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x y f(xs) 0.394525 0.015636 0.384614 0.725423 0.883720 1.367414 0.214386 0.109000 0.224628 0.357188 0.921254 1.100068 0.160378 0.451832 0.362428 0.285724 0.722295 0.780215 0.061012 0.528430 0.336597 0.300466 0.693221 0.758237 0.436095 0.264486 0.492299 0.737517 0.175565 0.703270 0.587999 0.919851 1.303415 ...
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clear; clc; disp("--------------Example 23.4----------------") buffer_size=5000; //bytes recieved_unprocessed = 1000; // bytes rwnd=buffer_size-recieved_unprocessed ; // formula printf("The value of rwnd = %d . Hence Host B can receive only %d bytes of data before overflowing its buffer.",rwnd,rwnd); // display r...
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clc clear //Inputs //The Values in the program are as follows: //Temperature in Celcius converted to Kelvin(by adding 273) //Pressure in bar converted to kPa (by multiplying 100) //Volume in m^3 //Value of R,Cp and Cv in kJ/kg Km=1; P1=6; V1=0.01; V2=0.05; P2=2; W1=(((P1+P2)/2)*100)*(V2-V1); printf('The ...
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curFig = scf(100001); clf(curFig,"reset"); demo_viewCode("membrane.sce"); drawlater(); xselect(); //raise the graphic window // set a new colormap //------------------- cmap= curFig.color_map; //preserve old setting curFig.color_map = jetcolormap(64); //The initial surface definition //--------------...
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//Example 7_7 clc(); clear; //To calculate how large a horizontal force must the pavement exert m=1200 //units in Kg v=8 //units in meters/sec r=9 //units in meters F=(m*v^2)/r //units in Newtons printf("The horizontal force must the pavement exerts is F=%d Newtons",F) //In text book the answer is...
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//Tested on Windows 7 Ultimate 32-bit //Chapter 14 Operational Amplifiers Pg no. 435 and 436 clear; clc; //Given //Figure 14.21 R=12D3;//resistances R1,R2,R3 in RC network in ohms C=0.001D-6;//capacitances C1,C2,C3 in RC network in ohms A=29;//gain for oscillator operation //Solution fr=1/(2*%pi*R*C*s...
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function[fx] = fungsiNormal(mu,sigma,x) konst = 1 / (sigma * sqrt(2 * %pi)) fx = konst * exp(-0.5 * (((x-mu)/sigma)^2)) endfunction function[P_Normal] = PNormal(mu,sigma,a,b) n = 1000 h = (b-a) / n fa = fungsiNormal(mu,sigma,a) fb = fungsiNormal(mu,sigma,b) jum = 0 for i = 1 : (n-1) a = a + h fa1n = fungsiNor...
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// To determine which of the circuits 7-1(a) or 7-2(b) has greater accuracy // example 7-3 in page 166 clc; //Data given V1=495; I1=0.5; // voltmeter and ammeter reading in volt and ampere respectively of circuit 7-1(a) V2=500; I2=0.5;// voltmeter and ammeter reading in volt and ampere respectively of circuit 7-1(...
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function [x, y, z]=sph2cart(theta, phi, r) x = r.*(cos(phi).*sin(theta)); y = r.*(sin(phi).*sin(theta)); z = r.*cos(theta); endfunction
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@relation coil2000 @attribute MOSTYPE integer[1,41] @attribute MAANTHUI integer[1,10] @attribute MGEMOMV integer[1,6] @attribute MGEMLEEF integer[1,6] @attribute MOSHOOFD integer[1,10] @attribute MGODRK integer[0,9] @attribute MGODPR integer[0,9] @attribute MGODOV integer[0,5] @attribute MGODGE integer[0,9] @attribute ...
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//example 6.4 clc; funcprot(0); // Initialization of Variable Vd=28;//V f=100;//frequency I=50;//current //calculation Rl=(Vd-.3)/I; disp(Rl*1000,"load resistance in ohm:") printf('thus pick Rl=560ohm') Rl=560; Vp=2.4; Ib=500;//microAmp Rb=(Vp-.9)/Ib; disp(Rb*1000,"max value of Rb is in kohm:") printf('t...
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I=20 w=2000 R=200 L=0.25 Xl=w*L*%i Ir=I*Xl/(Xl+R) Il=I-Ir Vl=Xl*Il t=1E-3 ir=sqrt(2)*real(Ir*exp(%i*w*t)) il=sqrt(2)*real(Il*exp(%i*w*t)) vl=sqrt(2)*real(Vl*exp(%i*w*t)) is=sqrt(2)*real(I*exp(%i*w*t)) vs=vl Pr=ir*ir*R Pl=vl*il Ps=is*ir*R Pr=ir*vl disp(Ps,Pl,Pr)
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clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 3.9 Page 145 \n'); //Example 3.9 // Heat conduction through Rod kc = 398; //[W/m.K] From Table A.1, Copper at Temp 335K kal = 180; //[W/m.K] From Table A.1, Aluminium at Temp 335K kst = 14; ...
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// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART II : TRANSMISSION AND DISTRIBUTION // CHAPTER 3: STEADY STATE CHARACTERISTICS AND PERFORMANCE OF TRANSMISSION LINES // EXAMPLE : 3.24 : // Page number 156-157 clear ;...
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// Copyright (C) 2015 - IIT Bombay - FOSSEE // // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_...
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//Problem 21.17: A six-pole lap-wound motor is connected to a 250 V d.c. supply. The armature has 500 conductors and a resistance of 1 ohm. The flux per pole is 20 mWb. Calculate (a) the speed and (b) the torque developed when the armature current is 40 A //initializing the variables: p = 1; // let c = 2*p; // fo...
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clc clear //Initialization of variables g=32.2 //ft/s^2 p1=100 //psig p2=29.0 //in of Hg //calculations BP=p2*0.491 AP=BP+p1 //results printf("Absolute pressure = %.2f psia",AP)
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clear; chdir('J:\Stage\Simulations') getf('Modele un facteur gaussien\Prime de risque\VarPrime_Function.sci');disp('getf done'); getf('Modele un facteur gaussien\Cas Constant\Const_Function.sci');disp('getf done'); /////////////////////////////////////////////////////////////////////////////////////////////...
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clc //initialisation of variables clear vo= 10 //ft/sec a= 0.5 //ft^-1 b= 1 //ft x= -2 //ft y= 2 //ft b1= 2 a1= 3/5 //ft //CALCULATIONS Vx= vo/(a*x^2+b) Vy= -2*a*b*vo*x*y/(a*x^2+b)^2 V= sqrt(Vx^2+Vy^2) fx= -2*a*b^2*vo^2*x/(a*x^2+b)^3 fy= 2*a*b^2*vo^2*y*(b-a*x^2)/(a*x^2+b)^4 f= sqrt(fx^2+fy^2) r= b1^2/a...
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//Page Number: 653 //Example 12.8 clc; //Given d=2.4;//cm lmbc=1.8; c=3*10^10; //cm/s lmbg=2*d; lmb=(lmbg*lmbc)/(sqrt(lmbg^2+lmbc^2)); //Operating frequency f=c/lmb; disp('GHz',f/10^9,'Operating frequency:');
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//Script to test the calculation of the type II based on Bassos book sec 4.2.9 mag=10^(-18/20); boost=68; fc=5e3; //Result must be fp=25.7e3, fp0=7.8e3, fz=972 [regulator,fz1,fp1,fp0]= calctypeII(mag,fc,boost) bode(regulator,fc/103,fc*100,0.1)
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global file_name path fname extension clk__x1 = [0.2 2.5]; data_x1 = [2.5 2.5]; zero_x1 = [0.2 0.2]; clk_sr = [clk__x1 clk__x1 clk__x1 linspace(0.2,0.2,21)]; data_sr = [data_x1 zero_x1 zero_x1 linspace(0.2,0.2,21)]; input_vmmwta=[ zero_x1 zero_x1 zero_x1 linspace(2.5,2.5,21); // "1" zero_x1 zero_x1 zero_x1...
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//clear// clear; clc; //Example 22.2 //Given Dp = 1; //[in.] vdot = 25000; //[ft^3/h] T = 68; //[F] P = 1; //[atm] ya = 0.02; Mair = 29; Mg = 17; //Solution //The average molecular weiht of the entering gas M = (1-ya)*Mair+ya*Mg; rho_y = M*492/(359*(460+68)); //[lb/ft^3] rho_x = 62.3; //[lb/ft^3] //...
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// circuito RL passa-faixa em dB R1=1000;R2=220;L1=1e-3;L2=1e-3; fc1=R1/(2*%pi*L1); fc2=R2/(2*%pi*L2); f = logspace(1,10,1e4); mod_H1=1./((1+(fc1./f).^2).^0.5); mod_H2=1./((1+(f./fc2).^2).^0.5); mod_H=mod_H1.*mod_H2; ang_H1=(180/%pi)*(atan(fc1./f)); ang_H2=-(180/%pi)*(atan(f./fc2)); ang_H=ang_H1+ang_H2; scf(...
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; asinf.tst ; ; Copyright 2009-2023, Arm Limited. ; SPDX-License-Identifier: MIT OR Apache-2.0 WITH LLVM-exception func=asinf op1=7fc00001 result=7fc00001 errno=0 func=asinf op1=ffc00001 result=7fc00001 errno=0 func=asinf op1=7f800001 result=7fc00001 errno=0 status=i func=asinf op1=ff800001 result=7fc00001 errno=0 sta...
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// Inverting and non inverting amplifier clc; clear all; close; m = 200000; vi = input("Enter input voltage= ") rf =input("Enter feedback resistor value= ") r1 =input("Enter input resistor value= ") vo = (-rf/r1)*vi; // vo = (1+(rf/r1)*vi; //for non inverting amp //vo = vo-(vo/m); // correction for vo...
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clc clear //input data D=8//Outer diameter of the turbine in m Db=3//Inner diameter of the turbine in m P=30000//Power developed by the turbine in kW nH=0.95//Hydraulic efficiency N=80//Speed of the turbine in rpm H=12//Head operated by the turbine in m Q=300//Discharge through the runner in m^3/s g=9.81//Acc...
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//Ex 2.9.2 clc;clear;close; format('v',6); //Given : IF=10;//mA VF=0.3;//volts T=27+273;//K Eta=1;//for Ge diode VT=T/11600;//V Io=IF/(exp(VF/Eta/VT)-1);//mA disp(Io*10^6,"Reverse saturation current in nA : ");
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errcatch(-1,"stop");mode(2);//332 ; ; A=[-1 2 2]'; disp(A,'A='); [U diagnol V]=svd(A); disp(U,'U='); disp(diagnol,'diagnol='); disp(V','V''='); disp(U*diagnol*V','A=U*diagnol*V''') //end exit();
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// Example 10.7 format('v',6) clc; clear; close; // given data R_C= 1*10^3;// in Ω r_desh_e= 2.5;//in Ω Zin= 1*10^3;// in Ω A2= 10;// unit less A3= 1;// unit less A1= (R_C*Zin/(R_C+Zin))/r_desh_e;// unit less // The overall voltage gain A= A1*A2*A3; disp(A,"The overall voltage gain is : ")
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// Test # 1 : No Input Arguments exec('./iirpowcomp.sci',-1); [b,p]=iirpowcomp(); //!--error 10000 //2/3 input arguments allowed //at line 33 of function iirpowcomp called by : //[b,p]=iirpowcomp()
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//Page Number: 250 //Example 4.24 clc; //Given //As it is perfectly matched S12=1/sqrt(2); S21=S12; s=[0 S12;S21 0]; disp(s,'Scattering matrix:');
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# towers_of_hanoi.tst >>> import sys >>> import pyke >>> import os >>> new_path = os.path.join(os.path.dirname(os.path.dirname(pyke.__file__)), ... 'examples/towers_of_hanoi') >>> sys.path.append(new_path) >>> import driver >>> driver.test(1) got 1: ((0, 2)...
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//Example 11.7 //Relaxation Method //Page no. 376 clc;clear;close; for i=0:4 for j=0:4 if i==0 | j==0 then U(5-i,j+1)=0 elseif i==4 | j==4 U(5-i,j+1)=(i*j)^2 else U(5-i,j+1)=0; end end end S=['A','B','C','D','E','F','G','H','I'...
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clc //initialisation of variables o= 90 //degrees H= 15.5 //in Cd= 0.6 g= 32.2 //ft/sec^2 //CALCULATIONS Q= 8*Cd*tand(o/2)*sqrt(2*g)*(H/12)^2.5/15 //RESULTS printf ('Total Discharge= %.2f cuses',Q)
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//check o/p for a given matrix r=[1 34 4]; [a,e,k]=levinson(r); disp(a); ////output /// 1. - 0.0883117 - 0.9974026
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stacksize('max'); t = read("t", -1, 1); tamount = size(t,1)*1; t = t(1:tamount); tamount = size(t,1); x = read("x", -1, 1); xamount = size(x,1); re=read("re", xamount, tamount); im=read("im", xamount, tamount); v = re+%i*im; xx = 1:xamount; tt = 1:tamount; xx = (x ./ (60*60)); tt = (t ./ (60*60)); ...
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// Scilab code Exa5.10 : : Page 206 (2011) clc; clear; h_kt = 1.05457e-34; // Reduced Planck's constant, joule sec e = 1.60218e-19; // Charge of an electron, coulomb l = 2; // Orbital angular momentum eps_0 = 8.5542e-12; // Absolute permittivity of free space, coulomb square per...
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//Problem 35.02: If the load impedance Z in Figure 35.2 of problem 35.01 consists of variable resistance R and variable reactance X, determine (a) the value of Z that results in maximum power transfer, and (b) the value of the maximum power. //initializing the variables: rv = 120; // in volts thetav = 0; // in deg...
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// Exa 6.4 clc; clear; close; // Given data V_DD = 15;// in V R = 3;// in kohm I_D = V_DD/R;// in mA R_D = 1;// in kohm V_D = V_DD - (I_D*R_D);// in V disp(V_D,"The drain voltage in V is");
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//=========================================================================== //chapter 7 example 15 clc; clear all; //variable declaration W1 = 300; //wattmeter reading in kW W2 = 100; //wattmeter reading in kW //calculations P = W1+W2; //input power in kW phi = atan(((W1-W2)/(W1+W2))*sqrt(3)); //pha...
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//To find axial thrust clc //Given: L=175/1000, d2=100/1000, r2=d2/2 //m theta=70 //degrees G=1.5, T2=80 Tf=75 //Torque on faster wheel, N-m funcprot(0) //Solution: //Spiral angles for each wheel: //Calculating the number of teeth on slower wheel T1=T2*G //Calculating the pitch circle diameter of the slower...
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//Section-10,Example-3,Page no.-CT.31 //To calculate q,W,dl_E,dl_H. clc; V_2=20 V_1=4 P=1 W=-(P*(V_2-V_1))*(8.314/0.08206) disp(W,'maximum work done in(J)') dl_E=0 dl_H=0 q=dl_E-W disp(q,'in J')
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clc;funcprot(0);//EXAMPLE 5.5 // Initialisation of Variables N=1;..........//N0. of atoms on one side of iron bar H=1;..........//No. of atoms onother side of iron bar d=3;.......//Diameter of an impermeable cylinder in cm l=10;.....//Length of an impermeable cylinder in cm A1=50*10^18*N;..........// No. of gaseo...
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clear; clc(); // To find the temperature of planes indicated by grid points using relaxation method t1=800; // inner surface temperature of wall in degF t4=200; // outer surface temperature of wall in degF //Grids are square in shape so delx =dely where delx,y sre dimensions of square gri...
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// Scilab Code Ex6.11 Time required for carburizing of steel: Page 209 (2010) C0 = 0.0018; // Intial carbon concentration of steel Cx = 0.0030; // Carbon concentration of steel at 0.60 mm below the surface of the gear Cs = 0.01; // Carbon concentration of steel at the surface x = 0.6e-03; // Diffusion d...
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clc //initialisation of variables P= 15 //psia sx= 1.7050 //Btu/lb R sg= 1.7549 //btu/lb R sfg= 1.4415 //Bru/lb R hg= 1150.8 //btu/lb hfg= 969.7 //Btu/lb vg= 26.29 //cu ft/lb vfg= 26.27 //cu ft/lb //CALCULATIONS n= (sg-sx)/sfg sx= sg-n*sfg hx= hg-n*hfg vx= vg-n*vfg //RESULTS printf ('Volume= %.2f cu ft/lb',vx) printf (...
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//Example 4.27 //Interconnectiuion of LTI systems n2=0:18; h1=[1 5 10 11 8 4 1]; h2=[1 1 zeros(1,5)]; h3=[1 1 zeros(1,5)]; a=convol(h1,h2); h=convol(a,h3); x=[1 -1]; n1=[0 1]; n3=0:19; y=convol(x,h); subplot(3,1,1) xtitle("input signal x(n)","....................n","x[n]"); plot(n1,x,'.'); subplot(3,1,2) xtitle("system...
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//Example 10.9// T1=290;//degree C //recrystallization temperature T2=920;// degree C //solidus temperature T3=273;//K //Kelvin T4=(T1+T3)/(T2+T3) disp(T4)
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//scilab 5.4.1 //windows 7 operating system //chapter 5:Semiconductor Junction Diodes clc clear Vz=3//Vz=breakdown voltage of zener diode Vi=12//Vi=input voltage V=[12;-3]//V=[Vi:-Vz] R1=1000 R2=1000 R3=500//R1,R2,R3=resistances R=[R1+R2 -R2;-R2 R2+R3] I1=inv(R)*V//solving this matrix on the basis of appli...
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//Chapter 14 //Example 14_11 //Page 372 clear;clc; v=400; ph_v=230; w1=100; w2=150; r1=ph_v^2/w1; r2=ph_v^2/w2; i=v/(r1+r2); v1=i*r1; v2=i*r2; printf("Resistance of lamp L1 = R1 = %.2f ohm \n\n", r1); printf("Resistance of lamp L2 = R2 = %.2f ohm \n\n", r2); printf("Curretn through lamps = %.3f A \n\n", i); printf...
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clc; clear all; clf; //UNIT IMPULSE x=[zeros(1,10),ones(1,1),zeros(1,10)]; t=-10:1:10; subplot(2,3,1) plot(t,x); title('unit impulse');xlabel('t');ylabel('amp'); //UNIT STEP x=[zeros(1,10),ones(1,11)]; t=-10:1:10; subplot(2,3,2) plot(t,x); title('unit step');xlabel('t');ylabel('amp'); //SINUSOIDAL FUNCTION t=0:0.0...
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clc,clear printf('Example 6.13\n\n') V_L=11*10^3 V_ph=V_L/sqrt(3) VA=700*10^3 I_FL=VA/(sqrt(3)*V_L) //full load current IR_a=(1.5/100)*V_ph //product of I and R_a R_a=IR_a/I_FL IX_s=(14/100)*V_ph // product of I and X_s X_s=IX_s/I_FL //synchronous reactance //at full load and 0.8 pf I=I_FL phi=acos(0.8)...
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clc; //page no 2-11 //example: 2.4 //Given carrier power=400 watt and modulation depth as 80% u=0.8; Pc=400; ptotal=Pc*(1+(u^2/2)); disp(+'watts',ptotal, 'Total power delivered is ')
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// Scilab Code Ex9.8: Page-9.28 (2004) clc;clear; Eg = 1; // Bandgap of silicon, eV e = 1.6e-19; // Electronic charge, C k = 1.38e-23; // Boltzman constant,joule per kelvin E_F = (0.6-0.5)*e; // Fermi energy, joules // E_F =((Ev+Ec)/2)+3/4*k*T1*(log(4)); // Ev & Ec= valance and conduction band energies...
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// PG (144) deff('[y]=f(x)','y=sqrt(x)') funcprot(0) deff('[y]=fp(x)','y=0.5/sqrt(x)') funcprot(0) deff('[y]=fpp(x)','y=-0.25*x^(-3/2)') funcprot(0) deff('[y]=fppp(x)','y=3*x^(-2.5)/8') deff('[y]=fpppp(x)','y=-15*x^(-7/2)/16') // f[2.0,2.1,.....2.4] = -0.002084 fpppp(2.3103)/factorial(4)
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clc clear //Initialization of variables h=2.5 //Btu/hr ft^2 F kc=0.1 //Btu/hr ft F r1=0.811/2 //calculations r2c=kc/h *12 //results if r2c>=r1 then printf("Thin layer of insulation would increase the heat dissipation from wire, r2c = %.2f in",r2c) else printf("Thin layer of insulation would decr...
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clc; // page no 26 // prob no 1_8_1 //High frequency transformer with identical primary and secondary circuits Lp=150*10^-6; Ls=150*10^-6; Cp=470*10^-12; Cs=470*10^-12; //Lp=Ls=150 uH,Cp=Cs=470 pF Q=85//Q-factor for each ckt is 85 c=0.01//Coeff of coupling is 0.01 Rl=5000//Load resistance Rl=5000 ohm r=7500...
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//Chapter-4,Example 12,Page 96 clc; close; q_rev= 19.14 //latent heat n= 18 //mols T= 273 //temperature in Kelvin dS= q_rev*n/T printf('the change of molar entropy is %.2f J/mol',dS)
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// Display mode mode(0); // Display warning for floating point exception ieee(1); clc; disp("Principles of Heat transfer, Seventh Edition, Frank Kreith, Raj M Manglik and Mark S Bohn, Chapter 9, Example 2") //Temperature at which sun is radiating as a blackbody in K T=5800; //Lower limit of wavelength f...
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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 10, Example 2") //Hot oil(specific heat,ch=2.09kJ/(kg*K)) flows through counter flow heat excahnger at the mass flow rate of mdoth=(0.7kg/s) ch=2.09*10^3; mdoth=...
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//Chapter 4, Example 4.5, Page 96 clc clear // Q value of the reaction mn = 1.0086649 MB = 10.0129370 MHe = 4.0026032 MLi = 7.0160040 C2 = 931.5 Q = (mn+MB-MHe-MLi)*C2 -0.48 printf("\n Q of the reaction = %f MeV",Q); //Answer may vary due to round off error
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clc; L1=10*1e-3; //in henry c3=10*1e-12; //in faraday pi=3.14; fr=1/(2*pi*sqrt(L1*c3)); disp(+'Hz',fr,'resonant frequency =')
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clc; E1=2500; // primary side voltage E2=250; // secondary side voltage P=10000; // rated VA of transformer // to achieve a voltage level of 2625, two equal parts of 125 V each of secondary winding are connected in parallel with each other and in series with primary winding Eo=E1+E2/2; // desired output of autot...
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//Load scripts getd('../scripts/') //Load image imgPos="../images/" img=readpbm(imgPos+'Gliese 667Cc_surface.pbm') // Do a normalisation on the image display_gray(normalisation(img))
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clc; disp("Example 9.2") density= 1000 // in kg/m^3 densitym=13600 // of mercury in kg/m^3 c=0.62 // orifice coefficient b=0.5 U=135.6/60 // velocity in m/s delP= ((U*((1-b^4)^0.5)/c)^2)*density/2 g=9.81 R=delP/(g*(densitym-density)) disp(R,"Reading on the manometer is ")
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clc; clear; printf("\t\t\tChapter4_example8\n\n\n"); rou=7817; c=461; k=14.4; alpha=.387e-5; L1=.03; L2=0.03; L3=0.04; x=0.04; T_i=95; T_inf=17; // for infinite plate L=L1/2; hc=50; reciprocal_Bi_plate=k/(hc*L); printf("\nThe value of 1/Bi for infinite plate is %.1f",reciprocal_Bi_plate); T=50; n=1; t=[3000 1500 700 40...
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//to find the weight of flywheel clc //given N=120//rpm k=3.5//ft Ef=2500//ft lb Ks=.01 g=32.2//ft/s^2 w=%pi*N/30//angular velocity W=g*Ef/(w^2*k^2*Ks*2240)//Weight of flying wheel printf("\nWeight of flying wheel, W = %.2f tons",W)
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clc //initialisation of variables d= 6 //in N= 120 //in Q= 5 //gpm //CALCULATIONS Vc= %pi*d^2*N/(4*231) //RESULTS printf ('minimum size of the reservoir = %.2f gpm',Vc)
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// // printf("\n chainage 15.5 and 27.5') a1=15.5,b1=27.5, //finding base and height of each triangle base=b1-a1 o1=0,o2=22.5, mo1=(o2+o1)/2 //calculating area ae1=base*mo1 ap1=0 an1=ae1 printf("\n area GAM= %0.3f sq meters",ae1) printf("\n chainage 15.5 and 50') a1=15.5,b1=50, base=b1-a1 o1=22.5,o2=30, mo...
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//Exa:6.3 clc; clear; close; //Given: BWt=100;//in kHz Fh=5;//in KHz n=BWt/(2*Fh); printf("\n\t number of stations = %f",n);
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// Data Reconciliation Benchmark Problems From Literature Review // Author: Edson Cordeiro do Valle // Contact - edsoncv@{gmail.com}{vrtech.com.br} // Skype: edson.cv // aux functions to weighted least squares functions function f = objfun ( x ) f = obj_user(x); endfunction function c = confun(x) //flowsheet_residu...
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clc //Initialzation of variables xN2=0.780 xO2=0.210 xAr=0.009 P=100 //kPa //Calculations PN2=xN2*P PO2=xO2*P PAr=xAr*P //Results printf('Partial pressure of Nitrogen(kPa) = %.1f',PN2) printf('\n Partial pressure of Oxygen(kPa) = %.1f',PO2) printf('\n Partial pressure of Argon(kPa) = %.1f',PAr)
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//Exa2.41 clc; clear; close; // given data t1=20;// in degree C t2=36;// in degree C alpha_20=0.0043;// in per degree C (Temperature Coefficient) InsulationResistance=480*10^6;// in ohm copper_cond_res=0.7;// in ohm (copper conductor resistance) l=500*10^-3;// in kilo meter (length) R1_desh=InsulationResis...
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//clc() x = [0.25,0.75,1.25,1.75,2.25]; y = [0.28,0.57,0.68,0.74,0.79]; a0 = 1; a1 = 1; sr = 0.0248; for i = 1:5 pda0(i) = 1 - exp(-a1 * x(i)); pda1(i) = a0 * x(i)*exp(-a1*x(i)); end Z0 = [pda0(1),pda1(1);pda0(2),pda1(2);pda0(3),pda1(3);pda0(4),pda1(4);pda0(5),pda1(5)] disp(Z0,"Z0 = ") R = Z0'*Z0; ...
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clear; clc; //Example 3.13 b=100; Vcc=12; Vbe=0.7; Icq=1;//mA Vceq=6; Rc=(Vcc-Vceq)/Icq; printf('\ncollector resistance=%.3f KOhms\n',Rc) Ibq=Icq/b; printf('\nbase current=%0.3f mA\n',Ibq) Rb=(Vcc-Vbe)/Ibq; printf('\nbase resistance=%.3f KOhms\n',Rb)
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Haha HelloWorld 123456799999 123456788888 18007328601
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// **** Purpose **** // This code is to help calculate PiLab_kif_esc in parallel way. // **** Variables **** // [Ek],[Vk]: data readed from /kif/Ek/Ek_x.sod, /kif/Vk/Vk_x.sod. // [EVal],[EWin],[StateProj]: kif_esc input data // [k_idx_start]: the start index of Ek and vk // **** Version **** // 05/18/2016 first built ...
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// This file is part of the materials accompanying the book // "The Elements of Computing Systems" by Nisan and Schocken, // MIT Press. Book site: www.idc.ac.il/tecs // File name: projects/07/MemoryAccess/BasicTest/BasicTestVME.tst load BasicTest.vm, output-file BasicTest.out, compare-to BasicTest.cmp, output-list R...
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//Ex8_10 clc Av = -200 Ri = 10*10^3 RL = 3*10^3 Ai = Av*Ri/RL disp("Av = "+string(Av))//voltage gain disp("Ri = "+string(Ri)+"ohm")//input resistance disp("RL = "+string(RL)+"ohm")//load resistance disp("Ai = Av*Ri/RL = "+string(Ai))//current gain // note : there are mis-printring in the textbook for the ab...
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clear; clc; // Illustration 3.4 // Page: 169 printf('Illustration 3.4 - Page: 169\n\n'); // solution //*****Data*****// // a-ammonia T = 300; // [K] P = 101.3; // [kPa] Kg = 2.75*10^-6; // [kmole/square m.s.kPa] m = 1.64; res = 0.85; // [gas phase resistance] xa_g = 0.115/100; // [mole fraction of...
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//Control of DC motors// //Example 11.1// //Since the speed control is required in both directions we will have to use a dual converter for the application.It would be prefarable to use six pulse dual converter with thyristors connected in antiparallel connection// //speed control from 20% rated speed to 100% rated ...
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function [y,err] = CRCDetector(in,polynomial,initialstate,chksumsperframe) y=[]; err=[] // Display mode mode(0); // Display warning for floating point exception ieee(1); //CRCDetector finds the message signal and also finds if error has occured in a subframe //Y = CRCGene...
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clc IP=35; // Power developed by the engine in kW S=284; //Steam combustion in kg/h p2=0.14; //Condenser pressure in bar p1=15; //bar h1=2923.3; //kJ/kg s1=6.709; //kJ/kg K h_f=220; //kJ/kg h_fg=2376.6; //kJ/kg s_f=0.737; //kJ/kg K s_fg=7.296; //kJ/kg K x2=(s1-s_f)/s_fg; disp("(i) Final condition of s...
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//example 6.4 clc; funcprot(0); Cc=0.28; Hc=18*12; e0=0.9; sigmao=11*100+40*(121.5-64)+18/2*(118-62.4); H2=5+40+18; H1=5+40; qo=3567; //from table IaH2=0.21; IaH1=0.225; Dsigma=qo*((H2*IaH2-H1*IaH1)/(H2-H1))*4; Scp=Cc*Hc/(1+e0)*log10(sigmao/sigmao+Dsigma/sigmao); disp(Scp,"settlement in inches");
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//Calculations on dual combustion cycle clc,clear //Given: r=18 //Compression ratio P1=1.01,P3=69 //Pressure at 1, 3 in bar T1=20+273 //Temperature at 1 in K cv=0.718 //Specific heat at constant volume in kJ/kgK cp=1.005 //Specific heat at constant pressure in kJ/kgK g=1.4 //Specific heat ratio(gamma) R=0.287 ...
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//EXAMPLE 26.26 //LONG SHUNT DYNAMO clc; funcprot(0); //Variable Initialisation N=1000;.....................//Speed of the generator in rpm Po=22;....................//Output power in Kilo Watts V=220;....................//Terminal voltage in Volts Ra=0.05;..................//Armature resisitance in Ohms R...
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clc; warning("off"); printf("\n\n example4.4 - pg101"); // given x1=0; //[cm] x2=30; //[cm] p1=0.3; //[atm] p2=0.03; //[atm] D=0.164; //[am^2/sec] R=82.057; //[cm^3*atm/mol*K] T=298.15; //[K] // using the formula Nax*int(dx/Ax)=-(D/RT)*int(1*dpa) a=integrate("1/((%pi/4)*(10-(x/6))^2)","x",x1,x2); b=...
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//Example7.13 // to determine the output voltage of the precision rectifier circuit clc; clear; close; Vi = 10 ; //V i/p volt R1 = 20 ; // K ohm R2 = 40 ; // K ohm Vd = 0.7 ; // V the diode voltage drop // the output of the half wave precision rectifier is defined as // Vo = -(R2/R1)*Vi ; for Vi < 0 // ...
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//ques-1.3 //Calculating temporary and permanent hardness of a sample of water clc A=7.3;//content of Magnesium hydrogencarbonate (in mg/L) B=16.2;//content of Calcium hydrogencarbonate (in mg/L) C=9.5;//content of Magnesium chloride (in mg/L) D=13.6;//content of Calcium sulphate (in mg/L) a1=(A/146)*100;//CaCO3...
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clc // Given that mu = 1.5 // refractive index of glass // Sample Problem 1 on page no. 3.23 printf("\n # PROBLEM 1 # \n") Ip = atan(mu) * (180 / %pi) // by brewster's law r = 90 - Ip // calculation for angle of refraction printf("Standard formula used \n mu=tan(Ip)\n") printf("\n Brewster angle = %f degree\n Angle of...
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PL/SQL Developer Test script 3.0 5 begin -- Call the procedure personas_por_usuario(pusuario_id => :pusuario_id, p_recordset => :p_recordset); end; 2 pusuario_id 1 gago 5 p_recordset 1 <Cursor> 116 0
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THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.305650D+00 ...
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function [p]=derivat(p) //pd=derivat(p) computes the derivative of the polynomial or rational //function marix relative to the dummy variable //! t=type(p) if t=1 then p=0*p,return,end if t=2 then [m,n]=size(p);var=varn(p); for i=1:m for j=1:n pij=p(i,j);nij=degree(pij); if nij=0 the...
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// PD control law from polynomial coefficients, as explained in Sec. 9.8. // 9.22 function [K,taud,N] = pd(Rc,Sc,Ts) // Both Rc and Sc have to be degree one polynomials s0 = Sc(1); s1 = Sc(2); r1 = Rc(2); K = (s0+s1)/(1+r1); N = (s1-s0*r1)/r1/(s0+s1); taudbyN = -Ts*r1/(1+r1); taud = taudbyN * N; endfunc...
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v=[5, -5, 9, 12, -1, 0, 4] v(v>0) = 10