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// Scilab code Exa4.8.1: To calculate the energy liberated during fusion reaction: Page 194 (2011) // 5*H(1,2)= He(2,3)+He(2,4)+H(1,2)+2*n(0,1)+25MeV is the given reaction N = 6.023e+026/2*10; // Number of atoms in 10Kg of H-2, atoms E = 25/5*1.6e-013; // Energy liberate during fusion of 1 atom of H-2, J E_l = E*N...
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//Exa 3.6 clc; clear; close; format('v',6); //Given Data : T1=23+273;//K COP_HP=2.5; HeatLost=60000;//KJ/hr HeatGenerated=4000;//KJ/hr Q1=HeatLost-HeatGenerated;//KJ/hr W=Q1/COP_HP;//KJ/hr W=W/3600;//KJ/s or KW disp(W,"Power input in KW : ");
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errcatch(-1,"stop");mode(2);//Chapter 6, Problem 11 ; C=12*10^-6; //Equivalent capacitance C1=30*10^-6; //Capacitance of capacitor1 C2=(C*C1)/(C1-C); //Calculating capacitance of capacitor2 printf("Unknown capacitance = %f uF",C2*10^6); exit();
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//Chapter 3: Thermodynamic and Chemical Equilibrium //Problem: 10 clc; //Declaration of Constant R = 8.314 //in J / K //Declaration of Variables V_O2 = 2.8 //in litres V_H2 = 19.6 //in litres // Solution na = V_O2 / 22.4 //in mol nb = V_H2 / ...
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//Example 12.9 Q=0.120;//Flow rate (cm^3/s) Q=Q*(10^-2)^3;//Flow rate (m^3/s) r=0.150*10^-3;//Radius of needle (m) eta=1*10^-3;//Viscosity of saline solution(N.s/m^2) //Above information from Exercise 12.8 ////////////////////////////////////// A=%pi*r^2;//Cross-sectional area of needle v=Q/A;//Fluid speed (m/s...
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argc:7 Dataset: ../datasets/converted/football.net Nodes Edges Com Mod NMI Time seq async 115 1226 10 0.605159 -1 0.000211709 par async 115 1226 12 0.52036 -1 0.068867
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// Chapter 10 example 13 //------------------------------------------------------------------------------ clc; clear; //Given data PRF_1 = 750; // pulse rep. freq in Hz PRF_2 = 1000; // pulse rep. freq in Hz PRF_3 = 1250; // pulse rep. freq in Hz Ra_1 = 100; // Apparent range for PRF_...
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function m = ifht(d, varargin) //Calculate the inverse Fast Hartley Transform of real input D //Calling Sequence //m= ifht (d) //m= ifht (d,n) //m= ifht (d,n,dim) //Parameters //d: real or complex valued scalar or vector //n: Similar to the options of FFT function //dim: Similar to the options of FFT function //Descr...
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//Example 2.5 page 28 //Given the DSP system shown in Figures 2.16 to 2.18, where a sampling //rate of 16,000 Hz is used and the anti-aliasing filter is a second-order //Butterworth lowpass filter with a cutoff frequency of 3.4 kHz, determine //the percentage of aliasing level at the cutoff frequency. clc,clear,cl...
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//Exa:1.32 clc; clear; close; f=50;//in hertz V=440;//in volts P_o=110*1000;//in watts P=24;//No.Of Poles N_s=120*f/P;//Synchronous Speed (in rpm) N=245;//in rpm s_f=(N_s-N)/N_s;//Full load Speed T_f=P_o/(2*%pi*N/60);//Full load Torque (in N-m) R=0.04;//in ohms R2=R/2;//Rotor resistance per phase (in ohms)...
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//Chapter 9 Ionic Equilibria and Buffer Action clc; clear; //Initialisation of Variables k1= 3.5*10**-7 k2= 4.4*10**-11 //CALCULATIONS c= sqrt(k1*k2) //RESULTS mprintf("Concentration of solution = %.2e mol per litre",c)
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---------------------------------------------------------------------------------------- DROP TABLE tblDeep; SELECT '***** EXECUTING SP_RegrDataPrep *****'; EXEC SP_RegrDataPrep(NULL, 'tblAutoMpg', 'ObsID', 'MPG', 'tblDeep', true, false, false, false,...
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//Ex17.7 I_max=700*10^-3; //in Amperes R_ext=0.7/I_max; disp(R_ext,'value of resistor in Ohms for which max current is 700mA')
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clear; clc; V_s=450; R=10; disp("for 180deg mode"); I_or=sqrt((V_s/(3*R))^2*2/3+(2*V_s/(3*R))^2*1/3); printf("rms value of load current=%.3f A",I_or); I_T1=sqrt((1/(2*%pi))*((V_s/(3*R))^2*2*%pi/3+(2*V_s/(3*R))^2*%pi/3)); printf("\nrms value of load current=%.0f A",I_T1); P=3*I_or^2*R; printf("\npower de...
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//clear// //Example 3.2:CTFS of a periodic signal x(t) //Expression of continuous time signal //using continuous time fourier series clear; close; clc; t = -3:0.01:3; //t1 = -%pi*4:(%pi*4)/100:%pi*4; //t2 =-%pi*6:(%pi*6)/100:%pi*6; xot = ones(1,length(t)); x1t = (1/2)*cos(%pi*2*t); xot_x1t = xot+x1t; x2t =...
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clear; clc; //Example 1.9 //circuit and diode parameters V_PS=5;//(V) R=5;//(KOhm) V_Y=0.6;//(V) V_T=0.026;//(Volt)terminal voltage v_i=0.1//*sin(wt)Volt //dc analysis I_DQ=(V_PS-V_Y)/R; printf('\ndc quiescent current=%f mA\n',I_DQ) V_O=I_DQ*R; printf('\ndc output voltage=%f V\n',V_O) //ac analysis V_P...
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java -ea trip.Main -m trip-tests/map05 trip-tests/request05
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clc; disp("Principles of Heat Transfer, 7th Ed. Frank Kreith et. al Chapter - 7 Example # 7.3 ") //Velocity of air in m/s Uinfinity = 0.5; //Length and breadth of square shaped array in m L = 2.5; //Surface temperature...
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//ques-34.7 //Calculating concentration of compound X in a solution clc EC=245;//molar extinction coefficient (in m^2/mol) Io=100;//original intensity percentage l=0.01;//length (in m) reduction=25;//percentage reduction in intensity I=Io-reduction; C=log10(Io/I)/(EC*l); printf("The concentratino o fcompound X...
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 7.1w //calculation of the maximum speed the car can take on the turn without skidding //given data R=45//radius(in m) of the turn mus=2.0//coefficient of static friction between the tyre and the road g=10//gravitatio...
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clc V=200; //Assigning values to parameters Z1=5*%i; Z2=5+%i*8.66; Z3=15; Z4=-10*%i; Y1=1/Z1; Y2=1/Z2; Y3=1/Z3; Y4=1/Z4; Yeq=Y1+Y2+Y3+Y4; Zeq=1/Yeq; I=V/Zeq; disp("Amperes",I,polar(I),"Total current");
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#************************************************************ # Scenario of humanTestEnv # # date : Thu Oct 16 14:15:52 2014 #************************************************************ p3d_sel_desc_name P3D_ENV humanTestEnv p3d_sel_desc_name P3D_ROBOT HERAKLES_HUMAN p3d_set_robot_steering_method Linear p3d_set_...
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//Example 18.1 //Extrusion Process //Page No. 629 clc;clear;close; Db=6; //in inches Df=2; //in inches L=15; //in inches v=2; //in inches/s alpha=60; //in degrees mu=0.1; //no unit R=Db^2/Df^2; ...
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//polynomial function[]=pol(n) x=poly([n^2,n+2,n^3+1],'x','c') disp(x) endfunction
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//Example 2.13 clear; clc; //Given R=8.314;// gas constant in J K^-1 mol^-1 p1=10;// initial pressure in atm p2=1;// final pressure in atm T=300;// temperature in K // To determine Wexp and Wcomp Wexp=R*T*(1-(p2/p1));//work done in J mol^-1 during expansion Wcomp=(-1)*R*T*(1-(p1/p2));//work done in J mol^...
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// scilab Code Exa 18.54 Maximum Height of Hydro Turbines H=52; // head developed in m c1=6.5; // exit velocity in m/s pa=1.0133; // ambient pressure in bar rho=1e3; n_d=0.75; // Efficiency of the Draft Tube g=9.81; // Gravitational acceleration in m/s^2 ta=20; // ambient temperature of water in degree C sigm...
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//Example 12.11 //Program to determine the average incident optical power required at //the receiver clear; clc ; close ; //Given data k=1.38*10^(-23); //J/K - BOLTZMANN's CONSTANT e=1.602*10^(-19); //Coulumbs - CHARGE OF AN ELECTRON SNR_dB=55; //dB - SIGNAL POWER TO RMS NOIS...
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// Author : Pierre Vuillemin (2017) // License : BSD function target = slr_fill_fields(target, ref, check_fields, tests) // fills a structure based on a reference structure. // // Syntax // out_stru = slr_fill_fields(in_stru, ref_stru) // out_stru = slr_fill_fields(in_stru, ref_stru, check_fields)...
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clear all; clc; disp("Scilab Code Ex 12.22 : ") //Given: l = 8; //m l_ab = l/2; //m l_bc = l/2; //m l_af = l_ab/2; //m b = 12/1000; //m w = 24; //kN/m E = 200*10^6; //Kn/m^2 I = 80*10^-6;// m^4 //Compatibility Equation: nu_b = (5*w*l^4)/(768*E*I); //nu_b = (5wl^4)/768EI nu_b_byBy = (l^3)/(48*E*I); //...
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// generated by builder.sce: Please do not edit this file // ------------------------------------------------------ libpremia_path=file('join',['.','../../lib/libpremia.dll']); libpnl_path=file('join',['.','../../lib/libpnl.dll']); link(libpnl_path); link(libpremia_path); libpremiamodel_path=file('join',['.','libpremi...
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// Exa Misc. 6.10 clc; clear; close; format('v',6) // Given data R_F = 60;// in ko hm R1 = 20;// in k ohm Vin1 = 2;// in V Vin2 = 0.1;// in V // The output voltage, by using super position theorm, Vo = ((-R_F/R1)*Vin1) + ((1+(R_F/R1))*Vin2);// in V disp(Vo,"The output voltage in V is");
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//Example 17.3 clc delta_v=120 I=6.4 R=(delta_v)/I disp(R,"The resistance in ohm=")
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//Exa 4.1 clc; clear; close; //Given data : f=50;//Hz d=1*100;//cm r=1.25/2;//cm r_dash=r*0.7788;//cm L=0.4*log(d/r_dash);//mH disp(L,"Loop inductance per km(mH)"); XL=2*%pi*f*L*10^-3;//ohm/Km disp(XL,"Reactance of transmission line(ohm/km)");
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clc clear printf("example 8.8 page number 368\n\n") //to find the coefficient of discharge for converging cone pipe_diameter=0.15; //in m venturi_diameter=0.05; //in m pressure_drop=0.12; //m of water flow_rate=3; //in kg/s density = 1000; //in kg/m3 v...
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//to calculate stator current,pf, net mech o/p, torque, motor performance clc; j=sqrt(-1); Vt=400; P=6; f=50; Inl=7.5; Pnl=700; disp('block rotor test results'); Vbr=150; Ibr=35; Pinbr=4000; R1=.55;disp(R1,'R1(ohm)'); k=1/.5; s=0.04; Zbr=Vbr/(sqrt(3)*Ibr); Rbr=Pinbr/(3*Ibr^2); Xbr=sqrt(Zbr^2-Rbr^2); ...
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before x, phead=0, pbody= - 14*y, ptail= - 448*y^2 - 576, mlead= + 4*x^2, flead=256, root2=16, widev=64 after x, phead=2*x^2, pbody= - 14*y, ptail= - 840*y^2 - 576, vmapt={x=> - 14*y+16*x,y=>y} zero y, pbody == 0, fbody=-840 after y, phead=2*x^2 - 840*y^2, pbody=0, ptail= - 576, vmapt={x=> - 14*y+16*x,y=>y} ...
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//example4.12 clc disp("f_H=1/(2*pi*R_eq*C_eq)") disp("and f_H''=2(f_H)") disp("1/(2*pi*R_eq*C_eq) = 2/(2*pi*R_eq*C_eq)") disp("R_eq'' = R_eq/2") disp("R_eq=(r_b''e)parallel to (r_bb''+R_s)") disp("= (r_b''e)=1000 ohm") disp("Therefore R_eq'' =500 ohm") disp("Therefore 500=((r_b''e)*(r_bb''+R_s))/((r_b''e)+(...
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clc // Given that lambda = 6000 // wavelength of first source in angstrom d = 2 // Spacing between sources in mm D = 0.1 // Distance between source and screen in meter t = 0.5 // Thickness of plate in mm shift = 5 // Shift of fringe in mm // Sample Problem 7 on page no. 95 printf("\n # PROBLEM 7 # \n") pr...
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13_4.sce
//chapter 13 //example 13.4 //page 410 clear all; clc ; //given Vi=50;//ip voltage in mV Vo=2;//op voltage volts Ibmax=500;//maximum base current nA I2=(100*Ibmax)/1000; printf("\noutput current(I2)=%d microA",I2); R3=Vi/I2; printf("\nR3=%d kohm,standard value",R3); R23=1000*Vo/I2; R2=R23-R3; R1=(R2*R3)/...
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35.sce
clc // Given that lambda = 5000 // wavelength of light in angstrom theta = 1e-3 // Separation between stars in degree // Sample Problem 35 on page no. 175 printf("\n # PROBLEM 35 # \n") printf(" Standard formula used \n") printf(" theta = 1.22*lambda/a \n") a = 1.22*lambda*1e-10/(theta*%pi/180) // Calculation...
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ex19_2.sce
clc;clear; //Example 19.2 //calculation of current //given values e=1.6*10^-19; kT=.026*e;//temp eqvlnt at room temp Io=2*10^-7;//current flowing at room temp in A V=.1;//forward bias voltage in volts //calculation I=Io*(%e^(e*V/kT)-1);//in Ampere disp(I*10^6,'current flowing when forward bias applied(in ...
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Ex16_7.sce
//Example 16_7 page no:758 clc; //given Z11x=3; Z12x=2; Z21x=2; Z22x=3; Z11y=15; Z21y=5; Z22y=25; Z12y=5; //calculating the parameters Z11=Z11x+Z11y; Z12=Z12x+Z12y; Z21=Z21x+Z21y; Z22=Z22x+Z22y; disp(Z11,"the parameter Z11 is (in ohm)"); disp(Z21,"the parameter Z21 is (in ohm)"); disp(Z22,"the paramet...
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// Estimate fracture strength clc c = 2 // crack of half length in micro meter Y = 70 // young’s modulus in GN m^-2 Gamma = 1 // specific surface energy printf("\n Example 12.1") sigma_f = sqrt(2*Gamma*Y*1e9/(%pi*c*1e-6))/1e6 r = Y*1e3/sigma_f printf("\n Fracture strength of %d MN m^-2 is 1/%dth of young’s mod...
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week-4-q1.sce
clear; clf; // constant variables dt1 = 0.02; // seconds dt2 = dt1 * 10; // seconds dt3 = dt1 * 100; // seconds function dMsdt = slopes(t, Ms) if t <= 5 K = [0.55, 0.5, 0.4, 0.1, 0.5, 0.55, 0.01]; // 1/s else K = [0.3, 0.5, 0.4, 0.1, 0.5, 0.3, 0.01]; end dMsdt(1) = K(7) * Ms(2) + K(2) * Ms(4) - K(1)...
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EX3_7_1.sce
//EXA:3.7.1 clc; clear; close; l=10*10^(-2) //circumference of ring(in m) A=20*10^(-4) //cross-sectional area(in m^2) u_r=500 //permeability of iron u_0=4*%pi*10^(-7) B=0.8 //flux density(in T) N=100 //no. of turns R=l/(u_0*u_r*A) phy=B*A mmf=phy*R I_e=mmf/N //exciting current disp(I_e,'Exciting current (...
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Ex1_37.sce
errcatch(-1,"stop");mode(2);//Example 1_37 ; ; //To find the separation between the slits lemda=5100 //units in angstroam lemda=5100*10^-8 //units in cm D=200 //units in cm betaa=0.01 //units in mts betaa=0.01*10^-3 //units in cm d=...
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moduloyurt.sce
function out=moduloluyurt(x,m) l=length(x); n=1; for i=1:1:l if modulo(x(i),e)==0 v(n)=x(i); n=n+1; end end out=v; endfunction
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5_6.sce
clc //initialisation of variables r=48//in A=450//gpd/acre B=8000//gpd/mile S=5280/350//manholes/mile //CALCULATIONS C=(B-S*100)/12//gpd/mile //RESULTS printf('the ground a quarter of it eventually =% f gpd/mile',C)
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TP.sce
clear; clc(); xdel(winsid()); //----------------------------------------------------------------------------// //Paramètres------------------------------------------------------------------// //----------------------------------------------------------------------------// m = 0.9; f0 = 30; w0 = 2*%pi*f0; K0 = 0.8; p =...
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Ex3_2.sce
//Example 3.2 //This code utilises dependency file 'Compute_Angle.sci' exec Compute_Angle.sci;//Execute dependency file //To find the position of the boat subplot(2,1,1) //To draw vector A x=0; y=0;//x and y are the starting point coordinates magnitude=27.5; theta=66; x_end=magnitude*cosd(theta); y_end=...
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hamming.sce
disp("Enter block length (n):"); n = input(""); k = 4; //Number of parity bits m = n-k; I = eye(k,k); //coefficient matrix P = zeros(k,m); disp("Enter elements of coefficient matrix P"); for i=1:k for j=1:m P(i,j)=input(""); end end disp(P, "Coefficient matrix P") // generator mat...
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solution12_3.sce
//Obtain path of solution file path = get_absolute_file_path('solution12_3.sce') //Obtain path of data file datapath = path + filesep() + 'data12_3.sci' //Clear all clc //Execute the data file exec(datapath) //Calculate the KE of the mass KEmass (J) KEmass = 0.5 * m * (v1^2 - v2^2) //Calculate the initial velocity of ...
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sce
16_12.sce
clear; clc; R1=500;R2=110; Ro=sqrt((2*2*R1*R1/4)+(2*R1*R2)); a=acosh(1+(2*R1/(2*R2))); Ri1=Ro;Ri2=150;Ai=round(a*10)/10; R3=(sqrt(Ri1*Ri2))/(fix(sinh(Ai)*100)/100); R2=(Ri2/(round(tanh(Ai)*10^4)/10^4))-(fix(R3*100)/100); R1=(Ri1/(round(tanh(Ai)*10^4)/10^4))-R3; printf("The desired elements of the attenuator ar...
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/1370/CH5/EX5.2/exp5_2.sce
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exp5_2.sce
//Example 5.2 clc disp("Given values are,") disp("P = 4, f = 50 Hz, %s = 4%") disp("s = Full load absolute slip = 0.04") ns=(120*50)/4 disp(ns,"N_s(in r.p.m) = 120*f / P =") disp("s = N_s-N / N_s where N_s = full load speed of motor") disp("0.04 = 1500-N_s / 1500") ns=1500-(1500*0.04) disp(ns,"Therefore, ...
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Chapter4_Example7.sce
//Chapter-4, Illustration 7, Page 170 //Title: Steam Nozzles and Steam Turbines //============================================================================= clc clear //INPUT DATA P1=10;//Pressure at point 1 in bar P2=0.5;//Pressure at point 2 in bar h1=3050;//Enthalpy at point 1 in kJ/kg h2s=2480;//Entha...
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116ex3.sce
clear; clc; close; x=poly(0,'x'); p1=x-1; p2=x^2-x-2; p=p1/p2; q1=x+2; q2=x^2+4*x+3; q=q1/q2; t=p-q; y=numer(t) //numerator of t z=factors(denom(t))//factors of denominator of t (more simplified form) disp("val=(1+2x)/(1+x)(-2+x)(3+x)")
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ch9_8.sce
//To determine the ratio of sheath loss to core loss of the cable clear clc; R=2*.1625; Rs=2*2.14; M=314; w=6.268*10^-4; r=Rs*M*M*w*w/(R*((Rs^2)+(M*M*w*w))); mprintf("ratio=%.4f \n",r);
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Ex1_6.sce
//Chapter 1: Structure and Bonding //Problem: 6 clc; //Declaration of Constant h = 6.626 * 10 ** -34 // Plank's constant,in J.sec // Variable lamda = 2 * 10 ** -10 // wavelength,in m // Solution p = h / lamda mprintf("The momentum of the particle is :%.2e kg m/s",p)
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//Chapter-1,Example1_14_15,pg 1-64 ro=2.7*10^3 //density of metal A=27 //atomic wt of metal N=6.023*10^26 //Avogadro's number a=4.05*10^-10 //lattice constant of metal n=(N*(a^3)*ro)/A pr...
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//========================================================================== //generates static table definitions // //Author : Rachid Djenidi, Alan Layec //Copyright INRIA // Modified for RT purposes by Roberto Bucher - RTAI Team // roberto.bucher@supsi.ch function txt=make_static_standalone42() txt=['']; //...
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x=1 // a real number s='2' //a string x+s // error since %s_a_c is undefined
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// // "vib-bode.sce" // clear; clf(); m=1; c=0.2; k=1; P=1; zeta = c/( 2*sqrt(m*k) ); //表3.1 omn = sqrt(k/m); //表3.1 function y = K(Om) global zeta; y=1/sqrt((1-Om^2)^2 +(2*zeta*Om)^2); endfunction function y = phase_diff(Om) global zeta; y=-atan((2*zeta*Om),(1-Om^2)); //atan2(y,x) endfunction om = lin...
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// Grob's Basic Electronics 11e // Chapter No. 19 // Example No. 19_7 clc; clear; // A coil L1 produces 80 uWb of magnetic flux. Of this total flux, 60 uWb arelinked with L2. How much is k between L1 and L2? // Given data lf1 = 80*10^-6; // Magnetic flux of coil L1=80 uWb lf2 = 60*10^-6; // Magnetic flux o...
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//Example 3.14. clc format(5) D=(((2*10^-2)*(1*10^-4)*(20*10^-2))/sqrt(800))*sqrt((1.6*10^-19)/(2*9.1*10^-31)) x1=D*10^2 disp("The deflection of the spot,") disp(x1,"D(cm) = (IBL/sqrt(Va))*sqrt(q/2m) =")
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// Ex5_7 clc; // Given: Q1=9.28;// in Mev Q2=0.21;// in Mev Q3=7.25;// in Mev Q4=3.63;// in Mev mn=1.008665; md=1.995311;// mass difference between Fe(56) & Fe(54) // Solution: E1=Q1+Q2+Q3+Q4;// part (a) E2=(2*mn-md)*931;// part (b) printf("The binding energy of last 2 neutron in part(a) and part(b)...
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//Chapter 13 //Example 13_12 //Page 323 clear;clc; Va=440; Ic=100; Id=200; Ie=250; If=300; Vb=430; Lac=150; Lcd=150; Lde=50; Lef=100; Lfb=150; r=0.01; l=600; //resistance for 100 m length of conductor R=2*r; Rac=R*Lac/100; Rcd=R*Lcd/100; Rde=R*Lde/100; Ref=R*Lef/100; Rfb=R*Lfb/100; //considering drop across variou...
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//Load required at point D //Refer fig. 2.22 (a),(b) and (c) //Using simple geometry we have alpha=acosd(1.3125/1.5) //degree Beta=acosd(2-1.3125) //degree //Applying Lami's Theorem at point C T1=1500*sind(90)/sind(180-alpha) //N T2=1500*sind(90+alpha)/sind(180-alpha) //N //Applying Lami's Theorem at point ...
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ScreenName String 'InGameMenu_Options_Control' ImplName String 'InGameMenu Screen' ElementChunkArray Int 6 ScreenElementType Int 0 ImplName String 'In Game Backdrop' TabIndex Int 1 Selectable Bool False Enabled Bool True ReferenceArea Rect( 53, 46, 747, 550 ) # left,top,right,bottom ScreenElementType Int 1 ImplName St...
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clear // // // //Variable declaration h=6.62*10^-34; //planck's constant(J-sec) m=9.1*10^-31; //mass of electron(kg) mp=1836*m; //mass of photon(kg) c=3*10^8; //velocity of light(m/sec) e=1.6*10^-19; //charge of electron(c) //Calculations E=m*c^2; //energy(J) v=sqrt...
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# Depeopanizer test (from old Hysim manual) units Field $thermo = VirtualMaterials.Peng-Robinson / -> $thermo thermo + Methane Ethane PROPANE thermo + ISOBUTANE n-BUTANE ISOPENTANE n-PENTANE n-Hexane thermo + n-Heptane n-Octane deprop = Tower.Tower() deprop.Stage_0 + 18 # twenty stages` cd deprop.Stage_0 v = Tower....
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//EXA:3.2.2 clc; clear; close; B=0.5 //magnetic field (in wb/m^2) S=.04 // area of square loop(in m^2) theta_1=0 //inclination flux_1=B*S*cosd(theta_1) disp(flux_1,'flux_1 (in wb)=') theta_2=60 flux_2=B*S*cosd(theta_2) disp(flux_2,'flux_2 (in wb)=') theta_3=90 flux_3=B*S*cosd(theta_3) disp(flux_3,'flux_3 ...
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clc horsepower=2.5 //rating of induction motor in horsepower at half load Vl=230 //terminal voltage of motor in volts Il=7 //load current of motor in amperes pf=0.8 //power factor of the machine Pin=sqrt(3)*Vl*Il*pf //input power in watts mprintf("Pin=%f W\n",Pin)//The answer may vary due to roundoff error ...
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//Exa 5.11 clc; clear; close; //Given data : A=800;//gain(unitless) f1=40;//in Hz f2=16;//in kHz Beta=2/100;//feedback fator (unitless) //Formula : Af=A/(1+A*Beta) Af=A/(1+A*Beta);//gain with feedback(unitless) disp(Af,"Voltage gin with feedback : "); BW=f2*10^3-f1;//Bandwidth of amplifier in Hz disp(BW*10...
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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/02/Inc16.tst load Inc16.hdl, output-file Inc16.out, compare-to Inc16.cmp, output-list in%B1.16.1 out%B1.16.1; set in %B0000000000000000, // in = 0 eva...
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clc clear //INPUT DATA p1=101;//Pressure in kPa t1=293;//temperature in K v12=7;//compression ratio Qs=1000;//heat added in kJ Ra=0.287;//gas constant m=1;//mass of air in kg/min cv=0.7243;//calorific value //CALCULATIONS v1=(Ra*t1)/p1;//specific volume in m^3/kg v2=v1/(v12);//specific volume in m^3/kg ...
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// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 2 : BESICS OF MAGNETIC CIRCUITS // Example : 2.12 clc;clear; // clears the console and command history // Given data l = 25*10^-2 // length of air-core coil in m A = 4*10^-4 // cross section...
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//Exa 6.7 clc; clear; close; format('v',8); //Given Data : mw=1;//Kg m_steam=39;//mass of dry steam in Kg ms=mw+m_steam;//Kg x=m_steam/ms;//dryness fraction disp(x,"Dryness fraction ; ");
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function y = g(x) if x < 2 then y = 2*x-3 else y = x-3+2 end endfunction plot(-5:5, g)
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//Problem 2.17: An electric heater consumes 3.6 MJ when connected to a 250 V supply for 40 minutes. Find the power rating of the heater and the current taken from the supply. //initializing the variables: V = 250; // in Volts E = 3.6E6; // energy in J t = 2400; // in sec //calculation: P = E/t I = P/V pri...
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clc //initialisation of variables k= 0.1200 t= 225 //sec d= 0.82 //CALCULATIONS v= t*k u= v*d //RESULTS printf ('kinematic viscosity = %.1f cP',u)
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//system specifications f_samp = 2000*10^6; fc = 800*10^6; Ts = 1/f_samp; n = linspace(1,2000,2000); n_points = 10000; I = 2*grand(n_points, 1, "bin", 1, 0.5) - 1; Q = 2*grand(n_points, 1, "bin", 1, 0.5) - 1; A = 0.001 R = 100*10^(-3) PLdB = 128.1 + 37.6*log10(R); PL = 10^(-PLdB/20); //fading HR = ...
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//Strength Of Material By G.H.Ryder //Chapter 2 //Example 1 //To Calculate diameter of Bolts clc(); //Initialization of Variables P=250; //Power transmitted by coupling, Unit in KW N=1000; //Rotations ,Unit in rpm n=6; //number fo bolts PCD=14; //Pitch circle diameter, Unit in cm ShearStress=75; //All...
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//Exa 5.3 clc; clear; close; //Evaluation at i=8% disp("Evaluation at i=8%"); //Alternative 1 : disp("Alternative 1 : "); P=500000;//in Rs A1=50000;//in Rs G=50000;//in Rs i=8;//in % per annum n=6;//in years //Formula : (F/P,i,n) : (1+i/100)^n //Formula : (F/A,i,n) : (((1+i/100)^n)-1)/(i/100) //Formula :...
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clear; clc; //Delta Function N=10; t1=-10:10; x1=[zeros(1,N), ones(1,1),zeros(1,N)]; plot(t1,x1); xlabel('Time'); ylabel('Delta function'); figure; //Unit Step Function N=100; t=-N:0.01:N; u=ones(t).*(t>=0); plot(t,u); set( gca(), "data_bounds",matrix([-100,100,-0.1,1.1],2,-1) ) xlabel("Time"); ylabe...
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[Bus1] EvtPluginINIFilePath = "RingdownEvtPlugin/RingdownEvtPlugin.ini" EvtParams.<size(s)> = "11 10" EvtParams 0 = "0" EvtParams 1 = "346.500E-3" EvtParams 2 = "696.700E-3" EvtParams 3 = "1.297E+0" EvtParams 4 = "1.457E+0" EvtParams 5 = "6.061E+0" EvtParams 6 = "7.229E+0" EvtParams 7 = "8.951E+0" EvtParams 8 = "11.705...
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410825 ( 250872 ( 498322 ( 673462 ( 276456 ( 698093 ( 344763 ( 254830 ( 808155 ( 0 ( ) ( ) ) ( 808914 ( ) ( ) ) ) ( 866745 ( 244865 ( ) ( ) ) ( 960740 ( ) ( ) ) ) ) ( 102613 ( 734874 ( 819695 ( ) ( ) ) ( 581100 ( ) ( ) ) ) ( 499849 ( 128381 ( ) ( ) ) ( 724909 ( ) ( ) ) ) ) ) ( 776348 ( 310022 ( 681415 ( 536908 ( ) ( ) ...
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// Calculating the heat conducted across the former from winding to core clc; disp('Example 4.11, Page No. = 4.17') // Given Data t = 2.5;// Thickness of former (in mm) t_air = 1;// Thickness of air space (in mm) lw = 150*250;// The inner dimentions of the former of field coil (in mm square) h = 200;// Winding h...
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load Mux4Way16.hdl, output-file Mux4Way16.out, compare-to Mux4Way16.cmp, output-list w%B1.16.1 x%B1.16.1 y%B1.16.1 z%B1.16.1 sel%B2.2.2 out%B1.16.1; set w 0, set x 0, set y 0, set z 0, set sel 0, eval, output; set sel 1, eval, output; set sel 2, eval, output; set sel 3, eval, output; set w %B0001001000110100, set ...
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//Chapter-4,Example4_15_9,pg 4-30 S1=220 //wall area a1=0.03 //absorption coefficient for the wall S2=120 //floor area a2=0.8 //absorption coefficient for the floor S3=120 ...
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clc T3=278; //K T2=350; //K T4=T2; T1=1350; //K Q1=100/[((T4/T1)*(T1-T2)/(T4-T3))+T2/T1]; //Q4+Q2=100; Q4=Q1*((T4/T1)*(T1-T2)/(T4-T3)); Q2=T2/T1*Q1; disp("Q1=") disp(Q1) disp("kJ")
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clear; clc; //Example - 7.4 //Page number - 240 printf("Example - 7.4 and Page number - 240\n\n"); //This problem involves proving a relation in which no numerical components are involved. //For prove refer to this example 7.4 on page number 240 of the book. printf(" This problem involves proving a relation ...
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clear;lines(0); A=rand(3,3); [al,be,Z] = gspec(A,eye(A));al./be clean(inv(Z)*A*Z) //displaying the eigenvalues (generic matrix) A=A+%i*rand(A);E=rand(A); roots(det(%s*E-A)) //complex case
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//Example_a_5_5 page no:204 clc; Vmag=100; Vang=0; f=50; L1=3+(%i*31.41); L2=5-(31.83*%i); L3=10+(%i*150.73); R1=3; R2=5; R3=10; L1mag=sqrt(real(L1)^2+imag(L1)^2); L1ang=atand(imag(L1)/real(L1)); I1mag=Vmag/L1mag; I1ang=Vang-L1ang; I1real=I1mag*cosd(I1ang); I1img=I1mag*sind(I1ang)*%i; I1=I1real+I1img; ...
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errcatch(-1,"stop");mode(2);//Example 4.21 //Program to Compute the DFT of given Sequence //x[n]=[1,-1,-1,-1,1,1,1,-1] using DIT-FFT Algorithm. ; ; ; x = [1,-1,-1,-1,1,1,1,-1]; //FFT Computation X = fft (x , -1); disp(X,'X(z) = '); exit();
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function [xm,fr]=frmag(num,den,npts) //[xm,fr]=frmag(num[,den],npts) //Calculates the magnitude of the frequency respones of //FIR and IIR filters. The filter description can be //one or two vectors of coefficients, one or two polynomials, //or a rational polynomial. //Case 1 (When den is not given): // num :Vector ...
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clc; clear; printf("\t\t\tChapter9_example3\n\n\n"); // Determination of the outlet temperature of the ethylene glycol for counterflow. // properties of air at (195 + 85)/2 = 140°F. from appendix table CII rou_1= 0.985*62.4; // density in lbm/ft^3 cp_1=0.9994; // specific heat BTU/(lbm-degree Rankine) v_1= 0.514e...
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// Exa 2.21.10 clc; clear; close; // Given data Rho = 9.6 * 10^-2;// in ohm-m Sigma_n = 1/Rho;// in (ohm-m)^-1 q = 1.6 * 10^-19;// in C Mu_n = 1300 * 10^-4;// in m^2/v-s N_D = Sigma_n / (Mu_n * q);// in atoms/m^3 A_D = N_D;// Atom density in atoms/cm^3 A_D = A_D * 10^6;// atoms/m^3 R_si = N_D/A_D;// ratio ...
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clc // initialization of variables clear // For 30 mm crack a=30/2 // mm crack S_30 =600 //MPa a=a*10^-3 C=S_30*sqrt(a) // For 120 mm crack a=120/2 a=a*10^-3 S_120=C/sqrt(a) printf('Sigma_120 = %d MPa',S_120)