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Ex7_1.sce
clear // // // //Variable declaration El=10**-2*50 //energy loss(J) H=El*60 //heat produced(J) d=7.7*10**3 //iron rod(kg/m**3) s=0.462*10**-3 //specific heat(J/kg K) //Calculation theta=H/(d*s) //temperature rise(K) //Result printf("\n temperature rise is %0.2f K",theta)
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printf("given G(s)=K/(s*(1+s*T)) \n Mp=20 percent \n resonant frequency=6 rad/sec\n we have to determine the value of K,T,resonant peak") printf(" H(s)=1 \n C(s)/R(s)=G(s)/(1+G(s)*H(s)) \n =(K/T)/(s^2+s/T+(K/T)"); printf("compare with w^2/(s^2+2*d*w*s+w^2)"); d1=log(0.2); d=sqrt(d1^2/(d1^2+%pi^2)); wr=6; w=wr/sqrt(1-(2*(d^2))); K=sqrt(4*d*w^2); T=sqrt(4*d/w^2); mr=1/(2*d*sqrt(1-d^2)); disp(w,"undamped natural frequency (in rad/sec)=") disp(d,"damping ratio=") disp(K,"value of K=") disp(T,"value of T=") disp(mr,"resonance peak=")
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Chapter8_Example14.sce
//Chapter-8, Example 8.14, Page 356 //============================================================================= clc clear //INPUT DATA d=0.0254;//Diatance between the plates in m Tl=60;//Temperature of the lower panel n degree C Tu=15.6;//Temperature of the upper panel in degree C //CALCULATIONS Tf=(Tl+Tu)/2;//Film temperature in degree C p=1.121;//Density in kg/m^3 k=0.0292;//Thermal conductivity in W/m.K v1=(0.171*10^-4);//Kinematic viscosity in m^2/s b=(3.22*10^-3);//Coefficient of thermal expansion in 1/K Pr=0.7;//Prantl number Gr=((9.81*b*d^3*(Tl-Tu))/(v1^2));//Grashof number Nu=(0.195*Gr^0.25);//Nussults number q=(Nu*k*(Tl-Tu))/d;//Heat flux across the gap in W/m^2 //OUTPUT mprintf('Free convection heat transfer is %3.1f W/m^2',q) //=================================END OF PROGRAM==============================
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clc // Given that f = 0.2 // Feed rate in cm/min l = 2.54 // Length of tool face in cm w = 2.54 // Width of tool face in cm T_b = 95 // Boiling temperature of electrolyte in °C Nita = 0.876e-3 // Viscosity of electrolyte in kg/m-sec D_e = 1.088 // Density of electrolyte in g/cm^3 c = .997 // Specific heat of electrolyte V = 10 // DC supply voltage in Volt k = 0.2 // Conductivity of electrolyte in ohm^-1-cm^-1 T = 35 // Ambient temperature in °C Vo = 1.5 // Total overvoltage in Volt F = 96500 // Faraday constant in coulombs per mole // Sample Problem 7 on page no. 355 printf("\n # PROBLEM 6.7 # \n") A = 55.85 // Atomic gram weight of iron in gm Z = 2 // Valency of dissolation of iron D = 7.86 // Density of iron in gm/cm^3 Ye = k*A*(V-Vo)*60/(D*Z*F*f) J = k*(V-Vo)/(Ye) D_T = T_b -T v = (J^2)*(l)/(k*D_T*D_e*c) Re = ((D_e*v*2*Ye)/Nita)*(0.1) p = 0.3164*D_e*(v^2)*l/(4*Ye*(Re^0.25))*(10^-4) A = l*w F = p*A*(10^-1)*(1/2) printf("\n Total force acting on the tool = %d N",F) // Answer in the book is given as 79 N
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void main() { int a = 100, b = -1, c = a; c = b; }
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// 11.05.28 // 11.08.24 // 15.05.03 function Setcolor(varargin) global Wfile FID; Nargs=length(varargin); Color='black'; Kosa=1; Color=varargin(1); if length(varargin)>1 Kosa=varargin(2); end; Iro=Ratiocmyk(Color); if type(Iro)==10 Str='\color{'+Iro+"}%"; else if size(Iro,2)>3 Str='\color[cmyk]{'; else Str='\color[rgb]{'; end; for J=1:size(Iro,2) Str=Str+string(Kosa*Iro(J)); if J<size(Iro,2) Str=Str+','; end; end; Str=Str+'}%'; end; if Wfile=='default' mprintf('%s\n',Str); else mfprintf(FID,'%s\n',Str); end endfunction
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//Chapter-9,Example9_32,pg 9_86 Po=7.46*10^3 V=250 Ilo=5 Ra=0.5 Rsh=250 Ish=V/Rsh Iao=Ilo-Ish Acl=(Iao^2)*Ra Fcl=(Ish^2)*Rsh Pi=V*Ilo FWl=Pi-Acl-Fcl//friction and windage loss //Pin=Eb*Ia=(V-Ia*Ra)*Ia //0.5*(Ia^2)-250*Ia+8452=0 b=-250 a=0.5 c=8452 Ia=(-b-sqrt((b^2)-4*a*c))/(2*a)//neglecting higher value TL=(Ia^2)*Ra+(Ish^2)*Rsh+FWl n=Po*100/(Po+TL) //for max. efficiency Ia=sqrt((FWl+Fcl)/Ra) Eb=V-Ia*Ra Pm=Eb*Ia //Po at nmax Po=Pm-FWl printf("maximum efficiency output\n") printf("Po=%.3f W",Po)
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// Given:- p1 = 3.0 // entry pressure in Mpa p2 = 0.5 // exit pressure in Mpa T1 = 320.0 // entry temperature in degree celcius T0 = 25.0 // in degree celcius p0 = 1.0 // in atm // From table A-4 h1 = 3043.4 // in kj/kg s1 = 6.6245 // in kj/kg.k h2 = h1 // from reduction of the steady-state mass and energy rate balances s2 = 7.4223 // Interpolating at a pressure of 0.5 MPa with h2 = h1, units in kj/kg.k // From table A-2 h0 = 104.89 // in kj/kg s0 = 0.3674 // in kj/kg.k // Calculations ef1 = h1-h0-(T0+273)*(s1-s0) // flow exergy at the inlet ef2 = h2-h0-(T0+273)*(s2-s0) // flow exergy at the exit // From the steady-state form of the exergy rate balance Ed = ef1-ef2 // the exergy destruction per unit of mass flowing is // Results printf( ' The specific flow exergy at the inlet is %.2f kJ/kg.',ef1) printf( ' The specific flow exergy at the exit is %.2f kJ/kg.',ef2) printf( ' The exergy destruction per unit of mass flowing is %.2f kJ/kg.',Ed)
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/* Nama : Mohamad Fahrio Ghanial Fatihah NPM : 140810190005 Deskripsi : Program Metode Trapezoida */ clear; clc; printf('\nProgram Integrasi Numerik Metode Trapezoida\n'); X = [0.10 0.12 0.14 0.16 0.18 0.20]; Y = [0.004 0.006 0.008 0.011 0.015 0.018]; printf('Diketahui Data Berikut:\n'); //Menampilkan data printf('x\tf(x)\n'); for i=1:6 printf('%.2f\t%.3f\n', X(i),Y(i)); end a=input('Masukkan batas awal x = '); b=input('Masukkan batas akhir x = '); h = 0.02; n=(b-a)/h; for i=1:6 if a == X(i) index=i; end if b == X(i) indexAkhir=i; end end I = Y(index) + Y(indexAkhir); sigma=0; for i=1 :(n-1) sigma=sigma+2*Y(index+1); index=index+1; end I=(I+sigma)*h/2; printf('Jadi luas daerahnya adalah : %.6f', I);
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clc; // plot for open circuit characteristics is given in fig 4.10 IF=[ 0 11.5 23 36.5 59.5 79 110 160]; EA=[0 40 80 120 160 180 200 220 ]; subplot(221); plot(IF,EA); xlabel('field ATs'); ylabel('voltage'); title('magnetising curve'); nf=800; // field winding turns rd=0.5; // total armature resistance along d-axis ifl=0.2; // field winding current d=10; // product of (difference between mmf of compensating winding and armature mmf along d-circuit)and load current nf1=nf*ifl; // field winding turns for field current of 200mA il=nf1/d; // maximum load current printf('Maximum field current is %d A\n',il); IL=[0 2 4 6 8 10 12 14 16]; // load currents ATD=nf1-d*IL; // net d-axis ATs disp('Net d-axis ATs is'); disp(ATD); // corresponding to each ATD open circuit EMF is obtained from magnetising curve EO=[220 213 204.7 194 180.5 161.4 128 70 0 ]; // open circuit EMF VRD=rd*IL; // d-axis resistance drop VO=EO-VRD; disp('Output voltage(V) is '); disp(VO); subplot(222); plot(IL,VO); xlabel('load current(A)'); ylabel('Output voltage(v)'); title('Output voltage vs Load current');
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clc //initialization of variables clear L=1 //m w=10 //kg h=50 //cm A=1 //cm^2 E=2*10^6 //kg/cm^2 Ar=1 //cm^2 Ec=3*10^4 //kg/cm^2 // For steel del=w*L*100/(A*E) P=w*(1+sqrt(1+(2*h/del))) printf('Stress in steeel = %d kg/cm^2 ',P) // for cloth laminate del=w*L*100/(A*Ec) P=w*(1+sqrt(1+(2*h/del))) printf('\n Stress in cloth laminate = %.1f kg/cm^2 ',P)
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COMMENT THE REDUCE INTEGRATION TEST PACKAGE Edited By Anthony C. Hearn The RAND Corporation This file is designed to provide a set of representative tests of the Reduce integration package. Not all examples go through, even when an integral exists, since some of the arguments are outside the domain of applicability of the current package. However, future improvements to the package will result in more closed-form evaluations in later releases. We would appreciate any additional contributions to this test file either because they illustrate some feature (good or bad) of the current package, or suggest domains which future versions should handle. Any suggestions for improved organization of this test file (e.g., in a way which corresponds more directly to the organization of a standard integration table book such as Gradshteyn and Ryznik) are welcome. Acknowledgments: The examples in this file have been contributed by the following. Any omissions to this list should be reported to the Editor. David M. Dahm James H. Davenport John P. Fitch Steven Harrington Anthony C. Hearn K. Siegfried Koelbig Ernst Krupnikov Arthur C. Norman Herbert Stoyan ; Comment we first set up a suitable testing functions; fluid '(gcknt!*); global '(faillist!* gcnumber!* inittime number!-of!-integrals unintlist!*); symbolic operator time; symbolic procedure initialize!-integral!-test; begin faillist!* := unintlist!* := nil; number!-of!-integrals := 0; gcnumber!* := gcknt!*; inittime := time() end; symbolic procedure summarize!-integral!-test; begin scalar totaltime; totaltime := time()-inittime; prin2t " ***** SUMMARY OF INTEGRAL TESTS *****"; terpri(); prin2 "Number of integrals tested: "; prin2t number!-of!-integrals; terpri(); prin2 "Total time taken: "; prin2 totaltime; prin2t " ms"; terpri(); if gcnumber!* then <<prin2 "Number of garbage collections: "; prin2t (gcknt!* - gcnumber!*); terpri()>>; prin2 "Number of incorrect integrals: "; prin2t length faillist!*; terpri(); prin2 "Number of unevaluated integrals: "; prin2t length unintlist!*; terpri(); if faillist!* then <<prin2t "Integrands of incorrect integrals are:"; for each x in reverse faillist!* do mathprint car x>>; if unintlist!* then <<prin2t "Integrands of unevaluated integrals are:"; terpri(); for each x in reverse unintlist!* do mathprint car x>> end; procedure testint(a,b); begin scalar der,diffce,res,tt; tt:=time(); symbolic (number!-of!-integrals := number!-of!-integrals + 1); res:=int(a,b); % write "time for integral: ",time()-tt," ms"; der := df(res,b); diffce := der-a; if diffce neq 0 then begin for all x let cot x=cos x/sin x, sec x=1/cos x, sin x**2=1-cos x**2, tan x=sin x/cos x, tan(x/2)=sin x/(1+cos x), tanh x= (e**(x)-e**(-x))/(e**x+e**(-x)), coth x= 1/tanh x; diffce := diffce; for all x clear cot x,sec x,sin x**2,tan x,tan(x/2), tanh x,coth x end; %hopefully, difference appeared non-zero due to absence of %above transformations; if diffce neq 0 then <<on combineexpt; diffce := diffce; off combineexpt>>; if diffce neq 0 then <<write " ***** DERIVATIVE OF INTEGRAL NOT EQUAL TO INTEGRAND *****"; symbolic(faillist!* := list(a,b,res,der) . faillist!*)>>; symbolic if smemq('int,res) then unintlist!* := list(a,b,res) . unintlist!*; return res end; symbolic initialize!-integral!-test(); % References are to Gradshteyn and Ryznik. testint(1+x+x**2,x); testint(x**2*(2*x**2+x)**2,x); testint(x*(x**2+2*x+1),x); testint(1/x,x); % 2.01 #2; testint((x+1)**3/(x-1)**4,x); testint(1/(x*(x-1)*(x+1)**2),x); testint((a*x+b)/((x-p)*(x-q)),x); testint(1/(a*x**2+b*x+c),x); testint((a*x+b)/(1+x**2),x); testint(1/(x**2-2*x+3),x); % Rational function examples from Hardy, Pure Mathematics, p 253 et seq. testint(1/((x-1)*(x**2+1))**2,x); testint(x/((x-a)*(x-b)*(x-c)),x); testint(x/((x**2+a**2)*(x**2+b**2)),x); testint(x**2/((x**2+a**2)*(x**2+b**2)),x); testint(x/((x-1)*(x**2+1)),x); testint(x/(1+x**3),x); testint(x**3/((x-1)**2*(x**3+1)),x); testint(1/(1+x**4),x); testint(x**2/(1+x**4),x); testint(1/(1+x**2+x**4),x); % Examples involving a+b*x. z := a+b*x; testint(z**p,x); testint(x*z**p,x); testint(x**2*z**p,x); testint(1/z,x); testint(1/z**2,x); testint(x/z,x); testint(x**2/z,x); testint(1/(x*z),x); testint(1/(x**2*z),x); testint(1/(x*z)**2,x); testint(1/(c**2+x**2),x); testint(1/(c**2-x**2),x); % More complicated rational function examples, mostly contributed % by David M. Dahm, who also developed the code to integrate them. testint(1/(2*x**3-1),x); testint(1/(x**3-2),x); testint(1/(a*x**3-b),x); testint(1/(x**4-2),x); testint(1/(5*x**4-1),x); testint(1/(3*x**4+7),x); testint(1/(x**4+3*x**2-1),x); testint(1/(x**4-3*x**2-1),x); testint(1/(x**4-3*x**2+1),x); testint(1/(x**4-4*x**2+1),x); testint(1/(x**4+4*x**2+1),x); testint(1/(x**4+x**2+2),x); testint(1/(x**4-x**2+2),x); testint(1/(x**6-1),x); testint(1/(x**6-2),x); testint(1/(x**6+2),x); testint(1/(x**8+1),x); testint(1/(x**8-1),x); testint(1/(x**8-x**4+1),x); testint(x**7/(x**12+1),x); % Examples involving logarithms. testint(log x,x); testint(x*log x,x); testint(x**2*log x,x); testint(x**p*log x,x); testint((log x)**2,x); testint(x**9*log x**11,x); testint(log x**2/x,x); testint(1/log x,x); testint(1/log(x+1),x); testint(1/(x*log x),x); testint(1/(x*log x)**2,x); testint((log x)**p/x,x); testint(log x *(a*x+b),x); testint((a*x+b)**2*log x,x); testint(log x/(a*x+b)**2,x); testint(x*log (a*x+b),x); testint(x**2*log(a*x+b),x); testint(log(x**2+a**2),x); testint(x*log(x**2+a**2),x); testint(x**2*log(x**2+a**2),x); testint(x**4*log(x**2+a**2),x); testint(log(x**2-a**2),x); testint(log(log(log(log(x)))),x); % Examples involving circular functions. testint(sin x,x); % 2.01 #5; testint(cos x,x); % #6; testint(tan x,x); % #11; testint(1/tan(x),x); % 2.01 #12; testint(1/(1+tan(x))**2,x); testint(1/cos x,x); testint(1/sin x,x); testint(sin x**2,x); testint(x**3*sin(x**2),x); testint(sin x**3,x); testint(sin x**p,x); testint((sin x**2+1)**2*cos x,x); testint(cos x**2,x); testint(cos x**3,x); testint(sin(a*x+b),x); testint(1/cos x**2,x); testint(sin x*sin(2*x),x); testint(x*sin x,x); testint(x**2*sin x,x); testint(x*sin x**2,x); testint(x**2*sin x**2,x); testint(x*sin x**3,x); testint(x*cos x,x); testint(x**2*cos x,x); testint(x*cos x**2,x); testint(x**2*cos x**2,x); testint(x*cos x**3,x); testint(sin x/x,x); testint(cos x/x,x); testint(sin x/x**2,x); testint(sin x**2/x,x); testint(tan x**3,x); % z := a+b*x; testint(sin z,x); testint(cos z,x); testint(tan z,x); testint(1/tan z,x); testint(1/sin z,x); testint(1/cos z,x); testint(sin z**2,x); testint(sin z**3,x); testint(cos z**2,x); testint(cos z**3,x); testint(1/cos z**2,x); testint(1/(1+cos x),x); testint(1/(1-cos x),x); testint(1/(1+sin x),x); testint(1/(1-sin x),x); testint(1/(a+b*sin x),x); testint(1/(a+b*sin x+cos x),x); testint(x**2*sin z**2,x); testint(cos x*cos(2*x),x); testint(x**2*cos z**2,x); testint(1/tan x**3,x); testint(x**3*tan(x)**4,x); testint(x**3*tan(x)**6,x); testint(x*tan(x)**2,x); testint(sin(2*x)*cos(3*x),x); testint(sin x**2*cos x**2,x); testint(1/(sin x**2*cos x**2),x); testint(d**x*sin x,x); testint(d**x*cos x,x); testint(x*d**x*sin x,x); testint(x*d**x*cos x,x); testint(x**2*d**x*sin x,x); testint(x**2*d**x*cos x,x); testint(x**3*d**x*sin x,x); testint(x**3*d**x*cos x,x); testint(sin x*sin(2*x)*sin(3*x),x); testint(cos x*cos(2*x)*cos(3*x),x); testint(sin(x*kx)**3*x**2,x); testint(x*cos(xi/sin(x))*cos(x)/sin(x)**2,x); % Mixed angles and half angles. int(cos(x)/(sin(x)*tan(x/2)),x); % This integral produces a messy result because the code for % converting half angle tans to sin and cos is not effective enough. testint(sin(a*x)/(b+c*sin(a*x))**2,x); % Examples involving logarithms and circular functions. testint(sin log x,x); testint(cos log x,x); % Examples involving exponentials. testint(e**x,x); % 2.01 #3; testint(a**x,x); % 2.01 #4; testint(e**(a*x),x); testint(e**(a*x)/x,x); testint(1/(a+b*e**(m*x)),x); testint(e**(2*x)/(1+e**x),x); testint(e**(2*x)*e**(a*x),x); testint(1/(a*e**(m*x)+b*e**(-m*x)),x); testint(x*e**(a*x),x); testint(x**20*e**x,x); testint(a**x/b**x,x); testint(a**x*b**x,x); testint(a**x/x**2,x); testint(x*a**x/(1+b*x)**2,x); testint(x*e**(a*x)/(1+a*x)**2,x); testint(x*k**(x**2),x); testint(e**(x**2),x); testint(x*e**(x**2),x); testint((x+1)*e**(1/x)/x**4,x); testint((2*x**3+x)*(e**(x**2))**2*e**(1-x*e**(x**2))/(1-x*e**(x**2))**2, x); testint(e**(e**(e**(e**x))),x); % Examples involving exponentials and logarithms. testint(e**x*log x,x); testint(x*e**x*log x,x); testint(e**(2*x)*log(e**x),x); % Examples involving square roots. testint(sqrt(2)*x**2 + 2*x,x); testint(log x/sqrt(a*x+b),x); u:=sqrt(a+b*x); v:=sqrt(c+d*x); testint(u*v,x); testint(u,x); testint(x*u,x); testint(x**2*u,x); testint(u/x,x); testint(u/x**2,x); testint(1/u,x); testint(x/u,x); testint(x**2/u,x); testint(1/(x*u),x); testint(1/(x**2*u),x); testint(u**p,x); testint(x*u**p,x); testint(atan((-sqrt(2)+2*x)/sqrt(2)),x); testint(1/sqrt(x**2-1),x); testint(sqrt(x+1)*sqrt x,x); % Examples from James Davenport's thesis: testint(1/sqrt(x**2-1)+10/sqrt(x**2-4),x); % p. 173 testint(sqrt(x+sqrt(x**2+a**2))/x,x); % Examples generated by differentiating various functions. testint(df(sqrt(1+x**2)/(1-x),x),x); testint(df(log(x+sqrt(1+x**2)),x),x); testint(df(sqrt(x)+sqrt(x+1)+sqrt(x+2),x),x); testint(df(sqrt(x**5-2*x+1)-sqrt(x**3+1),x),x); % Another such example from James Davenport's thesis (p. 146). % It contains a point of order 3, which is found by use of Mazur's % bound on the torsion of elliptic curves over the rationals; testint(df(log(1+sqrt(x**3+1)),x),x); % Examples quoted by Joel Moses: testint(1/sqrt(2*h*r**2-alpha**2),r); testint(1/(r*sqrt(2*h*r**2-alpha**2-epsilon**2)),r); testint(1/(r*sqrt(2*h*r**2-alpha**2-2*k*r)),r); testint(1/(r*sqrt(2*h*r**2-alpha**2-epsilon**2-2*k*r)),r); testint(r/sqrt(2*e*r**2-alpha**2),r); testint(r/sqrt(2*e*r**2-alpha**2-epsilon**2),r); testint(r/sqrt(2*e*r**2-alpha**2-2*k*r**4),r); testint(r/sqrt(2*e*r**2-alpha**2-2*k*r),r); testint(1/(r*sqrt(2*h*r**2-alpha**2-2*k*r**4)),r); testint(1/(r*sqrt(2*h*r**2-alpha**2-epsilon**2-2*k*r**4)),r); Comment many of these integrals used to require Steve Harrington's code to evaluate. They originated in Novosibirsk as examples of using Analytik. There are still a few examples that could be evaluated using better heuristics; testint(a*sin(3*x+5)**2*cos(3*x+5),x); testint(log(x**2)/x**3,x); testint(x*sin(x+a),x); testint((log(x)*(1-x)-1)/(e**x*log(x)**2),x); testint(x**3*(a*x**2+b)**(-1),x); testint(x**(1/2)*(x+1)**(-7/2),x); testint(x**(-1)*(x+1)**(-1),x); testint(x**(-1/2)*(2*x-1)**(-1),x); testint((x**2+1)*x**(1/2),x); testint(x**(-1)*(x-a)**(1/3),x); testint(x*sinh(x),x); testint(x*cosh(x),x); testint(sinh(2*x)/cosh(2*x),x); testint((i*eps*sinh x-1)/(eps*i*cosh x+i*a-x),x); testint(sin(2*x+3)*cos(x)**2,x); testint(x*atan(x),x); testint(x*acot(x),x); testint(x*log(x**2+a),x); testint(sin(x+a)*cos(x),x); testint(cos(x+a)*sin(x),x); testint((1+sin(x))**(1/2),x); testint((1-sin(x))**(1/2),x); testint((1+cos(x))**(1/2),x); testint((1-cos(x))**(1/2),x); testint(1/(x**(1/2)-(x-1)**(1/2)),x); testint(1/(1-(x+1)**(1/2)),x); testint(x/(x**4+36)**(1/2),x); testint(1/(x**(1/3)+x**(1/2)),x); testint(log(2+3*x**2),x); testint(cot(x),x); testint(cot x**4,x); testint(tanh(x),x); testint(coth(x),x); testint(b**x,x); testint((x**4+x**(-4)+2)**(1/2),x); testint((2*x+1)/(3*x+2),x); testint(x*log(x+(x**2+1)**(1/2)),x); testint(x*(e**x*sin(x)+1)**2,x); testint(x*e**x*cos(x),x); Comment the following set came from Herbert Stoyan; testint(1/(x-3)**4,x); testint(x/(x**3-1),x); testint(x/(x**4-1),x); testint(log(x)*(x**3+1)/(x**4+2),x); testint(log(x)+log(x+1)+log(x+2),x); testint(1/(x**3+5),x); testint(1/sqrt(1+x**2),x); testint(sqrt(x**2+3),x); testint(x/(x+1)**2,x); COMMENT The following integrals were used among others as a test of Moses' SIN program; testint(asin x,x); testint(x**2*asin x,x); testint(sec x**2/(1+sec x**2-3*tan x),x); testint(1/sec x**2,x); testint((5*x**2-3*x-2)/(x**2*(x-2)),x); testint(1/(4*x**2+9)**(1/2),x); testint((x**2+4)**(-1/2),x); testint(1/(9*x**2-12*x+10),x); testint(1/(x**8-2*x**7+2*x**6-2*x**5+x**4),x); testint((a*x**3+b*x**2+c*x+d)/((x+1)*x*(x-3)),x); testint(1/(2-log(x**2+1))**5,x); % The next integral appeared in Risch's 1968 paper. testint(2*x*e**(x**2)*log(x)+e**(x**2)/x+(log(x)-2)/(log(x)**2+x)**2+ ((2/x)*log(x)+(1/x)+1)/(log(x)**2+x),x); % The following integral would not evaluate in REDUCE 3.3. testint(exp(x*ze+x/2)*sin(pi*ze)**4*x**4,ze); % This one evaluates: testint(erf(x),x); % So why not this one? testint(erf(x+a),x); Comment some interesting integrals of algebraic functions; % The Chebyshev integral. testint((2*x**6+4*x**5+7*x**4-3*x**3-x*x-8*x-8)/ ((2*x**2-1)**2*sqrt(x**4+4*x**3+2*x**2+1)),x); % This integral came from Dr. G.S. Joyce of Imperial College London. testint((1+2*y)*sqrt(1-5*y-5*y**2)/(y*(1+y)*(2+y)*sqrt(1-y-y**2)),y); % This one has a simple result. testint(x*(sqrt(x**2-1)*x**2-4*sqrt(x**2-1)+sqrt(x**2-4)*x**2 -sqrt(x**2-4))/((1+sqrt(x**2-4)+sqrt(x**2-1))*(x**4-5*x**2+4)),x); % This used to reveal bugs in the integrator which have been fixed. % Since it takes a long time and doesn't have a closed form result, % it has been commented out. % testint(sqrt(-4*sqrt(2)+9)*x-sqrt(x**4+2*x**2+4*x+1)*sqrt(2),x); Comment here is an example of using the integrator with pattern matching; for all m,n let int(k1**m*log(k1)**n/(p**2-k1**2),k1)=foo(m,n), int(k1*log(k1)**n/(p**2-k1**2),k1)=foo(1,n), int(k1**m*log(k1)/(p**2-k1**2),k1)=foo(m,1), int(k1*log(k1)/(p**2-k1**2),k1)=foo(1,1), int(log(k1)**n/(k1*(p**2-k1**2)),k1)=foo(-1,n); int(k1**2*log(k1)/(p**2-k1**2),k1); COMMENT It is interesting to see how much of this one can be done; let f1s= (12*log(s/mc**2)*s**2*pi**2*mc**3*(-8*s-12*mc**2+3*mc) + pi**2*(12*s**4*mc+3*s**4+176*s**3*mc**3-24*s**3*mc**2 -144*s**2*mc**5-48*s*mc**7+24*s*mc**6+4*mc**9-3*mc**8)) /(384*e**(s/y)*s**2); int(f1s,s); factor int; ws; Comment the following integrals reveal deficiencies in the current integrator; %high degree denominator; %testint(1/(2-log(x**2+1))**5,x); %this example should evaluate; testint(sin(2*x)/cos(x),x); %this example, which appeared in Tobey's thesis, needs factorization %over algebraic fields. It currently gives an ugly answer and so has %been suppressed; % testint((7*x**13+10*x**8+4*x**7-7*x**6-4*x**3-4*x**2+3*x+3)/ % (x**14-2*x**8-2*x**7-2*x**4-4*x**3-x**2+2*x+1),x); symbolic summarize!-integral!-test(); end;
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clc,clear printf('Example 5.1\n\n') P=1000*10^3 //load power phi=acosd(0.8) //power factor lagging angle V_L=11*10^3 //rated terminal voltae R_a=0.4 //armature resistance per phase X_s=3//synchronous reactance per phase I_L=P/(sqrt(3)*V_L*cosd(phi)) I_aph=I_L //for star connected load I_a=I_L//current through armature V_ph=V_L/sqrt(3) //rated terminal volatge phase value E_ph= sqrt( (V_ph*cosd(phi)+I_a*R_a)^2+(V_ph*sind(phi)+I_a*X_s)^2 ) //emf generated phase value E_line=E_ph*sqrt(3) //line value of emf generated regulation=100*(E_ph-V_ph)/V_ph //pecentage regulation printf('Line value of e.m.f generated is %.2f kV\nRegulation is %.3f percent',E_line*10^-3,regulation)
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X= [100 110 120 130 140 150 160 170 180 190]; Y= [45 52 54 63 62 68 75 76 92 88]; plot2d(X, Y, -1); disp("A linear regression model seems appropriate")
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// Exa 11.10 clc; clear; close; format('v',6) // Given data Cs = 0.04;// in pF C_M = 2;// in pF Per = (1/2)*(Cs/C_M)*100;// in % disp("Parallel resonant frequency is greater than series resonant frequency by "+string(Per)+" %")
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//caption:find deflection sensitivity of CRO //Ex8.1 clc clear close l=20*10^-3//axial length of deflection plate(in meter) L=0.2//distance from the centre of the deflection plates to the screen(in meter) s=5*10^-3//spacing between two plates(in meter) V=2500//accelerating voltage(in Volt) S=(l*L)/(2*s*V) disp(S,'deflection sensitivity of CRO(in m/V)=')
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clc clear //Input data m1=100;//Air flow rate in kg/hr q1=600;//The heat generated by each person in kJ/hr h1=85;//The enthalpy of air entering the room in kJ/kg h2=60;//The enthalpy of air leaving the room in kJ/kg Q1=0.2;//The heat added by each lamp in the room in kW P1=0.2;//The power consumed by each fan in kW //Calculations q=(5*q1)/3600;//The heat generated by 5 persons in the room in kW Q=3*Q1;//The heat added by three lamps in the room in kW P=2*P1;//The power consumed by two fans in the room in kW m=m1/3600;//Mass flow rate of air in kg/s H=[q+Q+P]+[m*(h1-h2)];//Heat to be removed by the cooler in kW //Output printf('The rate at which the heat is to be removed by cooler X = %3.3f kJ/sec ',H)
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11_3_Crossflow_finned_tube_HeatX.sce
clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 11.3 Page 692 \n'); //Example 11.3 // Required gas side surface area //Operating Conditions Tho = 100+273 ;//[K] Hot Fluid outlet Temperature Thi = 300+273 ;//[K] Hot Fluid intlet Temperature Tci = 35+273 ;//[K] Cold Fluid intlet Temperature Tco = 125+273 ; //[K] Cold Fluid outlet Temperature mc = 1 ;//[kg/s] Cold Fluid flow rate Uh = 100 ;//[W/m^2.K] Coefficient of heat transfer //Table A.5 Water Properties T = 353 K cph = 1000 ; //[J/kg.K] Specific Heat //Table A.6 Saturated water Liquid Properties Tc = 308 K cpc = 4197 ; //[J/kg.K] Specific Heat Cc = mc*cpc; //Equation 11.6b and 11.7b Ch = Cc*(Tco-Tci)/(Thi-Tho); // Equation 11.18 qmax = Ch*(Thi-Tci); //Equation 11.7b q = mc*cpc*(Tco-Tci); e = q/qmax; ratio = Ch/Cc; printf("\n As effectiveness is %.2f with Ratio Cmin/Cmax = %.2f, It follows from figure 11.14 that NTU = 2.1",e,ratio); NTU = 2.1; A = 2.1*Ch/Uh; printf("\n Required gas side surface area = %.1f m^2",A); //END
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//Example 5.6 F_by_A=5.00*10^7;//Force per unit area at 5km depth (N/m^2) B=2.2*10^9;//Bulk modulus (N/m^2), See Table 5.3 v=(F_by_A)/B;//Fractional decrease in volume printf('Fractional decrease in volume (in percentage) = %0.1f%%',v*100) //Openstax - College Physics //Download for free at http://cnx.org/content/col11406/latest
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//i/p vector contains imaginary elements x=[1 0.2 0.3 0.5*%i]; y=vco(x,150,500); disp(y); //output // column 1 to 3 // // - 0.3090170 - 0.5358268 0.9510565 // // column 4 // // - 0.8687447 - 0.8804820i
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// exa 8.7 Pg 232 clc;clear;close; // Given Data dv=30;// mm Wv=10;// N Wl=25;// N lf=100;// mm del1=20;// mm p=3.5;// N/mm.sq. valve_lift=2;// mm C=6;// spring index tau=500;// N/mm.sq. G=0.84*10**5;// N/mm.sq. W=(%pi/4)*dv**2*p;// N (load on the valve at operating condition) W1=W-Wv;//N (Net load on the valve at operating condition) //W1*100=Wl*150+S1*200+P*300 // taking momens about the fulcrum //S1*200+P*300=W1*100-Wl*150 ...eqn(1) valve_lift=20*100/200;// mm //from figure (when spring is extended by 20 mm) spring_extension=2*200/100;// mm // from figure (when valve is lifted 2 mm) valve_load=W*12/10;// N // (when valve is lifted 2 mm) W2=valve_load-Wv;// N // (when valve is lifted 2 mm) del2=del1+4;// mm (when valve is lifted) //S2=S1*del2/del1;// spring force when valve is lifted //S1*del2/del1-s2=0 ... eqn(1) //W2*100=Wl*150+S2*200+P*300 // taking momens about the fulcrum //S2*200+P*300 =W2*100-Wl*150 ... eqn(2) //S1*200+P*300=W1*100-Wl*150 ...eqn(3) // solving above 3 eqn. by matrix method A=[del2/del1 -1 0;200 0 300;0 200 300]; B=[0;W1*100-Wl*150;W2*100-Wl*150]; X=A**-1*B;// solution matrix S1=X(1);// N S2=X(2);// N printf('\n Spring force when valve is lifted = %.1f N',S2) printf('\n\n Design of spring - ') k=(S2-S1)/(del2-del1);// N/mm (Spring stiffness) printf('\n Spring stiffness = %.2f N/mm',k) Kw=(4*C-1)/(4*C-4)+0.615/C;// Wahl's correction factor printf('\n Wahl''s correction factor = %.4f',Kw) // tau=Kw*8*S2*C/%pi/d**2 max. shear stress d=sqrt(Kw*8*S2*C/%pi/tau);// mm (spring diameter) printf('\n spring diameter = %.2f mm or %.f mm',d,d) d=ceil(d);// mm // k=G*d/(8*C**3*n) (Spring stiffness) n=G*d/(8*C**3*k);// no. of active coils printf('\n no. of active coils = %.2f. Use n=7',n) n=ceil(n);// rounding nt=n+1;// total no. of active coils printf('\n total no. of active coils = %.f',nt) p=lf/(n-1);// mm (pitch of coils) printf('\n pitch of coils = %.2f mm',p)
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#define ix0 0 #define lx0 0 #define ix1 1 #define lx1 1 #define ix2 2 #define lx2 2 #define ix4 4 #define lx4 4 #if __WORDSIZE == 32 # define ix7fe 0x7ffffffe # define ix7f 0x7fffffff # define ix80 0x80000000 # define iff 0xffffffff # define ife 0xfffffffe # define ifd 0xfffffffd # define ifc 0xfffffffc #else # define ix7fe 0x7ffffffffffffffe # define ix7f 0x7fffffffffffffff # define ix80 0x8000000000000000 # define iff 0xffffffffffffffff # define ife 0xfffffffffffffffe # define ifd 0xfffffffffffffffd # define ifc 0xfffffffffffffffc #endif /* check jumps are taken and result value is correct */ #define bopr_t(l, u, op, r0, r1, il, ir, iv) \ movi %r0 il \ movi %r1 ir \ b##op##r##u op##u##r##l##r0##r1 %r0 %r1 \ /* validate did jump */ \ movi %r0 0x5a5a5a5a \ op##u##r##l##r0##r1: \ beqi op##u##r##l##ok##r0##r1 %r0 iv \ calli @abort \ op##u##r##l##ok##r0##r1: #define bopi_t(l, u, op, r0, il, ir, iv) \ movi %r0 il \ b##op##i##u op##u##i##l##r0##r1 %r0 ir \ /* validate did jump */ \ movi %r0 0x5a5a5a5a \ op##u##i##l##r0##r1: \ beqi op##u##i##l##ok##r0##r1 %r0 iv \ calli @abort \ op##u##i##l##ok##r0##r1: #define bopr_f(l, u, op, r0, r1, il, ir, iv) \ movi %r0 il \ movi %r1 ir \ b##op##r##u op##u##r##l##r0##r1 %r0 %r1 \ beqi op##u##r##l##ok##r0##r1 %r0 iv \ op##u##r##l##r0##r1: \ calli @abort \ op##u##r##l##ok##r0##r1: #define bopi_f(l, u, op, r0, il, ir, iv) \ movi %r0 il \ b##op##i##u op##u##i##l##r0##r1 %r0 ir \ beqi op##u##i##l##ok##r0##r1 %r0 iv \ op##u##i##l##r0##r1: \ calli @abort \ op##u##i##l##ok##r0##r1: #define ccop(cc, l, u, op, r0, r1, il, ir, iv) \ bopr##cc(l, u, op, r0, r1, i##il, i##ir, i##iv) \ bopi##cc(l, u, op, r0, i##il, i##ir, i##iv) #define tadd(l, u, r0, r1, il, ir, iv) \ ccop(_t, l, u, oadd, r0, r1, il, ir, iv) \ ccop(_f, l, u, xadd, r0, r1, il, ir, iv) #define fadd(l, u, r0, r1, il, ir, iv) \ ccop(_f, l, u, oadd, r0, r1, il, ir, iv) \ ccop(_t, l, u, xadd, r0, r1, il, ir, iv) #define tsub(l, u, r0, r1, il, ir, iv) \ ccop(_t, l, u, osub, r0, r1, il, ir, iv) \ ccop(_f, l, u, xsub, r0, r1, il, ir, iv) #define fsub(l, u, r0, r1, il, ir, iv) \ ccop(_f, l, u, osub, r0, r1, il, ir, iv) \ ccop(_t, l, u, xsub, r0, r1, il, ir, iv) #define xopr6(l,op,r0,r1,r2,r3,r4,r5,llo,lhi,rlo,rhi,vlo,vhi) \ movi %r1 llo \ movi %r2 lhi \ movi %r4 rlo \ movi %r5 rhi \ op##cr %r0 %r1 %r4 \ op##xr %r3 %r2 %r5 \ beqi op##l##L##r0##r1##r2##r3##r4##r5 %r0 vlo \ calli @abort \ op##l##L##r0##r1##r2##r3##r4##r5: \ beqi op##l##H##r0##r1##r2##r3##r4##r5 %r3 vhi \ calli @abort \ op##l##H##r0##r1##r2##r3##r4##r5: #define xopr4_(l,op,r0,r1,r2,r3,llo,lhi,rlo,rhi,vlo,vhi) \ movi %r0 llo \ movi %r1 lhi \ movi %r2 rlo \ movi %r3 rhi \ op##cr %r0 %r0 %r2 \ op##xr %r1 %r1 %r3 \ beqi op##l##L_##r0##r1##r2##r3 %r0 vlo \ calli @abort \ op##l##L_##r0##r1##r2##r3: \ beqi op##l##H_##r0##r1##r2##r3 %r1 vhi \ calli @abort \ op##l##H_##r0##r1##r2##r3: #define xopr_4(l,op,r0,r1,r2,r3,llo,lhi,rlo,rhi,vlo,vhi) \ movi %r0 rlo \ movi %r1 rhi \ movi %r2 llo \ movi %r3 lhi \ op##cr %r0 %r2 %r0 \ op##xr %r1 %r3 %r1 \ beqi op##l##_L##r0##r1##r2##r3 %r0 vlo \ calli @abort \ op##l##_L##r0##r1##r2##r3: \ beqi op##l##_H##r0##r1##r2##r3 %r1 vhi \ calli @abort \ op##l##_H##r0##r1##r2##r3: #define xaddr(l,llo,lhi,rlo,rhi,vlo,vhi) \ xopr6(l,add,r0,r1,r2,v0,v1,v2,i##llo,i##lhi,i##rlo,i##rhi,i##vlo,i##vhi) \ xopr4_(l,add,r0,r1,r2,v0,i##llo,i##lhi,i##rlo,i##rhi,i##vlo,i##vhi) \ xopr_4(l,add,r0,r1,r2,v0,i##llo,i##lhi,i##rlo,i##rhi,i##vlo,i##vhi) #define xsubr(l,llo,lhi,rlo,rhi,vlo,vhi) \ xopr6(l,sub,r0,r1,r2,v0,v1,v2,i##llo,i##lhi,i##rlo,i##rhi,i##vlo,i##vhi) \ xopr4_(l,sub,r0,r1,r2,v0,i##llo,i##lhi,i##rlo,i##rhi,i##vlo,i##vhi) \ xopr_4(l,sub,r0,r1,r2,v0,i##llo,i##lhi,i##rlo,i##rhi,i##vlo,i##vhi) .data 16 ok: .c "ok\n" .code prolog tadd(__LINE__, , r0, r1, x7f, x1, x80) fadd(__LINE__, , r0, r1, x7fe, x1, x7f) tsub(__LINE__, , r0, r1, x80, x1, x7f) fsub(__LINE__, , r0, r1, x7f, x1, x7fe) tadd(__LINE__, _u, r0, r1, ff, x1, x0) fadd(__LINE__, _u, r0, r1, x7f, x1, x80) tsub(__LINE__, _u, r0, r1, x0, x1, ff) fsub(__LINE__, _u, r0, r1, x80, x1, x7f) /* 0xffffffffffffffff + 1 = 0x10000000000000000 */ xaddr(__LINE__, ff, ff, x1, x0, x0, x0) /* 1 + 0xffffffffffffffff = 0x10000000000000000 */ xaddr(__LINE__, x1, x0, ff, ff, x0, x0) /* 0xfffffffeffffffff + 1 = 0xffffffff00000000 */ xaddr(__LINE__, ff, fe, x1, x0, x0, ff) /* 1 + 0xfffffffeffffffff = 0xffffffff00000000 */ xaddr(__LINE__, x1, x0, ff, fe, x0, ff) /* 0xfffffffefffffffe + 2 = 0xffffffff00000000 */ xaddr(__LINE__, fe, fe, x2, x0, x0, ff) /* 2 + 0xfffffffefffffffe = 0xffffffff00000000 */ xaddr(__LINE__, x2, x0, fe, fe, x0, ff) /* 0xffffffffffffffff - 1 = 0xfffffffffffffffe */ xsubr(__LINE__, ff, ff, x1, x0, fe, ff) /* 1 - 0xffffffffffffffff = -0xfffffffffffffffe */ xsubr(__LINE__, x1, x0, ff, ff, x2, x0) /* 0xfffffffeffffffff - 1 = 0xfffffffefffffffe */ xsubr(__LINE__, ff, fe, x1, x0, fe, fe) /* 1 - 0xfffffffeffffffff = -0xfffffffefffffffe */ xsubr(__LINE__, x1, x0, ff, fe, x2, x1) /* 0xfffffffefffffffe - 2 = 0xfffffffefffffffc */ xsubr(__LINE__, fe, fe, x2, x0, fc, fe) /* 2 + 0xfffffffefffffffe = -0xfffffffefffffffc */ xsubr(__LINE__, x2, x0, fe, fe, x4, x1) prepare pushargi ok ellipsis finishi @printf ret epilog
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//example 21 clear emicur=1*10^-3;//ampere colcur=0.995*10^-3;//ampere alpha1=colcur/emicur; beta1=alpha1/(1-alpha1); disp("alpha = "+string((alpha1))); disp("beta = "+string((beta1)));
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//Problem 3 //calculate the energy falling on the target material per second and also calculate the cutoff wavelength of the X-rays clear clc V=20*10^3// potential difference in V e=1.6*10^(-19)//charge on an electron in C h=6.6*10^(-34)//planck's constant in J-s c=3*10^(8)//velocity of light in m/s i=1//current in mA E=i*V*10^(-3)//energy in j/s w=(h*c)/(e*V)//wavelength in nm printf('energy falling on the target material per second = %.1f j/s \n',E) printf('cutoff wavelength of the X-rays = %.3f nm',w*10^9)
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clc clear //DATA GIVEN D=0.2; //diameter of engine cylinder in m L=0.350; //length of stroke in m Pmico=6.5; //mean effective pressure on cover side in bar Pmicr=7; //mean effective pressure on crank side in bar N=420; //engine speed in R.P.M. Drod=0.02; //diameter of piston rod in m W=1370; //dead load on the brake in N S=145; //spring balance reading in N Db=1.2; //diameter of brake wheel in m d=0.02; //diameter of rope in m n=1; //no. of cylinders k=0.5; //for 4-stroke cylinder //INDICTED POWER ,I.P.=(n*Pmi*l*A*N*k*10)/6 kW Aco=(%pi/4)*(D^2); //area of cylinder om cover end in m^2 Acr=(%pi/4)*(D^2-Drod^2); //area of cylinder om crank end in m^2 IPco=(n*Pmico*L*Aco*N*k*10)/6; //IP on cover end side in kW IPcr=(n*Pmicr*L*Acr*N*k*10)/6; //IP on crank end side in kW IPtotal=IPco+IPcr; //IP total in kW //Brake Power, B.P.=(W-S)(pi)(Db+d)N/(60*1000) kW BP=(W-S)*(%pi)*(Db+d)*N/(60*1000); eta=BP/IPtotal; //mechanical efficiency printf('Mechanical efficiency is: %5.4f or %5.2f percent.\n',eta,(eta*100));
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// This file is part of the materials accompanying the book // "The Elements of Computing Systems" by Nisan and Schocken, // MIT Press. Book site: www.idc.ac.il/tecs // File name: projects/03/1/DFF1.tst // created by Xin Jia load DFF.hdl, output-file DFF1.out, //compare-to Bit.cmp, output-list time%S1.4.1 in%B2.1.2 out%B2.1.2; set in 0, tick, output; tock, output; set in 1, tick, output; tock, output; set in 0, tick, output; tock, output; set in 1, tick, output; tock, output; set in 0, tick, output; tock, output; set in 1, tick, output; tock, output; set in 0, tick, output; tock, output; set in 1, tick, output; tock, output;
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//Relating DFT and DTFT xn=[1 2 1 0]; XDFT=fft(xn,-1); //for F=k/4,k=0,1,2,3 for k=1:4 XF(k)=1+2*%e^(-%i*%pi*(k-1)/2)+%e^(-%i*%pi*(k-1)); end XF,XDFT disp(XF,'The DFT of x[n] is');
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clear all; clc; //This numerical is Ex 2_1S,page 29. Q1=18.2 N1=1000 N2=1500 delta_p1=10.3 P_s1=6 Q2=Q1*N2/N1 printf("\n The value of Q2 is equal to %g m^3/h",Q2) delta_p2=delta_p1*((N2/N1)^2) printf("\n The value of delta_p2 is equal to %0.1f bars",delta_p2) P_s2=P_s1*(N2/N1)^3 printf("\n The value of P_s2 is equal to %g kW",P_s2) E1=((Q1/3600)*delta_p1*10^2)/(P_s1) printf("\n The value of E1=E2 is equal to %g ",E1) disp("Thus the efficiency is equal to 86.8%")
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clc clear //input data M1=0.25 //Mach number at entrance f=0.04/4 //frictional factor D=0.15 //inner duct diameter in m p1=0.9 //Stagnation pressure ratio at exit to entry when loss in stagnation pressure is 10% ds=190 ///Change in entropy in J/kg-K k=1.3 //Adiabatic constant R=287 //Specific Gas constant in J/kg-K, wrong printing in question //calculation p2=2.4064 //Ratio of stagnation pressures at inlet to critical state from gas tables fanno flow tables @M1,k=1.3 X1=8.537 //frictional constant fanno parameter from gas tables,fanno flow tables @M1,k=1.3 p3=p1*p2 //Ratio of stagnation pressures at exit to critical state from gas tables fanno flow tables @M1,k=1.3 M2=0.28 //Mach number at p1=0.9 from gas tables @p3 X2=6.357 //frictional constant fanno parameter from gas tables,fanno flow tables @M2,k=1.3 X3=X1-X2 //overall frictional constant fanno parameter L1=(X3*D)/(4*f) //Length of the pipe in m p4=exp(ds/R) //Ratio of Stagnation pressure at entry to Stagnation pressure where ds=190 p5=p1/p4 //Ratio of Stagnation pressures where ds=190 to critical state M3=0.56 //Mach number where ds=190 X4=0.674 //frictional constant fanno parameter from gas tables,fanno flow tables @M3,k=1.3 X5=X1-X4 //overall frictional constant fanno parameter L2=(X5*D)/(4*f) //Length of the pipe in m //output printf('(A)Length of the pipe is %3.3f m\n (B)Length of the pipe would require to rise entropy by %3i J/kg-K is %3.5f m\n (C)Mach number is %3.2f',L1,ds,L2,M3)
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clear; clc; r1=0,r2=1,z1=-2,z2=2,q1=0,q2=2*%pi; Q=integrate('p^2','p',r1,r2)*integrate('(cos(Q)^2)','Q',q1,q2)*integrate('1','z',z1,z2); disp(Q,'Total charge is =');
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clc; Vpk=10; slewrate=0.5*10**6; fmax=slewrate/(2*%pi*Vpk); disp('HZ',fmax*1,"fmax=");//value of microamp 741 slewrate=13*10**6; fmax=slewrate/(2*%pi*Vpk); disp('kHZ',fmax*10**-3,"fmax=");//TLO 81 //value of microamp 741 is much lower than that of the input signal.And value of TLO81 is much higher than input signal,therefore TLO81 can be used
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clc //initialization of variables T1=69 //F T2=84 //F P=14.7 //lb/in^2 //calculations disp("from wet bulb n dry bulb temperature charts,") sh=82/7000 rh=47 Pwv=0.27 T=62 //F h=33.3 //results printf("Specific humidity = %.4f lbm/lbm",sh) printf("\n Relative humidity = %d ",rh) printf("\n Partial pressure = %.2f lb/in^2",Pwv) printf("\n Dew point = %d F",T) printf("\n Enthalpy per pound of air = %.1f V/lbm dry air",h)
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// The code was developed under Horizon2020 Framework Programme // Project: 748767 — SIMFREE function Out=SSSoEDFAgainNoise(In,G_dB,NF_dB) // Optical Amplifier // // Calling Sequence // Out=SSSoEDFAgainNoise(In,G_dB,NF_dB) // // Parameters // In : Optical Input // G_dB : Small Signal Gain [dB] // G_dB : Noise Figure [dB] // Out : Optical Output // // Description // A simple optical amplifier model with constant gain and white stationary noied added. // The amplifier gain is flat with wavelength. // [lhs,rhs]=argn(0); global MNT MNS MSR MLA select rhs case 0 then error("Expect at least one argument"); case 1 then G_dB=0; NF_dB=0; case 2 then NF_dB=0; end NF=10^(NF_dB/10); G=10^(G_dB/10); c=299792458; h=6.62607e-34; df=1e9*MSR/MNS; L=1e-9*MLA; sASE=sqrt(1e3*G*NF*df*h*c/L/4); x=complex(grand(MNT,1,'nor',0,sASE),grand(MNT,1,'nor',0,sASE)); y=complex(grand(MNT,1,'nor',0,sASE),grand(MNT,1,'nor',0,sASE)); Out=sqrt(G)*In; Out(:,1)= Out(:,1)+x; Out(:,2)= Out(:,2)+y; endfunction
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argc:7 Dataset: ../datasets/converted/karate.net Nodes Edges Com Mod NMI Time seq semisync 34 156 3 0.451841 -1 6.7428e-05 par semisync 34 156 5 0.4442 -1 0.06915
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s=%s; A=[-2 -3;4 2] B=[3;5] C=[1 1] TF=C*inv(s*eye(2,2)-A) *B disp(TF,"transfer function = ")
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[Read-the-docs-example-1-360-day] # Read the docs - example 1 indice_name: SU in_files: ['tasmax_day_HadGEM2-ES_historical_r1i1p1_19891201-19991130.nc', 'tasmax_day_HadGEM2-ES_historical_r1i1p1_19991201-20051130.nc'] dt1: 1990-01-01 dt2: 2004-12-31 slice_mode: month
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clear clc m=[0.01,0.02,0.05,0.10,0.20];// n=0.01;// mu=m+n;// E=[1.0495,1.0315,1.0073,0.9885,0.9694] E2=0.2225;//in V R=0.05913;//in V O=log10(m/n);// K=(E-E2)/R +O;// plot(mu,K,'mo-');// [m,c]=reglin(mu,K) Ksp=10^-c;// printf('Ksp=%.2f*10^-14',Ksp/10^-14) //There are some errors in the solution given in textbook //page 491
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//Section-14,Example-2,Page no.-PC.126 clc; CH_3COONa_1=0.01 CH_3COOH_1=0.1 K_a=1.75*10^-5 pK_a=-log10(K_a) pH1=pK_a +(log10(CH_3COONa_1/CH_3COOH_1)) disp(pH1,'pH of the given buffer solution') HCl=0.0002 CH_3COONa_2=0.01+0.0002 CH_3COOH_2=0.1-0.002 pH2=pK_a +(log10(CH_3COONa_2/CH_3COOH_2)) disp(pH2,'pH of the solution after addition of HCl') C_1=pH1-pH2 //change in pH CH_3COONa_3=0.01+0.002 CH_3COOH_3=0.1-0.002 pH3=pK_a +(log10(CH_3COONa_3/CH_3COOH_3)) pH4=pH1-pH3 //change in pH disp(pH4,'Required pH')
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disp('the GCD of the following numbers is:') V=int32([12,18]); [thegcd]=gcd(V) V=int32([12,-18]); [thegcd]=gcd(V) V=int32([12,-16]); [thegcd]=gcd(V) V=int32([29,15]); [thegcd]=gcd(V) V=int32([14,49]); [thegcd]=gcd(V)
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//Chapter 11 : Free Electron Theory Of Metals clear; //Variable declaration fE=0.01 //probability delE=8*10**-20 //ev to J //Calculations T=5797/log(99) //Result disp('K',T,"Temperature=")
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ScreenName String 'Arms Selection Screen' ImplName String 'NULL SCREEN' ElementChunkArray Int 14 ScreenElementType Int 0 ImplName String 'Assemble Fleet Backdrop' TabIndex Int 8 Selectable Bool False Enabled Bool True ReferenceArea Rect( 0, 0, 800, 600 ) # left,top,right,bottom ScreenElementType Int 1 ImplName String 'Fleet Select Button' TabIndex Int 11 Selectable Bool True Enabled Bool True ReferenceArea Rect( 27, 479, 459, 559 ) # left,top,right,bottom Font String 'UniversBold14' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_NULL' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'AutoAssign Button' TabIndex Int 1 Selectable Bool False Enabled Bool True ReferenceArea Rect( 168, 427, 320, 465 ) # left,top,right,bottom Font String 'BlackChancery16' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_AUTO_ASSIGN' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'Open Arms Selection Screen' TabIndex Int 5 Selectable Bool False Enabled Bool True ReferenceArea Rect( 308, 45, 434, 83 ) # left,top,right,bottom Font String 'BlackChancery16' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_ARMS' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'Open Crew Selection Screen' TabIndex Int 4 Selectable Bool False Enabled Bool True ReferenceArea Rect( 177, 45, 303, 83 ) # left,top,right,bottom Font String 'BlackChancery16' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_CREW' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'Open Ship Selection Screen' TabIndex Int 3 Selectable Bool False Enabled Bool True ReferenceArea Rect( 46, 45, 172, 83 ) # left,top,right,bottom Font String 'BlackChancery16' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_SHIPS' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 5 ImplName String 'Arms Types Available' TabIndex Int 7 Selectable Bool True Enabled Bool True ReferenceArea Rect( 516, 330, 770, 504 ) # left,top,right,bottom Font String 'Univers12' ScreenElementType Int 1 ImplName String 'Ship Victory Points' TabIndex Int 15 Selectable Bool False Enabled Bool True ReferenceArea Rect( 488, 79, 772, 109 ) # left,top,right,bottom Font String 'UniversLightBold14' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_BUTTON' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'Open Load Arms Profile Dialog Button' TabIndex Int 18 Selectable Bool False Enabled Bool True ReferenceArea Rect( 502, 45, 628, 83 ) # left,top,right,bottom Font String 'BlackChancery16' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_LOAD_ARMS' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'Open Save Arms Profile Dialog Button' TabIndex Int 19 Selectable Bool False Enabled Bool True ReferenceArea Rect( 634, 45, 759, 83 ) # left,top,right,bottom Font String 'BlackChancery16' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_SAVE_ARMS' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'Open Next Arrow Button' TabIndex Int 2 Selectable Bool False Enabled Bool True ReferenceArea Rect( 590, 510, 670, 590 ) # left,top,right,bottom Font String 'BlackChancery16' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_NULL' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'Arms Information Button' TabIndex Int 16 Selectable Bool True Enabled Bool True ReferenceArea Rect( 526, 121, 776, 280 ) # left,top,right,bottom Font String 'Univers10' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_BUTTON' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'Arms Upgrade Info Button' TabIndex Int 20 Selectable Bool True Enabled Bool True ReferenceArea Rect( 67, 163, 243, 294 ) # left,top,right,bottom Font String 'Univers12' Text String 'IDGS_TPFRONTENDTEXT_BUTTON_CREW' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1 ScreenElementType Int 1 ImplName String 'Modify Arms Position Button' TabIndex Int 17 Selectable Bool True Enabled Bool True ReferenceArea Rect( 46, 91, 438, 371 ) # left,top,right,bottom Font String 'Univers10' Text String 'IDGS_TPFRONTENDTEXT_SCREENS_BUTTON' Color Colour( 1.000000, 1.000000, 1.000000, 1.000000 ) HotKey Int -1
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//chapter-10,Example10_9,pg 503 err=0.5*10^-2//(+/-)0.5% R=100*10^6//test resistance //Re=((R*2*Rip)/(R+(2*Rip))) Re1=R-(err*R)//err=+0.5 Re2=R-(-err*R)//err=-0.5 Rip1=((R*Re1)/(2*(R-Re1)))//err=+0.5 Rip2=((R*Re2)/(2*(R-Re2)))//err=-0.5 printf("resistance of each insulating post-1\n") printf("Rip1=%.2f ohm\n",Rip1) printf("resistance of each insulating post-2\n") printf("Rip2=%.2f ohm",Rip2)
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clc // Given that lambda = 7620 // Wavelength of light in angstrom mu_r = 1.53914 // refractive index of quartz for right handed circularly polarized light mu_l = 1.5392 // refractive index of quartz for left handed circularly polarized light t = 0.5 // thickness of plate in mm // Sample Problem 19 on page no. 220 printf("\n # PROBLEM 19 # \n") theta = %pi*t*(mu_l-mu_r)/(lambda*1e-7)*180/%pi // Rotation of plane of polarization printf("\n Standard formula used \n theta = pi*t*(mu_l-mu_r)/(lambda). \n") printf("\n Rotation of plane of polarization is %f degree.",theta)
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nx = 500; ny = 500; x = linspace(-1,1,nx); y = linspace(-1,1,ny); [X,Y] = ndgrid(x,y); A = sin(20*X); A (find(A>0.0 ) ) = 1; A (find(A<0.0 ) ) = 0; f = scf(); grayplot(x,y,A); f.color_map = graycolormap(32); xs2png(gcf(),'grating.png');
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clc //Initialization of variables rv=8 k=1.4 Qa=1280 pa=14.7 //psia R=10.73/29 Ta=540 //R J=778 cv=0.17 //Btu/lbm R //calculations etat=1-1/rv^(k-1) W=etat*Qa va=R*Ta/pa vb=va/rv Tb=Ta*rv^(k-1) dt=Qa/cv Tc=Tb+dt pb=pa*(rv)^(k-1) pc= Tc*pb/Tb Td=Tc*(1/rv)^(k-1) pd=pa*Td/Ta imep = W*J/144/(va-vb) //results printf("Thermal efficiency = %.1f percent",etat*100) printf("\n Work done = %d btu/lbm air",W) printf("\n Imep = %d lbf/in^2",imep) printf("\n Pressure and temperature at A = %.1f psia and %d R",pa,Ta) printf("\n Pressure and temperature at B = %d psia and %d R",pb,Tb) printf("\n Pressure and temperature at C = %d psia and %d R",pc,Tc) printf("\n Pressure and temperature at D = %d psia and %d R",pd,Td) printf("\n The pressures given in textbook are wrong. Please check using a calculator")
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function [r]=%p_o_lss(s1,s2) //r=%p_o_lss(s1,s2) <=> r=(s1==s2) polynomail==state-space syslin list //! // Copyright INRIA r=%f
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//Solution 6-5 WD=get_absolute_file_path('6_05_solution.sce'); datafile=WD+filesep()+'6_05_example.sci'; clc; exec(datafile) //unit conversion V_1 = V_1 * 1000 / 3600 //from [km/hr] to [m/s] Wdot_act = Wdot_act * 1000 //from [kW] to [W] //(a) A_1 = %pi / 4 * D**2 //cross-sectional area of turbine blade span mdot = rho * V_1 * A_1 Wdot_max = mdot * V_1**2 / 2 eta_turbine = Wdot_act / Wdot_max printf("The efficiency of wind turbine generator is %1.2f percent", eta_turbine * 100) //(b) V_2 = V_1 * sqrt(1 - eta_turbine) F_R = mdot * (V_2 - V_1) //momentum equation in x direction printf("The Horizontal force exerted by wind in positive x direction on the supporting mast of wind turbine is %1.0f N", F_R)
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function [stk,txt,top]=%52sci() // genere le code relatif a la negation //! // Copyright INRIA txt=[] s2=stk(top) if s2(2)=='2' then s2(1)='('+s2(1)+')',end stk=list('~'+s2(1),s2(2),s2(3),s2(4),'4')
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// sum 3-13 clc; clear; R1=50; p=75; pmax=125; R2=sqrt((pmax+p)*R1^2/(pmax-p)); t=R2-R1; // printing data in scilab o/p window printf("t is %0.1f mm ",t);
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clc; clear; format('v',11); Er=-110; //as electric field has only radial component Re=6350000; Dr=8.85*10^-12*Er; Q=Dr*4*3.14*Re^2; disp(Q,"total charge dispersed on the earth(in coulomb)="); rho_s=Q/(4*3.14*Re^2); disp(rho_s,"surface charge density(in C/m^2)=");
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//Calcular raiz quadrada de C c = input ("Digitar valor de c = "); xa = 1; xf = (xa + c/xa)/2 while xf <> xa xa = xf; xf = (xa + c/xa)/2 end printf("raiz quadrada de %10.2f = %10.7f", c, xa);
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// Example 1.4: Contact difference of potential clc, clear N=5e22; // Number of acceptor or donor atoms per metres cube of step graded p-n junction ni=1.45e16; // Intrinsic carrier concentration in inverse metres cube VT=25e-3; // Voltage equivalent to temperatue at room temperature in volts Vo=VT*log(N^2/ni^2); // Contact difference of potential in volts Vo=Vo*1e3; // Contact difference of potential in milivolts disp(Vo,"Contact difference of potential (mV) = ");
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//AC Circuits : example 4.33 :pg(4.27) Vm=177; Im=14.14; phi=30; V=(Vm/sqrt(2)); I=(Im/sqrt(2)); pf=cosd(30); P=(V*I*pf); disp("v(t)=177sin(314t+10)");// value of 10 is in degrees disp("i(t)=14.14sin(314t-20)");//value of 20 is in degrees mprintf("\nCurrent i(t) lags behind voltage v(t) by 30degrees"); disp("phi=30degrees"); printf("Power factor pf=cos(30)=%.3f (lagging)",pf); printf("\nPower consumed P=V*I*cos(phi)=%.1f W",P);
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//Problem 14.08: A supply voltage has a mean value of 150 V. Determine its maximum value and its rms value //initializing the variables: Vmean = 150; // in Volts //calculation: //for a sine wave Vmax = Vmean/0.637 Vrms = 0.707*Vmax printf("\n\n Result \n\n") printf("\n peak value = %.1f V",Vmax) printf("\n rms value = %.1f V",Vrms)
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clc clear exec('tree_plot.sce',-1) function ordenacao=em_ordem(Arvore) n=length(Arvore) Arvore(n+1:2^ceil(log2(n))-1)=0 n=length(Arvore) ordenacao=zeros(1,n) potencia=floor(log2(n)) passo=2^potencia k=1 while k<n & passo>=1 for i=passo:passo:n if ordenacao(1,i)==0 then ordenacao(1,i)=k k=k+1 end end passo=passo/2 end endfunction function isBusca(Arvore) ordem = em_ordem(Arvore) EBusca = %T for i=1:length(Arvore)-1 if Arvore(i) > Arvore(i+1) printf('Falhou no indice %d, o valor %d é maior que %d\n', i, Arvore(i), Arvore(i+1)) EBusca = %F end end disp(EBusca) endfunction //Exemplo: //Arvore=[60 55 65 22 56 62 71 3 23 0 58] Arvore=[60 55 65 22 0 0 0 56 0 62 71 3 23 0 58] tree_plot(Arvore) isBusca(Arvore)
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//Ex:8.2 clc; clear; close; V_in=5*10^-3; R_in=2*10^6; I_in=V_in/R_in; printf("Input current = %e A",I_in);
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//Page Number: 95 //Example 2.5 clc; //Given, c=3D+8; //m/s a=4; //cm b=2; //cm //(i) Mode lamc=2*a; //cm lamcm=lamc/100; //m fc=c/lamcm; //20% above fc f=1.2*fc; //Hz //Operating wavelength lam1=c/f; //cm //For TE10 mode lamc10=2*b;//cm lamcm10=lamc10/100;//m fc10=c/lamcm10; disp('Hence mode of operation is TE10','Hz',fc,'Since guide is operating at'); //(ii)Guide wavelength lamm1=lam1*100;//cm lamg=lamm1/(sqrt(1-(lamm1/lamc)^2)); disp('cm',lamg,'Guide wavelength:'); //(iii) Phase velocity vp=f*lamg; disp('m/s',vp/100,'Phase velocity:'); //(iii) Group velocity vg=c^2/vp; disp('m/s',vg,'Group velocity:');
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clc z = 6 // number of steps n1 = 180 // rev/min n2 = 100 // rev/min Rn = n1/n2 phi = (Rn)^(1/(z-1)) // common ratio n3 = phi*n2 // rev/min n4 = (phi)^2*n2 // rev/min n5 = (phi)^3*n2 // rev/min n6 = (phi)^4*n2 // rev/min n7 = 225 // speed of input shaft in rev/min Ta=poly(0,'Ta') tb=n7/n5*Ta Ta=roots(tb+Ta-52) tb=horner(tb,Ta) tb = ceil(tb) Tc=poly(0,'Tc') td=n7/n6*Tc Tc=roots(td+Tc-52) td=horner(td,Tc) Tc = ceil(Tc) Te=poly(0,'Te') tf=n7/n1*Te Te=roots(tf+Te-52) tf=horner(tf,Te) tf = ceil(tf) Th=poly(0,'Th') tj=n2/n5*Th Th=roots(tj+Th-46) Th = ceil(Th) tj=horner(tj,Th) tj = floor(tj) Ti=poly(0,'Ti') tg=n5/n5*Ti Ti=roots(tg+Ti-46) tg=horner(tg,Ti) printf("\n Ta = %d Tb = %d \n Tc = %d Td = %d \n Te = %d tf = %d \n Th = %d Tj = %d \n Ti = %d Tg = %d" , Ta,tb,Tc,td,Te,tf,tj,Th,Ti,tg) // 'Answers vary due to round off error'
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Chap14_Ex10.sce
// Y.V.C.Rao ,1997.Chemical Engineering Thermodynamics.Universities Press,Hyderabad,India. //Chapter-14,Example 10,Page 500 //Title: Adiabatic reaction temperature //================================================================================================================ clear clc //INPUT T0=298.15;//temperature at the entrance (feed) in K P=0.1;//pressure (operating) in MPa //The reaction is given by: H2(g)+(1/2)O2(g)--->H20(g) n=[1;-1;-0.5];//stoichiometric coefficients of H2O(g),H2(g)and O2(g) respectively (no unit) n_r=[1;0.5];//stochiometric coefficients on the reactant side alone for computing the right hand side of Eq.(A) m=[0;1;0.5];//inlet mole number of H2O(g),H2(g) and O2(g) respectively //The isobaric molar capacity is given by Cp=a+bT+cT^2+dT^3+eT^-2 in J/molK and T is in K from Appendix A.3 //coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively) a=[28.850;27.012;30.255]; //coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively) b=[12.055*10^-3;3.509*10^-3;4.207*10^-3]; //coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively) c=[0;0;0]; //coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively) d=[0;0;0]; //coefficient in the expression for computing the isobaric molar heat capacity from Appendix A.3 (for H2O(g),H2(g),O2(g) respectively) e=[1.006*10^5;0.690*10^5;-1.887*10^5]; del_H=-241.997;//enthalpy of reaction at 298.15K in kJ del_G=-228.600;//Gibbs free energy of reaction at 298.15K in kJ R=8.314;//universal gas constant in J/molK //CALCULATION //Framing the isobaric molar heat capacity expression del_a=(n(1,:)*a(1,:))+(n(2,:)*a(2,:))+(n(3,:)*a(3,:)); del_b=(n(1,:)*b(1,:))+(n(2,:)*b(2,:))+(n(3,:)*b(3,:)); del_c=(n(1,:)*c(1,:))+(n(2,:)*c(2,:))+(n(3,:)*c(3,:)); del_d=(n(1,:)*d(1,:))+(n(2,:)*d(2,:))+(n(3,:)*d(3,:)); del_e=(n(1,:)*e(1,:))+(n(2,:)*e(2,:))+(n(3,:)*e(3,:)); mtot=m(1,:)+m(2,:)+m(3,:);//calculation of the total mole number of feed entering (no unit) del_n=n(1,:)+n(2,:)+n(3,:);//calculation of the total mole number (no unit) //Using Eq.14.21 to compute the value of del_H0 in kJ del_H0=((del_H*10^3)-((del_a*T0)+((del_b/2)*T0^2)+((del_c/3)*T0^3)+((del_d/4)*T0^4)-(del_e/T0)))*10^-3; //Using Eq.14.23 to compute the integration constant I=(1/(R*T0))*(((del_H0*10^3)-(del_a*T0*log(T0))-((del_b/2)*T0^2)-((del_c/6)*T0^3)-((del_d/12)*T0^4)-((del_e/2)*(1/T0))-(del_G*10^3))); //The conversion is computed by using Eq.(A) and by Eq.(B) and the two are plotted with respect to temperature. The point of intersection gives the adiabatic reaction temeperature and from that the conversion and the composition are determined.Let E_A denote the conversion obtained by using Eq.A and E_B denote the conversion obtained by using Eq.B (no unit) //For both the equations, conversion is determined for a temperature range of 2000 to 3800K, by incrementing temperature by 100K every time. T=2000:100:3800;//framing the temperature range in K l=length(T);//iteration parameter (no unit) i=1;//iteration parameter tol=1e-4;//tolerance limit for convergence of the system when using fsolve while i<l|i==l del_H_T(i)=((del_H0*10^3)+((del_a*T(:,i))+((del_b/2)*T(:,i)^2)+((del_c/3)*T(:,i)^3)+((del_d/4)*T(:,i)^4)-(del_e/T(:,i))))*10^-3; del_G_T(i)=((del_H0*10^3)-(del_a*T(:,i)*log(T(:,i)))-((del_b/2)*T(:,i)^2)-((del_c/6)*T(:,i)^3)-((del_d/12)*T(:,i)^4)-((del_e/2)*(1/T(:,i)))-(I*R*T(:,i)))*10^-3; Ka(i)=exp(-(del_G_T(i)*10^3)/(R*T(:,i)));//calculation of the equilibrium constant (no unit) //using Eq.A to determine the conversion(no unit) E_A(i)=(1/del_H_T(i)*10^-3)*(-((((n_r(1,:)*a(2,:))+(n_r(2,:)*a(3,:)))*(T(:,i)-T0))+((((n_r(1,:)*b(2,:))+(n_r(2,:)*b(3,:)))/2)*((T(:,i))^2-(T0^2)))+((((n_r(1,:)*c(2,:))+(n_r(2,:)*c(3,:)))/3)*((T(:,i))^3-(T0^3)))+((((n_r(1,:)*d(2,:))+(n_r(2,:)*d(3,:)))/4)*((T(:,i))^4-(T0^4)))+(((n_r(1,:)*a(2,:))+(n_r(2,:)*a(3,:)))*((1/T(:,i))-(1/T0))))); Eguess(i)=0.99;//taking a guess value for the conversion (no unit) function[fn]=solver_func(Ei) //Function defined for solving the system (Using Eq.B to determine the conversion (no unit)) fn=((((m(1,:)+(n(1,:)*Ei))/(mtot+(del_n*Ei)))^n(1,:))*(((m(2,:)+(n(2,:)*Ei))/(mtot+(del_n*Ei)))^n(2,:))*(((m(3,:)+(n(3,:)*Ei))/(mtot+(del_n*Ei)))^n(3,:)))-Ka(i); endfunction [E_B(i)]=fsolve(Eguess(i),solver_func,tol)//using inbuilt function fsolve for solving the system of equations i=i+1 end //plotting the conversions determined above (using Eqs.A and B respectively) against temperature to determine the adiabatic reaction temperature in K plot(T,E_A,T,E_B); legends(['Equation (A)';'Equation (B)'],[2,3],opt="lr"); xtitle('Plot of degree of conversion versus adiabatic reaction temperature','T(K)','E'); //From the above plot, it is determined that the point of intersection occurs around 3440K, which is taken as the reaction temperature, where the conversion=0.68(no unit). Therefore, the conversion at the adiabatic reaction temperature is 0.68 T_adiabatic=3440;//the adiabatic reaction temperature in K E_adiabatic=0.68;//conversion at the adiabatic reaction temperature (no unit) //Calculation of the composition of the burned gas (H2,O2 and H2O respectively) at the adiabatic reaction temperature (no unit) y_H2=((m(2,:)+(n(2,:)*E_adiabatic))/(mtot+(del_n*E_adiabatic))); y_O2=((m(3,:)+(n(3,:)*E_adiabatic))/(mtot+(del_n*E_adiabatic))); y_H2O=((m(1,:)+(n(1,:)*E_adiabatic))/(mtot+(del_n*E_adiabatic))); //OUTPUT mprintf('\n The adiabatic reaction temperature=%d K\n',T_adiabatic); mprintf('\n The composition of the burned gases is given by: y_H2=%0.4f \t y_O2=%0.4f \t y_H2O=%0.4f \n',y_H2,y_O2,y_H2O); //===============================================END OF PROGRAM===================================================
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// Exa 10.14 clc; clear; close; format('v',5) // Given data A = 200; BW = 10;// in kHz Beta = 10/100; Af =A/(1+(A*Beta)); disp(Af,"The gain with negative feedback is"); BWf = BW*(1+(A*Beta));// in kHz disp(BWf,"The bandwidth with negative feedback in kHz is");
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str='This is a character string' n=strindex(str,'i') // position of character 'i' in str part(str,n) // string in position n strsubst(str,'i','*') // replace 'i' with '*' tokens(str) // split according to the delimiter ' ' tokens(str,'i') // split according to the delimiter 'i' str='scilab' strsplit(str) // split characters strsplit(str,'i') // split according to delimiter 'i' str = ["hat" "cat" "chat" "tac" "dog"] grep(str,'a') str='aababbbaaabba' [first,last,match]=regexp(str,'/a(b)+/')
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//Chapter 12 //page no 431 //given clc; clear all; Pt=25; //in microW Prd=15; //in dBm Ptd=10*log10(Pt*10^-6/10^-3) //in dBm printf("\n Transmitter Power = %0.0f dBm",Ptd); Pm=Ptd-Prd; printf("\n Power margin= %0.0f dBm",Pm);
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//Example 7_5 clc(); clear; //To calculate number of reflections per meter and total distance covered n1=1.5 n2=1.49 phi=asin(n2/n1)*180/%pi //units in degrees a=25 //units in micro meters leng=2*a*tan(phi*%pi/180) //units inmicro meters totalnum=10^6/leng printf("Total number of reflections is %d\n",totalnum) l=1 //units in meters distance=l*(sin(phi*%pi/180)) printf("Total distance covered is %.4f Meters",distance) //in text book answer printed wrong as 1.006mcorrect answer is 0.9933meters
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clc; clear; f_s=1;//Signal Frequencies f_c=8;//Carrier Frequencies t=0:0.001:5; Phase_deviation_facotr=35 i_s=2*sin(2*%pi*f_s*t); // Information Signal c_s=15*sin(2*%pi*f_c*t);// Carrier Signal Modulated_signal=15.*cos((2*%pi*f_c*t)+(Phase_deviation_facotr.*sin(2*%pi*f_s*t))); subplot(3,1,1); plot(i_s) title('Information Signal'); subplot(3,1,2); plot(c_s) title('Carrier Signal'); subplot(3,1,3); plot(Modulated_signal) title('Phase Modulation Signal');
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alpha = [1, expm(%i*120*%pi/180), expm(%i*(-120)*%pi/180)] Vg = 13.8e3 VgL = Vg/(sqrt(3)*expm(%i*30*%pi/180))*alpha Comp12 = 15; Comp23 = 8; Z_eq = %i*1.15; Z_p = 0.2779 + %i*0.3921; I_max = 479;// # Corrente máxima que o condutor suporta Y_eq1 = 1/Z_eq; Ycond12 = 1/(Z_p*Comp12); Ycond23 = 1/(Z_p*Comp23); // ----------------------------------------- //2.1. Carga 1 //----------------------------------------- Z_cg_1a = 83.91 +%i*33.01; //Z_cg_1a = 0.01*Z_cg_1a/(0.01 + Z_cg_1a); Z_cg_1b = 117.47 + %i*46.21; Z_cg_1c = 73.42 + %i*28.89; Y_cg_1a = 1/Z_cg_1a; Y_cg_1b = 1/Z_cg_1b; Y_cg_1c = 1/Z_cg_1c; //----------------------------------------- // 2.2. Carga 2 // ----------------------------------------- Z_cg_2ab = 617.56 + %i*231.35; //Z_cg_2ab = Z_cg_2ab*0.01/(0.01 + Z_cg_2ab); Z_cg_2bc = 540.36 + %i*202.43; Z_cg_2ca = 385.97 + %i*144.59; Z_cg_2tot = Z_cg_2ab + Z_cg_2bc + Z_cg_2ca; Z_cg_2a = Z_cg_2ab*Z_cg_2ca/Z_cg_2tot; Z_cg_2b = Z_cg_2ab*Z_cg_2bc/Z_cg_2tot; Z_cg_2c = Z_cg_2ca*Z_cg_2bc/Z_cg_2tot; Y_cg_2a = 1/Z_cg_2a; Y_cg_2b = 1/Z_cg_2b; Y_cg_2c = 1/Z_cg_2c; // ----------------------------------------- // 2.3. Carga 3 // ----------------------------------------- Z_cg_3a = 165.26 + %i*60.83; Z_cg_3a = 0.01*Z_cg_1a/(0.01 + Z_cg_1a); Z_cg_3b = 198.32 + %i*73.00; Z_cg_3c = 123.95 + %i*45.62; Zn = 5.60 + %i*2.06; Yn = 1/Zn; Y_cg_3a = 1/Z_cg_3a; Y_cg_3b = 1/Z_cg_3b; Y_cg_3c = 1/Z_cg_3c; //A //B //C //A' //B' //C' Y = [Y_eq1 + Y_cg_1a + Ycond12, 0, 0, -Ycond12, 0, 0, 0, 0, 0, 0, 0; 0, Y_eq1 + Y_cg_1b + Ycond12, 0, 0, -Ycond12, 0, 0, 0, 0, 0, 0; 0, 0, Y_eq1 + Y_cg_1c + Ycond12, 0, 0, -Ycond12, 0, 0, 0, 0, 0; Ycond12, 0, 0, -Ycond12 - Ycond23 - Y_cg_2a, 0, 0, Ycond23, 0, 0, 0, Y_cg_2a; 0, Ycond12, 0, 0, -Ycond12 - Ycond23 - Y_cg_2b, 0,0,Ycond23,0,0,Y_cg_2b; 0,0,Ycond12,0,0,-Ycond12 - Ycond23 - Y_cg_2c,0,0,Ycond23,0,Y_cg_2c; 0,0,0,Ycond23, 0, 0, -Ycond23-Y_cg_3a,0,0,Y_cg_3a,0; 0,0,0,0,Ycond23,0,0,-Ycond23-Y_cg_3b,0,Y_cg_3b,0; 0,0,0,0,0,Ycond23,0,0,-Ycond23-Y_cg_3a,Y_cg_3c,0; 0,0,0,0,0,0,Y_cg_3a,Y_cg_3b,Y_cg_3c,-Y_cg_3a-Y_cg_3b-Y_cg_3c-Yn,0; 0,0,0,Y_cg_2a,Y_cg_2b,Y_cg_2c,0,0,0,0,-Y_cg_2a-Y_cg_2b-Y_cg_2c] //Vv = [Va1;Vb1;Vc1;Va2;Vb2;Vc2;Va3;Vb3;Vc3;Vn3;Vn2]; E = [VgL(1)*Y_eq1; VgL(2)*Y_eq1; VgL(3)*Y_eq1;0;0;0;0;0;0;0;0]; Vv = inv(Y)*E; //Vv(4) = Vv(4)*sqrt(3)*expm(%i*30*%pi/180); //Vv(5) = Vv(5)*sqrt(3)*expm(%i*30*%pi/180); //Vv(6) = Vv(6)*sqrt(3)*expm(%i*30*%pi/180); Z = [Z_cg_1a;Z_cg_1b;Z_cg_1c;Z_cg_2a;Z_cg_2b;Z_cg_2c;Z_cg_3a;Z_cg_3b;Z_cg_3c]; //Z = [Z_cg_1a;Z_cg_1b;Z_cg_1c;Z_cg_2a;Z_cg_2b;Z_cg_2c;0.01*Z_cg_3a/(0.01 + Z_cg_3a);Z_cg_3b;Z_cg_3c]; I = [Vv(1)/Z(1); Vv(2)/Z(2); Vv(3)/Z(3); Vv(4)/Z(4); Vv(5)/Z(5); Vv(6)/Z(6); Vv(7)/Z(7); Vv(8)/Z(8); Vv(9)/Z(9)]; As = [Vv(1)*conj(I(1));Vv(2)*conj(I(2));Vv(3)*conj(I(3));Vv(4)*conj(I(4));Vv(5)*conj(I(5));Vv(6)*conj(I(6));Vv(7)*conj(I(7));Vv(8)*conj(I(8));Vv(9)*conj(I(9))]; S = [As(1) + As(2) + As(3);As(4) + As(5) + As(6);As(7) + As(8) + As(9)]; P12 = (I(4) + I(7) + I(5) + I(8) + I(6) + I(9))/Ycond12; P23 = (I(7)+I(8)+I(9))/Ycond23; for i = 1:9 vp(i) = sqrt(real(Vv(i))^2 + imag(Vv(i))^2); vpf(i) = atand(imag(Vv(i))/real(Vv(i))); end for i = 1:3 sp(i) = sqrt(real(S(i))^2 + imag(S(i))^2); spf(i) = atand(imag(S(i))/real(S(i))); end pp12 = sqrt(real(P12)^2 + imag(P12)^2); ppf12 = atand(imag(P12)/real(P12)); pp23 = sqrt(real(P23)^2 + imag(P23)^2); ppf23 = atand(imag(P23)/real(P23));
d46190a0e04c7a4d1bcb778caea9fdb889e3851e
f42e0a9f61003756d40b8c09ebfe5dd926081407
/TP4/courbe.sci
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[]
no_license
BenFradet/MT09
04fe085afaef9f8c8d419a3824c633adae0c007a
d37451249f2df09932777e2fd64d43462e3d6931
refs/heads/master
2020-04-14T02:47:55.441807
2014-12-22T17:34:50
2014-12-22T17:34:50
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sci
courbe.sci
function[] = courbe() r = 2; a = linspace(0, 2 * %pi, 100); x = r * cos(a); y = r * sin(a); plot(x, y); x = linspace(-2, 2, 100); y = exp(x); gc = gca(); gc.isoview = 'on'; plot(x, y); endfunction
34cd14364928ee95265c7cb4a670772fca416fb8
e04f3a1f9e98fd043a65910a1d4e52bdfff0d6e4
/New LSTMAttn Model/.data/lemma-split/DEVELOPMENT-LANGUAGES/niger-congo/nya.tst
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davidgu13/Lemma-vs-Form-Splits
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refs/heads/master
2023-08-01T16:15:52.417307
2021-09-14T20:19:28
2021-09-14T20:19:28
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nya.tst
fulumira V;PL;3;PST fulumira V;NFIN fulumira V;PL;1;PRS fulumira V;PL;1;FUT fulumira V;SG;3;PST fulumira V;SG;2;FUT fulumira V;SG;3;FUT fulumira V;SG;1;PST fulumira V;PL;2;PST fulumira V;PL;2;FUT fulumira V;PL;3;FUT fulumira V;SG;1;PRS fulumira V;SG;2;PST fulumira V;PL;3;PRS fulumira V;SG;1;FUT fulumira V;SG;3;PRS fulumira V;PL;1;PST fulumira V;SG;2;PRS fulumira V;PL;2;PRS kola V;SG;3;PRS kola V;PL;2;PST kola V;PL;2;FUT kola V;PL;2;PRS kola V;NFIN kola V;PL;1;FUT kola V;SG;2;PRS kola V;SG;2;FUT kola V;SG;1;PST kola V;PL;3;PST kola V;SG;3;PST kola V;PL;3;PRS kola V;SG;2;PST kola V;SG;1;FUT kola V;SG;1;PRS kola V;SG;3;FUT kola V;PL;3;FUT kola V;PL;1;PRS kola V;PL;1;PST dikira V;SG;2;FUT dikira V;SG;1;FUT dikira V;SG;1;PST dikira V;PL;3;PRS dikira V;PL;1;PRS dikira V;SG;3;FUT dikira V;NFIN dikira V;SG;2;PST dikira V;PL;1;FUT dikira V;PL;1;PST dikira V;SG;3;PST dikira V;SG;3;PRS dikira V;PL;2;PRS dikira V;SG;2;PRS dikira V;PL;3;PST dikira V;PL;2;FUT dikira V;PL;3;FUT dikira V;PL;2;PST 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yamba V;PL;3;PST yamba V;SG;3;PRS yamba V;PL;2;PST yamba V;PL;1;FUT yamba V;SG;2;PRS yamba V;SG;2;PST yamba V;SG;1;PRS yamba V;PL;1;PRS yamba V;PL;2;FUT zindikira V;SG;1;PRS zindikira V;SG;3;FUT zindikira V;PL;2;FUT zindikira V;PL;1;FUT zindikira V;PL;3;PRS zindikira V;SG;3;PST zindikira V;SG;2;PRS zindikira V;PL;3;PST zindikira V;PL;3;FUT zindikira V;PL;1;PST zindikira V;NFIN zindikira V;PL;1;PRS zindikira V;SG;2;FUT zindikira V;PL;2;PRS zindikira V;SG;2;PST zindikira V;SG;1;PST zindikira V;SG;3;PRS zindikira V;PL;2;PST zindikira V;SG;1;FUT kalamba V;SG;1;PST kalamba V;SG;3;PRS kalamba V;SG;1;PRS kalamba V;SG;2;PRS kalamba V;SG;3;PST kalamba V;PL;2;PST kalamba V;SG;1;FUT kalamba V;PL;2;FUT kalamba V;SG;2;FUT kalamba V;PL;2;PRS kalamba V;PL;3;PST kalamba V;NFIN kalamba V;PL;1;PST kalamba V;SG;3;FUT kalamba V;PL;1;PRS kalamba V;SG;2;PST kalamba V;PL;3;FUT kalamba V;PL;3;PRS kalamba V;PL;1;FUT funsani V;SG;1;FUT funsani V;PL;1;FUT funsani V;PL;2;PRS funsani V;PL;3;FUT funsani V;SG;2;PRS funsani V;SG;2;PST funsani V;SG;1;PRS funsani V;SG;3;FUT funsani V;SG;1;PST funsani V;PL;2;PST funsani V;PL;3;PST funsani V;SG;3;PRS funsani V;PL;1;PST funsani V;PL;2;FUT funsani V;PL;3;PRS funsani V;SG;2;FUT funsani V;NFIN funsani V;PL;1;PRS funsani V;SG;3;PST bvoma V;SG;3;PRS bvoma V;SG;3;FUT bvoma V;SG;3;PST bvoma V;PL;3;PST bvoma V;PL;2;PRS bvoma V;PL;3;PRS bvoma V;SG;1;PST bvoma V;SG;1;FUT bvoma V;PL;1;PST bvoma V;SG;2;FUT bvoma V;SG;2;PRS bvoma V;SG;1;PRS bvoma V;PL;1;PRS bvoma V;NFIN bvoma V;PL;2;FUT bvoma V;PL;1;FUT bvoma V;PL;3;FUT bvoma V;SG;2;PST bvoma V;PL;2;PST tsata V;PL;1;FUT tsata V;SG;2;FUT tsata V;SG;1;PST tsata V;SG;3;PST tsata V;PL;3;FUT tsata V;SG;1;FUT tsata V;PL;1;PRS tsata V;NFIN tsata V;PL;3;PST tsata V;SG;3;PRS tsata V;SG;2;PST tsata V;PL;2;PST tsata V;PL;3;PRS tsata V;SG;3;FUT tsata V;PL;2;FUT tsata V;SG;1;PRS tsata V;SG;2;PRS tsata V;PL;1;PST tsata V;PL;2;PRS dandaula V;SG;3;PRS dandaula V;SG;2;FUT dandaula V;PL;3;PRS dandaula V;PL;3;PST dandaula V;SG;1;PST dandaula V;PL;1;PRS dandaula V;SG;2;PST dandaula V;SG;3;PST dandaula V;SG;1;FUT dandaula V;PL;1;PST dandaula V;PL;2;FUT dandaula V;SG;1;PRS dandaula V;SG;3;FUT dandaula V;PL;3;FUT dandaula V;SG;2;PRS dandaula V;PL;2;PST dandaula V;PL;2;PRS dandaula V;PL;1;FUT dandaula V;NFIN zungulira V;SG;1;PRS zungulira V;PL;2;FUT zungulira V;PL;3;PST zungulira V;SG;3;FUT zungulira V;PL;1;FUT zungulira V;PL;3;FUT zungulira V;SG;3;PST zungulira V;PL;1;PST zungulira V;SG;3;PRS zungulira V;SG;2;PST zungulira V;NFIN zungulira V;SG;1;FUT zungulira V;SG;1;PST zungulira V;SG;2;PRS zungulira V;PL;2;PST zungulira V;PL;1;PRS zungulira V;SG;2;FUT zungulira V;PL;3;PRS zungulira V;PL;2;PRS yang'ana V;SG;3;PRS yang'ana V;SG;3;FUT yang'ana V;SG;2;FUT yang'ana V;PL;1;FUT yang'ana V;SG;1;FUT yang'ana V;SG;1;PST yang'ana V;PL;3;PST yang'ana V;NFIN yang'ana V;SG;2;PRS yang'ana V;PL;3;PRS yang'ana V;SG;1;PRS yang'ana V;SG;3;PST yang'ana V;PL;2;FUT yang'ana V;PL;1;PST yang'ana V;PL;2;PST yang'ana V;PL;2;PRS yang'ana V;SG;2;PST yang'ana V;PL;3;FUT yang'ana V;PL;1;PRS gwa V;SG;2;PST gwa V;PL;3;FUT gwa V;SG;1;PST gwa V;SG;3;PST gwa V;SG;1;PRS gwa V;SG;2;FUT gwa V;PL;1;FUT gwa V;PL;2;FUT gwa V;PL;2;PRS gwa V;PL;3;PST gwa V;PL;1;PRS gwa V;PL;1;PST gwa V;NFIN gwa V;SG;3;FUT gwa V;SG;2;PRS gwa V;PL;3;PRS gwa V;PL;2;PST gwa V;SG;3;PRS gwa V;SG;1;FUT pempha V;PL;3;PST pempha V;PL;2;PRS pempha V;SG;1;FUT pempha V;PL;1;FUT pempha V;SG;2;PRS pempha V;SG;3;FUT pempha V;SG;3;PRS pempha V;PL;2;PST pempha V;PL;3;FUT pempha V;SG;3;PST pempha V;SG;1;PRS pempha V;SG;2;FUT pempha V;PL;1;PRS pempha V;SG;1;PST pempha V;PL;2;FUT pempha V;SG;2;PST pempha V;PL;1;PST pempha V;NFIN pempha V;PL;3;PRS dula V;SG;2;PRS dula V;SG;1;PST dula V;SG;2;FUT dula V;PL;2;PRS dula V;PL;1;PRS dula V;PL;1;PST dula V;SG;3;PRS dula V;NFIN dula V;PL;2;PST dula V;PL;3;FUT dula V;SG;2;PST dula V;SG;1;FUT dula V;PL;1;FUT dula V;SG;3;FUT dula V;PL;2;FUT dula V;PL;3;PRS dula V;SG;3;PST dula V;SG;1;PRS dula V;PL;3;PST patsa V;SG;1;PRS patsa V;PL;2;PST patsa V;PL;1;PRS patsa V;PL;1;FUT patsa V;PL;3;PST patsa V;PL;1;PST patsa V;PL;3;FUT patsa V;SG;1;PST patsa V;SG;2;PST patsa V;SG;2;FUT patsa V;PL;2;PRS patsa V;SG;1;FUT patsa V;SG;3;PRS patsa V;SG;3;PST patsa V;PL;2;FUT patsa V;PL;3;PRS patsa V;NFIN patsa V;SG;3;FUT patsa V;SG;2;PRS kwiya V;SG;3;PRS kwiya V;PL;2;PST kwiya V;PL;1;PRS kwiya V;NFIN kwiya V;PL;1;PST kwiya V;SG;3;PST kwiya V;SG;2;PST kwiya V;PL;3;PRS kwiya V;SG;1;FUT kwiya V;PL;2;PRS kwiya V;SG;1;PST kwiya V;SG;2;FUT kwiya V;PL;3;FUT kwiya V;PL;2;FUT kwiya V;PL;3;PST kwiya V;SG;1;PRS kwiya V;SG;3;FUT kwiya V;SG;2;PRS kwiya V;PL;1;FUT imba V;SG;3;PST imba V;SG;1;PST imba V;PL;1;PRS imba V;SG;1;PRS imba V;PL;1;FUT imba V;SG;1;FUT imba V;PL;3;FUT imba V;SG;2;PRS imba V;PL;3;PST imba V;SG;2;FUT imba V;SG;3;PRS imba V;PL;3;PRS imba V;PL;2;PRS imba V;SG;2;PST imba V;PL;2;FUT imba V;NFIN imba V;PL;1;PST imba V;PL;2;PST imba V;SG;3;FUT lamula V;PL;1;PST lamula V;SG;2;PST lamula V;SG;1;FUT lamula V;PL;3;FUT lamula V;PL;2;PRS lamula V;SG;3;PRS lamula V;SG;2;FUT lamula V;PL;2;PST lamula V;PL;2;FUT lamula V;PL;1;PRS lamula V;PL;1;FUT lamula V;SG;3;FUT lamula V;SG;1;PST lamula V;SG;1;PRS lamula V;SG;3;PST lamula V;NFIN lamula V;PL;3;PST lamula V;SG;2;PRS lamula V;PL;3;PRS peza V;PL;2;PRS peza V;PL;1;FUT peza V;PL;2;FUT peza V;SG;2;FUT peza V;PL;3;PST peza V;SG;2;PRS peza V;SG;2;PST peza V;NFIN peza V;SG;1;FUT peza V;PL;1;PRS peza V;SG;3;FUT peza V;PL;3;PRS peza V;SG;1;PST peza V;SG;1;PRS peza V;PL;1;PST peza V;SG;3;PST peza V;SG;3;PRS peza V;PL;3;FUT peza V;PL;2;PST mera V;PL;2;PRS mera V;SG;1;PRS mera V;PL;3;PST mera V;PL;2;FUT mera V;SG;1;PST mera V;SG;2;PRS mera V;PL;1;PRS mera V;PL;3;PRS mera V;SG;3;PST mera V;SG;3;FUT mera V;PL;3;FUT mera V;SG;3;PRS mera V;SG;2;FUT mera V;PL;1;FUT mera V;SG;2;PST mera V;NFIN mera V;PL;2;PST mera V;SG;1;FUT mera V;PL;1;PST tsiriza V;PL;3;PST tsiriza V;SG;1;PRS tsiriza V;SG;2;PST tsiriza V;PL;2;PRS tsiriza V;SG;1;FUT tsiriza V;SG;1;PST tsiriza V;PL;1;PRS tsiriza V;PL;1;PST tsiriza V;SG;2;FUT tsiriza V;PL;3;PRS tsiriza V;NFIN tsiriza V;PL;1;FUT tsiriza V;PL;2;FUT tsiriza V;SG;2;PRS tsiriza V;SG;3;PRS tsiriza V;SG;3;PST tsiriza V;PL;2;PST tsiriza V;SG;3;FUT tsiriza V;PL;3;FUT pitani V;SG;2;FUT pitani V;SG;1;PST pitani V;SG;3;PST pitani V;SG;3;FUT pitani V;PL;2;FUT pitani V;PL;3;FUT pitani V;PL;3;PRS pitani V;SG;2;PST pitani V;SG;3;PRS pitani V;NFIN pitani V;PL;1;FUT pitani V;PL;2;PST pitani V;PL;3;PST pitani V;PL;1;PST pitani V;PL;1;PRS pitani V;SG;1;PRS pitani V;PL;2;PRS pitani V;SG;1;FUT pitani V;SG;2;PRS khazika V;SG;2;PST khazika V;PL;2;FUT khazika V;SG;3;PRS khazika V;PL;1;FUT khazika V;PL;3;FUT khazika V;SG;2;PRS khazika V;PL;3;PST khazika V;SG;1;PST khazika V;PL;2;PRS khazika V;PL;1;PST khazika V;SG;3;FUT khazika V;SG;1;FUT khazika V;PL;2;PST khazika V;SG;2;FUT khazika V;SG;1;PRS khazika V;PL;1;PRS khazika V;PL;3;PRS khazika V;SG;3;PST khazika V;NFIN dalira V;SG;1;FUT dalira V;PL;1;PRS dalira V;PL;1;PST dalira V;SG;2;PRS dalira V;SG;1;PRS dalira V;PL;3;PRS dalira V;PL;2;PRS dalira V;PL;2;PST dalira V;SG;1;PST dalira V;PL;1;FUT dalira V;SG;3;FUT dalira V;SG;2;FUT dalira V;PL;3;PST dalira V;PL;2;FUT dalira V;PL;3;FUT dalira V;SG;2;PST dalira V;NFIN dalira V;SG;3;PST dalira V;SG;3;PRS lemba V;SG;1;PRS lemba V;PL;1;PST lemba V;PL;1;PRS lemba V;PL;2;FUT lemba V;SG;1;FUT lemba V;SG;3;PRS lemba V;SG;3;PST lemba V;PL;3;PST lemba V;SG;2;PRS lemba V;SG;2;PST lemba V;PL;2;PST lemba V;PL;1;FUT lemba V;PL;2;PRS lemba V;SG;2;FUT lemba V;NFIN lemba V;PL;3;FUT lemba V;SG;3;FUT lemba V;PL;3;PRS lemba V;SG;1;PST da V;SG;1;PRS da V;PL;3;FUT da V;NFIN da V;SG;2;PST da V;SG;2;FUT da V;PL;1;FUT da V;PL;3;PST da V;PL;1;PST da V;PL;2;PRS da V;SG;1;FUT da V;PL;3;PRS da V;SG;3;PST da V;SG;3;PRS da V;SG;1;PST da V;PL;2;FUT da V;SG;3;FUT da V;PL;2;PST da V;SG;2;PRS da V;PL;1;PRS pita V;PL;3;PST pita V;SG;2;PST pita V;SG;1;FUT pita V;PL;2;PST pita V;SG;1;PST pita V;PL;2;PRS pita V;PL;3;FUT pita V;PL;1;FUT pita V;SG;1;PRS pita V;PL;3;PRS pita V;PL;1;PRS pita V;SG;3;PST pita V;PL;1;PST pita V;SG;2;FUT pita V;NFIN pita V;PL;2;FUT pita V;SG;3;PRS pita V;SG;3;FUT pita V;SG;2;PRS sangalala V;SG;3;PRS sangalala V;PL;3;PRS sangalala V;PL;3;PST sangalala V;PL;2;FUT sangalala V;SG;1;PRS sangalala V;SG;2;FUT sangalala V;PL;3;FUT sangalala V;SG;3;PST sangalala V;SG;1;FUT sangalala V;PL;1;FUT sangalala V;SG;3;FUT sangalala V;SG;2;PRS sangalala V;PL;2;PRS sangalala V;NFIN sangalala V;PL;1;PRS sangalala V;PL;2;PST sangalala V;PL;1;PST sangalala V;SG;1;PST sangalala V;SG;2;PST taya V;SG;2;PRS taya V;PL;1;PRS taya V;PL;1;PST taya V;PL;3;PST taya V;PL;2;PRS taya V;SG;1;PST taya V;PL;3;PRS taya V;PL;2;FUT taya V;SG;3;FUT taya V;SG;3;PRS taya V;PL;3;FUT taya V;PL;2;PST taya V;SG;2;PST taya V;SG;3;PST taya V;SG;1;FUT taya V;SG;1;PRS taya V;PL;1;FUT taya V;NFIN taya V;SG;2;FUT tsegula V;SG;2;PRS tsegula V;PL;2;PRS tsegula V;SG;1;PRS tsegula V;PL;3;FUT tsegula V;SG;1;PST tsegula V;PL;2;PST tsegula V;PL;1;PST tsegula V;PL;3;PST tsegula V;SG;3;FUT tsegula V;PL;1;PRS tsegula V;SG;2;PST tsegula V;SG;2;FUT tsegula V;PL;3;PRS tsegula V;SG;3;PST tsegula V;SG;1;FUT tsegula V;NFIN tsegula V;PL;2;FUT tsegula V;PL;1;FUT tsegula V;SG;3;PRS
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/2534/CH8/EX8.4/Ex8_4.sce
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no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
7bc77cb1ed33745c720952c92b3b2747c5cbf2df
refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
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Ex8_4.sce
//Ex8_4 clc RL=5*10^3 Rs=1.2*10^3 hre=2.5*10^-4 hie=1.1*10^3 hfe=100 hoe=25*10^-6 disp("RL = "+string(RL)+"ohm")//load resistance disp("Rs = "+string(Rs)+"ohm")//source resistance //h-parameters for CE transistor amplifier are as follows: disp("hie = "+string(hie)+"ohm")//input resistance of CE transistor disp("hre = "+string(hre))//voltage gain of CE transistor disp("hfe = "+string(hfe))//current gain of CE transistor disp("hoe = "+string(hoe)+"mho")//output conductance of CE transistor //calculation for current gain: Ai=-hfe/(1+(hoe*RL)) disp("Ai = -hfe/(1+(hoe*RL)) = "+string(abs(Ai))) //calculation for input resistance: Ri = hie+(hre*Ai*RL) disp("Ri = hie+(hre*Ai*RL) = "+string(Ri)+"ohm") //calculation for voltage gain: Av = Ai*RL/Ri disp("Av = Ai*RL/Ri = "+string(Av)) //calculation for output resistance: Go=hoe-((hre*hfe)/(hie+Rs)) Ro = 1/Go disp("Ro = 1/Go") disp("Go = hoe-((hre*hfe)/(hie+Rs)) = "+string(Go)+"mho") disp("Ro = "+string(Ro)+"ohm") //note : in the textbook, above problem has given two values for "hfe" and no value for "hre"... // thus assuming value for "hre = 2.5*10^-4" as taken in previous example 8_2 // and "hfe=100" //note : in text LOAD RESISTANCE is noted as Rc in question, but RL in solution. // I have work with Load Resistance with notification RL.
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/1523/CH8/EX8.13/8_13.sce
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no_license
FOSSEE/Scilab-TBC-Uploads
948e5d1126d46bdd2f89a44c54ba62b0f0a1f5e1
7bc77cb1ed33745c720952c92b3b2747c5cbf2df
refs/heads/master
2020-04-09T02:43:26.499817
2018-02-03T05:31:52
2018-02-03T05:31:52
37,975,407
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555
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8_13.sce
//Transient analysis //pg no - 8.17 //example no - 8.13 a=((10*30)/(10+30)); d=5/a; b=0; c=5*(20/30); printf("iL(0-) = %.2f A", d); printf("\nvb(0-) = %.f", b); printf("\nva(0-) = %.2f V", c); disp("Applying Kcl equations at t=0+"); disp("((va(0+)-5)/10)+(va(0+)/10)+(va(0+)-vb(0+))/20 = 0"); //equation 1 disp("((vb(0+)-va(0+))/20)+((vb(0+)-5)/10)+(2/3) = 0"); //equation 2 //solving 1 and 2 M=[0.25, -0.05; -0.05, 0.15]; N=[0.5, -0.167]'; O=inv(M); X=O*N; disp(X); disp("va(0+)= 1.9 A"); disp("vb(0+)= -0.477 A");
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/1055/CH17/EX17.9/ch17_9.sce
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FOSSEE/Scilab-TBC-Uploads
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2020-04-09T02:43:26.499817
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ch17_9.sce
//Determine the reduced admittance matrices for prefault, fault and post fault conditions and determine the power angle characterstics for three conditions. clear clc; Y=[-%i*8.33 0 %i*8.33 0;0 -%i*28.57 0 %i*28.75;%i*8.33 0 -%i*15.67 %i*7.33;0 %i*28.57 %i*7.33 -%i*35.9]; YBB=[-%i*15.67 %i*7.33;%i*7.33 -%i*35.9]; YAA=[-%i*8.33 0;0 -%i*28.57]; YAB=[%i*8.33 0;0 %i*28.57]; YBA=YAB; Y=YAA-(YAB*(inv(YBB))*YBA); Y1=([-%i*8.33 0;0 -%i*28.57])-(([0;(%i*28.57/-%i*35.9)]*[0 %i*28.57])); disp(Y1,"Reduced admittance matrix during fault="); Yfull=[-%i*8.33 0 %i*8.33 0;0 -%i*28.57 0 %i*28.75;%i*8.33 0 -%i*12.33 %i*4;0 %i*28.57 %i*4 -%i*32.57]; YBB=[-%i*12.33 %i*4;%i*4 -%i*32.57]; Y=YAA-(YAB*(inv(YBB))*YBA); disp(Y,"(i) Post fault condition ,reduced matrix="); Y12=Y(1,2); E1=1.1; E2=1; printf("\n Power angle characterstics , Pe= %fsind",abs(Y12)*E1*E2);
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/branches/vac4.52mkg_24_06_2010/Idl/savevtk_xym.sci
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no_license
SpungMan/smaug-all
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01df12e98c734529ff984662badc26eaa3a9138b
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2021-11-29T14:09:47.094457
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savevtk_xym.sci
function [fd,err]=savevtk_xym(x,y,m,VarName) // Save Sci variables in VTK format // x is a list of points // y is a list of values in the x points // VarName is the Variable Name // Example: // // mtst=[11 12 13 14; 21 22 23 24]; // [f,c]=size(mtst); // savevtk_xym(1:c,1:f,mtst,'2DMatrix'); // // Coded by Sebastian Jardi Estadella // http://www.tinet.org/~sje/index_en.htm // // rotates the matrix -90 º. xaux=x; x=y; y=xaux; m=m'; nx=length(x); ny=length(y); [nf,nc]=size(m); if nx<>nc then disp('length(nx) and length(nc) have to be equals.'); disp(nx); disp(nc); abort; end if ny<>nf then disp('length(ny) and kength(nf) have to be equals.'); disp(ny); disp(nf); abort; end filename=sprintf('%s.vtk',VarName); mputl('# vtk DataFile Version 2.0',filename); // Delete previous content in filename [fd,err]=mopen(filename, 'a'); // Opens the file to Append. mfprintf(fd,'Structured Grid\n'); mfprintf(fd,'ASCII\n'); mfprintf(fd,'\n'); mfprintf(fd,'DATASET RECTILINEAR_GRID\n'); mfprintf(fd,'DIMENSIONS %d %d %d\n',nx,ny,1); //?? mfprintf(fd,'X_COORDINATES %d double\n',nx); for i=1:nx mfprintf(fd,'%e\n',x(i)); end mfprintf(fd,'\n'); mfprintf(fd,'Y_COORDINATES %d double\n',ny); for i=1:ny mfprintf(fd,'%e\n',y(i)); end mfprintf(fd,'\n'); mfprintf(fd,'Z_COORDINATES 1 double\n'); mfprintf(fd,'0 \n'); mfprintf(fd,'\n'); mfprintf(fd,'POINT_DATA %d\n',nx*ny); mfprintf(fd,'SCALARS '); mfprintf(fd,VarName); mfprintf(fd,' double 1\n'); mfprintf(fd,'LOOKUP_TABLE TableName\n'); for i_f=1:nf for i_c=1:nc mfprintf(fd,'%e ',m(i_f,i_c)); end mfprintf(fd,'\n'); end mfprintf(fd,'\n'); err=mclose(fd); endfunction
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/test/notoktests/maskjmp3.tst
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no_license
pilki/FPdNaCl
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54ed0e63fecb82ac163d153b1ffa17695cdb987d
refs/heads/master
2020-05-20T08:14:44.612852
2011-05-12T14:31:38
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maskjmp3.tst
;improperly masked indirect jump %2 <- %1 and 0xffffffe0; not twice the same register ; (should be accepted in a future version) ijmp %2
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clear; clc; Zo=2039.6;f=800; //value of Zo as taken in solution P=0.054* exp(%i*(%pi/(180/87.9))); w=2*%pi*f; Z=Zo*P; R=real(Z); printf('-Resistance R = %f ohms/km\n',R); L=(imag(Z))/w; printf('-Inductance L = %f mH/km\n',L*(10^3)); Y=P/Zo; G=real(Y); printf('-Conductance G = %f micromhos/km\n',G*(10^6)); C=((imag(Y))/w)*(10^6);c=round(C*10000)/10000 printf('-Capacitance C = %f microfarads/km\n',c);
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// Exa 6.42 format('v',7);clc;clear;close; // Given data Q = 1;// in k ohm Q = Q * 10^3;// in ohm S = Q;// in ohm P = 500;// in ohm r = 100;// in ohm C = 0.5;// in µF C = C * 10^-6;// in F //Using standard condition, Rx = (R2*R3)/R4; Rx = (P*Q)/S;// in ohm disp(Rx,"The value of Rx in Ω is"); //Lx = ((C*R2)/R4) * ( (R3*r) + (R4*r) + (R3*R4) ); Lx = ((C*P)/S) * ( (Q*r) + (S*r) + (Q*S) );// in H disp(Lx,"The value of Lx in H is");
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clc clear //Input data p=700;//pressure of fluid in kN/m^2 v1=0.28;//Initial volume of fluid in m^3 v2=1.68;//Final volume of fluid in m^3 //Calculations W=p*(v2-v1);//Work done in kJ //Output printf('The Work done W= %3.2f kJ',W)
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function [out]=erode(input_image ,kernel,actualkernel,anchorX,anchorY) input_image1=mattolist(input_image); a=opencv_erode(input_image1 ,kernel,actualkernel,anchorX,anchorY); dimension=size(a) for i = 1:dimension out(:,:,i)=a(i); end endfunction;
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//To Determine the Annual output of copper //Page 456 clc; clear; I=2000; //Current Passed NW=52; //Number of weeks in a Year T=100*3600; //Number of seconds per week TC=NW*T*I; //Total Charge supplied all over the year. ECu=31.8; //Equivalent Weight of Copper in grams F=96500; //One Farad of Charge // 1 F of charge gives 31.8 gms of copper W=(TC/F)*ECu/(1000*1000); //Weight of copper in tonnes printf('The Annual Output of Copper is %g Tonnes\n',W)
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// Scilab code Exa3.7: : Page 125(2011) clc; clear; Z = 82; // Atomic number E = 1; // Energy of the beta paricle, MeV I_l = 800; // Ionisation loss, MeV R = Z*E/I_l; // Ratio of radiation loss to ionisation loss E_1 = I_l/Z; // Energy of the beta particle when radiation radiation loss is equal to ionisation loss, MeV printf("\nThe ratio of radiation loss to ionisation loss = %5.3e \nThe energy of the beta particle = %4.2f MeV ", R, E_1); // Result // The ratio of radiation loss to ionisation loss = 1.025e-01 // The energy of the beta particle = 9.76 MeV
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function M=%hm_rand(M) // Copyright INRIA //creates a random hypermatrix with shape given by vector of dimensions or an //hypermatrix if type(M)==1 then dims=M else dims=M('dims') end M=mlist(['hm','dims','entries'],dims,rand(prod(dims),1))
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clear; clf; clc; ts = 0; te = 2; dt = 1/1000; t = te : dt : ts; x = 2*ones(1, (te-ts)/dt); z = sum(x)*dt; disp(z); t1 = 0; t2 = 2*%pi; T = t1 : dt : t2; y = sin(T); k = sum(y)*dt; disp(k); plot(T,y)
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clc; close(); clear(); //page no 456 //prob no. 13.9 //Data form ex13.8 Pe=10^-5; R=2*10^6; //bits/s (changed) k=1.38*10^-23; //Boltzmann cons Ti=475; //K Te=250; //K Tsys=Ti+Te; nsys=k*Tsys; //W/Hz function Eb=E(rhodb) //function for Eb rho=10^(rhodb/10); Eb=nsys*rho; endfunction function Pr=P(E) //function for Pr Pr=R*Eb; endfunction rhodb=9.6; Eb=E(rhodb); Pr=P(E); mprintf('\nBit energy , Eb=%.2f*10^-21 J \n',Eb*10^21); mprintf(' Required reciver carrier power , Pr=%.2f fW \n',Pr*10^15);
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clear // // //Initilization of Variables d=25 //mm //diameter of steel d2=18 //mm //Diameter at neck L=200 //mm //length of stee P=80*10**3 //KN //Load P1=160*10**3 //N //Load at Elastic Limit P2=180*10**3 //N //Max Load L1=56 //mm //Total Extension dell_l=0.16 //mm //Extension //Calculations A=%pi*d**2*4**-1 //Area of steel //mm**2 p=P1*A**-1 //Stress at Elastic Limit //N/mm**2 Y=P*L*(A*dell_l)**-1 //Modulus of elasticity //Let % elongation be x x=L1*L**-1*100 //Percentage reduction in area //Let % A be a a=((%pi*4**-1*d**2)-(%pi*4**-1*d2**2))*(%pi*4**-1*d**2)**-1*100 //Ultimate tensile stress sigma=P2*A**-1 //N/mm**2 //result printf("\n Stress at Elastic limit is %0.2f N/mm**2",p) printf("\n Youngs Modulus is %0.2f N/mm**2",Y) printf("\n Percentage Elongation is %0.2f ",a) printf("\n Percentage reduction in area is %0.2f ",P2) printf("\n Ultimate tensile stress %0.2f N/mm**2",sigma)
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V=120;F=60;N=2000;A=0.6; Ohm=20;L=0.25; V=120;F=60;N=2000;Ia=0.6; a=20;L=0.25; Edc=V-(Ia*Ra) X=2*%pi*F*L Eac=(-Ia*Ra)+sqrt(V^2-(Ia*X)^2) Nac=N*(Eac/Edc) Pf=(Eac+(Ia*Ra))/V Pmech=Eac*Ia Wm=(Nac*2*%pi)/F T=Pmech/Wm
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clc k= 8 // slope between points in degrees l=428 //measured length in m D1=l*cosd(k) printf('a)Horizontal distance between the points =%f m\n',D1) h=62 D2=sqrt(l^2-h^2) printf(' b)Horizontal distance between the points =%f m\n',D2) k= atan(0.25) D3=l*cos(k) printf(' c)Horizontal distance between the points =%f m',D3)
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//To calculate the de Broglie wavelength m = 1.674*10^-27; //mass of neutron, kg h = 6.626*10^-34; //planck's constant e = 1.6*10^-19; KE = 0.025; //kinetic energy, eV E = KE*e; //kinetic energy, J lamda = h/sqrt(2*m*E); //de Broglie wavelength, m lamda_nm = lamda*10^9; //de Broglie wavelength, nm printf("de Broglie wavelength is %5.3f nm",lamda_nm);
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clc // //Variable declaration G=90 // Modulus of rigidity(ksi) disp1=0.04 // Displacement of upper rod(in) Lda=2 // Height of bar(in) A=8*2.5 // Area of cross section(in**2) //Calculation Yxy=(disp1/Lda) // Shearing strain(rad) Txy=(90*((10**3)))*(0.020) // Shearing stress(psi) P=(Txy*A)/((10**3)) // Force exerted on the upper plate(kips) // Results printf("\n Shearing strain in rod=%1f rad' ,Yxy) printf("\n Force exerted on the upper plate=%1f kips' ,P)
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//Ex10_5 clc; // Given: ai=14000;// counts per min per 0.1 cm^3, initial activity of blood Si=1.4*10^5;// c min^-1 cm^-3, initial specific activity a=250;// 250 net counts in 10 min, this implies 25 net counts in a min // Formula: Si/Sr = V // Solution: V=Si/25;// total blood in the patient in cm^3 V1=V/1000;// volume in lit printf("The volume of blood in the patient is = %f lit",V1)
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errcatch(-1,"stop");mode(2);//page 238 ; ; A=[3 0;0 2]; eig=spec(A); [V,Val]=spec(A); disp(eig,'Eigen values:') x1=V(:,1); x2=V(:,2); disp(x1,x2,'Eigen vectors:'); //end exit();
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function [r]=%col(l1,l2) //%col(l1,l2) : l1==l2 //! r=%f
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//Example 5.4.3 page 5.14 clc; clear; Ttx= 8*10^-9; Tintra= 1*10^-9; Tmodal=5*10^-9; Trr= 6*10^-9; Tsys= sqrt(Ttx^2+(8*Tintra)^2+(8*Tmodal)^2+Trr^2); BWnrz= 0.7/Tsys; BWnrz=BWnrz/1000000;//converting in ns for dislaying... BWrz=0.35/Tsys; BWrz=BWrz/1000000;//converting in ns for dislaying... printf("Maximum bit rate for NRZ format is %.2f Mb/sec",BWnrz); printf("\n\nMaximum bit rate for RZ format is %.2f Mb/sec",BWrz);
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//Caption:Calculate (i)-required size of guide,(ii)-frequencies that can be used for this mode of propagation //Exa:4.10 clc; clear; close; wl_c=10;//in cm c=3*10^10;//in cm/s r=wl_c/(2*%pi/1.841);//in cm area=%pi*r^2;//in sq. cm f_c=c/wl_c; disp(r,'Radius of circular waveguide(in cm) ='); disp(area,'Area of cross-section of circular waveguide(in cm) ='); disp('Frequency above'); disp(f_c); disp('can be propagated');
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clc //initialisation of variables p1=1000//ft p2=50//ft g=20//ft/mile v=5280//ft q=7.5*10^-6//ft t=60//F k=2835//ft/days p=7.5//ft //CALCULATIONS S=g/v//ft W=k*(g/v)//ft/day Q=W*p1*p2*q//mgd P=k*p//ft P1=P*p2//mgd //RESULTS printf('the velocity of flow =% f mgd',Q) printf('the standard coefficient pf permeability=% f mgd',P1)
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function [pp,qq]=suivantFracCont(p,q) pp = 2*q + p qq = p + q endfunction N=5 p=1 q=1 valeursP = [] valeursQ = [] for k=1:N valeursP = [valeursP,p] valeursQ = [valeursQ,q] [p,q] = suivantFracCont(p,q) end A = [1 2 ; 1 1] [P,D]=spec(A) //N=5 //p=1 //q=1 //for k=1:N // disp(A^k) // [p,q] = suivantFracCont(p,q) // disp([p,q,p^2-2*q^2]) //end
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//achar pi através do método dos trapézios //achar a área A. Depois fazer pi=4*A t1=0; t2=1; h=0.0001; x=t1:h:t2; function y=caio(x) y=sqrt(1-x.^2); endfunction integral=0.0; y1=caio(x); for i=1:length(x)-1; integral= integral+((y1(i)+y1(i+1))/2)*h; end pi=4*integral; disp(pi)
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Lk_L = 1; Ck_C = 1; La_L = 1; Rs_R = 1; Rm_R = 1; Rk_R = 1; num_state = 3; num_invar = 1; num_outvar = 1; s = poly(0,'s'); polymat = ... [[Lk_L*s, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0]; [-1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, -1, 0, 0, 0, 0, 0]; [1, 0, -1, 0, 0, 0, 0, -1, 0, 0, 0, 1, 0, 0, 0, 0, 0]; [0, -Ck_C*s, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0]; [0, 1, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0]; [0, 0, La_L*s, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0]; [0, 0, -Rs_R, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0]; [0, 0, -Rm_R, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]; [0, 0, 0, 1, 0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]; [0, 0, 0, 0, 0, 1, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0]; [0, 0, 0, 0, 0, 1, 0, 0, -1, 0, 1, 0, 0, 0, 0, 0, 0]; [0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0]; [0, 0, 0, 0, 0, 0, 0, Rk_R, 0, 1, 0, 0, 0, 0, 0, 0, 0]; [0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, -1, 0]; [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 0, 1, 1] ];
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//clc() F = 100;//kg //F, D and W be the flow rates of the feed, the distillate and bottom product resp., xf , xd and xw be the mole fraction of methanol in the fee, distillate and the bottom product resp. xf = 0.20; xd = 0.97; xw = 0.02; //using, F = D + W and F*xf + D*xd + W*xw,we get D = 18.95;//kg/h W = 81.05;//kg/h R = 3.5; //R = L / D //for distillate = 1kg D1 = 1;//kg L = R*D1; //Taking balance around the condenser, G = L + D1; mcondensed = G * D / F; disp("kg",D,"(a)Amount of distillate = ") disp("kg",W," Amount of Bottom Product = ") disp("kg",G,"(b)Amount of vapour condensed per kg of distillate = ") disp("kg",mcondensed,"(c)Amount of vapour condensed per kg of feed = ")
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//Serie de Maclaurin de la funcion logaritmo natural //function [et, ea, aprox] = funcionLnMaclaurin(x, iter) //Entrada: //x : valor en que la serie sera evaluada //iter : cantidad de terminos de la serie //Salida: //et : vector de errores relativos //ea : vector de errores aproximados //aprox : vector con valores aproximados de la serie function [et, ea, aprox] = funcionLnMaclaurin(x, iter) verdad = log(1+x); aprox(1) = x; et(1) = ((verdad - aprox(1))/verdad)*100; ea(1) = %nan; for i = 2:iter aprox(i) = aprox(i-1) + ((-1)^(i+1))*((x^i)/i); et(i) = ((verdad - aprox(i))/verdad)*100; ea(i) = ((aprox(i) - aprox(i-1))/aprox(i))*100; end endfunction
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//Example 11.13 //Eigenvalue Problem //Page no. 387 clc;clear;close; h1=1/2;h2=1/3; lbd=poly(0,'lbd') mu=9*lbd/16; r1(1)=64 A=[2*lbd-324,81;243,lbd-324]; disp(determ(A),'Characteristic Equation = '); r=roots(determ(A)) disp(r,'Roots = ') r1(2)=r(2) Q=((h1/h2)^2*r1(2)-r1(1))/((h1/h2)^2-1) disp(Q,'Q12 = ')