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//Chapter 14 //Example 14_8 //Page 369 clear;clc; v=400; vl=230; ia=70; ib=84; ic=33; im=200; pf=0.2; //part 1 printf("LAMP LOAD ALONE: \n"); //Refering to the phasor diagram in the book hc=ib*cos(30*%pi/180)-ic*cos(30*%pi/180); vc=ia-ib*cos(60*%pi/180)-ic*cos(60*%pi/180); in=sqrt(hc^2+vc^2); printf("Resultant horizontal component = %.2f A \n", hc); printf("Resultant vertical component = %.2f A \n", vc); printf("Neutral component = %.2f A \n\n", in); //part 2 printf("BOTH LAMP AND MOTOR LOAD: \n"); ac=im*pf; rc=im*sin(acos(pf)); Ir=sqrt((ac+ia)^2+rc^2); Iy=sqrt((ac+ib)^2+rc^2); Ib=sqrt((ac+ic)^2+rc^2); printf("Nuetral current remains the same, ie In = %.2f A \n", in); printf("Active component of motor current = %.0f A \n", ac); printf("Reactive component of motor current = %.0f A \n", rc); printf("\t Ir = %.2f A \n", Ir); printf("\t Iy = %.2f A \n", Iy); printf("\t Ib = %.2f A \n\n", Ib); //part 3 printf("POWER SUPPLIED: \n"); pl=vl*(ia+ib+ic); pm=sqrt(3)*v*im*pf; printf("Power supplied to lamps = %.0f W \n", pl); printf("Power supplied to motor = %.0f W \n", pm);
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disp("Example 4.10") disp("Grade of Steel,fy = Fe250","Grade of Concrete,fck = M20","D=600mm","d=550mm","b=300mm","Bars used = 4 - 25 dia") b=300 d=550 D=600 fck=20 Ast=%pi*4*25*25/4 disp("mm^2",Ast,"Ast=") disp("For Fe415 Steel,") Es=2*10^5 fy=250 Est=0.87*fy/Es xumaxd=(0.0035/(0.0055+Est)) disp(xumaxd,"xumax/d") xumax=xumaxd*d disp("mm",xumax,"xu,max=") disp("Assuming, xu</xu,max and applying the force equilibrium condition Cu=Tu") xu= (0.87*fy*Ast)/(0.362*fck*b) disp("mm",xu,"xu") disp("mm",xu,"xu<xu,max, therefore xu=")
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// A Texbook on POWER SYSTEM ENGINEERING // A.Chakrabarti, M.L.Soni, P.V.Gupta, U.S.Bhatnagar // DHANPAT RAI & Co. // SECOND EDITION // PART IV : UTILIZATION AND TRACTION // CHAPTER 2: HEATING AND WELDING // EXAMPLE : 2.4 : // Page number 728 clear ; clc ; close ; // Clear the work space and console // Given data w_brass = 1000.0 // Weight of brass(kg) time = 1.0 // Time(hour) heat_sp = 0.094 // Specific heat fusion = 40.0 // Latent heat of fusion(kcal/kg) T_initial = 24.0 // Initial temperature(°C) melt_point = 920.0 // Melting point of brass(°C) n = 0.65 // Efficiency // Calculations heat_req = w_brass*heat_sp*(melt_point-T_initial) // Heat required to raise the temperature(kcal) heat_mel = w_brass*fusion // Heat required for melting(kcal) heat_total = heat_req+heat_mel // Total heat required(kcal) energy = heat_total*1000*4.18/(10**3*3600*n) // Energy input(kWh) power = energy/time // Power(kW) // Results disp("PART IV - EXAMPLE : 2.4 : SOLUTION :-") printf("\nAmount of energy required to melt brass = %.f kWh", energy)
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// Exa 2.10 clc; clear; close; // Given data ni= 1.8*10^15;// in /m^3 rho= 2*10^5;// in Ωm q=1.6*10^-19;// in C dopingConcentration= 10^25;// in /m^3 n=dopingConcentration; MCC= ni^2/dopingConcentration; // Minority carrier concentration per cube meter miu_n= 1/(2*rho*q*ni);// in m^3/Vs disp(miu_n,"The value of µn in m^3/Vs is : ") // Part (b) sigma= q*n*miu_n;//in (Ωm)^-1 rho= 1/sigma;// in Ωm disp(rho,"Resistivity in Ωm is : ") // Part(c) kT= 26*10^-3;//in V no= n;// in /m^3 Shift_inFermiLevel= kT*log(no/ni);// in eV disp(Shift_inFermiLevel,"Shift in Fermi level due to doping in eV is :") disp("Hence, E_F lies "+string(Shift_inFermiLevel)+" eV above Fermi level Ei") // Part (d) MCC= ni^2/dopingConcentration; // Minority carrier concentration per cube meter disp(MCC,"Minority carrier concentration per cube meter when its temperature is increased is : ")
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example9_3.sce
//clc() F1 = 6*1000;//L/s BOD1 = 3 * 10^-5;//g/L BOD2 = 5 * 10^-3;//g/L V = 16 * 10^3;//m^3/day v = V * 10^3 / (24 * 3600);//L/s //Let BOD of the effluent be BODeff, BODeff = (BOD2 * (F1 + v) - BOD1 * F1) / ( v ); disp("g/L",BODeff,"BOD of the effluent of the plant = ")
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//programa: P1S.sce //Semestre:1/2018 clear; format('e',10); //Constantes e = 1.602e-19; //[C] h = 6.6262e-34; //[J.s] c = 3e8; //[m/s] kb = 1.38e-23; //[J/K] //Questão-1 lambda = 500e-9; //[m] p = h/lambda; Ec = p*c; //[J] ec = Ec/e; //[eV] disp('Q1a: p = '); disp(p); //[kg.m/s] disp('Q1b: ec = '); disp(ec); //[eV] //Questão-2 Nd = 1e16; //[cm^-3] n0 = Nd; Rh = -1/(e*n0); //[cm^3/C] disp('Q2a: n0 = '); disp(n0); //[cm^-3] disp('Q2b: Rh = '); disp(Rh); //[cm^3/C] //Questão-3 L = 1; //[cm] W = 0.1; //[cm] t = 1e-4; //[cm] //T = 300;//[K] mi_n = 1350; //[cm^2/V.s] mi_p = 480; //[cm^2/V.s] //n = p = ni ni = 1.5e10; //[cm^-3] s = e*ni*(mi_n + mi_p); //condutividade [S/cm] disp('Q3a: s = '); disp(s); //[Siemens/cm] A = t*W; //[cm^2] R = L/(A*s); //[ohms] disp('Q3b: R = '); disp(R); //[ohms] //Questão-4 Vldr = [2 3]; //[volts] Ildr = [1e-3 30e-3]; //[amp] Iled = [0 10e-3]; //[amp] C2 = Ildr(1)/Vldr(1); //[Siemens] C1 = (Ildr(2) - C2*Vldr(2))/(Iled(2)*Vldr(2)); //C1 = [Siemens/A]ou[1/V]ou[A/W] disp('Q4a: C1 = '); disp(C1); disp('Q4b: C2 = '); disp(C2); //Questão-5 T1 = 300; Is1 = 1e-8; //[amp] T1 = 300 K Is2 = 3e-8; //[amp] T2 = ? Va = 1.2; //[volt] Vb = 0.7; //[volt] Vc = 0; I = Is1*(exp((e/(kb*T1))*(Va-Vb))-1);//[amp] T2 = (e*(Vb-Vc))/(kb*log(I/Is2 + 1)); //[K] disp('Q5a: T2 = '); disp(T2); //[K] disp('Q5b: I = '); disp(I); //[amp]
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//Introduction to Fiber Optics by A. Ghatak and K. Thyagarajan, Cambridge, New Delhi, 1999 //Example 8.8 //OS=Windows XP sp3 //Scilab version 5.5.2 clc; clear; //given lambda0=1.3e-6;//operating wavelength of single mode fiber in m thetah=2.74;//angle corresponding to 3 dB point in degrees k0=2*%pi/lambda0;//free space wave number in rad/m omega=sqrt(2*log(2))/(k0*sind(2.74));//corresponding spot size of fiber in m d=2*omega;//corresponding value of Gaussian mode field diameter in m mprintf("Corresponding mode field diameter=%f um",d/1e-6)//division by 1e-6 to convert in um //The answer provided in the textbook is wrong
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clc //initialisation of variables h=20.2 //lbf/in^2 T=40 //F //CALCULATIONS Cp=h/T//Btu/lbm F //RESULTS printf('The constant pressure specific heat of steam =% f Btu/lbm',Cp)
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<<<<<<< HEAD function y = sigmoid_train(t, ranges, rc) // Evaluate a train of sigmoid functions at T. //Calling Sequence //y = sigmoid_train(t, ranges, rc) //Parameters //t: integer //ranges: matrix //rc:timeconstant //Description //The number and duration of each sigmoid is determined from RANGES. Each row of RANGES represents a real interval, e.g. if sigmoid 'i' starts at 't=0.1' and ends at 't=0.5', then 'RANGES(i,:) = [0.1 0.5]'. The input RC is an array that defines the rising and falling time constants of each sigmoid. Its size must equal the size of RANGES. //Examples //sigmoid_train(0.1,[1:3],4) //Output : // ans = // // 0.2737470 funcprot(0); //************************************************************************************************** //______________________________________________version1 code (not working)_________________________ //__________________________________________________________________________________________________ //************************************************************************************************** //rhs=argn(2); //if (rhs<3 | rhs>3) then // error("Wrong number of input arguments"); //end // //select(rhs) //case 3 then // y=callOctave("sigmoid_train", t, ranges, rc) //end //************************************************************************************************** //______________________________________________version2 code ( working)____________________________ //__________________________________________________________________________________________________ //************************************************************************************************** nRanges = size (ranges, 1); if isscalar (rc) rc = rc * ones (nRanges,2); elseif or( size(rc) ~= [1 1]) if length(rc) ~= nRanges error('signalError','Length of time constant must equal number of ranges.') end if isrow (rc) rc = rc'; end rc = repmat (rc,1,2); end flag_transposed = %F; if iscolumn (t) t = t.'; flag_transposed = %T; end [ncol nrow] = size (t); T = repmat (t, nRanges, 1); RC1 = repmat (rc(:,1), 1, nrow); RC2 = repmat (rc(:,2), 1, nrow); a_up = (repmat (ranges(:,1), 1 ,nrow) - T)./RC1; a_dw = (repmat (ranges(:,2), 1 ,nrow) - T)./RC2; Y = 1 ./ ( 1 + exp (a_up) ) .* (1 - 1 ./ ( 1 + exp (a_dw) ) ) y = max(Y,'r'); if flag_transposed y = y.'; end ======= function y =sigmoid_train(t, ranges, rc) // Evaluate a train of sigmoid functions at T. //Calling Sequence //y = sigmoid_train(t, ranges, rc) //Parameters //t: integer //ranges: matrix //Description //The number and duration of each sigmoid is determined from RANGES. Each row of RANGES represents a real interval, e.g. if sigmoid 'i' starts at 't=0.1' and ends at 't=0.5', then 'RANGES(i,:) = [0.1 0.5]'. The input RC is an array that defines the rising and falling time constants of each sigmoid. Its size must equal the size of RANGES. //Examples //sigmoid_train(0.1,[1:3],4) //ans = // 0.27375 funcprot(0); rhs=argn(2); if (rhs<3 | rhs>3) then error("Wrong number of input arguments"); end select(rhs) case 3 then y=callOctave("sigmoid_train", t, ranges, rc) end >>>>>>> 6bbb00d0f0128381ee95194cf7d008fb6504de7d endfunction
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clc //initialisation of variables h= 6.6234*10^-27 //ergs sec m= 2.59 //gms v= 3.35*10^4 //cm sec ^-1 e= 4.8*10^-10 //ev V= 40000 //volts M= 300 //gms L= 1836 //A N= 6*10^23 //molecules //CALCULATIONS p= m*v l= h/p E= V*e/M P= sqrt(2*E*(1/(L*N))) L1= h*10^8/P //RESULTS printf (' wavelength = %.2e cm',l) printf (' \n wavelength = %.4f cm',L1)
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Example_10_5.sce
clear; clc; //To find Approx Value function[A]=approx(V,n) A=round(V*10^n)/10^n;//V-Value n-To what place funcprot(0) endfunction //Example 10.5 //Caption : Program to Find L V {xi} and {yi} for a System z1=0.45; z2=0.35; z3=0.2; P=110;//[KPa] T=353.15;//[K] P1_sat=195.75;//[KPa] P2_sat=97.84;//[KPa] P3_sat=50.32;//[KPa] //BUBL Calculation x1=z1; x2=z2; x3=z3; P_BUBL=(x1*P1_sat)+(x2*P2_sat)+(x3*P3_sat); //DEW Calculation y1=z1; y2=z2; y3=z3; P_Dew=1/((y1/P1_sat)+(y2/P2_sat)+(y3/P3_sat)); //Since P_Bubl<P<P_dew //Flash Calculation K1=P1_sat/P; K2=P2_sat/P; K3=P3_sat/P; //Finding V from Eqn(10.17) //E((zi*Ki)/(1+(V*(Ki-1))))=1 x=0; F_x=(((z1*K1)/(1+((K1-1)*x)))+((z2*K2)/(1+((K2-1)*x)))+((z3*K3)/(1+((K3-1)*x)))-1); F_a=F_x; x=0.9; F_x=(((z1*K1)/(1+((K1-1)*x)))+((z2*K2)/(1+((K2-1)*x)))+((z3*K3)/(1+((K3-1)*x)))-1); F_b=F_x; A=0; B=0.9; i=1; while(i==1) a=A; F_a=(((z1*K1)/(1+((K1-1)*a)))+((z2*K2)/(1+((K2-1)*a)))+((z3*K3)/(1+((K3-1)*a)))-1); b=B; F_b=(((z1*K1)/(1+((K1-1)*b)))+((z2*K2)/(1+((K2-1)*b)))+((z3*K3)/(1+((K3-1)*b)))-1); x1=((a*F_b)-(b*F_a))/(F_b-F_a); F_x1=(((z1*K1)/(1+((K1-1)*x1)))+((z2*K2)/(1+((K2-1)*x1)))+((z3*K3)/(1+((K3-1)*x1)))-1); if((F_a*F_x1)<0) then flag=1; A=a; B=x1; else((F_x1*F_b)<0) flag=2; A=x1; B=b; end x1_a=approx(x1,4); b_a=approx(b,4); a_a=approx(a,4); if(x1_a==b_a) V=approx(x1,4); i=0; break; elseif(x1_a==a_a) root=approx(x1,4); i=0; break; end end disp(V,'Hence By solving the polynomial V = ') L=1-V; //from eqn 10.16 //yi=(zi*Ki)/(1+(V*(Ki-1))) y1=approx((z1*K1)/(1+((K1-1)*V)),4); y2=approx((z2*K2)/(1+((K2-1)*V)),4); y3=approx((z3*K3)/(1+((K3-1)*V)),4); x1=approx(y1/K1,4); x2=approx(y2/K2,4); x3=approx(y3/K3,4); y=[y1 y2 y3]; x=[x1 x2 x3]; disp(L,'Moles Of liquid') disp(V,'Moles Of vapor') disp(x,'Mole fraction Of liquid') disp(y,'Mole fraction Of vapor') //End
7e9b015036bb3276b6f2417cd78f943a7d90f7f3
e04f3a1f9e98fd043a65910a1d4e52bdfff0d6e4
/New LSTMAttn Model/.data/form-split/DEVELOPMENT-LANGUAGES/oto-manguean/xty.tst
26aaa1ce03e523bbcbdcfaadc13cdca5c7886421
[]
no_license
davidgu13/Lemma-vs-Form-Splits
c154f1c0c7b84ba5b325b17507012d41b9ad5cfe
3cce087f756420523f5a14234d02482452a7bfa5
refs/heads/master
2023-08-01T16:15:52.417307
2021-09-14T20:19:28
2021-09-14T20:19:28
395,023,433
3
0
null
null
null
null
UTF-8
Scilab
false
false
14,275
tst
xty.tst
ta³ndaʔ³a⁴ V;PFV;LGSPEC1 ta¹xin¹ V;IRR ku³-ta⁴tan⁴ V;PFV;LGSPEC2 nda³-chaʔ⁴bi³ V;IRR ku³-ni³ni³ V;PFV;LGSPEC2 niʔ¹i⁴ V;IRR;NEG ka³ka³ V;PFV;LGSPEC1 ya³tan³ V;IPFV nda³-ya¹⁴kun² V;IRR;NEG ta¹⁴ni³ V;IPFV ta³xaʔ⁴a⁴ V;IRR;NEG ku³-xi⁴ni⁴ V;PFV;LGSPEC1 nda³-kwi³so³, ndi³-kwi³so³ V;PFV;LGSPEC2 ki³taʔ⁴an⁴ V;IRR;NEG nda³ndo³o³ V;IRR;NEG ku³un³ V;IRR ku³-i³tun⁴ V;PFV;LGSPEC1 nda³-tu¹u³ V;IRR ka³sun² V;PFV;LGSPEC2 ku³-naʔ³a⁴ V;IRR;NEG kaʔ³bi³ V;PFV;LGSPEC1 ku³ni² V;PFV;LGSPEC2 ku³-ndo³so⁴ V;IPFV ka³tu⁴ V;PFV;LGSPEC1 ka³na³ V;PFV;LGSPEC1 xo⁴kwi¹in¹ V;PFV;LGSPEC1 xi³kwiʔ⁴na⁴ V;IRR;NEG kaʔ¹a¹ V;IPFV xa³nda⁴ V;IRR nda³-ki³si³ V;PFV;LGSPEC1 ka³ndwaʔ¹a³ V;IRR tiʔ³bi³ V;IRR;NEG nda³kaʔ¹nu¹ V;PFV;LGSPEC1 xi¹ni³ V;IPFV ku³-ndo³so⁴ V;IRR ku³u² V;PFV;LGSPEC2 ka³ni² V;PFV;LGSPEC1 ko³o³ V;IPFV taʔ¹yu¹ V;IRR nda³-ta³ba⁴ V;IRR ku¹ndaʔ¹a³ V;PFV;LGSPEC1 kwi¹so³ V;IRR;NEG sa⁴-yuʔ³bi² V;PFV;LGSPEC1 kwi³nda²a² V;PFV;LGSPEC1 ke³-baʔ¹a³ V;IRR;NEG ta³xi³kwaʔ⁴a⁴ V;IRR ko¹ko³ V;PFV;LGSPEC2 chi³chin⁴ V;IPFV ku³-ndaʔ¹yu¹ V;IPFV nda³ka³ni³ni² V;PFV;LGSPEC2 ku³-xaʔ³an² V;PFV;LGSPEC2 ka³sa³ V;IRR nda³-ko³to³ V;PFV;LGSPEC1 ta¹ni¹ V;IRR chi³pa⁴chi¹ V;IRR xa¹a¹ V;PFV;LGSPEC1 nda³sa³ V;IRR kaʔ³ni⁴ V;PFV;LGSPEC1 tu³un² V;PFV;LGSPEC2 sa³ña⁴ V;IRR kwa³chi³ V;PFV;LGSPEC2 sa⁴-kwi³ko⁴ V;IPFV xi³kwe⁴nda² (alto es xo³kwe⁴nda²) V;IRR;NEG chi³nduʔ⁴u⁴ V;PFV;LGSPEC1 ndu³-ka⁴xi³ V;PFV;LGSPEC1 ndo¹o³ V;IRR taʔ³bi⁴ V;PFV;LGSPEC2 kwi¹i⁴ V;IRR;NEG tu³u⁴ V;PFV;LGSPEC1 a³sa³ V;PFV;LGSPEC1 ya¹ni³ V;IPFV ku³-xiʔ³na³ V;PFV;LGSPEC1 nda³kwi³in³ V;IPFV nda³kwi³in³ V;IRR ku³-nda³si⁴ V;PFV;LGSPEC2 chi³pa⁴chi¹ V;PFV;LGSPEC2 ta¹ni¹ V;IRR;NEG ku³-ndi³chi² V;PFV;LGSPEC1 ka³si² V;IPFV ndu³-tu¹u¹ V;PFV;LGSPEC1 niʔ¹i⁴ V;PFV;LGSPEC2 ndo¹o³ V;IRR;NEG ndu³-ku³na⁴ V;IRR ku³-ti¹ki³xin⁴ V;IPFV xi¹nu³ V;IRR;NEG nda³-ka³na³ V;IRR nda³kuʔ³u⁴ V;PFV;LGSPEC2 ka¹sun¹ V;PFV;LGSPEC2 ndo³ko³o⁴ V;PFV;LGSPEC2 nda³kwe¹ta³ V;PFV;LGSPEC2 ku³-ti³saʔ⁴ma³ V;IRR nu¹ma¹ V;PFV;LGSPEC2 naʔ¹na¹ V;PFV;LGSPEC1 ku³-ndi³xi³ V;IRR;NEG kaʔ³ma³ V;PFV;LGSPEC1 xa¹xa¹ V;IRR;NEG nda³chi³i⁴ V;PFV;LGSPEC1 kaʔ³a³ V;PFV;LGSPEC1 nda³ka³ba³ V;IPFV nda³i³ni² V;IPFV ku³-na³mi⁴ V;IPFV ta³kweʔ³e² V;IRR ndu³ku³xa³ V;IRR;NEG ka³kwi¹in¹ V;IPFV ka³ndi⁴xa³ V;IRR nda³-kwi¹in¹ V;PFV;LGSPEC1 nda³chi⁴ V;IPFV ki¹ni⁴ V;PFV;LGSPEC2 ku³-nda³si⁴ V;IRR tu³xi⁴ V;IPFV chioʔ¹o⁴ V;PFV;LGSPEC1 nda³kwaʔ¹a³ V;PFV;LGSPEC2 ndaʔ³a² V;IRR;NEG nda³ka³ba³ V;IRR;NEG chi³chin⁴ V;IRR ku³-nda⁴a⁴ V;PFV;LGSPEC1 tu¹xuʔ⁴u² V;IRR ku³-nda³si⁴ V;IPFV taʔ³bi⁴ V;PFV;LGSPEC1 sa⁴-na³na³ V;IPFV ka³ti¹in³ V;PFV;LGSPEC2 ndu¹xin¹ V;IRR ndo³o³ V;IPFV ka¹xan⁴ V;IPFV ku³-nda³tu³ V;IRR tiʔ³bi⁴ V;PFV;LGSPEC1 su³ku⁴ V;PFV;LGSPEC1 i¹chi¹ V;PFV;LGSPEC1 ki¹ni⁴ V;IRR ku³ndu³ndu² V;PFV;LGSPEC1 nda³-naʔ¹a¹ V;PFV;LGSPEC1 nda³chi⁴ V;IRR;NEG ki³ni⁴ V;PFV;LGSPEC2 ta¹xi⁴ V;IRR sa⁴-ndo³to³ V;IRR;NEG kaʔ¹a¹ V;PFV;LGSPEC2 ndiʔ³i³ V;IRR ku³u² V;IPFV ke¹e³ V;PFV;LGSPEC2 sa⁴-to¹o³ V;IPFV nda³ta⁴ V;IRR;NEG ndu³-ma¹ni¹ V;IRR nda³-taʔ³bi⁴ V;IPFV xi³kwiʔ⁴na⁴ V;IPFV ku³-nuʔ³ni² V;IRR ndu³-baʔ¹a³ V;IRR ki³si³ V;PFV;LGSPEC1 ndu³toʔ³ni³ V;IRR;NEG chuʔ³u⁴ V;IPFV ko¹yo³ V;PFV;LGSPEC2 ta³an⁴ V;IPFV ta³seʔ⁴e² V;IRR nda³tuʔ⁴un⁴ V;PFV;LGSPEC2 ku³-laʔ⁴la¹ V;IPFV ku³-ta³ka³a³ V;PFV;LGSPEC1 ka³ko¹o³ V;PFV;LGSPEC2 kaʔ³a³ V;IRR nde¹e³ V;IRR;NEG ta³xi³ V;PFV;LGSPEC1 ku³nda³si² V;PFV;LGSPEC1 ka³si² V;PFV;LGSPEC1 sa⁴-yuʔ³bi² V;IRR;NEG ndu³-ka⁴xi³ V;IRR;NEG ku³xi³ V;IRR su¹kun¹ V;IPFV kaʔ³ni⁴ V;IRR;NEG nda³-ki³si³ V;IPFV ta³xaʔ⁴a⁴ V;PFV;LGSPEC1 xiʔ¹⁴ñu³ V;PFV;LGSPEC2 sa³na³ V;IRR ke¹yu⁴ V;PFV;LGSPEC2 xa³xa⁴ V;IRR taʔ³bi⁴ V;IRR nda³ka³ni³ni² V;PFV;LGSPEC1 ku³-ñu³u⁴² V;PFV;LGSPEC2 nda³-ka³na³ V;PFV;LGSPEC2 ka¹ña¹ V;IRR;NEG nduʔ¹u⁴ V;IRR sa⁴-ka³xan⁴ V;IRR;NEG kaʔ³bi³ V;IRR sa⁴-kiʔ³in³ V;PFV;LGSPEC2 nda³ndi³i⁴ V;PFV;LGSPEC1 ya¹⁴kun² V;IPFV ndi³kweʔ³e² V;PFV;LGSPEC2 ndu³-i⁴ta⁴ V;PFV;LGSPEC2 ka³ba⁴ V;IRR nda³kwaʔ¹a³ V;IPFV kwi¹in¹ V;PFV;LGSPEC2 kwi¹in¹ V;IPFV ku³nduʔ⁴u⁴ V;IPFV sa⁴-ka³sun² V;PFV;LGSPEC1 ku³na¹⁴ni³ V;PFV;LGSPEC2 ndu³-si¹i⁴ V;IRR kaʔ³yu⁴ V;IRR ndu³ta³ V;IPFV ko³nda²a² V;IPFV ko¹ko³ V;IPFV ku³-laʔ⁴la¹ V;PFV;LGSPEC2 nda³-ka³a⁴ V;PFV;LGSPEC2 ko³ko³ V;PFV;LGSPEC1 kaʔ³ndi² V;PFV;LGSPEC1 ku³si⁴ki²⁴ V;IRR;NEG nda³-ki³si³ V;PFV;LGSPEC2 sa³-kwe⁴nda² V;IPFV ku³-nu¹mi⁴ V;IPFV ku³tu³ V;PFV;LGSPEC1 nda³keʔ³e⁴ V;IRR ndo³ko³o⁴ V;IRR;NEG kaʔ³bi³ V;IPFV sa⁴-ndaʔ³bi² V;IRR;NEG ndu³i³ko⁴ / ndu³kwi³ko⁴ V;IRR nda³-xa³a³ V;IRR;NEG kwe³e² V;IRR ndu³ku³xa³ V;IRR nda³xin³ V;IRR nda³xiʔ³i⁴ V;IRR;NEG tu¹un¹ V;PFV;LGSPEC2 kaʔ¹an¹ V;IRR;NEG ku³-chi³tu⁴ V;IPFV ku³u² V;PFV;LGSPEC1 ka¹kan¹ V;PFV;LGSPEC2 nda³xiʔ³i⁴ V;PFV;LGSPEC1 xi¹⁴ko³ V;IRR su³-kwe⁴nda² V;IRR;NEG ti³so⁴ V;PFV;LGSPEC2 nda³ndi³so³ V;IRR nda³kuʔ³u⁴ V;IPFV kwiʔ¹in³ V;IRR;NEG kwiʔ¹in³ V;IPFV ki³ni⁴ V;PFV;LGSPEC1 nda³i³ni² V;PFV;LGSPEC1 nda³-sa¹ka¹ V;PFV;LGSPEC2 taʔ¹bi⁴ V;IRR;NEG nda³ka¹tuʔ⁴un⁴ V;IPFV nda³a³ V;IRR cha³ka³ta⁴ V;IPFV ka³nda²a² V;PFV;LGSPEC2 ndu³-bi⁴chi⁴ V;PFV;LGSPEC1 nda³i³chi² V;PFV;LGSPEC2 ndu³-yaʔ⁴bi³ V;IRR ku³-na³a⁴ V;PFV;LGSPEC2 ndu³-ndi³i⁴ V;IRR nda³ta³a³ V;IRR sa¹a⁴ V;IPFV ndu³-si¹i⁴ V;PFV;LGSPEC2 ku³ma¹ni⁴ V;IRR;NEG nda³-ko¹o³ V;PFV;LGSPEC1 nda³-ya¹⁴kun² V;IRR ya¹⁴kun² V;PFV;LGSPEC1 chuʔ³u⁴ V;PFV;LGSPEC2 ya¹⁴kun² V;IRR;NEG kaʔ¹nu¹ V;IRR xi¹⁴ko³ V;PFV;LGSPEC1 nda³-ndu³ku⁴/ndu³ku⁴ V;IRR;NEG ka³ni² V;IRR;NEG xi¹⁴ko³ V;IRR;NEG ko³-nde³e⁴ V;PFV;LGSPEC2 ku³-yaʔ⁴bi³ V;IPFV nda³ka³ta³ V;PFV;LGSPEC1 ku³-xaʔ¹a¹ V;IRR;NEG ndu³-yaʔ⁴bi³ V;IRR;NEG ka³ni² V;IPFV sa⁴-sa¹a⁴ V;IRR;NEG ndo¹o³ V;PFV;LGSPEC2 ti¹⁴bi³ V;IPFV ndu³-kwa¹chi³ V;IPFV na³na³ V;IRR;NEG ya¹⁴kun² V;PFV;LGSPEC2 nda³kin² V;IRR;NEG chi³ka² V;IPFV ku³-niʔ³i³ V;IRR ndaʔ¹yu¹ V;PFV;LGSPEC1 nda³ko⁴ V;PFV;LGSPEC1 kuʔ¹un¹ V;IRR ka¹tu⁴ V;IPFV niʔ¹i⁴ V;IPFV ka¹sun¹ V;IRR;NEG ka¹kan¹ V;IRR;NEG na³na³ V;PFV;LGSPEC2 ka³ti¹in³ V;IPFV ta³ya² V;IRR;NEG xi³kwaʔ⁴a⁴ V;IPFV yuʔ¹⁴bi² V;IRR ko³ndoʔ³o³ V;IPFV ka³-ta³ni³ V;IRR;NEG kwaʔ³nu³ V;IRR;NEG ko³ndoʔ³o³ V;IRR;NEG kuʔ³u² V;IRR;NEG sa⁴-ndaʔ³bi² V;PFV;LGSPEC2 ti³chaʔ⁴ni³ V;IRR;NEG kwa¹xin³ V;IRR ko³ko³ V;IRR;NEG ku³-ti¹sun¹ V;PFV;LGSPEC2 ku¹un⁴ V;PFV;LGSPEC2 tiʔ³nu³ V;IRR chi³ndaʔ³a⁴ V;IPFV nda³-tu¹u⁴ V;PFV;LGSPEC1 xi¹ni³ V;PFV;LGSPEC1 ndu³-ndi³i⁴ V;IRR;NEG ku³-nde³ta³ V;IRR;NEG sa¹ña⁴ V;IRR su¹kun¹ V;PFV;LGSPEC2 ki³xa²a² V;PFV;LGSPEC2 ka³-taʔ³nu³ V;IRR;NEG ku³-nuʔ³ni² V;PFV;LGSPEC1 ka³ta³ V;IPFV kwi¹ya⁴ V;IRR tu¹u³ V;PFV;LGSPEC1 ndu³-ku³na⁴ V;PFV;LGSPEC1 kwi¹ko⁴ V;IRR;NEG ku¹nuʔ¹u⁴ V;IRR;NEG ku³-naʔ³a⁴ V;PFV;LGSPEC1 ku³-le³ke³ V;IRR;NEG kiaʔ³bi¹³ V;IPFV ka¹ba¹ V;IRR ko³seʔ⁴e² V;PFV;LGSPEC1 ka³ndiʔ³i³ V;IPFV nda³kwi³in³ V;PFV;LGSPEC1 ndu³-xi¹nu¹ V;IPFV ku³chi³ V;IPFV xa¹⁴ni² V;IRR;NEG ku³nduʔ⁴u⁴ V;IRR;NEG keʔ³e⁴ V;IRR ke³e³ V;PFV;LGSPEC2 kaʔ¹a¹ V;IRR;NEG nda³ka¹a¹ V;IPFV nda³-chi³kun² V;PFV;LGSPEC2 ku³-xi⁴ni⁴ V;IPFV ku³-ta³ni³ V;PFV;LGSPEC1 ta³xi³kwaʔ⁴a⁴ V;PFV;LGSPEC2 ke¹nu³ V;IRR;NEG sa⁴-ka³xan⁴ V;IPFV nda³ku³ni² V;PFV;LGSPEC1 ku³-ta³ni³ V;PFV;LGSPEC2 si⁴-kwe¹kun¹ (sa⁴-kwe¹kun¹) V;IRR ku³-yaʔ⁴bi³ V;PFV;LGSPEC2 ti³su⁴ku²⁴ V;IRR ka³si⁴ V;PFV;LGSPEC2 xiʔ¹⁴ni³ V;PFV;LGSPEC1 ku³-chaʔ¹mba⁴ V;IPFV koʔ³ni⁴ V;IRR;NEG ndu³-ndi³kun² V;IPFV cha³ka³ba³ V;PFV;LGSPEC2 sa⁴-ndo³to³ V;PFV;LGSPEC2 sa⁴-nda³ba³ V;IRR;NEG ko¹ko³ V;IRR tu¹u³ V;IPFV ka³kiʔ¹i³ V;IRR keʔ³e⁴ V;PFV;LGSPEC1 kwi³ko⁴ V;IPFV sa⁴-nda³ba³ V;PFV;LGSPEC2 xi³kwaʔ⁴a⁴ V;IRR;NEG ti³nda²a² V;IPFV ku¹ndaʔ¹a³ V;IRR ka³kwi¹in³ V;PFV;LGSPEC1 nda³ka³ni² V;PFV;LGSPEC1 ku³-ba³ta⁴ V;PFV;LGSPEC1 ndu³-tiʔ⁴bi³ V;IRR nda³i³ni² V;IRR ku³-ni³ni³ V;IPFV ndu³-bi³ta⁴² V;PFV;LGSPEC2 ka³ba⁴ V;IPFV kuʔ¹un¹ V;PFV;LGSPEC1 ku³un³ V;IRR;NEG taʔ¹nda¹ V;PFV;LGSPEC2 cha³ka³ta⁴ V;IRR;NEG ndu³-ku³tu⁴ V;PFV;LGSPEC1 ku³-yo⁴ko¹ V;IRR xi³i² V;IPFV chi³kun² V;IRR ndu³su⁴ku²⁴ V;IRR ka³xaʔ⁴an² V;IRR;NEG sa⁴-xi¹nu³ / ja⁴-xi¹nu³ V;PFV;LGSPEC1 ka³tu⁴ V;IPFV kwe³ta³ V;IRR ta¹ku¹ V;IRR ndo³ko³to² V;PFV;LGSPEC1 sa⁴-kwi³ko⁴ V;PFV;LGSPEC1 chi³kun² V;PFV;LGSPEC2 ka³nda³ba³ V;IRR ku³-ñu³u⁴² V;PFV;LGSPEC1 kwa³ku³ V;PFV;LGSPEC1 ka³ndi³chi² V;STAT ka¹sun¹ V;IRR ka³kiʔ¹i³ V;IPFV ka³xaʔ⁴ni³ V;PFV;LGSPEC2 ndu³-ku³nduʔ⁴u⁴ V;PFV;LGSPEC1 ndu³kwi³ko⁴ V;PFV;LGSPEC2 ku³-kaʔ³an³ V;IRR;NEG ku¹sun¹ V;IRR ku³-laʔ⁴la¹ V;PFV;LGSPEC1 ku³-chaʔ¹mba⁴ V;PFV;LGSPEC2 tu³un² V;PFV;LGSPEC1 xo⁴kwi¹in¹ V;IPFV ku³-xi⁴ni⁴ V;IRR nda³-na¹ma³ V;PFV;LGSPEC2 ya³tan³ V;IRR ndu³su⁴ku²⁴ V;IRR;NEG ndu³-ku³nduʔ⁴u⁴ V;IRR;NEG ndu³-ndi³xi³ V;IRR;NEG kaʔ³ma³ V;IRR;NEG ku³-baʔ¹a³ V;IRR ka¹sun¹ V;PFV;LGSPEC1 na³ma⁴ V;IPFV naʔ¹na¹ V;IRR;NEG ka³niʔ¹i³ V;PFV;LGSPEC1 ka³ndi⁴xa³ V;PFV;LGSPEC2 ndu³-baʔ¹a³ V;IRR;NEG sa⁴-ti¹⁴bi³ V;IPFV nda³-tiʔ³bi³ V;IRR;NEG cha³kwi³in³ V;STAT ka³si² V;IRR ko³nde³e³ V;PFV;LGSPEC1 chi³keʔ⁴le¹ V;IRR;NEG xa³a³ V;PFV;LGSPEC1 ku³-ñu³u³ V;IPFV nda³nuʔ³u³ V;PFV;LGSPEC2 chi³i³ V;IRR;NEG ki¹⁴tu³ V;IPFV ku³-nda¹a⁴ V;PFV;LGSPEC1 nda³kwe¹ta³ V;PFV;LGSPEC1 nda³-ndu³ku⁴/ndu³ku⁴ V;IPFV ka³sa³chiu⁴un⁴ V;PFV;LGSPEC1 ti³so⁴ V;IRR;NEG sa⁴-ndiʔ¹i³ V;IPFV ta³xi³kwaʔ⁴a⁴ V;PFV;LGSPEC1 sa¹ña⁴ V;PFV;LGSPEC1 nda³koʔ³ma⁴ V;IRR;NEG nda³-tu¹u⁴ V;PFV;LGSPEC2 ka³ta³ V;PFV;LGSPEC1 ka³xi⁴ V;IRR;NEG nda³ndi³i⁴ V;IRR taʔ¹bi⁴ V;PFV;LGSPEC1 xa³ka³ V;IPFV kwi¹ta¹ V;IPFV chi³kuʔ³ba² V;PFV;LGSPEC1 ka³ndi³so³ V;PFV;LGSPEC1 ki³ni⁴ V;IRR;NEG nda³kuʔ³u⁴ V;IRR;NEG tu³u⁴ V;IPFV ka³sun² V;IRR ko³ndo³ V;IRR sa⁴-kaʔ³a³ V;IPFV xi³kwaʔ⁴a⁴ V;PFV;LGSPEC1 ki³xa²a² V;IPFV taʔ¹yu¹ V;IPFV tu³tu⁴ V;PFV;LGSPEC1 ti¹in³ V;IPFV ta¹nda³² V;IRR;NEG ku³-su⁴ku²⁴ V;PFV;LGSPEC2 ka³-ta³ni³ V;PFV;LGSPEC2 tiʔ³bi³ V;PFV;LGSPEC2 ndi³ko³ V;IRR;NEG na³ma³ V;PFV;LGSPEC2 nda³ba³ V;IRR;NEG ndo³ko³o⁴ V;IPFV ta³chi⁴ñu³ V;IPFV nda³ndi³i³ V;PFV;LGSPEC2 kwa¹ñu¹ V;IRR;NEG nda³-nu¹na⁴ V;PFV;LGSPEC1 ke¹ta³ V;IRR;NEG xi¹⁴nda² V;PFV;LGSPEC2 nda³sa¹ma³ V;IPFV xiʔ¹⁴ñu³ V;IPFV sa⁴-chuʔ¹⁴ma¹ V;IRR;NEG ndu³naʔ³a² (tu⁴ni¹) V;PFV;LGSPEC1 xi³i² V;IRR;NEG sa⁴-ndu¹xin¹ V;IRR ndu³-ka⁴chi¹ V;IRR;NEG nda³ka¹xin³ V;PFV;LGSPEC1 sa⁴-chuʔ¹⁴ma¹ V;STAT ka³ko¹o³ V;IRR chuʔ¹⁴ma¹ V;PFV;LGSPEC1 ka³ndi³so³ V;IRR ku³-nda¹a⁴ V;IRR;NEG ka³nduʔ⁴u⁴ V;IPFV ke¹nu³ V;PFV;LGSPEC1 ku¹ndaʔ¹a³ V;IRR;NEG choʔ¹ma⁴ V;PFV;LGSPEC2 ta³ba⁴ V;IRR;NEG ndu³-yaʔ⁴bi³ V;PFV;LGSPEC2 nda³chi³i⁴ V;PFV;LGSPEC2 nda³-ka¹ya¹ V;PFV;LGSPEC2 nda³ndi³so³ V;PFV;LGSPEC2 sa⁴-nde³ta³ V;PFV;LGSPEC2 chi³nde³e⁴ V;PFV;LGSPEC2 ka³ni⁴nu³ V;IRR ndaʔ³ba² V;PFV;LGSPEC2 ku³-su⁴ku²⁴ V;IPFV tu¹xi⁴ V;IRR sa³ka⁴ V;PFV;LGSPEC1 nda³ka³ti³ V;PFV;LGSPEC2 chuʔ¹⁴ma¹ V;PFV;LGSPEC2 ndi³kweʔ³e² V;IRR ku³-yu⁴ma⁴ V;IPFV ku³-nde³ta³ V;PFV;LGSPEC2 xiʔ¹⁴ni³ V;IRR naʔ¹ma¹ V;PFV;LGSPEC1 xa³a³ V;IPFV nda³-xa³a³ V;IRR ku³i³ni² V;PFV;LGSPEC1 ndu³ku⁴ V;IRR nda³ka³ta³ V;IRR ta¹bi¹ V;PFV;LGSPEC1 kiʔ³in³ V;IRR;NEG ka¹ña¹ V;PFV;LGSPEC1 xa¹⁴bi² V;PFV;LGSPEC2 ka³kiʔ¹i³ V;IRR;NEG kwiʔ¹in³ V;IRR chi³ka² V;PFV;LGSPEC2 ku³-yo⁴ko¹ V;PFV;LGSPEC2 ka³ndiʔ³i³ V;PFV;LGSPEC1 ka³ba³ V;IPFV nda³ku³ V;IPFV ndi¹ko³ V;IPFV ka³ta³ V;PFV;LGSPEC2 ta³bi³ V;IRR;NEG chi³chin⁴ V;PFV;LGSPEC2 ka³na³nu³ V;IRR;NEG ku³-so³ko² V;IRR;NEG ki³taʔ⁴ni³ V;IRR;NEG ku³-nde³ta³ V;PFV;LGSPEC1 kwa³ku³ V;IPFV ku³-ti¹sun¹ V;IRR nda³ndi¹ka¹ V;IRR tu³xi⁴ V;PFV;LGSPEC1 sa³ka⁴ V;IRR;NEG ki³taʔ⁴an⁴ V;IPFV ndi³kwi¹kun¹ V;PFV;LGSPEC1 ku¹sun¹ V;PFV;LGSPEC1 nda³-sa¹ka¹ V;IRR nda³ndi¹ka¹ V;PFV;LGSPEC2 nda³ndo³o³ V;PFV;LGSPEC2 kaʔ¹nu¹ V;PFV;LGSPEC1 xi³kwaʔ⁴a⁴ V;IRR ku³-xaʔ³an² V;IPFV ka³sa³chiu⁴un⁴ V;PFV;LGSPEC2 xi³ka⁴ba¹³ V;PFV;LGSPEC2 ka¹xan⁴ V;PFV;LGSPEC1 nda³ka³ti³ V;IRR kaʔ³nda² V;IRR;NEG nda³keʔ³e⁴ V;IPFV tu³teʔ⁴e⁴ V;IRR ku³-nuʔ³ni² V;IRR;NEG tu³teʔ⁴e⁴ V;PFV;LGSPEC1 sa⁴-sa¹a⁴ V;IPFV nda³a³ V;IPFV nda³ko¹so⁴ V;IRR nda³-ki³nde³e⁴ V;IRR ka³ta⁴ V;PFV;LGSPEC2 ma¹ni⁴ V;IRR ku³tu⁴ V;IRR;NEG ndi³ki³taʔ⁴an⁴ V;PFV;LGSPEC2 ka¹ba¹ V;IRR;NEG ku³-xaʔ³an² V;IRR chi³kun² V;IPFV kwi¹in¹ V;IRR sa⁴-naʔ¹a¹ V;IRR;NEG tiʔ³nu³ V;IRR;NEG sa⁴-nda³ba³ V;IRR ta³kweʔ³e² V;PFV;LGSPEC2 nda³-ta³ku² V;PFV;LGSPEC1 nda³ndo⁴so²⁴ V;IRR ku¹un⁴ V;IRR choʔ¹ma⁴ V;IPFV nda³ke³e⁴ V;PFV;LGSPEC2 ndu³-kwa¹chi³ V;IRR;NEG nda³ko⁴ V;PFV;LGSPEC2 nda³ndo⁴so²⁴ V;IPFV ku³-naʔ¹a¹ V;IRR nda³xi⁴ V;IRR kaʔ¹yu¹ V;IRR ndaʔ¹ba¹ V;PFV;LGSPEC1 ko³-nde³e⁴ V;IRR sa³na³ V;PFV;LGSPEC1 ku³chi⁴ V;PFV;LGSPEC1 sa⁴-na³na³ V;IRR;NEG ka¹ku³ V;IRR ta³seʔ⁴e² V;PFV;LGSPEC1 koʔ¹ni⁴ V;IRR;NEG ku³-nu¹mi⁴ V;IRR ka³-ta³ni³ V;STAT ka¹nda¹ V;IRR kwiʔ¹ña¹ V;IRR;NEG nu¹ma¹ V;PFV;LGSPEC1 kwi¹in¹ V;PFV;LGSPEC1 ku³ndi³to³ V;IRR taʔ¹nda¹ V;PFV;LGSPEC1 ka³xi⁴ta³ V;PFV;LGSPEC1 nda³-ko³to³ V;IRR;NEG sa⁴-kaʔ³a³ V;PFV;LGSPEC1 ku³i³ni² V;IRR nda³koʔ³ma⁴ V;PFV;LGSPEC1 chi³ndaʔ³a⁴ V;PFV;LGSPEC2 ku³-kaʔ³an³ V;IRR sa⁴-naʔ¹a¹ V;IPFV nda³-chi³kun² V;IPFV ko¹so¹ V;PFV;LGSPEC1 ku³-tu³un³ V;PFV;LGSPEC1 chi³pa⁴chi⁴ V;IRR;NEG
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clc; p=6; // number of poles m=3; // number of phases f=50; // frequency of motor P=40000; // rated power of induction motor v=400; // rated voltage of induction motor // results of blocked rotor test vb=200; // applied voltage ib=110; // applied current pf=0.4; // power factor f1=45; // frequency at starting torque is to be determined e=380; // voltage at starting torque is to be determined vbp=vb/sqrt(3); // per phase voltage during blocked rotor test zb=vbp/ib; // total impedance referred to stator R=zb*pf; // net resistance referred to stator X=zb*(sqrt(1-pf^2)); // net reactance referred to stator X=X*(f1/f); // net reactance at frequency=45 Z=R+X*%i; // impedance at frequency=45 v1=e/sqrt(3); // per phase stator is=v1/(Z); // starting current ws=(4*%pi*f)/p; // synchronous speed T=(3/ws)*abs(is)^2*(R/2); printf('Starting torque is %f Nm',T);
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44_05.sce
//Problem 44.05: The voltages at the input and at the output of a transmission line properly terminated in its characteristic impedance are 8.0 V and 2.0 V rms respectively. Determine the output voltage if the length of the line is doubled. //initializing the variables: Vs = 8; // in Volts VR = 2; // in Volts x = 2; //calculation: // receiving end voltage VR = Vs*e^(-nr) //e^-nr = p p = VR/Vs //If the line is doubled in length, then VR = Vs*(p)^2 printf("\n\n Result \n\n") printf("\n Receiving end voltage If the line is doubled in length, VR is %.2f +(%.2f)i V",real(VR), imag(VR))
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// Exa 5.17 clc; clear; close; // Given data alpha = 1.732; k_f = 1.274; C1 = 1;// in F C2 = C1;// in F R1 = alpha/2;// in ohm R2 = 2/alpha;// in ohm R_F = R2;// in ohm f_3dB = 2;// in kHz f_3dB = f_3dB * 10^3;// in Hz f_c = f_3dB/k_f;// in Hz Omega_c = 2*%pi*f_c;// in rad/sec R1 = R1/Omega_c;// in ohm R1 = R1 * 10^8;// in ohm R2 = R2/Omega_c;// in ohm R2 = R2 * 10^8;// in ohm R_F = R2;// in ohm C1 = C1/10^8;// in F disp(R1*10^-3,"The value of R1 in kΩ is : ") disp(R2*10^-3,"The value of R2 and R_F in kΩ is : ") disp(C1*10^9,"The value of C1 and C2 in nF is : ")
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//Chapter 2,Ex2.58,Pg 2.74 clc; disp("Refer to the diagram shown in the figure") A=[7 -2;-2 10] B=[20;-12] I=A\B printf("\n I2= %.2f A \n",I(2)) printf("\n In=%.2f A \n",-I(2)) //Calculation of Rn Rn=(5*2/(5+2))+8 printf("\n Rn=%.2f ohms \n",Rn) //Calculation of Il Il=0.67*(Rn/(Rn+10)) //Current is short circuit current calculated printf("\n Il=%.2f A \n",Il)
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//Example 21.1. clc format(6) u=10*200 // in cm^2/V-s disp(u,"The electron mobility, un(cm^2/V-s) = sigma*RH =")
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//Initilization of variables F=[100;50;-150] //Force vector N a=2 //m b=2 //m c=3 //m d=2 //m e=4 //m f=8 //m //Calculations R=F(1,1)+F(2,1)+F(3,1) //N M_x=-F(1,1)*a+F(2,1)*b-F(3,1)*c //N-m M_z=F(1,1)*d+F(2,1)*e+F(3,1)*f //N-m C=sqrt(M_x^2+M_z^2) //N-m thetax=atand(-M_x/M_z) //degrees //result clc printf('The resultant is %f N \n',R) printf('The moment about x axis is %f N.m \n',M_x) printf('The moment about z axis is %f N.m\n',M_z) printf('The couple acting is %f N.m\n',C) printf('The trace makes an angle with x axis of %f degrees',thetax)
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codeblock readtextfile(ScriptDir+"\_TOOLS.sci"); codeblock readtextfile(ScriptDir+"\_SSYS.sci"); codeblock readtextfile(ScriptFilePath+"\_PlanetariumTools.sci"); codeblock readtextfile(ScriptFilePath+"\_Colors.sci"); #================================================================================= # SETUP #================================================================================= mydata=map; mydata.scfac=100; mydata.starsizefrac=0.004; mydata.longit=(3+43/60.0)/180*Pi; mydata.lattit=(51+3/60.0)/180*Pi; mydata.camh=0.1; InitPlanetarium(ref(mydata)); CreatePlanetariumClock; root.SC.Universe.ClockFrame.Clock1.size=0.1; root.SC.Universe.ClockFrame.Clock1.position=point(0.11,0.11); root.SC.Universe.ClockFrame.Clock2.visible=false; CreateBackDropGent(ref(mydata)); root.SC.Universe.SolarSystem.Earth.Inclin.Globe.ViewerFrame.Grid.visible=false; root.SC.Universe.SolarSystem.Earth.Inclin.Globe.ViewerFrame.Directions.visible=false; Planetarium_CreateViewPort_Earth(ref(mydata)); vpearth=mydata.viewport_earth; vpearth.EraseBackground=true; ratio=root.Viewports.main.aspectratio; sz=0.4;offset=0.03; vpearth.XMinFrac=1-offset-sz; vpearth.YMaxFrac=1.0-offset*ratio; vpearth.XMaxFrac=1-offset; vpearth.YMinFrac=vpearth.YMaxFrac-sz*ratio; vpearth.active=false; dist=14000000; dist=10000000; vpearth.FocalDistance=2.0*dist; vpearth.cameradir=vecnorm(vector(-1,0,-0.5)); vpearth.camerapos=point(0,0,0)-dist*vpearth.cameradir; vpearth.Aperture=60/180*Pi; vpearth.nearclipplane=0.4*dist; vpearth.FrameSize=0.015; vpearth.FrameColor=color(0.2,0.2,0.2); root.SC.Universe.SolarSystem.Earth.PlanetsIndicators.visible=false; root.SC.Universe.SolarSystem.Earth.MoonHalo.visible=false; root.SC.Universe.SolarSystem.Earth.StarGlobeFrame.StarglobeFront.linesize=0; #enhanced texture on Earth eglobe=GetPlanetBodyFrame("Earth"); etx2=eglobe.CreateTexture("Earth2",DataDir+"\textures\earth_3.jpg"); eglobe.GlobeRendering.Earth.Texture=etx2.name; ############################### Create Ptolemaeus viewport try { DelObject(root.Viewports.Ptol); } displayname=ReadSetting("DisplayName",""); if displayname=="" then displayname="\\.\DISPLAY1"; vp2=CreateNewViewPort(0.55,0.3,1,1); vp2.name="Ptol"; vp2.Framesize=0.003; vp2.start; vp2.setscene(root.SC); vp2.FocalDistance=90000000; vp2.cameradir=vecnorm(vector(1,0,0)); vp2.camerapos=point(0,0,0)-0.5*vp2.FocalDistance*vp2.cameradir; vp2.cameraupdir=vector(0,0,1); vp2.enableusernavigation=root.Viewports.main.enableusernavigation; vp2.EnableUserTimeControl=root.Viewports.main.EnableUserTimeControl; vp2.NearClipPlane=0.1*vp2.FocalDistance; vp2.FarClipPlane=40*vp2.FocalDistance; vp2.Aperture=40/180*Pi; vp2.XMinFrac=offset; vp2.YMaxFrac=1.0-offset*ratio; vp2.XMaxFrac=offset+sz; vp2.YMinFrac=vp2.YMaxFrac-sz*ratio; #vp2.EraseBackground=true; vp2.FrameSize=0.015; vp2.FrameColor=color(0.2,0.2,0.2); vp2.FrameSize=0; vp2.active=false; root.SC.Universe.StarBackFrame.addignoreviewport("Ptol"); root.SC.Universe.SolarSystem.Earth.Inclin.Globe.ViewerFrame.addignoreviewport("Ptol"); root.SC.Universe.ClockFrame.addignoreviewport("Ptol"); root.SC.Universe.SolarSystem.Earth.PlanetsIndicators.addignoreviewport("Ptol"); root.SC.Universe.SolarSystem.Earth.SunHalo.addignoreviewport("Ptol"); root.SC.Universe.SolarSystem.Earth.MoonHalo.addignoreviewport("Ptol"); root.SC.Universe.SolarSystem.Earth.Luna.addignoreviewport("Ptol"); root.SC.Universe.SolarSystem.Earth.StarGlobeFrame.addignoreviewport("Ptol"); globeframe=root.SC.Universe.SolarSystem.Earth.addsubframe("StarGlobe"); globeframe.addignoreviewport("main"); globeframe.addignoreviewport("Earth"); globeradius=4*root.SC.Universe.SolarSystem.Earth.Inclin.Globe.GlobeRendering.Earth.radius; #create halo halfcircle=FlatContourSet; haloframe=globeframe.addviewdirframe(point(0,0,0),"haloframe"); halfcircle.generate(functor("point("+str(globeradius)+"*sin(a),"+str(globeradius)+"*cos(a),0)","a"),Pi,0,40); halo=haloframe.add("SolidObject","Name":"Halo"); halo.Revolve(halfcircle,40); halo.BlendType=BlendTransparent; halo.RenderBack=true; halo.DepthMask=DepthMaskDisable; halo.EnableLight=false; halo.GenerateVertexProperty(functor("color(0,0,1/(1+5*sqr(p.z/"+str(globeradius)+")),0.2)","p"),VertexPropertyColor); halo.canbuffer=true; #Star globe tx=globeframe.createtexture("star",DataDir+"\textures\star4.bmp"); st1=globeframe.add("StarGlobe","Name":"StarglobeFront"); st1.radius=globeradius; st1.texture="star"; st1.StarSize=0.013*globeradius; st1.LineSize=0;#0.01*globeradius; st1.linecolor=color(0,0.5,1,0.4); st1.renderback=false; st2=globeframe.add("StarGlobe","Name":"StarGlobeBack"); st2.radius=globeradius; st2.texture="star"; st2.StarSize=0.007*globeradius; st2.LineSize=0;#0.01*globeradius; st2.renderfront=false; st2.linecolor=color(0,0.5,1,0.4); st2.color=color(1,1,1,0.15); #Create milky way #galactic pole glong=179.32095/180*Pi; glatt=29.811954/180*Pi; ez=-1*vector(cos(glong)*cos(glatt),sin(glong)*cos(glatt),sin(glatt)); #galactic center glong=266.14097/180*Pi; glatt=-5.52967943/180*Pi; ex=vector(cos(glong)*cos(glatt),sin(glong)*cos(glatt),sin(glatt)); ey=vecnorm(ez*ex); mwf=globeframe.addsubframe("MilkyWay"); mwf.transf.Xaxis=-1*ex; mwf.transf.Yaxis=-1*ey; mwf.transf.Zaxis=ez; tx=mwf.createtexture("MilkyWay",DataDir+"\textures\milkyway.png"); mw=mwf.add("sphere","EnableLight":false); mw.color=color(0.3,0.5,1,0.45); mw.texture=tx.name; mw.BlendType=BlendTransparent;mw.DepthMask=DepthMaskDisable; mw.renderback=false;mw.renderfront=true; mw.radius=globeradius; linew=25; crossframe=root.SC.Universe.addscreenframe("CrossFrame"); crossframe.addignoreviewport("main"); crossframe.addignoreviewport("earth"); ln1=crossframe.add("Curve","Color":GetColor("Red"),"Size":linew); csz=0.12;coffs=0.03; ln1.makeline(point(coffs,1-coffs,0),point(coffs+csz,1-coffs-csz,0)); ln1=crossframe.add("Curve","Color":GetColor("Red"),"Size":20); ln1.makeline(point(coffs,1-coffs-csz,0),point(coffs+csz,1-coffs,0)); crossframe.visible=false; ext=linew/5000; checkframe=root.SC.Universe.addscreenframe("CheckFrame"); checkframe.addignoreviewport("main"); checkframe.addignoreviewport("Ptol"); ln1=checkframe.add("Curve","Color":color(0,0.75,0),"Size":linew); csz=0.14;sz1=0.06;sz2=0.12; ln1.makeline(point(coffs,1-coffs-sz2+sz1,0),point(coffs+sz1+ext,1-coffs-sz2-ext,0)); ln1=checkframe.add("Curve","Color":color(0,0.75,0),"Size":20); ln1.makeline(point(coffs+sz1,1-coffs-sz2,0),point(coffs+sz1+sz2,1-coffs,0)); checkframe.visible=false; ############################### End Create Ptolemaeus root.time=time(2012,1,1,17,40,0); myviewport=T_getviewport; myviewport.FocalDistance=0.06*mydata.scfac; myviewport.Aperture=80/180*Pi; myviewport.cameradir=vecnorm(vector(0.92526291, 0.08590846, 0.369470278-0.05)); #================================================================================= # ANIMATION #================================================================================= root.TimeSpeedFactor=0; FadeViewportsIn; animate(4); Cam_RotateHor(myviewport,-0.6*Pi,14); animate(18); root.TimeSpeedFactor=700; animate(6); Cam_RotateHor(myviewport,-0.6*Pi,10); animate(12); PointHighlight_Start("Pole",color(1,0.5,0), 40.5,89.3); animate(6); ConstellHighlight_Stop("Pole"); animate(4); root.Viewports.Ptol.active=true; animate(10); root.Viewports.Ptol.EraseBackground=true; root.Viewports.Ptol.FrameSize=0.015; vpearth.active=true; animate(3); root.SC.Universe.CrossFrame.visible=true; root.SC.Universe.CrossFrame.blinkperiod=0.4;root.SC.Universe.CrossFrame.maxblinkcount=4; root.SC.Universe.CheckFrame.visible=true; root.SC.Universe.CheckFrame.blinkperiod=0.4;root.SC.Universe.CheckFrame.maxblinkcount=4; animate(10); FadeViewportsOut; stop;
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EX_2_27.SCE
// Example 2.27: Find R clc; clear; close; Vcc=24;// Colector voltage in volts Beta=45; Rc=10;// Collector resistance in killo ohms Re= 0.27;// in kilo ohms Vce=5;// Collector to emitter voltage in volts Vbe=0.6;// Base to emitter voltage in volts Ib=(Vcc-Vce)/((1+Beta)*(Rc+Re));//in milli ampere Ic=Ib/Beta;// in micro ampere R=(Vce-Vbe)/Ib;// Resistance in killo ohms disp (R,"Base resistance in killo ohms")
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Chapter5_Example4.sce
//Chapter-5, Illustration 4, Page 252 //Title: Air Compressors //============================================================================= clc clear //INPUT DATA x=0.05;//Ratio of clearance volume to swept volume P1=1;//Pressure at point 1 in bar T1=310;//Temperature at point 1 in K n=1.2;//Adiabatic gas constant P2=7;//Pressure at point 2 in bar Pa=1.01325;//Atmospheric pressure in bar Ta=288;//Atmospheric temperature in K //CALCULATIONS V1=1+x;//Ratio of volume of air sucked to stroke volume V4=((P2/P1)^(1/n))/20;//Ratio of volume delivered to stroke volume DV=V1-V4;//Difference in volumes nv1=DV*100;//Volumetric efficiency V=(P1*DV*Ta)/(T1*Pa);//Ratio of volumes referred to atmospheric conditions nv2=V*100;//Volumetric efficiency referred to atmospheric conditions W=(n*0.287*T1*((P2/P1)^((n-1)/n)-1))/(n-1);//Work required in kJ/kg //OUTPUT mprintf('Volumetric efficiency is %3.1f percent \n Volumetric efficiency referred to atmospheric conditions is %3.1f percent \n Work required is %3.1f kJ/kg',nv1,nv2,W) //==============================END OF PROGRAM=================================
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22_2.sce
//1 m radius wheel //refer fig. 22.4(a),(b),(c),(d),(e) and (f) vA=1*5 //m/sec aA=1*4 //m/sec^2 vBA=1*5 //m/sec vB=vA+vBA //m/sec aBA=1*4 //m/sec^2 an=5^2 //m/sec^2 aB=sqrt((8^2)+(25^2)) //m/sec^2 theta=atand(25/8) //degree //Consider rotation of point D vDx=5+3*sind(60) //m/sec vDy=3*cosd(60) //m/sec vD=7.745 //m/sec //inclination to horizontal theta2=atand(1.5/7.598) //degree vDA=0.6*5 //m/sec^2 aD=sqrt((14.190^2)+(1.422^2)) //m/sec^2 theta3=atand(14.190/1.422) //degree printf("\nAt B\naB=%.3f m/sec^2\ntheta=%.2f degree\nvB=%.3f m/sec\nAt D\nvD=%.3f m/sec^2\ntheta2=%.2f degree\naD=%.3f m/sec^2\ntheta3=%.2f degree",aB,theta,vB,vD,theta2,aD,theta3)
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Newton-Raphson.sci
// metodo de Newton-Raphson // "inicio" indica el valor x para el cual comenzar // ATENCION!!!!!! ANTES DE USAR EL METODO SE DEBE ACLARAR LA FUNCION A UTILIZAR function f=newtonRaphson(xcero,iteraciones) for i=1: iteraciones disp(" ^^^ITERACION ",i) f=xcero-(miFuncion(xcero)/miFuncionDerivada(xcero)) xcero=f disp(f) end endfunction function y=miFuncion(x) //aqui especificar FUNCION y=2*x^3+6*x^2+6*x-1 disp('valor f(x) ',y) endfunction function d=miFuncionDerivada(x) //aqui especificar FUNCION d=6*x^2+12*x+6 disp('valor df(x)',d) endfunction
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Ex6_13.sce
//=============================================================================================================================================== // chapter 6 example 13 clc; clear; //input data a = 110*10^-3; //area in m^2 d = 2; //thickness in mm er = 5; //relative permitivity E = 12.5*10^3; //electric field strength in V/mm e0 = 8.854*10^-12; //charge of electron in coulombs //calculations A = a*a; //area in m^2 C = e0*((er*A)/(d*10^-3)) //capacitance in F V = E*(d); Q = (C)*(V) //charge on capacitor in C // result mprintf('capacitance =%3.2e.F\n',C); mprintf(' charge=%3.4e C\n',Q); //==============================================================================================================================================
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Example10_1.sce
//Example 10.1. refer fig.10.8. clc format(6) hie=1600 hfe=60 hre=5*10^-4 hoe=25*10^-6 hic=1600 hfc=-61 hrc=1 hoc=25*10^-6 disp("The AC equivalent circuit of the CE-CC amplifier is shown in fig.10.9(a)") disp("The Second Stage :") disp("Current gain :") disp("The current gain of a particular stage is given by") disp(" AI = -hf / (1 + ho*ZL)") disp("For the second stage ZL = RE2 and the current gain of the second stage is") RE2=4000 AI2=-hfc/(1+(hoc*RE2)) disp(AI2," AI2 = -Ie2 / Ib2 = -hfc / (hoc*RE2) =") disp("The input impedance Ri of a particular stage is given by") disp(" Ri = hi + hf*AI*ZL") disp("For the second stage,") Ri2 = hic + (hrc*AI2*RE2) Ri22=Ri2*10^-3 disp(Ri22," Ri2(k-ohm) = hic + (hrc*AI2*RE2) =") disp("Thus, the CC stage has a high input impedance.") disp("The voltage gain of a particular stage is") disp(" AV = (AI*ZL) / Zi") disp("For the second stage,") Re2=4000 AV2=(AI2*Re2)/Ri2 disp(AV2," AV2 = Vo/V2 = (AI2*Re2) / Ri2") disp("The First Stage :") RC1=4000 format(5) RL1=(RC1*Ri2)/(RC1+Ri2) RL11=RL1*10^-3 disp(RL11," RL1(k-ohm) = RC1 || Ri2 =") disp("Current gain,") AI1= -hfe/(1+(hoe*RL1)) disp(AI1," AI1 = -IC1/Ib1 = -hfe/(1+(hoe*RL1)) =") disp("The input impedance of the first stage, which is also the input impedance of the cascaded amplifier is") Ri1=hie +(hre*AI1*RL1) // answer in textbook is wrong Ri11=Ri1*10^-3 disp(Ri11," Ri1(k-ohm) = hie + hre*AI1*RL1 =") disp("The voltage gain of the first stage is") format(7) AV1=(AI1*RL1)/Ri1 // answer in textbook is wrong disp(AV1," AV1 = V2/V1 = (AI1*RL1) / Ri1 =") disp("The output admittance of the first transistor Q1") RS=600 format(5) Yo1=hoe-((hfe*hre)/(hie+RS)) Yo0=Yo1*10^6 disp(Yo0," Yo1(uA/V) = hoe - ((hfe*hre) / (hie+RS)) =") disp("The output impedance of the first stage") format(6) Ro1=1/Yo1 Ro0=Ro1*10^-3 disp(Ro0," Ro1(k-ohm) = 1 / Yo1 =") disp("The output impedance taking RC1 into account is") format(5) Rot1=(Ro1*RC1)/(Ro1+RC1) Rott=Rot1*10^-3 disp(Rott," Rot1(k-ohm) = Ro1 || RC1 =") disp("This is the effective source resistance RS2 of the second stage") disp("The output admittance of the second stage") format(7) Yo2=hoc-((hfc*hrc)/(hic+Rot1)) disp(Yo2," Yo2(A/V) = hoc-((hfc*hrc) / (hic+Rot1)) =") disp("Output impedance,") format(4) RO2=1/(11.525*10^-3) disp(RO2," RO2(ohm) = 1 / Yo2 =") disp("The amplifier output impedance taking RE2 into account is RO2 || RE2") format(6) Ro2=(87*4000)/(87+4000) disp(Ro2,"Hence, Ro2(ohm) = (RO2*RE2) / (RO2+RE2) =") disp("Overall current gain :") disp("The output or total current gain of both the stages is") disp(" AI = -Ie2 / Ib1 = (-Ie2/Ib2)(Ib2/IC1)(IC1/Ib1)") disp(" = -AI2*(Ib2/Ic1)*AI1") disp("From fig.10.9(b),") disp(" Ib2 = (-IC1)(Rc1 / Rc1+Ri2)") Rc1=4000 format(7) x=(-Rc1)/ (Rc1+Ri2) disp(x," Ib2/Ic1 = -Rc1/ Rc1+Ri2 =") format(6) AI=-AI2*x*AI1 disp(AI," AI = -AI2*AI1*(Rc1 / Ri2+Rc1) =") disp("The overall voltage gain of the amplifier,") disp(" AV = Vo / V1 = (Vo/V2)(V2/V1)") AV=AV2*AV1 disp(AV," AV = AV2*AV1 =") // answer in textbook is wrong disp("The overall voltage gain taking the source impedance into account,") format(4) AVs=AV*(Ri1/(Ri1+RS)) disp(AVs," AVs = Vo/Vs = Av(Ri1 / Ri1+Rs) =") // answer in textbook is wrong
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Ex12_6.sce
clc // Fundamental of Electric Circuit // Charles K. Alexander and Matthew N.O Sadiku // Mc Graw Hill of New York // 5th Edition // Part 2 : AC Circuits // Chapter 12 : Three Phase Circuit // Example 12 - 6 clear; clc; close; // // Given data Vp_mag = 110.0000; Vp_angle = 0.0000; Ip_mag = 6.8100; Ip_angle = -21.8000; Z1 = complex(10,8); Z2 = complex(5,-2); // // Calculations Complex Power Absorbed by The Source S_s_mag = -3*Vp_mag*Ip_mag; S_s_angle = Vp_angle + (-1*Ip_angle); P_s = S_s_mag * cosd(S_s_angle); Q_s = S_s_mag * sind(S_s_angle); // Calculations Complex Power Absorbed By Load 1 Z1_mag = norm(Z1); Z1_real = real(Z1); Z1_imag = imag(Z1); Z1_angle = atand(Z1_imag,Z1_real) S_1_mag = 3*(Ip_mag)^2.00*Z1_mag S_1_angle = Z1_angle P_1 = S_1_mag * cosd(S_1_angle); Q_1 = S_1_mag * sind(S_1_angle); // Calculations Complex Power Absorbed By Load 2 Z2_mag = norm(Z2); Z2_real = real(Z2); Z2_imag = imag(Z2); Z2_angle = atand(Z2_imag,Z2_real) S_2_mag = 3*(Ip_mag)^2.00*Z2_mag S_2_angle = Z2_angle P_2 = S_2_mag * cosd(S_2_angle); Q_2 = S_2_mag * sind(S_2_angle); // disp("Example 12-6 Solution : "); printf(" \n S_s_mag = Magnitude of Complex Power Absorbed by The Source = %.3f VA",S_s_mag) printf(" \n S_s_Angle = Angle of Complex Power Absorbed by The Source = %.3f Degree",S_s_angle) printf(" \n P_s = Real Power Absorbed by The Source = %.3f Watt",P_s) printf(" \n Q_s = Reactive Power Absorbed by The Source = %.3f Var",Q_s) printf(" \n S_1_mag = Magnitude of Complex Power Absorbed by The Load 1 = %.3f VA",S_1_mag) printf(" \n S_1_Angle = Angle of Complex Power Absorbed by The Load 1 = %.3f Degree",S_1_angle) printf(" \n P_1 = Real Power Absorbed by The Load 1 = %.3f Watt",P_1) printf(" \n Q_1 = Reactive Power Absorbed by The Load 1 = %.3f Var",Q_1) printf(" \n S_2_mag = Magnitude of Complex Power Absorbed by The Load 2 = %.3f VA",S_2_mag) printf(" \n S_2_Angle = Angle of Complex Power Absorbed by The Load 2 = %.3f Degree",S_2_angle) printf(" \n P_2 = Real Power Absorbed by The Load 2 = %.3f Watt",P_2) printf(" \n Q_2 = Reactive Power Absorbed by The Load 2 = %.3f Var",Q_2)
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mode(1) //execution mode with echo A=[1 2;3 4];y=[3;5]; x1=linsolve(A,-y); //not displayed, even "with echo" x2=A^(-1)*y //displayed if "with echo" disp(x1,'x=') //displayed even "with no echo"
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$thermo = VirtualMaterials.Advanced_Peng-Robinson / -> $thermo thermo + n-BUTANE ISOBUTANE pfr = KineticReactor.PFR() pfr.In.T = 330 K pfr.In.P = 3000 kPa pfr.In.Fraction = 0.9 0.1 pfr.In.MoleFlow = 163 pfr.Length = 12.9 m pfr.Diameter = 0.6 m pfr.OutQ = 0 pfr.NumberSections = 40 pfr.NumberRxn = 1 pfr.Rxn0.Formula = theRxn0:1.0*ISOBUTANE-1.0*!'n-BUTANE' pfr.CustomEquationUnitSet = sim42 pfr.Rxn0.ReactionRateEq = """ #The following are plain Python lines with the final goal of defining a variable called r #which will be interpreted as the reaction rate in the CustomEquationUnitSet units. #Define some constants E = 65700.0 #J/mol = kJ/kmol T1 = 360.0 #K kRef = 31.1 #1/h #R is automatically loaded in sim42 units as kJ/kmolK k = kRef*exp( (E/R)*(1.0/T1 - 1/T) ) #1/h T2 = 60.0 + 273.15 #K KcRef = 3.03 Kc = KcRef*exp( (-6900.0/R)*(1/T2 - 1/T) ) #The unit set defined is sim42, hence concentration comes in kmol/m3 r = k*(rxnCmp['n-BUTANE'].Concentration - rxnCmp['ISOBUTANE'].Concentration/Kc) #The unit set defined is sim42, hence r has to be returned in kmol/(s*m3) r = r/3600.0 #kmol/(s*m3) """ pfr.DeltaP = 0.0 pfr.T pfr.f pfr.r pfr.Ignored = 1 #Now solve it by providing r in different units pfr.CustomEquationUnitSet = Field pfr.Rxn0.ReactionRateEq = """ #The following are plain Python lines with the final goal of defining a variable called r #which will be interpreted as the reaction rate in the CustomEquationUnitSet units. #Define some constants E = 65700.0 * 0.43 #Btu/lbmol T1 = 360.0 * 1.8 #R kRef = 31.1 #1/h #T came in F. Make it R T = T + 459.67 #R #R is automatically loaded in Field units as psia-ft3/lbmolR R = 1.987 #Btu/lbmolR k = kRef*exp( (E/R)*(1.0/T1 - 1/T) ) #1/h T2 = (60.0 + 273.15) * 1.8 #R KcRef = 3.03 Kc = KcRef*exp( ((-6900.0*0.43)/R)*(1/T2 - 1/T) ) #The unit set defined is sim42, hence concentration comes in lbmol/ft3 r = k*(rxnCmp['n-BUTANE'].Concentration - rxnCmp['ISOBUTANE'].Concentration/Kc) #The unit set defined is Field, hence r has to be returned in lbmol/(s*ft3) r = r/3600.0 #lbmol/(s*ft3) """ pfr.Ignored = None pfr.T pfr.f pfr.r copy /pfr paste / pfrClone.T pfrClone.f pfrClone.r
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function[res]=slaplacien(D,n) if n>1 then h=1/(n-1) //Définir T1 & Tn sous forme de matrice creuse T1=sparse(-3*eye(n,n)+diag(ones(n-1,1),1)+diag(ones(n-1,1),-1)) T1(1,1)=T1(1,1)+1 T1(n,n)=T1(n,n)+1 T1=(1/(h*h))*T1 //Définir Tk sous forme de matrice creuse Tk=sparse(-4*eye(n,n)+diag(ones(n-1,1),1)+diag(ones(n-1,1),-1)) Tk(1,1)=Tk(1,1)+1 Tk(n,n)=Tk(n,n)+1 Tk=(1/(h*h))*Tk //Définir A sous forme de matrice creuse A=sparse(zeros(n*n,n*n)+(1/(h*h))*diag(ones(n*(n-1),1),n)+(1/(h*h))*diag(ones(n*(n-1),1),-n)) k=1 //Affecter les T1 i= indice_i(k,n) j= indice_j(k,n) A([ i : j ],[ i : j ])=T1 k=k+1 //Affecter les Tk while(k<=n-1), i= indice_i(k,n) j= indice_j(k,n) A([ i : j ],[ i : j ])=Tk k=k+1 end //Affecter les T1 au bout de k=n => Tn if k==n then i= indice_i(k,n) j= indice_j(k,n) A([ i : j ],[ i : j ])=T1 end //Renvoyer la matrice creuse résultante de (n*n ; n*n) res=D.*A else res=0 end endfunction //Calcul de l'indice i de sous-matrice function[res]=indice_i(k,n) res=(n*(k-1))+1 endfunction //Calcul de l'indice j de sous-matrice function[res]=indice_j(k,n) res=k*n endfunction //-- L'Exemple --// n=3 D=2 res=slaplacien(D,n)
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//Example 7.22 // SRL design for satellite attitude control xdel(winsid())//close all graphics Windows clear; clc; //------------------------------------------------------------------ //Transfer function for satellite attitude control system s=poly(0,'s'); nums=1; dens=s^2; num_s=1; den_s=(-s)^2; G0s=syslin('c',nums/dens); //G0(s) G0_s=syslin('c',num_s/den_s); //G0(-s) //evans(G0s*G0_s) evans(1/s^4) zoom_rect([-3 -3 3 3]) f=gca(); f.x_location = "origin" f.y_location = "origin" xset("color",2); h=legend(''); h.visible = "off" //Title, labels and grid to the figure exec .\fig_settings.sci; //custom script for setting figure properties title('Symmetric root locus for the satellite','fontsize',3); //------------------------------------------------------------------ //Root locus design //choose rho=4.07 that places pole at -1+-j rho=4.07; chr_eqn=(1+rho*G0s*G0_s) p=[-1+%i, -1-%i]; sig=real(p); omega=imag(p); plot(sig,omega,'ro') xstring(-2.2,0.5,["pole locations at";"$\rho=4.07$"]) //------------------------------------------------------------------ //pole-placement design; sys=tf2ss(G0s); exec('acker_dk.sci', -1); K=acker_dk(sys.A,sys.B,p); syscl=syslin('c',(sys.A-sys.B*K),sys.B, sys.C, sys.D) disp(spec(syscl.A),"Closed loop eigen values"); //------------------------------------------------------------------
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// page no 629 // example no A.8 // SUBTRACTION OF SIGNED NUMBERS clc; printf('Part a \n \n') printf('Minuend= FAH \n \n'); printf('It is a negative no since D7= 1 for FAH, this must be represented in \n2s compliment form. \n'); // finding 2's complement of subtrahend (FAH); m=hex2dec(['FA']); x=hex2dec(['62']); y=bitcmp(m,8); // 1's compliment of FAH z=y+1; // 2's compliment of FAH printf('2s compliment of minuend is= '); disp(z); printf('\n \n Subtrahend= 62H \n'); printf('It is a positive no since D7= 0 for 62H. \n'); // subtraction can be represented as // FAH-62H= (-06H)-(+62H) s=-x-z; a=-s; d=dec2hex(a); printf('Subtraction= '); disp(s); disp(d); printf('in hexadecimal with a negative sign \n \n'); g=bitcmp(a,8); // 1's compliment of result q=g+1; // 2's compliment of result e=dec2hex(q); printf('2s compliment of result would be= '); disp(e);
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scenario = "Visual Oddball Instructions (German Version)"; scenario_type = trials; # sets the default text font default_font = "Arial"; default_font_size = 14; default_text_color = 0,0,0; # sets text to black # sets the background colour to white (default is black) default_background_color = 255,255,255; #center the text default_text_align = align_center; begin; bitmap { filename = "SleepLookCircle.bmp";} NoSleep; bitmap { filename = "SleepLook.bmp";} Sleep; bitmap { filename = "IncorrectLookCircle.bmp";} NoLookAway; bitmap { filename = "IncorrectLook.bmp";} LookAway; bitmap { filename = "CorrectLookCircle.bmp";} YesLook; bitmap { filename = "CorrectLook.bmp";} Look; bitmap { filename = "BlankSubject.bmp";} sub; bitmap { filename = "F.pcx";} F; bitmap { filename = "T.pcx";} T; bitmap { filename = "default.pcx";} blankPCX; wavefile { filename = "nvmmn_instructionsDE.wav"; } visInstruct; sound { wavefile visInstruct; attenuation = 0.2; } visualInstruction; picture { bitmap blankPCX; x = 0; y = 0; } default; trial { trial_duration = 58000; sound visualInstruction; time = 0; picture {bitmap sub; x = 0; y = 0; }; time = 1000; duration = 8000; picture {bitmap F; x = 0; y = 0; }; time= 9000; # 9 secs duration = 6000; picture {bitmap T; x = 0; y = 0; }; time = 16000; # 16 secs duration = 6000; picture {bitmap T; x = 0; y = 0; }; time = 28000; # 28 secs duration = 2000; picture {bitmap sub; x = 0; y = 0; }; time = 34000; # 34 secs duration = next_picture; picture {bitmap Look; x = 0; y = 0; }; time = 40200; #40.2 secs duration = next_picture; picture {bitmap YesLook; x = 0; y = 0; }; time = 40400; #40.4 secs duration = next_picture; picture {bitmap Look; x = 0; y = 0; }; time = 40600; #50.4 secs duration = next_picture; picture {bitmap YesLook; x = 0; y = 0; }; time = 40800; duration = next_picture; picture {bitmap Look; x = 0; y = 0; }; time = 41000; duration = next_picture; picture {bitmap YesLook; x = 0; y = 0; }; time = 41200; duration = next_picture; picture {bitmap sub; x = 0; y = 0; }; time = 44200; duration = next_picture; picture {bitmap LookAway; x = 0; y = 0; }; time = 444000; duration = next_picture; picture {bitmap NoLookAway; x = 0; y = 0; }; time = 44800; duration = next_picture; picture {bitmap LookAway; x = 0; y = 0; }; time = 46000; duration = next_picture; picture {bitmap NoLookAway; x = 0; y = 0; }; time = 46200; duration = next_picture; picture {bitmap LookAway; x = 0; y = 0; }; time = 46400; duration = next_picture; picture {bitmap NoLookAway; x = 0; y = 0; }; time = 46600; duration = next_picture; picture {bitmap sub; x = 0; y = 0; }; time = 46800; duration = next_picture; picture {bitmap Sleep; x = 0; y = 0; }; time = 55000; duration = next_picture; picture {bitmap NoSleep; x = 0; y = 0; }; time = 55200; duration = next_picture; picture {bitmap Sleep; x = 0; y = 0; }; time = 55400; duration = next_picture; picture {bitmap NoSleep; x = 0; y = 0; }; time = 55600; duration = next_picture; picture {bitmap Sleep; x = 0; y = 0; }; time = 55800; duration = next_picture; picture {bitmap NoSleep; x = 0; y = 0; }; time = 56000; duration = next_picture; picture {bitmap sub; x = 0; y = 0; }; time = 49000; duration = next_picture; };
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//i/p arg a is negative x=[12 3 4 5 6 7 8 9]; a=-5; y=polyscale(x,a); disp(y); //output // // column 1 to 5 // // 12. - 15. 100. - 625. 3750. // // column 6 to 8 // // - 21875. 125000. - 703125.
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clc // given data w=0.6 // in km h2=2.5 // in km p=5/100.0 // porosity rhor=3000.0 // density of sediment in kg/m^3 cr=750.0 // specific heat of sediment in J/kg-K rhow=1000.0 // density of water in kg/m^3 cw=4200.0 // specific heat of water in J/kg-K G=35.0 // temperature gradient in degree C/km T1=45.0 // temp 1 in degree celsius T0=12.0 // temp 2 in degree celsius Q=0.75 // water extraction rate in m^3/sec-km^2 tau=((p*rhow*cw+(1-p)*rhor*cr)*w*1000**3/(Q*rhow*cw))/(60*60*24*365) // in years printf( "Time constant is %.1f years",tau) // the answer is different in textbook due to wrong calculations
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<?xml version="1.0" encoding="utf-8"?> <test> <description>Linear stability with coupled solver (LM with Arpack and Real Shift): ChannelMax Ev = (0.00248682 -0.158348i) </description> <executable>IncNavierStokesSolver</executable> <parameters> -P nvec=4 -P kdim=32 -P imagShift=0.0 -I ArpackProblemType=LargestMag PPF_R15000_3D.xml</parameters> <files> <file description="Session File">PPF_R15000_3D.xml</file> </files> <metrics> <metric type="Eigenvalue" id="0"> <value tolerance="0.001">-0.000201531,0</value> </metric> </metrics> </test>
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// Chapter6 // Page.No-193, Figure.No-6.5(a) // Example_6_3 // Components of peak amplifier // Given clear;clc; fp=16*10^3; // Peak frequency Af=10; // Gain at peak frequency C=0.01*10^-6; // Assume L=1/(((2*%pi*fp)^2)*10^-8); // Simplifying fp=1/(2*pi*sqrt(L*C)) printf("\n Inductance is = %.4f H \n",L) L=10*10^-3; // Approximate R=30; // Assume the value of internal resistance of the inductor Xl=2*%pi*fp*L; // Inductive reactance Qcoil=Xl/R; // Figure of merit of the coil printf("\n Figure of merit of the coil is = %.1f \n",Qcoil) Rp=(Qcoil)^2*R; // Parallel resistance of the tank circuit printf("\n Parallel resistance of the tank circuit is = %.1f ohm \n",Rp) R1=100; // Assume the value of internal resisrance of the coil Rf=-Rp/(1-(Rp/(Af*R1))); // Simplifying Af=(Rf||Rp)/R1 printf("\n Feedback resistance is = %.1f ohm \n",Rf)
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function rectOutput = GetBoundingBox(plPoints) xMin = min(plPoints(:,1)); xMax = max(plPoints(:,1)); yMin = min(plPoints(:,2)); yMax = max(plPoints(:,2)); rectOutput = [xMin, yMax; xMax - xMin, yMax - yMin]; //[Left Top; Width Hieght] endfunction
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// Scilab Code Ex3.5: Page-89 (2006) clc; clear; k = 1.38e-023; // Boltzmann constant, J/K theta_D = 1440; // Debye temperature for Be, K h = 6.626e-034; // Planck's constant, Js f_D = k*theta_D/h; // Debye cut off frequency of Be, Hz printf("\nThe Debye cut off frequency of Be = %g per sec", f_D); // Result // The Debye cut off frequency of Be = 2.99909e+013 per sec
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//Ex:3.18 clc; clear; close; n=10;// number of isotropic elements // d=y/4 // Do=1.789(4n*(d/y)) // Do=1.789(4n*(y/4y)=2n(1/4)) Do=1.789*(4*n*(1/4)); D0=10*log(Do)/log(10);// Directivity in db printf("the Directivity = %f dB", D0);
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welcome to my activities: A1 - Flowcharts & Pseudocodes A2 - Flow-charting Exercise A3 - Pseudo-code Exercises
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//To calculate motor speed and its slip clc; clear; f=50; sf=3/2; s=sf/f; p=8; N=120*f/8; Nr=poly([0 1],'Nr','c'); // Actual Speed Variable x=(750*s)-(750-Nr); // Equation To find the Actual Speed Nr=roots(x); // Actual Speed Constant printf('The motor runs at a speed of %g rpm and has a slip of %g \n',ceil(Nr),s)
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test.txt.tst
This is the first test for github.
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// Example 6.14;//voltage gain ,input & output resistance clc; clear; close; A= 500;// open voltage gain Beta=0.01;// feedback ratio Ri=3;//input resistance without feedback in kiilo ohms Ro=20;//output resistance without feedback in kiilo ohms Af=(A/(1+A*Beta));//Voltage gain is Rif= (1+A*Beta)*Ri;//input RESISTANCE with feedback in kiilo ohms Rof=(Ro/(1+Beta*A));//output resistance with feedback in killo ohms disp(Rif,"input resistance with feedback in kiilo ohms is ") disp(Rof,"output resistance with feedback in killo ohms is ")
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clc; close; fls=loadfls('carro-maxmin'); //carrega o arquivo carro2.fls diretamente no workspace para a variável chamada 'fls' figure(1); plotvar(fls,"input",[1 2]); figure(2); plotvar(fls,"output",1); figure(3); plotsurf(fls,[1 2], 1); //scf();clf(); //plotsurf(fls,[1 2],1); //scf();clf(); //plotsurf(fls,1,1,[0 50]); //scf();clf(); //plotsurf(fls)
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//Exa3 clc; clear; close; //given data //atomic radius r=1.278; //in Angstrum //atomic weight aw=63.5; //Avogadro's number an=6.023*10^23; //copper has FCC structure for which a=(4*r)/sqrt(2);// in Angstrum a=a*10^-10;//in m //Mass of one atom m=aw/an;//in gm m=m*10^-3;//in kg //volume of one unit cell of copper crystal, V=a^3;//in meter cube //Number of atoms present in one unit cell of Cu(FCC Structure), n=4; //Density of crystal rho=(m*n)/V;//in kg/m^3 disp("Density of crystal is : "+string(rho)+"kg/m^3");
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clc //initialisation of variables T= 90 //C T1= 25 //C Cp= 6.9 //cal per mole per degree Cp1= 7.05 //cal per mole per degree Cp2= 18 //cal per mole per degree H= -68.37 //kcal //CALCULATIONS H1= H+(Cp2-Cp-0.5*Cp1)*((T-T1)/1000) //RESULTS printf (' heat of formation= %.2f cal',H1)
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// // 09.02.27 // 14.03.05 MARKLEN function Setmarklen(varargin) global MARKLEN MARKLENInit MARKLENNow; Nargs=length(varargin); if Nargs==0 Tmp=MARKLEN/MARKLENInit; Tmp=round(Tmp*100)/100; disp(Tmp); return; end; Size=varargin(1); MARKLEN=MARKLENInit*Size; MARKLENNow=MARKLEN; endfunction
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# start Dutch.X12 - 2*m + 6*m^2 - 14*m^3 + 4*m^4 + 1 # start Dutch.X12 2*m - 6*m^2 - 4*m^3 + 5*m^4 - 1 # start Dutch.X12 3*m + 3*m^4 # start Dutch.X12 - 3*m + 9*m^2 - 9*m^3 + 6*m^4 Dutch.X12 [0] 0 - 2*M + 6*M^2 - 14*M^3 + 4*M^4 + 1 Dutch.X12 [0] 1 2*M - 6*M^2 - 4*M^3 + 5*M^4 - 1 Dutch.X12 [0] 2 3*M + 3*M^4 Dutch.X12 [0] 3 - 3*M + 9*M^2 - 9*M^3 + 6*M^4 Dutch.X12 [1] 0 - 16*M - 12*M^2 + 2*M^3 + 4*M^4 - 5 Dutch.X12 [1] 1 - 2*M + 12*M^2 + 16*M^3 + 5*M^4 - 4 Dutch.X12 [1] 2 15*M + 18*M^2 + 12*M^3 + 3*M^4 + 6 Dutch.X12 [1] 3 12*M + 18*M^2 + 15*M^3 + 6*M^4 + 3 Dutch.X12 [-1] 0 - 72*M + 72*M^2 - 30*M^3 + 4*M^4 + 27 Dutch.X12 [-1] 1 - 18*M + 36*M^2 - 24*M^3 + 5*M^4 Dutch.X12 [-1] 2 - 9*M + 18*M^2 - 12*M^3 + 3*M^4 Dutch.X12 [-1] 3 - 72*M + 72*M^2 - 33*M^3 + 6*M^4 + 27 Dutch.X12 [2] 0 - 18*M + 18*M^2 + 18*M^3 + 4*M^4 - 27 Dutch.X12 [2] 1 90*M + 90*M^2 + 36*M^3 + 5*M^4 + 27 Dutch.X12 [2] 2 99*M + 72*M^2 + 24*M^3 + 3*M^4 + 54 Dutch.X12 [2] 3 117*M + 99*M^2 + 39*M^3 + 6*M^4 + 54 Dutch.X12 [-2] 0 - 322*M + 186*M^2 - 46*M^3 + 4*M^4 + 205 Dutch.X12 [-2] 1 - 182*M + 138*M^2 - 44*M^3 + 5*M^4 + 83 Dutch.X12 [-2] 2 - 93*M + 72*M^2 - 24*M^3 + 3*M^4 + 42 Dutch.X12 [-2] 3 - 339*M + 207*M^2 - 57*M^3 + 6*M^4 + 210 Dutch.X12 [3] 0 88*M + 96*M^2 + 34*M^3 + 4*M^4 - 5 Dutch.X12 [3] 1 398*M + 228*M^2 + 56*M^3 + 5*M^4 + 248 Dutch.X12 [3] 2 327*M + 162*M^2 + 36*M^3 + 3*M^4 + 252 Dutch.X12 [3] 3 456*M + 252*M^2 + 63*M^3 + 6*M^4 + 315 Dutch.X12 [-3] 0 - 848*M + 348*M^2 - 62*M^3 + 4*M^4 + 763 Dutch.X12 [-3] 1 - 610*M + 300*M^2 - 64*M^3 + 5*M^4 + 452 Dutch.X12 [-3] 2 - 321*M + 162*M^2 - 36*M^3 + 3*M^4 + 234 Dutch.X12 [-3] 3 - 948*M + 414*M^2 - 81*M^3 + 6*M^4 + 819 Dutch.X12 [4] 0 398*M + 222*M^2 + 50*M^3 + 4*M^4 + 217 Dutch.X12 [4] 1 1042*M + 426*M^2 + 76*M^3 + 5*M^4 + 935 Dutch.X12 [4] 2 771*M + 288*M^2 + 48*M^3 + 3*M^4 + 780 Dutch.X12 [4] 3 1173*M + 477*M^2 + 87*M^3 + 6*M^4 + 1092 Dutch.X12 [-4] 0 - 1746*M + 558*M^2 - 78*M^3 + 4*M^4 + 2025 Dutch.X12 [-4] 1 - 1422*M + 522*M^2 - 84*M^3 + 5*M^4 + 1431 Dutch.X12 [-4] 2 - 765*M + 288*M^2 - 48*M^3 + 3*M^4 + 756 Dutch.X12 [-4] 3 - 2043*M + 693*M^2 - 105*M^3 + 6*M^4 + 2268 Dutch.X12 [5] 0 1008*M + 396*M^2 + 66*M^3 + 4*M^4 + 891 Dutch.X12 [5] 1 2142*M + 684*M^2 + 96*M^3 + 5*M^4 + 2484 Dutch.X12 [5] 2 1503*M + 450*M^2 + 60*M^3 + 3*M^4 + 1890 Dutch.X12 [5] 3 2412*M + 774*M^2 + 111*M^3 + 6*M^4 + 2835 Dutch.X12 [-5] 0 - 3112*M + 816*M^2 - 94*M^3 + 4*M^4 + 4411 Dutch.X12 [-5] 1 - 2738*M + 804*M^2 - 104*M^3 + 5*M^4 + 3464 Dutch.X12 [-5] 2 - 1497*M + 450*M^2 - 60*M^3 + 3*M^4 + 1860 Dutch.X12 [-5] 3 - 3768*M + 1044*M^2 - 129*M^3 + 6*M^4 + 5115 Dutch.X12 [6] 0 2014*M + 618*M^2 + 82*M^3 + 4*M^4 + 2365 Dutch.X12 [6] 1 3818*M + 1002*M^2 + 116*M^3 + 5*M^4 + 5411 Dutch.X12 [6] 2 2595*M + 648*M^2 + 72*M^3 + 3*M^4 + 3906 Dutch.X12 [6] 3 4317*M + 1143*M^2 + 135*M^3 + 6*M^4 + 6138 Dutch.X12 [-6] 0 - 5042*M + 1122*M^2 - 110*M^3 + 4*M^4 + 8437 Dutch.X12 [-6] 1 - 4678*M + 1146*M^2 - 124*M^3 + 5*M^4 + 7115 Dutch.X12 [-6] 2 - 2589*M + 648*M^2 - 72*M^3 + 3*M^4 + 3870 Dutch.X12 [-6] 3 - 6267*M + 1467*M^2 - 153*M^3 + 6*M^4 + 10062 Dutch.X12 [7] 0 3512*M + 888*M^2 + 98*M^3 + 4*M^4 + 5083 Dutch.X12 [7] 1 6190*M + 1380*M^2 + 136*M^3 + 5*M^4 + 10352 Dutch.X12 [7] 2 4119*M + 882*M^2 + 84*M^3 + 3*M^4 + 7224 Dutch.X12 [7] 3 7032*M + 1584*M^2 + 159*M^3 + 6*M^4 + 11739 Dutch.X12 [-7] 0 - 7632*M + 1476*M^2 - 126*M^3 + 4*M^4 + 14715 Dutch.X12 [-7] 1 - 7362*M + 1548*M^2 - 144*M^3 + 5*M^4 + 13068 Dutch.X12 [-7] 2 - 4113*M + 882*M^2 - 84*M^3 + 3*M^4 + 7182 Dutch.X12 [-7] 3 - 9684*M + 1962*M^2 - 177*M^3 + 6*M^4 + 17955
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// Problem no 6.5,Page No.157 clc;clear; close; b=0.1 //m //width d=0.2 //m //depth L=2 //m //Length of beam L_1=1 //m //Length from free end E=210*10**9 W=1*10**3 //N //Concentrated Load w=2*10**3 //N/m //Calculations I=b*d**3*12**-1 //m**4 //M.I of the beam section //Slope at free end theta=W*L**2*(2*E*I)**-1+w*L**3*(6*E*I)**-1-w*(L-L_1)**3*(6*E*I)**-1 //Deflection at free end y_b=(W*L**3*(3*E*I)**-1+w*L**4*(8*E*I)**-1-w*(L-L_1)**4*(8*E*I)**-1-w*(L-L_1)**3*L_1*(6*E*I)**-1)*10**3 //Result printf("Slope at free end is %.5f radian",theta) printf("\n Deflection at free end is %.2f mm",y_b)
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clc clear //Input data A1=465.125 //Cross sectional area at entry in cm^2 T1=26.66+273 //Static temperature at section-1 in K P1=3.4473 //Static Pressure at section-1 in bar C1=152.5 //Velocity at section-1 in m/s P2=2.06838 //Static Pressure at section-2 in bar T2=277.44 //Static temperature at section-2 in K C2=260.775 //Velocity at section-2 in m/s Cp=1005 //Specific heat capacity at constant pressure in J/kg-K k=1.4 //Adiabatic constant R=287 //Specific gas constant in J/kg-k //Calculations To1=T1+(C1^2/(2*Cp)) //Stagnation temperature at entry in K To2=T2+(C2^2/(2*Cp)) //Stagnation temperature at exit in K //here To1=To2 from answers d1=(P1*10^5)/(R*T1) //Density at section-1 d2=(P2*10^5)/(R*T2) //Density at section-2 ar=(d2*C2)/(d1*C1) //Ratio of inlet to outlet area A2=A1/ar //Cross sectional area at exit in cm^2 C_max=sqrt(2*Cp*To1) //Maximum velocity at exit in m/s m=d1*A1*C1*10^-4 //Mass flow rate in kg/s F=((P1*10^5*A1*10^-4)-(P2*10^5*A2*10^-4))+(m*(C1-C2)) //Force acting on the duct wall between two sections in N //Output printf('(A)Maximum velocity and stagnation temperature at exit are %3.2f m/s and %3.2f K\n (B)Since Stagnation temperature %3i K at entry and %3i K at exit are equal, the flow is adiabatic\n (C)Cross sectional area at exit is %3.2f cm^2\n (D)Force acting on the duct wall between two sections is %3.2f N',C_max,To2,To1,To2,A2,F)
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// Exa 1.31 clc; clear; close; format('v',6) // Given data R = 1;// in ohm V = 5;// in V V1 = 0.5;// in V R1 = 1;// in k ohm R1 = R1 * 10^3;// in ohm // V-(I_D*R1)-(I_D*R) - V1 = 0; I_D = (V-V1)/(R1+R);// in A I_D = I_D * 10^3;// in mA V_D = (I_D*10^-3*R) + V1;// in V disp("The operating point of the diode is : "+string(V_D)+" V, "+string(I_D)+" mA")
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mode(-1) //to Check all the demos funcprot(0) clearfun('x_message') clearfun('x_dialog') clearfun('x_mdialog') clearfun('x_choose') clearfun('mode') clearfun('xclick') deff('[]=mode(x)','x=x') deff('[]=halt( )',' ') getf('SCI/macros/util/x_matrix.sci','c') getf('SCI/macros/util/getvalue.sci') names=read('SCI/macros/scicos/names',-1,1,'(a)') for k=1:size(names,'r') getf('SCI/macros/scicos/'+names(k)+'.sci') end lines(0) clearfun('lines') deff('x=lines(x)','x=0 ') getf('SCI/tests/dialogs.sci','c') I=file('open','SCI/tests/demos.dialogs','old') O=file('open','/dev/null','unknown') %IO=[I,O] lines(0) exec('SCI/demos/alldems.dem') file('close',I) file('close',O)
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clc clear //INPUT DATA ni=2.1*10^19//intrinsic charge carriers in m^-3 me=0.4//electron mobility in m^2 V^-1 s^-1 mh=0.2//hole mobility in m^2 V^-1 s^-1 d=4.5*10^23//density of boron in m^-3 e=1.6*10^-19//charge of electron in coulombs //CALCULATION C=(ni*e)*(me+mh)//conductivity before adding boron atoms in ohm^-1 m^-1 c=(d*e*mh)/10^4//conductivity after adding boron atoms in ohm^-1 m^-1 *10^4 //OUTPUT printf('Before adding boron atoms,the semiconductor is an intrinsic semiconductor \n conductivity before adding boron atoms is %3.3f ohm^-1 m^-1 \n Aefore adding boron atoms,the semiconductor becomes a P-type semiconductor \n conductivity after adding boron atoms is %3.2f*10^4 ohm^-1 m^-1',C,c)
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y=wavread("./154997700.wav"); //I/Q data broken apart f = 154997700 //f is the Am frequency (carrier) O = 2*%pi*f //O omega (angular frequency) Q = y(1,:); //Q is the first row of the data I = y(2,:); //I is the second row of the data n = size(I,2) //n is the size of I. 5986304 p = floor(log(n)/log(2)); //rounds down result of log(size of matrix)/log(2))=22 n = 2^p //n now is smaller. 419304 2^22 I = I(1:n); //make I only the elements up to n (smaller) Q = Q(1:n); //make Q only the elements up to n (smaller) dt = 0.5e-6 //change in time is very small .000005 (time samples)2000000 //Demodulation t = linspace(0,dt*(n-1),n) //split from 0 to how many samples there are by size. //creates n evenly spaced pts from 0 to dt*vectorlength-1 //wonder if should be just n. //t is period (T) E = I + %i*Q; //E is the whole signal. D = exp(%i*O*t); //D is e^(i*omega*time) Eulers form (whiteboard) why time? //D=e^iOt //D is carrier frequency //Removes carrier B = E./D; //B is whole signal divided by carrier frequency. Br = real(B); //Br is only real values of B. no imaginary stuff. //B1 = Br.*cos(O*t); //taking real part of Euler. //Eulers t=x //plot(abs(fE)); //plot fE //Fast Fourier Transform //f=1/t and period is 1/f //differentiation? df = 1/(n*dt); //change in f is 1 over number of elements times time. //amplitude? m = n //copy n to m fE = fft(E(1:m), -1); //fE is the fast fourier transform of entire signal,-1 idk //separate frequency into two halves for i = 1:m //for entire length of vector, if(i<(m/2)+1) then //if first half of vector fr(i) = (i-1)*df; //frequency of i is previous element times df else //second half of vector fr(i) = (i-m-1)*df; //subtract length of whole vector and 1 more if in tophalf. end //creating negative part. end //plot(fr,abs(fE)); //plot frequency of fast fourier of entire signal.doesnt work //plot(fr(1:2000000)',abs(fE(1:2000000))) //same plot but only 1 to 2000000th element playsnd(Br,2e6) //play sound for real whole signal //plot fr,t
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@relation balance @attribute Left-weight real[1.0,5.0] @attribute Left-distance real[1.0,5.0] @attribute Right-weight real[1.0,5.0] @attribute Right-distance real[1.0,5.0] @attribute Balance_scale{L,B,R} @inputs Left-weight,Left-distance,Right-weight,Right-distance @outputs Balance_scale @data R B R R R B B L R R R R R R R R B B R R R B R B R R L B R B L B R B L B L L R R L B R R R R L B L L B B L B R R R B L R B R R R L L L L L B R R L B L B L R B R L L L B R R R B R R L B L B R L R R R L R R R R L L L L L L L L L L L L L L L B L L L B
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// Example 8.5, Page No-371 clear clc Iadjmax=100*10^-6 R1=240 Vref=1.25 // First case: Vo=4 Vo=4 R2a1=(Vo-Vref)/(Vref/R1 + Iadjmax) R2a=R2a1/1000 printf('\nR2= %.2f kohm', R2a) // First case: Vo=12 Vo=12 R2b1=(Vo-Vref)/(Vref/R1 + Iadjmax) R2b=R2b1/1000 printf('\nR2= %.2f kohm', R2b)
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clc //initialisation of variables mr= 2.01474 //amu mH= 0.00237 //amu mD= 1.00814 //amu //CALCULATIONS mn= mr+mH-mD //RESULTS printf ('mass of neutron = %.5f amu',mn)
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clc h1=240; //kJ/kg h2=192; //kJ/kg dZ=20; //m g=9.81; //m/s^2 Q=(h2-h1)+dZ*g/1000; disp("heat transfer = ") disp(-Q) disp("kJ/kg")
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SET SERVEROUTPUT ON CREATE OR REPLACE PACKAGE nocopy_test IS PROCEDURE pass_by_value ( nums IN OUT number_varray); PROCEDURE pass_by_ref ( nums IN OUT NOCOPY number_varray); END; / CREATE OR REPLACE PACKAGE BODY nocopy_test IS PROCEDURE pass_by_value ( nums IN OUT number_varray) IS BEGIN FOR indx IN nums.FIRST .. nums.LAST LOOP nums(indx) := nums(indx) * 2; IF indx > 2 THEN RAISE VALUE_ERROR; END IF; END LOOP; END; PROCEDURE pass_by_ref ( nums IN OUT NOCOPY number_varray) IS BEGIN FOR indx IN nums.FIRST .. nums.LAST LOOP nums(indx) := nums(indx) * 2; IF indx > 2 THEN RAISE VALUE_ERROR; END IF; END LOOP; END; END; / DECLARE nums1 number_varray := number_varray (1, 2, 3, 4, 5); nums2 number_varray := number_varray (1, 2, 3, 4, 5); PROCEDURE shownums ( str IN VARCHAR2, nums IN number_varray) IS BEGIN DBMS_OUTPUT.PUT_LINE (str); FOR indx IN nums.FIRST .. nums.LAST LOOP DBMS_OUTPUT.PUT (nums(indx) || '-'); END LOOP; DBMS_OUTPUT.NEW_LINE; END; BEGIN shownums ('Before By Value', nums1); BEGIN nocopy_test.pass_by_value (nums1); EXCEPTION WHEN OTHERS THEN shownums ('After By Value', nums1); END; shownums ('Before NOCOPY', nums2); BEGIN nocopy_test.pass_by_ref (nums2); EXCEPTION WHEN OTHERS THEN shownums ('After NOCOPY', nums2); END; END; /
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COM_receive file enable testTemp\test_results.txt sleep 1000 COM_send string start sleep 900000 COM_receive file disable testTemp\test_results.txt
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clear; clc; // Stoichiometry // Chapter 5 // Energy Balances // Example 5.28 // Page 261 printf("Example 5.28, Page 261 \n \n"); // solution lv1 = 26694 // kj/kmol Tc = 466.74 lv2 = lv1*((Tc-298.15)/(Tc-307.7))^.38/1000 // kJ/mol Hf = -252 // kJ/mol Hf1 = Hf-lv2 // kJ/kmol printf("Heat of formation of liquid di ethyl ether = "+string(Hf1)+" kJ/mol.")
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//[r]=%lssor(s1,s2) //%lssor(s1,s2) effectue le test d'egalite entre systemes d'etat et transfert //correspond a l'operation s1==s2 //! r=%f //end
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x=input("enter sample x(n)"); h=input("enter semple of h(n)"); n_x=input("enter index of x(n)"); n_h=input("enter index of h(n)"); exec('conv.sce'); [y,n_y]=fn_conv(x,n_x,h,n_h); figure(1); subplot(3,1,1); plot2d3(n_x,x); xlabel("x"); ylabel("n"); title("x(n)"); figure(1); subplot(3,1,2); plot2d3(n_h,h); xlabel("h"); ylabel("n"); title("h(n)"); figure(1); subplot(3,1,3); plot2d3(n_y,y); xlabel("y"); ylabel("n"); title("y(n)");
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4_02.sce
//Problem 4.02: //initializing the variables: mdt = 0.15; // in kg/sec v = 420; // in m/sec //calculation: vxin = v vxout = 0 vyin = 0 vyout = v Fxgc = mdt*(vxout - vxin) Fygc = mdt*(vyout - vyin) Fres = (Fxgc^2 + Fygc^2)^0.5 theta = (atan(Fygc/Fxgc))*180/%pi + 180 printf("\n\nResult\n\n") printf("\n resultant supporting force is %.1f N and direction is %.0f degree",Fres,theta)
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2_8.sce
clear; clc; Zoc=2000*exp(%i*(-%pi/(180/80)));Zsc=20*exp(%i*(%pi/(180/20)));l=0.5;w=10000; //value of length of cable as taken in solution Zo=sqrt(Zoc*Zsc); C=real(Zo); D=imag(Zo); printf("-Zo = %f /_ %f ohms\n",abs(Zo),atan(D,C)*180/%pi); A=atanh(sqrt(Zsc/Zoc)); P=A/l; a=real(P); printf("-a = %f neper/km\n",fix(a*10000)/10000); b=imag(P); printf("-b = %f henry/km",round(b*10000)/10000);
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bode2freq.sci
function f=bode2freq(sys,val,fmin,fmax,typ) // Interpolation for bode values f=sqrt(fmin*fmax); repf=repfreq(sys,[fmin,f,fmax]); [db,phi]=dbphi(repf); if typ=='db' then valf=db; else valf=phi; end while(abs(val-valf(2))>1000*%eps) delta=val-valf; if delta(1)*delta(2) >=0 then fmin=f; else fmax=f; end f=sqrt(fmin*fmax); repf=repfreq(sys,[fmin,f,fmax]); [db,phi]=dbphi(repf); if typ=='db' then valf=db; else valf=phi; end end endfunction
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Ex8_18.sce
clc// // // //Variable declaration I=30; //current(A) B=1.75; //magnetic field(T) n=6.55*10^28; //electron concentration(/m^3) t=0.35*10^-2; //thickness(m) e=1.6*10^-19; //Calculation VH=I*B*10^6/(n*e*t); //hall voltage(micro V) //Result printf("\n hall voltage is %0.3f micro V",VH)
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Ex17_4.sce
//Example 17.4 //Torque and Horsepower //Page No. 614 clc;clear;close; w=12; //in inches hi=0.8; //in inches hf=0.6; //in inches D=40; //in inches N=100; //in rpm R=D/2; dh=abs(hf-hi); e1=log(hi/hf); r=(hi-hf)/hi; sigma=20*e1^0.2/1.2; Qp=1.5; //no unit P=2*sigma*w*(R*(hi-hf))^(1/2)*Qp/sqrt(3); a=0.5*sqrt(R*dh); a=a/12; //conversion to ft hp=4*%pi*a*P*N*1000/33000; printf('\nRolling Load = %g\nHorsepower = %g',P,hp);
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5_6_soln.sce
clc; pathname=get_absolute_file_path('5_6_soln.sce') filename=pathname+filesep()+'5_6_data.sci' exec(filename) // Solutions: // Theoretical flow rate, Qt=Qa/(eta_v/100); //gpm // Area of piston, A=(%pi/4)*(d^2); //in^2 // tan of offset angle, T_theta=(231*Qt)/(D*A*N*Y); // offset angle, theta=atand(T_theta); //deg // Results: printf("\n Results: ") printf("\n The offset angle of axial piston pump is %.1f deg.",theta)
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Ex17_5.sce
clc; clear; mprintf('MACHINE DESIGN \n Timothy H. Wentzell, P.E. \n EXAMPLE-17.5 Page No.388\n'); //Deflection D=0.75; E=30*10^6; L=15; F=96; I=%pi*D^4/64; delta=F*L^4/(48*E*I); delta=floor(100*delta)*10^-2; Nc=188/sqrt(delta); mprintf('\n Critical speed = %f rpm.',Nc);
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example49.sce
//to find transfer function using mason gain formula printf("syms R1 R2 C1 C2 \n //gains of forward path\n P1=1/(R1*R2*C1*C2*s^2);//forward path1 gain\n //gain of individual loops\n L1=-1/(R1*C1*s);\n L2=-1/(R2*C1*s);\n L3=-1/(R2*C2*s);\n //gain of two non touching loops\n g1=1/(s^2*R1*R2*C1*C2);\n //since all the loops touches the forward path1 so\n d1=1\n d=1-(L1+L2+L3)+g1;\n G=(P1*d1)/d;\n transfer function C/R=G")
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NormaP.sce
// José Augusto Câmara Filho - Matemática Industrial //ATENÇÃO EXECUTAR ESTÁ FUNÇÃO JUNTO COM A FUNÇÃO AUXILIAR "norma". function x= NormaP(A,p,m) [l, c] = size(A); //Armazena em l e c, o tamanho das linhas e das colunas v= zeros(1,m); //inicia um vetor com todos os elementos iguais a zero for i=1:m s(:,i)= rand(c,1); //gera m vetores randômicos de acordo com a entrada do usuário end for i=1:m v(i)= norma(A*s(:,i),p)/norma(s(:,i),p); //calcula a norma através da função auxiliar "norma" com o p definido pelo usuário e em seguida armazena o valor no vetor v(i) end x = max(v); //pega o maior elemento e armazena na variável x que será o retorno da função endfunction
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Example1_2.sce
//clear// //Caption: Program to find the phase angle between two vectors //Example1.2 //page 11 clc; Q = [4,5,2]; //point Q x = Q(1); y = Q(2); z = Q(3); G = [y,-2.5*x,3]; //vector field disp(G,'G(rQ) =') aN = [2/3,1/3,-2/3]; //unit vector- direction of Q G_dot_aN = dot(G,aN); //dot product of G and aN disp(G_dot_aN,'G.aN =') G_dot_aN_aN = G_dot_aN*aN; disp(G_dot_aN_aN,'(G.aN)aN=') teta_Ga = Phase_Angle(G,aN) //phase angle between G and unit vector aN disp(teta_Ga,'phase angle between G and unit vector aN in degrees =') //Result // G(rQ) = 5. - 10. 3. // G.aN = - 2. // (G.aN)aN = - 1.3333333 - 0.6666667 1.3333333 // phase angle between G and unit vector aN in degrees = 99.956489
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Ex6_10.sce
//OptoElectronics and Fibre Optics Communication, by C.K Sarkar and B.C Sarkar //Example 6.10 //OS=Windows 10 ////Scilab version Scilab 6.0.0-beta-2(64 bit) clc; clear; //given E=1.15*(1.6e-19);//band gap energy in V h=6.62e-34;//plank's constant in S.I units c=3e8;//velocity of light in m/s lamda_c=(h*c)/(E);//critical wavelength in meter mprintf("The critical wavelength is=%.2f um",lamda_c*1e6);//multiplication by 1e6 to convert unit from m to um //the answer vary due to roundingoff
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ex4_15.sce
//Find the capacitance of 3 phase line clear; clc; //soltion //given r=1;//cm//radius of the conductor d=250;//cm//spacing L=100000;//m//length of the line Eo=8.854*10^-12//permitivity of the air C=2*%pi*Eo*L/(log(d/r)); C_=C*10^6; printf("Capacitance of 100km line= %.4fµF",C_);
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Example3_5.sce
//Chapter-3,Example3_3_5,pg 3-7 angle_0=30 //acceptance angle n1=1.4 //refractive index of core n2=sqrt(n1^2-sind(angle_0)^2) //Numerical aperture is 'NA^2 = n1^2 - n2^2' also numerical aperture is 'NA=sin(angle_0)' printf("\nThe refractive index of cladding is n2 = %.4f\n",n2)
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EX4_44.sce
//EXAMPLE 4-44 PG NO 257 //6I1+14I2=20 I1-I2=-6 I1=-3.2; I2=2.8; disp('i) Current(I1) is = '+string (I1) +' A '); disp('Ii) Current (I2) is = '+string (I2) +' A ');
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example3_18.sce
clear; clc; // Stoichiometry // Chapter 3 // Material Balances Without Chemical Reaction // Example 3.18 // Page 79 printf("Example 3.10, Page 79 \n \n"); // solution // Overall balance // F=R1+P2 // Balance across Module I // F+R2 = R1+P1 ==> R1+P2+R2 = R1+P1 // balance across module II // P1 = P2+R2 P2 = 5 //[m^3/h] P1 = P2/.8 //[m^3/h] R2 = P1-P2 //[m^3/h] F = P1/.66 - R2//[m^3/h] R1 = F-P2 //[m^3/h] // Overall balance of DS in water xR1 = (F*4200-P2*5)/R1 //[mg/l] xP1 = (P2*5)/(.015*P1) // [mg/l] xR2 = (P1*xP1-P2*5)/R2 //[mg/l] m1 = F*4200+R2*xR2 //[g] DS mixeed in MF C1 = m1/(F+R2) // [mg/l] m2 = R1*xR1 //[g] DS in R1 r = m2*100/m1 // rejection in module in I m3 = m1-m2 //[g] DS in P1 C2 = m3/P1 // [mg/l] R = R2/F R1 = P2*100/F printf("F = "+string(F)+" m^3/h \nR1 = "+string(R1)+" m^3/h \nP = "+string(P1+P2)+" m^3/h \nR2 = "+string(R2)+" m^3/h \nrecycle ratio = "+string(R)+" \nrejection percentage of salt in module I = "+string(r)+"")
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Ex13_42.sce
//EX13_42 Pg-23 clc clear x=['1010']; y=['0011']; //binary to decimal conversion// x=bin2dec(x) y=bin2dec(y) z=x+y; a=dec2bin(z)//decimal to binary conversion// printf('the addition of given numbers is: ') printf("%s",a)
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s=%s ;//convert to state space TFcont=syslin ('c',2/(s^3+6*s^2+11*s+6)) SScont=tf2ss (TFcont) [Ac ,Bc ,U, ind ]=canon( SScont( 2 ) , SScont( 3 ) ) disp(Ac,"A=") disp(Bc,"B=") C=[1 0 0] p=cont_mat(Ac,Bc) disp (p," controllability matrix="); d=det(p) if d==0 printf ("matrix is singular, so the system is uncontrollable"); else printf ("system is controllable "); end g= obsv_mat (Ac,C); disp (g," Observability Matrix="); i= det(g) if i ==0 printf ("matrix is singular, so the system is unobservable"); else printf (" system is observable "); end
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// Calculates the GPC law given by Eq. 12.36 on page 446. // 12.6 function [K,KH1,KH2,Tc,dTc,Sc,dSc,R1,dR1] = ... gpc_col(A,dA,B,dB,C,dC,N,k,rho) D=[1 -1]; dD = 0; AD=convol(A,D); dAD=dA+1; zj=1; dzj=0; Nu = N+1; G=zeros(Nu,Nu); H1=zeros(Nu,2*k+N-2+dB); H2 = zeros(Nu,k+N+dA); for j = 1:Nu, zj = convol(zj,[0,1]); dzj = dzj + 1; [Fj,dFj,Ej,dEj] = ... xdync(zj,dzj,AD,dAD,C,dC); [Nj,dNj,Mj,dMj] = ... xdync(zj,dzj,C,dC,1,0); [Gj,dGj] = polmul(Mj,dMj,Ej,dEj); [Gj,dGj] = polmul(Gj,dGj,B,dB); [Pj,dPj] = polmul(Mj,dMj,Fj,dFj); [Pj,dPj] = poladd(Nj,dNj,Pj,dPj); j,Fj,Ej,Mj,Nj,Gj,Pj G(j,1:j) = flip(Gj(1:j)); H1(j,1:dGj-j+1) = Gj(j+1:dGj+1); H2(j,1:dPj+1) = Pj; end K = inv(G'*G+rho*eye(Nu,Nu))*G' // Note: inverse need not be calculated KH1 = K * H1; KH2 = K * H2; R1 = [1 KH1(1,:)]; dR1 = length(R1)-1; Sc = KH2(1,:); dSc = length(Sc)-1; Tc = K(1,:); dTc = length(Tc)-1; endfunction;
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l=0.5; f=60; i=0.25; p=5; disp("Part a"); r=p/i^2; disp("the ac resistance (in Ω) of the coil is"); disp(r); disp("part b"); q=2*%pi*f*l/r; disp("the Q of the coil is"); disp(q);
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clc clear //Initialization of variables p=[2.75 0.5 0.31 0.31 2.75] v=[116.17 654.8 654.8 597 110.65] t=[440 440 170 140 410] h=[3325 3356 2802.6 2738.5 3257.7] e=[3005.6 3028.6 2602.6 2553.6 2953.4] //calculations dh1=h(2) - h(1) de1=e(2) - e(1) q2=e(3) - e(2) dh2=h(3) - h(2) dh3=h(4) - h(3) de3=e(4) - e(3) W3= p(3) *(v(4) - v(3)) Q3= de3+W3 dh4=h(5) -h(4) de4=e(5) -e(4) dh5=h(1) - h(5) de5= e(1) - e(5) W5= p(5) *(v(1) - v(5)) q5 = de5+W5 //results printf("In case 1 , dH = %.1f kJ/kg dE = %.1f kJ/kg W= pDv kJ/kg Q= %.1f + W kJ/kg",dh1,de1,de1) printf("\n In case 2, W =0 kJ/kg Q = dE = %d kJ/kg dH = %.1f kJ/kg",q2,dh2) printf("\n In case 3, dH= %.1f kJ/kg dE = %.1f kJ/kg W= %.1f kJ/kg Q = %.1f kJ/kg",dh3,de3,W3,Q3) printf("\n In case 4, Q= 0 kJ/kg dH = %.1f kJ/kg dE = -W = %.1f kJ/kg",dh4,de4) printf("\n In case 5, dH = %.1f kJ/kg dE = %.1f kJ/kg W = %.1f kJ/kg Q = %.1f kJ/kg",dh5,de5,W5,q5) xlabel("Volume (m^3/kg)") ylabel("Pressure (Mpa)") plot(v,p)
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//Exa 2.41 clc; clear; close; //Given data : format('v',6); VL=440;//in volt f=50;//in Hz P=6;//no. of poles phase=3;//no. of phase Ns=120*f/P;//in rpm fr=120;//alternations per minute fr=fr/60;//in Hz S=fr/f;//slip disp(S,"Slip : "); Nr=Ns-S*Ns;//in rpm disp(Nr,"Rotor speed(in rpm) :"); Rotor_input=80;//in KW RotorCuLoss=S*Rotor_input;//in KW disp(RotorCuLoss*10^3/phase,"Rotor Cu Loss per phase(in watts) :"); P_Mechdev=Rotor_input*10^3-RotorCuLoss*10^3;//in watts P_Mechdev=P_Mechdev/735.5;//in H.P. disp(P_Mechdev,"Mechanical power devloped(in H.P.) :"); Ir=60;//in Ampere R2=(RotorCuLoss*10^3/phase)/Ir^2;//in ohm disp(R2,"Rotor resistance per phase(in ohm) :");
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chapter6_ex7.sce
clc clear //input z1=10+(%i*15);//first impedance in ohms z2=15-(%i*25);//second impeddance in ohms //impedances 1 and 2 are connected in parallel //calculations Z1=(((real(z1)^2)+(imag(z1)^2)))^0.5;//magnitude of impedance 1 in ohms Z2=(((real(z2)^2)+(imag(z2)^2)))^0.5;//magnitude of impedance 2 in ohms phi1=(180/%pi)*atan((imag(z1))/real(z1));//phase angle 1 in degrees phi2=(180/%pi)*atan((imag(z2))/real(z2));//phase angle 1 in degrees Z=z1+z2;//total impedance in ohms Zt=(((real(Z)^2)+(imag(Z)^2)))^0.5;//magnitude of total impedance in ohms PHIt=(180/%pi)*atan((imag(Z))/real(Z));//total phase angle in degrees ZT=(Z1*Z2)/Zt;//magnitude of equivalent impedance in ohms PHIT=phi1+phi2-PHIt;//phase angle of equivalent impedance in degrees p=(PHIT*%pi)/180;// phase angle in radians Zs=(ZT*cos(p))+(%i*(ZT*sin(p)));//series impedance in ohms R=real(Zs);//resistance of equivalent series circuit in ohms X=imag(Zs);//reactance of equivalent series circuit in ohms //output mprintf('the resistance and inductive reactance of equivalent series circuit are %3.1f ohm and %3.2f ohm',R,X)
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<?xml version="1.0" encoding="UTF-8"?> <Project Name="map1109" Width="13" Height="9" CellSize="40" BackgroundSize="1" Background="9plus.png"> <Cell Name="丛林图腾2" X="4" Y="1" /> <Cell Name="樱桃树" X="8" Y="1" /> <Cell Name="蘑菇" X="10" Y="1" /> <Cell Name="蘑菇" X="11" Y="1" /> <Cell Name="樱桃树" X="1" Y="2" /> <Cell Name="出生点" X="2" Y="2" /> <Cell Name="蘑菇" X="3" Y="2" /> <Cell Name="丛林图腾2" X="4" Y="2" /> <Cell Name="樱桃树" X="9" Y="2" /> <Cell Name="丛林图腾2" X="10" Y="2" /> <Cell Name="樱桃树" X="2" Y="3" /> <Cell Name="丛林图腾2" X="3" Y="3" /> <Cell Name="丛林图腾2" X="4" Y="3" /> <Cell Name="樱桃树" X="5" Y="3" /> <Cell Name="樱桃树" X="8" Y="3" /> <Cell Name="蜘蛛怪" X="6" Y="4" arg0="3" /> <Cell Name="池塘-左上" X="7" Y="4" /> <Cell Name="池塘-右上" X="8" Y="4" /> <Cell Name="木桩" X="1" Y="5" /> <Cell Name="木偶" X="3" Y="5" arg0="22" /> <Cell Name="丛林图腾2" X="5" Y="5" /> <Cell Name="池塘-左上" X="6" Y="5" /> <Cell Name="池塘-内角左上" X="7" Y="5" /> <Cell Name="池塘-右" X="8" Y="5" /> <Cell Name="通关点-1" X="9" Y="5" /> <Cell Name="猪怪" X="10" Y="5" arg0="5" /> <Cell Name="木桩" X="2" Y="6" /> <Cell Name="樱桃树" X="5" Y="6" /> <Cell Name="池塘-左下" X="6" Y="6" /> <Cell Name="池塘-下" X="7" Y="6" /> <Cell Name="池塘-右下" X="8" Y="6" /> <Cell Name="木桩" X="2" Y="7" /> <Cell Name="樱桃树" X="3" Y="7" /> <Cell Name="木桩" X="6" Y="7" /> <Cell Name="木偶" X="7" Y="7" arg0="22" /> <Cell Name="樱桃树" X="11" Y="7" /> </Project>
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//To find torque exerted clc //Given: TA=15, TB=20, TC=15 NA=1000 //rpm Tm=100 //Torque developed by motor, N-m //Solution: //Refer Fig. 13.26 and Table 13.21 //Calculating the number of teeth on gears E and D TE=TA+2*TB TD=TE-(TB-TC) //Speed of the machine shaft: //From the fourth row of the table, x+y = 1000, or y+x = 1000 .....(i) //Also, y-x*(TA/TE) = 0 .....(ii) A=[1 1; 1 -TA/TE] B=[1000; 0] V=A \ B y=V(1) x=V(2) //Calculating the speed of machine shaft ND=y-x*(TA/TB)*(TC/TD) //rpm //Calculating the torque exerted on the machine shaft Ts=Tm*NA/ND //Torque exerted on the machine shaft, N-m //Results: printf("\n\n Speed of machine shaft, ND = %.2f rpm, anticlockwise.\n\n",ND) printf(" Torque exerted on the machine shaft = %d N-m.\n\n",Ts)
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clc //Initialization of variables N=6 g=4 //calculations sig=factorial(g+N-1) /(factorial(g-1) *factorial(N)) //results printf("No. of ways of arranging = %d ",sig)
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clear; clc; printf("\t\t\tExample Number 2.10\n\n\n"); // rod with heat sources // illustration2.10 // solution // q_dot is uniform heat source per unit volume // h is convection coefficient // k is heat transfer coefficient // A is area of crossection // P is perimeter // Tinf is environment temperature // we first make an energy balance on the element of the rod shown in figure(2-10) // energy in left place + heat generated in element = energy out right face + energy lost by convection // or // -(k*A*dT_by_dx)+(q_dot*A*dx) = -(k*A(dT_by_dx+(d2T_by_dx2)*dx))+h*P*dx*(T-Tinf) // simlifying we have // d2T_by_dx2-((h*P)/(k*A))*(T-Tinf)+q_dot/k = 0 // replacing theta = (T-Tinf) and (square meter) = ((h*P)/(k*A)) // d2theta_by_dx2-(square meter)*theta+q_dot/k = 0 // we can make a further substitution as theta` = theta-(q_dot/(k*(square meter))) // so that our differential equation becomes // d2theta`_by_dx2-(square meter)*theta` // which has the general solution theta` = C1*exp^(-m*x)+C2*exp^(m*x) // the two end temperatures are used to establish the boundary conditions: // theta` = theta1` = T1-Tinf-q_dot/(k*(square meter)) = C1+C2 // theta` = theta2` = T2-Tinf-q_dot/(k*(square meter)) = C1*exp^(-m*L)+C2*exp^(m*L) // solving for the constants C1 and C2 gives // (((theta1`*exp^(2*m*L)-theta2`*exp^(m*L))*exp^(-m*x))+((theta2`exp^(m*L)-theta1`)exp^(m*x))/(exp^(2*m*L)-1)) printf("the expression for the temperature distribution in the rod is "); printf("\n theta` = (((theta1`*exp^(2*m*L)-theta2`*exp^(m*L))*exp^(-m*x))+((theta2`exp^(m*L)-theta1`)exp^(m*x))/(exp^(2*m*L)-1))"); printf("\n for an infinitely long heat generating fin with the left end maintained at T1, the temperature distribution becomes "); printf("\n theta`/theta1 = exp^(-m*x)");
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java -ea make.Main -f make-tests/autograder_make07.mk -D make-tests/autograder_file07 T1
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clc; pathname=get_absolute_file_path('8_5_soln.sce') filename=pathname+filesep()+'8_5_data.sci' exec(filename) // Solution: // capacity coefficient in English Units, Cv=Q/sqrt(del_p/SG_oil); //gpm/sqrt(psi) // capacity coefficient in Metric Units, Cv1=Q1/sqrt(del_p1/SG_oil); //Lpm/sqrt(kPA) // Results: printf("\n Results: ") printf("\n The capacity coefficient in English unit is %.2f gpm/sqrt(psi).",Cv) printf("\n The capacity coefficient in Metric unit is %.2f Lpm/sqrt(kPa).",Cv1)
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//Given that TH = 850 //in K TL = 300 //in K W = 1200 //in J t = 0.25 //in sec //Sample Problem 21-3a printf("**Sample Problem 21-3a**\n") eta = 1 - (TL/TH) printf("The efficiency of the cycle is equal to %f\n", eta) //Sample Problem 21-3b printf("\n**Sample Problem 21-3b**\n") P = W/t printf("The average power of the cycle is %fW\n", P) //Sample Problem 21-3c printf("\n**Sample Problem 21-3c**\n") QH = W/eta printf("The heat extracted from the reservoir is equal to %fJ\n", QH) //Sample Problem 21-3d printf("\n**Sample Problem 21-3d**\n") QL = W - QH printf("The heat delivered to the reservoir is equal to %fJ\n", QL) //Sample Problem 21-3e printf("\n**Sample Problem 21-3e**\n") S = QH/TH + QL/TL printf("The net entropy change for the cycle is %fJ/k", S)
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Ex2_1.sce
clear // // //Given //Variable declaration L=4*(10**3) //Length of the bar in mm b=30 //Breadth of the bar in mm t=20 //Thickness of the bar in mm P=30*(10**3) //Axial pull in N E=2e5 //Youngs modulus in N/sq.mm mu=0.3 //Poisson's ratio //Calculation A=b*t //Area of cross-section in sq.mm long_strain=P/(A*E) //Longitudinal strain delL=long_strain*L //Change in length in mm lat_strain=mu*long_strain //Lateral strain delb=b*lat_strain //Change in breadth in mm delt=t*lat_strain //Change in thickness in mm //Result printf("\n change in length = %0.3f mm",delL) printf("\n change in breadth = %0.3f mm",delb) printf("\n change in thickness = %0.3f mm",delt)
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//Ex 21 clc; clear; close; commonLossGain=16; lossPercent=(commonLossGain/10)^2; printf("The loss is %3.2f percent",lossPercent);
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clc clear printf("Example 5.8 | Page number 129 \n\n"); //Redo example 5.7 for heat loss 10% of heat transferred mh = 9.45 // kg/s // flow rate of steam h_h2 = 140 // kJ/kg // enthalpy of condensate h_h1 = 2570 // kJ/kg // inlet enthalpy of steam t1 = 25 // °C //inlet temperature of cooling water t2 = 36 // °C //exit temperature of cooling water c = 4.189 // kJ/kg deg // specific heat of water fractionalheatloss = 0.1 // fractional heat loss //Solution mc = -1*((1-fractionalheatloss)*mh*(h_h2-h_h1))/(c*(t2-t1)) // kg/s //mass flow rate of cooling water printf("Mass flow rate of cooling water = %.1f kg/s",mc)
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intersectMultIntervals.sci
function [bool,intersection] = intersectMultIntervals(intv1,intv2) //Author : Maxens ACHIEPI //Space Robotics Laboratory - Tohoku University //Description: // //INPUT //intv1: matrix of intervals //intv2: matrix of intervals //OUTPUT //intersect: matrix of intervals //----------------------------------------------------------------------------// l1 = size(intv1,1); l2 = size(intv2,1); intersection = []; bool = %F; k = 1; for i=1:l1 for j=1:l2 [dump,temp] = intersectTwoIntervals(intv1(i,:),intv2(j,:)); if dump then bool = %T; intersection(k,:) = temp'; k = k+1; end end if dump then k = k+1; end end endfunction
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imcsplit.sci
// Splitting a polynomial B(z) // 10.3 // Splits a polynomial B into good, nonminimum with // positive real & with negative real parts. // All are returned in polynomial form. // Gain is returned in Kp and delay in k. function [Kp,k,Bg,Bnmp,Bm] = imcsplit(B,polynomial) k = 0; Kp = 1; if(polynomial) rts = roots(B); Kp = sum(B)/sum(coeff(poly(rts,'z'))); else rts = B; end Bg = 1; Bnmp = 1; Bm = 1; for i = 1:length(rts), rt = rts(i); if rt == 0, k = k+1; elseif (abs(rt)<1 & real(rt)>=0) Bg = convol(Bg,[1 -rt]); elseif (abs(rt)>=1 & real(rt)>=0) Bnmp = convol(Bnmp,[1 -rt]); else Bm = convol(Bm,[1 -rt]); end end