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ex1_10.sce
// Exa 1.10 clc; clear; close; // Given data m = 18.2;//quantity of air supplied of coal in kg T1 = 200;// in degree C T2 = 18;// in degree C del_T = T1-T2;// in degree C Spe_heat = 1;// in kJ/kg-K Q_C = m*Spe_heat*del_T;// in kJ disp(Q_C,"The Quantity of heat supplied per kg of coal in kJ is");
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Name=Break Cover V2 PlayerCharacters=Aimer BotCharacters=BOT1.bot;hayabot.bot;osobot.bot;jumot.bot IsChallenge=true Timelimit=30.0 PlayerProfile=Aimer AddedBots=BOT1.bot;BOT1.bot;BOT1.bot;BOT1.bot PlayerMaxLives=0 BotMaxLives=0;0;0;0 PlayerTeam=1 BotTeams=2;2;2;2 MapName=BC22.map MapScale=3.0 BlockProjectilePredictors=...
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clear; clc; close; R1 = 80*10^(3); R2 = 20*10^(3); Ro = 10*10^(3); Rd = 10*10^(3); gm = 4000*10^(-6); Rl = Ro*Rd/(Ro+Rd); A = -gm*Rl; Beta = -R2/(R1+R2); Af = A/(1+Beta*A); disp(A,'Gain without feedback = '); disp(Af,'Gain with feedback = ');
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function[ret, erro] = AproximacaoQuadNivel()//relacao linear da nota com o quadrado do nivel e com a taxa usando minimos quadrados //nota = a*nivel^2 + b*taxa + c T = Entrada()//matriz do tipo [nivel, taxa, nota] N = size(T) n = N(1,1) A = zeros(n,3) b = zeros(n,1) for i=1:1:n ...
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//clear// clear; clc; //Example 7.1 // Given rho_p = 2800; // [kg/m^3] g = 9.80665; //[m/s^2] ac = 50*g; // [m/s^2] //(a) //From appendix 20 Dp_100 = 0.147; //[mm] Dp_80 = 0.175; //[mm] Dp = (Dp_100+Dp_80)/2;//[mm] //From Appendix 14 mu = 0.801; //[cP] rho = 995.7; //[kg/m^2] // Using Eq.(7.45) ...
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## Mieke van Holstein, Roshan Cools, Hanneke den Ouden: probabilistic reversal learning task (PRL). ## Based on Cools et al. 2001, Swainson et al. 2000; Lawrence et al. 1999; den Ouden et al 2013. ## started: October 6th 2011 ## finished: November 26 2014 ## Prompt format entered text: "subject_12_session_2". ## chec...
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PL/SQL Developer Test script 3.0 22 -- Created on 27.05.10 by Kravchenko A.V. declare -- Local variables here i integer; m xmltype; s xmltype; p xmltype; begin --m := xmltype.createXML(:p_xml); s := xmltype.createXML(:p_xsl); select xmlelement("ROOT",xmlagg(xmlelement("ROW", XMLForest(o.id,o....
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//example1.15 clc disp("The various node voltage are shown in the fig 1.73(a).") disp("THe various braanch currents are shown.Applying KCL at various nodes.") disp("Node 1: 9-I1-I2-I3=0 ..(1)") disp("Node 1: I3-I4+4=0 ..(2)") disp("Node 1: I2-4-I5=0 ..(3)") disp("Key Point: Nodes V1 and V3 from supernod...
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// Theory and Problems of Thermodynamics // Chapter 5 // Second law of Thermodynamics // Example 4 clear ;clc; //Given data TL = -25 //TL = Lower temperature of reservoir TH = 40 //TH = High temperature of reservoir W_R = 0.5 //W_R = power consumption of carnot refrigerator // Units conve...
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edo - rk4.sce
clear; clc; //Problema exemplo: dada a equação diferencial dy/dx = -1.2y + 7exp(-0.3x) //com condições iniciais y(0) = 3 //no intervalo de a = 0 à b = 2.5 com um passo h = 0.5 //Definindo a função function dy = f(x, y) dy = -1.2*y + 7*exp(-0.3*x) endfunction //Implementação de RK4 function [x,y] = RK4(y0, a, b,...
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Example37_3.sce
//Given that d = 32*10^-3 //in meter f= 24*10^-2 //in meter lam = 550*10^-9 //in meter //Sample Problem 37-3a printf("**Sample Problem 37-3a**\n") theta = 1.22*lam/d printf("Angular sepration should be equal to %erad\n", theta) //Sample Problem 37-3b printf("\n**Sample Problem 37-3b**\n") deltaX = f*th...
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Ex5_10.sce
// // aa=1.155 ab=2.595 ba=0.985 bb=2.415 td=((ab-aa)+(bb-ba))/2 rla=525.5 rlb=rla-td dab=500 printf("\n true RL of B %0.3f meters',rlb) dab1=dab/1000 correct=0.0673*dab1*dab1 printf("\n combined corrction for 500m= %0.3f meters',correct) sc=100 a=1.155 e=-(0.0118*sc)/(dab) printf("\n collimation error per 10...
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-chain [[0,-2,1,-2],[-2,0,-2,1],[2,1,2,0],[-2,-1,-1,1]] [6,-3,-4,-5] 4 0 [[0,-2,1,-2],[-2,0,-2,1],[2,1,2,0],[-2,-1,-1,1]],det=-2 [6,-3,-4,-5], chain 8 => [12,-9,1,-10] => [39,-36,17,-26] => [141,-138,76,-85] => [522,-519,296,-305] => [1944,-1941,1117,-1126] => [7251,-7248,4181,-4190] => [27057,-27054,15616,-15625] => [...
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Ex13_7.sce
clear; clc; l = 12;//feet w = 150;// lb per sq.foot //Live load LL = w*l;//lb-wt //Dead Load assuming the slab thickness to be 6 inches t = 6;//inches DL = t*l*12;//lb-wt //total load W = LL+DL;//lb-wt M = W*l*12/10;//lb-inches d = sqrt(M/(12*126)); printf('d = %.3f inches',d); //With about an inch to co...
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clc f=50 n2=50 //Assigning values to parameters n1=500 kva=25 e1=3000 k=n2/n1 i1=kva*1000/e1 i2=i1/k e2=k*e1 fm=e1/(4.44*f*n1) disp("Amperes",i1,"The primary full load current is"); disp("Amperes",i2,"The secondary full load current is"); disp("Volts",e2,"The secondary emf is"); disp("Wb",fm,"The maximum ...
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//Example 12_4 clc(); clear; //To describe the Temperature changes of the gas printf("This type of process is termed as throttling process and described by the equation Delta U=- Delta W\n") printf("Where Delta W is the work done")
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errcatch(-1,"stop");mode(2);//Example 3.8: Unknown resistance ; ; //given data : P=100;// in ohm Q=10;// in ohm S=46;// in ohm R=((P/Q)*S); disp(R," Unknown resistance,R(ohm) = ") exit();
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1_19_d.sce
//example 1.19<d> //determine whether the following signals are power or energy signal clc ; t=0:0.01:100; x=1; P=(integrate('1^2','t',0,1))/2; disp(P,'The power of the signal is:'); E=(integrate('1^2','t',0,1)); disp(E,'The energy is:'); disp('As t tends to infinity energy also tends to infinity but power rem...
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clear // //Given //variable declaration P=4000 //Load in N sigma=95 //Stress in N/sq.mm //Calculation D=(sqrt(P/((%pi/4)*(sigma)))) //Diameter of steel wire in mm //Result printf("\n Diameter of a steel wire = %0.3f mm",D)
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##################################################################################### # SCE File for Recognition Task # # associated files: Recognition_PCL, subroutines_PCL, wordlist_recognition.txt) # ##################################################################################### ###########################...
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Example9_28.sce
//Chapter-9,Example9_28,pg 9_82 Po=20*735.5//(in W) V=230 N=1150 P=4 A=P Z=882 Ia=73 Ish=1.6 T=60*Po/(2*%pi*N) phi=T*A/(0.159*Ia*P*Z)//flux per pole Il=Ia+Ish Pin=V*Il n=Po*100/Pin printf("electromagnetic torque\n") printf("T=%.3f Nm\n",T) printf("flux per pole\n") printf("phi=%.3f Wb\n",phi) printf("...
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//Example 2.13 x_0=0;//Position at the start of the ramp (m) x=200;//Position at the end of the ramp (m) v_0=10;//Initial velocity (m/s) a=2;//Acceleration (m/s^2) //Use the equation x=x_0+v_0*t+(1/2)*a*t^2 and rearrange to form a quadratic equation with t as the variable //(1/2)*a*t^2+v_0*t+(x_0-x)=0 p=[((1/2)*...
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dsimul.sci
function [y]=dsimul(sld,u) [a,b,c,d,x0]=sld(2:6); y=c*ltitr(a,b,u,x0)+d*u;
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13_17_3.sce
clc; //page no 515 //prob no. 13.17.3 Z0=600;Zl=73;//in ohm F=0.9; QF=(2*%pi*F)/4; //For matching, the effective load impedance on the main line must equal the characteristic impedance of the mail line Zl1=Zl; Z01=sqrt(Zl1*Zl); Tl=(Zl-Z01)/(Zl+Z01); VI=1;//reference voltage Vi=VI*%e^(%i*QF); Vr=Tl*VI*%e^-(%...
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GEARS_1.sce
// sum 25-1 clc; clear; Zp=25; Zg=60; m=5; dp=m*Zp; dg=m*Zg; CD=(dp+dg)/2; ha=m; hf=1.25*m; c=hf-ha; r=0.4*m; // printing data in scilab o/p window printf("dp is %0.0f mm ",dp); printf("\n dg is %0.0f mm ",dg); printf("\n CD is %0.1f mm ",CD); printf("\n ha is %0.0f mm ",ha); print...
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//Exa 2.17 clc; clear; close; //Given data : format('v',8) d=10;//in um a=d/2;//in um lambda_c=1.3;//in um n1=1.55;//unitless //Part (a) //for single mode transmission cut-off wavelength is lambda_c=2*%pi*a*n1*sqrt(2*delta)/2.405 delta=(lambda_c*2.405/(2*%pi*a*n1))^2/2;//unitless disp(delta,"Normalized ref...
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// PG (258) deff('[y]=f(x)','y=exp(x)*cos(x)') x0=0; xn=%pi; x=x0:xn; // True value I = integrate('exp(x)*cos(x)','x',x0,xn) // Using Simpson's rule N=2; h=(xn-x0)/N; x1=x0+h; x2=x0+2*h; I1 = h*(f(x0)+4*f(x1)+f(x2))/3 N=4; h=(xn-x0)/N; x1=x0+h; x2=x0+2*h; x3=x0+3*h; x4=x0+4*h; I2 = h*(f(x0)+4*f(x...
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11_2.sce
clc //initialisation of variables g=32.2//ft/sec^2 n=(0.01*30.5)/(453.6*32.2) H=2//ft D=1/12 //CALCULATIONS p=62.4/g v=n/p k=v/(D*sqrt(2*g*H)) //RESULTS printf (' value of non dimensional constant= %.2e',k)
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ieee-divs.tst
*Testcase IEEE LOAD ROUNDED (3), LOAD FP INTEGER (3) DIVIDE (5), DIVIDE TO INTEGER (2) 13 instr total # Divide adjacent pairs of values in the input set (five values means four # quotients). Test data: 1, 2, 4, -2, -2; expected quotients 0.5, 0.5, -2, 1 # Load Floating Point Integer of the above result set. Exp...
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testRLG.sce
clear;clc;close();getd(".");getd("./kNN");getd("./RLG");getd("./RLG/Geometry");getd("./Pruning_strategy");getd("./Local_Planner"); f1=struct('leg','HL','pos',[-2.3750e-1,-4.3146e-1,+1.9095e-2]); f2=struct('leg','FL','pos',[-2.3750e-1,+4.3146e-1,+1.9095e-2]); f3=struct('leg','HR','pos',[+2.3750e-1,-4.3146e-1,+1.9095e-2...
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14_8.sce
clc //initialisation of variables M= 250 //KNm Torquemax= 200 //KNm allowablestress= 180 //N/mm^2 L=250 //mm B= 500 //mm t= 10 //mm t1= 12 //mm //CALCULATIONS Stressmax= (Torquemax*10^6)/(2*B*L*t) I= (2*t1*L*L^2)+((2*t*B^3)/12) sigma= (M*10^6*B)/(2*I) Stressallowable= sqrt(sigma^2+3*(Stressmax^2)) if(Str...
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2019-02-06T10:10:20
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man-propsensor.tst
#Property Sensor Example #generate a table of enthalpy versus molar fractions $thermo = VirtualMaterials.Advanced_Peng-Robinson / -> $thermo thermo + WATER TRIETHYLENE_GLYCOL #generate WATER/TEG bubble temperature curve units Field s = Stream.Stream_Material() s.In.P = 1 atm s.In.VapFrac = 0.0 ps = Sensor.PropertySe...
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4_11.sce
clc //initialisation of variables ha=360.025 //kj/kg hfb=38.05 //kj/kg hb=264.2 //kj/kg h1=2963 //kj/kg h2=1974.6 //kj/kg h3=163 //kj/kg h4=1087 //kj/kg h=1714 //kj/kg //CALCULATIONS m=h/(hb-hfb) wo=7.58*(ha-hb)+(h1-h2) qs=7.58*(ha-hfb)+(h4-h3)+(h1-h) teff=(wo/qs) //RESULTS printf('thermal efficiency is...
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Ex11_2.sce
//Caption: Find (a)Speed of motor (b)%Slip //Exa:11.2 clc; clear; close; p=6//Number of poles f_s=50//Stator frequency(in c/s) f_r=2//Rotor frequency(in c/s) n_s=(120*f_s)/p n=(f_r*120)/p s=n_s-n disp(s,'(a)Speed of motor(in r.p.m)=') S=(n/n_s)*100 disp(S,'(b)%Slip(in %)=')
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// Scilab code Exa8.8 : : Page-351 (2011) clc; clear; S = 1; // Spin angular momentum(s1+-s2), whereas s1 is the spin of proton and s2 is the spin of neutron. m = 2*S+1; // Spin multiplicity j = 1; // Total angular momentum printf("\nThe possible angular momentum states with their parities are as follo...
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//find minimum force per pitch and find actual stresses developed clc //solution //given t=15//mm d=25//mm p=75//mm ftu=400//N/mm^2 tu=320//N/mm^2 fcu=640//N/mm^2 pi=3.14 n=2 FS=4//factor of safety //min foce per pitch which will rupture the joint Ptu=(p-d)*t*ftu//N//ultimate teraing reisistance Psu=n*(p...
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S8_5.sce
// sum 8-5A clc; clear; Di=15; Do=20; d=2.3; D=17.5; C=D/d; Ks=1+(0.5/C); Wmax=100; Tmax=Ks*8*Wmax*D/(%pi*(d^3)); G=81000; delmax=67.7/2.366; k=100/28; Na=G*(d^4)/(8*k*(D^3)); Ls=Na+1; //(for plain ends) delmax=28; //TL= total working length TL=Ls+delmax+(0.15*delmax); // printing data in scilab ...
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//Chapter-17,Example 8,Page 371 clc(); close(); strength = 10*0.85/9 //strength of EDTA //1000 ml EDTA solution == 1 g CaCO3 //for 20 ml EDTA solution amnt= 20*strength/1000 //50 ml smple of water contains amnt CaCO3 hard= amnt*10^6/50 //hardness of water printf("the hardness of water is =...
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clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 13.3 Page 826 \n')// Example 13.3 // Net rate of Heat transfer to the absorber surface L = 10 ;//[m] Collector length = Heater Length T2 = 600 ;//[K] Temperature of curved surface A2 = ...
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//Sci_Function_Demos //use Sci_Test_Function example for help files out1 = Sci_Test_Function(in1,in2);
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//To find the gyroscopic effects clc //Given: m=5*1000 //kg N=1000 //rpm k=0.5 //m //Solution: //Calculating the angular speed of the rotor omega=2*%pi*N/60 //rad/s //When the ship steers to the left: v=30*1000/3600 //m/s R=60 //m //Calculating the angular velocity of precession omegaP=v/R //rad/s //Calcu...
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function res = viscircles(varargin) select length(varargin) case 02 then res = il2mat(raw_viscircles(mat2il(varargin(01)),varargin(02))) else error(39) end endfunction
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Engineering Thermodynamics by Onkar Singh Chapter 4 Example 5") T1=(25+273.15);//temperature of inside of house in K T2=(-1+273.15);//outside temperature in K Q1=125;//heating load in MJ/Hr disp("COP_HP=Q1/W=Q1...
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//Exa 5.2 clc; clear; close; //Given Data : format('v',7); R=2;//in ohm X=6;//in ohm P=10000*10^3;//in watts cos_fir=0.8;//unitless VR=22*10^3;//in volt I=P/(sqrt(3)*VR*cos_fir);//in Ampere VR_phase=VR/sqrt(3);//in volt Vs=sqrt((VR_phase*cos_fir+I*R)^2+(VR_phase*sqrt(1-cos_fir^2)+I*X)^2); disp(Vs,"Sending...
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clc disp("the soln of eg 1.5--> Gauss Seidel Method"); for i=1:3,xnew(i)=2,e(i)=1 end x=1e-6 while e(1)>x&e(2)>x&e(3)>x do for i=1:3, xold(i)=xnew(i),end xnew(1)=(44-xold(2)-2*xold(3))/10 xnew(2)=(-2*xnew(1)+51-xold(3))/10 xnew(3)=(-2*xnew(2)-xnew(1)+61)/10 for i=1:3,e(i)=abs(xnew(i)-xold...
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//Page Number: 375 //Example 7.8 clc; //Given V0=1.8D+3; //V I0=1.3; //A Pin=70; //W n=0.22; //(i) Power generated Pgen=n*I0*V0; disp('W',Pgen,'Power generated:'); //(ii) Total RF power generated Pt=Pin+Pgen; disp('W',Pt,'Total RF power generated:'); //(iii) Power gain G=Pt/Pin; Gdb=10*log10(G); ...
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clc //initialization of variables //For window with two panes 3 cm apart k = 0.57*10^-4 //cal/cm-sec-K l = 3 //cm g = 980 // cm/sec^2 Nu = 0.14 // cm^2/sec DeltaT = 30 // Kelvin T = 278 // Kelvin L = 100 // cm //calculations h = (0.065*(k/l)*(((l^3)*g*DeltaT/((Nu^2)*T))^(1/3))*((l/L)^(1/9)))*10^4//for two p...
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int main(void) { int x[3]; &x; }
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//Scilab Code for Example 7.19 of Signals and systems by //P.Ramakrishna Rao //Convolution of two signals clc; clear; clear x y n; x=[2,-1,1,0,2]; y=[1,0,-1,2]; n=-1:3; c = gca(); c.y_location = "origin"; c.x_location = "origin"; plot2d3(n,x,-5); title('x(k)') xlabel('k') figure(1); n=0:3; c = gca(); ...
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clear // //given and derived rm=75 im=300*(10**-6) //case a i=5 rsh=(rm*im)/(i-im) printf("\n rsh= %e ohm",rsh) //case b i=7.5 rsh=(rm*im)/(i-im) printf("\n rsh= %e ohm",rsh) //case c i=10 rsh=(rm*im)/(i-im) printf("\n rsh= %e ohm",rsh)
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t=[1.00 0 0 0 0 0 0 0 0 0 1.00 1.00 0 0 0 0 0 0 0 0 1.00 1.00 1.00 0 0 0 0 0 0 0 1.00 1.00 1.00 1.00 0 0 0 0 0 0 1.00 1.00 1.00 1.00 0.67 0 0 0 0 0 1.20 1.20 1.00 1.00 1.00 0.86 0 0 0 0 1.64 1.50 1.50 1.64 1.64 1.64 1.38 0 0 0 2.21 2.47 2.36 2.41 2.47 2.41 2.36 2.26 0 0 2.82 3.06 3.10 3.10 3.10 3.10 3.09 3.06 3.02 0 2....
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True-Class positive negative positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1.0 0.0 positive 1...
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// Example 34_11 clc;funcprot(0); //Given data L=30;// Total load in MW L_12=20;//Capacity of two steam turbines in MW // S_1=2000+10*L_1-0.0001*L_1^2 // S_2=1000+7*L_1-0.00005*L_2^2 //Calculation //L_1+L_2=L*10^3; // For the most loading,the required condition is (dS_1/dL_1=dS_2/dL_2) function[X]=Load(y) ...
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//find.. clc //soltuion //given n=10 v=25//m/s P=115*1000//W q=%pi B=22.5//deg u=0.2 m=0.6//kg/m //let T1 and T2 be tension on tight and slag side //P=(T1-T2)*v*n//W //T1-T2=460...eq1 //log(T1/T2)=u*q*cosec(%pi/180*B)=0.714 T2=T1/5.18//....eq2 //from eq1 and eq2 T1=570//N T2=110//N Tc=m*v^2 Tt1=T1+T...
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//Finding the Snubber Values for Limiting dv/dt and di/dt Values of a BJT Switch //Example 4.6(Page No- 154) clc clear //given data fs = 10*10^3;//kHz Vs = 220;//V Il = 100;//A Vce_sat = 0;//V td = 0; tr = 3*10^-6//sec tf = 1.2*10^-6//sec //part(a) Ls = (Vs*tr)/Il; printf('(a)\t Ls: %1.1f uH',Ls*10^6) ...
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clc // Given that lambda1 = 6500 // wavelength of first source in angstrom lambda2 = 5200 // wavelength of second source in angstrom d = 2 // Spacing between sources in mm D = 1.2 // Distance between source and screen n = 3 // Order of bright fringe // Sample Problem 6 on page no. 95 printf("\n # PROBLEM 6 # \...
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//Exa 14(i) clc; clear; close; //given data : inINV=50000;//initial investment in Rs. and equal for all projects life=5;//in years salvage=0;//in Rs. TaxRate=55;//in % //depreciation type :Straight line D=inINV/life;//in Rs //cash flows before tax of 1st,2nd,3rd,4th and 5th years CBFT1=10000;//in Rs. CBFT...
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// Exa 1.15 clc; clear; close; format('v',6) // Given data R1= 2;// in ohm R2= 6;// in ohm R3= 4;// in ohm R4= 3;// in ohm R5= 5;// in ohm V1= 10;// in V V2= 6;// in V V3= 2;// in V //Applying KVL in ABEFA : I1*(R1+R2+R3) - R2*I2 = V1-V2 (i) //Applying KVL in BCDEB : I1*-R2+I2*(R2+R4+R5) =V2-V3 (ii) ...
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//Problem 24.05: A 200 V, 50 Hz supply is connected across a coil of negligible resistance and inductance 0.15 H connected in series with a 32 ohm resistor. Determine (a) the impedance of the circuit, (b) the current and circuit phase angle, (c) the p.d. across the 32 ohm resistor, and (d) the p.d. across the coil. ...
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//Ex4_9 // Illustration of Jaggies in Image Zooming //Version : Scilab 5.4.1 // Operating System : Window-xp, Window-7 //Toolbox: Image Processing Design 8.3.1-1 //Toolbox: SIVP 0.5.3.1-2 //Reference book name : Digital Image Processing //book author: Rafael C. Gonzalez and Richard E. Woods clc; close; cl...
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//Supercharged four stroke oil engine clc,clear //Given: T1=20+273 //Temperature of air enters the compressor in K Q1=1340 //Heat added to air in kJ/min T3=60+273 //Temperature of air leaves the cooler or enters the engine in K P3=1.72 //Pressure of air leaves the cooler or enters the engine in bar eta_v=0.70 //Volumet...
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// written by Aurelien Garivier, CNRS & Telecom Paristech // January 2012 // // Baum-Welch algorithm for discrete Hidden Markov models // see http://www.telecom-paristech.fr/~garivier/code/index.html function [x,y] = HMMsample(nu, Q, g, n) // HMMsample sample a trajectory from a hidden markov chain // // in : nu = in...
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load HackComputer.hdl, output-file Prog3.out, output-list time%S1.5.1 reset%B2.1.2 ARegister[]%D1.7.1 DRegister[]%D1.7.1 RAM64[16]%D1.7.1 RAM64[17]%D1.7.1; //finds sum of first 99 natural numbers and stores it in RAM64[17] ROM32K load Prog3.hack, repeat 1400 { tick, tock, output; }
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//Optoelectronics - An Introduction, 2nd Edition by J. Wilson and J.F.B. Hawkes //Example 7.7 //OS=Windows XP sp3 //Scilab version 5.5.2 clc; clear; //given A=100e-6*100e-6;//Junction area of the photodiode in m^2 Epsilonr=12;//Relative permittivity of InGaAs Epsilon0=8.84e-12;//Permittivity of free space in...
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Test CONTOURS Numero : 1 Largeur : 4.698 Hauteur : 417.612 Angle_Haut/Gauche : 73.301 Numero : 0 Largeur : 4.5 Hauteur : 500.0 Numero : 2 Largeur : 4.698 Hauteur : 104.403 Angle_Haut/Gauche : 106.699 Numero : 3 Lmin : 4.5 Lmax : 4.698 Hmin : 190.802 Hmax : 192.152 Angle_Haut/Gauche : 16.699 Numero : 4 Lmin : 4.5 Lmax ...
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// Example 8.2 format('v',5) clc; clear; close; // given data V_EE= 12;// in V V_BE= 0.7;// in V R_E= 5.6*10^3;// in Ω V_CC= 15;// in V R_C= 6.8*10^3;// in Ω // The emitter current, I_E= (V_EE-V_BE)/R_E;// in A I_C= I_E;// in A // The collector to base voltage V_CB= V_CC-I_C*R_C;// in V disp(V_CB,"The v...
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boardsize 4 play w a2 play b b3 play w b2 play b c3 play w c2 play b d4 play w d2 play b b4 play w e1 play b c5 havannah_winner
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//( debFon ) function y=score_de_Deux(p) y = 3*p^2 - 2*p^3 endfunction function y=deux_d_affilee(p) q = 1-p y = ___ endfunction function y=deux_d_avance(p) q = 1-p y = p^2 / (p^3+p^2*q+p*q^2+q^3) endfunction //( finFon ) scf(1) clf() //( debAff ) p = linspace(.5,1) plot(p,score_de_Deux,'--') plot(p,deux_d_...
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//author :Vamsi Krishna Koppala //function :firlp2lp //input argument : fir typeI filter coefficient vector(Hr(w)) //output argument :fir typeI filter of coefficients[1 - Hr(π-w)]. // source http://www.mathworks.com/help/dsp/ref/firlp2lp.html function a=firlp2lp(b); r = length(b); a =b; a(int16(r/2+...
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// Chapter 9 example 4 //------------------------------------------------------------------------------ clc; clear; // Given Data PW_tx = 10^-6; // Transmitted pulse width Rx_PW = 10^-6; // Received pulse width c = 3*10^8; // velocity of EM waves in m/s // Cal...
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clc; v=3*10^8; //velocity in m/sec n=500*10^3; //frequency in Hz l=v/n; //calculating wavelength disp(l,"Wavelength in m = "); //displaying result
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tam_mat = 12 a = zeros(tam_mat,tam_mat) b = zeros(tam_mat,tam_mat) c = zeros(tam_mat,tam_mat) for i = 1:tam_mat for j = 1:tam_mat if (i == j) | (i == j-1) | (i == j+1) then a(i,j) = 1 c(i,j) = 1 end if (i <= j) then a(i,j) = 1 c(i,j) = 1 ...
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//Chapter 1, Example 1.8 clc //Variable Declaration a1 = 0.00047 a2 = 0.002 b1 = 690000 b2 = 0.00000013 //Calculation a = a1/a2 b = b1/b2 //Results printf("(a) %.1f x 10^-2 \n",a*100) printf("(b) %.2f x 10^12",b/1000000000000)
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#define szof_c 1 #define szof_uc szof_c #define szof_s 2 #define szof_us szof_s #define szof_i 4 #if __WORDSIZE == 64 # define szof_ui szof_i # define szof_l 8 #endif #define szof_f 4 #define szof_d 8 #define FILL(T) \ name fill##T \ fill##T: \ prolog \ arg $argp \ get...
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clc clear //Input data V1=600//Velocity in m/s Vb=120//Mean blade velocity in m/s a=16//Nozzle angle in degrees b=[18,21,35]//Exit angles in degrees m=5//Steam flow rate in kg/s h=25//Nozzle height in mm v=0.375//Specific volume in m^3/kg p=25//Pitch in mm t=0.5//Thickness in mm kb=0.9//Constant //Cal...
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clc; clear; printf("\n Example 9.32\n"); d2=54e-3; //outer diameter of the tube d1=70e-3; // fin diameter w=2e-3; //fin thickness n=230;// number of fins per metre run T_s=370; //Surface temperature T=280; //Temperature of surroundings h=30; //Heat transfer coeffecient between gas and fin k=43; //Thermal ...
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clear //Given //We will divide this into three parts F = 8 //k - force applied d = 16 //inch -distance l_1 = 1 //in l_2 = 3 //in b_1 = 4 //in b_2 = 1 //in A_1 = l_1* b_1 //sq.in - area of part_1 y_1 = 0.5 //in com distance from ab A_2 =l_2*b_2 //sq.in - area of part_1 y_2 = 2.5 //in com distance from...
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// Example 4.15.6 page 4.39 clc; clear; L=5; //length of optical link n1=1.5 //refractive index c=3d8; //speed of light delta=1/100; //relative refractive index delTS=L*n1*delta/c; //computing delay difference delTS=delTS*10^12; sigmaS=L*n1*delta/(2*sqrt(3)*c); //computing...
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clc; eff_c=0.8; // Isentropic efficiency of compression each stage eff_CT=0.88; // Isentropic efficiency of compressor turbine eff_PT=0.88; // Isentropic efficiency of power turbine eff_trans=0.98; // Turbine to compressor transmission efficiency rp=3; // Pressure ratio in each stage of compression T08=297; // Te...
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function PontoFixo(f,x1,tol) // erro - tolerância máxima para x1 // Saída - x1 raíz da função f. // x0 solução inicial para a raíz de f deff ( ' y=g ( x ) ' , ' y=f(x)+x' ) erro = 1; printf ( '%i\t%.10f\t%.1e\n',1,x1,erro) for (k=1:500) x0=x1 x1 = g(x0); if(x1<>0) t...
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errcatch(-1,"stop");mode(2);// Example 18.12, page no-467 epsr=12 N=5*10^28 //atoms.m^-3 eps=8.854*10^-12//F.m^-1 alfe=eps*(epsr-1)/N printf("The electronic polarisability of given element is %.3f * 10^-39 F m^2",alfe*10^39) exit();
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//Ex 1.11 clc;clear;close; VBE3=0.7;//V VBE1=0.7;//V IREF1=100;//micro A IC1=IREF1;//micro A IREF2=1;//mA IC2=IREF2;//mA Beta=200;//unitless //IC2/IC1=(IS*exp(VBE2/VT))/(IS*exp(VBE1/VT)) VT=26;//mV deltaVBE=VT*10^-3*log(IC2/IC1);//V(deltaVBE=VBE2-VBE1) deltaVx=2*deltaVBE;//V IO=IREF1/(1+2/(Beta^2+Beta));//...
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//Example 4_7 clc(); clear; //To estimate the lower limit for the speed //In a practical situation u should be atleast 0.5 u=0.5 g=9.8 //units in meter/sec^2 x=7 //units in meters v0=sqrt(2*u*g*x) //units in meters/sec printf("The lower limit of the speed v0=%.1f meter/sec",v0)
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clear; clc; J=20*10^3 //in kA/cm^2 e=1.6*10^-19 //in C Nd=2*10^15 //in cm^-3 //Calculation vz=J/(e*Nd) format("e",9) disp(vz,"avalanche-zone velocity is (cm/s)= ")
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// // printf("\n at station 1') h=1,h1=2.55,h2=0.95,b=9,b1=7.5,b2=5.25, w1=b1+b2 a=(((h/2)*(b1+b2))+((b/4)*(h1+h2))) printf("\n area= %0.3f sq. meter",a) printf("\n at station 2') h=1.5,h1=2.8,h2=1.35,b=9,b1=8.1,b2=4.75, a1=(((h/2)*(b1+b2))+((b/4)*(h1+h2))) d=50 k=10.01 v=(d/2)*(a+a1) w2=b1+b2 printf("\n area=...
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Turbomachinery Design and Theory,Rama S. R. Gorla and Aijaz A. Khan, Chapter 5, Example 6") disp("Since the degree of reaction is 50%, the velocity triangle is symmetric as shown in Figure Ex56") disp("Using the ...
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clear; clc; printf("\t\t\tProblem Number 6.12\n\n\n"); // Chapter 6: The Ideal Gas // Problem 6.12 (page no. 255) // Solution //The molecular weight of oxygen is 32.Therefore, R=1545/32; //Unit:ft*lbf/lbm*R //constant of proportionality J=778; //conversion factor cp=0.24; //Unit:Btu/lbm*R //specific heat at...
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// BOTÃO EXCLUIR BARRAS frame_left_estr.enable ="off" sca(eixoEstr) [botao,x0,y0]=xclick(); // ponto clicado na janela de eixos // Enquanto não clicar no botão central e houver barras para apagar while and(botao~=[2 5]) && size(Barras,1)>0 // Cria lista com ponto inicial e final das barras Lista ...
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// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 8 : STARTING, CONTROL AND TESTING OF AN INDUCTION MOTOR // Example : 8.5 clc;clear; // clears the console and command history // Given data V = 210 // supply voltage in V f = 50 // su...
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//Scilab Code for Example 7.22 of Signals and systems by //P.Ramakrishna Rao //Convolution of two signals clc; clear; clear x y n; for n=0:10; x(n+1)=(3/4)^n*u(n); end c = gca(); c.y_location = "origin"; c.x_location = "origin"; n=0:10; plot2d3(n,x,-4); title('x(n)') xlabel('n') for n=0:10; y(n...
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<?xml version="1.0" encoding="utf-8"?> <test> <description>Project2D Quad Lagrange basis P=6 Q=7</description> <executable>LocProject</executable> <parameters>-s quadrilateral -b GLL_Lagrange GLL_Lagrange -o 6 6 -p 7 7 -c 0.0 0.0 1.0 0.0 1.0 1.0 0.0 1.0</parameters> <metrics> <metric type="L2" i...
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clc clear //Input data T=[50+273,345+273]//Temperatures at the beginning and end of compression in K g=1.4//ratio of specific heats IHP=25//Indicated horse power in h.p m=5.44//Mass of fuel consumed per hour in kg CV=10300//Calorific value in kcal/kg //Calculations na=(1-(T(1)/T(2)))*100//Air standard effic...
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// Chapter9 // Page.No-440 // Example_9_10 // Design of current source // Given clear;clc; Vr=5; // Voltage in volt Il=0.25; // Load current in ampere Rl=48; // Load resistance in ohm dropout_volt=2; // Constant for IC7805C R=Vr/Il; // Approximate result sice Iq is negligible in the eq. Il=(Vr/Il)+Iq where Iq...
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// Scilab Code Ex4.18: Page-134 (2006) clc; clear; // We have from Mattheissen rule, rho = rho_0 + alpha*T1 T1 = 300; // Initial temperature, K T2 = 1000; // Final temperature, K rho = 1e-06; // Resistivity of the metal, ohm-m delta_rho = 0.07*rho; // Increase in resistivity of metal, ohm-m alpha = de...
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//ques-18.5 //Calculating standard heat of formation of actylene clc h1=-1300;//heat of combustion of acetylene (in kJ) h2=-395;//heat of combustion of graphite(C) (in kJ) h3=-286;//heat of combustion of hydrogen (in kJ) H=2*h2+h3-h1;//heat of formation of actylene (in kJ) printf("Heat of formation of actylene i...
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Ex10_3.sce
//Predict whether the following reaction would proceed spontaneously as written ( Cd(s)+Fe++(aq)=Cd++(aq)+Fe(s) //Example 10.3 clc; clear; C1=0.15; //Concentration of Cadmium ion in M C2=0.68; //Concentration of Ferrus ion in M E1=-0.447; //Standard Electrode potential for cathode in V E2=-0.403;...
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/3516/CH18/EX18.2/Ex18_2.sce
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2018-02-03T05:31:52
2018-02-03T05:31:52
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Ex18_2.sce
printf("\t example 18.2 \n"); tav=500; // F Ts=1000; t0=100; c=0.12; // Btu/(lb)*(F) k=24; // Btu/(hr)*(ft^2)*(F/ft) row=488; // lb/ft^3 alpha=0.41; // alpha=(k/(c*row)), ft^2/hr x=0.333; // ft theta=4; printf("\t values are approximately mentioned in the book \n"); X=(x/(2*(alpha*theta)^(1/2))); printf("\...