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main var cs142; { let cs142 <- 2412; call outputnum(cs142); call outputnewline() }.
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function [y, dy] = moc_polyval (p, x, s, mu) // Evaluate the polynomial p at the specified values of x // Calling Sequence // y = moc_polyval (p, x) // y = moc_polyval (p, x, [], mu) // [y,dy] = moc_polyval (p, x) // [y,dy] = moc_polyval (p, x,s) // [y,dy] = moc_polyval (p, x,s,mu) // Description // // Evaluate the polynomial p at the specified values of x}. When // mu is present, evaluate the polynomial for // (x-mu(1))/mu(2). // If x is a vector or matrix, the polynomial is evaluated for each of // the elements of x. // // In addition to evaluating the polynomial, the second output // represents the prediction interval, y +/- dy which // contains at least 50% of the future predictions. To calculate the // prediction interval, the structured variable @var{s}, originating // from moc_polyfit, must be supplied. // // See also // moc_polyfit, moc_poly // // Authors // Tony Richardson, 1994 // Holger Nahrstaedt 2014 // Author: Tony Richardson <arichard@stark.cc.oh.us> // Created: June 1994 // Adapted-By: jwe // Copyright (C) 1994-2013 John W. Eaton // // This file is part of Octave. // // Octave 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 3 of the License, or (at // your option) any later version. // // Octave 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 Octave; see the file COPYING. If not, see // <http://www.gnu.org/licenses/>. [nargout,nargin]=argn(0); if (nargin < 2 | nargin > 4 | (nargout == 2 & nargin < 3)) error ("[y, dy] = moc_polyval (p, x, s, mu)"); end if (nargin == 2) s=[]; end; if (isempty (x)) y = []; return; elseif (isempty (p)) y = zeros ((x)); return; elseif (~ moc_isvector (p)) error ("polyval: first argument must be a vector"); end if (nargin > 3) x = (x - mu(1)) / mu(2); end n = length (p) - 1; y = p(1) * ones ((x)); for i = 2:n+1 y = y .* x + p(i); end if (nargout == 2) // Note: the F-Distribution is generally considered to be single-sided. // http://www.itl.nist.gov/div898/handbook/eda/section3/eda3673.htm // t = finv (1-alpha, s.df, s.df); // dy = t * sqrt (1 + sumsq (A/s.R, 2)) * s.normr / sqrt (s.df) // If my inference is correct, then t must equal 1 for polyval. // This is because finv (0.5, n, n) = 1.0 for any n. try k = length (x); A = (x(:) * ones (1, n+1)) .^ (ones (k, 1) * (n:-1:0)); dy = sqrt (1 + moc_sumsq (A/s.R, 2)) * s.normr / sqrt (s.df); dy = matrix (dy, size (x)); catch if (isempty (s)) error ("polyval: third input is required."); elseif (isstruct (s) & and (moc_ismember ({"R", "normr", "df"}, fieldnames (s)))) error (lasterr ()); elseif (isstruct (s)) error ("polyval: third input is missing the required fields."); else error ("polyval: third input is not a structure."); end end end endfunction // // %!fail ("polyval ([1,0;0,1],0:10)") // // %!test // %! r = 0:10:50; // %! p = poly (r); // %! p = p / max (abs (p)); // %! x = linspace (0,50,11); // %! y = polyval (p,x) + 0.25*sin (100*x); // %! [pf, s] = polyfit (x, y, numel (r)); // %! [y1, delta] = polyval (pf, x, s); // %! expected = [0.37235, 0.35854, 0.32231, 0.32448, 0.31328, ... // %! 0.32036, 0.31328, 0.32448, 0.32231, 0.35854, 0.37235]; // %! assert (delta, expected, 0.00001); // // %!test // %! x = 10 + (-2:2); // %! y = [0, 0, 1, 0, 2]; // %! p = polyfit (x, y, numel (x) - 1); // %! [pn, s, mu] = polyfit (x, y, numel (x) - 1); // %! y1 = polyval (p, x); // %! yn = polyval (pn, x, [], mu); // %! assert (y1, y, sqrt (eps)); // %! assert (yn, y, sqrt (eps)); // // %!test // %! p = [0, 1, 0]; // %! x = 1:10; // %! assert (x, polyval (p,x), eps); // %! x = x(:); // %! assert (x, polyval (p,x), eps); // %! x = reshape (x, [2, 5]); // %! assert (x, polyval (p,x), eps); // %! x = reshape (x, [5, 2]); // %! assert (x, polyval (p,x), eps); // %! x = reshape (x, [1, 1, 5, 2]); // %! assert (x, polyval (p,x), eps); // // %!test // %! p = [1]; // %! x = 1:10; // %! y = ones (size (x)); // %! assert (y, polyval (p,x), eps); // %! x = x(:); // %! y = ones (size (x)); // %! assert (y, polyval (p,x), eps); // %! x = reshape (x, [2, 5]); // %! y = ones (size (x)); // %! assert (y, polyval (p,x), eps); // %! x = reshape (x, [5, 2]); // %! y = ones (size (x)); // %! assert (y, polyval (p,x), eps); // %! x = reshape (x, [1, 1, 5, 2]); // // %!assert (zeros (1, 10), polyval ([], 1:10)) // %!assert ([], polyval (1, [])) // %!assert ([], polyval ([], [])) //
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//Print the alternate characters of the file output1.dat warning('off'); fstream1=mopen('output1.dat','r'); for a=1:13 ch=mfscanf(fstream1,"%c"); //Read data from file output.dat printf("%c ",ch); //Print the data mseek(1,fstream1,'cur'); //Skeep 1 character after every reading end mclose(fstream1);
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-- Fuzzy Logix, LLC: Functional Testing Script for DB Lytix functions on Netezza -- -- Copyright (c): 2014 Fuzzy Logix, LLC -- -- NOTICE: All information contained herein is, and remains the property of Fuzzy Logix, LLC. -- The intellectual and technical concepts contained herein are proprietary to Fuzzy Logix, LLC. -- and may be covered by U.S. and Foreign Patents, patents in process, and are protected by trade -- secret or copyright law. Dissemination of this information or reproduction of this material is -- strictly forbidden unless prior written permission is obtained from Fuzzy Logix, LLC. -- Functional Test Specifications: -- -- Test Category: String Functions -- -- Test Unit Number: FLConcatStr-Netezza-01 -- -- Name(s): FLConcatStr -- -- Description: Concatenates into one string. -- -- Applications: -- -- Signature: FLConcatStr(String STRING, -- Position INTEGER) -- -- Parameters: See Documentation -- -- Return value: VARCHAR -- -- Last Updated: 07-04-2017 -- -- Author: Kamlesh Meena -- -- BEGIN: TEST SCRIPT \time --.run file=../PulsarLogOn.sql --.set width 2500 --SELECT COUNT(*) AS CNT, -- CASE WHEN CNT = 0 THEN ' Please Load Test Data!!! ' ELSE ' Test Data Loaded ' END AS TestOutcome --FROM fzzlSerial a; -- BEGIN: POSITIVE TEST(s) ---- Positive Test 1: Returns Concated string. SELECT FLConcatStr(a.String1,'|') AS ConcatString FROM tblString a; --Positive test case 2 --use the same one but try to get an order in concat string SELECT FLConcatStr(a.String1,'|') AS ConcatString FROM ( SELECT String1 as String1 FROM tblString ORDER BY 1 LIMIT 10) AS a; --Positive test case 3 --select names and concat them for a large number of data SELECT FLConcatStr(a.companyname,'|') AS ConcatString FROM ( SELECT companyname as companyname FROM finPubco LIMIT 30) AS a; --Positive test case 4 --select the top names alphabetically and concat them for a large number of data SELECT FLConcatStr(a.companyname,'|') AS ConcatString FROM ( SELECT companyname as companyname FROM finPubco ORDER BY 1 LIMIT 30) AS a; --Positive test case 5 --(Resolution to JIRA TD-82) example similar to the test case listed there --SELECT SUBSTRING(DatabaseNameI FROM 1 FOR 1) || ':' (TITLE ''), --FLConcatStr(DatabaseNameI,'|') (TITLE 'Databases') --FROM DBC.DBase --ORDER BY 1 --GROUP BY 1 SELECT SUBSTRING(string1 from 1 for 1), FLConcatStr(string1,'|') FROM tblstring GROUP BY 1 ORDER BY 1; -- BEGIN: NEGATIVE TEST(s) ---- Negative Test 1:More parameter given SELECT FLConcatStr(a.String1,'|','A') AS ConcatString FROM tblString a; ---- Negative Test 2: Invalid Input For Input String Column ---- Negative Test 2a --- Return expected results, Good SELECT FLConcatStr('String1','|') AS ConcatString FROM tblString; ---- Negative Test 2b --- Return expected results, Good SELECT FLConcatStr('','|') AS ConcatString FROM tblString; ---- Negative Test 2c --- Return expected results, Good SELECT FLConcatStr(NULL,'|') AS ConcatString FROM tblString; ---- Negative Test 3: Invalid Input For Delimiter --- Return expected results, Good --- To be investigated SELECT FLConcatStr(String1,'') AS ConcatString FROM tblString; SELECT FLConcatString(String1,NULL) AS ConcatString FROM tblString; --Negative test case 4 :try using more than 1 character as delimiter --it should ignore the other characters after the 1st SELECT FLConcatStr(a.companyname,'|pppp') AS ConcatString FROM ( SELECT companyname as companyname FROM finPubCo ORDER BY 1 LIMIT 25 ) AS a; ---Negative case 5 --When the input causes the concat string exceeds the max allowded value for the output SELECT FLConcatStr(a.CustName, ',') AS ConcatString FROM ( SELECT 'abcdefghijklmnopqrstuvwxyz' AS CustName, a.SerialVal AS CustNo FROM fzzlserial a ) AS a; -- END: NEGATIVE TEST(s) \time -- END: TEST SCRIPT
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// Scilab Code Ex3.4 : Page-72 (2010) lambda = 5893e-010; // Wavelength of monochromatic lihgt used, m n = 1; // Number of fringe for the least thickness of the film r = 0; // Value of refraction angle for normal incidence, degrees mu = 1.42; // refractive index of the soap film // As for constructive interference, // 2*mu*t*cos(r) = (2*n-1)*lambda/2, solving for t t = (2*n-1)*lambda/(4*mu*cos(r)); // Thickness of the film that appears bright, m printf("\nThe thickness of the film that appears bright = %6.1f angstrom", t/1e-010); // As for destructive interference, // 2*mu*t*cos(r) = n*lambda, solving for t t = n*lambda/(2*mu*cos(r)); // Thickness of the film that appears bright, m printf("\nThe thickness of the film that appears dark = %4d angstrom", t/1e-010); // Result // The thickness of the film that appears bright = 1037.5 angstrom // The thickness of the film that appears dark = 2075 angstrom
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function C=seriesC(C1,C2) C=C1*C2/(C1+C2) endfunction Ceq=seriesC(seriesC(10,20),40) disp(Ceq) q=Ceq*280 disp(q) V1=q/10 V2=q/20 V3=q/40 disp(V3,V2,V1)
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//Chapter 4 Ex 9 clc; clear; close; //(i) x1=17.28/(2*3.6*0.2); mprintf("(i)The value of x is %.0f",x1); //(ii) x2=364.824/(3794.1696+36.4824-3648.24); mprintf("\n(ii)The value of x is %.0f",x2); //(iii) x3=poly(0,'x'); for x3=1:0.1:10 if round(8.5-(5.5-(7.5+(2.8/x3)))*(4.25/0.04))==306 break; end end mprintf("\n(iii)The value of x is %.1f",x3);
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// Exa 6.11 clc; clear; close; format('v',6) // Given data Va = 0.2;// in V Vb = -0.5;// in V Vc = 0.8;// in V Ra = 33;// in k ohm Rb = 22;// in k ohm Rc = 11;// in k ohm R_F = 66;// in k ohm // Using Superposition theorm, the output voltage Vo = (-((R_F/Ra)*Va)) -(((R_F/Rb)*Vb)) -(((R_F/Rc)*Vc));// in V disp(Vo,"The output voltage in V is");
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// Chargement du fichier X1. // le fichier est stocké dans la variale "imgT". load('X1.dat'); // Déclaration & initialisation d'une variable où est stocké la nouvelle image // qui résulte de la transformée de fourrier inverse de l'image source. img=ifft(imgT) // Affichage de l'image. display_gray(img)
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//EX3_13 PG-3.40 clc Ep=230; //rms value of primary voltage N2toN1=1/15;//turns ratio Rf=0;//diode is ideal Rs=0;//transformer is ideal Rl=50;//load resistance Es=Ep*N2toN1;//rms vaue of primary voltage Esm=sqrt(2)*Es;//peak value of input voltage Im=Esm/(Rs+2*Rf+Rl); Idc=2*Im/%pi; Edc=Idc*Rl;//load voltage printf("\n Therefore load voltage is %.1f V\n",Edc) Rf=.482;//ripple factor for full wave rectifier Vrip=Rf*Edc;//ripple voltage printf("\n ripple voltage is %.4f V",Vrip) //in the book the ripple voltage has been rounded off to //6.6516V but the actual ans is 6.6539V
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clear; clc; disp('Example 15.13'); // aim : To determine // (a) the pressure, volume and temperature at cycle process change points // (b) the net work done // (c) the thermal efficiency // (d) the heat received // (e) the work ratio // (f) the mean effective pressure // (g) the carnot efficiency // given values rv = 15;// volume ratio P1 = 97*10^-3;// initial pressure , [MN/m^2] V1 = .084;// initial volume, [m^3] T1 = 273+28;// initial temperature, [K] T4 = 273+1320;// maximum temperature, [K] P3 = 6.2;// maximum pressure, [MN/m^2] cp = 1.005;// specific heat capacity at constant pressure, [kJ/kg K] cv = .717;// specific heat capacity at constant volume, [kJ/kg K] // solution // taking reference Fig. 15.27 // (a) R = cp-cv;// gas constant, [kJ/kg K] Gama = cp/cv;// heat capacity ratio // for process 1-2 V2 = V1/rv;// volume at stage2, [m^3] // using PV^(Gama)=constant for process 1-2 P2 = P1*(V1/V2)^(Gama);// pressure at stage2,. [MN/m^2] T2 = T1*(V1/V2)^(Gama-1);// temperature at stage 2, [K] // for process 2-3 // since volumee is constant in process 2-3 , so using P/T=constant, so T3 = T2*(P3/P2);// volume at stage 3, [K] V3 = V2;// volume at stage 3, [MN/m^2] // for process 3-4 P4 = P3;// pressure at stage 4, [m^3] // since in stage 3-4 P is constant, so V/T=constant, V4 = V3*(T4/T3);// temperature at stage 4,[K] // for process 4-5 V5 = V1;// volume at stage 5, [m^3] P5 = P4*(V4/V5)^(Gama);// pressure at stage5,. [MN/m^2] T5 = T4*(V4/V5)^(Gama-1);// temperature at stage 5, [K] mprintf('\n (a) P1 = %f kN/m^2, V1 = %f m^3, t1 = %f C,\n P2 = %f MN/m^2, V2 = %f m^3, t2 = %f C,\n P3 = %f MN/m^2, V3 = %f m^3, t3 = %f C,\n P4 = %f MN/m^2, V4 = %f m^3, t4 = %f C,\n P5 = %fkN/m^2, V5 = %fm^3, t5 = %fC\n',P1*10^3,V1,T1-273,P2,V2,T2-273,P3,V3,T3-273,P4,V4,T4-273,P5*10^3,V5,T5-273); // (b) W = (P3*(V4-V3)+((P4*V4-P5*V5)-(P2*V2-P1*V1))/(Gama-1))*10^3;// work done, [kJ] mprintf('\n (b) The net work done is = %f kJ\n',W); // (c) TE = 1-(T5-T1)/((T3-T2)+Gama*(T4-T3));// thermal efficiency mprintf('\n (c) The thermal efficiency is = %f percent\n',TE*100); // (d) Q = W/TE;// heat received, [kJ] mprintf('\n (d) The heat received is = %f kJ\n',Q); // (e) PW = P3*(V4-V3)+(P4*V4-P5*V5)/(Gama-1) WR = W*10^-3/PW;// work ratio mprintf('\n (f) The work ratio is = %f\n',WR); // (e) Pm = W/(V1-V2);// mean effective pressure, [kN/m^2] mprintf('\n (e) The mean effective pressure is = %f kN/m^2\n',Pm); // (f) CE = (T4-T1)/T4;// carnot efficiency mprintf('\n (f) The carnot efficiency is = %f percent\n',CE*100); // End
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clear; clc; //page no.266 G = 20000;//cfs k = 1/15; Q_m = G*(k)^(2+ 1/2); printf('Qm = %d cfs',Q_m); //there is a minute error in the answer given in textbook
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// To determine the phase shifting network to be used. clear clc; Z=1000*(cosd(60) + %i*sind(60));//impedence X=tand(50)*1000*cosd(60); Xl=1000*sind(60); Xc=Xl-X; C=1000000/(314*Xc); //Answers don't match due to difference in rounding off of digits disp(X,"X="); disp(Xc,"Xc="); disp(C,"C(micro farads)=");
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clc; Zout=80; avAoL=1180; Zoutf=Zout/(1+avAoL); disp('mohm',Zoutf*1000 ,"Zoutf=");//The answers vary due to round off error
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//Problem 26.05:The circuit shown in Figure 26.9 dissipates an active power of 400 Wand has a power factor of 0.766 lagging. Determine (a) the apparent power, (b) the reactive power, (c) the value and phase of current I, and (d) the value of impedance Z. //initializing the variables: Pa = 400; // in Watts rv = 100; // in volts thetav = 30; // in degrees R = 4; // in ohm pf = 0.766; // power factor //calculation: V = rv*cos(thetav*%pi/180) + %i*rv*sin(thetav*%pi/180) //magnitude of apparent power,S = V*I S = Pa/pf phi = acos(pf) theta = phi*180/%pi // in degrees //Reactive power Q Q = S*sin(phi) //magnitude of current Imag = S/rv thetai = thetav - theta I = Imag*cos(thetai*%pi/180) + %i*Imag*sin(thetai*%pi/180) //Total circuit impedance ZT ZT = V/I //impedance Z Z = ZT - R printf("\n\n Result \n\n") printf("\n (a)apparent power is %.2f VA ",S) printf("\n (b)reactive power is %.2f var ",Q) printf("\n (c)the current flowing and Circuit phase angle is %.2f/_%.0f° A ",Imag,thetai) printf("\n (d)impedance, Z is %.2f + (%.2f)i ohm ",real(Z), imag(Z))
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//Chapter 7: Solid State //Problem: 11 clc; mprintf("AB remain in BCC structure if the edge length is a then body diagonal ,is root(3)a\n") mprintf(" root(3)a = 2(r+ + r-)\n") A = (sqrt(3) * 0.4123 - 2 * 0.81) / 2 mprintf(" A+ = %.2f nm",A)
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//This is an infeasible model //Ref:H.A. TAHA,"OPERATIONS RESEARCH AN INTRODUCTION",PEARSON-Prentice Hall New Jersey 2007,chapter 3.5 //Example: //Toolco produces three types of tools, T1,T2 and T3. The tools use two rawmaterials, M1 and M2, according to the data in the following table: //-------------------------------------------------------------------- // Number of units of raw materials per tool // ----------------------------------------- //Raw material T1 T2 T2 //--------------------------------------------------------------------- //M1 3 5 6 //M2 5 3 4 //--------------------------------------------------------------------- //The available daily quantites of raw materials M1 and M2 are 1000 units and 1200 units, respectively. The marketing department informed the production manager that according to their research , the daily demand for all three tools must be at least 500 units. will the manufacturing department be able to satisfy the demand? // Copyright (C) 2015 - IIT Bombay - FOSSEE // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt // Author:Debasis Maharana // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in //====================================================================== clc; Nraw = 2; Nproduct = 3; Product_mix = [3 5 6;5 3 4]; Raw_avail = [1000 1200]; Demand = [500 500 500]; mprintf('Data received') mprintf('Number of product %d',Nproduct); mprintf('Number of Raw materials %d',Nraw); FirstR = ['Product 1','Product 2','Product 3','Available'] Raw_mat = ['M1','M2']; table = [['Raw materisl' FirstR];[Raw_mat' string(Product_mix) string(Raw_avail') ]]; disp(table); input('Enter to proceed '); for i = 1:Nraw A(i,:) = Product_mix(i,:); b(i) = Raw_avail(i); end A = [A;-eye(Nproduct,Nproduct)]; b = [b;-500*ones(Nproduct,1)]; lb = zeros(1,Nproduct);//Redudant due to demand ub = []; C = ones(1,Nproduct); [xopt,fopt,exitflag,output,lambda] = linprog(-C',A,b,[],[],lb,ub); clc select exitflag //Display result case 0 then mprintf('Optimal Solution Found') input('Press enter to view results') disp('Paint produced') mprintf('\nExterior paint %d\t interior paint %d',xopt(1),xopt(2)); case 1 then mprintf('Primal Infeasible\n') y = A*xopt-b; if sum(y(1:Nraw))>0 mprintf('\nInsufficient raw material'); else mprintf('\nCan not meet market demand'); end case 2 then mprintf('Dual Infeasible') case 3 mprintf('Maximum Number of Iterations Exceeded. Output may not be optimal') case 4 mprintf('Solution Abandoned') case 5 mprintf('Primal objective limit reached') else mprintf('Dual objective limit reached') end
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//quadratic splines //example 5.2 //page 18 clc;clear;close; X=[1 2 3]; y=[-8 -1 18]; h=X(2)-X(1); m1=(y(2)-y(1))/(X(2)-X(1)); m2=(2*(y(2)-y(1)))/h-m1; m3=(2*(y(3)-y(2)))/h-m2; deff('y2=s2(x)','y2=(-(X(3)-x)^2*m1)/2+((x-X(2))^2*m3)/2+y(2)+m2/2'); a=poly(0,'x'); disp(s2(a)); printf('the value of function is 2.5: %0.2f',s2(2.5)); x=2.0; h=0.01; deff('y21=s21(x,h)','y21=(s2(x+h)-s2(x))/h'); d1=s21(x,h); printf('\n\nthe first derivative at 2.0 :%0.2f',d1);
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Ex19_4.sce
fck=15//in MPa fy=415//in MPa MF=1.4//modification factor //let a be span to depth ratio l=1//span, in m a=MF*7 D=l*1000/a//in mm D=105//assume, in mm //to calculate loading W1=25*(D/10^3)*1.5//self-weight, in kN/m W2=0.5*1.5//finish, in kN/m W3=0.75*1.5//live load, in kN/m W=W1+W2+W3//in kN/m Wu=1.5*W//in kN/m lef=l+0.23/2//effective span, in m Mu=Wu*lef/2//in kN-m //check for depth d=sqrt(Mu*10^6/(0.138*fck*1500))//in mm dia=12//assume 12 mm dia bars D=d+12/2+15//<105, hence OK D=100//assume, in mm d=D-dia/2-15//in mm //steel //Xu=0.87*fy*Ast/0.36/fck/b = a*Ast a=0.87*fy/0.36/fck/1.5/10^3 //using Mu=0.87 fy Ast (d-0.416 Xu), we get a quadratic equation p=0.87*fy*0.416*a q=-0.87*fy*d r=Mu*10^6 Ast=(-q-sqrt(q^2-4*p*r))/2/p//in sq mm //provide 8 mm dia bars dia=8//in mm s1=1500*0.785*dia^2/Ast//>3d=3x79=237 mm s1=235//in mm Ads=0.12/100*1000*D//distribution steel, in sq mm //assume 6 mm dia bars s2=1000*0.785*6^2/Ads//in mm s2=235//round-off, in mm Tbd=1.6//in MPa Ld=dia*0.87*fy/4/Tbd//in mm Ld=452//in mm Tv=Wu*10^3/1500/d//in MPa Ast=1500*0.785*8^2/235//in sq mm pt=Ast/1500/d*100//in % //for M15 and pt=0.26 Tc=0.35//in MPa //as Tc>Tv, no shear reinforcement required mprintf("Summary of design\nThickness of slab = %d mm\nCover = 15 mm\nMain steel = 8 mm dia @ %d mm c/c\nDevelopment length = %d mm\nDistribution steel = 6 mm dia @ %d mm c/c",D,s1,Ld,s2)
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Ex13_19.sce
clc clear //Initialization of variables t=90 //F ts=67.2 //F phi=0.3 per=0.8 //calculations dep=t-ts dt=dep*per tf=t-dt disp("from psychrometric charts,") phi2=0.8 //results printf("Dry bulb temperature = %.2f F",tf) printf("\n percent humidity = %.2f",phi2)
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Ex3_2.sce
clc T=300 //K k=8.617*10^-5 //eV/K q=1.6*10**-19 //C NA=10^19 //cm^-3 ND=10^16//cm^-3 ni=9.65*10^9 epsilonx=8.854*10^-12 //F/m Vbi=(k*T)*log(NA*ND/ni^2) disp(Vbi,"the built in potential in V=") W=sqrt(2*Vbi/q*ND) disp(W,"W in cm =") // ans in textbook is wrong epsilonm=((q*ND*W)) disp(epsilonm,"epsilonm in V/cm") // ans in textbook is wrong
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ex5_35.sce
//Chapter-5, Example 5.35, Page 198 //============================================================================= clc clear //INPUT DATA Q=250;//quality factor fr=1.5*10^6;//resonant freq in hertz //CALCULATIONS Bw=(fr)/(Q);//bandwidth in Hz hf1=fr+Bw;//half power freq 1 hf2=fr-Bw;//half power freq 2 mprintf("Thus bandwidth is %d hz\n",Bw); mprintf("Thus value of half-power frequencies are %g hz and %g hz",hf1,hf2); //=================================END OF PROGRAM======================================================================================================
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Ex3_9.sce
//CHAPTER 3- THREE-PHASE A.C. CIRCUITS //Example 9 clc; disp("CHAPTER 3"); disp("EXAMPLE 9"); //VARIABLE INITIALIZATION v_ab=400; //in Volts v_bc=400; //in Volts v_ac=400; //in Volts z_ab=100; //in Ohms z_bc=100; //in Ohms z_ac=100; //in Ohms //solution (a) //function to convert from polar to rectangular form function [x,y]=pol2rect(mag,angle); x=mag*cos(angle); y=mag*sin(angle); endfunction; I_AB=v_ab/z_ab; mag1=abs(real(I_AB)); ang1=0; //I_AB is represented as mag1∠ang1 I_BC=v_bc/z_bc; ang2=-210*(%pi/180); //I_BC is represented as mag1∠ang2 I_AC=v_ac/z_ac; ang3=210*(%pi/180); //I_AB is represented as mag1∠ang3 [x1,y1]=pol2rect(I_AB,ang1); [x2,y2]=pol2rect(I_BC,ang2); [x3,y3]=pol2rect(I_AC,ang3); //let us consider values X1, Y1, X2, Y2, X3 and Y3 for the ease of calculation (these are not mentioned in the book) X1=x1-x3; Y1=y1-y3; X2=x2-x1; Y2=y2-y1; X3=x3-x2; Y3=y3-y2; I_A=X1+(%i*Y1); I_B=X2+(%i*Y2); I_C=X3+(%i*Y3); //function to convert from rectangular to polar form function [z,angle]=rect2pol(x,y); z=sqrt((x^2)+(y^2)); //z is impedance & the resultant of x and y if(x==0 & y>0) then angle=90; //in case atan=∞ elseif(x==0 & y<0) then angle=-90 //in case atan=-∞ else angle=atan(y/x)*(180/%pi); //to convert the angle from radians to degrees end; endfunction; [mag4,ang4]=rect2pol(X1,Y1); [mag5,ang5]=rect2pol(X2,Y2); [mag6,ang6]=rect2pol(X3,Y3); disp(sprintf("(a) The line current I_A is %f∠%f A",mag4,ang4)); disp(sprintf("The line current I_B is %f∠%f A",mag5,(180+ang5))); disp(sprintf("The line current I_C is %f∠%f A",mag6,ang6)); //solution (b) //since power is consumed only by 100Ω resistance in the arm AB r1=100; p1=(I_AB^2)*r1; p2=160000; r2=p2/p1; disp(sprintf("(b) The star connected balanced resistance is %d Ω",r2)); //END
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Ex10_1.sce
//Ex10_1 clc x=72;//given value in Decimal disp("Decimal number="+string(x)) str=dec2bin(x) disp("Eqivalent Binary number="+string(str))//Binary value
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Example8_4a.sce
clc //Given that t1 = 1 // time period of satellite s1 in hours t2 = 8 // time period of satellite s2 in hour r1 = 1.2e4 // radius of orbit of satellite s1 in km // sample problem 4a page No. 300 printf("\n\n\n # Problem 4a # \n") printf("Standard formula r2/r1 = (t2/t1)^(2/3)") r2 = r1 * (t2/t1)^(2/3) // calculation of radius of orbit of satellite s2 in km v1 = 2 * %pi * r1 / t1 // calculation of speed of satellite s1 in km/h v2 = 2 * %pi * r2 / t2 // calculation of speed of satellite s2 in km/h del_v = v2 - v1 // calculation of relative speed of satellites in km/h printf (" \n Relative speed of satellite s2 wrt satellite s1 is %e km/h.", del_v)
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ex8_4.sce
clc; clear all; C=5.1*1e-12;//given capacitance fr=10*1e6;//given resonant frequency L=2*1e-6;//given L //part a Lt=1/(4*%pi*%pi*fr*fr*C);//new inductance disp(Lt,'new inductancetance for fr=10mhz is='); //part b L2=Lt-L;//required Ls disp(L2,'required series inductance is');
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Ex1_8.sce
//Tested on Windows 7 Ultimate 32-bit //Chapter 1 Introduction to Electronics Pg no. 33 //Solved Problem 1 clear; clc; //Given Data wattage=100;//wattage in watts voltage=220;//voltage in volts //Solution I=wattage/voltage;//current in amperes printf("I=%.3f Amp.",I);//Displaying upto 3 places of decimal //Error due to decimal approximations
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example14.sce
clc clear //input data dH=0.14//Rise in static pressure of the air by fan in m of water N=650//The running speed of the fan in rpm P=85*0.735//Power consumed by the fan in kW H1=0.75//The static pressure of the air at the fan in m of Hg T1=298//The static pressure at the fan of air in K m=260//Mass flow rate of air in kg/min dHg=13590//Density of mercury in kg/m^3 dw=1000//Density of water in kg/m^3 g=9.81//Acceleration due to gravity in m/s^2 R=287//The universal gas constant in J/kg.K //calculations P1=dHg*g*H1*10^-3//The inlet static pressure in kPa dP=dw*g*dH*10^-3//The total change in static pressures at inlet and outlet in kPa P2=P1+dP//The exit static pressure in kPa d1=(P1*10^3)/(R*T1)//The inlet density of the air in kg/m^3 Q=m/d1//The volume flow rate of air in fan in m^3/min //output printf('(a)The exit static pressure of air in the fan is %3.2f kPa\n(b)The volume flow rate of the air is %3.1f m^3/min',P2,Q)
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Ex10_3.sce
// Example 10_3 clc;funcprot(0); // Given data T=200;// °C P=1554;// kPa R=0.462;// kJ/kg.K // Calculation v_g=(R*(T+273))/P;// m^3/kg rho=1000;// kg/m^3 v_f=0.001;// m^3/kg dPbydT=(1906-1254)/(210-190);// kN/m^2.K h_fg=(T+273)*(v_g-v_f)*dPbydT;// kJ/kg h_fga=1941;// kJ/kg (From steam tables) error=((h_fg-h_fga)/h_fga)*100;// The percentage error printf("\nThe percent error=%2.1f percentage",error);
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Ex9_9.sce
// Exa 9.9 // To calculate- // a) The total bandwidth required, // b) The bandwidth efficiency, // c) The Eb/No required, and // d) No of carried bits per symbol. clc; clear all; M=8; //number of different signal elements Fc=250; //carrier frequency in kHz DelF=25; //kHz Pe=10^-6;//probability of error //solution TotalBW=2*M*DelF; nb=2*log2(M)/(M+3); //Pe=7*Q(z) and z=approx(5.08) z=5.08; Eb_No=(z)^2/log2(M); bits_sym=log2(M); printf('Total bandwidth required is %d kHz \n ',TotalBW); printf('The bandwidth efficiency is %.4f \n ',nb); printf('The required Eb/No is %.3f dB \n ',10*log10(Eb_No)); printf('Carried bits per symbol are %d \n ',bits_sym);
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clc; clear all; P=8.9*1e-3;//resistivity of doped sillicon Rh=3.6*1e-4;//hall coefficient e=1.6*1e-19;//charge of electron ne=3*%pi/(8*Rh*e);//carrier density of electron disp('m^-3',ne,'carrier density of electron is='); ue=1/(P*ne*e);//mobility of electon disp('m^2*V^-1*s^-1',ue,'mobility of electon is=')
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//Determine appropriate values of RS and RD. //Solved Example Ex4.8 page no 120 clear clc Rs=750 //kΩ printf("\n Rs = %0.2f K ohm",Rs) Vdd=24 //V Vdsq=15 //V Idq=0.002 //mA Rd=((Vdd-Vdsq-(Idq*Rs))/Idq)/1000 printf("\n Rd = %0.2f K ohm",Rd)
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Ex2_11.sce
//Variable Declaration JDref=2415020 //Reference Julian days JC=36525 JD=2451897.0417 //Julian days with reference from Example 2.10 //Calculation T=(JD-JDref)/JC //Time in julian Centuries //Result printf("The time for given date is : %.8f Julian Centuries",T)
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function[n] = cauchy(n) i = 1:n; // tail de l'échantillon u = rand(i, 'uniform'); x = tan(%i .* (u - 0.5)); // échantillon loi de Cauchy S = cumsum(x); y = S ./ i; // vecteur de composants Si / i xbasc(); plot2d(i', y', 2); // ligne polygonale de sommets (i, Si / i) en bleu xsegs([0 n], [0 0], 5); // trace en rouge l'axe des abscisses. endfunction
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<?xml version="1.0" encoding="utf-8"?> <test> <description>Euler, Subsonic Cylinder, WeakDG advection, Implicit solver</description> <executable>CompressibleFlowSolver</executable> <parameters>CylinderSubsonic_WeakDG_Implicit.xml</parameters> <files> <file description="Session File">CylinderSubsonic_WeakDG_Implicit.xml</file> </files> <metrics> <metric type="L2" id="1"> <value variable="rho" tolerance="1e-12">10.7744</value> <value variable="rhou" tolerance="1e-12">1.09375</value> <value variable="rhov" tolerance="1e-7">0.0768964</value> <value variable="E" tolerance="1e-12">2.228e+06</value> </metric> <metric type="Linf" id="2"> <value variable="rho" tolerance="1e-12">1.22501</value> <value variable="rhou" tolerance="2e-6">0.246672</value> <value variable="rhov" tolerance="2e-6">0.134329</value> <value variable="E" tolerance="1e-12">253314</value> </metric> </metrics> </test>
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clc; pathname=get_absolute_file_path('2_15_soln.sce') filename=pathname+filesep()+'2_15_data.sci' exec(filename) // Solution: // specific Weight of water, gamma1=9800; //N/m^3 // we know pressure, // p=(specific weight of liquid * liquid column height) p=(gamma1*H); //Pa pK=p/1000; //kPa // Results: printf("\n Results: ") printf("\n The pressure on skin diver is %.0f kPa.",pK)
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clc //initialisation of variables p=1/10//in v1=3/4//in v2=3/5//in m=1*1/2//in l=4//cranks a1=1.25//in a2=0.7//in //CALCULATIONS C=a1/a2//in A=l*a1/a2//in S=(A/2-a1)//in //RESULTS printf('the travel of the value =% f in',S)
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i = imread('test3.jpg'); corners = detectHarrisFeatures(i,'SensitivityFactor',0.08,'MinQuality',0.5); disp(corners);
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//Chapter 07: Discrete Probability clc; clear; //For sum to be 7 out of the total 36 equally likely possible outcomes there are 6 outcomes //(1,6) (2,5) (3,4) (4,3) (5,2) (6,1) total_outcomes=36 //total no of outcomes seven_sum_outcome=6 //no of outcomes where sum of numbers appearing on dice is 7 prob_seven=seven_sum_outcome/total_outcomes disp('Probability that 7 comes when 2 dice are rolled is') disp(prob_seven)
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clear; close; clc; A=[1 -1;1 1;1 2]; disp('A=',A); b=[1;1;3]; disp('b=',b); x=(A'*A)\(A'*b); disp('x=',x); C=x(1,1); D=x(2,1); disp('C=',C); disp('D=',D); disp('The line of best fit is b=C+Dt');
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clc(); clear; // To calculate the critical temperature M1=199.5; //isotopic mass M2=203.4; Tc1=4.185; //1st critical temp in K Tc2=Tc1*sqrt(M1/M2); printf("the critical temperature is %f K",Tc2);
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//Function to calculate the roots of a cubic equation function[r] = cubic(a, b, c) //Use Gerolamo Cardano's solution for cubic equations with real roots //roots function can also be used //Use auxiliary variables p and q p =((3 * b)-(a^2))/3 q =(c)+((2 * (a^3))/27)-((a * b)/3) //Calculate discrimant Di for the cubic equation Di = ((p/2)^2)+((q/3)^3) //Calculate values of cos(Fi) and Fi if(p<0) //For |p| p = p * -1 end Fi = acos((-1 * q)/(2 * (sqrt((p/3)^3)))) Fi1 = Fi/3 Fi2 = (Fi + (4 *%pi))/3 Fi3 = (Fi + (2 *%pi))/3 //x,y and z are the three real distinct roots of the cubic equation x = (-1 * (a/3)) + (2 * cos(Fi1) * sqrt(p/3)) y = (-1 * (a/3)) + (2 * cos(Fi2) * sqrt(p/3)) z = (-1 * (a/3)) + (2 * cos(Fi3) * sqrt(p/3)) printf('\nx,y and z are the three real roots of this cubic equation\n') printf('x = %f\ny = %f\nz = %f\n',x,y,z) //d should be maximum for cost and area of friction lining to be minimum //Assigning proper value to d if (x > y) then e = x if (e > z) then r = e else r = z end else e = y if (e > z) then r = e else r = z end end endfunction //Obtain path of solution file path = get_absolute_file_path('solution11_3.sce') //Obtain path of data file datapath = path + filesep() + 'data11_3.sci' //Clear all clc //Execute the data file exec(datapath) //Consider torque transmitted by one pair of contacting surfaces Mt = Mt / 2 printf('\nApplying uniform-wear theory\n') //Evaluate the inner diameter of the friction disk d (mm) printf('\nInner diameter of the friction disk is d (mm)\n') printf('\nUsing the torque transmitting capacity formula:\n') printf('\nMf = (pi * mu * Pa * d)*((D^2)-(d^2)))/8\n') printf('we get the following cubic equation in d\n') //Representing cubic equation in the form x3 + ax2 + bx + c = 0 (x=d) a = 0 b = -1 * (D^2) c = (8 * Mt * 1000)/(%pi * mu * Pa) printf('\n(d^3) + (%fd) + (%f)= 0\n',b,c) //Call the declared function d = cubic(a, b, c) //Evaluate spring force P (N) P = ((%pi * Pa * d)*(D - d))/2 //Print results printf('\nInner diameter of the friction disk(d)= %f(mm)\n',d) printf('\nSpring force for keeping the clutch engaged(P)= %f(N)\n',P)
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//Copyright INRIA //sharing common data files=G_make(['/tmp/ext13c.o'],'ext13c.dll'); link(files,['ext13ic','ext13oc'],'C'); //Must be linked together a=1:10; n=10;a=1:10; call('ext13ic',n,1,'i',a,2,'r','out',2); //loads b with a c=call('ext13oc',n,1,'i','out',[1,10],2,'r'); //loads c with b if norm(c-a) > %eps then pause,end
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THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.284770D+00 2 -0.369080D-02 0.227838D-02 3 0.244550D-01 0.232960D-03 0.215975D+00 4 0.910765D-03 0.174106D-03 -0.151360D-02 0.183820D-02 5 0.991548D-03 -0.154771D-03 -0.174586D-03 -0.165104D-03 0.319149D-02 6 -0.726687D-04 -0.611736D-04 0.189253D-03 -0.414567D-04 0.850608D-04 7 0.167675D-02 -0.716535D-04 0.833817D-03 -0.491210D-04 0.238074D-03 8 -0.240475D-03 0.855644D-04 -0.232679D-03 -0.483678D-06 0.331327D-04 9 -0.157011D+00 0.983733D-02 -0.462568D-01 0.258117D-02 0.716089D-01 10 -0.394571D-01 -0.987950D-02 0.284100D-01 -0.834195D-02 0.148007D+00 11 -0.398864D-01 -0.108460D-01 0.385973D-01 -0.161492D-01 -0.536659D-01 12 0.400197D+00 -0.231569D-01 -0.406789D+00 0.468647D-01 -0.232452D-01 13 0.542674D-01 -0.102676D-01 0.223784D-01 0.228816D-03 0.176188D-01 14 -0.781739D-01 0.212216D-01 -0.446439D+00 0.392101D-02 0.277814D-01 15 -0.156252D+01 -0.276406D-01 0.566307D-01 -0.471616D-01 -0.106681D+00 16 -0.352959D-01 -0.686802D-03 -0.748201D-02 0.188107D-02 -0.753144D-03 17 0.954149D-02 0.135226D-03 0.294264D-02 0.470496D-03 -0.449580D-03 18 0.146027D+00 -0.249561D-01 -0.191958D+00 -0.271624D-01 -0.997418D-03 19 0.500811D-01 0.836195D-03 0.348468D-01 0.114709D-02 -0.214203D-01 20 0.290234D-01 -0.156865D-01 -0.544628D+00 -0.239090D-01 -0.155341D-01 21 -0.693236D-01 0.181075D-02 -0.552495D-01 0.140363D-02 0.200155D-01 22 0.174575D-02 0.339154D-03 0.339636D-02 0.327185D-03 0.618357D-04 23 -0.544920D-02 0.558485D-03 -0.121371D-03 -0.490909D-02 -0.100688D-02 24 0.820514D-03 0.182814D-03 -0.131936D-02 0.102063D-03 0.187985D-03 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 0.704396D-03 7 0.578005D-03 0.245143D-02 8 0.639763D-04 -0.859763D-04 0.235342D-02 9 -0.276513D-02 0.121183D-01 0.222954D-01 0.287954D+02 10 0.907670D-02 0.205656D-01 0.124386D-02 0.185591D+01 0.168567D+02 11 0.224028D-01 0.352973D-01 0.358854D-01 0.400504D+00 -0.252165D+01 12 -0.338623D-01 0.907447D-02 0.665657D-01 0.482890D+01 -0.598304D+00 13 0.438219D-01 0.745585D-01 -0.180543D-01 -0.123435D+01 0.321543D+01 14 -0.886569D-02 0.991812D-02 0.170609D+00 0.221689D+01 0.300290D+01 15 0.310397D-02 0.292809D-01 -0.193964D-01 -0.227929D+00 -0.765885D+01 16 0.102796D-02 0.130594D-02 -0.213077D-02 0.469012D+00 -0.453616D-01 17 -0.266428D-03 -0.470747D-03 0.180956D-03 -0.103967D+00 -0.444775D-01 18 -0.396128D-01 -0.781016D-01 -0.222091D-02 0.132121D+01 -0.134512D+01 19 -0.144318D-02 0.111973D-01 -0.319188D-02 -0.881027D+00 -0.160716D+01 20 -0.130807D-01 -0.401790D-01 -0.111368D+00 0.224304D+00 0.208891D+00 21 0.895724D-03 -0.102740D-01 0.144233D-02 0.109372D+01 0.142192D+01 22 -0.893147D-05 -0.258949D-03 0.103462D-03 -0.142021D-01 0.298944D-02 23 -0.682454D-03 -0.337267D-03 0.118933D-02 -0.232208D-01 -0.359312D-01 24 0.215683D-03 0.911131D-04 -0.392245D-03 -0.228737D-02 -0.405070D-02 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 0.284762D+02 12 0.188092D+01 0.122856D+03 13 0.386815D+00 -0.177225D+00 0.991005D+01 14 0.318616D+01 0.102953D+02 0.581213D-01 0.436681D+02 15 -0.744726D+00 0.558334D+01 -0.950559D+00 -0.301393D+01 0.197539D+03 16 0.178779D+00 0.205098D-01 0.717761D-01 -0.330022D+00 0.144188D+01 17 -0.835460D-02 -0.256450D-01 -0.276934D-01 0.158541D-01 -0.101534D+01 18 -0.423119D+01 0.329681D+01 -0.477808D+01 -0.178897D+01 0.103901D+02 19 0.223611D+01 0.182088D+00 0.538229D-01 -0.121855D+01 0.202076D+01 20 0.202564D+01 -0.242088D+02 -0.707512D+00 -0.160885D+02 0.793236D+01 21 -0.176353D+01 0.380514D+00 -0.156781D+00 0.117068D+01 -0.200688D+01 22 -0.393122D-01 -0.243981D-01 -0.822828D-02 -0.332106D-02 -0.297461D-01 23 0.276137D+00 0.465272D+00 -0.530413D-01 0.183223D+00 0.990725D-01 24 -0.870117D-01 -0.420059D-01 0.469433D-02 -0.575096D-01 -0.226286D-01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 0.332804D+00 17 -0.245760D-01 0.125378D-01 18 0.717541D-02 -0.364001D-01 0.100185D+03 19 0.301077D-01 0.185460D-02 -0.476698D+00 0.280834D+01 20 0.148680D+00 -0.545857D-01 0.149623D+02 0.192384D+00 0.134699D+03 21 0.450223D-01 -0.190532D-02 0.229637D+01 -0.270709D+01 0.224711D+00 22 0.214359D-03 0.876983D-03 -0.489788D+00 -0.513335D-02 -0.128511D+00 23 0.939370D-03 -0.231637D-03 0.198759D+00 0.264013D-01 0.158406D+01 24 0.100774D-02 0.169844D-03 -0.107678D+00 -0.651802D-02 -0.650785D+00 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 0.330664D+01 22 -0.241383D-01 0.572791D-02 23 0.234585D-01 -0.421923D-02 0.240902D+00 24 -0.141642D-02 0.159257D-02 -0.206755D-01 0.804123D-02 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.145 1.000 3 0.099 0.011 1.000 4 0.040 0.085 -0.076 1.000 5 0.033 -0.057 -0.007 -0.068 1.000 6 -0.005 -0.048 0.015 -0.036 0.057 7 0.063 -0.030 0.036 -0.023 0.085 8 -0.009 0.037 -0.010 0.000 0.012 9 -0.055 0.038 -0.019 0.011 0.236 10 -0.018 -0.050 0.015 -0.047 0.638 11 -0.014 -0.043 0.016 -0.071 -0.178 12 0.068 -0.044 -0.079 0.099 -0.037 13 0.032 -0.068 0.015 0.002 0.099 14 -0.022 0.067 -0.145 0.014 0.074 15 -0.208 -0.041 0.009 -0.078 -0.134 16 -0.115 -0.025 -0.028 0.076 -0.023 17 0.160 0.025 0.057 0.098 -0.071 18 0.027 -0.052 -0.041 -0.063 -0.002 19 0.056 0.010 0.045 0.016 -0.226 20 0.005 -0.028 -0.101 -0.048 -0.024 21 -0.071 0.021 -0.065 0.018 0.195 22 0.043 0.094 0.097 0.101 0.014 23 -0.021 0.024 -0.001 -0.233 -0.036 24 0.017 0.043 -0.032 0.027 0.037 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 1.000 7 0.440 1.000 8 0.050 -0.036 1.000 9 -0.019 0.046 0.086 1.000 10 0.083 0.101 0.006 0.084 1.000 11 0.158 0.134 0.139 0.014 -0.115 12 -0.115 0.017 0.124 0.081 -0.013 13 0.524 0.478 -0.118 -0.073 0.249 14 -0.051 0.030 0.532 0.063 0.111 15 0.008 0.042 -0.028 -0.003 -0.133 16 0.067 0.046 -0.076 0.152 -0.019 17 -0.090 -0.085 0.033 -0.173 -0.097 18 -0.149 -0.158 -0.005 0.025 -0.033 19 -0.032 0.135 -0.039 -0.098 -0.234 20 -0.042 -0.070 -0.198 0.004 0.004 21 0.019 -0.114 0.016 0.112 0.190 22 -0.004 -0.069 0.028 -0.035 0.010 23 -0.052 -0.014 0.050 -0.009 -0.018 24 0.091 0.021 -0.090 -0.005 -0.011 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 1.000 12 0.032 1.000 13 0.023 -0.005 1.000 14 0.090 0.141 0.003 1.000 15 -0.010 0.036 -0.021 -0.032 1.000 16 0.058 0.003 0.040 -0.087 0.178 17 -0.014 -0.021 -0.079 0.021 -0.645 18 -0.079 0.030 -0.152 -0.027 0.074 19 0.250 0.010 0.010 -0.110 0.086 20 0.033 -0.188 -0.019 -0.210 0.049 21 -0.182 0.019 -0.027 0.097 -0.079 22 -0.097 -0.029 -0.035 -0.007 -0.028 23 0.105 0.086 -0.034 0.056 0.014 24 -0.182 -0.042 0.017 -0.097 -0.018 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 1.000 17 -0.380 1.000 18 0.001 -0.032 1.000 19 0.031 0.010 -0.028 1.000 20 0.022 -0.042 0.129 0.010 1.000 21 0.043 -0.009 0.126 -0.888 0.011 22 0.005 0.103 -0.647 -0.040 -0.146 23 0.003 -0.004 0.040 0.032 0.278 24 0.019 0.017 -0.120 -0.043 -0.625 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 1.000 22 -0.175 1.000 23 0.026 -0.114 1.000 24 -0.009 0.235 -0.470 1.000
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clc //initialisations po=101396.16//N/m^2 vo=22.4//l t=273 m=4*1000//gm //calculations R=po*vo/t c=R/m //results printf(' pressure of the gas= % 1f j',c)
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clc // Given that D=10//in inch Grinding Wheel diameter N=4000//in rpm w=1//in inch d=0.002//in inch depth of cut v=60//inch/min feed rate of the workpiece // Sample Problem on page no. 715 printf("\n # force in Surface Grinding # \n") Mrr=d*w*v//material removal rate //for low carbon steel , the specific energy is 15hp min/in3 u=15//in hp min/in3 P=u*Mrr*396000//in lb/min Fc = P/(2*3.14*N*(D/2)) printf("\n\n Cutting Force = %f lb",Fc) // Answer in the book is approximated to 5.7 lb // from the experimental data in book thrust force is taken as 30% higher than cutting force Fn = Fc+(30/100)*Fc printf("\n\n Thrust Force = %f lb",Fn) // Answer in the book is approximated to 7.4 lb
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; @Harness: disassembler ; @Result: PASS section .text size=0x00000046 vma=0x00000000 lma=0x00000000 offset=0x00000034 ;2**0 section .data size=0x00000000 vma=0x00000000 lma=0x00000000 offset=0x0000007a ;2**0 start .text: label 0x00000000 ".text": 0x0: 0x0e 0x90 ld r0, -X 0x2: 0x1e 0x90 ld r1, -X 0x4: 0x2e 0x90 ld r2, -X 0x6: 0x3e 0x90 ld r3, -X 0x8: 0x4e 0x90 ld r4, -X 0xa: 0x5e 0x90 ld r5, -X 0xc: 0x6e 0x90 ld r6, -X 0xe: 0x7e 0x90 ld r7, -X 0x10: 0x8e 0x90 ld r8, -X 0x12: 0x9e 0x90 ld r9, -X 0x14: 0xae 0x90 ld r10, -X 0x16: 0xbe 0x90 ld r11, -X 0x18: 0xce 0x90 ld r12, -X 0x1a: 0xde 0x90 ld r13, -X 0x1c: 0xee 0x90 ld r14, -X 0x1e: 0xfe 0x90 ld r15, -X 0x20: 0x0e 0x91 ld r16, -X 0x22: 0x1e 0x91 ld r17, -X 0x24: 0x2e 0x91 ld r18, -X 0x26: 0x3e 0x91 ld r19, -X 0x28: 0x4e 0x91 ld r20, -X 0x2a: 0x5e 0x91 ld r21, -X 0x2c: 0x6e 0x91 ld r22, -X 0x2e: 0x7e 0x91 ld r23, -X 0x30: 0x8e 0x91 ld r24, -X 0x32: 0x9e 0x91 ld r25, -X 0x34: 0xae 0x91 ld r26, -X ; undefined 0x36: 0xbe 0x91 ld r27, -X ; undefined 0x38: 0xce 0x91 ld r28, -X 0x3a: 0xde 0x91 ld r29, -X 0x3c: 0xee 0x91 ld r30, -X 0x3e: 0xfe 0x91 ld r31, -X 0x40: 0x0e 0x90 ld r0, -X 0x42: 0x0a 0x90 ld r0, -Y 0x44: 0x02 0x90 ld r0, -Z start .data:
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clear; clc; close; disp("system given is dy(t)/dt+2y(t)=x(t)+dx(t)/dt"); disp("taking fourier transform on both sides we get"); disp("H(w)=Y(w)/X(w)=1-(1/(2+%j*w))"); w=-10:0.1:10; dw=.1; Hw=1-ones(1,length(w))./(2+%i*w); t=0:0.1:10; d=gca() plot(w,Hw); poly1=d.children.children; poly1.thickness=3; poly1.foreground=2; xtitle('X(w)','w') for i=1:length(t) if t(i)==0 then delta(i)=1; else delta(i)=0; end end h=delta'-exp(-2*t); figure; d=gca() plot(t,(h)); poly1=d.children.children; poly1.thickness=3; poly1.foreground=2; xtitle('h(t)','t')
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EX17_24.sce
clc;funcprot(0);//EXAMPLE 17.24 // Initialisation of Variables v1=0.216;.....................//Gas consumption in m^3/min pw=75;........................//Pressure of gas in mm of water t1=290;......................//Temperature of gas in K ac=2.84;....................//Air consumption in kg/min br=745;......................//Barometer reading in m of Hg D=0.25;.....................//Engine bore in m L=0.475;......................//Engine stroke in m N=240;........................//Engine rpm R=287;......................//Gas constant for air in J/kgK //Calculations p1=br+(pw/13.6);...................//Pressure of gas in mm of mercury p2=760;t2=273;.....................//NTP conditions in mm of Hg and Kelvin v2=(p1*v1*t2)/(t1*p2);...............//Volume of gas used at NTP in m^3 gs=v2/(N/2);.........................//Gas used per stroke in m^3 v=(ac*R*t2)/(1.0132*10^5);...........//Volume occupied by air at NTP in m^3/min aps=v/(N/2);...........................//Air used per stroke Va=gs+aps;.....................//Actual volume of mixture in m^3 drawn per stroke at NTP Vs=(%pi/4)*D*D*L;...............//Swept volume in mm^3 etaV=(Va/Vs);...................//Volumetric efficiency disp(etaV*100,"Volumetric efficiency (in %):")
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//Example No. 3_20 //Floating Point Arithmetic //Pg No. 56 clear ;close ;clc ; fx = 0.875000 ; Ex = -18 ; fy = 0.200000 ; Ey = 95 ; fz = fx/fy Ez = Ex - Ey mprintf('\n fz = %f \n Ez = %i \n z = %f E%i \n',fz,Ez,fz,Ez) if fz > 1 then fz = fz/10 Ez = Ez + 1 mprintf('\n z = %f E%i (normalised) \n',fz,Ez) end
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PointerTestVME.tst
//PointerTestVME.tst load PointerTest.asm, output-file PointerTest.out, compare-to PointerTest.cmp, output-list RAM[256]%D1.6.1 RAM[3]%D1.6.1 RAM[4]%D1.6.1 RAM[3032]%D1.6.1 RAM[3046]%D1.6.1; set RAM[0] 256, repeat 15 { vmstep; } output;
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# f(x, y) = x + y 1.0 1.0 = 2.0 2.0 2.0 = 4.0 5.0 3.0 = 8.0 34.0 2.0 = 36.0 32.0 -2.0 = 30.0 3.0 3.0 = 6.0
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function y = quickSort(n,x) y = []; if x == [] then y = []; elseif ((size(x,'r') > 1) | (length(n) > 1)) then error ('quickSort(n,x): n deve ser escalar, e x deve ser matriz-linha') else select n case 1 then [lt,gt] = partition(x(1),x) y = [quickSort(1,lt) x(1) quickSort(1,gt)] else if n > length (x) y = quickSort(n-1,x) else ps = quickSort(1,x(1:n)); xs = x((n+1):$); for i = 1:n if i == 1 then [lt,gt] = partition(ps(i),xs); y = [quickSort(n,lt) ps(i)]; elseif i < n then [lt,gt] = partition(ps(i),gt); y = [y quickSort(n,lt) ps(i)]; else [lt,gt] = partition(ps(i),gt); y = [y quickSort(n,lt) ps(i) quickSort(n,gt)] end end end end end endfunction function varargout = partition(x,xs) lt=[]; gt=[]; for i = 1:length(xs) if xs(i)<x lt = [lt xs(i)]; elseif xs(i)>x gt = [gt xs(i)]; end end varargout = list(lt,gt) endfunction function b = isOrd(x) b = %T; for i = 2 : length(x) if x(i)<x(i-1) then b = %F; break end end endfunction
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clc; Cai=90; alpha=20*%pi/180; Cf=Cai*sin(alpha) Cb=4*Cf/3; v=0.6686;//m^3/kg m=9000/3600; A=m*v/Cf h=0.04; r=A/(2*%pi*h) N=Cb/(A/h) disp("rev/s",N,"Wheel speed is:") //partII Cw=2*Cai*cos(alpha)-Cb; DP=m*Cb*Cw; disp("kW",DP/1000,"diagram powar is:"); //part III R=Cb*Cw Cri=[(Cai^2)+(Cb^2)-(2*Cai*Cb*cos(alpha))]^0.5 Ei=Cai^2-(Cri^2/2) DE=R/Ei disp("%",DE*100,"diagram efficiency is:"); //part IV Ed=(Cai^2-Cri^2)/2; Td=2*Ed; disp("kJ/kg",Td/1000,"total enthalpy drop per stage:")
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clc //initialisation of variables clear T= 25 //C b= 0.785*10^-6 //atm^-1 a= 49.2*10^-6 //deg^-1 d= 8.93 //gm/cc aw= 63.57 //gms //CALCULATIONS dC= a^2*(273.2+T)*aw*0.0242/(b*d) //RESULTS printf ('Cp-Cv = %.3f cal deg^-1g atom^-1',dC)
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//shubham-nakaskar iirnotch i/p arg:fo,fs,q o/p arg:x,bw disp("enter the values of fo, fs and q(quality factor) in the fo.txt,fs.txt and q.txt file provided"); //reading the notch freq.ie fo from the user fo=read('fo.txt',1,1) //reading the signal freq. ie fs from the user fs=read('fs.txt',1,1) //reading the quality factor value ie q from user q=read('q.txt',1,1) //bw=bandwidth bw=w0/q w0=fo/(fs/2) //generating linearspace vector which resembles to w0 aw0=linspace(0,w0*2,3) //getting the difference between two consecutive elements of generated linspace vector d=aw0(1,2)-aw0(1,1) xtitle("notch filter","normaised frequency(x rad/sample)","magnitude(dB)") for x=aw0(1,1):w0/8:aw0($,$) if(x==w0) then continue else plot2d(x,bw,style=0) end end
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//Chapter-1, Example 1.12, Page 24 //============================================================================= clc; clear; //INPUT DATA m=20;//mass of aluminium in kg S=0.896;//specific heat of aluminium in KJ/Kg degree centigrade L=402;//latent heat of fusion of aluminium in KJ/Kg theta2=657;//final temperature theta1=20;//initial temperature(assumed) P=25;//power of furnace in Kw n=80;//efficiency of kettle in percentage //CALCULATIONS H=m*S*(theta2-theta1)+(m*L);//heat energy required to melt aluminium or energy output from the furnace in Kj H=H/4.186;//heat energy required to melt aluminium or energy output from the furnace in Kcal H=H/860;//heat energy required to melt aluminium or energy output from the furnace in KWh n=n/100; E=H/n;//electrical energy or input energy to kettle in Kwh t=E/P;//time taken to melt the aluminium in hr t=t*60;// time taken to melt the aluminium in min //OUTPUT mprintf("Thus the time taken to melt the aluminium is %2.2f min",t); //=================================END OF PROGRAM==============================
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function [slp]=projsl(sl,q,m) //slp= projected model of sl q*m is the full rank //factorization of the projection. //! slp=syslin(sl(7),m*sl(2)*q,m*sl(3),sl(4)*q,sl(5),m*sl(6))
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//to calculate generator terminal voltage,excitation emf,power angle clc; Xd=1.48; Xq=1.24; Xe=.1; Xdt=Xd+Xe; Xqt=Xq+Xe; MVA=1; Vb=1; pf=.9; phi=acosd(pf); //(Vt*cosd(phi))^2+(Vt*sind(phi)+Ia*Xe)^2=Vb^2; //after solving //Vt^2-.0870*Vt-.99=0; function [x]=quad(a,b,c) d=sqrt(b^2-4*a*c); x1=(-b+d)/(2*a); x2=(-b-d)/(2*a); if(x1<Vb) x=x2; else x=x1; end endfunction Vt=quad(1,-.0870,-.99);disp(Vt,'terminal voltage(V)'); //after solving phi=20; j=sqrt(-1); Ia=1; Iaa=Ia*complex(cosd(-phi),sind(-phi)); Ef=Vb+j*Iaa*Xqt; Eff=abs(Ef); dl=atand(imag(Ef)/real(Ef));disp(dl,'power angle'); w=dl+phi; Id=Ia*sind(w); Ef=Ef+Id*(Xdt-Xqt); disp(abs(Ef),'excitation emf(V)');
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//Example 6.8(d) clear; clc; Vs=13; A=10; ft=1*10^6; SR=0.5*10^6; f=2*10^3; Vommax=SR/(2*%pi*f); if Vommax>Vs then Vimmax=Vs/A; printf("Useful Input Amplitude Range is Vim<=%.2f V",Vimmax);
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// Example 8.4 // Impedance Matching with a Transformer omega=10^5; R_L=500 ; I_s_m=100*10^-3; Z_s=(400*(-%i*200))/(400-%i*200); // from figure 8.10(a) V_s_m=abs(Z_s)*I_s_m; // From figure 8.10(b),load impedance referred to the primary // Turn ratio N=sqrt(500/80); // from condition of impedance matching L=(160*N^2)/omega;// from condition of impedance matching P_max=(V_s_m/sqrt(2))^2/(4*real(Z_s)); // Load reactance will be X_L=%i*omega*L; disp(X_L,"Load reactance for maximum power transfer(Ohms)=") disp(N,"Turn ratio for maximum power transfer=") disp(P_max,"Maximum power transferred(Watts)=")
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//ques1 //Compression of High-Speed Air in an Aircraft clear clc //(a) the stagnation pressure at the compressor inlet (diffuser exit) can be determined from Eq. 17–5 in book //state 1 T1=255.7;//Temperature in K V1=250;//velocity in m/s Cp=1.005;//specifc heat at const pressure in kJ/kg/K T01=T1+V1^2/(2*Cp)/1000;//divide 1000 to convert it into K //now from eqn 17-5 P1=54.05;//pressure in kPa k=1.4; P01=P1*(T01/T1)^(k/(k-1)); printf('(a) Pressure P01 = %.2f kPa \n',P01); //(b) To determine the compressor work r=8//P02/P01 T02=T01*r^(1-1/k);//Temperature in K //Disregarding potential energy changes and heat transfer, the compressor work per unit mass of air is determined from Eq. 17–8 Win=Cp*(T02-T01);//Work input in kJ/kg printf(' (b) Work input = %.1f kJ/kg \n',Win);
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errcatch(-1,"stop");mode(2);//Caption:Find the efficiency of transformer at full load and 75 percent of full load (a)at unity power factor (b)0.8 power factor //Exa:3.16 ; ; I_2=200;//in amperes R_O2=0.008;//in ohms x=0.75; P_cu=x^2*I_2^2*R_O2;//in watts P_i=190;//in watts KVA=40; P_o=40*1000;//in watts Eff=P_o/(P_o+P_i+I_2^2*R_O2); Eff_2=KVA*1000*0.8/(KVA*1000*0.8+P_i+I_2^2*R_O2); disp(Eff*100,'(a)Efficiency of transformer at full load and at unity power factor (in %)='); disp(Eff_2*100,'(b)Efficiency of transformer at full load and at 0.8 power factor (in %)='); Eff_3=x*P_o/(x*P_o+P_i+P_cu); Eff_4=x*P_o*0.8/(x*0.8*P_o+P_i+P_cu); disp(Eff_3*100,'(a)Efficiency of transformer at 75% of load and at unity power factor (in %)='); disp(Eff_4*100,'(b)Efficiency of transformer at 75% of load and at 0.8 power factor (in %)='); exit();
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//Chapter-5,Example5_3_12,pg 5-12 h=6.63*10^-34 //Plancks constant m=6.68*10^-27 //mass of alpha particle E=1.6*10^-16 //energy asociated with alpha particle wavelength=h/sqrt(2*m*E) printf("\nThe de Broglie wavelength of an alpha particle\n") disp(wavelength) printf("meter\n") v=h/(m*wavelength) //velocity of an alpha particle printf("\nThe velocity of an alpha particle v = %.2f m/s\n",v)
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clc clear //input data Q=0.015//Discharge of water in pump in m^3/s D1=0.2//Internal diameter of the impeller in m D2=0.4//External diameter of the impeller in m b1=0.016//Width of impeller at inlet in m b2=0.008//Width of impeller at outlet in m N=1200//Running speed of the pump in rpm b22=30//Impeller vane angle at outlet in degree g=9.81//Acceleration due to gravity in m/s^2 d=1000//Density of water in kg/m^3 //calculations printf('From velocity triangles the following values have been deduced') a11=90//The absolute water angle at inlet in degree Cx1=0//Absolute whirl component at inlet in m/s A1=3.1415*D1*b1//Area of flow at inlet in m^2 Cr1=Q/A1//Flow velocity through impeller at inlet in m/s C1=Cr1//Absolute velocity at inlet in m/s A2=3.1415*D2*b2//Area of flow at outlet in m^2 Cr2=Q/A2//Flow velocity through impeller at outlet in m/s U2=(3.1415*D2*N)/60//Blade outlet speed in m/s Cx2=U2-(Cr2/tand(b22))//Absolute whirl component at outlet in m/s C2=(Cx2^2+Cr2^2)^(1/2)//Velocity at impeller exit in m/s Ihl=((Cx2*U2)/g)-((C2^2)/(2*g))+((C1^2)/(2*g))//Pressure rise in impeller in m //output printf('\n\nThe rise in pressure in the impeller is %3.3f m',Ihl)
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Example_9_4.sce
clc clear printf('Solve the equation y''+y=3X^2, with boundary points (0,0) and (2, 3.5)') printf('\n Subdivide into seven elements that join at x=0.4, 0.7, 1.1, 1.3 and 1.6') printf('\n Augmented matrix') P=[2.367 -2.567 0 0 0 0 0 0 -0.024 -2.567 5.6 -3.383 0 0 0 0 0 -0.160 0 -3.383 8.167 -5.033 0 0 0 0 -0.328 0 0 -5.033 9.867 -5.033 0 0 0 -0.492 0 0 0 -5.033 9.867 -5.033 0 0 -0.732 0 0 0 0 -5.033 8.167 -3.383 0 -1.378 0 0 0 0 0 -3.383 5.600 -2.567 -2.89 0 0 0 0 0 0 -2.567 2.367 -1.944] disp(P) printf('Matrix after ajjusting boundary condition') T=[ 5.6 -3.383 0 0 0 0 -0.160 -3.383 8.167 -5.033 0 0 0 -0.328 0 -5.033 9.867 -5.033 0 0 -0.492 0 0 -5.033 9.867 -5.033 0 -0.732 0 0 0 -5.033 8.167 -3.383 -1.378 0 0 0 0 -3.383 5.600 6.094] disp(T) A=[ 5.6 -3.383 0 0 0 0 -3.383 8.167 -5.033 0 0 0 0 -5.033 9.867 -5.033 0 0 0 0 -5.033 9.867 -5.033 0 0 0 0 -5.033 8.167 -3.383 0 0 0 0 -3.383 5.600 ] B=[-0.160 -0.328 -0.492 -0.732 -1.378 6.094]' S=A\B printf('The table showing the analytical solution and the errors of our computation') X=[0.4 0.7 0.9 1.1 1.3 1.6] for i=1:6 B(1,i)=6*cos(X(1,i))+3*(X(1,i).*X(1,i)-2) T=[X(1,i), S(i,1), B(1,i), B(1,i)-S(i,1)] disp(T) end
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clc clear //Initialization of variables p1=2.758 //Mpa p2=0.552 //Mpa T1=700 //K T2=700 //K n=1 R=8.3143 Cv=21 Cp=29.3 //calculations dsa=n*R*log(p1/p2) T3=437.5 //K dsb=Cv*log(T3/T2) T4=350 //K dsc=Cp*log(T4/T3) T5=634 //K dsd=0 T6=700 //K dse=Cp*log(T6/T5) dstotal=dsa+dsb+dsc+dsd+dse //results printf("Entropy change in case a = %.3f kJ/kmol K",dsa) printf("\n Entropy change in case b = %.3f kJ/kmol K",dsb) printf("\n Entropy change in case c = %.3f kJ/kmol K",dsc) printf("\n Entropy change in case d = %.3f kJ/kmol K",dsd) printf("\n Entropy change in case e = %.3f kJ/kmol K",dse) printf("\n Entropy change in total process = %.3f kJ/kmol K",dstotal)
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clc disp("Example 3.7") printf("\n") printf("Given") disp("Total resistance of three resistors is 50 ohm") R=50; disp("Output voltage is 10 percent of the input voltage") //Let v be input voltage and v1 be output voltage //Let v1/v=V V=0.1; //As V=R1/(Total resistance) //Solving for R1 R1=V*R; //As R=R1+R2 //Solving for R2 R2=R-R1; printf("R1=%dohm\n R2=%dohm\n",R1,R2)
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//Example 2-46// //Conversion to decimal// clc //clears the console// clear //clears all existing variables// p=1 q=1 z=0 b=0 w=0 f=0 //initialising// //bin=input (enter the binary number to be converted to its decimal form)// //accepting the input from the user// bin=1101.000101 d=modulo(bin,1) //separating the decimal part from the integer part// d=d*10^10 a=floor(bin) //removing the decimal part// while(a>0) //loop to enter the binary bits of the integer part into a matrix// r=modulo(a,10) b(1,q)=r a=a/10 a=floor(a) q=q+1 end for m=1: q-1 //multiplying each bit of the integer part with its corresponding positional value and adding// c=m-1 f=f+b(1,m)*(2^c) end while(d>0) //loop to take the bits of the decimal part into a matrix// e=modulo(d,2) w(1,p)=e d=d/10 d=floor(d) p=p+1 end for n=1: p-1 //multiplying each bit with its corresponding positional value and adding// z=z+w(1,n)*(0.5)^(11-n) end z=z*10000 z=round(z) //rounding off to 4 decimal places// z=z/10000 x=f+z disp('the decimal equivalent of the binary number is :') disp(x) //result is displayed//
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function distance = euclidianDistance(instance1, instance2, len) distance = 0 for x = 1:len distance = distance + (instance1(x) - instance2(x))^2 end distance = sqrt(distance) endfunction function sortedTable = msort(table, sortBy, direction) len = size(table, 1) sortedTable = [] [a, h] = gsort(table(:, sortBy), 'r', direction) // sorted distances for x = 1:len sortedTable = cat(1, sortedTable, table(h(x), :)) end endfunction function neighbors = getNeighbors(trainingSet, testInstance, k) distances = [] len = length(testInstance)-1 // type less instance for x = 1:size(trainingSet, 1) dist = euclidianDistance(testInstance, trainingSet(x,:), len) distances = cat(1, distances, [trainingSet(x,:), dist]) end distances = msort(distances, $, 'i') neighbors = [] for x = 1:k neighbors = cat(1, neighbors, distances(x,1:3)) end endfunction function response = classify(neighbors) len = size(neighbors, 1) votes = zeros(1, len) for x = 1:len guess = neighbors(x, $) votes(guess) = votes(guess) + 1 end [amount, index] = max(votes) response = index endfunction function accuracy = getAccuracy(testSet, predictions) correct = 0 len = size(testSet, 1) for x = 1:len if testSet(x, $) == predictions(x) then correct = correct + 1; end end accuracy = (correct/len) * 100 endfunction function [predictions, wrong] = knn(trainingSet, testSet, k) predictions = [] wrong = [] for x = 1:size(testSet, 1) neighbors = getNeighbors(trainingSet, testSet(x,:), k) result = classify(neighbors) predictions = cat(1, predictions, result) disp('> predicted='+string(result)+', actual='+string(testSet(x,$))) if testSet(x, $) <> predictions(x) then wrong = cat(1, wrong, testSet(x, 1:2)) end end accuracy = getAccuracy(testSet, predictions) disp('Accuracy='+string(accuracy)+'%') endfunction
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// Scilab code Ex3.8 : Pg:108 (2008) clc;clear; mu_v = 1.5230; // Refractive index of violet color mu_r = 1.5145; // Refractive index of red color R1 = 40; // Radius of curvature of first curvature of lens, cm R2 = -10; // Radius of curvature of second curvature of lens, cm // As 1/f_r = (mu_r - 1)*(1/R1 - 1/R2), solving for f_r f_r = 1/((mu_r-1)*(1/R1 - 1/R2)); // Focal length for red color, cm f_v = 1/((mu_v-1)*(1/R1 - 1/R2)); // Focal length for violet color, cm CA = f_r - f_v; // The longitudinal chromatic abberation, cm printf("\nThe longitudinal chromatic abberation for the object at infinity = %5.3f cm", CA); // Result // The longitudinal chromatic abberation for the object at infinity = 0.253 cm
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n=6;m=5; L=([1:n]')*ones(1,m); // row number C=ones(n,1)*[1:m]; // column number D=1+abs(L-C) // diagonal index
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h=0.35;//location of center of gravity from leading edge Hn=0.516;//Neutral point location
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(set-strategy depth) (unwatch all) ; misclns1.bat test (clear) (open "misclns1.rsl" misclns1 "w") (dribble-on "misclns1.out") (batch "misclns1.bat") (dribble-off) (load "compline.clp") (printout misclns1 "misclns1.bat differences are as follows:" crlf) (compare-files misclns1.exp misclns1.out misclns1) ; close result file (close misclns1)
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//chapter 20 //example 20.4 //page 649 clear all; clc ; //given Ip=1;//mA Vp=65;//mV Id=1;//mA Vd=65;//mV Iv=0.12;//mA Vv=350;//mV Vf=500;//mV Id=[0 Id Iv Iv Ip] Ed=[0 Vd Vv 450 Vf]; plot(Ed,Id,'-.*'); xtitle('piecewise linear characteristics','Ed in mV','Id in mA'); Rd=-(350-65)/(1-0.12); printf("\nValue of RD=%d ohm",round(Rd));
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@relation vehicle @attribute COMPACTNESS integer[73,119] @attribute CIRCULARITY integer[33,59] @attribute DISTANCECIRCULARITY integer[40,112] @attribute RADIUSRATIO integer[104,333] @attribute PRAXISASPECTRATIO integer[47,138] @attribute MAXLENGTHASPECTRATIO integer[2,55] @attribute SCATTERRATIO integer[112,265] @attribute ELONGATEDNESS integer[26,61] @attribute PRAXISRECTANGULAR integer[17,29] @attribute LENGTHRECTANGULAR integer[118,188] @attribute MAJORVARIANCE integer[130,320] @attribute MINORVARIANCE integer[184,1018] @attribute GYRATIONRADIUS integer[109,268] @attribute MAJORSKEWNESS integer[59,135] @attribute MINORSKEWNESS integer[0,22] @attribute MINORKURTOSIS integer[0,41] @attribute MAJORKURTOSIS integer[176,206] @attribute HOLLOWSRATIO integer[181,211] @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 bus bus van van van bus saab saab van van van van bus bus saab bus bus saab bus saab bus bus van van saab saab saab opel bus bus opel saab saab van saab saab van van bus saab saab saab van bus van van bus bus van van opel bus bus bus van van saab saab bus bus bus opel bus bus van van opel opel saab opel saab opel saab saab saab opel opel saab opel opel bus bus bus bus bus bus opel saab saab opel saab opel opel saab van van opel opel opel saab saab saab van van van van bus bus van van bus bus saab saab bus bus opel opel opel saab opel saab opel opel saab van saab opel bus opel opel saab opel saab opel saab bus bus saab saab van van opel saab van bus opel saab van van bus bus saab opel saab opel bus bus van van opel saab opel saab opel opel
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//ques3 //Entropy Change of a Substance in a Tank clear clc //specific volume remains constant during this process //state 1 P1=140//initial pressure in kPa T1=20 //initial temperature in C s1=1.0624//entropy in kJ/Kg.K from table v1=0.16544//specific volume in m^3/Kg //state 2 P2=100//pressure in kPa v2=0.16544//specific volume remains same ie v2=v1 //from table vf=0.0007259//specific volume of saturated water in m^3/kg vg=0.19254//specific volume of saturated vapor in m^3/kg //Final state-saturated liquid–vapor mixture x2=(v2-vf)/(vg-vf);//x-factor sf=0.07188//entropy of saturated water in kJ/Kg.K sfg=0.87995//entropy change in kJ/kg.K s2=sf+x2*sfg;//entropy at state 2 in kJ/kg.K m=5//mass in Kg S=m*(s2-s1);//entropy change in process in kJ printf('Entropy change = %.3f kJ',S);
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// Electric Machinery and Transformers // Irving L kosow // Prentice Hall of India // 2nd editiom // Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY // Example 13-5 clear; clc; close; // Clear the work space and console. // Given data P_o = 50 ; // Power rating of the WRIM in hp // Class F insulation T_hottest = 160 ; // Hottest-spot winding temperature recorded by the embedded // hot-spot detectors in degree celsius T_ambient = 40 ; // Standard ambient temperature recorded by the embedded // hot-spot detectors in degree celsius P_f_a = 40 ; // Power rating of load a in hp P_f_b = 55 ; // Power rating of load a in hp // Calculations // case a delta_T_o = T_hottest - T_ambient ; // Temperature rise for the insulation type // used in degree celsius // subscript a in delta_T_f ,P_f_a and T_f indicates case a delta_T_f_a = (P_f_a/P_o)*delta_T_o ; // final temperature rise in degree celsius T_f_a = delta_T_f_a + T_ambient ; // Approximate final hot-spot temperature in degree celsius // case b // subscript b in delta_T_f ,P_f and T_f indicates case b delta_T_f_b = (P_f_b/P_o)*delta_T_o ; // final temperature rise in degree celsius T_f_b = delta_T_f_b + T_ambient ; // Approximate final hot-spot temperature in degree celsius // Display the results disp("Example 13-5 Solution : "); printf(" \n a: ΔT_o = %d degree celsius ",delta_T_o); printf(" \n ΔT_f = %d degree celsius ",delta_T_f_a); printf(" \n T_f = %d degree celsius \n",T_f_a); printf(" \n b: ΔT_f = %d degree celsius ",delta_T_f_b); printf(" \n T_f = %d degree celsius \n",T_f_b); printf(" \n Yes,motor life is reduced at the 110 percent motor load because"); printf(" \n the allowable maximum hot-spot motor temperature for Class F"); printf(" \n insulation is 155 degree celsius.");
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pathname=get_absolute_file_path('17_13.sce') filename=pathname+filesep()+'17_13_data.sci' exec(filename) //Brake power(in kW) bp=(W*g*N*C)/60000 //Brake thermal efficiency nbth=(bp*60)/(mf*CV) //Indicated thermal effciency nith=(nbth/nm)*100 //Indicated mean effective pressure(in bar) imep=(bp*60000)/(nm*L*0.25*%pi*D^2*0.5*N*K)*10^-5 //Brake specific fuel consumption(in kg/kWh) bsfc=(mf*60)/bp printf("\n\nRESULTS\n\n") printf("\nBrake thermal effciency:%f\n",nbth*100) printf("\nIndicated thermal effciency:%f\n",nith) printf("\nIndicated mean effective pressure:%f\n",imep) printf("\nBrake specific fuel consumption:%f\n",bsfc)
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//EXAMPLE 27.4 //4-POLE GENERATOR clc; funcprot(0); //Variable Initialisation P=4;...........//Total number of poles Z=492;..........//Total number of conductors Ia=143+10;...........//Armature current in Amperes(Current supplied + Shunt field current) b=10;...........//Actual brush lead in degrees Al=P;............//Number of parallel paths in a lap wound generator Aw=2;...........//Number of parallel paths in a wave wound generator I1=Ia/Al;.....//Current per path for lap wound in Amperes I2 =Ia/Aw;.....//Current per path for wave wound in Amperes ATd1=Z*I1*b/360;.......//Demagnetizing ampere-turns per pole for lap wound y=round(ATd1);.....//Rounding of decimal places sft1=y/b;........//Extra shunt field turns for lap wound r=round(sft1);......//Rounding of decimal places ATd2=Z*I2*b/360;.....//Cross-magnetizinfampere-turns per pole for wave wound y1=round(ATd2);......//Rounding of decimal places sft2=y1/b;..........//Extra shunt field turns for wave wound r1=round(sft2);.......//Rounding of decimal places disp(y,"Demagnetizing ampere-turns per pole for lap wound:"); disp(r,"Extra shunt field turns for lap wound:"); disp(y1,"Demagnetizing ampere-turns per pole for wave wound:"); disp(r1,"Extra shunt field turns for wave wound:");
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//Exa 3.6 clc; clear; close; //given data : //l=lambda/15 meter //Assume %pi^2 = 10 Rl=2;//in Ohm //Gain : Gain=5.33/4;//Unitless //Directivity Rr=80*10*(1/15)^2;//in Ohm ETA=Rr/(Rr+Rl);//Unitless Directivity=Gain/ETA;//unitless //Beam solid angle BSA=4*%pi/Directivity;//in steradian disp(Directivity,"Directivity : "); disp(Gain,"Gain = Pt/Pr = "); //Effective aperture disp("Effective aperture = G*lambda^2/(4*%pi) "); disp(string(Gain/(4*%pi))+"lambda^2"); disp(BSA,"Beam Solid Angle in steradian : "); disp("Radiation Resistance :") disp("Rr=80*%pi^2*(dl/lambda)^2 in Ohm"); disp("dl/lambda = 1/15 : as l=lambda/10 "); Rr=80*10*(1/15)^2;//in Ohm disp(Rr,"Radiation Resistance in Ohm : "); disp("Pt = Area of sphere * (E^2/(120*%pi))"); disp("Pt = ((4*%pi^2)/(120*%pi))*((60*%pi*I/r)*(dl/lambda)^2)"); disp("Pt=120*%pi^2*(lambda*15/lambda)*I^2"); disp("Pt = "+string(120*10/225)+"I^2"); disp("Pr = I^2*Rr = 4*I^2");
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THE OPTIMIZATION ALGORITHM HAS CHANGED TO THE EM ALGORITHM. ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 0.446554D+00 2 -0.603996D-02 0.363568D-02 3 0.837217D-01 -0.372349D-03 0.442359D+00 4 -0.622759D-03 0.793500D-03 -0.615097D-02 0.378185D-02 5 -0.305237D-03 -0.246094D-04 -0.170908D-02 0.134610D-03 0.205052D-02 6 -0.108578D-02 0.537074D-04 -0.364014D-03 0.630398D-04 -0.223168D-03 7 -0.889490D-03 0.473331D-04 0.896038D-03 0.637948D-04 0.376215D-04 8 -0.627418D-03 0.452553D-04 -0.464877D-03 0.347503D-04 -0.878718D-04 9 0.325154D+00 -0.385277D-01 -0.590432D+00 0.396028D-01 0.422219D-01 10 0.133200D+00 -0.351990D-02 0.474724D-01 0.744427D-02 0.120970D+00 11 0.255654D+00 -0.629146D-02 0.346275D+00 0.984033D-02 -0.381709D-01 12 0.735149D+00 0.637678D-02 0.687794D+00 0.256038D-01 0.116853D-01 13 -0.671869D-01 0.108680D-02 -0.125609D+00 0.128068D-01 0.243475D-01 14 0.829790D-01 0.143607D-01 0.115582D+00 0.128044D-02 -0.214376D-01 15 0.940914D+00 0.623077D-01 -0.126411D+01 -0.267787D-02 -0.685590D-01 16 0.594936D-01 0.429654D-03 -0.286092D-01 0.277099D-02 0.993696D-05 17 -0.692297D-02 0.392527D-03 0.112269D-01 0.458140D-04 -0.719068D-03 18 -0.151737D+01 0.962871D-02 -0.554150D+00 0.168005D-02 0.178898D-01 19 -0.439703D-01 0.371198D-02 -0.142844D+00 0.798105D-02 0.648895D-02 20 -0.618679D+00 -0.277829D-01 0.187578D+01 0.573564D-02 -0.651139D-01 21 -0.205419D-01 0.312844D-02 0.193483D+00 -0.121943D-01 -0.782754D-02 22 0.125365D-01 -0.420259D-04 0.698571D-03 0.520523D-03 -0.247123D-03 23 0.191147D-01 -0.169779D-03 -0.795397D-02 0.113609D-01 -0.934878D-03 24 0.370882D-03 0.455086D-03 0.189536D-02 -0.331469D-03 0.496257D-03 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 0.795916D-03 7 0.570980D-03 0.259026D-02 8 0.279169D-03 -0.152490D-03 0.298008D-02 9 -0.242087D-02 0.441478D-01 0.228235D-01 0.984704D+02 10 -0.794684D-03 0.603489D-02 0.628557D-02 0.986879D+00 0.257852D+02 11 0.168744D-01 0.270577D-01 0.366372D-01 0.180892D+02 -0.214298D+01 12 -0.297580D-01 0.926513D-02 0.892463D-01 0.724764D+01 0.469692D+01 13 0.629330D-01 0.113286D+00 0.391062D-01 0.211233D+01 0.373333D+01 14 0.485139D-01 0.420116D-01 0.372821D+00 0.257675D+01 0.579227D+01 15 -0.770249D-01 -0.502082D-01 0.575960D-01 -0.109193D+02 -0.966543D+01 16 -0.191756D-02 -0.167917D-02 0.698638D-04 0.667973D+00 -0.866122D-01 17 0.454282D-03 0.165975D-04 0.861438D-04 -0.198093D+00 -0.908065D-01 18 -0.565370D-01 -0.105162D+00 0.187626D-01 -0.604064D+01 -0.157424D+01 19 -0.103472D-01 0.820964D-02 0.483248D-02 0.134724D+01 0.445507D+00 20 -0.607430D-02 0.340376D-01 -0.330731D+00 -0.986809D+01 -0.439097D+00 21 0.848293D-02 -0.968132D-02 0.130298D-03 -0.574501D+00 -0.819533D+00 22 -0.580133D-04 -0.144477D-03 -0.697134D-03 -0.578047D-01 -0.104476D-01 23 0.566478D-03 0.264940D-02 -0.482727D-03 -0.579904D+00 -0.282075D-02 24 -0.282500D-03 -0.740634D-03 0.225061D-03 0.136321D+00 -0.528932D-02 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 0.608260D+02 12 0.233308D+02 0.169371D+03 13 -0.486318D+01 0.147377D+01 0.231822D+02 14 0.767926D+01 0.107678D+02 0.154113D+02 0.111028D+03 15 0.384778D+01 0.235240D+02 -0.102193D+02 0.759680D+00 0.524717D+03 16 -0.161516D+00 -0.278476D+00 -0.182740D+00 0.568621D-01 0.608309D+01 17 0.184308D-02 -0.531531D-01 0.405498D-01 0.226358D-01 -0.273133D+01 18 0.132742D+01 0.922296D+00 -0.991093D+01 -0.651085D+01 0.742367D+02 19 -0.204176D+01 0.534299D+00 0.255254D+00 0.196965D+01 -0.461994D+01 20 -0.137007D+01 -0.272456D+02 -0.103353D+02 -0.696184D+02 0.394933D+02 21 0.226858D+01 -0.630683D+00 -0.564888D+00 -0.140295D+01 0.568510D+01 22 -0.830013D-01 -0.206698D-01 -0.104559D-01 -0.109552D+00 -0.131256D+00 23 0.811688D+00 0.121186D+01 0.723904D-01 -0.274308D+00 0.194913D+00 24 -0.617498D-01 -0.137265D+00 -0.277725D-01 -0.301075D-01 -0.209093D+00 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 0.832346D+00 17 -0.760177D-01 0.306241D-01 18 0.100575D+01 -0.300308D+00 0.338718D+03 19 -0.960888D-01 0.520588D-01 -0.431900D+00 0.799625D+01 20 -0.200353D-01 -0.250478D+00 0.745457D+02 -0.823336D+01 0.557948D+03 21 0.444971D+00 -0.641802D-01 0.697041D+01 -0.737750D+01 0.932490D+01 22 -0.725739D-02 0.295562D-02 -0.165666D+01 -0.282041D-01 -0.276063D+00 23 0.931465D-01 -0.971502D-02 -0.170718D+00 -0.364094D+00 0.605966D+01 24 -0.689998D-02 0.281357D-02 -0.133769D+00 0.544085D-01 -0.250155D+01 ESTIMATED COVARIANCE MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 0.918836D+01 22 -0.837541D-01 0.189302D-01 23 0.709126D+00 -0.769323D-02 0.919708D+00 24 -0.753576D-01 0.325364D-02 -0.840903D-01 0.267396D-01 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 1 2 3 4 5 ________ ________ ________ ________ ________ 1 1.000 2 -0.150 1.000 3 0.188 -0.009 1.000 4 -0.015 0.214 -0.150 1.000 5 -0.010 -0.009 -0.057 0.048 1.000 6 -0.058 0.032 -0.019 0.036 -0.175 7 -0.026 0.015 0.026 0.020 0.016 8 -0.017 0.014 -0.013 0.010 -0.036 9 0.049 -0.064 -0.089 0.065 0.094 10 0.039 -0.011 0.014 0.024 0.526 11 0.049 -0.013 0.067 0.021 -0.108 12 0.085 0.008 0.079 0.032 0.020 13 -0.021 0.004 -0.039 0.043 0.112 14 0.012 0.023 0.016 0.002 -0.045 15 0.061 0.045 -0.083 -0.002 -0.066 16 0.098 0.008 -0.047 0.049 0.000 17 -0.059 0.037 0.096 0.004 -0.091 18 -0.123 0.009 -0.045 0.001 0.021 19 -0.023 0.022 -0.076 0.046 0.051 20 -0.039 -0.020 0.119 0.004 -0.061 21 -0.010 0.017 0.096 -0.065 -0.057 22 0.136 -0.005 0.008 0.062 -0.040 23 0.030 -0.003 -0.012 0.193 -0.022 24 0.003 0.046 0.017 -0.033 0.067 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 6 7 8 9 10 ________ ________ ________ ________ ________ 6 1.000 7 0.398 1.000 8 0.181 -0.055 1.000 9 -0.009 0.087 0.042 1.000 10 -0.006 0.023 0.023 0.020 1.000 11 0.077 0.068 0.086 0.234 -0.054 12 -0.081 0.014 0.126 0.056 0.071 13 0.463 0.462 0.149 0.044 0.153 14 0.163 0.078 0.648 0.025 0.108 15 -0.119 -0.043 0.046 -0.048 -0.083 16 -0.075 -0.036 0.001 0.074 -0.019 17 0.092 0.002 0.009 -0.114 -0.102 18 -0.109 -0.112 0.019 -0.033 -0.017 19 -0.130 0.057 0.031 0.048 0.031 20 -0.009 0.028 -0.256 -0.042 -0.004 21 0.099 -0.063 0.001 -0.019 -0.053 22 -0.015 -0.021 -0.093 -0.042 -0.015 23 0.021 0.054 -0.009 -0.061 -0.001 24 -0.061 -0.089 0.025 0.084 -0.006 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 11 12 13 14 15 ________ ________ ________ ________ ________ 11 1.000 12 0.230 1.000 13 -0.130 0.024 1.000 14 0.093 0.079 0.304 1.000 15 0.022 0.079 -0.093 0.003 1.000 16 -0.023 -0.023 -0.042 0.006 0.291 17 0.001 -0.023 0.048 0.012 -0.681 18 0.009 0.004 -0.112 -0.034 0.176 19 -0.093 0.015 0.019 0.066 -0.071 20 -0.007 -0.089 -0.091 -0.280 0.073 21 0.096 -0.016 -0.039 -0.044 0.082 22 -0.077 -0.012 -0.016 -0.076 -0.042 23 0.109 0.097 0.016 -0.027 0.009 24 -0.048 -0.065 -0.035 -0.017 -0.056 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 16 17 18 19 20 ________ ________ ________ ________ ________ 16 1.000 17 -0.476 1.000 18 0.060 -0.093 1.000 19 -0.037 0.105 -0.008 1.000 20 -0.001 -0.061 0.171 -0.123 1.000 21 0.161 -0.121 0.125 -0.861 0.130 22 -0.058 0.123 -0.654 -0.072 -0.085 23 0.106 -0.058 -0.010 -0.134 0.268 24 -0.046 0.098 -0.044 0.118 -0.648 ESTIMATED CORRELATION MATRIX FOR PARAMETER ESTIMATES 21 22 23 24 ________ ________ ________ ________ 21 1.000 22 -0.201 1.000 23 0.244 -0.058 1.000 24 -0.152 0.145 -0.536 1.000
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//Chapter 6 //Example 6.5 //page 210 //To find bus voltages and Reactive power injected using GS iterations clear;clc; //Ybus matrix from the network Ybus=[3-9*%i -2+6*%i -1+3*%i 0; -2+6*%i 3.666-11*%i -0.666+2*%i -1+3*%i -1+3*%i -0.666+2*%i 3.666-11*%i -2+6*%i 0 -1+3*%i -2+6*%i 3-9*%i] //Case(i) ////////////////////////////////////////////////////// //Pi Qi Vi Remarks Bus no// P1=0; Q1=0; V1=1.04; //Slack bus 1 P2=0.5; Q2=0.2; V2=1.04; //PVbus 2 P3=-1.0; Q3=0.5; V3=1; //PQbus 3 P4=0.3; Q4=-0.1; V4=1; //PQbus 4 ///////////////////////////////////////////////////// printf('\nCase(i) When 0.2<Q2<1 pu and running for 1 iteration,we get \n\n'); Q2min=0.2;Q2max=1; n=1; for i=1:n if Q2<Q2min then Q2=Q2min; V2=(1/Ybus(2,2))*(((P2-%i*Q2)/conj(V2))-Ybus(2,1)*V1-Ybus(2,3)*V3-Ybus(2,4)*V4); elseif Q2>Q2max then Q2=Q2max; V2=(1/Ybus(2,2))*(((P2-%i*Q2)/conj(V2))-Ybus(2,1)*V1-Ybus(2,3)*V3-Ybus(2,4)*V4); else Q2=-imag(conj(V2)*Ybus(2,1)*V1+conj(V2)*(Ybus(2,2)*V2+Ybus(2,3)*V3+Ybus(2,4)*V4)); [mag,delta2]=polar((1/Ybus(2,2))*(((P2-%i*Q2)/(conj(V2)))-Ybus(2,1)*V1-Ybus(2,3)*V3-Ybus(2,4)*V4)); V2=abs(V2)*(cos(delta2)+%i*sin(delta2)); end V3=(1/Ybus(3,3))*(((P3-%i*Q3)/conj(V3))-Ybus(3,1)*V1-Ybus(3,2)*V2-Ybus(3,4)*V4); V4=(1/Ybus(4,4))*(((P4-%i*Q4)/conj(V4))-Ybus(4,1)*V1-Ybus(4,2)*V2-Ybus(4,3)*V3); end printf('Q2=');disp(Q2);printf('pu'); printf('\n\n\ndelta2=');disp(abs(delta2));printf('rad'); printf('\n\n\nV1=');disp(V1);printf('pu'); printf('\n\n\nV2=');disp(V2);printf('pu'); printf('\n\n\nV3=');disp(V3);printf('pu'); printf('\n\n\nV4=');disp(V4);printf('pu'); // case(ii) printf('\n\n\nCase(ii) When 0.25<Q2<1 pu and running for 1 iteration,we get \n\n'); ////////////////////////////////////////////////////// //Pi Qi Vi Remarks Bus no// P1=0; Q1=0; V1=1.04; //Slack bus 1 P2=0.5; V2=1.04; //PVbus 2 P3=-1.0; Q3=0.5; V3=1; //PQbus 3 P4=0.3; Q4=-0.1; V4=1; //PQbus 4 ///////////////////////////////////////////////////// Q2min=0.25;Q2max=1; n=1; for i=1:n if Q2<Q2min then Q2=Q2min; V2=(1/Ybus(2,2))*(((P2-%i*Q2)/conj(V2))-Ybus(2,1)*V1-Ybus(2,3)*V3-Ybus(2,4)*V4); elseif Q2>Q2max then Q2=Q2max; V2=(1/Ybus(2,2))*(((P2-%i*Q2)/conj(V2))-Ybus(2,1)*V1-Ybus(2,3)*V3-Ybus(2,4)*V4); else Q2=-imag(conj(V2)*Ybus(2,1)*V1+conj(V2)*(Ybus(2,2)*V2+Ybus(2,3)*V3+Ybus(2,4)*V4)); [mag,delta2]=polar((1/Ybus(2,2))*(((P2-%i*Q2)/(conj(V2)))-Ybus(2,1)*V1-Ybus(2,3)*V3-Ybus(2,4)*V4)); V2=abs(V2)*(cos(delta2)+%i*sin(delta2)); end V3=(1/Ybus(3,3))*(((P3-%i*Q3)/conj(V3))-Ybus(3,1)*V1-Ybus(3,2)*V2-Ybus(3,4)*V4); V4=(1/Ybus(4,4))*(((P4-%i*Q4)/conj(V4))-Ybus(4,1)*V1-Ybus(4,2)*V2-Ybus(4,3)*V3); end printf('Q2=');disp(Q2);printf('pu'); printf('\n\n\nV1=');disp(V1);printf('pu'); printf('\n\n\nV2=');disp(V2);printf('pu'); printf('\n\n\nV3=');disp(V3);printf('pu'); printf('\n\n\nV4=');disp(V4);printf('pu');
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clc clear printf("Example 8.11 | Page number 224 \n\n"); //This is a theoritical question.Refer textbook for solution printf("This is a theoritical question.Refer textbook for solution")
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//developed in windows XP operating system 32bit //platform Scilab 5.4.1 clc;clear; //example 16.16w //calculation of the speed,wavelength in the rod,frequency,wavelength in the air //given data l=90*10^-2//length(in m) of the rod rho=2600//density(in kg/m^3) of the aluminium Y=7.80*10^10//Young modulus(in N/m^2) vai=340//speed(in m/s) of the sound in the air //calculation v=sqrt(Y/rho)//speed of the sound in aluminium lambda=2*l//wavelength....since rod vibrates with fundamental frequency nu=v/lambda//frequency lambdaai=vai/nu//wavelength in the air printf('the speed of the sound in aluminium is %d m/s',v)//Textbook Correction : correct answer is 5477 m/s printf('\nthe wavelength of the sound in aluminium rod is %d cm',lambda*10^2) printf('\nthe frequency of the sound produced is %d Hz',nu)//Textbook Correction : correct answer is 3042 Hz printf('\nthe wavelength of the sound in air is %3.1f cm',lambdaai*10^2)
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clc //initialisation of variables h11= 2786.2 //kJ/kg h12= 340.5 //kJ/kg h7= 327.9 //kJ/kg h6= 169.0 //kJ/kg h10= 756.7 //kJ/kg h9= 480.9 //kJkg h14= 2818 //kJ.kg h15= 762.8 //kJ/kg h8= 462.7 //kJ/kg h13= 2974.5 //kJ/kg h5= 168.8 //kJ/kg P= 150 //kW v1= 0.02293 //m^3/kg v= 40 //m/s h1= 3448.6 //kJ/kg h3= 3478.5 //kJ/kg h2= 2818 //kJ/kg h4= 2527.1 //kJ/kg //CALCULATIONS y1= (h10-h9)/(h14-h15) y2= ((h8-h7)-y1*(h15-h7))/(h13-h7) y3= (h7-h6)*(1-y1-y2)/(h11-h12) qin= h1-h10+(1-y1)*(h3-h2) qout= (h5-h4)*(1-y1-y2)+y3*(h4-h12) wnet= qin+qout n= wnet*100/qin m1= P*1000/wnet A1= m1*v1/v D= sqrt(4*A1/%pi) //RESULTS printf (' quality= %.4f ',y1) printf (' \n quality= %.4f ',y2) printf (' \n quality= %.4f ',y3) printf (' \n efficiency= %.2f percent',n) printf (' \n mass flow rate= %.2f kg/s',m1) printf (' \n diameter= %.3f m',D)
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Auto Scaling - Self Healing/ Self Recovery Infrastructure - Auto Scaling Group (ASG) - Ability to monitor existing resource utilization and when certain threshold is breached, they are able to take action. - Implementation of elasticity from AWS. - Terminate EC2 if required. - Triggers - Upper limit - Lower Limit - Minimum Size (minimum contribution to target should be 1 instance) - Cloud Watch - monitoring capability which keeps an hawk eye on utiliztion and raises an alarm on which autoscaling group has to take action. - Custom AMI & Boot Strap Scripts controlledwhen Instance is added by Auto Scaling Group. - ASG - Launch Configuration or Template - these are used to spawn an EC2 instance and install required s/w to make it ready for ASG. - Launch Template - 5 teams and want to make sure Auto Scaing Guidance - we don't miss on mandatory parameters when Launch Template is being used. ASG - Network Interface option will not be used ASG and Lunch Template should not be looked in context of elasticity. It should also be very effective in maintaining the infrastructure. launch template is immutable. only option is copy template with new version. Launch Template is a regional resource and available only in a specific region. A single Launch Template can be used by multiple ASG. 1:Many Relationship. Configuration Dissodance/Configuration Drift - change in configuration in launch template than in production. On Demand is always prioritized. Min Size, Desired Capacity and Maximum Size (Artificial upper limit) where Desired Capacity >= Min Size. Instance Scale-in protection - * enable during my observation window. at no point, ASG will not remove any instance during my peak period. So, it will help me to justify my decision what is the max size of ASG. * No SqL Instance - Horizontal Scaling. That instance will hold some data. How can be arbitary remove the instance. * Instances where you want to avoid Scale-In by ASG. Target Tracking Scaling Policy. Denial of Service Attack Scaling Policy - Step scaling or simple scaling AMI are region centric. EFS - mounted across multiple EC2 instances. - in AZ, multiple subnets can exist. One Mount Target in any of the subnet for all EC2 instances in that particular AZ. - Mount Target will get an IP addres, EC2 has to open nfs port 2049 and atatch tht security group to EC2. - Unlike EBS, amout is charged with data stored and expand file system at point of time. - Lifecycle Management - moving the data to infrquently stored data. - Performance Modes - General Purpose & Max I/O :for big data, mny I/O operation, so mode is Max I/O. - Throughput Mode - Bursting or Provisioned. - Migration will be required if you want to change performance mode of EFS. - Credit Concept exist. - It is not a cross region service. S3 - Object Based storage. - 100 soft limit buckets/account. Globally unique names. - Minimum File Size is 0Bytes and Max 5 TB, chunk size is 5Gb for multi part upload. - Once enabled versioning, disabled can't be done, only suspend versioning is possible. - Content Creation (sunrise to sunset of content creation) Lifecycle of content - Current -> Reduced Redundancy -> Infrquently Accessed ->Archival (Glacier Service) -> Auto Delete the content - 99.999,999,999 (11 9s) durability. 10 copies of your data by default. - Multi Region Replication. - S3 source of data + CDN (Cloud Front) as cache. - Reverse Syncronisation (update Cache and document travels back to S3. Bi Directional Data flow) - Static Work of website goes to S3 which in front is CDN. If no dynamic content, then no application server required, no http server required. - Eventual Consistency (Strong Read after write for a new file but for a old file, it is eventual consistency) - Key Value (Key is file name) - Object Lock - Write Once Read Many (WORM) - Intelligent Tearing - Storage Classes - standard - replica >= 3 - infrequently accessed (not old enough to be archived) - min 30 days old - Standard -> One Zone ->Reduced Redundancy -> Intelligent Tiering -> Glacier -> Glacier Deep Archive - Transfer Acceleration (Data replicated. Customer can access and update document around the world) - Once we create replication rule, exisitng content in bucket will not be replicated over. - Delete Marker for soft delete of files deleted. Delete Marker + Delete => permanently deleting the file. Nice way of recovering the documents accidently deleted. - Deleting the deleting marker doesn't delete on replica. - CORS (Cross Origin Resource Sharing) - Storage Lens - View across all buckets , Matrix View - only for a bucket.
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// Example 2_1 clc;funcprot(0); //Given data R=6.2;//Rainfall in cm A=2346;// Area in km^2 //Calculation Tr=A*10^6*(R/100);// Total rainfall in m^2 V=(A*R*10^4)/86400;// Rainfall in day-sec-metre R_k=(A*R*10^4)/10^6;// Rainfall in km^2-m printf('\n Total rainfall=%0.4e m^3 \nVolume of rainfall=%0.0f day-sec-metre \nRainfall in km^2-m=%0.2f km^2-m',Tr,V,R_k); // The answer provided in the textbook is wrong
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// Display modeK mode(0); // Display warning for floating point exception ieee(1); clear; clc; disp("Engineering Thermodynamics by Onkar Singh Chapter 7 Example 13") T_min=(20+273);//minimum temperature reservoir temperature in K T_max=(500+273);//maximum temperature reservoir temperature in K n=0.25;//efficiency of heat engine disp("reversible engine efficiency,n_rev=1-(T_min/T_max)") n_rev=1-(T_min/T_max) disp("second law efficiency=n/n_rev") n/n_rev disp("in %") n*100/n_rev
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// Scilab code: Ex3.7: de-Broglie wavelength of electrons: Pg: 80 (2008) L_1 = 3e-07; // Wavelength of ultraviolet light, m L_0 = 4e-07; // Threshold wavelength of ultraviolet light, m m = 9.1e-031; // Mass of an electron, kg c = 3e+08; // Velocity of light, m/s h = 6.624e-034; // Plancks constant, joule-second U = h*c*(1/L_1-1/L_0); // Maximum Kinetic energy of emitted electrons, joule // since U = m*v^2/2, Kinetic energy of electrons, joule // Thus v = sqrt(2*U/m), so that L_2 becomes L_2 = h/sqrt(2*m*U); // wavelength of electrons, m printf("\nThe wavelength of the electrons = %3.1f angstorm", L_2/1e-010); // Result // The wavelength of the electrons = 12.1 angstorm
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clc h=1.05*10^-34 disp("h = "+string(h)+"Js") //initializing value of reduced plancks constant or dirac constant or h-bar mo = 9.1*10^-31 disp("mo = "+string(mo)+"kg") //initializing value of mass of electron me = 0.067*mo disp("me* = "+string(me)+"kg") //initializing value of effective mass of GaAs kbT = 4.16*10^-21 disp("kbT = "+string(kbT)+"J/K") //initializing value of kbT at 300K Nc=2*(((me*kbT)/(2*%pi*(h^2)))^(3/2)) disp("for GaAs conduction band case effective density of states is ,Nc = 2*(((me*kbT)/(2*%pi*(h^2)))^(3/2)) = "+string(Nc)+"m^-3")//calculation ml = 0.98*mo disp("ml* = "+string(ml)+"kg") //initializing value of longitudinal mass mt = 0.19*mo disp("mt*= "+string(mt)+"kg")//initializing value of transverse mass mdos = (((6)^(2/3))*((ml)*((mt)^2))^(1/3)) disp("The conduction band density of states mass is (mdos* = (((6)^(2/3))*((ml*)*((mt*)^2))^(1/3)))= "+string(mdos)+"kg")//calculation Nc1 = 2*((mdos*kbT)/(2*(%pi)*(h^2)))^(3/2) disp("for silicon conduction band case effective density of states is ,Nc = 2*((mdos*kbT)/(2*(%pi)*(h^2)))^(3/2) = "+string(Nc1)+"m^-3")//calculation // Note : due to different precisions taken by me and the author ... my answer differ disp(" for silicon ") mhh =0.5*mo disp("mhh* = "+string(mhh)+"kg") //initializing value of heavy hole mass for silicon mlh = 0.15*mo disp("mlh*= "+string(mlh)+"kg")//initializing value of light hole mass for silicon Nv1 =((kbT/(2*(%pi)*(h^2)))^(3/2))*2*(mhh^(3/2)+mlh^(3/2)) disp("for silicon valence band case effective density of states is ,Nv = 2*(mhh^(3/2)+mlh^(3/2))*(kbT/(2*(%pi)*(h^2)))^(3/2)= "+string(Nv1)+"m^-3")//calculation disp("for GaAs ") mhh1 =0.45*mo disp("mhh* = "+string(mhh1)+"kg") //initializing value of heavy hole mass mlh1 = 0.08*mo disp("mlh*= "+string(mlh1)+"kg")//initializing value of light hole mass Nv = 2*(mhh1^(3/2)+mlh1^(3/2))*((kbT/(2*(%pi)*(h^2)))^(3/2)) disp("for GaAs valence band case effective density of states is ,Nv = 2*(mhh1^(3/2)+mlh1^(3/2))*(kbT/(2*(%pi)*(h^2)))^(3/2)= "+string(Nv)+"m^-3")//calculation // Answer given in the book for valence band case is wrong
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clc //initialization of variables clear s=3 //m n=60 p=50 //kg // calculations W=n*p Rc=W*2/s Rb=W-Rc dx = 0.001; x = 0:dx:s n = s/dx +1; for i = 1:n Sx(i) = -Rb + Rc*x(i)^2/6; Mx(i) = Rb*x(i) - Rc*x(i)^3 /18; end //Results figure(1);plot(x,Sx);title("Shear force diagram");xlabel("X (in m)");ylabel("Shear force (in kg)"); figure(2);plot(x,Mx);title("Bending Moment diagram");xlabel("X (in m)");ylabel("Bending Moment (in kg-m)");
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clc //to find location of second fragment // GIVEN:: //refer to figure 7-11 fron page no. 145 //consider +ve x direction as our reference axis //mass of projectile M = 9.6//in kg //initial velocity of projectile v0 = 12.4//in m/s //angle of projectile above horizontal fi0 = 54//in degrees //mass of first piece after explosion m1 = 6.5//in kg //time after which first piece id observed t = 1.42//in seconds //vertical distance at which first piece is observed y1 = 5.9//in meters //horizontal distance at which first piece is observed x1 = 13.6//in meters //acceleration due to gravity g = 9.80//in m/s^2 // SOLUTION: //refer to figure 7-11 from page no. 145 //mass of second piece //by mass conservation m2 = M-m1//in kg //velocity of projectile in +ve x direction v0x = v0*cosd(fi0)//in m/s //velocity of projectile in +ve y direction v0y = v0*sind(fi0)//in m/s //using kinematic equation of motion //x coordinate of position of original projectile x = v0x*t//in m //y coordinate of position of original projectile y = (v0y*t)-(0.5*g*t^2)//in m //applying center of mass formula //x coordinate of posion of second piece x2 = (M*x - m1*x1)/m2//in meters //y coordinate of posion of second piece y2 = (M*y - m1*y1)/m2//in meters x = nearfloat("succ",10.4) y = nearfloat("pred",4.3) x2 = nearfloat("succ",3.7) y2 = nearfloat("pred",0.9) printf ("\n\n x coordinate of position of original projectile x = \n\n %.1f m",x); printf ("\n\n y coordinate of position of original projectile y = \n\n %.1f m",y); printf ("\n\n x coordinate of posion of second piece x2 = \n\n %.1f m",x2); printf ("\n\n y coordinate of posion of second piece y2 = \n\n %.1f m",y2);
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//scilab 5.4.1 clear; clc; printf("\t\t\tProblem Number 5.26\n\n\n"); // Chapter 5 : Properties Of Liquids And Gases // Problem 5.26 (page no. 212) // Solution //The Mollier chart has lines of constant moisture in the wet region which correspond to (1-x).Therefore,we read at 20% moisture(80% Quality) and 120 psia, printf("The enthalpy of a wet steam mixture at 120 psia having quality 80 percent is 1015 Btu/lbm\n"); //Which also agrees well with the calculated value in problem 5.7
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clear; clc; //Example 4.3 //Caption : Program To Find the Final Temperature with Heat Given function[Q]=MCPH(T0,T,A,B,C,D) t=T/T0; Q=(A+((B/2)*T0*(t+1))+((C/3)*T0*T0*((t^2)+t+1))+(D/(t*T0*T0))) funcprot(0); endfunction //Given values for Ammonia R=8.314; T0=533.15; A=3.578; B=3.020*(10^-3); C=0; D=-0.186*(10^5); Q=422*(10^3); n=11.3; del_H=Q/n; //Solution i=-1; a=round(T0);//Initial while (i==-1) b=R*MCPH(T0,a,A,B,C,D); c=b*(a-T0); flag=del_H-c; if(flag<=100) then T=a-1; i=1; else a=a+1; i=-1; end end disp('K',T,'Temperature Required(Approx)') //End
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clc I=0.62; //kg/m^2 N1=2500; //rpm w1=2*%pi*N1/60; //rad/s m=1.9; //kg; Water equivalent of shaft bearings cp=4.18; T0=293; //K t0=20; //0C disp("(i)Rise in temperature of bearings") KE=1/2*I*w1^2/1000; //kJ dT=KE/(m*cp); //rise in temperature of bearings disp("dT=") disp(dT) disp("0C") t2=t0+dT; disp("Final temperature of the bearings =") disp(t2) disp("0C") T2=t2+273; disp("(ii)Final r.p.m. of the flywheel") AE=integrate('m*cp*(1-T0/T)', 'T', T0, T2); UE=KE - AE; disp("Available energy =") disp(AE) disp("kJ") UAE=KE-AE; disp("Unavailable energy =") disp(UAE) disp("kJ") w2=sqrt(AE*10^3*2/I); N2=w2*60/2/%pi; disp("Final rpm of the flywheel =") disp(N2) disp("rpm")