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ex_4_11.sce
clc //solution //given lc=3000//mm//length of steel and copper bar lst=3000//mm//length of steel bar Ec=105//kN/mm^2//young's modulus of copper Est=210//kN/mm^2//young's modulus of steel b=25//mm//width t=12.5//mm//thickness P=50//kN//load applied //refer fig 4.14 in book //let dl be increace in length of compound bar Ast=b*t//mm^2//area of steel bar Ac=b*t//mm^2//area of copper bar Pc=(P*Ec)/(Ec+Est)//kN//load taken by copper bar Pst=P-Pc//kN//load taken by steel bar dl=(Pc*lc)/(Ac*Ec)//mm//change in length //stresses produced in individual bars are //since strain produced are same therefore //(Fst/Est)=(Fc/Ec)//since Est=2Ec,therefore Fst(stress in steel)=2*Fc(stress in copper) P=Pst+Pc//(Fst*Ast)+(Fc*Ac)//Ast=Ac//Fst=2Fc,therefore gievn equation can ve written as //50=2*Fc*Ac+(Fc*Ac) Fc=50/(3*Ac)//N/mm^2//stress in copper bar Fst=2*Fc//N/mm^2//stress in steel bar printf("the change in lentgth of compound bar is,%f mm\n",dl) printf("the stress in copper bar is ,%f kN/mm^2\n",Fc) printf("the stress in steel bar is , %f kN/mm^2",Fst)
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// Exa 3.7 clc; clear; close; // Given data n=2; V_T=26;// in mV Io= 30;// in mA // (i) when I_D= 0.1;// in mA V_D= n*V_T*log(I_D/Io);// in mV disp(V_D,"(i) When I_D is 0.1 mA, The junction forward-bias voltage in mV is : ") // (ii) when I_D= 10;// in mA V_D= n*V_T*log(I_D/Io);// in mV disp(V_D,"(ii) When I_D is 10 mA, The junction forward-bias voltage in mV is : ") // Note: There is calculation error in the book so answer in the book is wrong.
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Chapter3_Example33.sce
//Chapter-3, Example 3.33, Page 117 //============================================================================= clc clear //INPUT DATA k=200;//Thermal conductivity of aluminium in W/m.K t=0.001;//Thickness of fin in m L=0.015;//Width of fin in m D=0.025;//Diameter of the tube in m Tb=170;//Fin base temperature in degree C Ta=25;//Ambient fluid temperature in degree C h=130;//Heat transfer coefficient in W/m^2.K //CALCULATIONS Lc=(L+(t/2));//Corrected length of fin in m r1=(D/2);//Radius of tube in m r2c=(r1+Lc);//Corrected radius in m Am=t*(r2c-r1);//Corrected area in m^2 x=Lc^(3/2)*sqrt(h/(k*Am));//x for calculating efficiency n=0.82;//From fig. 3.18 on page no. 112 efficiency is 0.82 qmax=(2*3.14*(r2c^2-r1^2)*h*(Tb-Ta));//Maximum heat transfer in W qactual=(n*qmax);//Actual heat transfer in W //OUTPUT mprintf('Heat loss per fin is %3.2f W',qactual) //=================================END OF PROGRAM==============================
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//variable initialization RestEnergy=0.51 //rest energy of electron (Mev) //calculation of minimum energy of a gamma ray photon which is required to produce an electron positron pair E=2*RestEnergy; //minimum energy of gamma ray photon (Mev) printf("\nMinimum energy required = %.2f Mev",E);
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// Exa 6.25 clc; clear; close; // Given data I_DSS = 20;// in mA V_P = 4;// in V I_D = I_DSS;// in mA disp(I_D,"The maximum drain current in mA is"); V_GS = -V_P;// in V disp(V_GS,"The gate source cut off voltage in volts is"); R_DS = V_P/I_DSS;// in kΩ disp(R_DS*10^3,"The value of ohmic resistance in Ω is");
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//Page Number: 11.21 //Example 11.16 clc; //(b)I(X;Y) //Given a=0.5; p=0.1; //As we know //P(Y)=P(X)*P(Y/X) //We have PX=[a (1-a)]; PYbyX=[(1-p) p;p (1-p)]; PY=PX*PYbyX; //As H(Y)=-Sum of[P(yi)log2P(yi)] //Where i=0 to n; HofY=0; for i=1:2 HofY=HofY+(PY(i)*log2(PY(i))); end //For BSC, I(X;Y)=H(Y)+plog2(p)+(1-p)log2(1-p) IXY=-HofY+[(p*log2(p))+((1-p)*log2(1-p))]; disp(IXY,'I(X;Y) for a=0.5 and p=0.1:'); //(c)I1(X;Y) //Given a1=0.5; p1=0.5; //As we know //P(Y)=P(X)*P(Y/X) //We have PX1=[a1 (1-a1)]; PYbyX1=[(1-p1) p1;p1 (1-p1)]; PY1=PX1*PYbyX1; //As H(Y)=-Sum of[P(yi)log2P(yi)] //Where i=0 to n; HofY1=0; for i=1:2 HofY1=HofY1+(PY1(i)*log2(PY1(i))); end //For BSC, I(X;Y)=H(Y)+plog2(p)+(1-p)log2(1-p) IXY1=-HofY1+(p1*log2(p1))+((1-p1)*log2(1-p1)); disp(IXY1,'I(X;Y) for a=0.5 and p=0.5:');
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[-63,16,-65]*42 [[1,2,0,0,0],[0,2,2,0,0],[1,2,2,0,0]] [3,4,5] [4,3,5]*8 [15,8,17]*9 [12,5,13]*32 [35,12,37]*25 [24,7,25]*72 [63,16,65]*49 [[-1,0,1,0,0],[0,2,0,0,0],[1,0,1,0,0]] [0,1,1]*2 [-3,4,5]*4 [-4,3,5]*18 [-15,8,17]*16 [-12,5,13]*50 [-35,12,37]*36 [-24,7,25]*98 [[-1,2,0,0,0],[0,-2,2,0,0],[-1,2,-2,0,0]] [1,0,-1] [0,-1,-1]*8 [-3,-4,-5]*9 [-4,-3,-5]*32 [-15,-8,-17]*25 [-12,-5,-13]*72 [-35,-12,-37]*49 [[-1,-1,1,1,0],[0,2,2,0,0],[1,1,1,1,0]] [0,1,1]*4 [-3,4,5]*6 [-4,3,5]*24 [-15,8,17]*20 [-12,5,13]*60 [-35,12,37]*42 [-24,7,25]*112 [[-1,-2,0,-2,1],[0,2,0,-2,0],[-1,-2,-2,2,-1]] [-1,0,-1]*4 [-35,12,-37] [-35,12,-37]*4 [-391,120,-409] [-221,60,-229]*4 [-1739,420,-1789] [-775,168,-793]*4 [[-1,0,1,-2,1],[0,-2,2,0,0],[-1,0,-1,2,-1]] [-1,0,-1] [-15,-8,-17] [-77,-36,-85] [-247,-96,-265] [-609,-200,-641] [-1271,-360,-1321] [-2365,-588,-2437] [[-1,1,0,-1,1],[0,-2,2,-2,0],[-1,1,-2,1,-1]] [0,-1,-1]*2 [-3,-4,-5]*3 [-4,-3,-5]*14 [-15,-8,-17]*13 [-12,-5,-13]*42 [-35,-12,-37]*31 [-24,-7,-25]*86 [[-1,-2,-1,0,1],[0,0,2,2,0],[1,2,1,0,1]] [-3,4,5] [-35,12,37] [-143,24,145] [-399,40,401] [-899,60,901] [-1763,84,1765] [-3135,112,3137] [[-1,0,0,0,1],[0,2,0,2,0],[1,0,2,0,1]] [0,1,1]*4 [-3,4,5]*5 [-4,3,5]*20 [-15,8,17]*17 [-12,5,13]*52 [-35,12,37]*37 [-24,7,25]*100 [[-1,2,-1,0,1],[0,0,2,-2,0],[-1,2,-1,0,-1]] [1,0,-1] [-3,4,-5] [-35,12,-37] [-143,24,-145] [-399,40,-401] [-899,60,-901] [-1763,84,-1765] [[-1,0,0,0,1],[0,0,2,0,0],[1,0,0,0,1]] [0,1,1]*2 [-15,8,17] [-40,9,41]*2 [-255,32,257] [-312,25,313]*2 [-1295,72,1297] [-1200,49,1201]*2 [[-1,-1,0,1,1],[0,2,2,2,0],[1,1,2,1,1]] [0,1,1]*6 [-3,4,5]*7 [-4,3,5]*26 [-15,8,17]*21 [-12,5,13]*62 [-35,12,37]*43 [-24,7,25]*114 [[-1,2,0,2,1],[0,2,0,-2,0],[-1,2,-2,-2,-1]] [1,0,-1]*4 [5,12,-13] [-5,12,-13]*4 [-119,120,-169] [-91,60,-109]*4 [-851,420,-949] [-425,168,-457]*4 [[-1,0,1,2,1],[0,2,2,0,0],[1,0,1,2,1]] [3,4,5] [-7,24,25] [-65,72,97] [-231,160,281] [-589,300,661] [-1247,504,1345] [-2337,784,2465] [[0,0,0,2,2],[0,0,1,2,0],[0,0,1,2,2]] [4,3,5] [3,4,5]*2 [8,15,17] [5,12,13]*2 [12,35,37] [7,24,25]*2 [16,63,65] [[0,0,2,0,-2],[0,1,-1,-2,0],[0,-1,1,0,-2]] [0,-1,-1]*2 [1,0,-1]*6 [4,3,-5]*4 [3,4,-5]*10 [8,15,-17]*6 [5,12,-13]*14 [12,35,-37]*8 [[0,0,2,0,-2],[0,1,1,-2,0],[0,1,1,0,-2]] [0,0,0] [3,4,5]*2 [8,15,17]*2 [5,12,13]*6 [12,35,37]*4 [7,24,25]*10 [16,63,65]*6 [[0,0,0,2,-2],[0,0,1,-2,0],[0,0,-1,2,-2]] [0,-1,-1] [1,0,-1]*2 [4,3,-5] [3,4,-5]*2 [8,15,-17] [5,12,-13]*2 [12,35,-37] [[0,0,0,2,1],[0,0,2,2,0],[0,0,2,2,1]] [3,4,5] [5,12,13] [7,24,25] [9,40,41] [11,60,61] [13,84,85] [15,112,113] [[0,0,2,-1,-1],[0,2,0,-2,0],[0,2,0,-1,-1]] [0,0,0] [5,12,13] [7,24,25]*2 [9,40,41]*3 [11,60,61]*4 [13,84,85]*5 [15,112,113]*6 [[0,1,1,-1,-1],[0,0,2,2,0],[0,1,1,1,1]] [0,1,1]*4 [3,4,5]*3 [4,3,5]*8 [15,8,17]*5 [12,5,13]*12 [35,12,37]*7 [24,7,25]*16 [[0,0,1,0,-1],[0,0,0,2,0],[0,0,1,0,1]] [0,1,1]*2 [3,4,5] [4,3,5]*2 [15,8,17] [12,5,13]*2 [35,12,37] [24,7,25]*2 [[0,-1,1,1,-1],[0,0,2,-2,0],[0,1,-1,1,-1]] [0,0,0] [-3,4,5] [-4,3,5]*4 [-15,8,17]*3 [-12,5,13]*8 [-35,12,37]*5 [-24,7,25]*12 [[0,0,2,1,-1],[0,2,0,-2,0],[0,-2,0,1,-1]] [1,0,-1]*2 [3,4,-5]*3 [5,12,-13]*4 [7,24,-25]*5 [9,40,-41]*6 [11,60,-61]*7 [13,84,-85]*8 [[0,0,0,2,-1],[0,0,2,-2,0],[0,0,-2,2,-1]] [1,0,-1] [3,4,-5] [5,12,-13] [7,24,-25] [9,40,-41] [11,60,-61] [13,84,-85] [[0,0,2,2,0],[0,1,2,0,0],[0,1,2,2,0]] [4,3,5] [3,4,5]*4 [8,15,17]*3 [5,12,13]*8 [12,35,37]*5 [7,24,25]*12 [16,63,65]*7 [[0,0,2,-2,0],[0,-1,2,0,0],[0,-1,2,-2,0]] [0,1,-1] [1,0,-1]*4 [4,-3,-5]*3 [3,-4,-5]*8 [8,-15,-17]*5 [5,-12,-13]*12 [12,-35,-37]*7 [[0,0,2,1,0],[0,2,2,0,0],[0,2,2,1,0]] [3,4,5] [5,12,13]*2 [7,24,25]*3 [9,40,41]*4 [11,60,61]*5 [13,84,85]*6 [15,112,113]*7 [[0,0,2,-1,0],[0,-2,2,0,0],[0,-2,2,-1,0]] [1,0,-1] [3,-4,-5]*2 [5,-12,-13]*3 [7,-24,-25]*4 [9,-40,-41]*5 [11,-60,-61]*6 [13,-84,-85]*7 [[0,1,0,-1,0],[0,0,2,0,0],[0,1,0,1,0]] [0,1,1]*2 [3,4,5]*2 [4,3,5]*6 [15,8,17]*4 [12,5,13]*10 [35,12,37]*6 [24,7,25]*14 #---> reslines=58
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clear // // // //Variable declaration h=6.6*10^-34; //planck's constant(J-sec) m=9.1*10^-31; //mass of electron(kg) c=3*10^8; //velocity of light(m/sec) lamda=0.82*10^-10; //wavelength(m) //Calculations E=h*c/lamda; //energy(J) lamda=h*10^10/sqrt(2*m*E); //wavelength of photo-electron(angstrom) //Result printf("\n wavelength of photo-electron is %0.1f angstrom",lamda)
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//Chapter 6 //Example 6.10 //Page 159 //impedance //run clear command then execute dependancy file and then the source file //dependency file is pucalc.sci clc; //Given P_g = 300e6; V_g = 20e3; X11_g = 0.20; l = 64; V_m = 13.2e3; P_m1 = 200e6; P_m2 = 100e6; X11_m = 0.20; T1_P = 350e6; T1_vht = 230e3; T1_vlt = 20e3; x_T1 = 0.10; T2_1_P = 100e6; T2_1_vht = 127e3; T2_1_vlt = 13.2e3; x_T2 = 0.10; x_line = 0.5;//ohm per km V_base = V_g; P_base = P_g; //Calculations T2_P = 3*T2_1_P; T2_vht = sqrt(3)*T2_1_vht; T2_vlt = T2_1_vlt; V_base_line = (T1_vht/T1_vlt)*V_base; V_base_m = V_base_line * (T2_vlt/T2_vht); x_T1_base = x_T1 * (P_base/T1_P); x_T2_base = x_T2 * (T2_vlt/V_base_m); z_line_base = (V_base_line)^2/P_base; x_line_pu = x_line * l / z_line_base; X11_m1_pu = pucalc(X11_m,V_m,V_base_m,P_base,P_m1); X11_m2_pu = pucalc(X11_m,V_m,V_base_m,P_base,P_m2); //Reactance diagram is given in xcos file disp('Base Voltages in different parts of circuit') printf("\n Generator voltage = %.0f kV",V_g/1e3) printf("\n Line voltage = %.0f kV",V_base_line/1e3) printf("\n Motor voltage = %.1f kV \n\n\n",V_base_m/1e3) disp('Base reactance in different parts of circuit') printf("\n Transformer 1 reactance = %.4f per unit",x_T1_base) printf("\n Transformer 2 reactance = %.4f per unit",x_T2_base) printf("\n Line reactance = %.4f per unit",x_line_pu) printf("\n Motor 1 reactance = %.4f per unit",X11_m1_pu) printf("\n Motor 2 reactance = %.4f per unit",X11_m2_pu) //impedance diagram is shown in the xcos file
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// Example 9_3 clc;funcprot(0); // Given data m=0.800;// kg/s V_1=93.0;// m/s // Station 1 p_1=97.0;// kPa T_1=80.0;// °C // Station 2 p_2=101.3;// kPa g_c=1;// The gravitational constant c_p=523;// J/(kg.K) R=208;// J/(kg.K) // Calculation T_2=(T_1+273.15)+((V_1^2)/(2*g_c*c_p));// K S_p=m*((c_p*log(T_2/(T_1+273.15)))-(R*log(p_2/p_1)));// The rate of entropy production within the diffuser in W/K printf("\nThe rate of entropy production within the diffuser,S_p=%1.2f W/K",S_p);
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// Example 2.14.b:Maximum Core Readius clc; clear; close; n1=1.48;//Waveguide Refractive Index d= 0.01;// Cange in core-cladding refractive index a=2;// parabolic refractive index h=1.3;//wavelngth in micro meters v= 2.4*sqrt(1+(2/a));//maximum value of normalised frequence a= (v*h)/(2*%pi*n1*sqrt(2*d));//Core Radius disp(a,"maximum core radius in micro meters")
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# careful. t1 m1 t2 m1 m2 t3 m2 =T:3 M:2 t:1 m:1
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stepdetails.sci
function stepdetails(resp) // This program is free software; you can redistribute it and/or modify // it under the terms of the GNU General Public License as published by // the Free Software Foundation; either version 2 of the License, or // (at your option) any later version. // // This program is distributed in the hope that it will be useful, // but WITHOUT ANY WARRANTY; without even the implied warranty of // MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the // GNU General Public License for more details. // // You should have received a copy of the GNU General Public License // along with this program; if not, write to the Free Software // Foundation, Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA // Authors // Holger Nahrstaedt - 2010 // Ishan Pendharkar - 2001-2007 // //RLTOOL for scilab (c) Ishan Pendharkar. //function calculates response parameter of step response. // resp is the response vector computed by csim [m,n]=size(resp); flag=0; finval=resp(n); //final value ess=string(clean(1-abs(finval),1e-5)); // steady state erorr [overshoot,tpeak]=max(resp); // peak overshoot (if at all!!) if( clean(abs(overshoot-finval),1e-4) >0 & (finval*overshoot)>0 ) then flag=1; //YES there is overshoot! perovershoot=string((overshoot-finval)/finval*100); //percentage over shoot tpeak=string(tpeak*tstep); //peak time else perovershoot='No Overshoot'; tpeak= 'None'; end; //****************damping ration calculations if flag==1 then //there is overshoot tmp=(%pi/(log(overshoot-finval)))^2+1; tmp=1/tmp; zeta=string(sqrt(tmp)); else zeta='overdamped' end //*********Settling time calculations************************ band=0.03*abs(resp(n)) // 3% tolerance band for i=1:n-2 if (abs( resp(n)-resp(n-i) )) >band then, ts=string((n-i)*tstep); tmp=n-i; break else ts='Couldnot resolve settling time.' tmp=0; end; end; if tmp>n-5 then ts='not settled. Select a larger time.'; tmp=[]; end; if tmp<>[] then, // if not unstable then display! messagebox([ 'Final Value (on the plot): '+string(finval); 'Error (as seen on plot): '+ess; 'Percent Peak Overshoot : '+perovershoot; 'Damping ratio (2nd order approx): '+zeta; 'Peak Time (sec): '+tpeak ; 'Setting Time (sec): '+ts ; ' '],"modal"); else messagebox(['System is unstable or output has not settled';'Increase simulation time in Settings->Step Response'],"modal"); end; //return endfunction
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5_1.sce
clc //initialisation of variables t1=5.25//yr t2=10.00//yr yi=171000//in ye=111000//in yt=5.23300//in yl=5.04532//in yn=31500//in ym=0.09853//in tm=9.25//yr tn=10.00//yr //CALCULATIONS T=t1/t2//yr T1=tm/tn//yr Y=yi-ye//in Yt=yt-yl//in //RESULTS printf('the fifth intercensal year =% f yr',T) printf('the ninth postcensal year =% f yr',T1)
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Afcam/Materiais-Eletricos-Magneticos
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aphi_sup.sce
//programa: aphi_sup.sce //Superposição em Poço 1D de Potenciais Infinitos clear; N = 100; //Número de Pontos L = 1e-9;//[m] x = linspace(0,L,N);//[m] h = 6.626e-34; //[J.s] hb = h/(2*%pi); m = 9.1e-31; //[kg] A = sqrt(2/L); n1 = 1; //Nível de Energia-1 n2 = 2; //Nível de Energia-2 kn1 = n1*%pi/L; //[rad/m] kn2 = n2*%pi/L; //[rad/m] wn1 = (hb*(kn1^2))/(2*m); //[rad/s] wn2 = (hb*(kn2^2))/(2*m); //[rad/s] fn1 = wn1/(2*%pi); Tn1 = 1/fn1; dx = x(2)-x(1); dt = Tn1/100 //[s] scf(1) plot(x,0*x); f=gcf(); a=get("current_axes"); a.data_bounds=[0,-1.2*A;L,1.2*A]; t = 0; while( t < 10*Tn1 ) t = t + dt; phi1 = ((A*sin(kn1*x))*exp(-%i*wn1*t)); phi2 = ((A*sin(kn2*x))*exp(-%i*wn2*t)); phi = (1/sqrt(2))*(phi1 + phi2); p = abs(phi).^2; pn = (A/max(p))*p; //Normalizada drawlater; scf(1); clf(1); plot(x,real(phi),'r',x,imag(phi),'g'); plot(x,pn,'b'); a=get("current_axes"); a.data_bounds=[0,-1.2*A;L,1.2*A]; drawnow; end
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Ex13_10.sce
clc clear //DATA GIVEN P=400; //maximum value of force that can be developed in N mu=0.25; //coefficient of friction d=0.6; //diameter of drum in m //Refer the figure theta=180+45; //angle of contact in degrees theta=theta*(%pi)/180; //theta converted into radians //moments about A, Ma=0, T1=P*1/0.5; //(i)Drum is rotating anticlockwise //T1>T2 (T1/T2)=e^(mu*theta) T2=T1/(%e^(mu*theta)); Mcac=(T1-T2)*(d/2); //maximum braking torquethat can be developed in N //(i)Drum is rotating clockwise //T2>T1 (T2/T1)=e^(mu*theta) T2=T1*(%e^(mu*theta)); Mcc=(T2-T1)*(d/2); //maximum braking torquethat can be developed in N printf(' (i) The Maximum braking torque that can be developed in anticlockwise direction is: %3.0f Nm. \n',Mcac); printf(' (ii) The Maximum braking torque that can be developed in clockwise direction is: %3.1f Nm. \n',Mcc);
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ex5_16.sce
clc; p1=3; // Pressre of air at state 1 in bar p2=p1; // constant pressure process T1=450; // Temperature of air at state 1 in kelvin T2=1250; // Temperature of air at state 2 in kelvin T3=1000; // Temperature of air at state 3 in kelvin V3=50; // Velocity of air at state 3 in m/s T4=800; // Temperature of air at state 4 in kelvin Cpo=1.0035; // Specific heat at constant pressure in kJ/kg K // (a).Combustion chamber q=Cpo*(T2-T1); // Heat added to air disp ("kJ/kg (round off error)",q,"Heat added to air = ","(a).Combustion chamber"); // (b).Turbine k=1.4; // Index of adiabatic process w=Cpo*(T2-T3)-V3^2/2000; // Work done disp ("kJ/kg (round off error)",w,"Work done = ",("(b).Turbine)")); // (c).Nozzle V4=sqrt (2*Cpo*10^3*(T3-T4)+V3^2); // Velocity of air leaving the nozzle disp ("m/s (round off error)",V4,"Velocity of air leaving the nozzle = ","(c).Nozzle"); // (d).Pressure drop p3=p2*(T3/T2)^(k/(k-1)); // Pressure of air leaving turbine p4=p3*(T4/T3)^(k/(k-1)); // Pressure of air leaving nozzle disp ("bar ",p4,"Pressure of air leaving nozzle = ","bar",p3,"Pressure of air leaving turbine = ","(d).Pressure drop");
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Exa1_20.sce
//Exa 1.20 clc; clear; close; //given data cos_fi=0.8;//unitless fi=acosd(cos_fi); tan_fi=tand(fi);//unitless //For Alternator A : cos_fi_A=0.9;//unitless fi_A=acosd(cos_fi_A); tan_fi_A=tand(fi_A);//unitless //Formula : Active load, KW=V*I*cos_fi //Formula : Reactive load, KVAR=V*I*sin_fi ActiveLoad=8000;//in KW ReactiveLoad=ActiveLoad*tan_fi;//in KVAR //For A: ActiveLoadA=5000;//in KW ReactiveLoadA=ActiveLoadA*tan_fi_A;//in KVAR //For B : ActiveLoadB=ActiveLoad-ActiveLoadA;//in KW ReactiveLoadB=ReactiveLoad-ReactiveLoadA;//in KVAR tan_fi_B=ReactiveLoadB/ActiveLoadB;//unitless fi_B=atand(tan_fi_B);//in degree cos_fi=cosd(atand(tan_fi_B));//unitless disp("Power factor of the other machine : "+string(cos_fi));
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ex_3_24.sce
errcatch(-1,"stop");mode(2);//Example 3.24 : concentration of iron ; ; format('v',9) //given data : d=7.87; N=6.023*10^23; // avogadro's number A=55.85;// atomic weight I=A/N;// mass of iron atom atom=d/I; disp(atom,"number of atoms(atoms/cm^3) = ") exit();
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result2s0.tst
@relation unknow @attribute COMPACTNESS real[73.0,119.0] @attribute CIRCULARITY real[33.0,59.0] @attribute DISTANCECIRCULARITY real[40.0,112.0] @attribute RADIUSRATIO real[104.0,333.0] @attribute PRAXISASPECTRATIO real[47.0,138.0] @attribute MAXLENGTHASPECTRATIO real[2.0,55.0] @attribute SCATTERRATIO real[112.0,265.0] @attribute ELONGATEDNESS real[26.0,61.0] @attribute PRAXISRECTANGULAR real[17.0,29.0] @attribute LENGTHRECTANGULAR real[118.0,188.0] @attribute MAJORVARIANCE real[130.0,320.0] @attribute MINORVARIANCE real[184.0,1018.0] @attribute GYRATIONRADIUS real[109.0,268.0] @attribute MAJORSKEWNESS real[59.0,135.0] @attribute MINORSKEWNESS real[0.0,22.0] @attribute MINORKURTOSIS real[0.0,41.0] @attribute MAJORKURTOSIS real[176.0,206.0] @attribute HOLLOWSRATIO real[181.0,211.0] @attribute class{van,saab,bus,opel} @inputs COMPACTNESS,CIRCULARITY,DISTANCECIRCULARITY,RADIUSRATIO,PRAXISASPECTRATIO,MAXLENGTHASPECTRATIO,SCATTERRATIO,ELONGATEDNESS,PRAXISRECTANGULAR,LENGTHRECTANGULAR,MAJORVARIANCE,MINORVARIANCE,GYRATIONRADIUS,MAJORSKEWNESS,MINORSKEWNESS,MINORKURTOSIS,MAJORKURTOSIS,HOLLOWSRATIO @outputs class @data van van van van bus bus van van bus bus van van saab van saab opel opel van van van saab opel van van bus bus bus bus opel opel van van saab bus bus bus bus bus bus bus van van bus bus opel opel van van van van van van saab saab bus bus van van saab opel bus bus bus bus opel van opel opel saab saab saab saab van van van van saab saab bus bus saab opel van van opel opel bus bus van van van van opel saab bus bus van van opel saab bus bus bus bus opel van bus bus saab saab van van saab van van van van van bus bus bus bus bus bus van van opel opel van van saab opel saab saab saab saab saab saab bus bus bus bus opel saab saab opel saab saab saab opel opel opel opel van opel van opel opel bus bus van van bus bus bus bus bus bus bus van bus bus opel saab saab saab van van bus bus saab opel saab saab van van opel opel bus bus bus bus van van saab opel opel opel saab opel opel saab saab saab van van opel opel van van van van opel opel bus bus van van bus van saab opel bus bus opel opel saab saab van van saab opel opel opel opel saab opel van opel van opel opel opel saab saab van van van saab opel bus bus opel saab opel saab opel opel opel opel bus bus saab opel saab opel saab opel bus bus bus bus van van saab saab opel opel opel opel van van saab saab saab saab bus bus saab saab opel opel van van bus bus opel opel saab opel saab opel saab saab bus bus van van bus bus saab saab van van opel opel saab opel bus bus opel saab van van opel van opel opel bus bus opel opel opel opel saab saab van van
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Woumpousse/KHL
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permutaties.sce
function r=swap(s,x,y) // wissel in vector s element x en y van plaats r=s; r(y)=s(x); r(x)=s(y); endfunction function permutatie(n) // vind alle permutaties van 1...n // in lexicografische volgorde s=1:n; disp(s); for i=2:factorial(n) m=n-1; while s(m)>s(m+1) // zoek het eerste element dat kleiner is dan het // tweede te beginnen van rechts m=m-1; end k=n; while s(m)>s(k) // vind het meest rechtse element s(k) // waarbij s(m)<s(k) k=k-1; end s=swap(s,m,k); p=m+1; q=n; while p<q s=swap(s,q,p); p=p+1; q=q-1; end disp(s) end endfunction permutatie(3)
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Example_10_5.sce
//Caption: Forecasting //Simple Moving Average Method //Example10.5 //Page381 clear; clc; Dt = [24,30,27,24,39,45,42,51];//Demand Di n = length(Dt);//Month (t) //Three months moving average for i = 3:n Mt(i-2) = mean(Dt([(i-2):i])); end disp(Mt,'Three Months moving average Mt=') for i = 1:length(Mt)-1 Ft(i) = Mt(i); et(i) = Dt(i+3)-Ft(i); end disp(Ft,'Forecast Ft=') disp(et,'Error et=') MAD = sum(abs(et(:)))/length(et); disp(MAD,'Mean Absolute Deviation MAD=') MFE = sum(et(:))/length(et); disp(MFE,'Mean Forecast Error MFE=') //Result // Three Months moving average Mt= // // 27. // 27. // 30. // 36. // 42. // 46. // // Forecast Ft= // // 27. // 27. // 30. // 36. // 42. // // Error et= // // - 3. // 12. // 15. // 6. // 9. // // Mean Absolute Deviation MAD= // // 9. // // Mean Forecast Error MFE= // // 7.8
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clc h1=2776.4; //kJ/kg h2=h1; h_f1=884.6; //kJ/kg h_fg1=1910.3; //kJ/kg x1=(h1-h_f1)/h_fg1; disp("Initial dryness fraction = ") disp(x1)
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ch_4_eg_10.sce
clc disp("the solution of eg 4.10 -->Series of Stirred Tanks with Coil Heaters") Cp=2000,A=1,U=200,m=1000,mdot=2,Ts=250 //given data T0=20, T1=0, T2=0, T3=0 //from energy balances for the tanks we have accumulation=inlet-outlet T1_steady=(mdot*Cp*(T0)+U*A*(Ts))/(mdot*Cp+U*A) disp(T1_steady,"the steady state temperature of tank 1 is"); T2_steady=(mdot*Cp*(T1_steady)+U*A*(Ts))/(mdot*Cp+U*A) disp(T2_steady,"the steady state temperature of tank 2 is"); T3_steady=(mdot*Cp*(T2_steady)+U*A*(Ts))/(mdot*Cp+U*A) disp(T3_steady,"the steady state temperature of tank 3 is"); final_T3=.99*T3_steady function dT1_by_dt=f1(t,T1,T2,T3), dT1_by_dt=(mdot*Cp*(T0-T1)+U*A*(Ts-T1))/(m*Cp), endfunction function dT2_by_dt=f2(t,T1,T2,T3), dT2_by_dt=(mdot*Cp*(T1-T2)+U*A*(Ts-T2))/(m*Cp), endfunction function dT3_by_dt=f3(t,T1,T2,T3), dT3_by_dt=(mdot*Cp*(T2-T3)+U*A*(Ts-T3))/(m*Cp), endfunction T1=20,T2=20,T3=20 //solving by Newton's Method for t=0:1:10000, h=1 //step increment of 1 k1=h*f1(t,T1,T2,T3) l1=h*f2(t,T1,T2,T3) m1=h*f3(t,T1,T2,T3) k2=h*f1(t+h/2,T1+k1/2,T2+l1/2,T3+m1/2) l2=h*f2(t+h/2,T1+k1/2,T2+l1/2,T3+m1/2) m2=h*f3(t+h/2,T1+k1/2,T2+l1/2,T3+m1/2) k3=h*f1(t+h/2,T1+k2/2,T2+l2/2,T3+m2/2) l3=h*f2(t+h/2,T1+k2/2,T2+l2/2,T3+m2/2) m3=h*f3(t+h/2,T1+k2/2,T2+l2/2,T3+m2/2) k4=h*f1(t+h,T1+k3,T2+l3,T3+m3) l4=h*f2(t+h,T1+k3,T2+l3,T3+m3) m4=h*f3(t+h,T1+k3,T2+l3,T3+m3) T1=T1+(k1+2*k2+2*k3+k4)/6 T2=T2+(l1+2*l2+2*l3+l4)/6 e1=abs(T3-final_T3) if e1<1e-3 then disp(t,"the approx. time when Temperature in 3rd tank is 99% of steady value is"); break end T3=T3+(m1+2*m2+2*m3+m4)/6 end
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Example6_8_a.sce
errcatch(-1,"stop");mode(2);//Example 6.8(a) ; ; Vs=15; A=10; Vim=0.5; SR=0.5*10^6; Vom=A*Vim; fmax=SR/(2*%pi*Vom); printf("fmax=%.f kHz",fmax*10^(-3)); exit();
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ex4_7.sce
clc; clear all; V = 120000; // Volume of hall in cubic meters T = 1.5; // Reverberation time TSA = 25000; // Total absorbing surface area in square meters A = (0.163*V)/T TA = A/TSA;//The average absorbing power of the surface disp('Sabine',TA,'The average absorbing power of the surface is ') // Slight variation in answer compared to textbook.. there is mistake in book.. checked in calculator also..
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Ex4_5.sce
clc //Variable Initialisation Ea=220//Input Voltage to armature in volts N1=1000//Rated Speed of Motor in rpm N2=500//Speed of Motor in rpm Ia=24//Armature Current in Ampere Ra=2//Armature resistance in ohm Es=230//Source voltage in Volts //Solution Eb1=Ea-(Ia*Ra) Eb2=(N2/N1)*Eb1 E0=Eb2+(1.2*Ia*Ra) d=E0/Es printf('\n\n The Duty Ratio=%0.1f\n\n',d) //The answers vary due to round off error
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4_13.sci
//4.13 clc; d_rate=100; fc= 0.5*d_rate; printf("cutt off frquency =%.1f kHz ",fc)
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PA_Ex_11_2.sce
clc //Chapter 11:Power amplifiers //example 11.2 page no 466 //given Po=5//max power in watts Rl=50//load resistance in ohm //asumme'1:1 truns ratio transformer coupled push pull amplifier each supllying 2.5 watt' disp('since a push pull amplifier is used, each class B amplifier will supply 2.5W') Pomax=2.5 Vcc=sqrt(4*Rl*Po)//supply voltage Ptmax=Pomax*(4/%pi^2)//maximum power handling requriment of the transistor I=sqrt((4*Pomax)/Rl)//peak output current mprintf('maximum power handling requriment of the transistor is %d W \n peak output current is %f A ',Ptmax,I)
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C2P3.sce
clear clc //to find magnitude and direction of resultant of a and b and c vector // GIVEN:: //coefficient in x direction for vector a ax = 4.3 //coefficient in y direction for vector a ay = -1.7 //coefficient in x direction for vector b bx = -2.9 //coefficient in y direction for vector b by = 2.2 //coefficient in x direction for vector c cx = 0 //coefficient in y direction for vector c cy = -3.6 //we can write a,b and c in vector form a = [4.3 -1.7] b = [-2.9 2.2] c = [0 -3.6] // SOLUTION: //coefficient in x direction for resultant vector sx = ax + bx + cx //coefficient in y direction for resultant vector sy = ay + by + cy //direction of resultant vector fi = atand(sy/sx)+360 printf ("\n\n Coefficient of resultant vector in x direction sx = \n\n %.1f",sx); printf ("\n\n Coefficient of resultant vector in y direction sy =\n\n %.1f",sy); printf ("\n\n Resultant vector s =\n\n %.1fi + %.1fj',sx,sy); printf ("\n\n Direction of resultant vector with positive x axis measured counterclockwise fi =\n\n %3i degrees",fi);
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ok_bv.tst
; checks bit vector sorts (set-logic QF_BV) (declare-fun x () (_ BitVec 4)) (declare-fun y () (_ BitVec 1)) (declare-fun z () (_ BitVec 4)) (assert (= x #b0101 )) (assert (= z #xa ))
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14_5.sce
//To find gyroscopic couple and direction clc //Given: N=1500 //rpm m=750 //kg omegaP=1 //rad/s k=250/1000 //m //Solution: //Calculating the angular speed of the rotor omega=2*%pi*N/60 //rad/s //Calculating the mass moment of inertia of the rotor I=m*k^2 //kg-m^2 //Calculating the gyroscopic couple transmitted to the hull C=I*omega*omegaP/1000 //kN-m //Results: printf("\n\n Gyroscopic couple transmitted to the hull, C = %.3f kN-m.\n\n",C) printf(" When the pitching is upward, the relative gyroscopic couple acts in the clockwise direction.\n\n")
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ch4_22.sce
clear; clc; V_r=1000; //rating of SCR I_r=200; //rating of SCR V_s=6000; //rating of String I_s=1000; //rating of String disp("when DRF=.1"); DRF=.1; n_s=V_s/(V_r*(1-DRF)); printf("number of series units=%.0f",ceil(n_s)); n_p=I_s/(I_r*(1-DRF)); printf("\nnumber of parrallel units=%.0f",ceil(n_p)); disp("when DRF=.2"); DRF=.2; n_s=V_s/(V_r*(1-DRF)); printf("number of series units=%.0f",ceil(n_s)); n_p=I_s/(I_r*(1-DRF)); printf("\nnumber of parrallel units=%.0f",ceil(n_p));
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Ex6_9.sce
clear ; clc; // Example 6.9 printf('Example 6.9\n\n'); printf('Page No. 157\n\n'); // given P = 10;// Boiler pressure in bar Ts = 180;// Steam temperature in degree celcius Tf = 80;// Feed water temperature in degree celcius X = 0.95;// Steam dryness fraction m_s = 4100;// steam rate in kg/h m_f = 238;// Gas rate in kg/h G_CV = 53.5*10^6;// In J/kg N_CV = 48*10^6;//in J/kg //from steam table,AT 10 bar and at temperature T = Ts h2 = (763+(X*2013))*10^3;//Specific enthalpy of steam in J/kg //At temperature T = Tf h1 = 335*10^3;//Specific enthalpy of feed steam in J/kg E_G = ((m_s*(h2-h1)*100)/(m_f*G_CV));// printf('The gross efficiency percentage is %.0f \n',E_G) E_N = ((m_s*(h2-h1)*100)/(m_f*N_CV));// printf('The net efficiency percentage is %.0f',E_N)
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Example_4_3.sce
//Example 4.3 (b) clear; clc; //Given a=1.24;//alpha at 290K and 1 atm in 10^-3 K^-1 b=9.3;//beta at 290K and 1 atm in 10^-5 atm^-1 T=290;//temperature in K delS=2.1;//entropy change in J K^-1 mol^-1 //to calculate the change in molar volume delV=(delS*b)/(a*100*101.325);//change in molar volume in dm^3 mol^-1 mprintf('change in molar volume = %f dm^3 mol^-1',delV); //end
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42.sce
clc cp=1.11; T3=883; //K T2a=529; //K W_turbine=290.4; //kJ/kg W_net=48.2; //kJ/kg Qs=cp*(T3-T2a); n_thermal=W_net/Qs*100; disp("Thermal efficiency =") disp(n_thermal) disp("%") W_ratio=W_net/W_turbine; //Work ratio=net work output/Gross work output disp("Work ratio =") disp(W_ratio)
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example_6_1.sce
//Chapter 6 //Example 6-1 //ProbOnThresholdVoltage //Page 149,151, Figure 6-1 clear;clc; //Given R1 = 100*10^3 ; R2 = 86*10^3 ; Vsatp = 15 ; Vsatm = -15 ;//Saturation voltages Vut = (R2 * Vsatp)/(R1 + R2); Vlt = (R2 * Vsatm)/(R1 + R2); printf ( "\n\n Upper Threshold Voltage = %.4f V ", Vut ) printf ( "\n\n Lower Threshold Voltage = %.4f V ", Vlt )
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example9_8.sce
//clc() F = 100;//kg xf = 0.15; P1 = 80;//% ( Carbonate recovered ) M1 = 106;//(Molecular weight of Na2CO3) M2 = 286;//(Molecular weight of Na2CO3.10H2O) x1 = M1 / M2;//(Weight fraction of Na2CO3 in crystals) Mrecovered = P1 * F * xf / 100; Wcrystal = Mrecovered / x1; disp("kg",Wcrystal,"(a)quantity of crystals formed = ") //Na2CO3 balance gives, F*xf = Wcrystal*x1 + W2*x2 //W2 weight of mother liquor remaining after crystallization //let M = W2 * x2,therefore M = F * xf - Mrecovered; x2 = 0.09; W2 = M/x2; W3 = F - Wcrystal - W2;//weight of water evaporated disp("kg",W3,"(b)Weight of water evaporated = ")
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Ex9_1.sce
//Example number 9.1, Page number 202 clc;clear; close; //Variable declaration e=1.6*10**-19; //charge(c) ni=2.4*10**19; //particle density(per m**3) mew_e=0.39; //electron mobility(m**2/Vs) mew_h=0.19; //hole mobility(m**2/Vs) //Calculation rho=1/(ni*e*(mew_e+mew_h)); //resistivity(ohm m) //Result printf("resistivity is %.5f ohm-m",rho)
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Example7_4.sce
// Electric Machinery and Transformers // Irving L kosow // Prentice Hall of India // 2nd editiom // Chapter 7: PARALLEL OPERATION // Example 7-4 clear; clc; close; // Clear the work space and console. // Given data P1 = 300 ; // Power rating of generator 1 in kW P2 = 600 ; // Power rating of generator 2 in kW V = 220 ; // Voltage rating of generator 1 and 2 in volt V_o = 250 ; // No-load voltage applied to both the generators in volt // Assume linear characteristics V_1 = 230 ; // Terminal voltage in volt (case a) V_2 = 240 ; // Terminal voltage in volt (case b) // Calculations // case a kW1_a = (V_o - V_1)/(V_o - V) * P1 ; // kW carried by generator 1 kW2_a = (V_o - V_1)/(V_o - V) * P2 ; // kW carried by generator 2 // case b kW1_b = (V_o - V_2)/(V_o - V) * P1 ; // kW carried by generator 1 kW2_b = (V_o - V_2)/(V_o - V) * P2 ; // kW carried by generator 2 // case c frac_a = (V_o - V_1)/(V_o - V); // Fraction of rated kW carried by each generator frac_b = (V_o - V_2)/(V_o - V); // Fraction of rated kW carried by each generator // Display the results disp("Example 7-4 Solution : "); printf(" \n a: At 230 V, using Eq.(7-3) below : "); printf(" \n Generator 1 carries = %d kW ", kW1_a ); printf(" \n Generator 2 carries = %d kW \n", kW2_a ); printf(" \n b: At 240 V, using Eq.(7-3) below : "); printf(" \n Generator 1 carries = %d kW ", kW1_b ); printf(" \n Generator 2 carries = %d kW \n", kW2_b ); printf(" \n c: Both generators carry no-load at 250 V; "); printf(" \n %f rated load at %d V; ", frac_b , V_2 ); printf(" \n %f rated load at %d V; ", frac_a , V_1 ); printf(" \n and rated load at %d V. ", V );
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EX3_16.sce
//Calculating secondary voltage and voltage regulation //Chapter 3 //Example 3.16 //page 218 clear; clc; disp("Example 3.16") kVA=10; //rating of the transformer V1=2000; //primary voltage in volts V2=400; //secondary voltage in volts R1=5.5; //primary voltage in ohms R2=0.2; //secondary voltage in ohms X1=12; //primary reactance in ohms X2=0.45; //secondary reactance in ohms //assuming (V1/V2)=(N1/N2) Re=R2+(R1*(V2/V1)^2); printf("equivalent resistance referred to the secondary=%fohms",Re); Xe=X2+(X1*(V2/V1)^2); printf("equivalent reactance referred to the secondary=%fohms",Xe); Ze=sqrt(Re^2+Xe^2); printf("equivalent impedance referred to the secondary=%fohms",Ze); phi=acosd(0.8); Vl=374.5; printf("\nVoltage across the full load and 0.8 p.f lagging=%fV",Vl); reg=((V2-Vl)*100)/Vl; printf("\npercentage voltage regulation=%f percent",reg);
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Example5_4_4.sce
//Example 5.4.4 page 5.15 clc; clear; Ts= 10*10^-9; Tn=9*10^-9; Tc=2*10^-9; Td=3*10^-9; BW= 6*10^6; Tsyst= 1.1*sqrt(Ts^2+(5*Tn)^2+(5*Tc)^2+Td^2); Tsyst=Tsyst*10^9;//converting in ns for displying... Tsyst_max = 0.35/BW; Tsyst_max=Tsyst_max*10^9;//converting in ns for displying... printf("Rise system of the system is %.2f ns",Tsyst) printf("\n\nMaximum Rise system of the system is %.2f ns",Tsyst_max) printf("\n\nSpecified components give a system rise time which is\n adequate for the bandwidth and distance requirements of the optical fibre link.");
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Ex9_12.sce
//Example number 9.12, Page number 208 clc;clear; close; //Variable declaration n=5*10**17; //concentration(m**3) vd=350; //drift velocity(m/s) E=1000; //electric field(V/m) e=1.6*10**-19; //charge(c) //Calculation sigma=n*e*vd/E; //conductivity(per ohm m) //Result printf("conductivity is %.3f per ohm-m",sigma)
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RLG_Euler.sci
function [P,Q,THETA,SUCCESS,NB_TRY] = RLG_Euler(STANCE,NORMALS,PARAMS) //Author : Maxens ACHIEPI //Space Robotics Laboratory - Tohoku University //Description: // Cxy,WS_proj_R0,footPlane_Rmat,zFinalInterval,psiInter,thetInter,phiInter //[] //INPUT //STANCE: Row array of the current footholds. Contains struct describing // footholds: // *foothold: struct. // *foothold.leg: string identifying the leg (FR,FL,HR,HL); // *foothold.pos: row vector. Position of the foot in R0 //PARAMS: a struct containing all the parameters relating to the robot // geometry, as well as problem-specific parameters: // *PARAMS.extRad; // *PARAMS.distApiOb: distances between the leg attachment and the EoF CoM // *PARAMS.intRad; // *PARAMS.halfAngle; // *PARAMS.shellPtsNb; // *PARAMS.shrink // *PARAMS.kpxy; // *PARAMS.kpz; // *PARAMS.kRz; // *PARAMS.kRx; // *PARAMS.tInc; // *PARAMS.aInc; // *PARAMS.baseDimensions: (1) on x, (2) on y; // *PARAMS.legLength: [l1,l2,l3] // *PARAMS.verbose: %T or %F //OUTPUT //P: the base position //Q: the quaternion defining the rotation //THETA: the leg's joint angles //RMAT: the rotation matrix //SUCCESS: boolean for benchmarking purposes (atm) //NB_TRY: number of tries to sample a configuration //TODO : put psi/theta/phi range finding in function // could you put bounds on phi so that it doesn't flip over? NAH(?) // put IK in function // Remove RMAT output // Change rotation parametrization to full quaternion //----------------------------------------------------------------------------// P = 0;Q = 0;THETA = 0;SUCCESS = %F;RMAT = 0;NB_TRY = 1; //Compute LS-fit plane by ACP stance_pos_list = STANCE(:).pos; stance_pos_array = []; for i=1:size(STANCE,2) stance_pos_array(i,:) = stance_pos_list(i); end foot_nb = size(stance_pos_array,1); [footPlane_z,footPlane_d,footPlane_or] = plane_ACP(stance_pos_array); footPlane_z = footPlane_z/norm(footPlane_z); FL_present = %f;HL_present = %f;FR_present = %f;HR_present = %f; for i=1:foot_nb select STANCE(i).leg case 'FL' then FL = i; FL_present = %t; case 'HL' then HL = i; HL_present = %t; case 'FR' then FR = i; FR_present = %t; case 'HR' then HR = i; HR_present = %t; end end if HR_present&FR_present then footPlane_x = (STANCE(HR).pos-footPlane_or) + 0.5*(STANCE(FR).pos-STANCE(HR).pos); footPlane_x = projectionPlan(footPlane_x,footPlane_or,footPlane_z); footPlane_x = footPlane_x - (footPlane_z*footPlane_or')*footPlane_z footPlane_x = footPlane_x/norm(footPlane_x); footPlane_y = cross(footPlane_z,footPlane_x); elseif HL_present&FL_present then footPlane_x = (STANCE(HL).pos-footPlane_or) + 0.5*(STANCE(FL).pos-STANCE(HL).pos); footPlane_x = projectionPlan(footPlane_x,footPlane_or,footPlane_z); footPlane_x = footPlane_x/norm(footPlane_x); footPlane_y = cross(footPlane_z,footPlane_x); end footPlane_Rmat = [footPlane_x;footPlane_y;footPlane_z]; [footPlane_angle,footPlane_vector] = angle_vector_FromMat(footPlane_Rmat); footPlane_Q = createQuaternion(footPlane_angle,footPlane_vector); //Compute leg approximate workspaces. Project them on footPlane. for i = 1:foot_nb //leg workspace, all points in R0 WSmi_R0 = []; WSmi_proj_RP = []; //shell descriptions shellDesc_i = struct('origin',stance_pos_array(i,:),'extRad',PARAMS.extRad(i),'intRad',PARAMS.intRad(i),'axis',NORMALS(i,:),'halfAngle',PARAMS.halfAngle); shellDesc(i) = shellDesc_i; shellDesc_AUG_i = struct('origin',stance_pos_array(i,:),'extRad',PARAMS.extRad(i)+PARAMS.distApiOb(i),'intRad',PARAMS.intRad(i),'axis',NORMALS(i,:),'halfAngle',PARAMS.halfAngle); shellDesc_AUG(i) = shellDesc_AUG_i; WSmi_alpha = linspace(0,2*%pi,PARAMS.shellPtsNb); WSmi_theta = linspace(%pi/2-shellDesc_AUG_i.halfAngle,%pi/2,PARAMS.shellPtsNb); [x1,y1,z1] = halfSph(shellDesc_AUG_i.origin,shellDesc_AUG_i.extRad,2*WSmi_alpha,WSmi_theta,shellDesc_AUG_i.axis); WSmi_R0 = [x1',y1',z1']; if shellDesc_i.intRad then [x2,y2,z2] = halfSph(shellDesc_AUG_i.origin,shellDesc_AUG_i.intRad,2*WSmi_alpha,WSmi_theta,shellDesc_AUG_i.axis); WSmi_R0 = [WSmi_R0;x2' y2' z2']; end WS_R0(:,:,i) = WSmi_R0; //projection, all points in RP for j=1:size(WSmi_R0,1) v = projectionPlan(WSmi_R0(j,:),footPlane_or,footPlane_z); WSmi_proj_R0(j,1) = v(1);WSmi_proj_R0(j,2) = v(2);WSmi_proj_R0(j,3) = v(3); v = footPlane_Rmat*(v'-footPlane_or'); WSmi_proj_RP(j,1) = v(1);WSmi_proj_RP(j,2) = v(2); end WS_proj_RP(:,:,i) = WSmi_proj_RP; WS_proj_R0(:,:,i) = WSmi_proj_R0; end //Compute Cxy Cxy = computeCxy(WS_proj_RP,[1 0;0 1]); if isnan(Cxy.origin) then if PARAMS.verbose then mprintf('Could not compute intersection of workspaces! Stance is probably unreachable...\n'); end return; end //Sample pxy_RP, transform into pxy_R0 kpxy = 0; while kpxy<PARAMS.kpxy kpxy = kpxy+1; kpz = 0; pxy_RP = sampleInBBox(Cxy,PARAMS.shrink); pxy_R0 = footPlane_Rmat'*[pxy_RP 0]'+footPlane_or'; if PARAMS.verbose then mprintf("XY - At iteration %d of %d:\nBase xy_R0 position: [%.4f, %.4f]\n",kpxy,PARAMS.kpxy,pxy_R0(1),pxy_R0(2)); end zInterval = cell(1,foot_nb); //Compute intersections of the line perpendicular to footPlane, going through pxy_R0, with the WSmi line_z = struct('origin',pxy_R0','direction',footPlane_z); for i=1:foot_nb [boolInterT_i,tMultiple_i,tInterval_i,d_i]=intersectLineWS(WS_R0(:,:,i),shellDesc_AUG(i),line_z,PARAMS.tInc); if boolInterT_i then tInterval(i).entries = createZInterval(tInterval_i,d_i); if PARAMS.verbose & tMultiple_i then mprintf(" T - For leg %d, t lies in %d different intervals", i, size(tInterval(i).entries,1)); elseif PARAMS.verbose then mprintf(" T - For leg %d, t range is: %.4f to %.4f\n",i,tInterval(i).entries(1),tInterval(i).entries(2)); end else if PARAMS.verbose then mprintf(" T - No intersection with leg %d workspace! Resampling pxy_RP...\n",i); NB_TRY = NB_TRY+1; end break; end end if ~boolInterT_i then continue; end //Sample pz_R0 [tFinalBool,tFinalInterval] = intersectSetIntervals(tInterval); if ~tFinalBool then if PARAMS.verbose then mprintf(" T - t valid intervals do not intersect! Resammpling pxy_RP...\n"); NB_TRY = NB_TRY+1; end continue; end while kpz<PARAMS.kpz kpz = kpz+1; kRz = 0; t_R0 = sampleFromMultInterval(tFinalInterval); if PARAMS.verbose then mprintf('T - At iteration %d of %d:\n Base t_R0 : %.4f\n",kpz,PARAMS.kpz,t_R0); end //Compute intersections of Api arcs and WSmi for each rotation parameters //Rotations are represented by Euler angles (norm ZXY): (psi,Z0);(thet,X1);(phi,Y2) base_R0 = pxy_R0'+t_R0*line_z.direction; P = base_R0; offset_i = []; xOff = [1 0 0]*PARAMS.baseDimensions(1)/2; yOff = [0 1 0]*PARAMS.baseDimensions(2)/2; R_0_EF = footPlane_Rmat; //Transformation matrix between end-eff frame and R0. At first is R_0_plane. psiInter=cell(1,foot_nb); arcDesc_psi = struct('origin',base_R0,'normal',[0 0 1]) //rotation around Z0 //First start with (psi,Z0) for i=1:foot_nb select STANCE(i).leg case 'FR' then offset_i = xOff + yOff; case 'FL' then offset_i = - xOff + yOff; case 'HR' then offset_i= + xOff - yOff; case 'HL' then offset_i = - xOff - yOff; else if PARAMS.verbose then mprintf("Error in the definition of foothold %d : leg name does not exist!\n",i); end return; end [boolInterPsi_i,psiMultiple_i,psiInter_i] = intersectArcWS(WS_R0(:,:,i),offset_i,R_0_EF,shellDesc(i),arcDesc_psi,PARAMS.aInc); if boolInterPsi_i then psiInter(i).entries = createAngleInterval(psiInter_i); if PARAMS.verbose & psiMultiple_i then mprintf(" PSI - For leg %d, psi lies in %d different intervals\n", i, size(psiInter(i).entries,1)); elseif PARAMS.verbose then mprintf(" PSI - For leg %d, psi range is: %.4f to %.4f\n",i,psiInter(i).entries(1),psiInter(i).entries(2)); end else if PARAMS.verbose then mprintf(" PSI - No intersection with leg %d workspace! Resampling pz_0...\n",i); NB_TRY = NB_TRY+1; end break; end end if ~boolInterPsi_i then continue; end //Sample (psi,Z0) [psiBoolFinal,psiFinalInterval] = intersectSetIntervals(psiInter); if ~psiBoolFinal then if PARAMS.verbose then mprintf(" PSI - psi valid intervals do not intersect! Resampling z_R0...\n"); NB_TRY = NB_TRY+1; end continue; end while kRz<PARAMS.kRz kRz = kRz +1; kRx = 0; psi = sampleFromMultInterval(psiFinalInterval); if PARAMS.verbose then mprintf("PSI - At iteration %d of %d:\n Base psi: %.4f\n",kRz,PARAMS.kRz,psi); end //Rotate base Rz0 = [cos(psi), -sin(psi), 0;sin(psi), cos(psi) 0;0 0 1]; [Rz0_angle,Rz0_vector] = angle_vector_FromMat(Rz0); Rz0_Q = createQuaternion(Rz0_angle,Rz0_vector); R_0_EF = R_0_EF*Rz0; thetInter=cell(1,foot_nb); arcDesc_thet = struct('origin',base_R0,'normal',[1 0 0]) //rotation around X1 //Then (thet,X1) for i=1:foot_nb select STANCE(i).leg case 'FR' then offset_i = xOff + yOff; case 'FL' then offset_i = - xOff + yOff; case 'HR' then offset_i= + xOff - yOff; case 'HL' then offset_i = - xOff - yOff; else if PARAMS.verbose then mprintf("Error in the definition of foothold %d : leg name does not exist!\n",i); end return; end [boolInterThet_i,thetMultiple_i,thetInter_i] = intersectArcWS(WS_R0(:,:,i),offset_i,R_0_EF,shellDesc(i),arcDesc_thet,PARAMS.aInc); if boolInterThet_i then thetInter(i).entries = createAngleInterval(thetInter_i); if PARAMS.verbose & thetMultiple_i then mprintf(" THET - For leg %d, theta lies in %d different intervals\n", i, size(thetInter(i).entries,1)); elseif PARAMS.verbose then mprintf(" THET - For leg %d, theta range is: %.4f to %.4f\n",i,thetInter(i).entries(1),thetInter(i).entries(2)); end else if PARAMS.verbose then mprintf(" THET - No intersection with leg %d workspace! Resampling psi...\n",i); NB_TRY = NB_TRY+1; end break; end end if ~boolInterThet_i then continue; end //Sample (theta,X1) [thetBoolFinal,thetFinalInterval] = intersectSetIntervals(thetInter); if ~thetBoolFinal then if PARAMS.verbose then mprintf(" THET - theta valid intervals do not intersect! Resampling psi...\n"); NB_TRY = NB_TRY+1; end continue; end while kRx<PARAMS.kRx kRx = kRx +1; theta = sampleFromMultInterval(thetFinalInterval); if PARAMS.verbose then mprintf("THETA - At iteration %d of %d:\n Base theta: %.4f\n",kRx,PARAMS.kRx,theta); end //Rotate base Rx1 = [1, 0, 0;0, cos(theta), -sin(theta);0 sin(theta) cos(theta)]; R_0_EF = R_0_EF*Rx1; phiInter=cell(1,foot_nb); arcDesc_phi = struct('origin',base_R0,'normal',[0 1 0]) //rotation around Y2 //Then (phi,Y2) for i=1:foot_nb select STANCE(i).leg case 'FR' then offset_i = xOff + yOff; case 'FL' then offset_i = - xOff + yOff; case 'HR' then offset_i= + xOff - yOff; case 'HL' then offset_i = - xOff - yOff; else if PARAMS.verbose then mprintf("Error in the definition of foothold %d : leg name does not exist!\n",i); end return; end [boolInterPhi_i,phiMultiple_i,phiInter_i] = intersectArcWS(WS_R0(:,:,i),offset_i,R_0_EF,shellDesc(i),arcDesc_phi,PARAMS.aInc); if boolInterPhi_i then phiInter(i).entries = createAngleInterval(phiInter_i); if PARAMS.verbose & phiMultiple_i then mprintf(" PHI - For leg %d, phi lies in %d different intervals\n", i, size(phiInter(i).entries,1)); elseif PARAMS.verbose then mprintf(" PHI - For leg %d, phi range is: %.4f to %.4f\n",i,phiInter(i).entries(1),phiInter(i).entries(2)); end else if PARAMS.verbose then mprintf(" PHI - No intersection with leg %d workspace! Resampling theta...\n",i); NB_TRY = NB_TRY+1; end break; end end if ~boolInterPhi_i then continue; end //Sample (phi,Y2) [phiBoolFinal,phiFinalInterval] = intersectSetIntervals(phiInter); if ~phiBoolFinal then if PARAMS.verbose then mprintf(" PHI - phi valid intervals do not intersect! Resampling theta...\n"); NB_TRY = NB_TRY+1; end continue; end phi = sampleFromMultInterval(phiFinalInterval); if PARAMS.verbose then mprintf("PHI - Base phi: %.4f\n",phi); end Ry2 = [cos(phi), 0, sin(phi);0, 1 0;-sin(phi) 0 cos(phi)]; R_0_EF = R_0_EF*Ry2; RMAT = R_0_EF; [angle,vector] = angle_vector_FromMat(RMAT); Q = createQuaternion(angle,vector); if PARAMS.verbose then mprintf("\nBase state sampled! Now using closed form IK for the legs...\n"); end for i=1:foot_nb select STANCE(i).leg case 'FR' then offset_i = xOff + yOff; R_Leg_EF = [0 1 0;1 0 0;0 0 -1]; factor_t2 = -1; factor_t3 = -1; factor_elbow = +1; case 'FL' then offset_i = - xOff + yOff; R_Leg_EF = [0 1 0;-1 0 0;0 0 1]; factor_t2 = +1; factor_t3 = +1; factor_elbow = -1; case 'HR' then offset_i= + xOff - yOff; R_Leg_EF = [0 -1 0;1 0 0;0 0 1]; factor_t2 = +1; factor_t3 = +1; factor_elbow = -1; case 'HL' then offset_i = - xOff - yOff; R_Leg_EF = [0 -1 0;-1 0 0;0 0 -1]; factor_t2 = -1; factor_t3 = -1; factor_elbow = +1; end IK_target_RLeg = -R_Leg_EF*offset_i' + R_Leg_EF*R_0_EF'*(STANCE(i).pos'-base_R0'); //the foothold for the ith leg, in the leg base frame IK_target_array(:,i) = IK_target_RLeg; THETA(i,1) = atan(IK_target_RLeg(2),IK_target_RLeg(1)); rem = sqrt(IK_target_RLeg(1)**2+IK_target_RLeg(2)**2)-PARAMS.legLength(1); nc3 = IK_target_RLeg(3)**2+rem**2-PARAMS.legLength(2)**2-PARAMS.legLength(3)**2; dc3 = 2*PARAMS.legLength(2)*PARAMS.legLength(3); c3 = nc3/dc3; bool_ik = abs(c3)>1; if bool_ik then if PARAMS.verbose then mprintf("\nIK - NO SOLUTION FOR LEG %s INVERSE KINEMATICS\n",STANCE(i).leg); NB_TRY = NB_TRY+1; end // return; break; end s3 = factor_elbow*sqrt(1-c3**2); //ELBOw UP THETA(i,3) = factor_t3*atan(s3,c3); THETA(i,2) = factor_t2*(atan(IK_target_RLeg(3),rem)-atan(PARAMS.legLength(3)*s3,PARAMS.legLength(2)+PARAMS.legLength(3)*c3)) end if bool_ik then continue; end SUCCESS=%T; // NB_TRY = kRx*kRz*kpz*kpxy; if PARAMS.verbose then mprintf("\nSUCCESS!\n"); end return; end if PARAMS.verbose then mprintf("THET - Reached maximum number of trials, resampling psi...\n"); end end if PARAMS.verbose then mprintf("PSI - Reached maximum number of trials, resampling z...\n"); end end if PARAMS.verbose then mprintf("Z - Reached maximum number of trials, resampling XY...\n"); end end if PARAMS.verbose then mprintf("XY - Reached maximum number of trials, aborting...\n"); end endfunction
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Example5_19.sce
//Exa 5.19 clc; clear; close; //Given data : L=80;//km f=50;//Hz Z=(0.15+%i*0.78)*L;//ohm Y=(%i*5*10^-6)*L;//mho A=1+1/2*Y*Z;//parameter of 3-phase line D=A;//parameter of 3-phase line B=Z*(1+1/4*Y*Z);//parameter of 3-phase line C=Y;//parameter of 3-phase line disp(A,"Parameter A : "); disp(B,"Parameter B : "); disp(C,"Parameter C : "); disp(D,"Parameter D : "); //Answer of B is wrong in the book.
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clc clear //Inputs V1=0.028; P1=1; T1=27+273; n=1.3; V2=0.0046; T3=T1; T2=T1*((V1/V2)^(n-1)); printf('Temperature after compression: %1.2f K',T2); printf('\n'); P2=P1*((V1/V2)^n); W=((P1*100*V1)-(P2*100*V2))/(n-1); printf('Work Done: %1.2f kJ',W); printf('\n'); P3=(T3*P2)/T2; printf('Final Pressure: %1.2f bar',P3); printf('\n');
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example12_4.sce
//example 12.4 //page 454 clc; funcprot(0); //initialisation of variable Q=500/449; D=8/12; pi=3.14; g=32.2; N=1800;//rpm A=pi*D^2/4; V=Q/A; f=0.022//from chart HL=V^2/2/g*(12.1+224.9*f); hs=HL+119.4; Ns=N*sqroot(Q*449)/hs^0.75; disp(Ns,"specific speed (rpm)"); clear
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clear u u1 u2 u3 tr x x1 nu col=['-+b'; '-+g'; '-+r'; '-+c']; iter=1; flag=1; //lambda=input('lambda : '); while(flag) //h=input('Enter space step h : '); h=1/20; k=lambda*h; ax=-3; bx=3; at=0; bt=2; n=(bt-at)/k + 1; m=(bx-ax)/h + 1; //True Solution for comparison //t=2; //x=ax; //for i=1:m // tx=x-t; // if(abs(tx)<.5) // tr(i)=cos(%pi*tx); // else // tr(i)=0; // end // x=x+h; //end u=zeros(n,m); w=zeros(n,m); x=ax; for i=1:m u(1,i)=1-abs(x); w(1,i)=1-2*abs(x); if u(1,i)<0 u(1,i)=0; end if w(1,i)<0 w(1,i)=0; end x=x+h; end for j=1:n-1 for i=2:m-1 //Lax-Friedrichs u(j+1,i)=(1-lambda)/2*u(j,i+1)+(1+lambda)/2*u(j,i-1) end u(j+1,m)=u(j+1,m-1); end x1=ax:h:bx; plot(x1,u(n,:),col(iter)) //err3=abs(u(n,:)-tr'); //if abs(err3)<5 then // nratio3(iter)=err3(m); // nerr3=norm(err3); // disp(nerr3) //else disp('u is usless'); //end iter=iter+1; lambda=input('lambda : '); if lambda == 0 then flag = 0; end end plot(x1,tr,'k')
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Ex12_1.sce
//Part B Chapter 4 Example 1 clc; clear; close; R=75;//mm G=75;//GN/m^2 L=3;//m tau_s=75;//MN/m^2 theta=tau_s*L/R/G*180/%pi;//degree disp("Angle of twist is "+string(theta)+" degree."); r=50;//mm tau=tau_s*r/R;//MN/m^2 disp("Shear stress at inside surface is "+string(tau)+" MN/m^2");
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clc //initialisation of variables Vm= 0.6 //in^3 N= 2400 //rpm Qa= 6.5 //gpm p= 50 //CALCULATIONS ev= Vm*N*100/(Qa*231) Tf= (100-ev)*Qa/100 Cl= p*Tf/100 //RESULTS printf ('Case drain loss = %.3f gpm',Cl)
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clc; f=50 //Frequence in hertz Irms=20 //Rms current in amperes Im=Irms*sqrt(2) disp("(i)") printf("\n Im=%02f A \n",Im) t=0.0025 //Time in seconds i=Im*sin(2*%pi*f*t) disp("(ii)") printf("\n i=%.0f \n",i) t=0.0125 i=Im*sin(2*%pi*f*t) disp("(iii)") printf("\n i=%.0f \n",i) i1=14.14/Im disp(i1) i2=asin(i1) i2=i2*180/%pi disp(i2) i=i2/(2*180*f) printf("\n i=%.2f \n ms",i*10^3)
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Ch13Ex11.sce
// Scilab Code Ex13.11: Page-652 (2011) clc;clear; mu = 1.5;....// Optical index of refraction of NaCl crystal K = 5.6;....// Static dielectric constant of NaCl crystal P_IP = (1-((mu^2-1)*(K+2))/((mu^2+2)*(K-1)))*100; printf("\nThe percentage of ionic polarizibility in NaCl crystal = %4.1f percent ", P_IP); // Result // The percentage of ionic polarizibility in NaCl crystal = 51.4 percent
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max_hold_plots.sce
// plot max hold graphs // change directory to the desired location to save these plot files cd('C:\Documents and Settings\pflynn\Desktop\Max hold plots'); clf() plot([x],evstr(A)) //threshold line plot([x],threshold,'r--') // plots a dash-dotted line // give it titles xset("font size", 3) // sets font size for axis values xtitle('MAX HOLD','Frequency (MHz)','Power (dBm)'); a = gca() //gets handle of current axis a.title.font_size = 3 // sets title fontsize to 5 a.x_label.font_size = 3 // sets x lable font size to 4 a.y_label.font_size = 3 // sets y lable font size to 4 a.x_label.font_style = 2; a.title.font_style = 4; a.tight_limits = "on" xs2jpg(0,'mh plot '+ band_name'); xs2jpg(gcf(),'/mh plot '+ band_name + 'MHz.jpg') //end
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9_1_1.sce
clc //initialisation of variables d= 1.6 //lb/ft^3 vk= 6.2*10^-6 //ft^2/sec R= 1.8 //lbf v= 100 //ft/sec d1= 64 //lb/ft^3 vk1= 1.7*10^-5 //ft62/sec l= 10 //ft //CALCULATIONS u= v*vk1/(vk*l) u1= v*vk1/(vk*l*1.98) r= d1*l^2*(u/100)^2/d F= r*R //RESULTS printf (' resistance= %.f lbf ',F)
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result8.tst
@relation vowel @attribute TT integer[0,1] @attribute SpeakerNumber integer[0,14] @attribute Sex integer[0,1] @attribute F0 real[-5.211,-0.941] @attribute F1 real[-1.274,5.074] @attribute F2 real[-2.487,1.431] @attribute F3 real[-1.409,2.377] @attribute F4 real[-2.127,1.831] @attribute F5 real[-0.836,2.327] @attribute F6 real[-1.537,1.403] @attribute F7 real[-1.293,2.039] @attribute F8 real[-1.613,1.309] @attribute F9 real[-1.68,1.396] @attribute Class{0,1,2,3,4,5,6,7,8,9,10} @inputs TT,SpeakerNumber,Sex,F0,F1,F2,F3,F4,F5,F6,F7,F8,F9 @outputs Class @data 0 0 2 2 5 6 6 6 10 10 7 7 2 2 7 7 0 9 4 4 5 5 10 10 9 9 2 2 3 3 1 1 5 5 8 10 0 0 5 4 3 5 8 8 3 3 7 6 2 2 1 1 8 8 2 2 6 7 3 3 4 4 4 4 9 9 5 10 10 10 1 1 5 4 6 6 10 5 8 9 10 5 4 4 2 2 3 3 6 6 6 6 10 10 0 0 9 9 6 6 9 9 8 8 10 10 0 0 10 10 1 9 8 8 1 9 5 3 5 5 9 9 1 1 6 7 2 2 4 4 7 7 9 9 4 6 3 3 7 7 6 6 8 8 10 10 3 3 6 6 2 5 5 5 7 7 0 0 9 9 0 0 7 7 9 9 1 2 3 3 8 8 0 0 7 7 8 1 2 2 7 7 1 1 3 3 9 8 4 4 0 0 1 1 4 4 4 4
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clear; clc; V=400; V_ph=V/sqrt(3); N_s=1000; N=800; a=.7; I_d=110; R=2; k=1-((1-N/N_s)*(2.339*a*V_ph)/(I_d*R)); printf("value of duty cycle=%.3f",k); P=I_d^2*R*(1-k); I1=a*I_d*sqrt(2/3); r1=.1; r2=.08; Pr=3*I1^2*(r1+r2); P_o=20000; P_i=P_o+Pr+P; eff=P_o/P_i*100; printf("\nefficiency=%.2f",eff); I11=sqrt(6)/%pi*a*I_d th=43; P_ip=sqrt(3)*V*I11*cosd(th); pf=P_ip/(sqrt(3)*V*I11); printf("\ninput power factor=%.4f",pf);
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clear; clc; Y1=1;Y2=1;Y3=2;V3=3; Z1=1/Y1;Z2=1/Y2;Z3=1/Y3; V1=1;I1=-1; z11=V1/I1; V2=1;I2=3; z22=V2/I2; z21=V2/I1; printf("z11 = %f ohms\n",z11); printf(" z22 = %f ohms\n",z22); printf(" z21 = %f ohms\n",z21); printf(" z11 = %f ohms\n",0);
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clc //initialisation of variables h1= 182.07 //kJ/kg h4= 76.26 //kJ/kg h2= 217.97 //kJ/kg Q= 10^6 //kJ/h Tc= -5 //C Th= 32 //C //CALCULATIONS COP= (h1-h4)/(h2-h1) W= Q/(COP*3600) COPcarnot= (273.15+Tc)/(Th-Tc) //RESULTS printf (' COP= %.2f ',COP) printf (' \n power= %.1f kW ',W) printf (' \n COP= %.3f ',COPcarnot)
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c13_15.sce
//(13.15) Devise and evaluate an exergetic efficiency for the internal combustion engine of Example 13.4. For the fuel, use the chemical exergy value determined in Example 13.12(a). //solution mFdot = 1.8e-3 //fuel mass flow rate in kg/s ech = 47346 //in kj/kg, from example 13.12(a) Wcvdot = 37 //power developed by the engine in kw Efdot = mFdot*ech //rate at which exergy enters with the fuel in kw epsilon = Wcvdot/Efdot //exergetic efficiency printf('the exergetic efficiency is: %f',epsilon)
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// Scilab Code Ex12.3 Binding energy of helium nucleus: Pg: 247 (2008) e = 1.6e-019; // Energy equivalent of 1 eV, J/eV amu = 931; // Energy equivalent of 1 amu, MeV m = 2*1.007825+2*1.008665-4.002603; // Mass difference in formation of He, amu E = m*amu; // Energy equivalent of mass difference for He nucleus, MeV printf("\nThe minimum energy required to break He nucleus = %5.2f MeV", E); // Result // The minimum energy required to break He nucleus = 28.28 MeV
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// Test #10 : Valid input test case #2 exec('./allpasslp2bsc.sci',-1); [n,d]=allpasslp2bsc(0.786,[0.549,0.8746]); disp(d); disp(n); // //Scilab Output //d=1. 0.0888982 - 0.1132783i //n=0.1439960 0.6173655 - 0.7866765i // //Matlab Output //n= 0.1440 + 0.0000i 0.6174 - 0.7867i //d= 1.0000 + 0.0000i 0.0889 - 0.1133i
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clc,clear printf('Example 1.21\n\n') Pole=4 Z=32 //no of conductors coil_sides=Z segments=16 pole_pitch=Z/Pole slots=16 slots_per_pole=slots/Pole //for Simplex lap winding y_b=pole_pitch+1 //back pitch y_f=pole_pitch-1 //front pitch y_c=1 //Commutator pitch; Note that it is positive and it is progressive type of Simplex lap winding printf('WINDING TABLE:\n\n 1<- 10-> 3<- 12-> 5<- 14\n-> 7<- 16-> 9<- 18-> 11<- 20\n->13<- 22-> 15<- 24-> 17<- 26\n->19<- 28-> 21<- 30-> 23<- 32\n->25<- 2-> 27<- 4-> 29<- 6\n->31<- 8->1 ') printf('\nNote that <- indicates back connection with y_back=%.0f and -> indicates front connection with y_front=%.0f\n',y_b,y_f) printf('\nAnother form of winding table:') printf('\n BACK CONNECTIONS FRONT CONNECTIONS') printf('\n\n 1 to (1+9) =10 -> 10 to (10-7) =3') printf('\n 3 to (3+9) =12 -> 12 to (12-7)= 5') printf('\n 5 to (5+9) =14 -> 14 to (14-7)= 7') printf('\n 7 to (7+9) =16 -> 16 to (16-7)= 9') printf('\n 9 to (9+9) =18 -> 18 to (18-7)=11') printf('\n 11 to (11+9)=20 -> 20 to (20-7)=13') printf('\n 13 to (13+9)=22 -> 22 to (22-7)=15') printf('\n 15 to (15+9)=24 -> 24 to (24-7)=17') printf('\n 17 to (17+9)=26 -> 26 to (26-7)=19') printf('\n 19 to (19+9)=28 -> 28 to (28-7)=21') printf('\n 21 to (21+9)=30 -> 30 to (30-7)=23') printf('\n 23 to (23+9)=32 -> 32 to (32-7)=25') printf('\n 25 to (25+9)=34=(34-32)=2 -> 2 to (34-7)=27') printf('\n 27 to (27+9)=36=(36-32)=4 -> 4 to (36-7)=29') printf('\n 29 to (29+9)=38=(38-32)=6 -> 6 to (38-7)=31') printf('\n 31 to (31+9)=40=(40-32)=4 -> 8 to (40-7)=33 -32= 1')
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function imgout=contours(imgsrc) filterx = [-1 -2 -1;0 0 0;1 2 1] imgfx = convolution(imgsrc,filterx,6) filtery = [-1 0 1;-2 0 2;-1 0 1] imgfy = convolution(imgsrc,filtery,6) [wd,he]=size(imgfx); //Create an empty image imgout = zeros(wd,he); //For each lines for i=1:he //For each columns for j=1:wd pix1 = imgfx(j,i); pix2 = imgfy(j,i); imgout(j,i) = sqrt(pix1^2 + pix2^2); end end endfunction
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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 //Rao, R Ramesh, and Shankar Narasimhan. 1996. //“Comparison of Techniques for Data Reconciliation of Multicomponent Processes.” //Industrial & Engineering Chemistry Research 35:1362-1368. //http://dx.doi.org/10.1021/ie940538b. //Bibtex Citation //@article{Rao1996, //author = {Rao, R Ramesh and Narasimhan, Shankar}, //isbn = {0888-5885}, //journal = {Industrial \& Engineering Chemistry Research}, //month = apr, //number = {4}, //pages = {1362--1368}, //publisher = {American Chemical Society}, //title = {{Comparison of Techniques for Data Reconciliation of Multicomponent Processes}}, //url = {http://dx.doi.org/10.1021/ie940538b}, //volume = {35}, //year = {1996} //} // 12 Streams // 7 Equipments clear xm var jac nc nv i1 i2 nnzeros sparse_dg sparse_dh lower upper var_lin_type constr_lin_type constr_lhs constr_rhs getd('../functions'); // In the original paper, all streams for this problem are unmeasures, //theses values are estimates givem by the paper's original author. xm =[691.67 727.54 699.36 687.15 35.87 12.51 27.88 23.36 22.67 4.79 4.52 9.31 ]; //the variance proposed by the original author //var = (0.0001*ones(12,1)).^2; //the variance proposed by this work var = (0.03*xm).^2; // gross error gerror = zeros(length(xm),1); // to setup gross errors, select the stream and magnitude as the line bellow //gerror(2) = 9*sqrt(var(2)); xm = xm + gerror; //The jacobian of the constraints // 1 2 3 4 5 6 7 8 9 10 11 12 jac = [ 1 -1 0 0 1 0 0 0 0 0 0 0 0 1 -1 0 0 0 -1 0 0 0 0 0 0 0 1 -1 0 -1 0 0 0 0 0 0 0 0 0 0 -1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 -1 1 0 0 -1 0 0 0 0 0 0 1 0 -1 1 0 0 0 0 0 0 0 0 0 0 0 -1 -1 1 ]; // 1 2 3 4 5 6 7 8 9 10 11 12 //observability/redundancy tests umeas_P11 = []; [red_P11, just_measured_P11, observ_P11, non_obs_P11, spec_cand_P11] = qrlinclass(jac,umeas_P11) // reconcile with all measured. To reconcile with only redundant variables, uncomment the "red" assignments measured_P11 = setdiff([1:length(xm)], umeas_P11); red = measured_P11;// // to reconcile with all variables, comment the line above and uncomment bellow //red = [1:length(xm)]; // to run robust reconciliation,, one must choose between the folowing objective functions to set up the functions path and function parameters: //WLS = 0 // Absolute sum of squares = 1 //Cauchy = 2 //Contamined Normal = 3 //Fair = 4 //Hampel = 5 //Logistic = 6 //Lorenztian = 7 //Quasi Weighted = 8 // run the configuration functions with the desired objective function type obj_function_type = 0; exec ../functions/setup_DR.sce // to run robust reconciliation, it is also necessary to choose the function to return the problem structure if obj_function_type > 0 then [nc_eq, n_non_lin_eq, nv, nnzjac_ineq, nnzjac_eq, nnz_hess, sparse_dg, sparse_dh, lower, upper, var_lin_type, constr_lin_type, constr_lhs, constr_rhs] = robust_structure(jac, 0, xm, objfun, res_eq, res_ineq); else // for WLS, only the line bellow must be choosen and comment the 3 lines above [nc, nv, i1, i2, nnzeros, sparse_dg, sparse_dh, lower, upper, var_lin_type, constr_lin_type, constr_lhs, constr_rhs] = wls_structure(jac); end params = init_param(); // We use the given Hessian params = add_param(params,"hessian_approximation","exact"); params = add_param(params,"derivative_test","second-order"); params = add_param(params,"tol",1e-8); params = add_param(params,"acceptable_tol",1e-8); params = add_param(params,"mu_strategy","adaptive"); params = add_param(params,"journal_level",5); [x_sol, f_sol, extra] = ipopt(xm, objfun, gradf, confun, dg, sparse_dg, dh, sparse_dh, var_lin_type, constr_lin_type, constr_rhs, constr_lhs, lower, upper, params); //Q = 2*hessf ( xm ); //p=-4*(xm./var)'; //C=jac; //me=nc; //b=zeros(nc,1); //ci=lower; //cs=upper; // //[x,iact,iter,f_sol]=qpsolve(Q,p,C,b,ci,cs,me) //[x_solqp,lagr,info]=qld(Q,p,C,b,ci,cs,me, 1.0e-8) //status = info; //x_sol = x'; //f_sol=0; mprintf("\n\nSolution: , x\n"); for i = 1 : nv mprintf("x[%d] = %e\n", i, x_sol(i)); end mprintf("\n\nObjective value at optimal point\n"); mprintf("f(x*) = %e\n", f_sol);
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errcatch(-1,"stop");mode(2); //example 9.4 //page 318 ; funcprot(0); //initialisation of variable L=500; S=0.004;//slope of slope line hf=S*L; disp(hf,"head loss (ft)="); exit();
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//////// Probleme de Poisson Dirichlet par differences finies sur (0,1) clear /////// Parametres M = 100 // nb de subdivisions h = 1.0/M // pas k = 1.0 // raideur vect_M = [10,20,50,100,200] for l=1:length(vect_M) M = vect_M(l) h=1.0/M // sous-divisions de l'espace X = linspace(0,1,M+1) // f : terme source function z = f(x) //z = 1.0 //z = x z=x.*x endfunction // u : solution exacte function z = u(x) // z = x.*(1.0-x) ./ (2.0*k) //z = x.*(1.0-x.*x) ./ (6.0*k) z = x.*(1.0-x.*x.*x) / (12.0*k) endfunction // sous-divisions de l'espace X = linspace(0,1,M+1)' // calcul de la matrice de Poisson Dirichlet A = zeros(M-1,M-1) A(1,1) = 2 A(1,2) = -1 for i=2:M-2 A(i,i-1) = -1 A(i,i) = 2 A(i,i+1) = -1 end A(M-1,M-2) = -1 A(M-1,M-1) = 2 A = k.*A./(h*h) // calcul du second membre F for i=1:M-1 F(i) = f(X(i+1)) end // calcul de la solution num U U_temp = A \ F U(1) = 0 U(2:M) = U_temp U(M+1) = 0 // affichage de la solution exacte Uex = u(X) figure(1,"Figure_name",'probleme de Dirichlet homogene') plot(X,Uex,'o-b') // affichage de la solution approchee plot(X,U,'o-r') // calcul de l'erreur err = abs(Uex - U) disp(max(err)) vect_err(l) = max(err) vect_h(l) = h end figure(2,"Figure_name",'etude de la convergence') plot(log10(vect_h),log10(vect_err),'o-b') xtitle("Erreur du schema en fonction de h (echelle logarithmique)","log(h)","log(erreur)") //legend("k=5","k=1","k=0.5") //xtitle("Solution du problème de Poisson-Dirichlet","Position (x)","Solution (u)")
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function v = algoritmo(m, xf) x = 0 k = 0.046 v2 = 0 incx = xf / 10 R = 0 flag = 0 while (x <= xf) if flag == 2 incx = incx * 2 end var = 19.6 - (2000 / m) * R v2 = v2 + incx * var R = k * v2 flag = flag + 1 x = x + incx printf("x = %f, R = %f, var = %f, v2 = %f \n", x, R, var, v2) end v = sqrt(v2) endfunction
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//To find linear and angular velocity and acceleration clc //Given: OC=150/1000, PC=600/1000, CD=150/1000 //m N=450 //rpm //Solution: //Refer Fig. 15.6 //Calculating the angular speed of the crank omega=2*%pi*N/60 //rad/s //By measurement, OM=145/1000, CM=78/1000, QN=130/1000, NO=56/1000 //m //Velocity and acceleration of alider: //Calculating the velocity of the slider P vP=omega*OM //m/s //Calculating the acceleration of the slider P aP=omega^2*NO //m/s^2 //Velocity and acceleration of point D on the connecting rod: //Calculating the length od CD1 CD1=CD/PC*CM //m //By measurement, OD1=145/1000, OD2=120/1000 //m //Calculating the velocity of point D vD=omega*OD1 //m/s //Calculating the acceleration of point D aD=omega^2*OD2 //m/s^2 //Angular velocity and angular acceleration of the connecting rod: //Calculating the velocity of the connecting rod PC vPC=omega*CM //m/s //Calculating the angular velocity of the connecting rod omegaPC=vPC/PC //rad/s //Calculating the tangential component of the acceleration of P with respect to C atPC=omega^2*QN //m/s^2 //Calculating the angular acceleration of the connecting rod PC alphaPC=atPC/PC //rad/s^2 //Results: printf("\n\n Velocity of the slider P, vP = %.3f m/s.\n\n",vP) printf(" Acceleration of the slider P, aP = %.1f m/s^2.\n\n",aP) printf(" Velocity of point D, vD = %.3f m/s.\n\n",vD) printf(" Acceleration of point D, aD = %.2f m/s^2.\n\n",aD) printf(" Angular velocity of the connecting rod, omegaPC = %.3f rad/s.\n\n",omegaPC) printf(" Angular acceleration of the connecting rod PC, alphaPC = %.2f rad/s^2.\n\n",alphaPC)
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clc clear //input v=230;//voltage of a shunt generator in volts ra=0.2;//armature resistance of the shunt generator in ohms rf=115;//feild resistance of the shunt generator in ohms n=0.85;//overall effeciency in per units il=37;//load current in amperes //calculations inp=(v*il)/n;//input in watts inp1=inp/1000;//input power in kilo watts fi=v/rf;//feild current in amperes ai=il+fi;//armature current in amperes e=v+(ai*ra);//generated e.m.f. in volts ap=e*ai;//armature power in watts ml=inp-ap;//mechanical losses in watts nm=ap/inp;//mechanical effeciency in per units Nm=nm*100; ne=(v*il)/ap;//electrical effeciency in per units Ne=ne*100; //output mprintf('the input power will be %3.0f kW and the mechanical losses are %3.0f W \n the mechanical and electrical effeciecies are %3.1f per cent and %3.1f per cent',inp1,ml,Nm,Ne)
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//LU DECOMPOSITION //FACTORIZING A INTO L AND U (A = LU) clc;clear; function lu_decomposition(A) [r,c]=size(A); u=A; l=eye(r,c); for i=1:(r-1) m=det(u(i,i)); for j=i+1:c n=det(u(j,i)) a=n/m; l(j,i)=a; u(j,:)=u(j,:)-u(i,:)/(m/n); end end disp(l,'The lower triangular matrix L is'); disp(u,'The upper triangular matrix U is'); endfunction disp('Factorization of A into L and U'); A=input('Enter elements of matrix: '); disp(A,'The given matrix is A='); lu_decomposition(A); //SOLVING SYSTEM OF EQUATIONS BY LU DECOMOSITION clc;clear; format('v',5); function lu_decomposition(a, b) [r,c]=size(a); b=b'; l=eye(r,c); for i=1:r for j=1:c s=0; if j>=i for k=1:i-1 s=s+l(i,k)*u(k,j); end u(i,j)=a(i,j)-s; else for k=1:j-1 s=s+l(i,k)*u(k,j); end l(i,j)=(a(i,j)-s)/u(j,j); end end end c=l\b; x=u\c; disp(l,'The lower triangular matrix L is'); disp(u,'The upper triangular matrix U is'); disp(x,'Solution of system of equation is '); endfunction disp("Solving system of equation by LU decomposition"); a=input('Enter elements of matrix A: '); b=input('Enter elements of matrix B: '); disp(a,'The coefficient matrix A is'); disp(b,'The constant matrix b is'); lu_decomposition(a,b);
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//Chapter 16, Problem 8 clc; L=0.20; //inductance R=60; //resistance C=20e-6; //capacitance V=20; //supply voltage fr=(2*%pi)^-1*sqrt((1/(L*C))-(R^2/L^2)); Xl=2*%pi*fr*L; //inductive reactance Rd=L/(R*C); Ir=V/Rd; Q=Xl/R; printf("(a) Resonant frequency of the circuit = %f Hz\n\n",fr); printf("(b) Dynamic resistance Rd = %f ohm\n\n",Rd); printf("(c) Current at resonance Ir = %f A\n\n",Ir); printf("(d) Q factor of circuit = %f",Q);
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I=35 VR=25 Vrl=40 VRrl=50 Vc=45 C=50E-6 Xc=Vc/I w=1/(Xc*C) theta=acos((VR^2+VRrl^2-Vrl^2)/(2*VR*VRrl)) x=VRrl*cos(theta)-25 y=VRrl*sin(theta) r=x/I L=y/(I*w) Vappl=sqrt((VR+x)^2+y^2) R=VR/I disp(L,r,R)
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clc //ex11.7 R_i=1*10^3; R_o=100; A_voc=100; //V_ooc=A_voc*V_i and I_i=V_i/R_i gives R_moc=V_ooc/I_i R_moc=A_voc*R_i; disp('The resulting transconductance model is with an:') disp(R_i,'input resitance in ohms') disp(R_o,'output resistance in ohms') disp(R_moc,'and transresistance in ohms')
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clc;clear; printf("\nBalanço de Massa\nMétodos Decomposição LU e Jacobi\n\n") /*ordem = input("Qual a ordem da matriz? ") printf("Preencha os valores da matriz dos coeficientes:\n") for i = 1:ordem printf("Linha %d\n", i) for j = 1:ordem printf("Coluna %d", j) A(i, j) = input("Valor: ") end end printf("Preencha os valores dos termos independentes:\n") for i = 1:ordem B(i) = input("Termo: ") end*/ n = 3; A = [0.8 0.8 0.9; 0.05 0.9 0.9; 0.02 0.02 0.05]; b = [500; 300; 200]; L = zeros(n,n); // Matriz triangular inferior com os coeficientes U = zeros(n,n); // Matriz que resta da eliminação de Gauss // Decomposição LU for j = 1:n L(j,j) = 1; // diagonal principal com 1 for i = 1:j soma = 0.0; for k=1:i-1 soma = soma + L(i,k)*U(k,j); end U(i,j) = A(i,j) - soma; // Cálculo da matriz U end for i = j+1:n soma = 0.0; for k=1:j-1 soma = soma + L(i,k)*U(k,j); end L(i,j) = (A(i,j)-soma)/U(j,j); // Cálculo da matriz L end end printf('Matriz L: \n') disp(L) printf('\nMatriz U: \n') disp(U) // resolve L*y = b: substituicao progressiva y = zeros(1,n); y(1) = b(1)/L(1,1); for i=2:n soma = 0.0; for j=1:i-1 soma = soma + L(i,j)*y(j); end y(i) = (b(i)-soma)/L(i,i); end // resolve U*x = y: substituicao retroativa x(n) = y(n)/U(n,n); for i=n-1:-1:1 soma = 0.0; for j=i+1:n soma = soma + U(i,j)*x(j); end x(i) = (y(i)-soma)/U(i,i); end // Eliminação de Gauss function y = eliminacaoGauss(A, n) for i = 1:n w = abs(A(i,i)) // Pegando o primeiro elemento como o maior r = i for j = i:n // Analisando o maior elemento da coluna if abs(A(j,i)) > w then w = abs(A(j,i)) r = j end end troca = A(i, :) // Fazendo a troca da linha do pivô com a maior em módulo A(i, :) = A(r, :) A(r, :) = troca for k = i+1:n // Fazendo a eliminação de Gauss mult = -A(k,i)/A(i,i) for p = i:n+1 A(k,p)=A(k,p)+mult*A(i,p) end end end y = A endfunction function y=subRetroativa(A, n) pos = 2 x(n) = A(n,n+1) / A(n,n) // Cálculo do último x for i = n-1:-1:1 // Cálculo dos demais x soma = 0 for j = i+1:n // Somatório da substituição retroativa soma = soma + A(i,j) * x(j) end x(i) = (A(i,i+pos) - soma) / A(i,i) pos = pos + 1 end y = x endfunction matrizAumentada = [A b] matrizAumentada = eliminacaoGauss(matrizAumentada,n) printf("\nO sistema triangular superior resultante é:\n") disp(matrizAumentada) printf('\nResultados pelo método da Decomposição LU: \n') disp(x) printf("\nResultados pelo método da Eliminação de Gauss com pivotação parcial: \n") x2 = subRetroativa(matrizAumentada, n) disp(x2)
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function [a,b,c]=foo(x,y,z) a=x+y b=x*y c=z endfunction foo(2,3,4) // only the value of a is displayed [a,b]=foo(2,3,4) // a and b are retrieved as outputs [a,b,c]=foo(2,3,4) // a,b,c are retrieved as outputs
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clc clear //Input data T1=263;//Minimum temperature at which ammonia refrigerating machine works in K T2=303;//Maximum temperature at which ammonia refrigerating machine works in K x1=0.6;//Dryness fraction of ammonia during suction stroke sf1=0.5443;//Liquid entropy at 263 K in kJ/kg K hfg1=1297.68;//Latent heat at 263 K in kJ/kg sf2=1.2037;//Liquid entropy at 303 K in kJ/kg K hfg2=1145.8;//Latent heat at 303 K in kJ/kg hf1=135.37;//Liquid enthalpy at 263 K in kJ/kg hf2=323.08;//Liquid enthalpy at 303 K in kJ/kg //Calculations s1=sf1+((x1*hfg1)/T1);//Entropy at point 1 in kJ/kg K x2=(s1-sf2)/(hfg2/T2);//Entropy at point 2 in kJ/kg K h1=hf1+(x1*hfg1);//Enthalpy at point 1 in kJ/kg h2=hf2+(x2*hfg2);//Enthalpy at point 2 in kJ/kg COP=(h1-hf2)/(h2-h1);//Theoretical COP of ammonia refrigerating machine //Output printf('The theoretical COP of a ammonia refrigerating machine working between given temperatures is %3.2f',COP)
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clear $th = VirtualMaterials.Advanced_Peng-Robinson / -> $th th + WATER pipe = PipeSegment.PipeSegment() cd /pipe.In P = 300 kPa T = 300.0 K Fraction = 1.0 MoleFlow = 1000.0 cd /pipe.Out #P = 200.0 #T = 300.0 cd / pipe.Diameter = 0.1 pipe.Length = 20.0 pipe.Roughness = 0.0001 pipe.Elevation0 = 0.0 pipe.Elevation1 = 0.0 pipe.OutQ = 0 pipe.In pipe.Out #Calculate flow pipe.In.MoleFlow = pipe.Out.P = 280 kPa pipe.Out #Calculate from deltaP pipe.Out.P = pipe.DeltaP = 10 pipe.Out #Back to pipe.In.P = pipe.Out.P = 270 pipe.In pipe.Out #Negative deltaP -> Negative flow calc pipe.DeltaP = -10 pipe.Out #Calculate with negative flow pipe.DeltaP.DP = pipe.Out.P = pipe.In.P = 300 kPa pipe.In.MoleFlow = -5000.0 pipe.In pipe.Out #Play with elevation now pipe.In.MoleFlow = 9000.0 pipe.Out pipe.Elevation1 = 10 pipe.Out pipe.Elevation0 = 10 pipe.Out pipe.Elevation0 = 20 pipe.Out #Play with roughness pipe.Elevation1 = 0 pipe.Elevation0 = 0 pipe.Roughness = 0.0001 pipe.Out pipe.Roughness = 0.000001 pipe.Out pipe.Roughness = 0.0 pipe.Out #ignore kinetic and potential energy calcs /pipe.In.T = /pipe.In.P = /pipe.In.H = -34353.018 /pipe.In /pipe.Out /pipe.IgnoreKineticAndPotential = 1 #The enthalpy should be passed directly from the In to Out /pipe.Out #Solve /pipe.In.P = 300 copy /pipe paste / /pipeClone.In /pipeClone.Out #Resolve /pipe.In.H = /pipe.In.T = 90 /pipe.Out #Get rid of this IgnoreKineticAndPotential = 0 #Remove energy /pipe.OutQ.Energy = 1.0e7 /pipe.Out /pipe.Energy /pipe.T #Solve with T out as a spec /pipe.In.MoleFlow = /pipe.Out.T = 38 /pipe.Out #Did not work, solve with other numerical method /pipe.SolutionMethod = Secant /pipe.Out /pipe.Energy /pipe.T #Worked, now try again newton raphson but don't minimize error /pipe.SolutionMethod = NewtonRaphson /pipe.MinimizeError = 0 /pipe.TryLastConverged = 0 /pipe.Out /pipe.Energy /pipe.T #Different solve scheme /pipe.In.T = /pipe.In.MoleFlow = 9000 /pipe.Out /pipe.Out /pipe.Energy /pipe.T #Flip specs around /pipe.In.P = /pipe.Out.T = /pipe.Out.P = 150 /pipe.In.T = 90 /pipe.Out /pipe.Energy /pipe.T /pipe.u #Change energy model /pipe.EnergyLossModel = LinearTemperature #Didn't finish. Add iterations /pipe.MaxNumIterations = 50 /pipe.Out /pipe.Energy /pipe.T #add sections NumberSections = 5 /pipe.Out /pipe.Energy /pipe.T #all u are equal /pipe.EnergyLossModel = EqualU /pipe.Out /pipe.Energy /pipe.T /pipe.u #Try secant method /pipe.SolutionMethod = Secant /pipe.Out /pipe.Energy /pipe.T /pipe.u #Now solve for energy /pipe.OutQ.Energy = /pipe.Out.T = 38 /pipe.Out /pipe.Energy /pipe.T /pipe.u #Change numerical method /pipe.SolutionMethod = NewtonRaphson /pipe.Out /pipe.EnergyLossModel = LinearTemperature /pipe.Out /pipe.T /pipe.EnergyLossModel = LinearEnergy /pipe.Out /pipe.Energy /pipe.Out.P = /pipe.In.P = 300 /pipe.Out #Now spec u and change the energy models (nothing should change since u as a spec implies all u equal) /pipe.Out.T = /pipe.U.U = 4.8846261 /pipe.u /pipe.EnergyLossModel = LinearTemperature /pipe.u /pipe.EnergyLossModel = LinearEnergy /pipe.u #Back to energy spec but change nergy models /pipe.U.U = /pipe.OutQ.Energy = 9931444.4 /pipe.Out /pipe.Energy /pipe.T /pipe.u /pipe.SolutionMethod = Secant /pipe.EnergyLossModel = LinearTemperature /pipe.T /pipe.EnergyLossModel = EqualU /pipe.SolutionMethod = NewtonRaphson /pipe.T /pipe.u
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// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART II : TRANSMISSION AND DISTRIBUTION // CHAPTER 7: UNDERGROUND CABLES // EXAMPLE : 7.16 : // Page number 222-223 clear ; clc ; close ; // Clear the work space and console // Given data V = 33.0*10**3 // Line Voltage(V) f = 50.0 // Frequency(Hz) l = 4.0 // Length(km) d = 2.5 // Diameter of conductor(cm) t = 0.5 // Radial thickness of insulation(cm) e_r = 3.0 // Relative permittivity of the dielectric PF = 0.02 // Power factor of unloaded cable // Calculations // Case(a) r = d/2.0 // Radius of conductor(cm) R = r+t // External radius(cm) e_0 = 8.85*10**-12 // Permittivity C = 2.0*%pi*e_0*e_r/log(R/r)*l*1000 // Capacitance of cable/phase(F) // Case(b) V_ph = V/3**0.5 // Phase voltage(V) I_c = V_ph*2.0*%pi*f*C // Charging current/phase(A) // Case(c) kVAR = 3.0*V_ph*I_c // Total charging kVAR // Case(d) phi = acosd(PF) // Φ(°) delta = 90.0-phi // δ(°) P_c = V_ph*I_c*sind(delta)/1000 // Dielectric loss/phase(kW) // Case(e) E_max = V_ph/(r*log(R/r)*1000) // RMS value of Maximum stress in cable(kV/cm) // Results disp("PART II - EXAMPLE : 7.16 : SOLUTION :-") printf("\nCase(a): Capacitance of the cable, C = %.3e F/phase", C) printf("\nCase(b): Charging current = %.2f A/phase", I_c) printf("\nCase(c): Total charging kVAR = %.4e kVAR", kVAR) printf("\nCase(d): Dielectric loss/phase, P_c = %.2f kW", P_c) printf("\nCase(e): Maximum stress in the cable, E_max = %.1f kV/cm (rms)", E_max)
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pathname=get_absolute_file_path('6_16c.sce') filename=pathname+filesep()+'6_16c_data.sci' exec(filename) A=2*Tf/W;B=W/S;C=1/L_Dmax^2;E=sqrt(A^2-C) Vmax=sqrt((A*B+B*E)/(D*Cdo)) printf("\Answer:\n") printf("\n\Maximum Velocity for CJ-1: %f m/s\n\n",Vmax)
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//// compute_bd // find boundary of mesh, returned bd will be in ccw consective order. For // multiple boundary mesh, return a cell, each cell is a closed boundary. // For single boundary mesh, return an array. // //// Syntax // bd = compute_bd(face) // //// Description // face: double array, nf x 3, connectivity of mesh // // bd: double array, n x 1, consective boundary vertex list in ccw order // cell, n x 1, each cell is one closed boundary // //// Contribution // // Author : Wen Cheng Feng // Created: 2014/03/06 // Revised: 2014/03/14 by Wen, add document // Revised: 2014/03/23 by Wen, revise doc // // Copyright 2014 Computational Geometry Group // Department of Mathematics, CUHK // http://www.lokminglui.com function bd = compute_bd(face) // amd stores halfedge information, interior edge appear twice in amd, // while boundary edge appear once in amd. We use this to trace boundary. // // currently, there is problem for multiple boundary mesh. Some boundary may // be missing. [am,amd] = compute_adjacency_matrix(face); md = am - (amd>0)*2; [I,~,~] = find(md == -1); [~,Ii] = sort(I); bd = zeros(size(I)); k = 1; for i = 1:size(I) bd(i) = I(k); k = Ii(k); end
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clear; clc; disp('Example 8.18'); // aim : To determine // the actual mass of air supplied/kg coal // the velocity of flue gas // given values mc = 635;// mass of coal burn/h, [kg] ea = .25;// excess air required C = .84;// mass composition of carbon H2 = .04;// mass composition of hydrogen O2 = .05;// mass composition of oxygen ash = 1-(C+H2+O2);// mass composition of ash P1 = 101.3;// pressure, [kJn/m^2] T1 = 273;// temperature, [K] V1 = 22.4;// volume, [m^3] T2 = 273+344;// gas temperature, [K] P2 = 100;// gas pressure, [kN/m^2] A = 1.1;// cross section area, [m^2] aO2 = .23;// composition of O2 in air mCO2 = 44;// moleculer mass of carbon mH2O = 18;// molecular mass of hydrogen mO2 = 32;// moleculer mas of oxygen mN2 = 28;// moleculer mass of nitrogen // solution mtO2 = 8/3*C+8*H2-O2;// theoretical O2 required/kg coal, [kg] msa= mtO2/aO2;// stoichiometric mass of air supplied/kg coal, [kg] mas = msa*(1+ea);// actual mass of air supplied/kg coal, [kg] m1 = 11/3*C;// mass of CO2/kg coal produced, [kg] m2 = 9*H2;// mass of H2/kg coal produced, [kg] m3 = mtO2*ea;// mass of O2/kg coal produced, [kg] m4 = mas*(1-aO2);// mass of N2/kg coal produced, [kg] mt = m1+m2+m3+m4;// total mass, [kg] x1 = m1/mt*100;// %age mass composition of CO2 produced x2 = m2/mt*100;// %age mass composition of H2O produced x3 = m3/mt*100;// %age mass composition of O2 produced x4 = m4/mt*100;// %age mass composition of N2 produced vt = x1/mCO2+x2/mH2O+x3/mO2+x4/mN2;// total volume v1 = x1/mCO2/vt*100;// %age volume composition of CO2 v2 = x2/mH2O/vt*100;// %age volume composition of H2O v3 = x3/mO2/vt*100;// %age volume composition of O2 v4 = x4/mN2/vt*100;// %age volume composition of N2 Mav = (v1*mCO2+v2*mH2O+v3*mO2+v4*mN2)/(v1+v2+v3+v4);// average moleculer mass, [kg/kmol] // since no of moles is constant so PV/T=constant V2 = P1*V1*T2/(P2*T1);//volume, [m^3] mp = mt*mc/3600;// mass of product of combustion/s, [kg] V = V2*mp/Mav;// volume of flowing gas /s,[m^3] v = V/A;// velocity of flue gas, [m/s] mprintf('\n The actual mass of air supplied is = %f kg/kg coal\n',mas); mprintf('\n The velocity of flue gas is = %f m/s\n',v); // End
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<?xml version="1.0" encoding="utf-8"?> <test> <description>desc P=10</description> <executable>AcousticSolver</executable> <parameters>LEE_2DPulseAdv_WeakDG_MODIFIED.xml</parameters> <files> <file description="Session File">LEE_2DPulseAdv_WeakDG_MODIFIED.xml</file> </files> <metrics> <metric type="L2" id="1"> <value variable="p" tolerance="1e-4"> 6.77196</value> <value variable="rho" tolerance="1e-12"> 1.04459e-06</value> <value variable="rhou" tolerance="1e-7">0.00191276</value> <value variable="rhov" tolerance="1e-7">0.00195208</value> </metric> <metric type="Linf" id="2"> <value variable="p" tolerance="1e-4"> 30.1611</value> <value variable="rho" tolerance="1e-12"> 7.98548e-06</value> <value variable="rhou" tolerance="1e-7">0.00713893</value> <value variable="rhov" tolerance="1e-7">0.0098122</value> </metric> </metrics> </test>
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clear //Given D=0.13*10**-2 R=3.4 //ohms l=10.0 //Calculation // A=(%pi/4.0)*D**2 a=R*A/l b=1/a //Result printf("\n Conductivity of a material is %0.1f *10**6 S/m",b*10**-6)
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clear;lines(0); driver("Pos") xinit("foo.ps") plot2d() xend() driver("X11")
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//Initialisation of variables clc h=1 k=1 l=1 dhkl=1.75e-8// a=dhkl*sqrt(h^2+k^2+l^2) printf('inter atomic spacing is %e cms \n',a)
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//4*x^3*y/(6*x*y^3) clear; clc; close; d=int32([4,6]); m=4/gcd(d); n=6/gcd(d); x=poly(0,'x');y=poly(0,'y'); p1=x^3;p2=x;p=p1/p2; q1=y;q2=y^3;q=q1/q2; //val=m/n*p*q disp('val=') mprintf("%i/%i*x^2/y^2",m,n)
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function y = upsamplefill (x, w, cpy) //This function upsamples a vector interleaving given values or copies of the vector elements. //Calling Sequence //y = upsamplefill (x, w) //y = upsamplefill (x, w, cpy) //Parameters //x: scalar, vector or matrix of real or complex numbers //w: scalar or vector of real or complex values //cpy: can take in "true" or "false", default is false //Description //This is an Octave function. //This function upsamples a vector interleaving given values or copies of the vector elements. //The second argument has the values in the vector w that are placed in between the elements of x. //The third argument, if true, means that w should be scalar and that each value in x repeated w times. //Examples //upsamplefill([0.4,0.5],7) //ans = // 0.4 7. 0.5 7. funcprot(0); rhs = argn(2) if(rhs<2 | rhs>3) error("Wrong number of input arguments.") end select(rhs) case 2 then y = callOctave("upsamplefill", x, w) case 3 then y = callOctave("upsamplefill", x, w, cpy) end endfunction
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[profile regional_s3_endpoint] s3_us_east_1_regional_endpoint=regional
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clc; funcprot(0); //Example 10.7 //Initializing the variables d1 = 0.140; d2 = 0.250; DpF_DpR = 0.6; //Difference in head loss when in forward and in reverse direction K = 0.33 ;//From table g = 9.81; //Calculations ratA = (d1/d2)^2; v = sqrt(DpF_DpR*2*g/((1 - ratA)^2 - K)); disp(v,"Velocity (m/s):");
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function W=cijfers(V) // output initialiseren W=[]; // controleren of V enkel cijfers 0:9 bevat for i=1:length(V) if (V(i)<0)|(V(i)>9) disp('input mag enkel cijfers tussen 0 en 9 bevatten') abort // foute input, dus functie afbreken end end // cijfer per cijfer controleren of het voorkomt in V for i=0:9 for j=1:length(V) if V(j)==i W=[W,i] // cijfer komt voor, dus plakken aan output-vector break // cijfer is gevonden, dus niet meer verder zoeken // voor dit cijfer end end end endfunction
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// ==================================================================== // Allan CORNET // INRIA 2008 // Template toolbox_skeleton // This file is released into the public domain // ==================================================================== // // function s = scilab_sum(valA,valB) s = valA + valB; endfunction // ====================================================================
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function [xnew,ynew,znew] = translate3d(x,y,z,tx,ty,tz) xnew = x + tx ynew = y + ty znew = z + tz endfunction
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//Chapter-9,Example9_17,pg 9_66 Ra=0.08 Eb1=242 V=250 Ia=87 Vt=V//generator supply Nm=1500 Ia1=(V-Eb1)/Ra //at start N=0, Eb=0 Ias=V/Ra//Ia(start) Ia2=120 Eb2=V-Ia2*Ra Eg=Vt+Ia*Ra//generator e.m.f Ng=Nm*Eg/Eb1//speed as generator printf("speed as generator\n") printf("Ng=%.2f r.p.m",Ng)
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IDsmb.sci
0 fbegin 100 internode 101 internode-bone 200 default 202 cottonleaf 203 sunflowerleaf 204 wheatleaf 205 leaflob1 206 leaflob2 207 leaf_5lob 208 leaf_5lob2 209 leaf_aescu2 210 maizeleaf2 211 leafaecu 212 leafpen11 213 leafpen11_2 214 leafplatan 215 leafplatan2 216 leafsquare1 217 leafsquare2 218 needle 219 needle2 220 needle4 221 cube 222 yinxing_win 223 test1 224 test2 225 test3 226 test4 227 test5 228 test6 229 test7 230 test8 231 test9 232 test10 233 tomatoleaf 234 maizeleaf6 235 cottonleafvert2 236 cotilydonver 237 e1 238 leaf_maize_vert 239 leaf_wheat_vert 240 maizeleaf 241 maizeleaf_vert 242 petal1 243 spike 244 spike_a 245 spike_b 247 maizeleaf_yell 248 leaf_wheat_yell 249 leaf_tomato2 250 leaf_tomato2d 251 leaf_maize_yell 252 cottonleafyel2 300 femaleorgan 301 flower_1 302 cube2 303 maizecone 304 sunflower 305 wheatspike 306 sunflower2 307 tomatobunch 308 flower_chrys_a_glab 309 flower 310 ball 311 cob2 312 fcoton2 313 fruit1 314 flower_2 315 r3 316 r4 317 sunflower0 318 sunflower_a 319 sunflower_b 320 sunflower_c 400 male3 401 maizetassel 402 malemaize2 403 male3a 404 malemaizeok2 405 male_yell2 406 male_yell 500 fend %index of smb files. 100-200 internode, 200-300 leaf, 300-400 female organ, 400-500 male organ
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//Ex:3.4 clc; clear; close; b=1/2;// propagtion constant printf("normalised propagtion constant"); printf("\n B=((b/k)^2-n2^2)/(n1^2-n2^2)"); printf("\n thus when b=1/2"); printf("\n B=k*sqrt(n2^2+b*(n1^2-n2^2))"); printf("\n B=k*sqrt((n1^2-n2^2)/2)"); printf("\n which gives its rms value");
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//Chapter 03: Algorithms clc; clear; function []= binarysearch (arr ,n ,i) last =1; h=n; while (last <= h ) mid = int (( last + h ) /2) ; if ( arr ( mid ) == i ) printf ( "\nElement:%d found at position %d",i ,mid) ; break ; else if ( arr ( mid ) >i ) h = mid -1; else last = mid +1; end end end endfunction //Note:input array has to be sorted ar =[1 2 3 5 6 7 8 10 12 13 15 16 18 19 20 22] l=length(ar) disp (ar , " Given array " ) ; binarysearch (ar ,l ,19) //Note:input format for function is (array,length,element to be searched)
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clc //Initialization of variables P1=10 //psia Q=0.6 //cfs A1=0.0491 //ft^2 g=32.2 V=39.2//fps A0=0.0218 //ft^2 d1=2 //in d2=3 //in //calculations Phead=P1*144/62.4 V1=Q/A1 V2i= sqrt(2*g*(Phead + V1^2 /(2*g))) Cv=V/V2i A2=Q/V Cc=A2/A0 Cd=Cc*Cv hL=(1/Cv^2 -1)*(1- (d1/d2)^4)*V^2 /(2*g) //results printf("Cc = %.2f ",Cc) printf("\n Cd= %.2f",Cd) printf("\n Cv= %.2f",Cv) printf("\n Head loss = %.2f ft",hL)
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clc //Example 10.6 //Calculate the pump head rho=62.3//lbm/ft^3 g=32.2//ft/s^2 v=18.46//ft/s //1 lbf/s^2 = 32.2 lbm.ft h=(v^2/2)*32.2/rho/g//ft printf("The pump head is %f ft",h);