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//Example 5.2 clc; clear; close; format('v',9); //Given data : v1=400*10^-3;//m/s d1=300/1000;//meter d2=450/1000;//meter A1=%pi*d1^2/4;//m^2 A2=%pi*d2^2/4;//m^2 Q1=A1*v1*1000;//litres/sec(1m^3=1000litres) disp(Q1,"Discharge of pipe in litres/sec : "); v2=(Q1/1000)/A2;//m/s(Q1=Q2) disp(v2,"Mean velocity o...
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//Problem 3.09: //initializing the variables: T = 230; // in deg celcius P = 2500; // in psia Pa = 14.7; // in psia //calculation: //critical values Tc = 417 // in K Pc = 76 // in atm w = 0.074 // acentric factor Tr = (T + 273)/Tc Pr = P/(Pa*Pc) printf("\n\nResult\n\n") printf("\n the reduced tempera...
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function [varargout] = ForegroundDetector(varargin) // This function creates a ForeggroundDetector struct which is used to detect Foreground with Gaussian Mixture Models. // // Calling Sequence // result = ForegroundDetector() // result = ForegroundDetector(Name, Value, ..) // // Parameters // results: ForegroundDet...
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// 08.08.21 // 09.10.27 function Out=Phrawparadata(VL,FaceL) Out=Facesdata(list(VL,FaceL),'rawpara'); endfunction
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//example 2.8// //multiplication in hexadecimal form// clc //clears the screen // clear //clears all the existing variables // x=hex2dec('2B') //first number to be multiplied is x // //hexadecimal to decima l conversion // y=hex2dec ('3' ) //second number to be multiplied is y // z=x*y //multiplication// a...
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Совместная жизнь 1 Должен ли мужчина платить за кафе? 1 Приготовление пищи — женская обязанность? 1 Должны ли вы делать друг другу сюрпризы? 1 За порядок в доме ответствены двое? 1 Бюджет должен быть общим? 1 Допустимо ли совершать крупные покупки не посоветовавшись? 1 Стульчак должен быть опущен? 1 Должна ли женщина р...
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//check o/p when no i/p args are passed to the function a=shiftdata(); //output //!--error 10000 //Incorrect number of input arguments. //at line 62 of function shiftdata called by : //a=shiftdata();
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a = 1,b = 2 c= a+b d = a-b e = a*b f = a/b g = a^b exec("foo.sci")
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// Chapter 6 example 18 //------------------------------------------------------------------------------ clc; clear; // Given data from fig R_a = 1000; // resistance shown in fig a W1 = 0.15*10^-3 // width of geometry fig 6.72a L1 = 3*10^-3 // Length of geometry fig 6.72a W2 = 75*10^-6 // width of g...
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clc //initialisation of variables d= 4 //in v= 30 //ft/sec a= 22.5 //degrees w= 62.4 //lbs/ft^3 g= 32.2 //ft/sec^2 //CALCULATIONS P= w*(%pi*(d/12)^2/4)*v^2*sqrt(2*(1-cosd(a)))/g //RESULTS printf ('Resultant force tending to move the pipe= %.f lbs',P) //ANSWER GIVEN IN THE TEXTBOOK IS WRONG
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clc //initialisation of variables z1= 10 ///m h1= 10 //m v1= 12 //ft/sec v2= 4 //m/sec k= 0.6 w= 62.4 //lb/in^2 g= 32.2 //ft/sec^2 //CALCULATIONS p= (w/144)*(z1+h1+(v1^2/(2*g))-(v2^2/(2*g))-(k*(v1-v2)^2/(2*g))) //RESULTS printf ('pressure at bottom end = %.2f lb/in^2',p)
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//exa 1.17 clc;clear;close; format('v',8); pf=0.8;//power factor delf=1;//%//drop in frequency(delf/f) //delP=-2*(sind(theta))^2*delf theta=acosd(pf);//degree delP_BY_delf=-2*sind(theta)^2;//increase in load wrt frequency disp(-delP_BY_delf,"1% drop in frequency, Increased in Load(%)");
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//Checking if error message pops up when only 12 inputs are given instead of 13 //Reading first stereo image I1 = imread("stereo1.png"); //Reading second stereo image I2 = imread("stereo2.png"); // Number of disparities numofDisparities = 144; // Minimum of disparity value minDisparity = 0; // Matched blocked size sad...
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v=15; i=50*10^(-3); disp("Part a"); r=v/i; disp("the resistance (in Ω) of R is"); disp(r); disp("Part b"); p=v*i; disp("The average power dissipated (in mW) in R is"); disp(p*1000);
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// Example 9.4.5 page 9.16 clc; clear; output=22; //laser output sensitivity=-35; //APD sensitivity sl=0.05; //loss correspond to one splice in dB fl=0.4; //fiber loss in dB/km connector_loss=2; saftey_margin=6; penalties=1.5 allowed_loss=output-sensitivity; Length = (allowed_loss-(connector_...
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clc //Initialization of variables W=14.5 //B/lbm Q=141.7 //B/lbm Tr=520 //R Ta=1040 //R //calculations n1=W/Q n2=1-Tr/Ta Wc=n2*Q //results printf("Thermal efficiency = %.2f percent",n1*100) printf("\n Work done in carnot cycle = %.1f Btu/lbm",Wc)
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clc; p=20000; // rated power of transformer vbh=2500; // base voltage in volts for h. v. side vbl=250; // base voltage in volts for l. v. side ibh=p/vbh; // base current in Ampere for h. v. side zbh=vbh/ibh; // base impedance in ohm ze=2.6+4.3*%i; // equivalent leakage impedance referred to h. v. side in ohm zep...
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clear all; t=-%inf:%inf; w=2*%pi*%i K=4 k=0:0.004:4; sigma=0 Wmax=w.*k./K; function x=f(t); x=sign(t).*exp(-sigma.*abs(t)).*exp(-%i.*w.*t) endfunction Y=intg(-10,10,f) // for the CTFS of x(t)=u(t) f=1 // Analog frequency f=1 Hz U=(1/%i*2*%pi*f)+(1/2).*[ones(1,1)] disp(Y,'The CTFS of sgn(t)=')...
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//Example 5_18 clc(); clear; //To calculate the glancing angle h=1 k=1 l=0 lamda=0.065*10^-9 //units in m n=2 a=0.26*10^-9 //units in nm sintheta=(n*lamda*sqrt(h^2+k^2+k^2))/(2*a) theta=asin(sintheta)*180/%pi //units in degrees printf("Theta=%.2f degrees",theta) //the answer in the textbook is given ...
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//To find turning moment clc //Given: D=300/1000, L=450/1000, r=L/2, d=50/1000, l=1.2 //m N=200 //rpm mR=225 //kg theta=125 //degrees p1=30*1000, p2=1.5*1000 //N/m^2 //Solution: //Calculating the angular speed of the crank omega=2*%pi*N/60 //rad/s //Calculating the area of the piston A1=%pi/4*D^2 //m^2 //C...
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// residue function //RESIDUE Partial-fraction expansion (residues). // [R,P,K] = RESIDUE(B,A) finds the residues, poles and direct term of // a partial fraction expansion of the ratio of two polynomials B(s)/A(s). // If there are no multiple roots, // // B(s) R(1) R(2) R(n) // -...
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function scs_m=do_color(scs_m) // do_block - edit a block icon while %t [btn,xc,yc]=xclick(0); pt=[xc,yc] [n,pt]=getmenu(datam,pt); if n>0 then n=resume(n),end K=getblock(scs_m,[xc;yc]) if K<>[] then break,end end o=scs_m(K) graphics=o(2) gr_i=graphics(9) if type(gr_i)==10 then,gr_i=list(gr_i,[]),end if gr_...
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// Example 15_19 clc;funcprot(0); // Given data T=25;// °C p=0.1;// MPa // Calculation n_H2=1;// The stoichiometric coefficient of the reaction n_O2=0.5;// The stoichiometric coefficient of the reaction n_H2O=1;// The stoichiometric coefficient of the reaction g_f_H2O=-237.178;// MJ/kgmole // [(abar_f)_i]_c...
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clc //Chapter3: Modulation //Example3.28, page no 177 //Given R=1//Antenna Resistance assumed to be 1 ohm for ease of calculation Ic=10.8// current with no modulation Pc=Ic^2*R//power with no modulation It=12.15//modulated current Pt=It^2*R// modulated power ma=(sqrt(2*(((It/Ic)^2)-1)))//modulation depth) m...
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//Problem 7.12: A silicon iron ring of cross-sectional area 5 cm2 has a radial air gap of 2 mm cut into it. If the mean length of the silicon iron path is 40 cm, calculate the magnetomotive force to produce a flux of 0.7 mWb. //initializing the variables: A = 5E-4; // in m2 l = 0.4; // in m r = 2E-3; // in m u0 ...
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//Graphical// //Example 3.2.9 //Convolution Property Proof clear; clc; close; function [Ztransfer]= ztransfer_new (sequence,n) z = poly (0, 'z' , 'r' ); Ztransfer = sequence *(1/z )^n' endfunction x1 = [1,-2,1]; n1 = 0:length(x1)-1; X1 = ztransfer_new(x1,n1) x2 = [1,1,1,1,1,1]; n2 = 0:length(x2)-1; X2 = ztransfer_ne...
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//solution //given P=50*10^3//N d=75//mm ft=25//N/mm^2 t=20//N/mm^2 //let B1 be the width of strap //B1=d B1=75//mm //t11=B1/4//mm printf("the thickness is,%f mm\n",B1/4) printf("the thickness can be taken as 20mm\n ") t11=20//mm //let t1 be thickness of strap at thnner side //P=2*B1*t1*ft//N //t1=P/(2*B1*ft)//mm prin...
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//pagenumber 285 example 2 clear ic=1.3*10^-3;//ampere colres=2*10^3;//ohm re=500;//ohm v1=0.026;//volt beta1=100; vcc=15;//volt c1=10*10^-6;//farad ib=ic/beta1; ri=0.01/ib; volgai=beta1*colres*ib/0.01; disp("voltage gain = "+string((volgai))+"<180"); disp("voltage gain reduced ce removed"); disp("whe...
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clc Nd=6*10^16 disp("Nd = "+string(Nd)+" cm^-3") //initializing the value of donor concentration. Na=10^17 disp("Na = "+string(Na)+" cm^-3")//initializing the value of acceptor concentration. no=1.5*10^10 disp("no = "+string(no)+" cm^-3")//initializing the value of electron and hole per cm^3. p=((Na-Nd)+sqrt((Na...
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// Test avec la matrice suivante : K3 = [ 5 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 -1 0 0 0 -1 0 0 0 -1 -1 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 0 0 0 0 0 0 ; 0 8 0 -1 0 0 0 0 0 0 0 0 0 -1 -1 0 0 0 0 0 -1 -1 0 0 0 -1 -1...
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clear; clc; //Example 7.7 Vbe=0.7; b=100; Re=.5; Rc=5; Rl=10; R1=40; Cc=10; R2=5.7; Rs=.1; Vt=0.026; Icq=0.99; gm=Icq/Vt; printf('\ntransconductance=%.3f mA/V\n',gm) r=b*Vt/Icq; printf('\ndiffusion resistance=%.2f KOhm\n',r) Ri=r+(1+b)*Re; printf('\ninput resistance=%.2f KOhm\n',Ri) x=Rc*Rl/(Rc+Rl);...
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// Exa 5.14 clc; clear; close; format('v',6) // Given data V_L = 400;// in V f = 50;// in Hz I_L = 17.32;// in A phi = acosd(0.8); I_Ph =I_L/sqrt(3);// in A disp(I_Ph,"The phase current in A is"); V_Ph=V_L;// in V Z_Ph = V_Ph/I_Ph;// in ohm Z_Ph= Z_Ph*expm(phi*%i*%pi/180);// in ohm R= real(Z_Ph);// in oh...
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// Example 2.7: Calculate the collector to emitter voltage(Vce) and Collector current (Ic) clc; clear; Vcc= 12 ;// as Ic=0 so Vce=Vcc (In volts) Rc= 3;// Collector Resistance in killo oms Ic=Vcc/Rc; // Collector Current in Amperes Vce=Vcc; disp(Vce,"Colletor to emitter voltage (in volts)") disp(Ic," Collector c...
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errcatch(-1,"stop");mode(2);//Caption:Find the ratio of the speed as a generator to speed as a motor //Exa:2.19 ; ; V=250;//in volts I_L=80;//in amperes R_a=0.12;//in ohms R=100;//in ohms I_f=V/R;//in amperes I_a1=I_L+I_f;//in amperes (generator) E_1=V+(I_a1*R_a);//in volts (generator) I_a2=I_L-I_f;//in am...
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clc clear //Input data p=[5,20]//Pressure limits in atm T3=650+273//Temperature in K T1=60+273//Temperature in K T2=T1//Temperature in K Cp=0.24//Specific heat at constant pressure in kJ/kg.K g=1.4//Ratio of specific heats R=29.27//Characteristic gas constant in kg.m/kg.K J=427//Mechanical equivalent of heat ...
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clc //Initialization of variables T=298 //K m=32*1.66054*10^-27 //kg k=1.38066*10^-23 //j/k V=10^-4 //m^3 h=6.62608*10^-34 //J/s //calculations q=(2*%pi*m*k*T)^1.5 *V/h^3 //results printf("Translational partition function = %.2e",q)
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//caption:Find magnitude of limiting error fot R1 and R2 //Ex2.5 clc clear close R1=36//resistance(in ohm) R2=75//resistance(in ohm) er=0.005//limiting error(in ohm) dR1=R1*er disp(dR1,'magnitude of limiting error for R1(in ohm)=') dR2=R2*er disp(dR2,'magnitude of limiting error for R2(in ohm)=')
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clc disp("Example 7.3") printf("\n") t1=-10:0.05:10 v=5*cos (%pi*t1/6+%pi/6) figure a= gca (); plot(t1,v) xtitle ('v vs %pi*t/6','%pi*t/6','v '); a. thickness = 2;
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deff('d=f(x)','d=x^3-100') a=input("Enter the value of a:") b=input("Enter the value of b:") n=input("Enter the number of iterations n:") for i=1:n c=(a*f(b)-b*f(a))/(f(b)-f(a)) disp([i,c]) if f(a)*f(c)<0 then b=c end if f(b)*f(c)<0 then a=c end c1=(a*f(b)-b*f(a))/(f(b)-f(a)) if abs(c1-c)<...
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// Example 11.13: (a) Amplifier type // (b) Input resistance, Output resistance, Transfer ratio clc, clear r_pi=1e3; // in ohms gm=0.1; // in mho disp("Part (a)"); disp("Q1 is a common collector and Q2 is common emitter stage. Hence the given circuit is cascade of cc and CE stages. As the Rin of ...
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clc clear //Data taken from Ex.No.1 BHP=51//Brake horse power in h.p N=1000//Speed in r.p.m FHP=17//Friction horse power in h.p //Input data O1=BHP/2//Half of b.h.p output in h.p O2=10//Brake horse power in h.p //Calculations //Case(i) IHP1=(O1+FHP)//Indicated Horse power in h.p mn1=(O1/IHP1)*100//Mechani...
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spLiTter eunG {} fILter V { Not ::D:eBE:e:e:BD:cB << m OR noT N ( BItAnD ( ::C:f:3:aE:cc0, 252.132.19.216/17, ), ) nOT MLz ( ) << 251.224.7.234 or noT 255.113.2.4 > S } fIlTeR MjH {noT zP oR NL or sD OR n Or ui Or MtiyF I OR NOt RDWF } n braNCh hy grOUper p {ModuLE c{ } ModulE s{ } agGREgATE Nt } UngRoU...
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clear// //Variables hfe = 99.0 //hfe hie = 2.0 * 10**3 //hie (in ohm) Rc = 22.0 * 10**3 //Load resistor of frist stage (in ohm) R4 = 100.0 //Emitter resistance of first stage (in ohm) R1 = 220.0 * 10**3 ...
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int f(int, int b); int f(int a, int b); int a, b, c, d; int f(int c, int d) { c = d; } void main(void) { f(a=b,c=d); }
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ваймамс V;AC2;PL;1;CAUS перямс V;NEG;TR грек N;HUM;DAT;PL;PSS1S;LGSPEC1 вайгяль N комитет N;GEN;DEF;SG;LGSPEC1 альбом N;IN+ESS;SG;PSS1S;LGSPEC1 лектор N;HUM;TRANS;SG;LGSPEC1 конюх N;HUM;TRANS;SG;LGSPEC1 азкс N;NOM;SG;PSS3P пиче N;GEN;DEF;SG единоросс N;HUM;DAT;DEF;PL;LGSPEC1 дурак N;HUM;NOM;SG;LGSPEC1 шавомс V.PTCP;GEN...
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//16.10 clc; P_consumed=3000/3; E_per_phase=440/(3^0.5); IL=P_consumed/E_per_phase; printf("\nCurrent in each line=%.2f A",IL) R=E_per_phase/IL; printf("\nResistance of resistor=%.2f ohm",R)
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//example7.21 clc disp("The given values are,") disp("R_f=0.1 ohm, I_DC=10 A, R_s= 0 ohm, E_s(RMS)=30 V") e=30*sqrt(2) format(8) disp(e,"Now, (E_sm)=E_sm(RMS)*sqrt(2) =") disp("(I_DC)=(2*I_m)/pi") i=(%pi*10)/2 disp(i,"I_m(in A)=(pi*I_DC)/2=") disp("Now, (I_m)=(E_m)/(2*R_f+R_s+R_L)") disp("Therefore, 15.7079 ...
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5.sce
clc clear //input f=0.5 //focal length u=0.8 //distance of object f1=0.2 //focal length of converging lens d=1 //distance behind the first lens //calculation x=(1/f)-(1/u)//lens formula y=1/x u1=-(y-d)//second lens a=1/f1 +(1/-u1)//lens formula b=1/a //output printf("the image lies %3.3f m behind secon...
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clc clear //Input data n=6//Number of cylinders d=9//Bore in cm l=10//Stroke in cm N=2500//Speed in r.p.m Ta=25+273//Temperature of air entering the compressor in K q=16800//Heat rate in kcal/hour T=60+273//Temperature of air leaving the cooler in K p=1.6//Pressure of air leaving the cooler in kg/cm^2 t=14.5...
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//page 107 clear; close; clc; A=[1 2;3 6]; disp(A,'A='); [m,n]=size(A); disp(m,'m='); disp(n,'n='); [v,pivot]=rref(A); r=length(pivot); disp(r,'rank=') cs=A(:,pivot); disp(cs,'Column space='); ns=kernel(A); disp(ns,'Null space='); rs=v(1:r,:)'; disp(rs,'Row space=') lns=kernel(A'); disp(lns,'Left null sapce=');
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//chapter 13 //example 13.9 //page 568 printf("\n") printf("given") hic=2*10^3;hie=hic;hfe=100;hfc=100;Av=25000;B=1/33.4;R8=10*10^3;R5=R8; Ze=(hic+R8)/hfc Zo=Ze/(1+Av*B) Zout=(R5*Zo)/(R5+Zo); printf("output impedance is %3.2fohm\n",Zout)
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//Example 7.1 clc s=%s; tau1=0.5; tau2=1; R2=1; //From Eq.(7.8) g=R2/((tau1*s+1)*(tau2*s+1)) disp(g,'H2(s)/Q(s)=') Qs=1/s; H2s=g*Qs; disp(H2s,'H2(s)=') syms t; H2t=ilaplace(H2s,s,t); disp(H2t,'H2(t)=')
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 2 //First Law of Thermodynamics //Example 5 clear; clc; //Given: n_water = 10^3; //moles of water T = 373; //tempearture(K) P = 101.3; //pressure(kPa) sv_liquid = 0.00104; //specific volume of liquid(m^3/kmol) sv_vapour = 1.675; //specific...
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clear; clc; disp("--------------Example 4.5---------------") data_rate=1; // 1 Mbps frac= 0.25 // 4B/5B coding adds 25% to the baud rate add=data_rate*frac; N = (data_rate+add)*10^6; // Hz NRZI_B= N/2; // minimum bandwidth using NRZ-I Manchester_B = data_rate; // minimum bandwidth using Manchester scheme // d...
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//Exa 1.9 clc; clear; close // given data K_A=0.5;// in W/m degree C K_B=0.8;// in W/m degree C Ti_A=600;// inside temp. of slab A in degree C To_B=100;// outside temp. of slab B in degree C t_A=4*10^-2;// thickness of slab A t_B=6*10^-2;// thickness of slab B // Heat transfer rate per square meter through t...
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// Exa 5.28 format('v',6); clc; clear; close; // Given data A=2000;//V/V Bita= 0.1;// inV/V Ri= 1;// in kohm Ri= Ri*10^3;// in ohm Ro= 1;// in kohm Ro= Ro*10^3;// in ohm Af= A/(1+A*Bita); disp(Af,"The gain Af in volt is : ") Rif= Ri*(1+A*Bita);// in ohm disp(Rif*10^-3,"The input resistance in kohm is : "...
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errcatch(-1,"stop");mode(2);//caption:Find limiting error in percentage //Ex2.2 V=150//maximum range of voltmeter(in V) A=0.02//magnitude of accuracy(in V) Vm=83//voltage measured dA=A*V %er=(dA/Vm)*100 disp(%er,'limiting error(in %)=') exit();
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//Initilization of variables F3=500 //N alpha=60 //degree //angle made by F3 with F2 beta=40 //degree //angle made by F1 with F3 theta=80 //degree //angle made by F1 with F2 //Calculations // Solving by using law of sines F1=(F3*sind(alpha)/sind(theta)) //N //by law of sines F2=(F3*sind(beta)/sind(theta)) //N /...
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clc(); clear; //Given : Eg1= 0.72; //Energy gap for Germanium in eV Eg2= 1.10; //Energy gap for Silicon in eV Eg3= 5.6; //Energy gap for diamond in eV //Fraction of electron : n/N = exp(-Eg/(2*k*T)) , k*T = 0.026 eV F1 = exp(-Eg1/(2*0.026)); // For Germanium F2 = exp(-Eg2/(2*0.026)); // For Silicon F3 = exp(-E...
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clc; clear; printf("\n Example 9.20\n"); e=0.75; //Emissivity of grey surface r=1-e; //reflectivity of surface Ts=400; //Temperature of surface T_amb=295; sigma=5.67e-8; q1=3e3; //Rate of radiation arriving at grey surface //From equation 9.118 Eb=sigma*Ts^4; printf("\n Emissive Power = %.0f W/m^2",Eb); ...
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//SHUBHAM SHARMA //Roll no:18i190002 //Msc PhD OR clc,clear function[TR_busy, TR_total, TR_idle] =JobsDoneInOneArrival(m,u,X) IS = X TR = [0] i = 1 S = grand(1,1, "exp", 1/u) while (IS > 0) A = grand(1,1, "exp", 1/m) TR(i+1) = TR(i) + A JC =0 while TR(i+1) > S &IS >0 ...
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//Kunii D., Levenspiel O., 1991. Fluidization Engineering(II Edition). Butterworth-Heinemann, MA, pp 491 //Chapter-9, Example 2, Page 222 //Title: Horizontal Drift Of Solids //========================================================================================================== clear clc //INPUT Lmf=0....
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errcatch(-1,"stop");mode(2); //Initialization of variables C=1/4000 F=125 //lb n=3500 //rpm //calculations Bhp=F*n*C //results printf("Bhp developed by the engine = %.1f",Bhp) exit();
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//function dlyap //lyapunov A = [3.0 1.0 1.0 1.0 3.0 0.0 0.0 0.0 3.0]; B = [25.0 24.0 15.0 24.0 32.0 8.0 15.0 8.0 40.0]; X = dlyap (A, B); // Sylvester A = [1.0 2.0 3.0 6.0 7.0 8.0 9.0 2.0 3.0]; B = [7.0 2.0 3.0 ...
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//Renvoie la première valeur en nuance de où l'histogramme est différent de 0 en partant de la fin. //Pour le futur, inclure des pourcentages de pertes pour l'image, par exemple, des valeurs qui n'influe que très peu seront ignorées. function indexHisto=finHistogramme(histogram) for indexHisto = 1:256 if(h...
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clc // initialization of variables clear E=72 //GPa t=60 //mm M=24 //kN.m //part(a) ro=100 //mm r1=150 //mm A=t*r1 Am=t*log((ro+r1)/ro) R=ro+r1/2 E=E*10^3 Mx=M*10^6 Phi=Am*Mx*%pi/(A*(R*Am-A)*E) printf('part(a)') printf('\n Phi = %.5f rad',Phi) //part(b) //Mx=Mx+P*R*sin(th) delta_P=Am*Mx*R*2/(A*(R*Am-...
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clc; t=40*1e-2; //pulse covers 2x distance in arriving back //so, 30*1e-6=2*x/v //and, 2nd pulse will cover a distance of 2*40 cm in 80*1e-6 seconds //therfore, 80*1e-6=(2*40*1e-2)/v //compare both equation e1=30; e2=40*2 x=e1*t*2/(2*e2); disp(+'m',x,'distanc of the flow from near end =')
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// Ex2.9.2 //calculation of transition capacitance clc; clear; //given W=2.38*10^-6; //width of depletion layer for V=-1 W1=4.8*10^-6;//width of depletion layer for V=-5 A=0.8*10^-6;//area of junction epslnR=16;//for Ge epslnO=8.854*10^-12;//permittivity of free space epsln=epslnR*epslnO;//permittivity of sem...
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clc; VCC=18; Rx=2*10**3; VBE=0.7; Ix=(VCC-VBE)/Rx; I=Ix; disp('mA',I*10**3,"I=");
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//Exa 5.3 clc; clear; close; //Evaluation at i=8% disp("Evaluation at i=8%"); //Alternative 1 : disp("Alternative 1 : "); P=500000;//in Rs A1=50000;//in Rs G=50000;//in Rs i=8;//in % per annum n=6;//in years //Formula : (F/P,i,n) : (1+i/100)^n //Formula : (F/A,i,n) : (((1+i/100)^n)-1)/(i/100) //Formula :...
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clc;funcprot(0);//Example 3.10 //Initilisation of Variables Ti=20;...//Uniform temparature of the wood piece in degrees celcius To=120;...//Rised Surface temparature in degrees celcius t1=2*60;....//time after exposure of surface in degrees celcius t2=20*60;....//time after exposure of surface in degrees celcius ...
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clc clear x=poly(0,'x'); p=x^3-4*x-9 disp("Finding roots of this equation by bisection method"); disp('f(2) is -ve and f(3) is +ve so a root lies between 2 and 3'); l=2; m=3; function y=f(x) y=x^3-4*x-9; endfunction for i=1:4 k=1/2*(l+m); if(f(k)<0) l=k; else m=k; end end disp(k)
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clc //initialisation of variables g=9.8//m/s^2 h=20//m //CALCULATIONS V=sqrt(2*g*h)//m//s //RESULTS printf('The velocity is=% f m/s',V)
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T=300;//initial temperature in kelvin// T1=692;//temperature in kelvin// T2=1180;//temperature in kelvin// T3=1500;//final temperature in kelvin// Lf=1800;//Latent heat of fusion in cal per mol// Tf=692;//temperature of fusion in kelvin// Lv=27700;//Latent heat of vapourization in cal per mol// Tv=1180;//tempera...
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clc //initialisation of variables P0=100 //lbf/in^2 T0=660 //R Mx=1.5//lbf/in^2 P0x=0.272//lbf/in^2 T0x=0.690 //R Ax=1.176 //lbm Px=P0x*P0//lbf/in^2 Tx=T0x*T0//R Mx=0.701//lbf/in^2 Px1=2.458 //lbf/in^2 Tx1=1.320 //R P0x1=0.9298 //lbf/in^2 Ax=1.094 //lbf/in^2 P0y=0.721 //lbf/in^2 T0y=0.911 //R A=2.0 /...
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clear ; clc; // Example 5.6 printf('Example 5.6\n\n'); printf('Page No. 120\n\n'); // given x = 0.90;// dryness fraction m = 0.25;// Mass flow rate in kg/s P = 0.7;// pressure in bar T1 = 10;// in degree celcius //from steam table h_10= 42*10^3;// Specific enthalpy of water at 10 degree celcius in J/kg ...
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//A Textbook of Chemical Engineering Thermodynamics //Chapter 7 //Properties of Solutions //Example 22 clear; clc; //Given: C_water = 4.18*10^3; //heat capacity of water (J/kg K) C_ethanol = 2.58*10^3; //heat capacity of ethanol (J/kg K) G1 = -758; //heat of mixing 20 mol percent ethanol water at 298 K...
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// sempre eh bom colocar função na extensão .sci // vou colocar as variáveis iguais ao do caderno...se tu não gostar, pode mudar. Eh so pra me nao se perder. function [y, x]= MPotencia(A) [l,c] = size(A); y = ones(l,1); err = 1; while (err > 1E-4) z = A*y ; maximo =...
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clear// //Variables hie = 1.0 * 10**3 //hie (in ohm) hre = 1.0 * 10**-4 //hre hoe = 100.0 * 10**-6 //hoe (in mho) RC = 1.0 * 10**3 //Collector resistance (in ohm) RS = 1000.0 //Source resistance (in oh...
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clear; clc; close; matriz = input("Digite a matriz: "); erroAdmitido = input("Digite o erro admitido: "); aproxAnterior = input("Digite aproximação inicial: ") linha = size(matriz, "r"); coluna = size(matriz, "c"); diferenca = ones(linha); while erroAdmitido for i = 1:(linha) diagonal = 1/matriz(i,i); ...
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errcatch(-1,"stop");mode(2);; ; D2 = 0.1; D3 = 0.04; D4 = 0.02; THD = sqrt((D2^2)+(D3^2)+(D4^2))*100; disp(THD,'Total harmonic distortion(Percentage) = '); exit();
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//Example 3_4 clc; clear; close; format('v',5); //given data : //Let T=1 for calculation T=1; //i=5*t/T+5;//A Iav=1/T*integrate('5*t/T+5','t',0,T); disp(Iav,"Average value(A)"); Irms=sqrt(1/T*integrate('(5*t/T+5)^2','t',0,T));//V disp(Irms,"rms value(A)"); //Answer is not accurate in the book.
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//Example 5_2 clc(); clear; //To show d100:d110:d111=sqrt(6):sqrt(3):sqrt(2) //For d100 h=1 k=0 l=0 d100=1/sqrt(h^2+k^2+l^2) //Units in terms of a //For d110 h=1 k=1 l=0 d110=1/sqrt(h^2+k^2+l^2) //Units in terms of a //For d111 h=1 k=1 l=1 d111=1/sqrt(h^2+k^2+l^2) //Units in terms of a printf("...
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clc clear //Input data At=24 //Throat area in cm^2 A2=50 //Exit area in cm^2 Po=700 //Stagnation pressure in kPa To=100+273 //Stagnation temperature in K Ax=34 //Area before the shock in cm^2 Ay=34 //Area after the shock in cm^2 //Calculation a1=Ax/At //Ratio of areas Mx=1.78 //Mach number upstream of sh...
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clc //initialisation of variables v=25*10^-2//m a=8//m^3/min t=1/60//sec/min f=0.015 p=100//m p1=1.2//kg/m^3 p2=2.7//m/s g=9.8//m/s^2 q=998//kg/m^3 v1=9802//N/m^3 //CALCULATIONS V=(4*a*t)/((%pi)*(v)^2)//m/s hf=(f)*((p)*(p2)^2*(p1))/(v*2*g*v1)//m^-3 P=v1*hf*10//Pa //RESULTS printf('The pressure drop is=...
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function [ez]=ez(m,n,k,rho,phi,z) amn=1; bmn=1; gammamn=1; ez=besselj(m,gammamn*rho).*exp(%i*m*phi).*(amn*sin(k*z)+bmn*cos(k*z)); endfunction
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// Geovani Pereira dos Santo - IFNMG - 2019 function [x, y, yDerivado] = rungeKutta(funcao, funcaoDerivada, limiteInferior, limiteSuperior, passo, valorInicial, valorInicialDerivada) x(1) = limiteInferior; y(1) = valorInicial; yDerivado(1) = valorInicialDerivada; quantidadePassos = (limiteSup...
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//Scilab Code for Example 2.21 of Signals and systems by //P.Ramakrishna Rao //Z- transform of a^n u(n) clear; clc ; close ; syms a n z; x1 =1/2; x2=1/3; X1= symsum (x1*(z^(-n)),n ,0, %inf ); X2= symsum (x2*(z^(-n)),n ,0, %inf ); X=X1+X2; //Display the result disp (X,"Z-transform of u(n) is:"); disp('ROC...
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//CHAPTER 7- SINGLE PHASE TRANSFORMER //Example 25 disp("CHAPTER 7"); disp("EXAMPLE 25"); //VARIABLE INITIALIZATION va=25000; v1=2200; //primary voltage in Volts v2=110; //secondary voltage in Volts f=50; R1=1.75; X1=2.6; R2=0.0045; X2=0.0075; //SOLUTION //...
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// Example 4.2 clear; clc; close; format('v',7); // Given data Tst=1; // in N-m Ist=5;//in Ampere V=28;//in volt //Calculations kf=Tst/Ist;//in Nm/A omega_m=V/kf//in radian/sec No=omega_m*60/(2*%pi)//in rpm disp(No,"No load speed in rpm = ");
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declare sentconst \=\=\+\=\> A\+\-\1; taut \=\=\+\=\> and A\+\-\1; declare sentconst \1AB; assume \1AB;
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//To find MI and size clc //Given: a1=5*10^-5, a2=21*10^-5, a3=85*10^-5, a4=8*10^-5 //m^2 N2=98, N1=102 //rpm rho=8150 //kg/m^3 sigma=7.5*10^6 //N/m^2 //Solution: //Refer Fig. 16.20 //Calculating the net area a=a3-(a1+a2+a4) //Net area, m^2 //Calculating the value of 1 m^2 on the turning moment diagram in te...
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function mm=g_aranum(g) [lhs,rhs]=argn(0), if rhs=0 then g=the_g, end mm=g(6)
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clear ; clc; // Example 8.8 printf('Example 8.8\n\n'); printf('Page No. 236\n\n'); // This question doesnot contain any calculation part. //By refering figure 8.7 which shows Poer factor-load curve for a motor with a capacitor and one without a capacitor. printf('there is no computational part in the problem...
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label 5650 1 1122,112,188,170,14186,1105,1510,12759,5012,119 0 8362,2087,2836,7520,1193,26481,1105,7127 1 3643,1366,1106,2810,1115,1185,4371,1110,25156,1106,9712,24063,170,1558,1578,1112,170,2595,1870,1107,14850,2109,13140,119 1 1103,3176,117,13452,117,1390,117,7678,22556,1105,1839,1132,1155,1112,2430,6775,1158,154...
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considerNonPrimitive Expanding for base=2, level=2, reasons+features=base,primitive,same,similiar,accel igtriv,norm Refined variables=a,b,c,d [0+1a,0+1b,0+1c,0+1d]: unknown -> [1] [0,0,0,0] a³+2b³+4c³-9d³ ---------------- level 0 expanding queue[0]^-1,meter=[2,1,1,2]: a³+2b³+4c³-9d³ [0+2a,0+2b,0+2c,0+2d]: unknown accel...
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clc; //page no 52 //problem 2.8 //All frequencies in kHz //The time is in ms //Power in WATTS //All voltage in volts //following values are copied from Ex2-7 T=0.2; //ms f1=1/T; //There are only odd numbered harmonics Ap2p=40; // peak to peak R=50; //ohm A=Ap2p/2; C1=4*A/%pi; C3=4*A/(3*%pi); C5=4*A/...
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// ELECTRIC POWER TRANSMISSION SYSTEM ENGINEERING ANALYSIS AND DESIGN // TURAN GONEN // CRC PRESS // SECOND EDITION // CHAPTER : 4 : OVERHEAD POWER TRANSMISSION // EXAMPLE : 4.15 : clear ; clc ; close ; // Clear the work space and console // GIVEN DATA D_12 = 26 ; // distances in feet D_23 = 26 ; // dist...