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Ex12_2.sce
// chapter 12 // example 12.2 // fig. 12.8 // Determine resonant frequency, maximum operating frequency,Peak thyristor current, average thyristor current, rms thyristor current, rms load current and average supply current // page-760-762 clear; clc; // given C1=4; // in uF C2=4; // in uF Lr=40; // in uH R=2; // in ohm ...
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Ex12_7.sce
// Initilization of variables A= 50 // cm^2 // area of the shaded portion J_A=22.5*10^2 // cm^4 // polar moment of inertia of the shaded portion d=6 // cm // Calculations J_c=J_A-(A*d^2) // substuting the value of I_x from eq'n 2 in eq'n 1 we get, I_y=J_c/3 // cm^4 // M.O.I about Y-axis // Now from eq'n 2, I_...
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Ex4_3.sce
//Example 4.3, Page Number 158 //The Function fpround(dependency) is used to round a floating point number x to n decimal places clc; d=0.2*(10**-3) //Chip Diameter in meter d1=1 //Distance in Meter l=550*(10**-9 ) //Wavelength in Meter q=0.001 //External Quantum Efficiency i=50*(10**-3) //Operational Current ...
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Example3_7.sce
//Example 3.7 //Program to estimate rms pulse broadening per kilometer for the fiber clear; clc ; close ; //Given data lambda=0.85*10^(-6); //metre - WAVELENGTH L=1; //km - DISTANCE MD=0.025;//MATERIAL DISPERSION = mod(lamda^2*[del^2(n1)/del(lamda)^2) c=2.998*10^8; //m/s - ...
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clc //initialisation of variables d1=0.67//ft h1=2.00//ft h2=4.04//ft hv1=0.062//ft hv2=0.254//ft d=0.19//ft h=0.2//ft h1=0.04//ft q=0.644//ft q1=0.65//ft v=0.92//ft d2=6.5//ft v1=3.69//ft d3=0.542//ft hv3=0.21//ft delv=0.15//ft d4=0.02//ft //CALCULATIONS H=d1+hv1//ft H1=d1+hv2//ft he=h*d//ft hi=...
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Example42_3.sce
//Given that E = 7 //in ev V = 2*10^-9 //in m^3 density = 2*10^28 //in m^3/ev deltaE = 3*10^-3 //in ev //Sample Problem 42-3a printf("**Sample Problem 42-3a**\n") n = density*V printf("The number of states are equal to %1.2e per ev\n", n) //Sample Problem 42-3b printf("\n**Sample Problem 42-3b**\n") ...
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//Ex10_6 clc x='110'; disp("Octal number="+string(x))// octal value str=oct2dec(x)//octal to decimal disp("Eqivalent Decimal number="+string(str))
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clc //initialisation of variables W= -10 //KN/m Yac= 7 //m xad= -7.5 //m xac= -15 //m xcb= 10 //m //CALCULATIONS k= Yac/((xac)^2) yb= k*(xcb)^2 hb= Yac-yb yd= k*(xad)^2 hd= Yac-yd A=[(xcb-xac),(hb);(xcb),(-yb)] b=[-W*(-xac)*(-xad);0] c= A\b Rbv= c(1,1) Rbh= c(2,1) Rah= Rbh Rav= -Rbv-W*(-xac) dybydx=...
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Culvert_design.sce
disp("15CS101L"); disp("Programming Laboratory"); disp("Internet Programming lab"); disp("Culvert Design and Analysis"); disp("Mr. M. Mohamed Rabik"); disp("Aryaman Dhanda , RA1511003010481"); disp("Naman Maheshwari , RA1511003010471"); disp("Sidharth Suresh , RA1511003010477"); disp("Select Pipe material and i...
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3_37.sce
clc; clear; //Example 3.37 k=0.03 //W/(m.K) Npr=0.697 //Prandtl number v=2.076*10^-6 //m^2/s Beta=0.002915 //K^-1 D=25 ; //[Diameter in cm] D=D/100 //[m] Tf=343 //Film temperature in [K] A=%pi*(D/2)^2 //Area in [m^2] P=%pi*D //Perimeter [m] T1=293 //[K] T2=393 //[K] g=9.81 //[m/s^2] //C...
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//exa 2.20 clc;clear;close; format('v',6); //F1=0.004*P1^2+2*P1+80;//Rs./hr //F2=0.006*P2^2+1.5*P2+100;//Rs./hr P=250;//MW P1=poly(0,'P1');P2=poly(0,'P2'); dF1bydP1=2*0.004*P1+2; dF2bydP2=2*0.006*P2+1.5; //Let loads are P1 & P-P1 //Economical loading lambda1=lambda2 eqn=2*0.004*P1+2-2*0.006*(P-P1)-1.5; P1=r...
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Ex9_3.sce
//Chapter 9 : Eigenvalues and Eigenvectors //Example 9.3 //Scilab 6.0.1 //Windows 10 clear; clc; A=[-3 1 -1;-7 5 -1;-6 6 -2]; disp(A,'A=') eig=spec(A) disp(eig,'eigen values are:') e4=A-4*eye(3,3) mprintf('\n(A-4I3)x=') disp('*',e4) mprintf(' [x\n y\n z]=') z=zeros(3,1) disp(z) mprintf('\nthis reduc...
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17_02.sce
//pathname=get_absolute_file_path('17.02.sce') //filename=pathname+filesep()+'17.02-data.sci' //exec(filename) //Indicator diagram area & length(in m^2 & m): A=40*10^(-4) l=0.08 //Bore(in m): D=0.15 //Stroke(in m): L=0.20 //Rpm of motor: N=100 //Spring constant(in Pa/m): k=1.5*10^8 //Mep(in Pa): mep=A*k/...
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Ch06Exa12.sce
// Scilab code Exa6.12: : Page-244(2011) clc; clear; h_kt = 1.05457e-34; // Reduced planck's constant, joule sec c = 3e+08; // velocity of light, metre per sec m_e = 9.1e-31; // Mass of the electron, Kg ft_O = 3162.28; // Comparative half life for oxygen ft_n = 1174.90; // Com...
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//chapter-9 page 412 example 9.9 //============================================================================== clc; clear; //For an IMPATT diode Lp=0.5*10^(-9);//Inductance in Henry Cj=0.5*10^(-12);//Capacitance in Farad Ip=0.8;//RF peak current in A Rl=2;//Load Resistance in ohms Vbd=100;//Breakdown Volt...
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EX14_2.sce
//Book - Power System: Analysis & Design 5th Edition //Authors - J. Duncan Glover, Mulukutla S. Sarma, and Thomas J. Overbye //Chapter - 14 ; Example 14.2 //Scilab Version - 6.0.0 ; OS - Windows clc; clear; MVAtr1=40; //MVA FOA rating of transformer 1 MVAtr2=40...
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//========================================================================== // chapter 6 example 3 clc; clear; //input data t1 = 20; // temperature in °C alpha = 5*10^-3; //average temperature coefficient at 20°C R1 = 8; //resistance in ohm R...
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Ex7_5.sce
clc clear Vc=5*(10^-4); D=0.15; L=0.2; Vs=(22/7)*D*D*L*(1/4); r=(Vc+Vs)/Vc; G=1.4; Ea=[1-(1/(r^(G-1)))]; Eith=0.3; Erel=Eith/Ea; printf('Erel= %3.2f Percent',Erel*100); printf('\n'); Pm=500; //in kPa n=1000/2; IP=(Pm*Vs*n)/60; printf('IP= %3.2f kW',IP); printf('\n');
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6_17.sce
clc //initialisation of variables f= 0.008 l= 2000 //ft p1= 34 //ft p2= 8 //ft p3= 4 //ft g= 32.2 //ft/sec^2 d= 18 //in P= 140 //ft l1= 9500 //ft //CALCULATIONS v= sqrt((p1-p2-p3)*2*g/((d/12)+(4*f*l/(d/12)))) Q= %pi*(d/12)^2*v/4 v1= sqrt(P*2*g/((d/12)+(4*f*l1/(d/12)))) Q1= %pi*(d/12)^2*v1/4 //RESULTS ...
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//difference in 3 significant figures //example 1.8 //page 11 clc;clear;close; X1=sqrt(6.37); X2=sqrt(6.36); d=X1-X2;//difference between two numbers printf('the differencecorrected to 3 significant figures is %0.3g',d);
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ex2_10.sce
//Ex2.10 //let input wave be V_in=V_p_in*sin(2*%pi*f*t) f=1; //Frequency is 1Hz T=1/f; R_1=100; //Resistances in ohms R_L=1000; //Load V_p_in=10; //Peak input voltage V_th=0.7; //knee voltage of diode clf(); V_p_out=V_p_in*(R_L/(R_L+R_1)); //peak output voltage disp(V_p_out,'peak output vol...
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example4_5.sce
//clc() V = 250;//L T = 300;//K V1 = 1000;//L P1 = 100;//kPa T1 = 310;//K P = T * P1 * V1 /(T1 * V); disp("kPa",P,"Original pressure in the cylinder = ")
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Ex15_2.sce
clc // initialization of variables clear d=250 //mm c=30 //mm t=25 //mm // part (a) a=5 //mm lambda=a/(2*c) f1l=1.22 //from the tble f2l=1.02 //We don't know P yet so say P=1 P=1 Sfl=P/(t*2*c)*f1l+3*280*P*f2l/(2*t*c^2) K_IC=59*sqrt(1000) P=K_IC/(Sfl*sqrt(a*%pi)) printf('part (a)') printf('\n P = %.1f ...
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exa1_6_4.sce
//Caption:solution_of_differential_equation // example 1.6.4 //page 10 //after taking laplace transform and applying given condition, we get : //Y(s)=(6*s+6)/((s-1)*(s+2)*(s+3)) s=%s; syms t [A]=pfss((6*s+6)/((s-1)*(s+2)*(s+3))) F1 = ilaplace(A(1),s,t) F2 = ilaplace(A(2),s,t) F3 = ilaplace(A(3),s,t) F=F1+F2+...
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clc p1=1.02*10^5; //Pa T1=295; //K V1=0.015; //m^3 p2=6.8*10^5; //Pa y=1.4; disp("(i) Final temperature") T2=T1*(p2/p1)^((y-1)/y); t2=T2-273; disp("t2=") disp(t2) disp("°C") disp("(ii) Final volume :") V2=V1*(p1/p2)^(1/y); disp("V2=") disp(V2) disp("m^3") disp("(iii)Work done") R=287; m=...
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//chapter 6 //example 6.9 //Calculate average drift velocity of electrons //page 149 clear; clc; //given I=4; // in A (current in the conductor) e=1.6E-19; // in C (charge of electron) A=1E-6; // in m^2 (cross-sectional area) N_A=6.02E23; // in atoms/gram-atom (Avogadro's number) p=8.9; // in g/cm^3 (density...
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example_7_14.sce
s=%s; T=20/(s+10) syms t s; y=ilaplace(T,s,t); T1=20/((s+10)*s) c1=ilaplace(T1,s,t) T2=20/((s+10)*s^2) c2=ilaplace(T2,s,t)
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EX11_8.sce
//Example11.8 // determine the feed back current If and analog output voltage clc; clear; close; Vref = 5 ; BI = 101 ; BI = 011 ; BI = 100 ; BI = 001 ; Rf = 25*10^3 ; R = 0.2*Rf ; // The output current of given R-2R ladder D/A converter is defined as // If = -(Vref/2*R)*(2^0*b0+2^-1*b1+2^-2*b2) ;...
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//Chapter 21, Problem 10 clc; f=50; //frequency n1=25; //primary turns n2=300; //secondary turns A=300e-4; //cross-sectional area of the core v1=250; //primary voltage phim=v1/(4.44*f*n1); //flux Bm=phim/...
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11_04.sce
//pathname=get_absolute_file_path('11.04.sce') //filename=pathname+filesep()+'11.04-data.sci' //exec(filename) //Height of chimney(in m): H=60 //Ambient air temperature(in K): Ta=17+273 //Temperature of burnt gases(in K): Tg=300+273 //Temperature of the artificial burnt gases(in K): Tga=150+273 //Mass per kg...
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Ex1_10.sce
//Ex1.10 clc B = 2*10^-6 //magnetic flux density V = 4*10^6 //electron velocity e= 1.6*10^-19//elcetron charge disp("B ="+string(B)+"ax wb/m.sq") disp("V ="+string(V)+"az m/s") disp("e = "+string(e)+ "C") disp("F = e[VxB] ="+string(e*V*B)+"ay N")//force
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9_4.sce
clc //initialisation of variables M1= 208.3 //gms g= 2.69 //gms R= 0.08205 //l-atm mole^-1 deg^-1 T= 250 //C P= 1 //atm V= 1 //lit //CALCULATIONS M2= g*R*(273.1+T)/(P*V) a= (M1-M2)/M2 Kp= a^2*P/(1-a^2) //RESULTS printf ('Kp= %.2f ',Kp)
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Ex7_7_u1.sce
//Example 7_7_u1 clc(); clear; //To calculate the fiber length alpha=0.5 //units in db/KM it=2*10^-6 //units in W i0=1.5*10^-3 //units in W l=-1*(10/alpha)*log10(it/i0) //units in KM printf("The length of the fiber is L=%.1f KM",l)
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EX6_4_C.sce
//example 6.4.C //check the signal is periodic or not clc ; n=-15:0.01:15; y =(1+cos(2*(%pi)*n/8)/2); xlabel('n') ylabel('x(n)') plot(n,y); disp ( 'Plot shows that given signal is periodic of fundamental period=4 samples' ) ;
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funcprot(0); // Initialization of Variable function[dms]=degtodms(deg) d = int(deg) md = abs(deg - d) * 60 m = int(md) sd = (md - m) * 60 sd=round(sd*100)/100 dms=[d m sd] endfunction Long=30.0;//longitude in degrees GAT=13+15.0/60+10.0/3600;//GAT in hr ET=6.0/60+15.35/3600+0.3/360...
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//check o/p when the i/p is a cosine value t = 0:0.001:1-0.001; X = cos(2*%pi*100*t); r = rssq(X); disp(r); //output // 22.36068
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ex8_1.sce
// Exa 8.1 clc; clear; close; // Given data C= 85;// in % H= 12.5;// in % H1 = 35000;// heat liberated by carbon in kJ H2 = 143000;// heat liberated by hydrogen in kJ HCV = (C*H1+H*H2)/100;// Higher calorific value in kJ/kg disp(HCV,"Higher calorific value in kJ/kg is"); ms = 9; LCV= HCV -(ms*H*2442)/100 ;/...
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// Exa 3.2 clc; clear; close; format('v',7) // Given data Im = 141.4;// in A t = 3;// in ms t = t * 10^-3;// in sec disp(Im,"The maximum value of current in A is"); omega = 314;// in rad/sec // omega = 2*%pi*f; f = round(omega/(2*%pi));// in Hz disp(f,"The frequency in Hz is"); T = 1/f;// in sec disp(T,"...
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clc disp("the soln of eg 3.5-->Flash calc. using Modified Raoult law"); a12=292.66*4.18, a21=1445.26*4.18, v1=74.05*10^-6, v2=18.07*10^-6, R=8.314 t=100,z1=.3, z2=1-z1 a1=14.39155, a2=16.262, b1=2795.82, b2=3799.89, c1=230.002, c2=226.35 e1=1,e2=1,e3=1,e4=1,e5=1,e6=1,vnew=0 ...
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//chapter-12,Example12_2,pg 383 fc=10^6//carrier frequency m=0.4//modulation index fs=100//signal frequency V=2//(+/-)2V range delfc1=m*fc//frequency deviation for FS(full scale) //(+/-) 2V corresponds to delfc Hz deviation assuming linear shift, for (+/-)1V delfc2=delfc1/V//frequency deviation for...
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Ex4_2_2.sce
clc //initialization of variables d = 10 //cm s = 3 // km v = 500 //cm/sec nu = 0.15 // cm^2/sec //Calculations E = 0.5*d*v // cm^2/sec c1 = 1000 // m/km c2 = 1/100 // m/cm z = sqrt(4*E*c1*c2*s/v) percent = z*100/(s*c1) //Results printf(" The percent of pipe containing mixed gases is %.1f percent",percent...
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example18_13.sce
clc // Given that N = 6.5e25 // no. of atom per m^3 T = 300 // room temperature in K mu_ = 4 * %pi * 1e-7 // magnetic permittivity of space k = 1.38e-23 // Boltzmann's constant in J/K m = 9.1e-31 // mass of electron in kg e = 1.6e-19 // charge in an electron in C h = 6.62e-34 // Planck constant in J sec // Sample Pro...
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clc // Given that lambda = 6000 // wavelength of light in angstrom N = 200 // Grating element n = 3 // order d = 0.025 // diameter of wire in mm // Sample Problem 21 on page no. 165 printf("\n # PROBLEM 21 # \n") printf(" Standard formula used \n") printf(" n*lambda= sin(theta)/N \n") theta = 180/%pi*asin(N*...
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EX20_48.sce
clc;funcprot(0);//EXAMPLE 20.48 // Initialisation of Variables rp=4;........//Stagnation pressure ratio etaisen=0.85;.....//Stagnation isentropic efficiency p1=1;.............//Inlet stagnation pressure in bar t1=300;...........//Inlet stagnation temperature in K Rd=0.5;............//Degree of reaction Cu=180;.....
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2_6.sce
clc clear //Initialization of variables gam=62.4 x1=4 //ft x2=6 //ft y1=6 //ft z=8 //ft dy=1 //ft angle=60 //degrees //calculations A1=x1*x2 A2=1/2 *y1^2 yc = (A1*(x1+x2+dy) + A2*(x1+x2))/(A1+A2) hc=yc*sind(angle) F=hc*gam*(A1+A2) ic1=1/12 *x1*y1^3 ic2=1/36*y1*x2^3 ad1=A1*(x1+x2+dy-yc)^2 ad2=A2*(x1+...
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Bilinear_trans_using_butterworth.sce
clc; F1=input("Enter the pass band edge (Hz)="); F2=input("Enter the stop band edge (Hz)="); kp=input("Enter the pass band attenuation (-dB)="); ks=input("Enter the stop band attenuation (-dB)="); Fs=input("Enter the sampling rate (Hz)="); w1=2*%pi*F1*1/Fs; w2=2*%pi*F2*1/Fs; o1=2*Fs*tan(w1/2); o2=2*Fs*tan(w2/2); num=lo...
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ex4_23.sce
// Exa 4.23 clc; clear; close; format('v',6) // Given data V=230;// in V f= 50;// in Hz Z1= 10*expm(-30*%i*%pi/180);// in ohm Z2= 20*expm(60*%i*%pi/180);// in ohm Z3= 40*expm(0*%i*%pi/180);// in ohm Y1= 1/Z1;// in S Y2= 1/Z2;// in S Y3= 1/Z3;// in S Y= Y1+Y2+Y3;// in S phi= atand(imag(Y),real(Y));// in °...
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ex_24_14.sce
//find.. clc //solution //given P=15000//W N=900//rpm n=4 R=0.15//m u=0.25 //let m be the mass w=2*%pi*N/60//rad/s w1=(3/4)*w//rad/s r=0.12//m //Pc=m*w^2*r=1066*m//N //Ps=m*w1^2*r=600m//N T=P*60/(2*%pi*N)//N-m //T=u*(Pc-Ps)*R*n=70m m=T/70//kg printf("mass of shoes is,%f kg\n",m) a=%pi/3 l=R*a*1000//...
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c9_12.sce
//(9.12) Air enters a turbojet engine at 0.8 bar, 240K, and an inlet velocity of 1000 km/h (278 m/s). The pressure ratio across the compressor is 8. The turbine inlet temperature is 1200K and the pressure at the nozzle exit is 0.8 bar. The work developed by the turbine equals the compressor work input. The diffuser,...
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Example7_15.sce
////Example 7.15 clc; clear; close; format('v',6); //Given data : g=9.81;//gravity constant D1=50/1000;//meter D2=100/1000;//meter l1=100;l2=100;//meter hf1=10;//meter(level difference) f=0.008;//coeff. of friction Q2BYQ1=sqrt((l1/l2)*(D2/D1)^5);//as hf1=hf2 Q1=sqrt(hf1/f/l1*(3.0257*D1^5));//m^3/sec Q2=Q2...
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ex1_20.sce
clc; p=200; //power in Watt v=12; //voltage in volt i=p/v; //calculating current in Ampere I=p/6; //calculating disp(i,"Current in Ampere = "); //displaying disp(I,"Current in Ampere if voltage were 6V = "); //displaying result
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Ex4_9.sce
// Problem no 4.4.9,Page No.99 clc;clear; close; M_C=40 //KNM //Moment at Pt C w=20 //KNm //u.d.l on L_AD L=10 //m //Length of beam L_CB=5 //m //Length of CB L_DC=1 //m //Length of DC L_AD=4 //m //Length of AD //Calculations //Let R_A & R_B be the reactions at A & B //R_A+R_B=80 //Taking Moment at A //M_A=0=R_B*L-M...
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Example5_1.sce
//Example 5.1 //Program to Calculate Collector and Base Currents clear; clc ; close ; //Given Circuit Data alpha=0.98; //alpha(dc) Ico=1*10^(-6); //Ampere Ie=1*10^(-3); //Ampere //Calculation Ic=alpha*Ie+Ico; //Collector Current Ib=Ie-Ic; //Base Current //Displaying The Results in Command Window printf("\...
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Example11_3.sce
//Example 11.3 clc; clear; close; format('v',9); //Given data : l=2;//meter d0=0;//meter d1=0.3;//meter d2=1.0;//meter d3=1.2;//meter d4=1.6;//meter d5=2.0;//meter d6=1.4;//meter d7=1.0;//meter d8=0.4;//meter d9=0.3;//meter d10=0.2;//meter V0=0;//meter V1=0.5;//meter V2=0.7;//meter V3=0.8;//meter ...
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Ex3_12.sce
//EX3_12 PG-3.39 clc Rl=5e3; N1toN2=2;//transformer turns ratio Ep=460;//rms value of primary voltage Es=Ep/N1toN2; Esm=sqrt(2)*Es;//peak value of the secondary voltage Im=Esm/Rl;//We neglect forward diode resistance Idc=2*Im/%pi; printf("\n Therefore DC load current is %f A \n",Idc) Edc=Idc*Rl; printf("\n ...
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B8_Salva_Dados.sce
Arquivo = uiputfile(["*.dinam","Arquivos de Análise Dinâmica"], ... fileparts(pwd()),"Salvar dados da análise"); if ~isempty(Arquivo) then [Path,Name,Extension] = fileparts(Arquivo); Arquivo = Path + Name + ".dinam" EditBoxData = [] Campos = {MaterialData SectionDat...
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E11_1.sce
clc //Initialization of variables S=[10 20 40 80 120 180 300] v=[0.32 0.58 0.9 1.22 1.42 1.58 1.74] //calculations bys=1000/S byv=1/v n=size(S) x=bys y=byv disp("From graph,") m=26.17 c=0.476 //Sx =sum(x); //Sxx =sum(x.*x); //Sy =sum(y); //Syy =sum(y.*y); //Sxy =sum(x.*y); //m =(n*Sxy-(Sx*Sy))/(n*Sxx-...
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M_Ex_3_18.sce
clc //Chapter3: Modulation //Example3.18, page no 172 //Given deltaF=1e6// max freq deviation fm=10e3//modulating freq mf=(2*deltaF)/fm// modulation coefficient BW=mf*fm// bandwidth mprintf('The approximate bandwidth is: %d MHz',BW/1e6)
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//Example 6.27b //x(t)=(cos(3t))^3 clc; syms t; x=(cos(3*t))^3; X=laplace(x);
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function [rs] = solve(ns,ds) // Reken de startpositie uit van de wagens rs = ns a = length(ns) b = max(ns) for i = 1:a for u = 1:b if ds(1,i) == u then x = rs(1,i) rs(1,i) = rs(1,i+u) rs(1,i+u) = x end ...
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Chapter4_Example20.sce
clc clear //INPUT k=5.64*10^-14;//kinetic energy of the hydrogen molecule ergs t=273;//temperature of the oxygen molecule in K r=8.32*10^7;//universal gas constant in ergs //CALCULATIONS N=(3/2)*(r*t/k);//avagadro number //OUTPUT mprintf('the avagadro number is %3.2f',N)
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errcatch(-1,"stop");mode(2);//Example 5_16 ; ; //To calculate the distane between (110) planes a=0.38 //units in nm h=1 k=1 l=0 d=a/sqrt(h^2+k^2+l^2) printf("Distance between (110) planes d = %.2f nm",d) exit();
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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 10, Example 5") //Acceleration due to gravity in m/s^2 g=9.81; //Length of the tube in meters L=1.5; //Temperatu...
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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 2, Example 09") //A thin walled copper tube of outside metal radius r=0.01m carries steam at temprature, T1=400K.It is inside a room where the surrounding air tempr...
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//========================================================================= // chapter 9 example 5 clc; clear; // Variable declaration dr = 12.8 // original diameter of steel wire in mm df = 10.7; // diameter at fracture in mm // Calculations percent_red = (((%pi*dr*dr) - (%pi*df*df))/(%pi*...
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clc; clear all; //chapter 3 //page no 90 //example 3.8 mprintf('(a) The RF burst frequency is 500 MHz\n'); mprintf(' (b) The pulse repetition rate is 1 MHz\n'); f0=10*10^6; //Zero crossing frequency in Hz tau=1/f0; //in second mprintf(' (c) The pulse width is %.1f micro second\n',tau*10^6);
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function y=g(x) //y=x^4+2*x^2-x-3; //y=(-2*x^2+x+3)^(1/4); //y=2 ./(3*x)+(2/3)*x; // Si pongo 2/x oscila y no termina ojo! //y=%pi+0.5*sin(x/2); //y=3/(x^3-3) //y=(x+((3*x+3)/x^3))/2 //y=(x+1)./x^2; //y=(3*x+3)^(1/4) //y=(4*x+1)^(1/4) //y=(3+x-2*x^2)^(1/4) y=((3*x^2.+3)/x)^(1...
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clc; eff_C=0.85; // Isentropic efficiency of the compressor rp=4; // Pressure ratio r=1.4; // specific heat ratio eff_pc=(((r-1)/r)*log (rp))/log (((rp^((r-1)/r)-1)/eff_C)+1); disp ("%",eff_pc*100,"Polytropic efficiency = "); disp ("variation of compressor efficiency with compression ratio is shown in window1"); ...
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Example_27_1.sce
//clear// clear; clc; //Example 27.1 //Given T = 60; //[F] wA = 0.30; //[MgSO4] wB = 0.70; //[H2O] //Solution //From Fig. 27.3 it is noted that the crystals are MgSO4.7H2O //and that the concentration of the mother liquid is xA = 0.245; //[anhydrous MgSO4] xB = 0.755; //[H2O] //Bases: F_in = 1000; //...
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ex5_2a.sce
//Ex 5.2a clc; syms T; disp('x(t)=1+cos(20%pi*t)'); w=20*%pi; f=w/(2*%pi); T=1/(2*f); disp(T,'minimun sampling interval');
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ztransfer.sce
//function// function [Ztransfer]=ztransfer(sequence) z = poly(0, 'z', 'r') Ztransfer=sequence*(1/z)^[0:(length(sequence)-1)]' endfunction
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//Exa 5.9 clc; clear; close; //Given data : A=200;//gain without feedback(unitless) Ri=2;//in kOhm Ro=12;//in kOhm Beta=0.02;//feedbak ratio(unitless) //Part (i) : Af=A/(1+A*Beta);//gain with feedback(unitless) disp(Af,"(i) Gain with Negative Feedback :"); //Part (ii) : Rif=Ri*(1+A*Beta);//in kOhm disp(Ri...
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Ex8_7.sce
clc;clear; //Example 8.7 //given data P1=1000; V=200; T1=300; T0=T1; P0=100; //constants used R=0.287;//in kPa m^3/kg K //calculations m1=P1*V/(R*T1); O1=R*T0*(P0/P1-1)+R*T0*log(P1/P0);// O refers to exergy X1=m1*O1/1000;//factor of 1000 for converting kJ into MJ X1=round(X1); disp(X1,'work obtaine...
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main.sce
clc; clear();close(); exec("catenaria3.sce"); exec("quadrotor.sce"); exec("descenso.sce"); exec("TensVerticales.sci"); exec("quadrotor_izq.sce"); exec("quadrotor_izqmov.sce"); exec("quadrotor_dcha2.sce"); exec("seguidores.sce"); Circuito_x=linspace(0,500,600); Circuito_y=zeros(1,400); Circuito_y(1:150)=l...
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exa_3_14.sce
// Exa 3.14 clc; clear; close; // Given data bita=100; hFE= 100; VCEsat= 0.2;// in V VBEsat= 0.8;// in V VBEactive= 0.7;// in V VBB= 5;// in V VCC= 10;// in V R_C= 3;// in kΩ R_C=R_C*10^3;// in Ω R_B= 50;// in kΩ R_B=R_B*10^3;// in Ω // Formula VCC= ICsat*R_C+VCEsat ICsat= (VCC-VCEsat)/R_C;//A disp(I...
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//scilab 5.4.1 //Windows 7 operating system //chapter 9 Basic Voltage and Power Amplifiers clc clear Vorms=2//Vorms=rms output voltage in the midband region of an amplifier Pa=42//Pa=power gain in dB Pol=0.4//Pol=power output in W at the lower cut-off frequency 100Hz Ri=10^3//Ri=input resistance in ohms VOrms=...
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clc m1=1000 //mass of wet steam in kg vg1=0.06663 vf1=0.0012163 V=(2*m1)/((1/vf1)+(1/vg1)) P1=3*10^5 mprintf("V=%fmetre-cube\n",V)//ans vary due to roundoff error\n mf=V/(2*vf1) mg=V/(2*vg1) mprintf("mass of liquid=%fkg\n",mf)//ans vary due to roundoff erorr mprintf("mass of steam=%fkg\n",mg)//ans vary due t...
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6_3_1.sce
clc //initialisation of variables p= 60 //lbf/in^2 w= 62.4 //lbf/ft^3 l= 1 //ft g= 32.2 //ft/sec^2 //CALCULATIONS i= p*144/(w*l) a= i*g //RESULTS printf ('accelaration of fluid = %.f ft/sec^2',a)
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1_7.sce
// To determine circuit impedance and current in a parallel connection of a resistor and capacitor. clc; clear; R=4700; V=240; f=60; w=2*%pi*f; C=2*(10^-6); Xc=-(1/(C*w))*%i;// Reactance in polar form Ir=V/R; Ic=V/Xc; I=Ir+Ic;// Total current Z=V/I; theta=atand(imag(Z)/real(Z)); mprintf('Imp...
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//Exa 5.12 clc; clear; close; //Given Data : format('v',8); R=2;//in ohm X=3;//in ohm VR=10*1000;//in volt P=1000*10^3;//in watt(power delivered) cos_fir=0.8;//unitless I=P/(VR*cos_fir);//in Ampere Vs=sqrt((VR*cos_fir+I*R)^2+(VR*sqrt(1-cos_fir^2)+I*X)^2);//in volt Reg=(Vs-VR)*100/VR;//in % disp(Reg,"% Reg...
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//Ex 10 clc; clear; close; n1=6;c=10; n2=4;s=6; n=double(lcm(int32([4,6]))); //Number of bananas cp=(c/n1)*n; sp=(s/n2)*n; lossPercent=(cp-sp)/cp*100; printf("The loss is %d percent",lossPercent);
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 11.10\n\n\n"); // Chapter 11 : Heat Transfer // Problem 11.10 (page no. 566) // Solution //A bare steel pipe ro=90; //Outside diameter //Unit:mm ri=75; //inside diameter //Unit:mm Ti=110; //Inside temperature //Unit:Celcius To=40; //Outside temperat...
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//Example 14.2 //Install Symbolic toolbox //Find the Laplace transform syms t s clc z=integ(2*exp(-s*t),t,3,%inf) //The second term will result in zero disp(z,'F(s)=')
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// define a grid x=[-1:0.1:1];y=x; // surface computation [X,Y]=meshgrid(x,y); Z=X.^2-Y.^2; //surface display clf; F=gcf();F.color_map=jetcolormap(64); surf(x,y,Z) colorbar(min(Z),max(Z))
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clear; clc; //Example - 1.9 //Page number - 26 printf("Example - 1.9 and Page number - 26\n\n"); //Given //Antoine equation for water ln(Psat)=16.262-(3799.89/(T_sat + 226.35)) P = 2;//[atm] - Pressure P = (2*101325)/1000;//[kPa] P_sat = P;// Saturation pressure T_sat = (3799.89/(16.262-log(P_sat)))-2...
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//Problem 5 //calculate the glancing angle for third order reflection clear clc w=0.842*(10)^(-10)//wavelength in m x=8.5833//glancing angle(in degrees) for the first order reflection a=1,b=3//a=1 for 1st order b=3 for 3rd order reflection d=(a*w)/(2*sind(x))//inerplanar spacing for first order reflection y=a...
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//Chapter-12, Example 12.19, Page 367 //============================================================================= clc clear //INPUT DATA Rh=200;//Hall-coefficient in cubiccentimeter/C a=10;//conductivity in s/m //CALCULATIONS un=a*Rh;//electron mobility in cm^2/V-s mprintf("electron mobility is %d cm^2/V-s...
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// Example 18.2, page no-460 clear clc r=0.12*10^-9//m eps=8.854*10^-12 alf=4*%pi*eps*r^3 printf("The electronic polarisability of an isolated Se is %.4f * 10^-40 F m^2",alf*10^40)
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Example_4_15_2.sce
// Example 4.15.2 page 4.36 clc; clear; L=10; //fiber length in km Pin=100d-6; //input power Pout=5d-6; //output power len=12; //length of optical link interval=1; //splices after interval of 1 km l=0.5; //loss due to 1 splice attenuation=-10*log10(Pin/Pout); //compu...
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//Scilab Code for Example 4.4 of Signals and systems by //P.Ramakrishna Rao clear; clc; //X(f)=A*T/1+j*2*pi*f*T syms f w; A=1; T=1; X=(A^2*T^2)/(1+4*%pi^2*f^2*T^2) disp('Putting f = tan @'); disp('Total Energy:'); Ex=integrate('(A^2*T)/(2*%pi)','w',-%pi/2,%pi/2) disp('Energy Contained in the Output Signal')...
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errcatch(-1,"stop");mode(2);//Example 7_2_u1 ; ; //To find the fraction of initial intensity alpha=-2.2 //units in db/Kilo meters //When l=2 Kilo meters l=2 //units in Kilo meters //Case (a) when L=2 Kilo meters It_I0=10^(alpha*l/10) printf("The fraction of initial intensity left when L=2 It/I0=%...
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EX17_8.sce
clc;funcprot(0);//EXAMPLE 17.8 // Initialisation of Variables T=175;.......................//Torque due to brake load in Nm N=500;.........................//Engine speed in rpm //calcuations BP=(2*%pi*N*T)/(60*1000);.......................//Brake power developed by engine in kW disp(BP,"Brake power developed by e...
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clear; clc; close; Vo = 1.25*(1+ (1.8*10^3/240)) + (100*10^(-6))*(1.8*10^3); disp(Vo,'Regulated Output voltage = ');
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exa4_5.sce
//Example 4-5// //Reduce a given expression// clc //clears the window// clear //clears all existing variables// //the given expression is as follows// disp(' Given Expression- ((AB''+ABC)''+A(B+AB''))'' ') disp('factorise') disp(' ((A(B''+BC))''+A(B+AB''))'' ') //reduce using laws 18 and 20// disp(' ((A(B''+...
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clear; clc; printf("\t\t\tExample Number 7.11 \n\n\n"); // reduction of convection in ar gap // Example 7.11 (page no.-347) // solution Tm = 300;// [K] mean temperature of air dT = 20;// [degree celsius] temperature difference R = 287;// [] universal gas constant g = 9.81;// [m/s^(2)] acceleration due to gr...
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//Example 10.3 //Taylor Method //Page no. 304 clc;clear;close; deff('y=f1(x,y)','y=1') deff('y=f2(x,y)','y=x*y') deff('y=f3(x,y)','y=x*f1(x,y)+y') deff('y=f4(x,y)','y=x*f2(x,y)+2*f1(x,y)') deff('y=f5(x,y)','y=x*f3(x,y)+3*f2(x,y)') h=0.5;y=0; x=[0.5,1] for i=1:2 if i==1 then k=...
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Exa2_12.sce
//Exa 2.12 clc; clear; close; //Given Data : format('v',8); //Vcon=V;//in volt //pf=cosfi;//unitless //Rcon=R;//in ohm //Part (i) : single phase system disp("Single phase system :"); P1=5*10^6;//in watt //I1=P1/(V*cosfi);//in Ampere disp("Line current,I1="+string(P1)+"/V*cosfi"); //W1=2*I1^2*R;//in Wats(L...
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//Caption:In a transformer find all day efficiency //Exam:3.33 clc; clear; close; KVA=15;//Rating of the transformer(in KVA) E_f=0.98;//Efficiency of the transformer P_F=1;//for unity power factor O_P=KVA*P_F;//Output of the transformer at unity power factor(in KW) I_P=O_P/E_f;//Input to the transformer(in KW)...
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clc clear close b=[0 1 1 0 1 1 1 0] one=ones(1,100) zero=-ones(1,100) //generation of bodd signal bodd=[] for i=1:2:length(b) if(b(i)==1) bodd=[bodd one] bodd=[bodd one] else bodd=[bodd zero] bodd=[bodd zero] end end //generation of beven signal beven=[] for i=2:2:lengt...
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