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// Scilab Code Ex1.7: Page-1.6 (2009) clc; clear; m = 9.1e-031; // Mass of the electron, kg e = 1.6e-019; // Energy equivalent of 1 eV, J/eV h = 6.626e-034; // Planck's constant, Js E = 20e+03*e; // Energy of the electron, J // As 1/2*m*v^2 = E, solving for v v = sqrt(2*E/m); // Velocity of the electron, m/s lambda = h/(m*v); // de Broglie wavelength of the electron, m n = 1; // First order diffraction d = 9.8e-011; // Atomic spacing for thin gold foil, m // Using Bragg's equation, 2*d*sin(theta) = n*lambda and solving for theta theta = asind(n*lambda/(2*d)); // Angle of deviation for first order diffraction maxima, degree printf("\nThe angle of deviation for first order diffraction maxima = %4.2f degrees", theta); // Result // The angle of deviation for first order diffraction maxima = 2.54 degrees
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// Terme de croissance de Monod qui intervient dans l'équation bilan des légionnelles function y = monod1 (N,l,a) // N pour nutriment, l légionnelle, et a amibes // terme de croissance qui apparaît dans l'équation des légionnelles y = ((k_1* ones(size(N,1),1) ./ (k_2*ones(size(N,1),1) + N)) + (k_3*ones(size(N,1),1) ./ (k_4*ones(size(N,1),1) + rho_A * a))) .* l; endfunction function y = monod2 (N,l,a) // terme de croissance qui apparaît dans l'équation de croissance des amibes y = (k_5 * ones(size(N,1),1) ./ (k_6 * ones(size(N,1),1) + N)) .* a - (k_3 * ones(size(N,1),1) ./ (k_4 * ones(size(N,1),1) + rho_A * a)) .* l ; endfunction function y = monod3(N,l,a) // terme de croissance qui apparaît dans le calcul de l'intégrale (vitesse) y = (k_1*ones(size(N,1),1) ./ (k_2*ones(size(N,1),1) + N)) .* l + (k_5*ones(size(N,1),1) ./ (k_6*ones(size(N,1),1) + N)) .*a ; endfunction function y = monod4(N,l,a) // terme de croissance qui apparaît dans l'équation de diffusion des nutriments y = ( (k_1 * ones(size(N,1),1) ./ (k_2*ones(size(N,1),1) + N)) .* (l .* (rho_L*ones(size(N,1),1))) ) + ( (k_5*ones(size(N,1),1) ./ (k_6*ones(size(N,1),1) + N)) .* (a .* (rho_A*ones(size(N,1),1))) ); endfunction function y = trouveDernierNonNul(xx) i = 1; while(i <> size(xx,1) & xx($-i+1) == 0 ) i = i+1; end y = i; endfunction
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//Example number 2.7, Page number 2.34 clc;clear;close // Variable declaration m=1 // unitless lamda_l=6000*10**-10 // in m theta=0.046*(%pi/180) // radian n=2*10**6// unitless // Calculation lamda_s=(m*lamda_l)/(sin(theta)) // in m v=n*lamda_s // in m/s // Result printf("Ultrasonic wavelength,lamda s =%0.2e m",(lamda_s)) printf("\nVelocity of ultrasonic waves in liquid = %0.f ms^-1",v) // Answer varies due to rounding of numbers
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clear; clc; printf("\n Example 6.1"); //Calculating minimum fluidisation velocity //Calculating Galileo number function[Ga]=Galileo_number() d = 3*10^(-3); //particle size is in meters p = 1100; //density of liquid is in kg/m^3 ps = 4200; //density of spherical particles is in kg/m^3 g = 9.81; //acceleration due to gravity is in m/sec^2 u = 3*10^(-3); //viscosity is in Ns/m^2 Ga = d^3*p*(ps-p)*g/u^2; funcprot(0); endfunction Ga = Galileo_number(); printf("\nGalileo number = %f*10^5",Ga*10^(-5)); //Calculating Re mf Remf = 25.7*(sqrt(1+5.53*10^(-5)*(1.003*10^5))-1); printf("\nValue of Remf is %d",Remf); umf = Remf*(3*10^(-3))/(3*10^(-3)*1100); printf("\nminimum fluidisation velocity is %.1f mm/sec",umf*1000);
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// Exa 4.21 clc; clear; close; // Given data q = 1.6 * 10^-19;// in C Mu_p = 500;// in cm^2/V-sec Rho_p = 3.5;// in ohm-cm Mu_n = 1500;// in cm^2/V-sec Rho_n = 10;// in ohm-cm N_A = 1/(Rho_p * Mu_p * q);// in /cm^3 N_D = 1/(Rho_n * Mu_n * q);// in /cm^3 V_J = 0.56;// in V n_i = 1.5 * 10^10;// in /cm^3 V_T = V_J/log((N_A * N_D)/(n_i)^2);// in V // V_T = T/11600 T = V_T * 11600;// in K T = T - 273;// in °C disp(T,"The Temperature of junction in °C is");
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//Given that R = 9.0 //in Ohm L = 2*10^-3 //in Henery E = 18 //in Volts //Sample Problem 31-5a printf("**Sample Problem 31-5a**\n") //As soon as switch is closed the inductor will act like current barrier Io = E/R printf("The current as soon as qwitch is closed is equal to %1.2fA\n", Io) //Sample Problem 31-5b printf("\n**Sample Problem 31-5b**\n") //After long time inductor will act like short circuit Req = R/3 If = E/(R/3) printf("The current through the battery after long time will be %1.2fA", If)
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// Example 4.8, page no-153 clear clc la=0.5 //length efficiency in azimuth direction le=0.7 //length efficiency in elevation direction A=10 // Actual projected area of an antenna Ae=la*le Aee=Ae*A printf("Aperture efficiency = %.2f \n Effective Aperture = %.1f m^2",Ae,Aee)
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//Example 3 // Velocity and wavelength clc; clear; close; //given data : Y=7*10^10;// in N/m^2 p=2.8*10^3;// in kg/m^3 v=sqrt(Y/p); disp(v,"(1). The velocity,v(m/s) = ") f=500;// in vibration/sec lamda=v/f; disp(lamda,"(2). The wavelength,(m/s) = ")
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clear ; clc; // Example 2.5 printf('Example 2.5\n\n'); //Page no. 54 // Solution m_wt=192 ;//[kg] d_sol=1.024*1000 ;//[kg/cubic metre] // 1000L=1 cubic metre c_sol=d_sol/1000 ;//[kg/L] c_drug=c_sol*.412 ;//[kg/L] printf('Concentration of drug in solution is %.3f kg/L .\n',c_drug); Q=10.5 ;//[L/min] Qmol=10.5*c_drug/m_wt ;//[kg mol/min] printf(' Flow rate of drug is %.3f kg mol/min. \n',Qmol);
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clc //Initialization of variables ps=0.64 //psia p=14.7 //psia M=29 M2=46 //calculations xa=ps/p mb=xa*9/M *M2/(1-xa) //results printf("percentage = %d percent",mb*100)
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//Example 31.5 N_C12=(6.02*10^23)/12*1000;//Number of carbon nuclei (C-12) in a kg N_C14=N_C12*(1.3*10^-12);//Number of carbon nuclei (C-14) in a kg t_half=5730;//Half-life of C-14 (y), See Appendix B R=0.693*N_C14/t_half;//Activity (y^-1 or decays per year) R=R*1/(3.16*10^7);//Activity (Bq or decays per second) printf('\nActivity R = %0.1f Bq',R) R=R/(3.7*10^10);//Activity (Ci) printf('\nActivity R = %0.2f nCi',R*10^9) //Answers vary due to round off error //Openstax - College Physics //Download for free at http://cnx.org/content/col11406/latest
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//Example 1.68// voltmeter reading clc; clear; close; //given data : f1=25; // in Hz f2=100; // in Hz R=300; // in ohm L=0.12; // in H XL1=2*%pi*f1*L; V_ac=15; // in volts Z1=sqrt(R^2+XL1^2); Vr1=V_ac*(R/Z1); XL2=2*%pi*f2*L; Z2=sqrt(R^2+XL2^2); Vr2=V_ac*(R/Z2) disp(Vr1,"the voltmeter reading at f1,Vr1(V) = ") disp(Vr2,"the volt meter reading at f2, Vr1(V) = ")
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#light position 0.000000 0.000000 1.000000 0.000000 #arball transformation 0.804307 0.147733 0.575663 0.000000 -0.593149 0.138834 0.793109 0.000000 0.037245 -0.979297 0.199280 0.000000 0.000000 0.000000 0.000000 1.000000 #translation parameter 0.170000 0.840000 -2.850000 #viewport 0 0 1152 678 #modelview 0.804307 0.031967 0.593456 0.000000 -0.593149 -0.019404 0.804935 0.000000 0.037245 -0.999361 0.003354 0.000000 0.170000 1.382619 -7.728937 1.000000 #projection 1.420865 0.000000 0.000000 0.000000 0.000000 2.414214 0.000000 0.000000 0.000000 0.000000 -1.000004 -1.000000 0.000000 0.000000 -0.020000 0.000000
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####### INITIATION ####### sce scenario = "mirror"; pcl_file = "mirrorMAIN.pcl"; active_buttons = 7; # na do o arithmos ton kato button_codes = 1,2,3,4,5,6,7; # na do WAS BUTTON BOX SPECIFIC default_background_color = 0, 0, 0; default_font = "arial"; default_font_size = 20; default_text_color = 235, 235, 235; default_text_align = align_left; begin; # den ksero giati ######## PICTURES ######## picture { } p_Default; #black screen picture { text { caption = "+"; font_size = 20; font = "Courier"; }; x = 0; y = 0; } p_Fixation; # fixation picture { text { caption = "Welcome to our experiment!"; } t_Welcome1; x = 0; y = 100; text { caption = "You will first have 10 Practice Trials"; } t_Welcome3; x = 0; y = 0; text { caption = "Please press [ENTER] when you are ready to start"; } t_Welcome2; x = 0; y = -100; } p_Welcome; picture { text { caption = "That was the end of the Practice"; } t_Practice1; x = 0; y = 100; text { caption = "You can now ask any questions you may have!"; } t_Practice3; x = 0; y = 0; text { caption = "Please press [ENTER] when you are ready to start"; } t_Practice2; x = 0; y = -100; } p_Practice; picture { text { caption = " "; } t_Info1; x = 0; y = 0; text { caption = " "; } t_Info2; x = 0; y = -40; text { caption = " "; } t_info3; x = 0; y = -150; } p_Info; picture { text { caption = "Please wait"; } t_Info4; x = 0; y = 0; } p_wait; ### trial ### ## na dokimaso afto picture { bitmap { filename = " "; preload = false; width = 1000; scale_factor = scale_to_width; } p_1; x = 0; y = -100; } pic1; # afto prepei na vgei picture { bitmap { filename = " "; preload = false; width = 1000; scale_factor = scale_to_width; } p_2; x = 0; y = -100; } pic2; picture { bitmap { filename = " "; preload = false; width = 1000; scale_factor = scale_to_width; } p_3; x = 0; y = -100; } pic3; #picture { bitmap { filename = "SAMpleasure.png"; } p_S_V; x = 0; y = -300; } p_sam_V; #picture { bitmap { filename = "SAMarousal.png"; } p_S_A; x = 0; y = -500; } p_sam_A; trial { picture pic1; # pos to vriskei time = 0; picture pic2; # pos to vriskei time = 100; picture pic3; # pos to vriskei time = 200; }main_trial; ### apo emotion #trial { #picture{ # bitmap p_sti_all; x = 0; y = 150; # allazei thesi? # } stimuli; deltat = 1000; duration = 6000; # duration of PICTURE PRESENTATION !! #} main_trial; #picture{ # bitmap p_sti_all; x = 0; y = 150; # allazei thesi? # } p_stimuli; # san digma #picture { # bitmap #{ filename = "1_1.jpg"; # afto prepei na vgei #width = 800; # 300 pixel width #scale_factor = scale_to_width; # with same aspect ratio # } p_sti_all; x = 0; y = 7000; } p_sti_final; # picture { bitmap { filename = "SAMpleasure.png"; } p_S_V; x = 0; y = -300; } p_sam_V; #trial { # nomizo pos oxi #picture{ # bitmap p_sti_all; x = 0; y = 150; # allazei thesi? # } stimuli; deltat = 1000; duration = 6000; # duration of PICTURE PRESENTATION !! #} main_trial;
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//Exa 1.15 clc; clear; close; //Given data W1=2.5;//in eV W2=1.9;//in eV ContactPotential=W1-W2;//in Volt disp(ContactPotential,"Contact potential in Volts : ");
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clear; off=fscanfMat("off.txt"); on=fscanfMat("on.txt"); reinit=fscanfMat("reinit.txt"); off=off/100; on=on/100; reinit=reinit/100; clf; scf(0); plot([off,on,reinit]); title("eTr with round time 8 ms (green=feedforward, red=feedforward+reinit)"); xlabel("rounds"); ylabel("eTr in milliseconds");
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clc TL=250 TH=291 COPR=TL/(TH-TL) mprintf("COPR=%f\n",COPR)//ans vary due to roundoff error QL=4*10^4 W=QL/COPR mprintf("W=%fkJ/d\n",W)//ans vary due to roundoff error CW=200 //compressor work in watts mprintf("Fraction of time compressor runs=%f\n",W/((CW*3600*24)/1000))//ans vary due to roundoff error TH=310 COPR=TL/(TH-TL) mprintf("COPR=%f\n",COPR)//ans vary due to roundoff error W=QL/COPR mprintf("W=%fkJ/d\n",W)//ans vary due to roundoff error mprintf("fraction of time the compressor runs=%f\n",W/((CW*3600*24)/1000))//ans vary due to roundoff error
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// Example 5.4 // The program selects and prints the largest of the three numbers //using nested if...else statement disp("Enter three values"); A=input("A="); B=input("B="); C=input("C="); disp("Largest value is:"); if(A>B) , //Test for largest between A&B if(A>C) , //Test for largest between A&C disp(A); else disp(C); end else if(C>B) , //Test for largest between C&B disp(C); else disp(B); end end
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// Example 2.6 format('v',6) clc; clear; close; // given data Vin= 15;// in V V_K= 0.7;// in V Vout=0;// in V R_L= 10;// in kΩ R_L= R_L*10^3;// in Ω // The peak output voltage V_P= Vin-V_K;// in V // The maximum forward current I_P= V_P/R_L;// in A // The peak inverse voltage PIV= Vin-Vout;// in V I_P= I_P*10^3;// in mA disp(V_P,"The peak output voltage in volts is : "); disp(I_P,"The maximum forward current in mA is : "); disp(PIV,"The peak inverse voltage in volts is : ")
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clc clear //Initalization of variables cr=9 p1=14 //psia t1=80+460 //R n=1.4 heat=800 //Btu c=0.1715 R=53.35 J=778 //calculations p2=p1*(cr)^n t2=t1*cr^(n-1) t3=heat/c +t2 p3=p2*t3/t2 eff=(1-1/cr^(n-1))*100 t4=t3/cr^(n-1) Qr=c*(t4-t1) cyclework=heat-Qr eff2= cyclework/heat *100 V1=R*t1/(144*p1) pd=(1-1/cr)*V1 mep=cyclework*J/(pd*144) //results printf("Max. temperature = %d R",t3) printf("\n Max. pressure = %d psia",p3) printf("\n In method 1,Thermal efficiency = %.1f percent",eff) printf("\n In method 2,Thermal efficiency = %.1f percent",eff2) printf("\n Mean effective pressure mep = %.1f psia",mep)
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//Chapter 5, Problem 6, Figure 5.13 clc; //Potential difference across R1 is the same as the supply voltage V R1=5; R3=20; I=11; I1=8; //Hence supply voltage is V=R1*I1; I3=V/R3; //Reading on ammeter, printf("Reading on ammeter = %f A\n\n\n",I3); I2=I-I1-I3; R2=V/I2; //Current flowing through R2 printf("Resistance R2 = %f ohm\n\n\n",R2);
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//Part A Chapter 7 Example 8 clc; clear; close; P1=1;//MPa V1=0.05;//m^3 x1=80/100;//dryness fraction P2=1;//MPa V2=0.2;//m^3 W=P1*1000*(V2-V1);//kJ vf=0.001127;//m^3/kg//at 1 MPa vg=0.19444;//m^3/kg//at 1 MPa uf=761.68;//kJ/kg//at 1 MPa ufg=1822;//kJ/kg//at 1 MPa vfg=vg-vf;//m^3/kg v1=vf+x1*vfg;//m^3/kg ms=V1/v1;//kg(mass of steam) v2=V2/ms;//m^3/kg T1=1000;T2=1100;//degree C(as v2>vg(1MPa)) T=T1+(T2-T1)*(v2-0.5871)/(0.6355-0.5871);//degree C u2=4209.6;//kJ/kg(at 1MPa & T degree C) u1=uf+x1*ufg;//kJ/kg Q=W+ms*(u2-u1);//kJ disp("Heat added is "+string(Q)+" kJ");
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clear;lines(0); //We use the 'levin' macro for solving the normal equations //on two examples: a one-dimensional and a two-dimensional process. //We need the covariance sequence of the stochastic process. //This example may usefully be compared with the results from //the 'phc' macro (see the corresponding help and example in it) // // //1) A one-dimensional process // ------------------------- // //We generate the process defined by two sinusoids (1Hz and 2 Hz) //in additive Gaussian noise (this is the observed process); //the simulated process is sampled at 10 Hz (step 0.1 in t, underafter). // t1=0:.1:100;rand('normal'); y1=sin(2*%pi*t1)+sin(2*%pi*2*t1);y1=y1+rand(y1);plot(t1,y1); // //covariance of y1 // nlag=128; c1=corr(y1,nlag); c1=c1';//c1 needs to be given columnwise (see the section PARAMETERS of this help) // //compute the filter for a maximum order of n=10 //la is a list-type variable each element of which //containing the filters of order ranging from 1 to n; (try varying n) //in the d-dimensional case this is a matrix polynomial (square, d X d) //sig gives, the same way, the mean-square error // n=15; [la1,sig1]=levin(n,c1); // //verify that the roots of 'la' contain the //frequency spectrum of the observed process y //(remember that y is sampled -in our example //at 10Hz (T=0.1s) so that we need to retrieve //the original frequencies (1Hz and 2 Hz) through //the log and correct scaling by the frequency sampling) //we verify this for each filter order // for i=1:n, s1=roots(la1(i));s1=log(s1)/2/%pi/.1; // //now we get the estimated poles (sorted, positive ones only !) // s1=sort(imag(s1));s1=s1(1:i/2);end; // //the last two frequencies are the ones really present in the observed //process ---> the others are "artifacts" coming from the used model size. //This is related to the rather difficult problem of order estimation. // //2) A 2-dimensional process // ----------------------- //(4 frequencies 1, 2, 3, and 4 Hz, sampled at 0.1 Hz : // |y_1| y_1=sin(2*Pi*t)+sin(2*Pi*2*t)+Gaussian noise // y=| | with : // |y_2| y_2=sin(2*Pi*3*t)+sin(2*Pi*4*t)+Gaussian noise // // d=2;dt=0.1; nlag=64; t2=0:2*%pi*dt:100; y2=[sin(t2)+sin(2*t2)+rand(t2);sin(3*t2)+sin(4*t2)+rand(t2)]; c2=[]; for j=1:2, for k=1:2, c2=[c2;corr(y2(k,:),y2(j,:),nlag)];end;end; c2=matrix(c2,2,128);cov=[]; for j=1:64,cov=[cov;c2(:,(j-1)*d+1:j*d)];end;//covar. columnwise c2=cov; // //in the multidimensional case, we have to compute the //roots of the determinant of the matrix polynomial //(easy in the 2-dimensional case but tricky if d>=3 !). //We just do that here for the maximum desired //filter order (n); mp is the matrix polynomial of degree n // [la2,sig2]=levin(n,c2); mp=la2(n);determinant=mp(1,1)*mp(2,2)-mp(1,2)*mp(2,1); s2=roots(determinant);s2=log(s2)/2/%pi/0.1;//same trick as above for 1D process s2=sort(imag(s2));s2=s2(1:d*n/2);//just the positive ones ! // //There the order estimation problem is seen to be much more difficult ! //many artifacts ! The 4 frequencies are in the estimated spectrum //but beneath many non relevant others. //
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//1.5 clc; V=100; R=10; i=50*10^-3; t=-0.5*log(1-((i*R/V)))*10^3 printf("t= %.1f ms", t) disp('So the minimum width of the gate pulse is 2.5 ms')
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clc //Intitalisation of variables clear T= 556 //K E= 44300 //cal R= 2 //cal /mole K //CALCULATIONS k= 10^8*T*%e^(-E/(R*T)) //RESULTS printf ('Specific rate of reaction = %.1e litre mole^-1 sec^-1',k)
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//chapter-5,Example5_6,pg 493 Tz=0.6*10^-3//discharge time Vref=1//ref. voltage t=4*10^-3//integrator time const. Vk=((Vref*Tz)/t)//rise in output integrator printf("rise in integrator output\n") printf("Vk=%.2f V\n",Vk) Vi=0.2//input voltage Tu=Vref*(Tz/Vi)//charging time printf("charging time\n") printf("Tu=%.4f sec",Tu)
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clc clear mprintf('Mechanical vibrations by G.K.Grover\n Example 1.6.2\n') //given data //Z=r*e^(i*theta) is represented as Z=r*cos(theta) + i*r*sin(theta)= x +i*y //where r*cos(theta)=x and r*sin(theta)=y //case 1 //V=5*e^(j*0.10) r1=5 theta1=0.1 x1=r1*cos(theta1) y1=r1*sin(theta1) v1=complex(x1,y1) //case 2 //V=17*e^(-j*3.74) r2=17 theta2=-3.74 x2=r2*cos(theta2) y2=r2*sin(theta2) v2=complex(x2,y2) //output mprintf('case(i):V=5*e^(j*0.10) is represented as') disp(v1) mprintf('\ncase(ii):V=17*e^(-j*3.74) is represented as') disp(v2) mprintf('\nNOTE:complex number is represented as x+y*i in SCILAB')
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interpolacionLagrange.sci
// Problema 6 //T = [-40 0 20 50]; //d = [1.52 1.29 1.2 1.09]; //density = interpolacionLagrange(T,d,15) function yint = interpolacionLagrange(x,y,xx) //Lagrange: Lagrange interpolating polynomial //yint = Lagrange(x,y,xx): Uses an (n - 1)-order //Lagrange interpolating polynomial based on n data points //to determine a value of the dependent variable (yint) at //a given value of the independent variable, xx. //input: //x = independent variable //y = dependent variable //xx = value of independent variable at which the //interpolation is calculated //output: //yint = interpolated value of dependent variable n = length(x); if length(y)~=n then yint = %nan; //error('x and y must be same length'); else s = 0; for i = 1:n product = y(i); for j = 1:n if i ~= j product = product*(xx-x(j))/(x(i)-x(j)); end end s = s+product; end yint = s; end endfunction
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Example11_18.sce
//Example 11.18 clear; clc; VI=5; Vo=12; Io=1; fs=100*10^3; IL=(Vo/VI)*Io; deliL=0.2*IL; L=(VI*(1-(VI/Vo)))/(fs*deliL); Ip=IL+(deliL/2); Irms=[(IL^2)+((deliL/(sqrt(12)))^2)]^(1/2); Iomin=deliL/2; printf("L=%.f uH",L*10^6); printf("\nAt full load the coil must withstand Ip=%.2f A",Ip); printf(" and Irms=%.1f A",Irms); printf("\nMinimum Load Current (Iomin)=%.1f A",Iomin-0.1);
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Ex2_12.sce
// A Textbook of Fluid Mecahnics and Hydraulic Machines - By R K Bansal // Chapter 2 - Pressure and its measurements // Problem 2.12 //Given Data Set in the Problem h2=20/100 SG2=13.6 SG1=1 dens1=1000 dens2=13.6*dens1 g=9.81 //Calculations //equating pressure above the datum line; function [f]=F(h1) f=(dens2*g*h2)-(dens1*g*h1) endfunction h1=10; H1=fsolve(h1,F) //When vessel is completely filled with wter; //Equating pressure in the two limbs function [g]=G(y) g=(dens2*g*(0.2+2*y/100))-(dens1*g*(3+H1+y/100)) endfunction y=10; Y=fsolve(y,G) mprintf("The difference in the mercury level in the two limbs is %f cm\n",(20+2*Y))
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function [stk,txt,top]=sci_load() // Copyright INRIA txt=[] if rhs<=0 then stk=list('load(''scilab.save'')','0','0','0','0') return end args=[] for k=1:rhs args=[args,stk(top-rhs+k)(1)] end stk=list('mtlb_load'+rhsargs(args),'0','0','0','0')
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6_4_2.sce
//CAPTION: Characteristics_Of_a_MOSFET //chapter_no.-6, page_no.-262 //Example_no.6-4-2 clc; //(a)Calculate_the_insulation_capacitance eir=3.9; ei=8.854*(10^-12)*eir; d=.05*(10^-6); Ci=ei/d; disp(Ci,'the_insulation_capacitance(in F/m^2)'); //(b)Calculate_the_saturation_drain_current Z=12*(10^-12); Vg=5; Vth=.10; vs=1.70*(10^7); Idsat=Z*Ci*(Vg-Vth)*vs; Idsat=Idsat*10^7; disp(Idsat,'the_saturation_drain_current(in mA)'); //(c)Calculate_the_transconductance_in_the_saturation region Z=12*(10^-12); vs=1.70*(10^7); gmsat=Z*Ci*vs; gmsat=gmsat*10^7; disp(gmsat,'the_transconductance_in_the_saturation region(in_millimhos)'); //(d)Calculate_the_maximum_operating_frequency_in_the_saturation_region vs=1.70*(10^7); L=4*(10^-6); fm=vs/(2*%pi*L); fm=fm/(10^2); fm=fm/(10^9); disp(fm,'the_maximum_operating_frequency_in_the_saturation_region(in GHz)');
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//Problem 2 //Calculate the maximum speed of electron striking the anti-cathode clear clc V=18// Potential difference in kV e=1.6*10^(-19)//charge on an electron in C m=9.1*10^(-31)//mass of an electron in kg v=(2*e*V/m)^(0.5)//maximum speed of electron in m/s printf('maximum speed of electron striking the anti-cathode = %.1f m/s',v)
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clc; //page no 199 //prob no. 5.8 //LSB transmitter refering fig.5.14 with new carrier freq 9.0015 MHz & local oscillator freq 12.5015MHz fco=9.0015;//carrier oscillator freq flo=12.5015;//local oscillator freq //Determination of new o/p freq fo=fco+flo; disp('MHz',fo,'The o/p carrier freq');
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//Ex 6.8 clc;clear;close; format('v',5); n=8;//no. of bits Range=0:10;//V Vin=5.2;//V oneLSB=max(Range)/2^n;//V disp(oneLSB*1000,"(a) Minimum voltage for 1 LSB in mV"); Vifs=max(Range)-oneLSB;//V disp(Vifs,"(b) For all ones input voltage should be (V)"); D=Vin/oneLSB;//Digital output in decimal D=dec2bin(round(D));//Digital output in binary disp(D,"(c) Digital Output");
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Ex7_3.sce
//Example number 7.3, Page number 147 clc;clear; close; //Variable declaration epsilonr=1.0000684; //dielectric constant N=2.7*10**25; //number of atoms epsilon0=8.85*10**-12; //Calculation alpha_e=epsilon0*(epsilonr-1)/N; //polarisability(Fm**2) //Result printf("polarisability is %.3e Fm^2",alpha_e) //answer varies due to rounding off errors
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//example 5.15 //calculate formation constant of acquifer using Chow's method clc;funcprot(0); //given Q=2500; //discharge(l/min) r=60; //distance of observation well from acquifer tmin=[1 1.5 2 2.5 3 4 5 6 8 10 12 14 18 24 30 40 50 60 80 100 120 150 180 210 240]; //time in minutes s=[0.2 0.26 0.3 0.33 0.36 0.41 0.45 0.48 0.53 0.56 0.59 0.62 0.66 0.71 0.75 0.80 0.83 0.86 0.91 0.95 0.98 1.03 1.05 1.08 1.10]; //drawdown for i=1:25 tday(i)=tmin(i)/(60*24); end //graph is plotted between s and t //point P is choosen on it whose ordinate is: s1=0.45; t=0.00347; ds=0.38; //for one log cycle of time Fu=s1/ds; //from fig 5.43 //or using relation Wu=2.303*Fu; u=0.035; //from table 5.2 Q=3600; //discharge in cumec/day T=Q*Wu/(4*%pi*s1); S=4*u*t*T/r^2; mprintf("formation constant of acquifer:"); mprintf("\nT=%i cubic metre/day/m.\nS=%f.",T,S);
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clc a = 8000 // annual requirement of parts c = 60 // unit cost of part in Rs r = 150 // ordering cost per lot in Rs i = 30 // annual carrying charges of average inventory i = 30/100 k = i*c // carrying cost per unit per year n = sqrt(2*r*a/k) // most economical order quantity printf("\n Most economical ordering quantity = %d units" , n)
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function cmap = hotcolormap(n) //graycmap yellow to red color map. // Copyright INRIA if size(n,'*')<>1|or(n<3) then error('hotcolormap : n must be an integer greater than 3') end n1=fix(3/8*n); n2=n1 n3=n-(n1+n2) // cmap=[(1:n1)'/n1 zeros(n1,1) zeros(n1,1); ones(n2,1) (1:n2)'/n2 zeros(n2,1); ones(n3,1) ones(n3,1) (1:n3)'/(n3)]
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exec("swigtest.start", -1); // Test constructors for std::deque<int> intDeque = new_IntDeque(); intDeque2 = new_IntDeque(3); intDeque3 = new_IntDeque(4, 42); //intDeque4 = new_IntDeque(intDeque3); // Test constructors for std::deque<double> doubleDeque = new_DoubleDeque(); doubleDeque2 = new_DoubleDeque(3); doubleDeque3 = new_DoubleDeque(4, 42.0); //doubleDeque4 = new_DoubleDeque(doubleDeque3); // Test constructors for std::deque<Real> realDeque = new_RealDeque(); realDeque2 = new_RealDeque(3); realDeque3 = new_RealDeque(4, 42.0); //realDeque4 = new_RealDeque(realDeque3); // average() should return the average of all values in a std::deque<int> IntDeque_push_back(intDeque, 2); IntDeque_push_back(intDeque, 4); IntDeque_push_back(intDeque, 6); avg = average(intDeque); checkequal(avg, 4.0, "average(intDeque)"); // half should return a deque with elements half of the input elements RealDeque_clear(realDeque); RealDeque_push_front(realDeque, 2.0); RealDeque_push_front(realDeque, 4.0); halfDeque = half(realDeque); checkequal(halfDeque, [2., 1.], "half(realDeque)"); // same for halve_in_place //DoubleDeque_clear(doubleDeque); //DoubleDeque_push_front(doubleDeque, 2.0); //DoubleDeque_push_front(doubleDeque, 4.0); //halfDeque2 = halve_in_place(doubleDeque); //checkequal(halfDeque2, [2., 1.], "halve_in_place(doubleDeque)"); delete_IntDeque(intDeque); delete_DoubleDeque(doubleDeque); delete_RealDeque(realDeque); exec("swigtest.quit", -1);
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//caption:Design of phase shifter //example9.1 disp("Given frequency fo=10 KHz,Vrms=5 V,phi=10 degrees\n"); disp("Taking A=C3*R4\n"); phi=10;//in degrees fo=1000;//in Hz disp("phi=180-2(atan(2*%pi*f*A))"); A=tan((180-10)*%pi/(180*2))/(2*fo*%pi); printf("Therefore A=C3*R4=%f sec.\n",A); R4=10000;//let (in ohms) printf("C3 and R4 values are selected such that their product equals or greater than %f, The above values are preferable for low cost and bias compensation",A); C3=A/R4; printf("\nC3=%f uF",C3*10^6); disp("To lower the cost of design,the preferred value is C31=0.22uF"); C31=0.22;//let Such that C31>C3 disp("since,C31*R4>A,C31 can be preferred") printf("Similarly, R1 and R2 values should be of Good matching to obtain accurate unity gain modulus "); printf("RESULTS:\n"); printf("(i)Resistors, R1=R2=10Kohms\n"); printf("(iii)R4=%d Kohms\n",R4/1000);//divided by 1000 to display in Kohms printf("(iii)Capacitor, C3=%1.2f uF\n",C31);
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function GetPicInfo(currentfolder,picname) { rs=Map; rs.tpe='side'; rs.col=color(1,1,1); filecontent=''; if FileIsPresent(currentfolder+"\"+picname+".txt") then filecontent=readtextfile(currentfolder+"\"+picname+".txt"); if filecontent.length==0 then if FileIsPresent(currentfolder+"\_settings.txt") then filecontent=readtextfile(currentfolder+"\_settings.txt"); if filecontent.length>0 then { while filecontent.length>0 do { line=filecontent.split("~n"); id=line.split('='); if id=='TYPE' then rs.tpe=line; if id=='COLOR' then rs.col=color(ToScalar(line.split(',')),ToScalar(line.split(',')),ToScalar(line.split(','))); } } return(rs); }
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clc; //page no 8-62 //Example 8.23_ii //To calculate fi such that alpha at 30MHz is 44.9 dB or 176.48 alpha=176.48; Q=125; rho=sqrt((alpha^2-1)/Q^2) disp(rho,'rho='); //rho=(fsi'/fs')-(fs'/fsi') //1.412=(fsi'/fs')-(fs'/fsi')=(1.93/1)-(1/1.93) //fs'/fsi'=1/1.93 //fs'/(fs'+2*fi)=1/1.93 fi=[(30*1.93)-30]/2;//Answer was slightly wrong disp(+'MHz',fi,'IF required=');
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function [flt] = get_filter(a, b) flt = struct('a', a, 'b', b, 'buff_in', zeros(1, size(b, '*')), 'buff_out', zeros(1, size(a, '*'))); endfunction function [flt_out, signal] = push_signal(x, flt_in) flt_in.buff_in(2:$) = flt_in.buff_in(1:$-1) flt_in.buff_in(1) = x; out_sum = flt_in.buff_in*flt_in.b'; out_sum = out_sum + flt_in.buff_out*flt_in.a'; flt_in.buff_out(2:$) = flt_in.buff_out(1:$-1) flt_in.buff_out(1) = out_sum; flt_out = flt_in; signal = out_sum; endfunction
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Mp=1.00728;//mass of proton in amu// Mn=1.00866;//mass of neutronin amu// MH=2.01355;//isotopic mass of H atom in amu// dM=((1*Mp)+(1*Mn)-MH);//dM value of H atom in amu// printf('dM value of H atom=dM=%famu',dM); BE=dM*931;//binding energy of H atom in MeV// printf('\nBinding energy of H atom=BE=%fMeV',BE);
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//Calculate frequency of oscillations clear; clc; //soltion //given R=22*10^3;//ohm C=100*10^-12;//F fo=1/(2*%pi*R*C); printf("The frequency of oscillations= %.2f KHz\n",fo/1000);
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//example 4.6 //ceck whether data at station X is consistence //year in which regime is indicated //compute the adjusted rainfall atX clc;funcprot(0); //given X=[69 55 62 67 87 70 65 75 90 100 90 95 85 90 75 95]; //annual rainfall at X Y=[77 62 67 68 86 90 65 75 70 70 70 75 65 70 55 75]; //average rainfall at 10 base stations cx(1)=69; //accumulated annual values at station X for i=2:16 cx(i)=cx(i-1)+X(i); end cy(1)=77; for i=2:16 cy(i)=cy(i-1)+Y(i); //accumulated annual values at ten stations end //since curve is not having unform slope mprintf("Record at X is not consistent."); mprintf("\nFrom the curve regime is observed in the year 1978.") Q=[1970 1971 1972 1973 1974 1975 1976 1977]; O=[95 75 90 85 95 90 100 90]; for i=1:8 A(i)=0.7051*O(i); end mprintf("\n\nYear Observed rainfall Adjusted rainfall"); for i=1:8 mprintf("\n%i %i %i",Q(i),O(i),A(i)); end //graph is plotted between cx and cy
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printf(" Given C(s)/R(s)=14/(s^2+1.4*s+14)") printf("characterstic equation of the given system is s^2+1.4*s+14=0"); printf("compare it with the standard second order characterstic equation s^2+2*d*w*s+w^2=0"); w=sqrt(14); d=1.4/(2*w); d1=.7; t=2*(d1-d)/w; pt1=%pi/(w*sqrt(1-d^2)); mo1=exp((-%pi*d)/sqrt(1-d^2))*100; rt=(%pi-atan(sqrt((1-d)/d)))/(w*sqrt(1-d^2)); disp(t,"Td=") disp(pt1,"peak time(in sec)for sytem without derivative control is") disp(rt,"rise time(in sec)for sytem without derivative control is") disp(mo1,"maximum overshoot(in %)for sytem without derivative control is") printf("overal transfer function with derivative control is C(s)/R(s)=14*(1+0.274*s)/(s^2+5.236*s+14)") printf("c(t)=1-e^(-2.618*t)*cos(2.673*t)+0.455*e^(-2.618t)*sin(2.673t)") tp=(%pi-atan(2.58))/2.673; tr=(1/2.673)*atan(1/0.455); mo2=1-%e^(-2.618*tp)*cos(2.673*tp)+0.455*%e^(-2.618*tp)*sin(2.673*tp); mp=(mo2-1)*100; disp(tp,"peak time(in sec)for sytem with derivative control is") disp(tr,"rise time(in sec)for sytem with derivative control is") disp(mp,"maximum overshoot(in %)for sytem with derivative control is")
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//DASSL // PROBLEM 1: LINEAR DIFFERENTIAL/ALGEBRAIC SYSTEM // //X1DOT + 10.0*X1 = 0 //X1 + X2 = 1 //X1(0) = 1.0, X2(0) = 0.0 // t=1:10;t0=0;y0=[1;0];y0d=[-10;0]; info=list([],0,[],[],[],0,0); // Calling Scilab functions deff('[r,ires]=dres1(t,y,ydot)','r=[ydot(1)+10*y(1);y(2)+y(1)-1];ires=0') deff('pd=djac1(t,y,ydot,cj)','pd=[cj+10,0;1,1]') // scilab function, without jacobian yy0=dassl([y0,y0d],t0,t,dres1,info); // scilab functions, with jacobian yy1=dassl([y0,y0d],t0,t,dres1,djac1,info); // fortran routine dres1 in dir. routines/default, without jocabian yy2=dassl([y0,y0d],t0,t,'dres1',info); //=yy0 norm(yy2-yy0,1) // fortran routines dres1 and djac1 in dir. routines/default, with jacobian yy3=dassl([y0,y0d],t0,t,'dres1','djac1',info); //=yy1 norm(yy3-yy1,1) yy3bis=dassl([y0,y0d],t0,t,'dres1',djac1,info); // call fortran dres1 and scilab's djac1 yy3ter=dassl([y0,y0d],t0,t,dres1,'djac1',info); // // with specific atol and rtol parameters atol=1.d-6;rtol=0; yy4=dassl([y0,y0d],t0,t,atol,rtol,dres1,info); yy5=dassl([y0,y0d],t0,t,atol,rtol,'dres1',info); //=yy4 norm(yy5-yy4,1) yy6=dassl([y0,y0d],t0,t,atol,rtol,dres1,djac1,info); yy7=dassl([y0,y0d],t0,t,atol,rtol,'dres1','djac1',info); //==yy6 norm(yy7-yy6,1) // yy7=dassl([y0,y0d],t0,t,atol,rtol,'dres1','djac1',info); //==yy6 norm(yy7-yy6,1)
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clear // // // //Variable declaration rho=2.48; //density(gm/cc) M=58; //molecular weight N=6.023*10^23; //avagadro number n=4; //number of atoms //Calculations a=(n*M/(rho*N))^(1/3); //lattice constant(cm) r=a*sqrt(2)*10^8/4; //radius of atom(angstrom) d=2*r; //distance between ions(angstrom) //Result printf("\n distance between ions is %0.1f angstrom",d)
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//Exa 2.26 clc; clear; close; //Given data : format('v',6); Pin_rotor=20;//Power input of rotor in KW phase=3;//no. of phase P=6;//no. of poles f=50;//in Hz N=960;//in rpm(Actual speed of motor) Ns=120*f/P;//synchronous speed in rpm S=(Ns-N)/Ns;//unitless RotorCuLoss=S*Pin_rotor*10^3;//in watts RotorCuLoss=RotorCuLoss/phase;//in watts/phase R2=1/3;//in ohm(Rotor resistance per phase) I2=sqrt(RotorCuLoss/R2);//in Ampere disp(I2,"Motor current per phase(in Ampere) :");
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function [x,y,z] = cylinder(r,n) xdel(winsid()); pi=22/7; Radius = r; Height = 1; SideCount = n+1; // Vertices n_side = SideCount; for i_ver=1:n_side VertexData(i_ver,:) = [Radius*cos(2*%pi/n_side*i_ver),Radius*sin(2*%pi/n_side*i_ver),0]; VertexData(n_side+i_ver,:) = [Radius*cos(2*%pi/n_side*i_ver),Radius*sin(2*%pi/n_side*i_ver),Height]; end // Side Patches for i_pat=1:n_side-1 Index_Patch1(i_pat,:) = [i_pat,i_pat+1,i_pat+1+n_side,i_pat+n_side]; end Index_Patch1(n_side,:) = [n_side,1,1+n_side,2*n_side]; for i_pat=1:n_side // Side patches data PatchData1_X(:,i_pat) = VertexData(Index_Patch1(i_pat,:),1); PatchData1_Y(:,i_pat) = VertexData(Index_Patch1(i_pat,:),2); PatchData1_Z(:,i_pat) = VertexData(Index_Patch1(i_pat,:),3); end x=PatchData1_X; y=PatchData1_Y; z=PatchData1_Z; // Draw side patches figure(1); plot3d(PatchData1_X,PatchData1_Y,PatchData1_Z); h1_fac3d = gce(); h1_fac3d.color_mode = 4; h1_fac3d.foreground = 1; h1_fac3d.hiddencolor = 4; // Bottom Patches Index_Patch2(1,:) = [1:n_side]; Index_Patch2(2,:) = [n_side+1:2*n_side]; for i_pat=1:2 // Bottom patch data PatchData2_X(:,i_pat) = VertexData(Index_Patch2(i_pat,:),1); PatchData2_Y(:,i_pat) = VertexData(Index_Patch2(i_pat,:),2); PatchData2_Z(:,i_pat) = VertexData(Index_Patch2(i_pat,:),3); end // Draw bottom patches figure(1); plot3d(PatchData2_X,PatchData2_Y,PatchData2_Z); h2_fac3d(i_pat) = gce(); h2_fac3d(i_pat).color_mode = 4; h2_fac3d(i_pat).foreground = 1; h2_fac3d(i_pat).hiddencolor = 4; // Axes settings xlabel("x",'fontsize',2); ylabel("y",'fontsize',2); zlabel("z",'fontsize',2); h_axes = gca(); h_axes.font_size = 2; h_axes.isoview = "on"; h_axes.box = "off"; h_axes.rotation_angles = [63.5,-127]; h_axes.data_bounds = [-0.2,-0.2,0;0.2,0.2,0.4]; xgrid;
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clear clc // Given //t1=30 min ;T1=336 k; //t2=15 sec ;T2=347 k; // Converting t2 in min t1=30;T1=336;t2=0.25;T2=347 R=8.314; //log(t1/t2)=E(1/T1-1/T2)/R E=(log(t1/t2)*R)/(1/T1-1/T2); printf("\nRESULT\n") printf("E(J/mol) is %f",E)
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// To Compute the number of electrons. clc; clear; I=(25)*(10^-3); t=(30)*(10^-3); C=I*t; // 1C = 6.242*(10^18) n= 6.242*(10^18); e_s=C*n; disp(e_s,'The Number Of Electrons passing through the person is' )
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//Chapter 9 //Example 9_3 //PAge 215 clear;clc; d=200; r=1.24/2; loop_l=(1e-7)*(0.5+2*log(d/r)); loop_ls=loop_l*1000; printf("Loop inductance/phase/m = %d*10^-7 H \n\n", loop_l*1e7 ); printf("Loop indictance per km = %.1f mH \n\n", loop_ls*1e3 );
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<?xml version="1.0" encoding="utf-8"?> <test> <description>StdProject3D Prism Modified basis P=6 Q=7</description> <executable>StdProject</executable> <parameters>-s prism -b Modified_A Modified_A Modified_B -o 6 6 6 -p 7 7 7</parameters> <metrics> <metric type="L2" id="1"> <value tolerance="1e-11">2.08237e-12</value> </metric> <metric type="Linf" id="2"> <value tolerance="1e-11">4.65228e-12</value> </metric> </metrics> </test>
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x = [1 2 3 4; 5 4 3 6] clf() Matplot(x) xs2pdf( gcf(), "images/05_matplot_v1.pdf" ) if getscilabmode() ~= "STD" quit() end
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clc //initialisation of variables w=28740000//kg-m s=6000//m //CALCULATIONS P=w/s//kg //RESULTS printf('the average tractive force=% f kg',P)
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//Example12.3 // to determine the binary output of the 8-bit dual slope A/D converter clc; clear; close; Vin = 8.5 ; VR = 10 ; f = 2 ; //MHz N = 8 ; C = 0.1*10^-6 ; R = 2*10^3 ; // the output of integrator is defined as // Viao(T1) = -(Vin/R*C)*T1 ; // charging time of capacitor T1 = 2^N/f ; disp('charging time of capacitor is = '+string(T1)+ ' u sec'); // the integrator output T1 = T1*10^-6 ; Viao =-(Vin/(R*C))*T1; disp('the integrator output is = '+string(Viao)+ ' V'); // the binary output of a dual slope A/D converter Bn = (2^N*Vin)/VR; disp('the decimal output of a dual slope A/D converter is = '+string(Bn)+ ' = 218' ); Bn=218; Bn = dec2bin(Bn) ; disp(' The binary output of a dual slope A/D converter is = '+string(Bn)+ ' ' );
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clear; clc; printf("Example 10.3\n"); P=101.3e3; //pressure of the operating column T=295; //Temperature of the operating column P_A=7e3; //partial pressure of ammonia x=1e-3; //=(y1-y2)Thickness of stationary gas film D=2.36e-5; //Diffusivity of ammonia C_A=(1/22.4)*(273/T)*(P_A/P);//=(C_A1-C_A2)Concentration of ammonia gas //X=C_T/C_BM X=P*log(P/(P-P_A))/(P-(P-P_A)); //From equation 10.33 N_A_=(D/x)*X*(C_A); printf("\n The transfer rate per unit area = %.2f *10^-5 kmol/m^2*s",N_A_*1e5)
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clear; clc; D = 7;// inches t = 3/4;// inches l = 16;// feet P = 12;// tons e = 3/4;// inches E = 6000;// tons/in^2 d = D-2*t;// inches A = 0.25*%pi*(D^2 - d^2);// in^2 I = (%pi/64)*(D^4 - d^4);// in^4 p_0 = P/A;// tons/in^2 Z = 2*I/D;// in^3 M = P*e*sec(0.25*l*12*sqrt(P/(E*I)));// ton-inches p_b = M/Z;// tons/in^2 p_max = p_0+p_b;// tons/in^2 p_min = p_0-p_b;// tons/in^2 //if tension is just on the point being induced in the section, p_b = p_0 e = p_0*t*Z/M;// inches printf('Stress intensities, p_max = %.3f tons/in^2.,compressive\n p_min = %.3f tons/in^2., compressive',p_max,p_min); printf('\n Maximum possible eccentricity, e = %.2f inches',e);
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clear; clc; //Example 17.5 b=50; V2=-5.2; Vbe=0.7; Rc2=0.240; Vor=-0.75; Re=1.180; iE=(Vor-Vbe-V2)/Re; printf('\nemitter current=%.2f mA\n',iE) iB=iE/(1+b); iB=iB*1000;//micro A printf('\ninput base current=%.2f microA\n',iB) R3=1.500; i3=(Vor-V2)/R3; printf('\ni3=%.2f mA\n',i3) iB=iB*0.001;//mA N=(-(Vor+Vbe)*(1+b)/(Rc2)-i3)/iB; printf('\nN=%.f\n',N)
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FOR_DECL_KO.tst
void main() { for(int i = 0; i < 10; i = i + 1) { print("ERROR_FOR"); } }
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exa_7_7.sce
// Exa 7.7 clc; clear; close; // Given data f=50;//in Hz R2=330*10^3;//in ohm R3=15*10^3;// in ohm R4=22*10^3;//in ohm C2=0.12*10^-6;//in F omega=2*%pi*f; R1= R2*R3/R4;// in ohm disp(R1*10^-3,"Resistive component of unknown resistance in kohm") C1= C2*R4/R3;// in F disp(C1*10^6,"Capacitive component of unknown capacitor in micro F") D=1/(omega*C1*R1); disp(D,"Dissipation factor is ")
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Example39_30.sce
// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART IV : UTILIZATION AND TRACTION // CHAPTER 1: INDUSTRIAL APPLICATIONS OF ELECTRIC MOTORS // EXAMPLE : 1.30 : // Page number 710 clear ; clc ; close ; // Clear the work space and console // Given data J = 1270.0 // Moment of inertia of fly-wheel(kg-m^2) N = 500.0 // Speed(rpm) hp = 50.0 // Motor rating(hp) // Calculations g = 9.81 T = hp*746*60/(2*%pi*N*g) // Full load torque of motor(kg-m) T_m = 2*T // Accelerating torque(kg-m) alpha = T_m*g/J // Angular acceleration(rad/sec^2) t = 2*%pi*N/(60*alpha) // Time taken to accelerate a fly-wheel(sec) // Results disp("PART IV - EXAMPLE : 1.30 : SOLUTION :-") printf("\nTime taken to accelerate a fly-wheel, t = %.1f sec", t)
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Ex21_3_2.sce
clc //intialization of variables d = 100*10^-4 // cm v = 10^-3// cm/sec nu = 0.2 // cm^2/sec DS = 0.3 // cm^2/sec DG = 3*10^-7 // cm^2/sec H = 4.3*10^-4 // at 60 degree centigrade //Calculations kG = (2+(0.6*((d*v/nu)^0.5)*((nu/DS)^(1/3))))*DS/d// cm/sec k = kG*H t = 30*DG/k^2 //Results printf("The mass transfer coefficient is %.5f cm/sec",k) printf("\nTHe time needed to dry the particle is %.6f sec",t) //Answer wrong in textbook starting from kG
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//problem 2 pagenumber 2.86 //given format(6); v1=5;//volt v2=2;//volt rf1=10e3;//ohm r1=10e3;//ohm //determine output voltage v0=-((-v1*rf1/r1)-(-v2*rf1/r1)); disp('Output voltage = '+string(v0)+' V');
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//a) Find the range of IDQ that can be expected if R1 ¼ 1M and R2 ¼ 3M. (b) Find the range of IDQ that can be expected if R1 ¼ 1M //and R2 = 7M. (c) Discuss the significance of the results of parts a and b. //Example 5.28 page no 159 clear clc Vdd=24 Idqmax=4 Idqmin=2.8 Rs=2 //MΩ Rd=1 //MΩ Vdsqmax=Vdd-Idqmax*(Rs+Rd) Vdsqmin=Vdd-Idqmin*(Rs+Rd) printf("\n The value of Vdsqmax=%0.3f V" ,Vdsqmax) printf("\n The value of Vdsqmin=%0.3f V" ,Vdsqmin)
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plot_graph.sce
////<first_task> //// //importXcosDiagram("/home/evgeniy/Рабочий стол/Учеба/Магистратура/Адаптивное и робастное управление/Lab5/scilab/for_output.zcos"); //xcos_simulate(scs_m, 4); //plot2d(Y.time, [Y.values Y_lin.values]); //legend("Выход модели ВСВ", "Выход регрессионной модели" ,4) //a = gca(); //a.x_label.text = "$t\text{, с}$" //a.x_label.font_size = 4; //line2 = a.children(2).children(1); //line2.thickness = 3; //line2.foreground = 1; //a.children(1).font_size = 2; //// ////<\first_task> //<second_task> // importXcosDiagram("/home/evgeniy/Рабочий стол/Учеба/Магистратура/Адаптивное и робастное управление/Lab5/scilab/for_state.zcos"); xcos_simulate(scs_m, 4); for i = 1:2 subplot(2,1,i) plot2d(X.time, [X.values(:,i) X_lin.values(:,i)]); legend("Модель ВСВ", "Наблюдатель" ,4) a = gca(); a.x_label.text = "$t\text{, с}$" a.x_label.font_size = 4; str = "$x_{\small" + string(i)+ "}$" a.y_label.text = str; a.y_label.font_size = 4; line2 = a.children(2).children(1); line2.thickness = 3; line2.foreground = 1; a.children(1).font_size = 2; end // //<\second_task>
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clc //given that del_t = 2e-12 // lifetime of exited state in sec h = 6.63e-34 // Plank constant printf("Example 2.18") h_bar = h / (2*%pi) // constant del_E = h_bar/(1.6e-19*2*del_t) // calculation of uncertainty in momentum printf("\n Minimum error in measurement of energy of this state is %e eV.\n\n\n",del_E) // Answer in book is 1.65e-4 eV
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jobname='tbg2'; niters=4; resarray=cell(1,niters); nx=6; ny=6; nz=6; %Load aeach data set in turn create cell array %Do animation for currentiter=1:niters outname=sprintf('results/%s_%d.mat',jobname, currentiter); load(outname); end
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//Chapter-1, Example 1.57, Page 67 //============================================================================= clc; clear; //INPUT DATA I1=25;//current source in A I2=6;//current source in A I3=5;//current source in A RAB=5;//Resistance in ohms RAC=10;//Resistance in ohms RBC=2;//Resistance in ohms //let currents across AC and BC and AB are Ix,Iy and Iz respectively //applying kirchoff's current law at node A //-I1+Ix+I3+Iz=0------eqn(1) //applying kirchoff's current law at node B //-Iz-I3+Iy+I2=0------eqn(2) //CALCULATIONS [a]=[((1/RAC)+(1/RAB)),(-1/RAB);(-1/RAB),((1/RAB)+(1/RBC))]; [b]=[20;-1]; [c]=inv(a)*(b) VA=c(1);//voltage at node A VB=c(2);//voltage at node B //OUTPUT mprintf("Thus voltages at node A and B are %2.1f V and %2.1f V",-VA,VB); //=================================END OF PROGRAM==============================
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clc; clear; hp=0.15;//from diagram F=9000; V=80;//Working voltage Lmt=1.25;//Mean length of the turn Vp=4;//voltage per pole disp('For a copper winding at 75 deg cel:') a=0.021*(10^(-6))*Lmt*(F/Vp); disp(a,'The conductor area is:') Vp=4;//voltage per pole S=Lmt*hp; C=0.019;//Assumed value disp('For a temp rise of 65 deg cel:') theta_m=65;//temperature rise p=(theta_m*S)/C; disp(p,'The power dissipated is:') I=p/Vp; disp(I,'The field current is:') N=F/I; disp(N,'The number of turns per pole is:') J=I/N; disp(J,'The current density is:')
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// Example 7-13 // Nyquist Plot of MIMO system clear; clc; xdel(winsid()); //close all windows A = [-1 -1 ; 6.5 0]; B = [1 1; 1 0]; C = [1 0; 0 1]; D = [0 0; 0 0]; G = syslin('c',A,B,C,D); P = clean(ss2tf(G)); subplot(2,2,1); nyquist(P(1,1),-100,100); xgrid(color('gray')); xtitle('Nyquist plot: From U1','Real Axis','To Y1'); subplot(2,2,2); nyquist(P(2,1),-100,100); xgrid(color('gray')); xtitle('Nyquist plot: From U1','Real Axis','To Y2'); subplot(2,2,3); nyquist(P(1,2),-100,100); xgrid(color('gray')); xtitle('Nyquist plot: From U2','Real Axis','To Y1'); subplot(2,2,4); nyquist(P(2,2),-100,100); xgrid(color('gray')); xtitle('Nyquist plot From U2','Real Axis','To Y2');
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//example 5.1, Page 98 clc; IL=(100*1000)/(400*.9*.8*sqrt(3))//Full load current FS=(1000-950)/1000//Full load slip //disp("hhhhhhhhhhh") //Writing loop x=[1 0.577350269 .7 0.333333333] for i=1:1:4 T=1.8*(x(i))^2 printf("\n The value of Tst/Tf when x=%f is %f ",x(i),T) end
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#************************************************************ # Scenario of Ikea # # date : Fri Aug 6 12:16:20 2010 #************************************************************ p3d_sel_desc_name P3D_ENV Ikea p3d_sel_desc_name P3D_ROBOT HUMAN_ACHILE p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 1.800000 -1.520000 0.750000 0.000000 0.000000 76.320000 4.292683 0.000000 -1.327433 -1.283183 0.219512 12.951220 65.780000 8.370000 -14.860000 91.170000 20.930000 -39.760000 -5.120000 -27.600000 14.650000 -43.080000 -34.208450 0.880000 -107.640000 8.620000 -11.570000 -90.000000 -5.490000 103.630000 0.000000 0.000000 3.260000 5.980000 -90.000000 8.840000 96.870000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_sel_desc_name P3D_ROBOT ROBOT_JUSTIN p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 2.800000 -1.880000 74.120000 25.240000 -33.380035 36.969028 -13.274338 -1.327433 3.871682 -66.863319 -95.575218 -15.044244 115.634224 1.671586 -16.914946 39.690266 31.719982 -90.792702 -19.723872 32.761383 40.727967 -13.935005 -27.958518 2.080000 -1.140000 1.110000 -4.195717 -10.580000 40.360000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 2.800000 -1.880000 74.120000 78.010000 -33.380035 36.969028 -13.274338 -1.327433 3.871682 -66.863319 -95.575218 -15.044244 115.634224 1.671586 -16.914946 39.690266 19.545673 -84.999603 -19.723872 52.556530 34.691275 -9.582022 -1.049950 1.880000 -1.880000 1.110000 -4.195717 -10.580000 61.450000 p3d_set_robot_traj /Users/jmainpri/workspace/BioMove3DDemos/CostHriFunction/ICRA_Justin_Trajectory/baseFixedWithVisib.traj p3d_constraint p3d_kuka_arm_ik 6 18 19 21 22 23 24 1 29 0 3 20 -1 1 p3d_set_cntrt_Tatt 0 -0.982797 0.018403 -0.183754 -0.036039 -0.003530 0.992963 0.118341 -0.346482 0.184640 0.116955 -0.975819 -0.017708 p3d_constraint p3d_min_max_dofs 0 2 4 3 2 0.000000 135.000000 0 p3d_constraint p3d_lin_rel_dofs 1 5 2 3 4 3 -1.000000 -1.000000 0.000000 0 p3d_constraint p3d_fixed_jnt 1 1 0 3 2.800000 -1.880000 74.120000 0 p3d_constraint p3d_fixed_jnt 1 3 0 1 -33.380035 0 p3d_constraint p3d_fixed_jnt 1 4 0 1 36.969028 0 p3d_constraint p3d_fixed_jnt 1 5 0 1 -13.274338 0 p3d_constraint p3d_fixed_jnt 1 7 0 1 -1.327433 0 p3d_constraint p3d_fixed_jnt 1 8 0 1 3.871682 0 p3d_constraint p3d_fixed_jnt 1 10 0 1 -66.863319 0 p3d_constraint p3d_fixed_jnt 1 11 0 1 -95.575218 0 p3d_constraint p3d_fixed_jnt 1 12 0 1 -15.044244 0 p3d_constraint p3d_fixed_jnt 1 13 0 1 115.634224 0 p3d_constraint p3d_fixed_jnt 1 14 0 1 1.671586 0 p3d_constraint p3d_fixed_jnt 1 15 0 1 -16.914946 0 p3d_constraint p3d_fixed_jnt 1 16 0 1 39.690266 0 p3d_constraint p3d_fix_jnts_relpos 1 29 1 24 0 0 p3d_set_cntrt_Tatt 16 -0.982797 -0.003530 0.184640 -0.033372 0.018403 0.992963 0.116955 0.346778 -0.183754 0.118341 -0.975819 0.017101 p3d_set_object_base_and_arm_constraints 29 1 0 1 0 p3d_sel_desc_name P3D_ROBOT Lampe p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 -1.966568 1.622419 0.983776 0.000000 0.000000 0.000000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_sel_desc_name P3D_ROBOT Assiette p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.884956 -1.573254 0.787611 0.000000 0.000000 0.000000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_sel_desc_name P3D_ROBOT Pommes p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.442478 -1.622419 0.762537 0.000000 0.000000 0.000000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_sel_desc_name P3D_ROBOT Verre p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_sel_desc_name P3D_ROBOT Tabouret p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 -0.360000 0.220000 0.000000 0.000000 0.000000 0.000000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_sel_desc_name P3D_ROBOT sailLamp1 p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.029268 0.263415 3.000000 0.000000 0.000000 0.000000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_sel_desc_name P3D_ROBOT sailLamp2 p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.029268 0.556098 3.000000 0.000000 0.000000 0.000000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_sel_desc_name P3D_ROBOT sailLamp3 p3d_set_robot_steering_method Linear p3d_set_robot_current 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.029268 0.848780 3.000000 0.000000 0.000000 0.000000 p3d_set_robot_goto 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 p3d_set_camera_pos 1.692213 -1.665001 0.468306 5.227740 1.316250 0.728750 0.000000 0.000000 1.000000 0.000000
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/2438/CH6/EX6.1/Ex6_1.sce
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2020-04-09T02:43:26.499817
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Ex6_1.sce
//============================================================================== // chapter 6 example 1 clc; clear; //input data R75 = 57.2; //resistance at 75 C in ohm R25 = 55; //resistance at 25 C in ohm t1 = 25; //temperature in C t2 = 75 // temperature in C //formula //Rt = R0*(1+(alpha*t)) //calculation alpha = (R25-R75)/((25*R75)-(75*R25)); //temperature cofficient //result mprintf('temperature coefficient =%3.5f.K^-1',alpha); //=====================================================================================
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True-Class Iris-setosa Iris-versicolor Iris-virginica Iris-setosa 1.0 0.0 0.0 Iris-setosa 1.0 0.0 0.0 Iris-setosa 1.0 0.0 0.0 Iris-setosa 1.0 0.0 0.0 Iris-setosa 0.0 0.9772727272727273 0.022727272727272728 Iris-versicolor 0.0 0.9772727272727273 0.022727272727272728 Iris-versicolor 0.0 0.9772727272727273 0.022727272727272728 Iris-versicolor 0.0 0.9772727272727273 0.022727272727272728 Iris-versicolor 0.0 0.9772727272727273 0.022727272727272728 Iris-versicolor 0.0 0.9772727272727273 0.022727272727272728 Iris-virginica 0.0 0.023809523809523808 0.9761904761904762 Iris-virginica 0.0 0.023809523809523808 0.9761904761904762 Iris-virginica 0.0 0.9772727272727273 0.022727272727272728 Iris-virginica 0.0 0.023809523809523808 0.9761904761904762 Iris-virginica 0.0 0.023809523809523808 0.9761904761904762
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 5.8\n\n\n"); // Chapter 5 : Properties Of Liquids And Gases // Problem 5.8 (page no. 193) // Solution //Using Table 2 ans a quality of 85%(x=0.85),we have //at 1.0 MPa x=0.85; sf=2.1387; //saturated liquid entropy //Unit:kJ/kg*K sfg=4.4487; //Evap. Entropy //Unit:kJ/kg*K hf=762.81; //saturated liquid enthalpy //Unit:kJ/kg hfg=2015.3; //Evap. Enthalpy //Unit:kJ/kg uf=761.68; //saturated liquid internal energy //Unit:kJ/kg ufg=1822.0; //Unit:kJ/kg //Evap. internal energy vf=1.1273; //Saturated liquid specific volume //Unit:m^3/kg vfg=(194.44-1.1273); //evap. specific volume //Unit:m^3/kg sx=sf+(x*sfg); //entropy //kJ/kg*K printf("Entropy of a wet steam mixture at 1.0 MPa is %f kJ/kg*K\n",sx); hx=hf+(x*hfg); //enthalpy //kJ/kg*K printf("Enthalpy of a wet steam mixture at 1.0 MPa is %f kJ/kg\n",hx); ux=uf+(x*ufg); //internal energy //kJ/kg*K printf("Internal energy of a wet steam mixture at 1.0 MPa is %f kJ/kg\n",ux); vx=(vf+(x*vfg))*(0.001); //specific volume //m^3/kg printf("Specific Volume of a wet steam mixture at 1.0 MPa is %f m^3/kg\n",vx); //As a check, px=10^6; //psia //pressure ux=hx-((px*vx)/10^3); //1 ft^2=144 in^2 //internal energy printf("As a check,\n") printf("Internal energy of a wet steam mixture at 120 psia is %f kJ/kg\n",ux); printf("Which agrees with the values obtained above");
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# possible rows[0] for target 19 5 1 0 0 1 4 0 0 2 2 1 0 3 0 2 0 0 1 3 0 3 1 0 1 0 2 1 1 1 0 2 1 1 1 0 2 # rows in subarray[0]: 9 # possible rows[1] for target 18 6 0 0 0 2 3 0 0 3 1 1 0 0 2 2 0 1 0 3 0 4 0 0 1 0 3 0 1 1 1 1 1 2 0 0 2 0 0 0 3 # rows in subarray[1]: 10 # possible rows[2] for target 21 7 0 0 0 3 3 0 0 4 1 1 0 0 4 1 0 1 2 2 0 2 0 3 0 5 0 0 1 1 3 0 1 2 1 1 1 0 0 3 1 3 0 0 2 0 1 1 2 1 0 0 3 # rows in subarray[2]: 13 # possible rows[3] for target 28 8 1 0 0 4 4 0 0 0 7 0 0 5 2 1 0 1 5 1 0 6 0 2 0 2 3 2 0 3 1 3 0 0 2 4 0 1 0 5 0 6 1 0 1 2 4 0 1 3 2 1 1 4 0 2 1 0 3 2 1 1 1 3 1 4 1 0 2 0 4 0 2 1 2 1 2 2 0 2 2 2 1 0 3 0 0 2 3 0 1 0 4 # rows in subarray[3]: 23 # possible combinations: 26910 # Vector-div: unimodular matrix preserved 8 2 2 1 0 1 1 1 1 1 2 2 0 2 3 2 0
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clear //Initialisation OG=2*10**5 //Open Loop Gain CG=20 //Closed Loop Gain OR1=75 //Output Resistance IR1=2*10**6 //Input Resistance R1=20*10**3 //Resistnce in Ohm R2=10**3 //Resistnce in Ohm //Calculation AB=OG*CG**-1 //factor (1+AB) OR2=OR1*AB**-1 //Output Resistance //the input is connected to a virtual earth point by the resistance R2, //so the input resistance is equal to R 2 , IR2=R2 //Input Resistance //Result printf("\n Output Resistance = %.1f mOhm\n",OR2*1000) printf("\n Input Resistance = %d KOhm",IR2*10**-3)
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//Initialiser C0 = 25 R0 = 5 V0 = [C0;R0] //la matrice de coefficients A = [0.5 0.4;-0.104 1.1] //nombre de mois n = 20 M = zeros(2,2) M(1,1) = C0 M(2,1) = R0 V=V0 for i = 2:n V(1:2,i) = A*V(1:2,i-1) scf(0) clf(0) a=get("current_axes"); a.x_location="origin"; a.y_location="origin"; xtitle("rats et chouettes en cours du temps","mois","effectif") plot(V(1,:),'bp:') plot(V(2,:),'gs-') legend(['chouettes','rats (en milliers)']) //2e façon // scf(1) // clf(1) // b=get("current_axes") // b.x_location="origin"; // b.y_location="origin"; // plot2d(1:n,V(1,:),color("red")) // plot2d(1:n,V(2,:),color("blue")) end
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clc; Rc=8000; //Ohm Zin1=1330; //Ohm rc1=(Rc*Zin1)/(Rc+Zin1);//Ohm re=25; //Ohm Zin2=1730; //Ohm rc2=(Rc*Zin2)/(Rc+Zin2);//Ohm re=25;//Ohm Av1=rc1/re; Av2=rc2/re; disp(' ',Av1,"Av1=");//The answers vary due to round off error disp(' ',Av2,"Av2=");//The answers vary due to round off error
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//Ex7_1 clc A=60000 disp("A= "+string(A)) //Amplifier gain Af=10000 disp("Af= "+string(Af)) //Feedback gain N_dB=20*log10(Af/A) disp("N_dB=20*log10(Af/A)= "+string(N_dB)+"dB") //Negative feedback gain B=[1/(Af)]-(1/A)// formulae using [Af=A/(1+A*B)] disp("B=[1/(Af)]-(1/A)= "+string(B)) //Feedback factor
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# blinky2_b.tst Test the blinky2 example: >>> import examples >>> test2 = examples.test_compile('blinky2', False) Traceback (most recent call last): ... NameError: name 'run' is not defined
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; @Harness: disassembler ; @Result: PASS section .text size=0x00000054 vma=0x00000000 lma=0x00000000 offset=0x00000034 ;2**0 section .data size=0x00000000 vma=0x00000000 lma=0x00000000 offset=0x00000088 ;2**0 start .text: label 0x00000000 ".text": 0x0: 0x0f 0x37 cpi r16, 0x7F ; 127 0x2: 0x1f 0x37 cpi r17, 0x7F ; 127 0x4: 0x2f 0x37 cpi r18, 0x7F ; 127 0x6: 0x3f 0x37 cpi r19, 0x7F ; 127 0x8: 0x4f 0x37 cpi r20, 0x7F ; 127 0xa: 0x5f 0x37 cpi r21, 0x7F ; 127 0xc: 0x6f 0x37 cpi r22, 0x7F ; 127 0xe: 0x7f 0x37 cpi r23, 0x7F ; 127 0x10: 0x8f 0x37 cpi r24, 0x7F ; 127 0x12: 0x9f 0x37 cpi r25, 0x7F ; 127 0x14: 0xaf 0x37 cpi r26, 0x7F ; 127 0x16: 0xbf 0x37 cpi r27, 0x7F ; 127 0x18: 0xcf 0x37 cpi r28, 0x7F ; 127 0x1a: 0xdf 0x37 cpi r29, 0x7F ; 127 0x1c: 0xef 0x37 cpi r30, 0x7F ; 127 0x1e: 0xff 0x37 cpi r31, 0x7F ; 127 0x20: 0x0f 0x3f cpi r16, 0xFF ; 255 0x22: 0x00 0x30 cpi r16, 0x00 ; 0 0x24: 0x0f 0x37 cpi r16, 0x7F ; 127 0x26: 0x0f 0x33 cpi r16, 0x3F ; 0x63 0x28: 0x0f 0x31 cpi r16, 0x1F ; 0x31 0x2a: 0x0f 0x30 cpi r16, 0x0F ; 0x15 0x2c: 0x07 0x30 cpi r16, 0x07 ; 7 0x2e: 0x03 0x30 cpi r16, 0x03 ; 3 0x30: 0x01 0x30 cpi r16, 0x01 ; 1 0x32: 0x00 0x3f cpi r16, 0xF0 ; 240 0x34: 0x08 0x37 cpi r16, 0x78 ; 120 0x36: 0x0c 0x33 cpi r16, 0x3C ; 0x60 0x38: 0x0e 0x31 cpi r16, 0x1E ; 0x30 0x3a: 0x0c 0x3c cpi r16, 0xCC ; 204 0x3c: 0x06 0x36 cpi r16, 0x66 ; 102 0x3e: 0x03 0x33 cpi r16, 0x33 ; 0x51 0x40: 0x09 0x31 cpi r16, 0x19 ; 0x25 0x42: 0x0c 0x30 cpi r16, 0x0C ; 0x12 0x44: 0x06 0x30 cpi r16, 0x06 ; 6 0x46: 0x0a 0x3a cpi r16, 0xAA ; 170 0x48: 0x05 0x35 cpi r16, 0x55 ; 0x85 0x4a: 0x0a 0x32 cpi r16, 0x2A ; 0x42 0x4c: 0x05 0x31 cpi r16, 0x15 ; 0x21 0x4e: 0x0a 0x30 cpi r16, 0x0A ; 0x10 0x50: 0x05 0x30 cpi r16, 0x05 ; 5 0x52: 0x02 0x30 cpi r16, 0x02 ; 2 start .data:
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/solvers/IncNavierStokesSolver/Tests/PrismHex_channel_m4.tst
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<?xml version="1.0" encoding="utf-8"?> <test> <description>3D channel flow, Hexahedral and Prismatic elements, P=4</description> <executable>IncNavierStokesSolver</executable> <parameters>PrismHex_channel_m4.xml</parameters> <files> <file description="Session File">PrismHex_channel_m4.xml</file> </files> <metrics> <metric type="L2" id="1"> <value variable="u" tolerance="1e-12">1.79242e-15</value> <value variable="v" tolerance="1e-12">1.76764e-15</value> <value variable="w" tolerance="1e-12">7.97721e-15</value> <value variable="p" tolerance="1e-12">1.11992e-13</value> </metric> <metric type="Linf" id="2"> <value variable="u" tolerance="1e-12">9.44016e-15</value> <value variable="v" tolerance="1e-12">9.24349e-15</value> <value variable="w" tolerance="1e-12">4.39926e-14</value> <value variable="p" tolerance="1e-12">6.00853e-13</value> </metric> </metrics> </test>
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clear; clc; close; Tj = 200; Ta = 40; Qjc = 0.5; Qcs = 0.6; Qsa = 1.5; Pd = (Tj-Ta)/(Qjc+Qcs+Qsa); disp(Pd,'Maximum power dissipated(Watts) = ');
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//ques2 //The Ideal Otto Cycle clear clc //the temperature and pressure of air at the end of the isentropic compression process (state 2), using data from Table A–17 T1=290;//initial temp in K u1=206.9;//initial internal energy in kJ/kg vr1=676.1;//initial reduced volume //Process 1-2 (isentropic compression of an ideal gas) //vr2/vr1=v2/v1=1/r r=8;//ratio vr2=vr1/r;//reduced volume at state 2 //using table corresponding to vr2 T2=652.4;//final temp in K u2=475.11;//final internal energy in kJ/kg P1=100;//initial pressure in kPa P2=P1*T2/T1*r;//final pressure in kPa //Process 2-3 (constant-volume heat addition) Qin=800;//heat input in kJ/kg u2=1275.11;// intenal energy at state 2 in kJ/kg u3=Qin+u2;//internal energy at state 3 in kJ/kg //using tables corresponding to u3 T3=1575.1;//temperature at state 3 in K vr3=6.108;//reduced volume at state 3 printf('(a) T3,Temperature at state 3 = %.1f K \n',T3); vr3=6.108;//reduced volume at state 3 P3=P2*(T3/T2)*1;//1 for v2/v3 printf(' Pressure P3 = %.3f MPa \n',P3/1000); //(b) vr3=r*vr3; //now from table T4=795.6;//temp at state 4 in K u4=588.74;//internal energy at state 4 in kJ/kg //Process 4-1 (constant-volume heat rejection) Qout=u4-u1;//heat output in kJ/kg w=Qin-Qout;//work done in kJ/kg printf(' (b) Net work done = %.2f kJ/kg \n',w); //(c) nth=w/Qin;//efficiency of heat engine k=1.4;//constant no=1-r^(1-k);//thermal efficiency printf(' (c) The thermal efficiency of the cycle is determined from its definition = %.3f \n',nth); printf(' Under the cold-air-standard assumptions thermal efficiency would be = %.3f \n',no); //(d) //The mean effective pressure is determined from its definition R=0.287//gas constant for water v1=R*T1/P1//specific volume at state 1 MEP=w/(v1*(1-1/r));//mean effective pressure in kPa printf(' (d) Mean effective pressure = %.0f kPa \n',MEP);
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v1=8; v2=20; r1=1; r2=2; r3=3; r4=4; r5=5; i1=2; vx=(i1-v2/r5+v1*r4*(1/r4+1/r5)/(r1+r2))/((1/r3+1/r4+1/(r1+r2))*(1/r4+1/r5)*r4-1/r4); vy=r4*((vx*(1/r3+1/r4+1/(r1+r2)))-v1/(r1+r2)); i4=(vx-vy)/r4; disp("the current (in mA) flowing through R4 is"); disp(i4); disp("and the direction is left to right");
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function [R_sys] = reliabilityCalc(lambda, c, n) //open a file fid = mopen("data.txt", "w"); if (fid==-1) then error("cannot open the file") end t = [0:0.0001:1]; //calc of reliability R = %e^(lambda*-t); //calc of reliability without redundancy R_nr = R .^ ((2^n)-1); //write file sR = size(R_nr); mfprintf(fid, "%ld\n",-255) ; for i = 1:sR(2) mfprintf(fid, "%f ", R_nr(i)); end //calc Reliability of system with redundance for different c // figure(); // set(gca(),"auto_clear","off"); //xlabel("time"); // ylabel("Reliability"); //plot(t, R_nr, "black"); R_sys = null; r_int = 1; for j = 0:(n-1) k = j; r_int = r_int .* ((2^k * c + 1) - 2^k * c * R); end R_sys = R_nr .* r_int; //size of R_sys sR = size(R_sys); mfprintf(fid, "%d\n", -255); for k = 1 : sR(2) mfprintf(fid, "%f", R_sys(k)); end plot (t, R_sys, "r-"); xlabel("Tiempo adimensional") ylabel("Confiabilidad") f=get("current_figure") f.background = 8; legend(['R(sys) con c = 1']); mclose(fid) return R_sys; ////////////////////////////////////////////////////////////////////// l = size(c); for i = 1:l(2) R_sys =null; r_int=1; for j = 0:(n-1) k = j; r_int = r_int .* (( 2^k * c(i) + 1) - 2^k * c(i) * R); end R_sys = R_nr .* r_int; //size of R_sys sR = size(R_sys); mfprintf(fid, "%d\n", -255); //c = 0.98 if i==1 then for k=1:sR(2) mfprintf(fid, "%f ", R_sys(k)); end plot(t, R_sys,"r--"); end //c = 0.99 if i==2 then for k=1:sR(2) mfprintf(fid, "%f ", R_sys(k)); end plot(t, R_sys,"b-."); end //c = 1 if i==3 then for k=1:sR(2) mfprintf(fid, "%f ", R_sys(k)); end plot(t, R_sys,"cyan-+"); end end legend(['R Not redundant';'C=0.98'; 'c=0.99'; 'c=1']); mclose(fid) endfunction
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//Graphical// //Example 4.4.2 //Frequency Response of Three point Moving Average System //y(n)= (1/3)[x(n+1)+x(n)+x(n-1)] //h(n) = [1/3,1/3,1/3] clear; clc; close; //Calculation of Impulse Response n =-1:1; h = [1/3,1/3,1/3]; //Discrete-time Fourier transform K = 500; k = 0:1:K; w = %pi*k/K; H = h * exp(-sqrt(-1)*n'*w); //phasemag used to calculate phase and magnitude in dB [Phase_H,m] = phasemag(H); H = abs(H); subplot(2,1,1) plot2d(w/%pi,H) xlabel('Frequency in Radians') ylabel('abs(H)') title('Magnitude Response') subplot(2,1,2) plot2d(w/%pi,Phase_H) xlabel('Frequency in Radians') ylabel('<(H)') title('Phase Response')
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clear mclose ( 'all' ); szPathToData = 'C:\ARGilyazeev\github\stud\data\'; previous_driver = driver ( 'PDF' ); function rDataPlot ( szFileName, g ) [fdData, err] = mopen ( szPathToData + szFileName ); kI_Magic = mget( 1, 'ul', fdData ); kN_FrameCount = mget( 1, 'u', fdData ); kN_GridX = mget( 1, 'u', fdData ); kN_ItersIdentMi = mget( 1, 'u', fdData ); kN_ItersSkips = mget( 1, 'u', fdData ); kR_ErrorMax = mget( 1, 'd', fdData ); kR_Alpha = mget( 1, 'd', fdData ); kR_K_Mu = mget( 1, 'd', fdData ); kR_dT = mget( 1, 'd', fdData ); kR_S_t = mget( 1, 'd', fdData ); kR_S_x = mget( 1, 'd', fdData ); kR_dX = 1.0 / ( kN_GridX - 1.0 ); listX = 0:kR_dX:1; function [ s, p, v ] = rDataLoadFrame ( i ) mseek ( 72 + i*(kN_GridX*2+1)*8, fdData ); s = mget( kN_GridX, 'd', fdData ); if i < kN_FrameCount then p = mget( kN_GridX, 'd', fdData ); v = mget( 1, 'd', fdData ); end; endfunction xinit ( szPathToData + szFileName + '.S.pdf' ); xtitle ( '', 'X', 'S' ); for i = 1:floor(0.2/kR_dT):kN_FrameCount plot ( listX, rDataLoadFrame(i) ); end; xend(); xinit ( szPathToData + szFileName + '.U.pdf' ); xtitle ( '', 't', 'u' ); listTimes = 0:floor(0.01/kR_dT):kN_FrameCount; list_dP = listTimes; for i = 1:length(listTimes) [ s, p, v ] = rDataLoadFrame(listTimes(i)); list_dP(i) = v; end; plot ( listTimes*kR_dT, list_dP ); xend(); endfunction rDataPlot ( 'St.bin', 0 ); rDataPlot ( 'Sx.bin', 0 ); rDataPlot ( 'S2.bin', 0 ); rDataPlot ( 'original.bin', 0 ); driver(previous_driver); mclose ( 'all' );
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Example3_11.sce
//Example 3.11 clc disp("Step 1: Identify topology") disp(" The feedback voltage is applied across the resistance R_e1 and it is in series with input signal. Hence feedback is voltage series feedback.") disp("") disp("step 2 and Step 3: Find input and output circuit.") disp(" To find input circuit, set Vo = 0 (connecting C2 to ground), which gives parllel combination of Re with Rf at E1. To find output ciruit, set Ii = 0 (opening the input node E1 at emitter of Q1), which gives series combination od Rf and R_e1 across the output. The resultant circuit is shown in fig.3.57") disp("") disp("Step 4: Find open loop voltage gain (Av)") rl2=(4.7*3.42)/(4.7+3.42) // in k-ohm format(5) disp(rl2," R_L2(in k-ohm) = R_c2 || (Rs+R) =") disp(" A_i2 = -hfe = -50") disp("R_i2 = hie = 1000 ohm = 1 k-ohm") av2=-50*1.98 format(3) disp(av2," A_v2 = A_i2*R_L2 / R_i2 =") disp(" A_i1 = -hfe = -50") format(7) rl1=((10*100*22*1)/((100*22)+(10*22)+(10*100)+(10*100*22)))*10^3 // in ohm disp(rl1," R_L1(in ohm) = R_c1 || R3 || R4 || R_i2 =") disp(" R_i1 = h_ie + (1+h_fe)*R_e1eff") re1=1+(51*((3.3*0.12)/(3.42))) // in k-ohm format(4) disp(re1,"where R_e1eff(in k-ohm) = Rs || R =") av1=(-50*865.46)/6900 format(5) disp(av1," A_v1 = A_i1*R_L1 / R_i1 =") disp("The overall voltage gain,") av=-6.27*-99 format(7) disp(av," Av = A_v1 * A_v2 =") disp("") disp("Step 5: Calculate beta") beta=120/(120+3300) format(6) disp(beta," beta = Vf / Vo = Rs / Rs+R =") disp("") disp("Step 6: Calculate D, A_vf, R_if, R_of and R''_of") d=1+(0.035*620.73) format(7) disp(d," D = 1 + Av*beta =") avf=620.73/22.725 format(5) disp(avf," A_vf = Av / D =") rif=6.9*22.725 // in k-ohm format(6) disp(rif," R_if(in k-ohm) = R_i1 * D =") disp(" R_of = Ro / D = infinity") rof=(1.98*10^3)/22.725 // in ohm disp(rof," R''_of(in ohm) = R''o / D = R_L2 / D =")
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/884/CH15/EX15.6/Example15_6.sce
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Example15_6.sce
// Computation of pH of solutions for solutions of given concentrations clear; clc; printf("\t Example 15.6\n"); //for HCL solution ConcHCl=1*10^-3;//Concentration of HCl solution, M H=ConcHCl;//Concentration of [H+] ion after ionisation of HCl pH=-log10(H); printf("\t the pH of the HCl solution is : %4.2f \n",pH); //for Ba(OH)2 solution ConcBaOH2=0.02;//Concentration of Ba(OH)2 solution, M OH=ConcBaOH2*2;//Concentration of [OH-] ion after ionisation of Ba(OH)2 as two ions are generated per one molecule of Ba(OH)2 pOH=-log10(OH); pH2=14-pOH; printf("\t the pH of the Ba(OH)2 solution is : %4.2f \n",pH2); //End
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/278/CH22/EX22.10/ex_22_10.sce
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ex_22_10.sce
//find... clc //solution //given P=180*1000//W N=240//rpm ft=5.2*10^6//N/m^2 //N1-N2=0.03 rho=7220//kg/m^3 tf=40//N/mm^2 Tmean=(P*60)/(2*%pi*N)//N-m printf("mean torque acig is,%f N-m\n",Tmean) //ref fig 22.18 q=4*%pi Wdpc=Tmean*q Wp=1.33*Wdpc//work done in power stroke....eq1 //Wp1=(0.5*%pi)*Tmax...eq2 Tmax=Wp/(%pi/2)//N-m printf("max torque is,%f N-m\n",Tmax) //BG=BF-FG BG=Tmax-Tmean//N-m BF=Tmax dE=Wp*(BG/BF)^2//N-m printf("dE is,%f N-m\n",dE) //let D be mean dia //let v be peripheral velo v=sqrt(ft/rho)//m/s D=(v*60)/(N*%pi)//m R=D/2 printf("the dia of wheel is,%f m\n",D) //N1-N2=0.03*N Cs=0.03 w=2*%pi*N/60//rad/s //dE=E*2*Cs m=dE/(R^2*w^2*Cs) printf("mass of wheel is,%f kg\n",m) //let t be thickness and b be width of rim //b=2t //A=b*t=2*t^2 t=sqrt(m/96730)//mm printf("the thicknes and width is,%f m\n,%f m\n",t,2*t) //let d be dia of hub ,d1 be dia of shaft,l be length of hub //let Tmax1 be max torque on shaft Tmax1=2*Tmean*1000//N-mm //d1=(Tmax1*16/(%pi*tf))^(1/3) printf("dia od shaft is,%f mm\n",(Tmax1*16/(%pi*tf))^(1/3)) printf("the dia of shaft is say 125mm\n") d1=125//mm d=2*d1//mm l=2*t//mm printf("the dia of hub and length of hub is,%f mm\n,%f m\n",d,l) //let a1 be major and b1 be minor axis //a1=2*b1 n=6 fb=15//N/mm^2 M=Tmax1*(D*1000-d)/(D*n*1000)//N-mm printf("bending moment is,%f N-mm\n",M) //Z=(%pi/32)*b1*a1^2=0.05*a1^3 //fb=M/Z a1=(M/(fb*0.05))^(1/3)//mm b1=0.5*a1 printf("major and minor axis is,%f mm\n,%f mm\n",a1,b1) printf("corrsponding to shaft of dia 125 mm,width is 36 mm and thicknss ofkey is 20 mm\n") //let L be length of key L=Tmax1/(36*tf*d1/2)//mm printf("length of key is,%f mm\n",L)
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/3760/CH4/EX4.45/Ex4_45.sce
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Ex4_45.sce
clc; clc; v=200; // rated voltage of dc shunt motor ra=0.1; // armature resistance n=1000; // running speed of motor ia=50; // armature current at n=1000 rpm re=0.1; // reduction in field flux ia2=(1/(1-re))*ia; // armature current when transients are over Ea1=v-ia2*ra; // counter EMF when transients are over // with sudden increase from 0.9*f to f (f=flux), counter EMF rises to Ea2=Ea1*(1/(1-re)); i=(v-Ea2)/ra; printf('Armature current is %f A',i); disp('Since armature current is negative, machine acts as a generator. Speed reduces till counter EMF becomes less than supply voltage,so that motor action takes place and torque balance is obtained')