) H_s=num/den; // Transfer function of given filter h1=syslin('c',H_s); bode(h1);
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argc:7 Dataset: ../datasets/converted/football.net Nodes Edges Com Mod NMI Time seq async 115 1226 9 0.602529 -1 0.000248959 par async 115 1226 13 0.50171 -1 0.071628
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clc;//clears the command window clear;//clears all the variables format('v',18);//changing the default prcision to 20 significant digits i=1;x=1;//flag bits dec=432.354;//given decimal number which should be expressed in base 2 temp2=floor(dec);//separating integer part from the given number temp4=modulo(dec,1);//separating decimal part from the given number while(temp2>0)//storing each integer digit in vector for convenience p(i)=(modulo(floor(temp2),2)) temp2=floor(temp2)/2; i=i+1; end temp2=0;//clearing temporary variable 'temp2' for j=1:length(p) //multipliying bits of integer part with their position values and adding temp2=temp2+(p(j)*10^(j-1)); end while(temp4~=0) //storing each decimal digit in vector for convenience temp4=temp4*2; d(x)=floor(temp4); x=x+1; temp4=modulo(temp4,1); end temp5=0; //clearing temporary variable 'temp5' for j=1:length(d) //multipliying bits of decimal part with their position values and adding temp5=temp5+(10^(-1*j)*d(j)) end temp3=temp2+temp5; //finally adding both the integer and decimal parts to get total output. disp(temp3);//displays output
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//Percentage error calculation in a wattmeter clc; clear; //Rated Parameters I=50; V=230; R=61;// No. of revolutions t=37/3600; // Time in hours C=520; // Normal Disc Speed Pfl=I*V;// Power at full load Ps=Pfl*t/1000; // Power Supplied in kWhr Pr=R/C; //Power recorded in kWhr err=(Ps-Pr)*100/Ps; printf('The Percentage Error = %g percent slow \n',err)
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main array [3] foo,boo; var a,b,c; { let c <- foo[2*a-6]; while 2 < 3 do let b <- foo[2*a-6]; while 4 < foo[2*a-6] do if 3 < 4 then let boo[c] <- 3 else let foo[2] <- 5 fi od; let c <- foo[2*a-6] od; let a <- foo[2*a-6] } .
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<?xml version="1.0" encoding="utf-8"?> <test> <description>desc P=10</description> <executable>APESolver</executable> <parameters>APE_2DVariableC_WeakDG_MODIFIED.xml</parameters> <files> <file description="Session File">APE_2DVariableC_WeakDG_MODIFIED.xml</file> </files> <metrics> <metric type="L2" id="1"> <value variable="p" tolerance="1e-4">145.343</value> <value variable="u" tolerance="1e-7">0.329227</value> <value variable="v" tolerance="1e-7">0</value> </metric> <metric type="Linf" id="2"> <value variable="p" tolerance="1e-4">119.917</value> <value variable="u" tolerance="1e-7">0.260865</value> <value variable="v" tolerance="1e-7">0</value> </metric> </metrics> </test>
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clc //initialisation of variables w= 120 //KN D= 30 //m L= 300 //m sigmamax= 600 //N/mm^2 h= 50 //m beta= 45 //degrees //CALCULATIONS Tmax= ((w*L)/2)*(sqrt(1+(L/(4*D))^2)) d= sqrt((4*Tmax*10^3)/(sigmamax*%pi)) H= (w*L^2)/(8*D) alpha= atand((w*L)/(2*H)) Mt= Tmax*(cosd(alpha)-cosd(beta))*h Vt= Tmax*(sind(alpha)+sind(beta)) Wa= Tmax*cosd(beta) //RESULTS printf ('Tmax= %.1f KN',Tmax) printf (' \n d=%.1f mm',d) printf (' \n H=%.0f KN',H) printf (' \n alpha=%.1f degrees',alpha) printf (' \n Mt=%.0f KNm',Mt) printf (' \n Vt=%.0f KN',Vt) printf (' \n Wa=%.0f KN',Wa)
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c = -1*[ 504 803 667 1103 834 585 811 856 690 832 846 813 868 793 .. 825 1002 860 615 540 797 616 660 707 866 647 746 1006 608 .. 877 900 573 788 484 853 942 630 591 630 640 1169 932 1034 .. 957 798 669 625 467 1051 552 717 654 388 559 555 1104 783 .. 959 668 507 855 986 831 821 825 868 852 832 828 799 686 .. 510 671 575 740 510 675 996 636 826 1022 1140 654 909 799 .. 1162 653 814 625 599 476 767 954 906 904 649 873 565 853 1008 632]'; //Constraint Matrix A = [ //Constraint 1 42 41 523 215 819 551 69 193 582 375 367 478 162 898 .. 550 553 298 577 493 183 260 224 852 394 958 282 402 604 .. 164 308 218 61 273 772 191 117 276 877 415 873 902 465 .. 320 870 244 781 86 622 665 155 680 101 665 227 597 354 .. 597 79 162 998 849 136 112 751 735 884 71 449 266 420 .. 797 945 746 46 44 545 882 72 383 714 987 183 731 301 .. 718 91 109 567 708 507 983 808 766 615 554 282 995 946 651 298; //Constraint 2 509 883 229 569 706 639 114 727 491 481 681 948 687 941 .. 350 253 573 40 124 384 660 951 739 329 146 593 658 816 .. 638 717 779 289 430 851 937 289 159 260 930 248 656 833 .. 892 60 278 741 297 967 86 249 354 614 836 290 893 857 .. 158 869 206 504 799 758 431 580 780 788 583 641 32 653 .. 252 709 129 368 440 314 287 854 460 594 512 239 719 751 .. 708 670 269 832 137 356 960 651 398 893 407 477 552 805 881 850; //Constraint 3 806 361 199 781 596 669 957 358 259 888 319 751 275 177 .. 883 749 229 265 282 694 819 77 190 551 140 442 867 283 .. 137 359 445 58 440 192 485 744 844 969 50 833 57 877 .. 482 732 968 113 486 710 439 747 174 260 877 474 841 422 .. 280 684 330 910 791 322 404 403 519 148 948 414 894 147 .. 73 297 97 651 380 67 582 973 143 732 624 518 847 113 .. 382 97 905 398 859 4 142 110 11 213 398 173 106 331 254 447 ; //Constraint 4 404 197 817 1000 44 307 39 659 46 334 448 599 931 776 .. 263 980 807 378 278 841 700 210 542 636 388 129 203 110 .. 817 502 657 804 662 989 585 645 113 436 610 948 919 115 .. 967 13 445 449 740 592 327 167 368 335 179 909 825 614 .. 987 350 179 415 821 525 774 283 427 275 659 392 73 896 .. 68 982 697 421 246 672 649 731 191 514 983 886 95 846 .. 689 206 417 14 735 267 822 977 302 687 118 990 323 993 525 322; //Constrain 5 475 36 287 577 45 700 803 654 196 844 657 387 518 143 .. 515 335 942 701 332 803 265 922 908 139 995 845 487 100 .. 447 653 649 738 424 475 425 926 795 47 136 801 904 740 .. 768 460 76 660 500 915 897 25 716 557 72 696 653 933 .. 420 582 810 861 758 647 237 631 271 91 75 756 409 440 .. 483 336 765 637 981 980 202 35 594 689 602 76 767 693 .. 893 160 785 311 417 748 375 362 617 553 474 915 457 261 350 635 ; ]; nbVar = size(c,1) b=[11927 13727 11551 13056 13460 ]; // Lower Bound of variables lb = repmat(0,1,nbVar) // Upper Bound of variables ub = repmat(1,1,nbVar) // Lower Bound of constrains intcon = []; for i = 1:nbVar intcon = [intcon i]; end options = list("MaxIterations",1); // The expected solution : // Output variables xopt = [0 1 1 0 0 1 0 1 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 0 1 1 0 1 1 0 1 .. 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 .. 0 0 1 0 0 1 0 1 0 0 1 0 0 1 0 1 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 1 0 0 1 0] // Optimal value fopt = [ 24381 ] // Calling Symphony [x,f,status,output] = intqpipopt(zeros(100,100),c,intcon,A,b,[],[],lb,ub,zeros(100,1),options);
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Ex2_12.sce
//Fluid system - By - Shiv Kumar //Chapter 2 - Impact of Jet //Example 2.12 clc clear //Given Data:- Vi=22; //Absolute velocity of Jet at Inlet of Vane, m/s u=11; //Velocity of Vane, m/s ui=u; uo=u; alpha_i=25; //Angle made by Jet at Inlet, degrees alpha_l=135; //Angle made by Jet at leaving, degrees alpha_o=180-alpha_l; //degrees //Data Used:- g=9.81; //Acceleration due to gravity, m/s^2 //Computations:- //(a) Vwi=Vi*cosd(alpha_i); //m/s Vfi=Vi*sind(alpha_i); //m/s Vrwi=Vwi-ui; //m/s beta_i=atand(Vfi/Vrwi); //degrees Vri=Vfi/sind(beta_i); //m/s Vro=Vri; beta_o=alpha_o-asind(uo*sind(180-alpha_o)/Vro); //degrees Vwo=Vro*cosd(beta_o)-uo; //degrees //(b) W=(Vwi+Vwo)*u/g; //N-m/N //Results:- printf("(a)Vane angle at Inlet, beta_i=%.2f degrees \n", beta_i) //The answer vary due to round off error printf(" Vane angle at Outlet, beta_o=%.2f degrees \n", beta_o) //The answer vary due to round off error printf("(b)Work done per second per unit weight of water striking the vane per second=%.2f N-m/N", W) //The answer vary due to round off error
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y = [] coefficients = [] function [value] = calculate(i,count) k=0 new = 0.0 value = 0 l = i - count for j = l:(l+count-1) new = new + (-1 * coefficients($-k) * y(j)) k = k + 1 end value = new / coefficients(1) endfunction count = input("Enter Highest Degree of the Differential Equation : "); for i = 0:count printf("Enter coefficient of y[n-%d] :", i) coefficients(i+1) = input(" "); end for i = 1:count printf("Enter value of y[%d] : ", i-1); y(i) = input(" "); disp(i, count) end n = input("Enter upper limit : ") if n <= count then disp("Wrong value of n given") exit() end for i = count+1:n+1 y($+1) = calculate(i,count) end disp("Final Y values : ") for i = 1:length(y) printf("\ty[%d] = %d\n", i-1, y(i)) end
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{ "name":"Test1", "questions": [ { "question_text":"Why the sky is blue?", "answers": [ "Text", "Gray", "Bank" ], "correct_answer" : 2 } ] }
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