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//example 4.36 //calculate //risk of failure of cofferdam //return period clc;funcprot(0); //given T=30; //deign for period n=6; //period of construction R=(1-(1-(1/T))^n)*100; R1=0.1; //reduced risk T1=1/(1-(1-R1)^(1/6)); R=round(R*10)/10; T1=round(T1*100)/100; mprint...
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// Fehlermeldung bei neudefinition vermeiden funcprot(0); function globalPlot(x,y,style,doch) // actual inputs arguments (rhs) and output arguments (lhs) [lhs,rhs]=argn(0) ; // does the variable PROCESS_PLOTS _NOT_ exist? if exists("PROCESS_PLOTS") ~= 1 then error(" ### ERROR ### S...
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function xyp=f1(t,Z); // M is mass of the Earth // SUN is the mass of the Sun // G is the gravitational constant M = 5.9737 * 10^24; SUN = 1.989 * 10^30; G = 6.67259 * 10^(-11); //here's the problem setup just like the previous examples shown x(1) = Z(1); x(2) = Z(2); y(1) = Z(3)...
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clc clear //Input data P=100//Power in MW T=550//temperature in degree C p=0.1//Pressure in bar m=500000//Mass flow rate in kg/h at rated load mo=25000//Mass flow rate in kg/h at zero load x=[1/4,1/2,3/4,1]//Fraction of load //Calculations b=(m-mo)/(P*10^3)//Steam rate in kg/kWh y1=(x(1)*(P*10^3))//For on...
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** File Info Version: 1.0 (encrypted) Num Logs = 0 Num Trans = 0 Num Writers = 0 Init Tranlog = 0 Total Entries = 11 Tranlog Offset = 0 Transaction Id = 10 Index Free List = 8 Total Size of Data = 13638 Data Transformation Id = 8 Index Transformation Id = 43 ** Entry Info for: 0-5 num: 0000000000000000 pos: 0...
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// Chapter 6_The pn junction //Caption_Space charge width //Ex_2//page 224 T=300 Na=10^16 //acceptor ion concentration Nd=10^15 //donor ion concentration eps=11.7*8.85*(10^-14) e=1.6*(10^-19) Vbi=0.635 //built in potential barrier W=(2*eps*Vbi/e*(Na+Nd)/(Na*Nd))^0.5 Emax=-e*Nd*W/eps printf('The s...
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# ATWM1 MEG Experiment scenario = "ATWM1_Working_Memory_MEG_salient_cued_run2"; #scenario_type = fMRI; # Fuer Scanner #scenario_type = fMRI_emulation; # Zum Testen scenario_type = trials; # for MEG #scan_period = 2000; # TR #pulses_per_scan = 1; #pulse_code = 1; pulse_width=6; default_monitor...
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clear; clc; // A Textbook on HEAT TRANSFER by S P SUKHATME // Chapter 2 // Heat Conduction in Solids // Example 2.2 // Page 31 printf("Example 2.2, Page 31 \n\n") d_i=0.02; // [m] inner radius d_o=0.04; // [m] outer radius r_i=d_i/2; // [m] inner radius r_o=d_o/2; // [m] outer radius k=0.58; // [w...
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// Example 1.52 clear; clc; close; format('v',5); // Given data P=6;//no. of poles f=50;//in Hz Sf=3;//in % R2=0.2;//in ohm per phase //Calculations Sf=Sf/100;//Slip Ns=120*f/P;//in rpm N1=Ns*(1-Sf);//in rpm N2=N1*90/100;//in rpm S2=(Ns-N2)/Ns;//new slip //Formula : T=K*S*E2^2*R2/R2^2;//S*X2 is ne...
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أ ب ا ا ل ح س ن ' a b a a l h a s a n أ ب ا ا ل ح س ن ' a b a l h a s a n أ ب ا ا ل ح س ن ' a b a l h a s s a n أ ب ا ا ل ح س ن ' a b e l h a s a n أ ب ا ا ل ح س ن a b a - a l - h a s a n أ ب ا ا ل ح س ن a b a a l h a s a n أ ب ا ا ل ح س ن a b a l h a s a n أ ب ا ا ل ح س ن a b a l h a s s a n أ ب ا ا ...
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//chapter 3 Ex 8 clc; clear; close; //let value to be found is x, y and z x=2.61*1.3; y=2.1693*1.4; z=0.4*.04*.004*40; mprintf("(i) x=%.3f",x); mprintf("\n(ii) y=%.5f",y); mprintf("\n(iii)z=%.5f",z)
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//Part A Chapter 7 Example 2 clc; clear; close; h2=2682.5;//kJ/kg(For 0.05 MPa & 100 degree C) h1=h2;//kJ/kg(for throttling) hf=1407.56;//kJ/kg(For 10 MPa) hfg=1317.1;//kJ/kg(For 10 MPa) x1=(h1-hf)/hfg;//dryness fraction disp("Dryness fraction = "+string(x1));
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clear; clc; //Example 1.2(Heating of water in an Electric Teapot) //Mass of liquid water m1=1.2,m2=0.5;//[Kg] //Initial Temperature t1=15;//[Degree Celcius] //Final Temperature t2=95;//[Degree Celcius] //Specific heat of water cp1=4.186;//[kJ/kG.K] //Specific heat capacity of teapot cp2=.7;//[] Em=(m1*cp...
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clear; clc; // Example: 11.2 // Page: 459 printf("Example: 11.2 - Page: 459\n\n"); // Solution //*****Data******// Tf = 5 + 273;// [K] Lf = 9830;// [J/mol] R = 8.314;// [J/mol K] M1 = 78;// [kg/kmol] //**************// Kf = R*Tf^2*M1/(1000*Lf);// [kg/kmol] printf("Molal Freezing point is %.2f kg/...
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clc; funcprot(0); //Example 25.8 //Initializing the variables f = 0.008; T = 290; L = 750; Dt = 9; // Diameter Tunnel Df = 0.63; // Diameter fan K1 = 0.98; K2 = 0.92; Vo = 27.9; n = 10; //Calculations alpha = (Df/Dt)^2; // equation 25.20 becomes when E = 1 nad C = 0 W = poly(0,'W' ); ...
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//Example No. 3_08 //-ve Integer to binary //Pg No. 50 clear ; close ; clc ; negint = -13 posbin = dec2bin(abs(negint)) posbin = strcat(['0',posbin]) compl_1 = strsubst(posbin,'0','d') compl_1 = strsubst(compl_1,'1','0') compl_1 = strsubst(compl_1,'d','1') compl_2 = dec2bin(bin2dec(compl_1) + 1) disp(com...
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//(Welded and Riveted Joints) Example 8.26 //Refer Fig.8.69 on page 323 //Number of rivets n n = 4 //Eccentric force P (kN) P = 5 //Eccentricity e (mm) e = 200 //Permissible shear stress tau (N/mm2) tau = 60 //Radial distance of rivets from the C.G. r (mm) r = 100
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clc //Initialization of variables p1=600 //psia p2=0.2563 //psia t1=486.21 //F t2=60 //F //calculations disp("from steam tables,") h1=1203.2 hf1=471.6 hfg1=731.6 h2=1088 hf2=28.06 hfg2=1059.9 s1=1.4454 sf1=0.6720 sfg1=0.7734 s2=2.0948 sf2=0.0555 sfg2=2.0393 xd=(s1-sf2)/sfg2 hd=hf2+xd*hfg2 xa=0.302...
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clc clear //Input data dw=1000;//Density of water in kg/m^3 dh=13590;//Density of Hg in kg/m^3 Pa=400;//Pressure at A in kPa g=9.81;//Gravity in N/m^2 Zw1=2.5;//First level of water in m Zw2=0.4;//Second level of water in m Zh=0.6;//Level of Hg in m //Calculations Pw1=dw*g*Zw1;//First level of water press...
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function I = trap(a,b,fun,n) // a - limite inferior da integração // b - limete superior da integração // n - número de segmentos // fun- função literal caso exista o argumento n // - vetor com os pontos caso não seja passado n if argn(2)< 4 then fx = fun; n = length(...
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29 11 ~~~~~~~~~~~~~~~~~~~~~~~~~~ 2
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//i/p arg filter order is negative a=0.6; f=intfilt(1,-2,a); disp(f); //output //!--error 4 //Undefined variable: y //at line 13 of function sinf called by : //at line 164 of function intfilt called by : //f=intfilt(1,-2,a);
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//Book - Power system: Analysisi & Design 5th Edition //Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J.Overbye //Chapter-9 ;Example 9.4 //Scilab Version - 6.0.0; OS - Windows clc; clear; Sb=100 //Base value of system in MVA Vb=13.8 //Base voltage...
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//Example 1.3 stories=39;//Number of stories in the building height_man=2;//Approximate height of an adult man (m) height_storey=2*height_man;//Approximate height of a single storey (m) height_building=2*height_man*stories;//Approximate height of building assuming height of 1 storey=2*height_man (m) printf('Approx...
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clc //initialisation of variables w= 62.4 //lb/ft^3 f= 0.005 Q= 100 //cuses m= 40 //Rs n= 0.75 n1= 0.065 K= 15 //Rs //CALCULATIONS d= ((5*w/(1.5*550*10))*n*f*Q^3*m/(K*n1))^(1/6.5) //RESULTS printf ('economical diameter of pipe line = %.3f ft ',d)
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function block=tanh_func(block,flag) if flag==1 block.outptr(1)=block.x(1) elseif flag==0 block.xd(1)=tanh((block.inptr(1)(1)-block.x(1)))/block.rpar; end endfunction
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//consumer voltage and energy loss //Example 15.1(pg 392) clc clear R=0.2//total resistance of cable in ohms I=200//current in A t=100//time in hours V=240//voltage in volts c=0.8//cost of electrical energy in Rs per unit V1=I*R//voltage drop in the cable //(i)consumer voltage Vc=V-V1 //(ii)Power loss in th...
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// Problem no 2.6,Page no.34 clc;clear; close; b_1=10 //cm //Breadth of the triangle h=9 //cm //Height of triangle b_2=2 //cm //width of rectangle d=3 //cm //Depth of rectangle //Triangle ABC-1 a_1=45 //cm**2 //Area of triangle y_1=3 //cm //C.G of triangle //Rectanglar hole-2 a_2=6 //cm**2 //Area of rectangle y_2=4...
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//Title:-Circular Convolution || Sonu Sharma,Roll No-B630 clc x = input("Enter sequence x(n)=") h = input("Enter sequence h(n)=") N = length(x) n = 0:1:N-1 k = 0:1:N-1 w = exp(-1*%i*2*%pi/N) z = n'*n TF = w.^z X = x*TF H = h*TF disp(X,"X(k) = ") disp(H,"H(k) = ") Y = X.*H disp(Y,"Y(k) = ") TFI = (...
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// Scilab code Ex4.9: Pg 121-122 (2008) clc; clear; // Part (a) I = 0.2; // Electric current, A l = 5e-02; // Effective length, m A = 7e-04; // Cross-sectional area, metre-square d = 0.5e-03; // Diametre, m mew_r = 1; //Relativ...
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//************************** INF Neuron ******************************** if (blk_name.entries(bl) =='infneuron') then //FIX For Multiple blocks mputl("# INF Neuron",fd_w); mputl(".subckt INFneuron in[0]=net"+string(blk(blk_objs(bl),2))+"_1"+ " in[1]=net"+string(blk(blk_objs(bl),3))+"_1"+ " in[2]=net"+string(b...
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clear; clc; l=300 R=.4 *3 X=.8*3 Vs=11e3/sqrt(3); P=3000; pf=.8 pfa=acos(pf) VIr=P/(3*pf) a=1; b=-Vs c=VIr * 1e3 * ((R*cos(pfa))+(X*sin(pfa))) vr=(-b+sqrt((b*b )- (4*a*c)))/(2*a) Ir=VIr*1e3/vr; Vr=vr*sqrt(3)/1000; mprintf("\nReceiving End Voltage = %.2f kV",Vr) Pl=3* (Ir)^2 * R/ 1000; eff=P*10...
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// Theory and Problems of Thermodynamics // Chapter 8 // Power and Refrigeration Cycles // Example 10 clear ;clc; //Given data r0 = 8 // Compression ratio of Otto cycle T1 = 300 // initial temperature of air in K P1 = 0.1 // initial pressure of air in MPa q1 = 25 ...
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//Chapter 20, Problem 15 clc; P1=8; //power 1 in watt P2=4; //power 2 in watt P=P1+P2; //total input power phi=atan(sqrt(3)*((P1-P2)/(P1+P2))); pf=cos(phi); //load power factor printf("(a) Total input power = %d kW\n\n",P); printf("(b) Power fact...
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//============================================================= //Chapter 5 example 29 clc; clear all; //variable declaration V1 = 1000; //potential of vane in volts //calculations //v = VA-VB mprintf("theta 10 S D"); mprintf("\ntheta praportional to Tt praportional to 2*V1*V") mprint...
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load main.asm, output-file main.out, compare-to main.cmp, output-list RAM[0]%D1.6.1 RAM[1]%D1.6.1 RAM[2]%D1.6.1 RAM[3]%D1.6.1 RAM[4]%D1.6.1 RAM[5]%D1.6.1; set RAM[0] 256, // initializes the stack pointer set RAM[1] 300, // initializes the stack pointer set RAM[2] 400, // initializes the stack point...
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clear; clc; // Example: 5.12 // Page: 167 printf("Example: 5.12 - Page: 167\n\n"); // Solution //*****Data*****// Vol_O2 = 5.6;// [L] Vol_H2 = 16.8;// [L] R = 1.987;// [cal/mol K] //*************// xA = Vol_O2/22.4;// [mole fraction O2] xB = Vol_H2/22.4;// [mle fraaction H2] N = xA + xB;// [total...
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//CHAPTER 8- DIRECT CURRENT MACHINES //Example 25 disp("CHAPTER 8"); disp("EXAMPLE 25"); //VARIABLE INITIALIZATION slot=24; //number of slots P=2; //number of poles N=18; //number of turns per coil B=1; //in Webers l=20/10...
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clc clear //Initialization of variables Gf=11.57 //lb per lb of fuel H=4.4/100 M=13.5/100 mr=700 mf=10000 mc=1 //lb //calculations pro=M+9*H mrf=mr/mf Aa=Gf+pro+mrf-mc At=8.83 ea=(Aa-At)/At *100 //results printf("Excess air = %.1f percent",ea)
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//Chapter-5, Illustration 11, Page 261 //Title: Air Compressors //============================================================================= clc clear //INPUT DATA V1=7*(10^-3);//Volume of air in (m^3)/s P1=1.013;//Pressure of air in bar T1=288;//Air temperature in K P2=14;//Pressure at point 2 in bar n=...
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//calculation of rate constant from time and pressure data clear; clc; printf("\t Example 13.5\n"); t=[0,100,150,200,250,300];//time(data given), s P=[284,220,193,170,150,132];//pressure(data given) in mmHg corresponding to time values lnP=log(P);//lnP values corresponding to P [slope]=reglin(t,lnP);//ln...
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Ex2_7.sce
//Chapter 2, Exmaple 7, page 69 //Electric feild induced at x clc clear e0 = 8.85418*10**-12 //Epselon nought q = 1 // C/m C = (q/(2*%pi*e0)) //Based on figure 2.33 E = C-(C*(1/3+1/7))+(C*(1+1/5+1/9))+(C*(1/5+1/9))-(C*(1/3+1/7)) printf("Electric Feild = %e V/m \n",E) //Answers might vary due to round off error
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Name=Vertical Switching Sparky PlayerCharacters=VS Challenger BotCharacters=VS Target.bot IsChallenge=true Timelimit=60.0 PlayerProfile=VS Challenger AddedBots=VS Target.bot;VS Target.bot;VS Target.bot;VS Target.bot;VS Target.bot;VS Target.bot;VS Target.bot;VS Target.bot PlayerMaxLives=0 BotMaxLives=0;0;0;0;0;0;0;0 Pla...
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Example_2.sce
//Chapter-5,Example 2,Page 122 clc(); close(); t=[7.18 18 27.05] //time in minute r=[ 21.4 17.7 15] //rotation in degrees r_0=24.09 r_a=-10.74 k=(1 ./t).*log10((r_0-r_a)./(r-r_a)) printf('values of k') disp(k) printf('since k values are fairly constant by putting in 1nd order rate equat...
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////////////////////////////////////////////////////// // Неделя 7. // Комплексная обработка двух измерителей с использованием байесовского подхода. // Исследование точности с помощью матрицы ковариаций. // Нерекуррентный метод. ////////////////////////////////////////////////////// clear; deff('[numd] = roundd(num,n)'...
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// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART II : TRANSMISSION AND DISTRIBUTION // CHAPTER 2: CONSTANTS OF OVERHEAD TRANSMISSION LINES // EXAMPLE : 2.9 : // Page number 108-109 clear ; clc ; close ; // Clear the...
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clc; //e.g 26.5 fo=2*10**6; BW=50*10**3; Qo=fo/BW; disp(Qo);
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clc disp("Example 14.10") printf("\n") s=%s; //Applying KVL equation to the two loops we get //V1=3*I1+3*(I1+I2) //V2=7*I1+3*(I1+I2)+2*I2 //On solving we get disp("6*I1+3*I2=V1 (1)"); disp("10*I1+5*I2=V2 (2)"); //The equations which contain Z parameters are //V1=Z11*I1+Z12*I2 //V2=Z21*I...
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8.sce
clc N_s=0.14; // m^(3/4)s^(-3/2) H=30; // m p_v=7.38*10^3; // N/m^2 p_l=50*10^3; // N/m^2 rho=992; // kg/m^3 g=9.81; // m/s^2 H_L=0.2; // m NPSH=2.8*N_s^(4/3)*H; z1=NPSH+(p_v-p_l)/rho/g+H_L; disp("The minimum level of the alarm =") disp(z1) disp("m")
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Ex1_3.sce
clear; clc; disp('Example 1.3'); // Given values P1 = 2070; // initial pressure, [kN/m^2] V1 = .014; // initial volume, [m^3] P2 = 207; // final pressure, [kN/m^2] n=1.35; // polytropic index // solution // since gas is following PV^n=constant // hence V2 = V1*(P1/P2)^(1/n); // final volume, [m^3...
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Generation Of QPSK signal: clc;clear; //Generating QPSK signal fc = input('Enter the frequency'); //2 n = input('Input Message sequence'); //[1 0 0 1 1 1 0 1] for i=1:length(n) if n(i) == 1 then pol(1,i) = 1; else pol(1,i) = -1; end end // Multiplexing // NRZ polar line coded signal gen...
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// Example 8_13 clc;funcprot(0); // Given data T=20.0+273.15;// K mu=0.700;// N.s/m^2 L=0.100;// m R_1=0.0500;// m R_2=0.0510;// m n=1000;// rev/min // Solution omega=(2*%pi*n)/60;// rad/s S_P_W=((2*%pi*L*omega^2*R_1^4*mu)/((R_2^2-R_1^2)^2*T))*((2*R_2^2*(log(R_2/R_1)))+((R_2^4)/(2*R_1^2))-(R_1^2/2));// W/K...
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// Theory and Problems of Thermodynamics // Chapter 12 // Statistical Thermodynamics // Example 4 clear ;clc; //Given data U = 10 // Total energy units N = 4 // number of non-degenerate energy states n = 1:4 // energy levels g = [1 2 2 1]; // dege...
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成功例.sce
// constant matrix // k1=1 k2=2 d1=1 d2=2 AF = [0 0 1 0;0 0 0 1;-k1 k1 -d1 d1;k1 -(k1+k2) d1 -(d1+d2)]; BF = [0;0;1;1]; S = [4 3 0 1]; // Discretized system // h = 0.02; // sampling time // cont = syslin('c',AF,BF,S); disc = dscr(cont,h); // Discretized system // [A,B,Sd] = abcd(disc); // initail conditions // X=[0....
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//FK - Based on: function [xsol,z] = AlgebricSimplexe(A,b,c) [m,n] = size(A); Ac = [A,eye(m,m)]; B = [n+1:n+m]; N = [1:m]; z = 0; l = 0; while(1) l = l + 1; disp ('Iteration' + string(l) + ':..'); disp (B, "Les indices de la variable de base"); disp (N, "Les indices des variables hors base"); xbase = SolutionDe...
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2_3.sce
clc //initialisation of variables h= 76 //cmhg g= 32.2 //ft/s^2 h= 76.0 //cmHg Dhg= 847 //lbm/ft^3 //CALCULATIONS Pa= Dhg*g*h*0.33 Pa1= Pa/1 //RESULTS printf ('Pa= %.f poundal/ft^2',Pa1)
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clc clear //Declaring Values m=600; //Mass in kg z=50000; //Distance in meters V=2500000; //Velocity in m/hr g=7.9; //Gravitational Field in m/s^2 Vel=V/3600; KE=(0.5*m*Vel*Vel)/1000000; //Kinetic Energy in MJ PE=(m*g*z)/1000000; //Potential Energy in MJ //Displaying ...
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clc(); clear; //Given: deltalambda1 = 0.01; // The line width of the orange line of krypton,Kr^86 in A lambda = 6058; // Wavelength in angstroms = 6058*10^-10 m deltalambda2 = 0.00015; // The line width of a laser source in A c = 3*10^8 ;// Velocity of light in vacuum in m/s nu0 = c/(lambda*10^-10);// lambda ...
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test3-4.tst
main array[3][4] a; { let a[0][0] <- 1; let a[0][1] <- 2; let a[0][2] <- 3; let a[0][3] <- 4; let a[1][0] <- 2; let a[1][1] <- a[0][0] + a[0][1] + a[1][0]; let a[1][2] <- a[0][1] + a[0][2] + a[1][1]; let a[1][3] <- a[0][2] + a[0][3] + a[1][2]; let a[2][0] <- 3; let a[2][1] <- a[1][0] + a[1][1] + a...
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chan3D_part_only.tst
<?xml version="1.0" encoding="utf-8"?> <test> <description> Process part-only option </description> <executable>FieldConvert</executable> <parameters> -f --part-only 2 chan3D.xml </parameters> <files> <file description="Session File">chan3D.xml</file> <file description="Session File">chan3D.fld...
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clc clear disp('example 15.10') tc=2100 //total capacity of plant n=60 //number of generaed p=35 //power of generated by each generator h=10 //head of water d=12 //duration of generation cee=2.1 //cost of electrical energy per kWh efft=0.85 //efficiency of turbine effg=0.9 //efficiency of g...
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clc;clear;close // independent vectors stored in A A=input("enter the maxtrix [INDEPENDENT VECTORS]") disp(A,'A='); [m,n]=size(A); for k=1:n V(:,k)=A(:,k); for j=1:k-1 R(j,k)=V(:,j)'*A(:,k); V(:,k)=V(:,k)-R(j,k)*V(:,j); end R(k,k)=norm(V(:,k)); V(:,k)=V(:,k)/R(k,k); end ...
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%Generated from '../examples/shi/inverse.dig'. query(instances(aconcept('Okos')), [o1]). query(instances(aconcept('Boldog')), []). implies(some(inv(arole(gyereke)), aconcept('Okos')), aconcept('Boldog')). subrole(arole(gyereke), arole(utodja)). subrole(arole(hasParent), arole(szuloje)). subrole(arole(szul...
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function [y] = Leaky_ReLU(alpha,x) for i=1:length(x) if x(i) < 0 then y(i) = alpha*x(i) else y(i) = x(i) end; end; endfunction
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clc disp("Example 2.20") printf("\n") disp("Calculate peak,RMS,DC load current, DC in each diode,DC output voltage,% regulation,PIV,RMS current,DC load voltage") printf("Given\n") Rf=500 RL=2000 V2=280 //Secondary voltage is Vm=sqrt(2)*V2 //Peak load current Im=Vm/(Rf+RL) //DC load current Idc=2*Im/(%pi) ...
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s = poly(0, 's'); t = 0:0.01:5; Wn1 = 3;
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//These "parameters" will be set by calling readinput and passed to // the main H3 code: // nx,ny,nz : grid sizes of the 3d cube. // dx,dy,dz : grid spacings // iterations: the number of iterations to evolve // par1,par2 : Parameters for the initial data. // //[iterations,nx,ny,nz,d...
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//To find load, thrust, reaction and speed clc //Given: N=1800 //rpm r=50/1000, l=200/1000, D=80/1000, x=10/1000 //m mR=1 //kg p=0.7 //N/mm^2 //Solution: //Calculating the angular speed of the crank omega=2*%pi*N/60 //rad/s //Net load on the gudgeon pin: //Calculating the load on the piston FL=round(%pi/4*(...
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clear; clc; exec( get_absolute_file_path('main.sce') + "support.sce", -1 ); exec( get_absolute_file_path('main.sce') + "variables.sce", -1 ); disp( "Determening the root isolation interval..." ); [ a, b ] = askBorders( functionToInspect ); disp( "Starting solving with bisection method..." ); x = BisectionSolve( fun...
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//To find line current and pf and powers of a balanced delta load clc; clear; Z=8+6*%i; // Load V=230; // Voltage supply iR=V/Z; theta= atand(imag(iR)/real(iR)); Il= iR*sqrt(3); // Line current Pa=sqrt(3)*V*abs(Il)*cosd(theta); // Active Power Pr=sqrt(3)*V*abs(Il)*sind(theta); // Reactive Power Pt...
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//solve for x, a*(x-2)=5*x-(a+b) clear; clc; close; //removing brackets. "a*x-5*x=a-b" x=string('(a-b)/(a-5)')
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clc; clear; //Example 6.6 mf_dot=5000 //[kg/h] ic=0.01 //Initial concentration [kg/h] fc=0.02 //Final concentration [kg/h] T=373 //Boiling pt of saturation in [K] Ts=383 //Saturation temperature of steam in [K] mdash_dot=ic*mf_dot/fc //[kg/h] mv...
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function x=Cramer(A,b) //Saída- x(N) com solução para A(N,N)*x(N)=b(N). [N N]=size(A); C=[A b]; D=det(A) printf("Matriz Aumentada [C=A|b]\n") printf("det(A)=%f",D) disp(C) for(k=1:N) Ak=A; Ak(:,k)=b(:,1) //substitui coluna k por vetor b Dk=det(Ak) ...
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clc; pathname=get_absolute_file_path('5_5_soln.sce') filename=pathname+filesep()+'5_5_data.sci' exec(filename) // Solutions: // Hydraulic Power lost in Fixed Displacemnt pump, HP_f=(p*Q_f)/1714; //HP // Hydraulic Power lost in Pressure Compensated pump, HP_p=(P*Q_p)/1714; //HP // Therefore, Hydraulic Power saved, HP=H...
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clc //initialisation of variables mc=0.1//kg vl1=150//cc vl2=150//cc hl1=600 gl1=1200 hl2=400 gl2=900 t1=50//c t2=40//c sc=100 r1=2 //CALCULATIIONS m1=vl1*gl1/(10^6) rc1=(m1*hl1+mc*sc)*r1 k= -rc1/t1 m2=vl2*gl2/(10^6) b=(m2*hl2+mc*sc) j=-k*t2 //results printf(' rate of cooling= % 1f cal/min',j)
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//Problem 8.06: //initializing the variables: Tc = 647; // in K Tn = 100 + 273; // in K Pc = 217.6; // in atm dHe = 970; // in Btu/lb //calculation: Tm = Tn/Tc dH = 2.17*Tn*((log(Pc)) - 1.0)/(0.930 - Tm) dHn = dH*454/(252*18) perdiff = (dHn - dHe)*100/dHe printf("\n\nResult\n\n") printf("\n the enthal...
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// Theory and Problems of Thermodynamics // Chapter 8 // Power and Refrigeration Cycles // Example 12 clear ;clc; //Given data T1 = 300 // minimum temperature of air in K P1 = 0.1 // minimum pressure of air in MPa T3 = 2200 // maximum temperature of air in K P3 = 5.0...
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clc //initialisation s=8.2*10^4 si=5.67*10^-8//j m^-2 sec^-1 deg ^-4 a=32 //CALCULATIONS r2=a/2 r1=(r2*3.14)/(60*180) r=r1^2 t=s/(r*60*si) T=t^0.25 //results printf(' \n surface temperature of sun= % 1f k',T)
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clc; //page 163 //problem 3.4 //Transmission power effiency n = {(m^2)/[2+(m^2)]}*100% where m is modulated index //Given modulated indices are m1 = 0.25, m2 = 0.5 & m3 = 0.75 //Transmission power effiencies are n1, n2 & n3 respectively for m1, m2 & m3 n1 = {(0.25^2)/[2+(0.25^2)]}*100 n2 = {(0.5^2)/[2+(0.5...
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argc:7 Dataset: ../datasets/converted/4adjnoun.net Nodes Edges Com Mod NMI Time seq semisync 112 850 1 0.0162048 -1 0.000330236 par semisync 112 850 1 0.016205 -1 0.096944
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// Converts the points from homogeneous to Euclidean space // // Calling Sequence // pts_euclidean = convertPointsToHomogeneous(pts_homogenous); // // Parameters // Input // pts_homogenous : vector of N points in homogenous space // Output // pts_euclidean : vector of N-1 points in Euclidean space // // Description ...
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PL/SQL Developer Test script 3.0 7 begin -- Call the procedure cux_plsql_autocreate.form_view_iud(p_block_name => :p_block_name, p_table_name => :p_table_name, p_owner => :p_owner, p_primary_key => :p_prim...
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//Example 8.35 clc disp("A binary counter with a modulus greater than 16 can be built by cascading 74X163s. When counters are cascaded, CLK, CLR'' and LD'' of all the 74X163s are connected in parallel, so that all of them count or are cleared or loaded at the same time. The RCO signal drives the ENT input of the next...
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//Page Number: 332 //Example 6.3 clc; //Given c=3D+8; //m/s d=2D-3; //m p=50D+2; //turns per m e=1.6D-19; //J m=9.1D-31; // Axial phase velocity vp=c/(%pi*p*d); disp('m/s',vp,'Axial phase velocity:'); //Anode voltage V0=(m*vp*vp)/(2*e); disp('V',V0,'Anode voltage:');
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//Chapter 4 Ex 8 clc; clear; close; //(i) expr1=12.05*5.4/0.6; mprintf("The value of expression is %.2f",expr1); //(ii) expr2=(0.6*0.6)+(0.6/6); mprintf("\n The value of expression is %.2f",expr2);
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// This file is part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. // File name: projects/05/CPU-external.tst load CPU.hdl, output-file CPU-external.out, compare-to CPU-external.cmp, output-list time%S0.4.0 inM%D0.6.0 instruction%B0.16.0 reset%B2.1.2 ou...
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//calculating Ecell and energy //Example 6.11 clc clear Ec=0.4 Ea=-0.87 Ecell=Ec-Ea F=96500 Wmax=(2*F*Ecell)/1000 printf('Thus (i)Ecell = %2.2f V',Ecell) printf('\n (ii)Wmax= %3.0f kJ',Wmax)
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 module App { $Enums(*Enum)[ export enum $Name { $Values[ $Name = $Value][,] }] }
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// Grob's Basic Electronics 11e // Chapter No. 33 // Example No. 33_20 clc; clear; // Vin is 5 V, R is 1 kOhms, and Rl is 100 Ohms. Calculate the output current, Iout. // Given data Vin = 5; // Input votage=5 Volts Ri = 1*10^3; // Input resistance=1 kOhms Rl = 100; // Load resistance=10...
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//Transport Processes and Seperation Process Principles //Chapter 3 //Example 3.1-3 //Principles of Momentum Transfer and Applications //given data basis=1;//taking basis as 1 m3 of packed bed rho=962;//bulk density of packed bed m=rho*basis;//total mass rho2=1600;//density of solid cylinders V=m/rho2;//volum...
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clc // machine A ic1 = 50000 // initial cost hoc1 = 10 // hourly operating charges pp1= 5 // pieces produced per hour i = 15 // interest rate i = i/100 oh = 2000 // operating hours fc1 = ic1*i // fixed cost vc1 = oh*hoc1 // variable cost tc1 = fc1+vc1 // total charges ao1 = oh*pp1 // annual output c1 = tc1/...
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clear;lines(0); A=1; isdef('A') clear A isdef('A')
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clc //initialisation of variables p1=120//lb/in^2 p2=15//lb/in^2 //CALCULATIONS v=1.65//lb D=sqrt(v)//lb //RESULTS printf('The above pressure are by gauge=% f lb',D)
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close ; clear all; clc; p=100; t=1:1:p*10; for n=1:(p/2) a(n)=n; end for n=(p/2)+1:p a(n)=n-p; end a1=a for i=1:9 a1=[a1 a]; end plot2d(t,a1); axis( [ 0 1000 -100 100]); xtitle('sawtooth sequence', 'amplitude’, 'time');
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//Chapter 6, Problem 9 clc; n=19; //No of interleaved plates n=n-1; A=(75*10^-3)*(75*10^-3); //Calculating area of plates er=5; e0=8.85*10^-12; d=0.2*10^-3; //Distance between plates C=(e0*er*A*n)/d; //Calculating capacitance of the capacitor print...
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chdir("C:\Users\matve\Desktop\Code These\") //TRAD SANS CHOC load(string(fichier)+'\Blim_sans_choc'+string(cas),'Blim'); load(string(fichier)+'\Pro_sans_choc'+string(cas),'Pro'); load(string(fichier)+'\Foodssect_sans_choc'+string(cas),'Foodssect'); load(string(fichier)+'\nb_boats_Post_BAU_t_sans_choc'+string(cas),'nb_b...
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x = linspace(0,3,100) y = sin(x) xlfont("reset") xlfont("Times New Roman",10) plot(x,y) xstring(0.5,0.5,"A Text from ffr") figure_entity = gcf(); axes_entity = figure_entity.children title_entity = axes_entity.children title_entity.font_style = 10
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//example 7 //Phase Equilibrium for a Saturated Mixture clear clc hf=503.81 // in kJ/kg, sf=1.5279 // in kJ/kg-K hg=2706.0 //in kJ/kg sg=7.1292 //in kJ/kg-K T=393.15 //Given temp. in K gf=hf-T*sf //in kJ/kg gg=hg-T*sg //in kJ/kg disp('Since, two results are in close agreement, Therefore mixture of saturated ...
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//find safe tensile load clc //soltuion //given d=30//mm ft=42//N/mm^2 //using table 11.1,area corresponding to d=30mm is A=561//N/mm^2 A=561//mm^2 F=A*ft//N printf("the value of force is,%f N",F)
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clc Hb = 200 // brinell hardness d = 12.7 // diameter in mm f = 0.254 // feed in mm/rev. N = 100 // rpm M = (Hb*(d)^2*f)/8 //moment in kgf-mm k = 1.1 //material factor p = (1.25*(d)^2*k*N*(0.056+1.5*f))/(10)^5 // power in kW T1a = (1.7*M)/d // thrust force kgf T1b = (3.5*M)/d // kgf T1 = (T1a+T1b)/2 // averag...