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//example 11.3 clc; funcprot(0); K=0.25; Ap=16*16/12/12; phi=30*%pi/180; Nq=25; q=110*50/1000; sigmao=q/2; p=4*16/12; L=50; FS=4; Qu=q*Nq*Ap+K*sigmao*tan(0.8*phi)*p*L; Qall=Qu/FS; disp(Qall,"allowed load in kip");
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Vl = 220 Vf = Vl/(sqrt(3))*expm(-30*%pi/180*%i) Snom = 820+%i*500 Zcd = -%i*400 Zcy = Zcd/3 Zl = 3.5 + (8 - 6)*%i Zcarga = (3*abs(Vf)^2/(Snom))' Zto = Zl + (Zcy*Zcarga/(Zcy + Zcarga)) I = Vf/Zto Vcarga = Vf - I*Zl Scarga = 3*abs(Vcarga^2)/(Zcarga)' Vcarga = 220/sqrt(3)*expm(-30*%pi/180*%i) Vcarga = Vf - Zl*(Vcarg...
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clear //Given Ev=200 //V Iv=10.0 f=50 //Hz //Calculation z=Ev/Iv R=z*cos(30*3.14/180.0) Xc=z*sin(30*3.14/180.0) C=1/(2.0*%pi*f*Xc) //Result printf("\n (i) Value of resistance is %0.2f ohm",R) printf("\n (ii) Capacitive reactance is %0.0f ...
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// Exa 3.4 format('v',6) clc; clear; close; // Given data R_L = 18;// in ohm Vz = 18;// in V V1 = 22;// in V V2 = 28;// in V // Minimum voltage across R, V = V1-Vz;// in V Izmin = 200;// in mA I_Lmax = Vz/R_L;// in A I = I_Lmax+Izmin*10^-3;// in A R =V/I;// in ohm disp(R,"The value of R in ohm is : ") ...
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exec('scilab-base-program-make_point.sce',-1) //to delete exec('scilab-base-program-check_point.sce',-1) //to delete M=make_point(0,1) // creates point M function %point_p(P) check_point(P) printf('(x=%f,y=%f)\n',P.x,P.y) endfunction M // point M's display is modified
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function resp = fat(n) resp = 1; for i = 2:n resp = resp * i; end endfunction
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clear; clc; close; Ri_stage2 = 15*(10^(3))*4.7*(10^(3))*1300/(15*(10^(3))*4.7*(10^(3))+4.7*(10^(3))*1300+15*(10^(3))*1300); Rd1 = 2.4*10^(3); Rd2 = 2.2*10^(3); gm = 2.6*10^(-3); Vi1 = 20*10^(-3); Vi2 = 1*10^(-3); Av1 = -gm*(Rd1*Ri_stage2/(Rd1+Ri_stage2)); Av2 = -338.46; Av = Av1*Av2; Vo1 = Av*Vi1; Vo2 = ...
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clear; close; clc; n=-10:10; N=2; for i=1:length(n) if n(i)>=-N & n(i)<=N then x(i)=1; else x(i)=0; end end x=x'; figure subplot(2,1,1) plot2d3(n,x); title("x[n]") plot(n,x,'r.') w=-10:0.1:10; Xw=x*exp(-%i*n'*w); subplot(2,1,2) plot2d(w,Xw); title("X[w] fourier transfor...
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/////////////////////////////////////////////////////////////////////////////// // Author: Jia Wu // Version: 0.1 // Date: Nov 2009 // Description: Kernel Principal Component Analysis // // Copyright (C) 2009 OpenPR // All rights reserved. // // Redistribution and use in source and binary forms, with or witho...
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//ques-34.23 //Calculating quantum yield of a reaction clc c1=0.0506; c2=0.0394;//initial and final concentration of oxalic acid (in M) q=8.81*10^8;//(in ergs) w=245;//wavelength (in nm) n=(c1-c2)/100;//moles of oxalic acid decomposed QY=(n*1.196*10^15)/(q*w); printf("The quantum yield of the reaction is %.3f."...
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// Scilab code Ex8.1: Pg 253 (2008) clc; clear; C = 8e-06; // Value of capacitance of capacitor, farad R = 0.5e+06; // Value of series resistor, ohm E = 200; // Value of d.c. voltage supply, volt // Part (a) tau = C*R; // Time constant of the R-C circuit while charging, s printf("\nThe circuit ti...
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clc clear //INPUT DATA ni=2*10^16//intrinsic charge carriers in m^-3 Na=5*10^23//density of acceptor concentration of silicon with arsenic in atoms Nd=3*10^23//density of donor concentration of silicon with arsenic in atoms //CALCULATION nh=(Na-Nd)//density of hole in m^-3 ne=(ni^2/(nh))/10^9//The electron co...
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optimizecode 1 maxversions 0 units Field /LiquidPhases = 2 /StdLiqVolRefT = 288.15 /StdLiqVolRefT = 60 F /RecycleDetails = 1 displayproperties displayproperties VapFrac T P MoleFlow MassFlow VolumeFlow StdLiqVolumeFlow StdGasVolumeFlow Energy H S MolecularWeight MassDensity Cp ThermalConductivity Viscosity molarV ZFa...
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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 // Fictitious but realistic mineral processing plant //Alhaj-Dibo, Moustapha, Didier Maquin, and José Ragot. 2008. //Data reconciliation: A robust approa...
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//Example 2.10 the overall percentage of failure clc; clear; function value = binomial(n, k, p) value = factorial(n)*(p^k)*((1-p)^(n-k))/(factorial(k)*factorial(n-k)); endfunction q=40/100; n=6; p=1-q; disp(n,"No. of candidates =",p,"Proabab. of(success) in a single trial",q,"Probab. of failure of a candid...
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// Chapter 9_The bipolar transistor //Caption_Breakdown voltage //Ex_8//page 387 Wb=0.5*10^-4 //metallurgical base width NB=10^16 eps=11.7*8.85*10^-14 e=1.6*10^-19 Vpt=25 //punch through voltage x=Vpt*2*eps/(e*Wb^2*NB) y=x-1 NC=NB/y xn=(2*eps*(Vpt)*NB/(e*NC*(NB+NC)))^0.5*10000 printf('The collector d...
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//water and its treatment// //example 2.18.21// clc Purity_Lime=.85 Purity_soda=.95 W1=55.5;//amount of CaCl2 in ppm// W2=20;//amount of SiO2 in ppm// W3=12.6;//amount of NaHCO3 in ppm// W4=250;//amount of KCl in ppm// W5=48;//amount of MgSO4 in ppm// W6=2.2;//amount of CO2 in ppm// W7=43.8;//amount of ...
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//divide -5.3716 by 3 clear; clc; close; //characteristic=-5=-6+1 or the log as -6+1.3716 characteristic=-6/3; mantissa=1.3716/3; characteristic-mantissa
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//example 5.3 clc; funcprot(0); // Initialization of Variable h1=3213.6;//kJ/kg V1=10.0; V2=665.0; mdot=2.0; h2=h1+(V1^2/2-V2^2/2)/1000; //using table with given h2 values v2=0.1627;//specific volume V2=665; A2=mdot*v2/V2; disp(A2,"Area in m^2"); clear()
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//Example_a_8_15 page no:335 clc; V=220; f=50; Vr=550; Ir=1; R=V/Ir; C=1/(Vr*2*%pi*f); C=C*10^6; L=1/((C*10^-6)*(100*%pi)^2); disp(R,"the resistance is (in ohm)"); disp(L,"the inducatance is (in H)"); disp(C,"the capacitance is (in microFarad)");
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//Chapter 10, Problem 20, figure 10.35 clc; fr=400e3; //resonant frequency Qf=100; //Q factor C=400e-12; //capacitance L=((2*%pi*fr)^2*C)^-1; //calculating inductance R=2*%pi*fr*L/Qf; //calculating resistance printf("(a) Inductance = %f mH\n\n\n",L*1000...
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// Example 6.7 //The program illustrate the use of continue statement disp("Enter 9999 to STOP"); count=0; negative=0; while(count<=100) number=input("Enter a number:"); if(number==9999) then break; //EXIT FROM THE LOOP end if(number<0), disp("Number is ...
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8 5:4.0 6:0.17391304347826086 34:0.1111111111111111 42:0.4 54:1.0 56:0.25 82:0.25 143:0.16666666666666666 150:0.017543859649122806 180:1.0 251:0.14285714285714285 273:0.3333333333333333 286:1.0 327:0.16666666666666666 345:1.0 407:1.0 473:0.5 494:1.0 583:1.0 644:1.0 664:1.0 672:1.0 768:0.5 1037:0.5 1067:1.0 1183:0.5 156...
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// Example 10.6 format('v',6) clc; clear; close; // given data V_CC= 30;// in V PP= V_CC;// in V R_L= 100;// in Ω // The value of P_Lmax P_Lmax= PP^2/(8*R_L);// in W disp(PP,"The value of PP in volts is : ") disp(P_Lmax,"The value of P_Lmax in W is : ")
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-- Fuzzy Logix, LLC: Functional Testing Script for DB Lytix functions on Teradata -- -- Copyright (c): 2014 Fuzzy Logix, LLC -- -- NOTICE: All information contained herein is, and remains the property of Fuzzy Logix, LLC. -- The intellectual and technical concepts contained herein are proprietary to Fuzzy Logix, LLC. ...
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clear; clc; //Example13.10[Heat Transfer through a Tubular Solar Collector] k=0.02588;//[W/m.degree Celcius] Pr1=0.7282,Pr2=0.7255;//Prandtl no nu1=1.608*(10^(-5)),nu2=1.702*10^(-5);//[m^2/s] T1=20,T2=40;//[degree Celcius] Tavg=((T1+T2)/2)+273;//[K] Do=0.1,L=1;//Dimensions of glass tube[m] Di=0.05;//Inner diameter of ...
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clear //Given l=6000*10**-8 D=254.0 //Calculation a=(1.22*l)/D //Result printf("\n Limt of resolution of a telescope is %0.1f *10**-7 Radian",a*10**7)
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Plan Carat 8 16,Ensemble 15-24 ans 101 -1 1 1 0 1,B198,101 f:\source\SFR01 7070000 1 1 1,1410,1,1,0,1.9, 02/02/98,1 1,2140,1,1,0,5.5, 02/02/98,1 2,0810,1,1,0,0.5, 02/02/98,1 3,2057,1,1,0,3.3, 02/02/98,1 16,1830,1,1,0,1.6, 02/02/98,1 2,0930,2,1,0,0.4, 03/02/98,1 16,1340,2,1,0,3.5, 03/02/98,1 16,2005,2,1,0,5.5, 03/02/...
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Lab4.sce
clear function [grade] = get_grade(x) if(x > = 76) then grade = "H" elseif (x >= 60) then grade = "S" else grade = "U" end endfunction printf("-----------------------------------------------------\n") printf("Score\t\tGrade\n") printf("----------------------------------------...
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err_defineSort3.tst
; define-sort with duplicate parameters (set-logic QF_UF) (define-sort A (X X) Bool)
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EXP7_9.sce
clc;funcprot(0);//EXAMPLE 7.9 // Initialisation of Variables n=7;.............//The total number of specimens F=n+1;...........//The probability of failure of ceramic sigma1=52;........//the maximum allowed stress level on ceramic at one point in MP. sigma2=23.5;.......//the maximum allowed stress level on ceramic...
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q2.sce
function k=Map_Matrix(A) k=zeros(A); N=unique(A); disp("The Total Unique are : ") disp(length(N)) for i =1:1:length(N) indexes = find(A==N(i)); for j = 1:1:length(indexes) k(indexes(j))=i; end end endfunction A=imread('D:\example.png'); //disp("The image contents of the image are as follows : ") disp(A)...
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Exa_1_22.sce
//Exa 1.22 clc; clear; close; format('v',6); //Given Data : Rdegree=8314.3;//Universal Gas Constant M=32;//Molecular weight of gas p1=3*10^6;//N/m^2 V1=250*10^-3;//m^3 T1=20+273;//K p2=1.8*10^6;//N/m^2 V2=V1;//m^3 T2=16+273;//K R=Rdegree/M;//Nm/KgK m1=p1*V1/R/T1;//Kg m2=p2*V2/R/T2;//Kg mass_used=m1-m...
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clear; clc; //Example 11.3 V1=10; V2=-10; Iq=0.8*10^-3; Rc=12000; Ro=25000; b=100; Vt=0.026; Ad=Iq*Rc/(4*Vt); printf('\ndifferential gain=%.3f\n',Ad) Acm=-(Iq*Rc/(2*Vt))/(1+(1+b)*Iq*Ro/(Vt*b)); printf('\ncommon mode gain=%.3f\n',Acm)
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//clear// //Caption:Duobinary Encoding //Example6.2: Precoded Duobinary coder and decoder //Page 256 clc; b = [0,0,1,0,1,1,0];//input binary sequence:precoder input a(1) = bitxor(1,b(1)); if(a(1)==1) a_volts(1) = 1; end for k =2:length(b) a(k) = bitxor(a(k-1),b(k)); if(a(k)==1) a_volts(k)=1; else a_volt...
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scores=[284, 280, 277, 282, 279, 285, 281, 283, 278, 277]; new_scores = scores - 280; final_mean = mean(new_scores)+ 280; disp(final_mean)
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clear //Given m=900 w=100.0 c=1 a=80 b=4.2 V=210 //V x=12 y=60 //Calculation Hout=(m+w)*c*a Hin=(V*x*y)/b Hin1=90/w*Hin I=Hout/Hin1 //Result printf("\n Strength of the current is %0.3f A",I)
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//Example 2_17 clc(); clear; //To find the slit width theta=15 //units in degrees lemda=6500 //units in angstrom lemda=6500*10^-8 n=1 a=(n*lemda)/sin(theta*%pi/180) printf("slit width of white light is %f",a)
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clc //to find final speed of larger craft // GIVEN:: //refer to diagram 6-14 from page no. 127 //we consider +ve x direction as original motion of spaceship(and also that of final velocity of smaller craft) //total mass of spaceship //M = m//in kg //let us consider m = 1 M = 1//in kg //mass of smal...
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//Chapter-1,Example1_3_6,pg 1-18 A=132.91 //atomic weight of chromium N=6.023*10^26 //Avogadro's number p=1900 //Density a=6.14*10^-10 //lattice constant //step 1 : type of structure n=(p*N*a^3)/A...
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// Example for stdaq_get_adc() as from the Reference Manual in docs/refman stdaq_open("COM0"); chseq = [0]; // [ch0] clkdiv = 0; // 1 MHz stdaq_set_adc(chseq,clkdiv); stdaq_set_dac(0); stdaq_enable_dac(); tapsperperiod = 20; periods = 10; out = []; value = floor(2047.5*(1 + sin(2*%pi*(1:tapsperperiod*periods)...
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// Estimate value of capacitance needed // Basic Electronics // By Debashis De // First Edition, 2010 // Dorling Kindersley Pvt. Ltd. India // Example 3-16 in page 161 clear; clc; close; // Given data Vrms=230; // RMS voltage in V f=50; // Frequency in Hz gamma_hwr=0.003; // Ripple factor assumed I=0.5; ...
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//pagenumber 520 example 9 clear induct=0.33;//henry c=0.065*10^-12;//farad c1=1*10^-12;//farad r=5.5*10^3;//ohm //(1) series resonant frequency freque=(1/(2*(3.14)))*sqrt(1/((induct)*c)); disp("frequency = "+string((freque))+"hertz"); //(2)exceed of frequency ratio1=sqrt((1+(c/c1))) disp("ratio parallel...
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//Ex12_3 Pg-588 clc L=0.33 //inductance in henry C=0.065*10^(-12) //capacitance in farad Cm=10^(-12) //capacitance in farad R=0.55*10^(3) //resistor R in ohm disp("Series resonant frequency, fs = 1/2*pi*sqrt(L*C)") fs=1/(2*%pi*sqrt(L*C)) printf(" = %.2f MHz \n",fs*1e-6) disp("Q of the crystal ...
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module App { $Classes(Bacchus.Models*)[ export class $Name { $Properties[ public $name: $Type;] }] }
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// Example 3.12 clear all; clc; // Given data N = 120; // Number of fuel rods P = 100; // Reactor power in MW t = 1; // Estimation time of fuel rod after removal in days T = 365; // Tim...
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//To Calculate the Number of Turns in the Primary Coil //Example 39.5 clear; clc; E1=220;//Input Voltage to the Transformer in Volts E2=6;//Output Voltage by the Transformer in Volts N2=18;//Number of Turns in the Secondary Coil N1=(E1/E2)*N2;//Number of Turns in the Primary Coil printf("Number...
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clc; clear; v1=100;//mi/hr ht=10000;//ft //from standard table for static pressure at an altitude p1=1456//lb/ft^2(abs) P1=1456*0.006947;//psi d=0.001756;//slugs/ft^3 //1 mi/hr = 1.467 ft/s p2=p1+(d*(v1*1.467)^2/2);//lb/ft^3 //in terms of gage pressure p2g p2g=p2-p1;//lb/ft^2 //1lb/ft^2 = 0.006947 psi P2=...
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//Forces in various segments of cable //Refer fig. 2.21 (a) and (b) //Apply Lami's theorem at point D T1=250*sind(180-60)/sind(60+45) //N T2=250*sind(90+45)/sind(60+45) //N //Now consider system of forces acting at B //Resolving the forces we get T3=(T2*cosd(60)+200)/cosd(30) //N T4=T3*sind(30)+T2*sind(60) ...
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// To determine the short circuit capacity of each station clear clc; X=1200*100/800;// percent reactance of other generating station Xc=.5*1200/(11*11); Sc=1200*100/86.59;// short circuit MVA of the bus Xf=119.84;// equivalent fault impedence between F and neutral bus MVA=1200*100/Xf; mprintf("short circuit ...
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// Exa 5.13.12 clc; clear; close; // Given data V_CC = 9;// in V V_BE = 0.8;// in V V_CE = 0.2;// in V R_B = 50;// in kΩ R_C=2;// in kΩ R_E = 1;// in kΩ bita=70; // Applying KVL to input loop, V_CC= I_B*R_B +V_BE +I_E*R_E // V_CC- V_BE= (R_B+R_E)*I_B + R_E*I_C (i) // Applying KVL to output loop, ...
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//Example 19.8 clc mo=4*%pi*10^-7//Tm/A d=0.1//in m x=1*10^-4//F/l I=sqrt((x*2*%pi*d)/mo) disp(I,"Current in A=")
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//Variable Declaration El=50 //Elevation Angle(degrees) h0=0.6 //Earth station altitude(km) hr=3 //Rain height(km) R01=10 //Point Rain Rate(mm/hr) f=12 //frequency(GHz) ah=0.0188 bh=1.217 av=0.0168 bv=1.2 //Calculation Ls=(hr-h0)/sin(El*3.142/180) //Slant path length(km) LG=Ls*cos(El*3.142/180) //Horizonta...
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//Ex10_3 Pg-517 clc R1=10*10^(3) //resistor R1 in ohm Rf=50*10^(3) //feedback resistor in ohm Vin=10*10^(-3) //input voltage in V Ro=5000 //load resistor in ohm disp("A'' = Vo/Vi = (-1)*Rf/R1*(1+1/A*(1+Rf/R1))^-1 ") A=5000 Vo=Vin*(Rf/R1)/(1+1/A*(1+Rf/R1)) //output voltage printf("\n When gain A=%.0f",A) ...
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//Fiber-optics communication technology, by Djafer K. Mynbaev and Lowell L. Scheiner //Example 8.1.1 //windows 8 //Scilab version-6.0.0 clc; clear ; //given //case 1 d1=65.5E-6;//diameter of the core considering 62.5+3 in m d2=59.5E-6;//diameter of the core considering 62.5-3 in m Losscore=-10*log10((d2/...
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// Theory and Problems of Thermodynamics // Chapter 8 // Power and Refrigeration Cycles // Example 5 clear ;clc; //Given data P = 10 // condenser pressure in kPa P1 = 2.5 // leaving pressure of steam in MPa P2 = 3.5 // leaving pressure of steam in MPa T1 = 5...
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clc //initialisation of variables c= 0.1 //M p= 1.34 //per cent T= 25 //C //CALCULATIONS C1= c*p/100 C2= c*p/100 C3= c-C1 Ka= C1*C2/C3 //RESULTS printf (' ionization constant = %.2e ',Ka)
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//chapter 14 //example 14.4 //page 441 clear all; clc ; //given Av=1;//voltage follower printf("\nc1=500 pF\nc2=2000 pF\nc3=1000 pF")
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Df=100//in mm bf=1500//in mm bw=300//in mm d=700//in mm Ast=4510//in sq mm fy=250//in MPa fck=15//in MPa Asf=round(0.36*fck*bf*Df/0.87/fy)//area of steel required for flange, in sq mm //as Ast>Asf, Xu>Df Xu=round((0.87*fy*Ast-0.446*fck*(bf-bw)*Df)/0.36/fck/bw)//in mm Xc=0.531*d//Xc>Xu; hence OK a=0.43*Xu//as...
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// FUNDAMENTALS OF ELECTICAL MACHINES // M.A.SALAM // NAROSA PUBLISHING HOUSE // SECOND EDITION // Chapter 4 : DIRECT CURRENT GENERATORS // Example : 4.10 clc;clear; // clears the console and command history // Given data Pw = 12 // power in kW P = 4 // number of poles Z = 500 ...
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<?xml version="1.0" encoding="UTF-8" standalone="yes"?> <AutoTestC version="2.0.0"> <Pulse>CUSTOM WAVE</Pulse> <Title>Waveform</Title> <Organization>Ford</Organization> <Standard>ES-XW7T-1A278-AC 2003</Standard> <Item>16.0 Immunity to Voltage Offset CI 250</Item> <voltage>13.5</voltage> <cou...
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clc;funcprot(0);//Example 5.24 //Initilisation of Variables Tm=45;.....//Inlet Temperature of water in degrees celcius a=0.02;......//Length of the tube in m b=0.03;.......//width of the tube in m Tw=85;........//Temperatube of tube wall in degrees celcius //Properties of water at 55 d c rho=985.5;......//Densit...
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// Example 4.3 // A program that reads a character from the keyboard and then print in reverse //case ,that is,if input is in upper case,the output will be lower case and vice-versa disp("Enter an alphabet"); alphabet=scanf("%c"); //Reading character if((ascii(alphabet))>=9...
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; testing loading QF_RDL (set-logic QF_RDL )
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clc clear //input ep=150*10^3 //electric energy to primary e=0.69 //efficieny t=70 //time //calculation es=e*ep//transformer equation ps=es/t//power //output printf("the power output is %3.3e W",ps)
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data node { int val; node next; }. pred ll<n,u,sm,lg> == self=null & n=0 & sm<=lg & u=1 or self::node<v, null> & n=1 & sm<=v<=lg & u=2 //or self::node<v, null> & n=1 & sm<=v<=lg & u=1 or self::node<v, q> * q::ll<n-1,_,sm,lg> & q!=null & sm<=v<=lg & u=2 inv n>=0 & sm<=lg. pred ll2<n,u,...
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clc; //e.g8.3 a=0.967; Ie=10*10**-3; Ic=Ie*a;//a=Ic/Ie disp('mA',Ic*10**3,"Ic="); Ib=Ie-Ic; disp('mA',Ib*10**3,"Ib=");
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clear(); stacksize(100000000); function im = quantification(img,n) quantificateur = 2^n im = img ./(256/quantificateur) im = im.*(256/quantificateur) endfunction function im = surEch(img,n) img = im2double(img); colonne = size(img,1)*n ligne = size(img,2)*n im = zeros(colonne,ligne) for i = 1 : c...
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clc //initialisation of variables T=2000 //F T1=1000//F T2=300//F T3=500//F s1=0.4369 s2=1.7085 p1=100 //lbf/in^2 T0=537 //F M=1009.5 h=0.26 Me=M/(h*(T-T1)) h1=269.6//lbf/in^2 h2=1279.1//lbf/in^2 p=2460/1460//lbf/in^2 //CALCULATIONS We=(h1-h2)-T0*(0.4369-1.7085)//Btu/lbm Wre=Me*(h*(T-T1)-T0*(h*log(p...
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clc //initialisation of variables d=163*48.5//cfs a=48.5//ft q=100//cfs Q=45.5*a//cfs c=0.57//cfs v=1.8//cfs p=0.45//ft //CALCULATIONS P=d/(q*sqrt(a))//percent C=Q/(a^0.8*(1+2*a^-0.3))//cfs d1=2.6//cfs T=(1-p*c+v*c*2)//cfs //RESULTS printf('the meyers rating =% f percent',P) printf('the magnitude of the...
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//Problem 5.09: //initializing the variables: nCO2 = 7.5 nCO = 1.3 nO2 = 8.1 nN2 = 83.1 //calculation: //Determine the amount of oxygen fed for combustion. Since nitrogen does not react (key component), using the ratio of oxygen to nitrogen in air will provide the amount of oxygen fed: O2f = (21/79)*83.1 /...
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b=250//column width, in mm D=500//column depth, in mm lex=4//in m ley=4//in m Pu=300//in kN Asc=1472//in sq mm Ast=1472//in sq mm fck=15//in MPa fy=250//in MPa c=50//cover, in mm Max=Pu*10^3*D/2000*(lex/(D/10^3))^2/10^6//in kN-m May=Pu*10^3*b/2000*(ley/(b/10^3))^2/10^6//in kN-m Puz=(0.45*fck*(b*D-(Asc+Ast))...
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load MRAM8.hdl, output-file MRAM8.out, output-list time%S1.4.1 in%B1.16.1 address%B1.3.1 load%B2.3.2 out%B1.16.1; set in %B1101101010101101, set address %B100, set load 1, tick, output; tock, output; set in %B0110000100001001, set address %B010, set load 0, tick, output; tock, output; set in %B00000101011101...
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// Accessing a Scilab string inside an interface Mystr='My string'; ex15c()
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//CHAPTER 8- DIRECT CURRENT MACHINES //Example 7 disp("CHAPTER 8"); disp("EXAMPLE 7"); //VARIABLE INITIALIZATION p_o=20*746; //output power from H.P. to Watts (1 H.P.=745.699 or 746 W) v_t=230; //in Volts N1=1150; //speed in rpm P=4; //...
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//STACK USING CIRCULAR LINKED LIST funcprot(0) function[link2]=append(ele,link1) link2=list(0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,,0,0); if(link1(1)(1).add==0) link1(1)(1).data=ele; link1(1)(1).add=1; link1(1)(1).nexadd=1; link2(1)=link1(1)(1); else if(link1(1)(1)...
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function []=Verification(q,z,f,p) /////////////////////////////////////////////////////////////////////////////// // // // VERIFICATION DES EQUATIONS D'EQUILIBRE D'UN RESEAU DE DISTRIBUTION D'EAU // // ...
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//Calculating the rotor current at slip 3 precent and when the rotor develops maximum torque //Chapter 4 //Example 4.33 //page 351 clear; clc; disp("Example 4.33") E20=100;...............................//induced emf between slip terminals in volts E20p=E20/sqrt(3);.......................//induced emf per phas...
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clc //initialization of variables c = 0.92 F = 93 // ft^-1 nu = 2 // cs dl = 63 // lb/ft^3 dg = 2.8 // lb/ft^3 G = 23 //lb/sex //Calculations G11 = c*((dl-dg)^0.5)/(((F)^0.5)*(nu^0.05))// lb/ft^2-sec A = G/G11// ft^2 d = sqrt(4*A/%pi)//ft //Results printf("The diameter of the tower is %.1f ft",d)
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//Example 27_12 clc(); clear; //To estimate the age of the axe handle n_no=0.034 t1=5730 //Units in Years t=-(log(n_no)*t1)/0.693 //Units in Years printf("The age of the axe handle is t=%d years",t) //In textbook answer is printed wrong as t=28000 years correct answer is t=27958 years
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load LogicCircuit.hdl, output-file LogicCircuit.out, compare-to LogicCircuit.cmp, output-list in%B1.16.1 zin%B2.1.2 nin%B2.1.2 out%B1.16.1; set in %B0000000000000000, set zin %B0, set nin %B0, eval, output; set nin %B1, eval, output; set nin %B0, set zin %B1, eval, output; set nin %B1, eval, output; set in %B11111...
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function y = f(m) g = 9.8 cD = 0.25 t = 4 v = 36 y=sqrt(g*m/cD)*tanh(sqrt(g*cD/m)*t)-v endfunction
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clc //initialisation of variables c=0.005//cm v=2*10^-1/2//N.s/m^2 l=7.5//cm mu=2*10^-2//N.s/m^2 d=5*10^-2//m N=1500*(1/60)//rev/s L=7.5*10^-2//m //CALCULATIONS F=(2*(mu)*(%pi)^2*(d)^2*N*L)/(c*10^-2)//N HP=(F*(%pi)*d*N)/746//hp //RESULTS printf('The friction horsepower loss is=% f hp',HP)
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function p=parallel(r1,r2) p=r1*r2/(r1+r2) endfunction R=parallel(60,120) //Mesh Analysis A=[6,-4;-4,12] I=inv(A)*[2.4;-3.6] I3=I(1)-I(2) I60=I3*120/(120+60) disp(I60)
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// chapter 8 // example 8.27 // Determine range of duty cycle, peak-to-peak choke ripple current and average supply current // page-523-524 clear; clc; // given Edc=24; // in V (dc source) f=50; // in KHz (switching frequency) L=500; // in uH E0=15; // in V (average output voltage) Edc_max=26; // in V (maximum voltage ...
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//Example 7.10 //Calculate the view factors of the surfaces. //Given ds=0.3 //m, diameter of shell r1=0.1 //m, distance from the centre //Calculation //by the defination of view factor F12=1 printf("The view factor from surface 1 to 2 is %f\n",F12) //F21 R=ds/2 //m, radius of sphere ...
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@relation distaa1828a1828 @attribute 5.3 real[0.19,16.0] @attribute 5.5 real[0.0,12.0] @attribute green{green,red} @inputs 5.3,5.5 @outputs green @data red red green green green green green green red red green green green green green green green green green green green green green green green green
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// Example 2.2.b:Numerical Aperture clc; clear; close; n1=1.50;//Waveguide Refractive Index n2=1.47;//Cladding Refractive Index NA=sqrt(n1^2-n2^2);// Numerical Aperture disp(NA,"Numerical Aperture is")
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// Data Reconciliation Benchmark and GED Problems From Lietrature Review // Author: Edson Cordeiro do Valle // Contact - edsoncv@{gmail.com}{vrtech.com.br} // Skype: edson.cv // we have 4 functions in this file: // -generate_data // -generate_data_multiple // -generate_data_random_err // -generate_data_errors function ...
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// Initilization of variables a=10 // m/s^2 // acceleration of the particle S_5th=50 // m // distance travelled by the particle during the 5th second t=5 // seconds // Calculations // The distance travelled by the particle in time t is given by, S=(u*t)+(1/2)*a*t^2.....(consider this as eq'n 1) // Here, The dista...
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//Book Name:Fundamentals of Electrical Engineering //Author:Rajendra Prasad //Publisher: PHI Learning Private Limited //Edition:Third ,2014 //Ex10_9.sce clc; clear; P_in_HP=10; eta=0.9; pf=0.8; Vl=400; Vsc=160; Isc=7.2; P_in_watt=P_in_HP*735.5; If=P_in_watt/(sqrt(3)*Vl*pf*eta); Isc_400=Isc*Vl/Vsc; I...
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Ex1_3.sce
//Example 1_3 page no:3 clc clear E=50;//Energy in joules t=2.5;//Time in second P=E/t; disp(P,"Power(in watts):")
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b0=1; //m=pi.*a.^2.*I/c ring current I radius a l=1; x=-1.0:0.1:1.0; y=0.1:0.1:2.0; [X,Y]=meshgrid(x,y); bx=b0.*cos(%pi*X./l).*exp(-%pi.*Y./l); by=-b0.*sin(%pi*X./l).*exp(-%pi.*Y./l); bmag=sqrt(bx.^2+by.^2); contour(x,y,bmag',[0.005 0.006 0.007 0.008 0.009 0.01 0.05 0.1 0.15 0.2 0.25 0.3 .3...
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//1.20 clc; T=.5*10^-3; V=10; Vp=0.6*V+0.5; Ip=5*10^-3; Rmax=(V-Vp)/Ip; printf("Rmax=%.0f ohm", Rmax) C=1*10^-6; R=T/(C*log(1/(1-0.6))); printf("\nR=%.1f ohm", R) disp('since the value of R is less than Rmax so the value is suitable')
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 10.3\n\n\n"); // Chapter 10 : Refrigeration // Problem 10.3 (page no. 505) // Solution T1=70+460; //70F=70+460 R //Energy flows into the system at reservoir at constant temperature T1(unit:R) T2=20+460; //20F=20+460 R //Heat is rejected to the constant...
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n=6; lambda = 0.2; alpha = 50; x=0.5:.1:50; l = length(x); s=zeros(1,l); //for i = 1:l // s(i) = (n-1)*lambda*x(i)/((n-1)*lambda +1); //end for i = 1:l s(i) = sigma(x(i),n,alpha,lambda); end clf() plot(x,s,color='black', 'Linewidth', 2) //plot(x,0.5*x, color = 'blue', 'Linewidth', 2)
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clc; //for part (i) hf=89.8;//kJ/kg x=0.95; h_fg=(1420-89.8);//kJ/kg hi=hf+x*h_fg;//kJ/kg disp("enthalpy of part (i)"); disp("kJ/kg",hi); //for part (ii) //ammonia heated by (60-20) K x=40/50; hf=1462.6;//kJ/kg h_fg=(1597.2-1462.6);//kJ/kg hii=hf+x*h_fg; disp("enthalpy of part (ii)"); disp("kJ/kg",hi...
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clear //Given q=1.6*10**-19 //C r=0.6 //m m=1.67*10**-27 //Kg f=10**7 //Calculation // B=(2*%pi*m*f)/q K=((B**2*q**2*r**2)/(2.0*m))/1.6*10**-13 //Result printf("\n Kinetic energy of the protons is %0.1f Mev",K*10**26)
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clc; // (a).Restoring to initial state by throttling process T1=303; //Temperature of air at state 1 in kelvin p1=1; //Pressure of air at state 1 in bar p2=5; //Pressure of air at state 2 in bar p3=1;//Pressure of air at state 3 in bar T3=303; //Temperature of air at state 3 in kelvin Cpo=1.0035; // Specific he...