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// Display mode mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Introduction to heat transfer by S.K.Som, Chapter 12, Example 1") //The pressure in the pipeline that transports helium gas at a rate of 4kg/s is maintained at pressure(p)=1 atm or 101*10^3 pascal. //The internal...
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// Theory and Problems of Thermodynamics // Chapter 4 // Energy Analysis of Process // Example 4 clear ;clc; //Given data m = 0.1 // mass wet steam in kg X1 = 0.8 // wet steam quality P1 = 0.3 // Pressure in MPa d = 0.8 // distance upto presence of latc...
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build_help.sce
help_lang_dir = get_absolute_file_path("build_help.sce"); TOOLBOX_TITLE="Communications Toolbox" tbx_build_help(TOOLBOX_TITLE, help_lang_dir); ok=add_help_chapter("Demo",get_absolute_file_path("build_help.sce")); clear help_lang_dir;
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//Example 17.12// n=(1.4*10^12);//m^-3 //density of charge carrier q=(0.16*10^-18);//C // Coulomb of Charge ue=0.720;//m^2 /(V s) //Electron mobility of GaAs uh=0.020;//m^2 /(V s) //Hole mobility of GaAs s=n*q*(ue+uh) mprintf("s = %e ohm^-1 m^-1",s) Eg=1.47;//eV //band gap k=86.2*10^-6;//eV/K //Boltzmann cons...
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//Caption:overall_transfer_function_of_given_system //example 5.9.9 //page 107 syms Ka Ke Kt J f N1 N2 s=%s; Ke=10;//error_detector_gain Ka=100;//amplifier_transconductance Kt=.0005;//motor_torque_const J=.0000125;//moment_of_inertia f=.0005;//coeff_of_viscous_friction g=N1/N2; g=1/20; a=(Ka*Ke); b=(a*Kt);...
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clear μ = 0.03 k = (1/6)*10^4 β = 10^-5 δ = 1 n = 101 function f = f(M) f1 = α.*M./(M+k) f2 = 1./(1+β*n*M) f = f1.*f2 endfunction function g = g(M,γ) g = γ.*M endfunction function eqn = eqnM1(M,γ,α)//1 cohort eqn = (α*k-α*β*n*M.^2-(γ+μ)*(1+β*n*M).^2.*(M+k).^2)./((1+β*n*M).^2.*(M+k).^2) endfunction...
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clear; clc; close; disp(log10(0.5),'ans for part a :- '); disp(log10(4000/250),'ans for part b :- '); disp(log10(0.6*30),'ans for part c :- ');
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// Exa 3.11 clc; clear; close; format('v',5) //Given data Beta = 100; V_CC = 10;// V R1 = 9.1;// in k ohm R_C = 1;// in k ohm R_E = 560*10^-3;// in k ohm R2 = 4.7;// in k ohm V_BE = 0.7;// in V V_Th = (V_CC/(R1+R2))*R2;// in V R_B = (R1*R2)/(R1+R2);// in k ohm // V_Th - I_B*R_B - V_BE - I_E*R_E = 0 or ...
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//Exa 5.4 clc; clear; close; //given data ND=8*10^14;//in cm^-3 NA=8*10^14;//in cm^-3 ni=2*10^13;//in cm^-3 T=300;//in Kelvin k=8.61*10^-5;//in eV/K e=1.6*10^-19;//coulamb Vo=k*T*log(ND*NA/ni^2);//in Volts disp(Vo,"Potential barrier in volts : ");
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clc // Variable Initialization Vm=240//Supply Voltage in Volts Ra=0.9//Combined Field and Armature circuit resistance in Ohm N=900 //Motor speed in Rpm V=220//Rated voltage of motor in Volts a=45//firing angle in Degree Kaf=0.035 //Constant in N-m/A^2 //Solution //For semi-converter controlled Dc Drive Va=...
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clear // // // //Variable declaration lamda=589.3*10**-9 //wavelength(m) mewe=1.65833 //refractive index of e-ray mew0=1.48640 //refractive index of o-ray //Calculation t1=lamda/(2*(mewe-mew0)) //thickness of half wave plate(m) t2=lamda/(4*(mewe-mew0)) //thickness of quarter wave pl...
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 11.20\n\n\n"); // Chapter 11 : Heat Transfer // Problem 11.20 (page no. 588) // Solution //The upper temperature is given as 120 F and the temperature difference is Ti=120; //Inside temperature //unit:fahrenheit To=70; //Outside temperature //unit:fa...
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 3.10w //calculation of height of balloon when stone reaches ground //given data x=-50//height(in m) of the ballon when the stone was dropped u=5//velocity(in m/s) of the ballon a=-10//acceleration(in m/s^2) of the ba...
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clc; R=0.287; ch_ent=R*log(2);//V2=2*V1 disp("increase in entropy is:"); disp("kJ/kg K",ch_ent);
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1000100000001101 LIC 13 1000110000000000 STC 0 1000100000000111 LIC 7 1000110000000001 STC 1 1110000000000000 RDM 0 1110000000000001 RDM 1 1000000000000000 LDA 0 1000010000000001 LDB 1 0000100000000000 SUB 1000110000000011 STC 3 1000110000000100 STC 4 1000000000000011 LDA 3 1000010000000100 LDB 4 000001000...
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//Chapter 10 : Crystallography and Crystal Imperfections clear; //Variable declaration r=1.278 //atomic weigth No=6.02*10**26 //Avagadro's No. //Calculations a=2*sqrt(2)*r //Result mprintf("Lattice constant a= %.3f Armstrong",a)
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clear; clc; //Example 16.3 Vdd=5; Vtnd=0.8; Vtnl=0.8; Kn=35; Vo=0.1; Vi=4.2; //W/L=Y yl=0.5; //Kd/Kl=x x=(Vdd-Vo-Vtnl)^2/(2*Vo*(Vi-Vtnd)-Vo^2); printf('\nKd/Kl=%.2f\n',x) //Kd/Kl=yd/yl yd=12.6 yl=0.5 iD=Kn*yl*(Vdd-Vo-Vtnl)^2/2; printf('\ndrain current =%.2f microA\n',iD) P=iD*Vdd; printf('\npower di...
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//Ex:57 clc; clear; close; d_ap=30000;//Difference of apogee and perigee dis in km s_ap=50000;//sum of apogee and perigee dis in km e=d_ap/s_ap; printf("The orbital eccentricity=%f",e);
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//1.21 clc; X_mean=(15+20+25+30+35+45)/6; printf("The sample mean of the temperature=%.2f degree C",X_mean) Y_mean=(1.9+1.93+1.97+2+2.01+2.01+1.94+1.95+1.97+2.02+2.02+2.04)/12*10^-6; printf("\nThe sample mean of the faliure=%.6f failures/hour",Y_mean) disp('from these values we get') a=1.80*10^-6; b=0.00226; d...
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style.fontSize=12; style.displayedLabel="<table> <tr> <td><b>G<br>S</b></td> <td align=center>pFET0</td> <td align=left><b>D</b></td> </tr> </table>"; pal11 = xcosPalAddBlock(pal11,"macrocab_pfet0",[],style);
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//Fatorização LU sem escolha de pivot function[L,U]=gauss(A) [n,m]=size(A); L=eye(n,n); for i=1:n-1 for j=i+1:n mu=A(j,i)/A(i,i); for k=1:n A(j,k)=-A(i,k)*mu+A(j,k); ...
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Datatype tag for DS1 is: "Character array" DS1[0]: OPAQUE0 DS1[1]: OPAQUE1 DS1[2]: OPAQUE2 DS1[3]: OPAQUE3
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clc; v=400; // rated voltage of motor zs=0.13+%i*1.3 ; // per phase synchronous impedance p=100000; // VA rating of motor l=4000; // stray losses pl=75000; // power delivered to load disp('case a'); il=p/(sqrt(3)*v); // line current vt=v/sqrt(3); // per phase rated voltage pd=pl+l ; // power developed poh=3*i...
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errcatch(-1,"stop");mode(2);//Ex 6.2 ;; format('v',5); step=10.3;//mV reading='101101111';//reading Vo=step*bin2dec(reading)/1000;//V disp(Vo,"Output voltage(V)"); exit();
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//Variable declaration: D = 0.0833 //Diameter of tube (ft) L = 2.0 //Length of tube (ft) h = 2.8 //Heat transfer coefficient (Btu/h.ft^2.°F) Ta1 = 1500.0+460.0 //Temperature of hot air in furnace (°R) Ta2 = 1350.0+460.0 ...
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function displayalldetails() // This program is free software; you can redistribute it and/or modify // it under the terms of the GNU General Public License as published by // the Free Software Foundation; either version 2 of the License, or // (at your option) any later version. // // This program is distri...
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clc V1=0.25; //m^3 p1=500; //kPa p2=100; //kPa V2=V1*(p1/p2)^(1/1.25) n=1.25 dU=3.64*(p2*V2 - p1*V1) disp("(i) If the expansion is quasi-static") W=(p1*V1-p2*V2)/(n-1); Q=dU+W disp("Heat transfered=") disp(Q) disp("kJ") disp("(ii) In another process") Q=32; //kJ W=Q-dU; disp("Work done=") disp(...
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clc // // // //Variable declaration theta1=35 //rotation of plane s=100 //Specific rotation c=0.1 //Concentration //Calculations l=((theta1)/(s*c))*10 //Result printf("\n The Length will be %i cm",l)
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3 3 0 5 4 1 2 2 3 1 3 4 5 6 6 1 2 2 5 5 3 4 3 4 6 6 1 ~~~~~~~~~~~~~~~~~~~~~~~~~~ NO YES YES
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//chapter12 //example12.10 //page246 V_BE=0.7 // V gain_beta=100 I_C=1 // mA V_CE=2 // V I_B=I_C/gain_beta // since V_CE=V_BE+V_CB we get V_CB=V_CE-V_BE R_B=V_CB/I_B printf("base resistance=%.3f kilo ohm \n",R_B)
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// Data Reconciliation Benchmark Problems From Lietrature Review // Author: Edson Cordeiro do Valle // Contact - edsoncv@{gmail.com}{vrtech.com.br} // Skype: edson.cv //Martins, Márcio A.F., Carolina A. Amaro, Leonardo S. Souza, Ricardo A. Kalid, and Asher Kiperstok. 2010. //New objective function for data reconciliat...
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// Case Study:-Chapter 6 // 4.Plotting of two Functions i.e. y1=exp(-ax) and y2=exp(-ax^2/2) a=0.4; printf(" y-------> \n"); printf("0 ---------------------------------------------------------------\n"); for x=0:0.25:4 //Evaluation of functions ...
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// problem 2.4 s1=13.6 s2=7.8 s3=1 // by archimede principle // weight of body = weight of liquid displaced // s2=s1*x+s3*(1-x) x=(s2-s3)/(s1-s3) disp(x,"fraction of steel below surface of mercury")
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clear // // Variable Declaration T_d=21// The dry bulb temperature in °C Q=14// Internal load in kW H=50// % saturation Q_l=1.5// Latent heat gain in kW T_ain=12// The inlet air temperature in °C C_p=1.02// The specific heat capacity of air in kJ/kg.K // Calculation deltaT=T_d-T_ain// Air temperature rise through roo...
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clc clear //Input data d=0.5//Diameter of the ring in cm n=4//number of bands w=5893//Wavelength of light in Angstrom q=30//Angle at which light enters in degrees //Calculations R=((d^2*cosd(q))/(2*(2*n+1)*w*10^-8))//Radius of curvature of lens in cm //Output printf('Radius of curvature of lens is %3.1f c...
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default_font_size = 48; response_matching = simple_matching; active_buttons = 5; button_codes = 1, 2, 0, 11, 22; begin; TEMPLATE "words.tem"; TEMPLATE "semantic_block.tem"; trial { trial_duration = forever; trial_type = specific_response; terminator_button = 3; picture { text { ...
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// Example A-8-12 // Computing optimal solution clear; clc; xdel(winsid()); //close all windows s = %s; t = 0:0.1:5; u = ones(1,length(t)); t1 = 0:0.01:5;N =length(t1); u1 = ones(1,N); k = 0; mprintf('Processing ...\n'); for K = 50:-1:2 for a = 2:-0.05:0.05 num = K * ((s + a)^2) ; den = s * s * (s^2 + 6*...
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//Example 3.6 v_river=1.2;//Velocity of river current (m/s) v_x=v_river;//x-component of velocity (m/s) v_boat=0.75;//Velocity of boat (m/s) v_y=v_boat;//y-component of velocity (m/s) v_tot=sqrt(v_x^2+v_y^2);//Magnitude of resultant velocity (m/s) printf('Magnitude of the boat''s velocity relative to the observer...
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a=0.28; b=0.29; c=0.028; d=0.029; t=[0:0.1:400]; x0y0=[21;7]; function dxdy=x_der(t,x) dxdy(1)=b*x(1)-d*x(1)*x(2); dxdy(2)=-a*x(2)+c*x(1)*x(2); endfunction x=ode(x0y0,0,t,x_der); n=size(x,"c"); for i=1:n x_der1(i)=x(1,i); x_der2(i)=x(2,i); end //plot(t,x_der1); //plot(t,x_der2); //plot(x_der1,x_der...
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function f=%s_q_r(s,f) // %s_q_r(s,f) f= s./f //! // Copyright INRIA if size(s,'*')==0 then f=[],return,end f=tlist(['r','num','den','dt'],f(2)./s,f(3).*ones(s),f(4)),
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<<<<<<< HEAD // Given a time series (vector) Y, return a matrix with ones in the first column and the first K lagged values of Y in the other columns. //Calling Sequence //autoreg_matrix(Y, K) //Parameters //Y: Vector //K: Scalar or Vector ======= function y = autoreg_matrix(Y, varargin) // Given a time series (vect...
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clc; //Example 31.3 //page no 493 printf("Example 31.3 page no 493\n\n"); //friction factor for smooth tubes can be approximated by //f = 0.079*R_e^(-1/4),if 2000< R_e<2e-5 // average velocity in the system ,involving the flow of water at 60 deg F is given by //v =sqrt(2180/(213.4R_e^(-1/4) + 10), flow of water...
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clear; clc; l=5;f=5000/(2*%pi);Rs=175;Ls=10*(10^-3);Rsh=270;Csh=0.2*(10^-6); w=2*%pi*f; Z1=(Rs+(%i*w*Ls)); //Z1=Z1/2 Z2=Rsh-(%i/(w*Csh)); t=sqrt((Z1)/((Z1)+(2*Z2))); P=2*(atanh(t))/l; A=real(P); B=imag(P); printf("Propagation constant = % f + %f per loop km\n",round(A*100)/100,round(B*100)/100); Zo=Z1/(tanh(...
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 9.14\n\n\n"); // Chapter 9 : Gas Power Cycles // Problem 9.14 (page no. 489) // Solution //A Brayton cycle rc=7; //Compression Ratio Rc=v2/v3 k=1.4; //It is apparent incerease in compression ratio yields an increased cycle efficiency cp=0.24; //Unit...
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clear; clc; //Example12.4[Emissivity of a surface and emissive Power] e1=0.3;//For 0<= lambda <= 3micron e2=0.8;//3micron<=lambda<=7micron e3=0.1;//7micron<=lamda<infinity lambda1=3,lambda2=7;//[micron] T=800;//[K] //Solution:- p=lambda1*T;//[micron.K] q=lambda2*T;//[micron.K] //Hence blackbody radiation ...
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errcatch(-1,"stop");mode(2);//caption:Find deviation //Ex2.14 x1=25.65//first reading(in W) x2=24.39//second reading(in W) x3=23.75//third reading(in W) x4=26.42//fourth reading(in W) x5=24.92//fifth reading(in W) n=5//number of readings X=(x1+x2+x3+x4+x5)/5 d1=x1-X d2=x2-X d3=x3-X d4=x4-X d5=x5-X D...
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// Grob's Basic Electronics 11e // Chapter No. 16 // Example No. 16_3 clc; clear; // A constant current of 2 uA charges a capacitor for 20 s. How much charge is stored? Remember I=Q/t or Q=I*t. // Given data I = 2*10^-6; // Current=2 uAmps t = 20; // Time=20 Sec Q = I*t disp (Q,'The Ch...
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//Example 4.7 clc printf("Enter alphabets:") [n,line] = mscanf(" %[ A-Z]"); //[A-Z] =[ABCDEFGHIJKLMNOPQRSTUVWXYZ] disp(line, "line is:");
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[[1],[4,6],[8,17,15],[8,22,27,18]] / [[1],[2,3]] = quot[2,2] = 6, remd = [[1],[4,6],[8,17,15],[8,22,27,18]], prod = [[0],[0,0],[0,0,6],[0,0,12,18]] quot[2,1] = 5, remd = [[1],[4,6],[8,17,9],[8,22,15], prod = [[0],[0,0],[0,5,0],[0,10,15] quot[2,0] = 4, remd = [[1],[4,6],[8,12,9],[8,12], prod = [[0],[0,0],[4,0,0],[8,...
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H = 40 ; // Power in hp s = 6000 ; // allowable shear stress in steel in psi // Part (a) n = 500 ; // rpm T = ((33000*H)/(2*%pi*n))*(5042/420); // Torque in lb-in d = ((16*T)/(%pi*s))^(1/3); // diameter in inch disp("inch",d,"Diameter of the shaft at 500 rpm") // Part (b) n1 = 3000 ; // rpm T1 = ((33000*H)/(2*...
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function [output] = imrect(img,pstDataX,pstDataY) image = mattolist(img); a = opencv_imrect(image,pstDataX,pstDataY) d = size(a); for i=1:size output(:,:,i) = a(i); end endfunction
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R1=4; //Assigning values to parameters V1=7; R2=2; R3=4; I1=8; R4=6; R5=9; V2=12; R6=10; I2=V1/R1; //Performing source transformation V3=I1*R2; I3=V2/R5; R7=R2+R3; I4=V3/R7; R=1/((1/R1)+(1/R7)+(1/R4)+(1/R5)); I=I2+I3-I4; V=I*R; IR6=V/(R+R6) disp("Amperes",IR...
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//Ex3_15 Pg-190 clc disp("(a) When the diode is forward biased figure 3.55(b),it offers zero resistance. It is like shorted switch.This shorted switch across AB also short-circuits the resistance R2.Obviously,a parallel combination of the diode and R2 is equivalent to a resistance of zero ohms.") R1=100 //reisitor...
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function [x,iter,ea] = jacobi(A,b,x0,tol,imax) [m,n]=size(A) iter=1 while iter <= imax for i=1:m su=0 for j=1:n if j~=i then su=su+A(i,j).*x0(j) end end x(i)=1 ./A(i,i) .* (b(i)-su) ...
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//Example_3 //Chapter 47 h=6.63*10^(-34)// in joule/sec v=4.39*10^(14)// cycles/sec E_o=h*(v) disp(E_o,"Energy in joule=")
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clear; clc; disp("--------------Example 19.2---------------") // a) 111.56.45.78 n=8; //number of bits i.e 1 byte a3=dec2bin(111,n); // convert decimal numbers to binary numbers a2=dec2bin(56,n); a1=dec2bin(45,n); a0=dec2bin(78,n); disp("a) Binary notation :- "+a3+" "+a2+" "+a1+" "+a0) //result i...
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clc; y=1.135; p1=10; pc=p1*[2/(y+1)]^[(y/(y-1))]; h1=2778; hc=2675; xc=0.962; vg=0.328; vc=xc*vg; Cc=(2*[h1-hc]*10^3)^0.5; A_m=vc/Cc*10^6; disp(A_m);
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//Chapter 4 //Example 4.1 //page 103 // to draw the per unit reactance diagram clear;clc; mvab=30; kvb=33; //MVA base and KVA base are selected gen1_mva=30; gen1_kv=10.5; gen1_x=1.6; //Generator No.1 details gen2_mva=15; gen2_kv=6.6; gen2_x=1.2; //Generator No.2 details gen3_mva=25; gen3_kv=6.6; gen3_x=0.56; //Gene...
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// Example 1.7 // Computation of eddy current loss if the apparatus is connected to a 60 Hz //source // Page No. 29 clc; clear; close; // Given data V=240; // Rated voltage F1=25; // Rated frequency Pe1=642; // Eddy current loss F2=60; // So...
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eval: A000027 1,2,3,4,5,6,7,8,9,10 eval: A001263 1,1,1,1,3,1,1,6,6,1 eval: A001263 1,1,1,1,3,1,1,6,6,1 eval: A001283 6,12,15,20,24,28,30,35,40,45 eval: A001614 1,2,4,5,7,9,10,12,14,16 eval: A003056 0,1,1,2,2,2,3,3,3,3 eval: A003506 1,2,2,3,6,3,4,12,12,4 A003506 result [[1],[-4/3,-4/3]] eval: A003989 1,1,1,1,2,1...
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Wt=712000;//total weight of plane including fuel (unit N) D=1.225;//density at sea level(Kg/m^3) S=153.29;//wing area in m^2 Clm=3;//maximum lift coefficient at subsonic speed
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<scriptConfig name="pvsim_comm_test" script="pvsim_comm"> <params> <param name="pvsim.terrasas.vmp" type="float">460.0</param> <param name="pvsim.terrasas.pmp" type="float">3000.0</param> <param name="pvsim.terrasas.channel" type="string">1</param> <param name="pvsim.terrasas.ipaddr" type="string">192...
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clear;lines(0); //create a simple m_file write(TMPDIR+'rot90.m',['function B = rot90(A,k)' '[m,n] = size(A);' 'if nargin == 1' ' k = 1;' 'else' ' k = rem(k,4);' ' if k < 0' ' k = k + 4;' ' end' 'end' 'if k == 1' ' A = A.'';' ' B = A(n:-1:1,:);' 'elseif k == 2' ' B = A(m:-1:1,n:...
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clc clear //input data b2=10//Rotor blade air angle at exit in degree Dt=0.6//The tip diameter in m Dh=0.3//The hub diameter in m N=960//The speed of the fan in rpm P=1//Power required by the fan in kW pi=0.245//The flow coefficient P1=1.02//The inlet pressure in bar T1=316//The inlet temperature in K R=287/...
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i=linspace(0,2.5,6) V=[0 50 84 105 120 131] plot(i,V) xtitle("Magnetization curve for example 18.3","Field Current","Generated emf") //refer Fig.18.6 in the textbook //OE is the field resistance line of critical resistance Rc=100 //solving (iii) Rsh=70//field resistance N=750//speed in rpm Nc=Rsh/Rc*N mp...
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pathname=get_absolute_file_path('1_5.sce') filename=pathname+filesep()+'1_5_data.sci' exec(filename) //fuel consumption(in kg/s) mf=bp/(nbth*CV) //air consumption(in m^3/s) A=(mf*Raf)/Da //air flow rate per cylinder(in m^3/s) A1=A/n //indicated power(in kW) Ip=bp/nm //indicated thermal efficiency nith=Ip/(m...
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clear; clc; //Example13.9[Radiation Heat Transfer in a Triangular Furnace] //Given:- A1=1,A2=1,A3=1;//Area of each side[m^2] T1=600,T2=1000;//[K] e=0.7; F12=0.5,F13=0.5,F23=0.5;//Symmetry //Solution:- Eb1=5.67*10^(-8)*(T1^4);//[W/m^2] Eb2=5.67*10^(-8)*(T2^4);//[W/m^2] Q=(Eb2-Eb1)/(((1-e)/(A1*e))+((((A1*F12...
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//Given that n = 1 //in mole T = 310 //in K Vi = 12 //in L Vf = 19 //in L R = .0821 //in atm.lit/(mol.K) //Sample Probelm 20-2 printf("**Sample Problem 20-2**\n") W = n*R*T*log(Vf/Vi) //in atm.lit printf("The work done by the gas is equal to %fJ", W*1.0125*10^5*10^-3)
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//each solution to the equation can be viewed as a combination of objects r=18; //number of objects M=3; //kinds of object m=factorial(r+(M-1))/(factorial(r+(M-1)-(M-1))*factorial(M-1)); disp(m,'number of non negative integer solutions of the given equation x+y+z=18')
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//Ex_2_16 clc; clear; close; format('v',5); //given data : LA=75/100;///m LB=25/100;///m lg=2/100;//m(airgap) mu_r1=1000;///relative permeability mu_r2=1500;///relative permeability mu0=4*%pi*10^-7;//permeability of air A=10*10^-4;//m^2//Area of core N=1000;//turns I=5;//A S=LA/(mu0*mu_r1*A)+LB/(mu0*mu_r...
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Pu=1500//in kN fck=15//in MPa fy=250//in MPa l=2.75//unsupported length, in m //assume 1% steel Ag=Pu*10^3/(0.4*fck*0.99+0.67*fy*0.01)//in sq mm L1=225//assuming a square column L2=Ag/L1//in mm L2=880//in mm Asc=0.01*L1*L2//in sq mm e1=l*10^3/500+L1/30//in mm e2=l*10^3/500+L2/30//in mm ep1=0.05*L1//<e1 ep2...
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//Efficinc of pump: Effp=0.8; //Design specific speed(in rpm): Nscu1=2000; //Impeller diameter(in inches): D1=8; //Opertion sped at esign point flow condition(in rpm): N1=1170; //Flow rate at design point flow condition(in gpm): Q1=300; //Density of water (in slug/ft^3): d1=1.94; //Acceleration due to gravi...
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// Element-wise division of threea vectors function vdiv3(V1, V2, V3) R = (V1 ./ V2) ./ V3; endfunction
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clc clear //input data P1=5000*10^3//The initial power produced in W H1=250//The initial head produced in m N1=210//The initial speed of turbine in rpm n0=0.85//Overall efficiency of the turbine H2=160//The final head produced in m d=1000//density of the water in kg/m^3 g=9.81//Acceleration due to gravity in ...
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// Example 6.12 clear all; clc; // Given data rdist = 25.4; // Distance between the rods in cm a = 1.02; // Radius of a rod in cm // From the Figure 6.9 b = rdist/sqrt(%pi); // Radius of equivalent cell in cm // Using the data from Table 5.2 L_F = 1.55;...
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clc //initialisation of variables w= 4000 //lb/ft l= 20 //ft y= 0.96 A= 4.18 //in^2 Icq= 5.6 //in^4 d= 28 //in b= 0.5 //in T= 8000 //psi d1= 0.75 //in //CALCULATIONS V= w*l/2 Ay= 2*A*((d/2)-y) I= b*d^3/12+4*(Icq+A*((d/2)-y)^2) p= (2*T*(%pi/4)*d1^2*I)/(V*Ay) //RESULTS printf ('Rivet spacing= %.2f in',p)...
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f_addNew = figure('dockable', 'off', 'infobar_visible', 'off', 'toolbar_visible', 'off', 'toolbar', 'none', 'menubar_visible', 'on', 'menubar', 'none', 'default_axes', 'off', ... 'layout', 'border', 'figure_name', gettext(prodName), 'visible', 'on','icon','jagannath.png','position',[230 40 850 630]); ...
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//Caption: Poisson distribution //Example6.4 //Page179 clear; clc; //(a): Exactly 0 customer will arrive in 10 minutes interval X1= 0; //no. of customer Mean = 4;//mean arrival rate of 4 per 10 minutes [P1,Q1]=cdfpoi("PQ",X1,Mean) disp(P1,'Exactly 0 customer will arrive P(X=0,4) is =') //(b): Exactly 2 custom...
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// find average LED current,power dissipation in series resistor // Electronic Principles // By Albert Malvino , David Bates // Seventh Edition // The McGraw-Hill Companies // Example 5-14, page 169 clear;clc; close; // Given data V=20;// ac source rms voltage in volts Rs=680;// series resistance in ohms // Calcula...
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S=54 P=6 m=S/3/P gammaa=%pi/3/m Kb1=sin(m*gammaa/2)/m/sin(gammaa/2) Kb3=sin(m*gammaa/2*3)/m/sin(gammaa/2*3) Kb5=sin(m*gammaa/2*5)/m/sin(gammaa/2*5) disp(Kb1) disp(Kb3) disp(Kb5)
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//Example 1.2 clc(); clear; //To calculate the wavelength //First case to calculte the wavelengths of the light source to obtain fringes 0.46*10^-2 mts lamda1=4200 //units in armstrongs lamda1=lamda1*10^-10 //units in mts betaa=0.64*10^-2 //units in mts D_d=betaa/lamda1 //units in mts //Second ca...
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//Network Theorem 1 //page no-2.25 //example2.17 disp("from the fig,"); disp("V1=-5*I1");....//equation 1 disp("V2=2*I2");....//equation 2 disp("Applying Kvl to mesh 1:"); disp("20*I1+3*I2=-5");....//equation 3 disp("Applying Kvl to mesh 2:"); disp("11*I1-3*I2=10");...//equation 4 disp("Solving equations 3 an...
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18 21:0.1111111111111111 70:0.25 91:0.25 104:0.3333333333333333 257:2.0 991:1.0 1100:1.0 1444:1.0 1556:1.0 18 7:0.6666666666666666 22:0.09090909090909091 45:0.1 70:0.25 91:0.25 107:0.14285714285714285 257:1.0 396:1.0 1051:1.0 1100:1.0 1827:1.0 18 7:0.6666666666666666 21:0.2222222222222222 22:0.09090909090909091 36:0.25...
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//subtraction using 1's complement// //Example 25.d// //subtraction in one's complement// clc //clears the window// clear //clears all the existing variables// x=85 y=32 c=bitcmp(y,7);//finding 1's complement// a=x+c+1 a=a-bin2dec('10000000') res=dec2bin(a,7) //binary conversion// disp('binary form of the...
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//Example 2.10 clear; clc; //Given R=8.314;// gas constant in J K^-1 mol^-1 t1=298;// temperature in K p1=30000;// initial pressure in N m^-2 p2=10000;// final pressure in N m^-2 Cv=20.8;// heat capacity of CO at constant volume in J K^-1 mol^-1 W=0.1;// weight of CO taken in kg // To determine t2,w,delH ...
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pathname=get_absolute_file_path('24_2.sce') filename=pathname+filesep()+'24_2data.sci' exec(filename) clear P=[2*L11 -2*L11 0;0 -L11 L11;2*A2 2*(A1+A2) 0],X=[L1;L2;440000]; q=inv(P)*X; //actual X is X=[L1;L2;-L2*L11],it leads to wrong answers than book; Sy1=-(q(2))*L11; Px4= 2*A1*-(q(2))/L11; Py2= Px4*tan(thet...
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//Calulate Total no. of created vacancies //Ex:6.5 clc; clear; close; r=1.7*10^-10;//atomic radius in m n1=10^-3;//1mm=10^-3m a=2*r;//in m n=n1/a; ed=2*10^-6;//edge dislocation in m ns=ed/a; nv=n*ns; disp(nv,"Total no. of created vacancies = ");
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clc //initialisation of variables p= 10 //bar P= 40 //percent x= 0.4 H1= 16 //kcal/kg H2= 31 //kcal/kg H3= 64 //kcal/kg H4= 140 //kcal/kg T= 157 //C He= 580 //kcal/kg //RESULTS printf (' Enthalpy= %.f kcal/kg',H1) printf (' \n Enthalpy= %.f kcal/kg',H2) printf (' \n Enthalpy= %.f kcal/kg',H3) printf (' \...
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function h=%p_v_s(p,s) //! // Copyright INRIA h=p/(1+p*s)
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//ques12 //Compressing a Substance in the Liquid versus Gas Phases clear clc // (a)steam as a saturated liquid v1=0.001043;//=vf(specific volume of fluids) @ 100kPa in m^3/kg P2=1000;//final pressure in kPa P1=100;//initial pressure in kPa w=integrate('v1*P^0','P',P1,P2);//work done in kJ/kg printf('(a) Work ...
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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_monito...
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// Scilab code Exa15.1 : : Page-652 (2011) clc; clear; N_0_235 = 1; // Number of uranium atom N_0_c = 10^5; // Number of graphite atoms per uranium atom sigma_a_235 = 698; // Absorption cross section for uranium, barns sigma_a_c = 0.003; // Absorption cross section for graphite, barns f = N_0_23...
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clc; clear; format('v',11) E=1; d=1*10^-3; A=(%pi*d^2)/4; //A=area of the cross-section. sigma=58.14*10^6; //conductivity of copper. J=sigma*E; disp(J,"The current density in copper(in A/m^2)="); I=J*A; disp(I,"The current in the wire I(in ampere)=");
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//Example 5_9 page no:199 clc function [r,th]=rect2pol(x,y) //rectangle to polar coordinate conversion r=sqrt(x^2+y^2); th=atan(y,x)*180/3.14; endfunction Z1=5+(%i*10) Z2=2-(%i*4) Z3=1+(%i*3) Zt=Z1+((Z2*Z3)/(Z2+Z3)) disp(Zt,"the equivalent impedence is(in ohm)")//imaginary term is rounded off [mag,theta]=re...
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//Example 1_6 clc; clear;close; //Given data Vdc=100;//V R=10;//ohm L=5;//H i=50*10^-3;//A //Solution : //i=Vdc/R*(1-exp(-R*t/L)) t=-log(1-i/Vdc*R)/R*L;//s disp("Minimum width of gate pulse is "+string(t*1000)+" milli-seconds.");
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function [port, configs, limit, sleep_time] = get_configs() // Description of get_configs() port = input("Seleccione el puerto: ") bauds = input("Indique los baudios: ") limit = input("Indique la duración de la sesión de medida (en minutos): ") * 60 sleep_time = input("Indique el intervalo de trazado: "...
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clc; us=0.25;// Coeffiecient of static friction //Applying equillibrium equation we get relation in x printf("Apply equillibrium equations. It is theoritical part. \n"); x=12-(.75*2)+1.5//in, Distance at which the applied load can be supported printf("Minimum distance at which the applied load can be supported is ...
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// File name: projects/02/FullAdder.tst load FullAdder.hdl, output-file FullAdder.out, compare-to FullAdder.cmp, output-list a%B3.1.3 b%B3.1.3 c%B3.1.3 sum%B3.1.3 carry%B3.1.3; set a 0, set b 0, set c 0, eval, output; set c 1, eval, output; set b 1, set c 0, eval, output; set c 1, eval, out...
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//Chapter2 //Page.No-45, Figure.No-2.9 //Example_2_1_a //Output voltage for open-loop differential amplifier //Given: clear;clc; vin1=5*10^-6;vin2=-7*10^-6; // Both input voltages are in volts A=200000; // Voltage gain vo=A*(vin1-vin2); //Output voltage in volts printf("\n Output voltage is vo = %.1f V dc \n"...