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# Use this file to test your implementation of the gogui-rules_legal_moves # command. boardsize 3 clear_board 1 gogui-rules_legal_moves #?[a1 a2 a3 b1 b2 b3 c1 c2 c3] play b a1 play b a2 play b a3 play b b1 play b b2 play b b3 2 gogui-rules_legal_moves #?[c1 c2 c3] play w c1 play w c2 play w c3 # if board is full 3 gogui-rules_legal_moves #?[] boardsize 5 clear_board play b a1 play b b1 play b c1 play b d1 play b e1 # if somebody has won 4 gogui-rules_legal_moves #?[]
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//Chapter-4, Illustration 4, Page 165 //Title: Steam Nozzles and Steam Turbines //============================================================================= clc clear //INPUT DATA P1=2.2;//Pressure at entry in MN/(m^2) T1=533;//Temperature at entry in K P2=0.4;//Pressure at exit in MN/(m^2) m=11;//mass flow rate of steam in kg/s n=0.85;//Efficiency of expansion h1=2940;//Enthalpy at entrance in kJ/kg from Moiller chart ht=2790;//Enthalpy at throat in kJ/kg from Moiller chart h2s=2590;//Enthalpy below exit level in kJ/kg from Moiller chart vt=0.16;//Throat volume in (m^3)/kg v2=0.44;//Volume at exit in (m^3)/kg //CALCULATIONS Ct=(2000*(h1-ht))^0.5;//Throat velocity in m/s h2=ht-(0.85*(ht-h2s));//Enthalpy at exit in kJ/kg C2=(2000*(h1-h2))^0.5;//Exit velocity in m/s At=((m*vt)/Ct)*(10^6);//Area of throat in (mm^2) A2=((m*v2)/C2)*(10^6);//Area of exit in (mm^2) //OUTPUT mprintf('Throat velocity is %3.0f m/s \n Exit velocity is %3.0f m/s \n Throat area is %3.0f (mm^2) \n Exit area is %3.0f (mm^2) \n',Ct,C2,At,A2) //==============================END OF PROGRAM=================================
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//Example 1.6 //Pg No. 18 choice = 'a' while choice ~= 'T' choice = input('Type T and press enter to terminate') end
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COM_receive file enable testTemp\test_results.txt sleep 10000 COM_send string start sleep 500000 COM_receive file disable testTemp\test_results.txt
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// Copyright (C) 2015 - IIT Bombay - FOSSEE // // This file must be used under the terms of the CeCILL. // This source file is licensed as described in the file COPYING, which // you should have received as part of this distribution. The terms // are also available at // http://www.cecill.info/licences/Licence_CeCILL_V2-en.txt // Author: Vinay Bhat,Gursimar Singh // Organization: FOSSEE, IIT Bombay // Email: toolbox@scilab.in function [dstMat] = imhmin(srcImg, Hmin) // This fucntion is used to get H-minima transform in the form of an image. // // Calling Sequence // I2 = imhmin(I,h) // // Parameters // I: image matrix of the source image. // h: h-maxima transform, specified as a nonnegative scalar. // I2: Transformed image, returned as a nonsparse numeric array of any class, the same size as I. // // Description // I2 = imhmin(I,h) suppresses all minima in the intensity image I whose depth is less than h, where h is a scalar. Regional minima are connected components of pixels with a constant intensity value, t, whose external boundary pixels all have a value greater than t // // Examples // i = imread('images/lena.jpeg'); // i2 = imhmin(i,200); // imshow(i2); // //Authors //Vinay Bhat //Gursimar Singh // //See also //imhmax //imhistmax [lhs, rhs] = argn(0); if rhs>2 then error(msprintf("Too many input arguments")); end if rhs<2 then error(msprintf("Not enough input arguments")); end if lhs >1 error(msprintf("Too many output arguments")); end srcMat = mattolist(srcImg) out = raw_imhmin(srcMat, Hmin) channel = size(out) for i = 1: channel dstMat(:,:,i) = out(i) end endfunction
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//Example 14.4 //Program to calculate: //(a)3 dB Pulse Broadening in ns/km //(b)Fiber Bandwidth-Length product clear; clc ; close ; //Given data tau_o=12.6; //ns - 3 dB width of Output Pulse tau_i=0.3; //ns - 3 dB width of Input Pulse L=1.2; //km - LENGTH //(a)3 dB Pulse Broadening in ns/km tau=sqrt(tau_o^2-tau_i^2)/L; //(b)Fiber Bandwidth-Length product Bopt=0.44/tau; //Displaying the Results in Command Window printf("\n\n\t (a)3 dB Pulse Broadening is %0.1f ns/km.",tau); printf("\n\n\t (b)Fiber Bandwidth-Length product is %0.1f MHz km.",Bopt*10^3);
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clear; clc; printf('FUNDAMENTALS OF HEAT AND MASS TRANSFER \n Incropera / Dewitt / Bergman / Lavine \n EXAMPLE 8.4 Page 506 \n'); //Example 8.4 // Length of tube for required heating // Surface temperature Ts at outlet section //Operating Conditions m = .01; //[kg/s] mass flow rate of water Ti = 20+273; //[K] Inlet temp To = 80+273; //[K] Outlet temperature D = .06; //[m] Diameter q = 2000; //[W/m^2] Heat flux to fluid //Table A.4 Air Properties T = 323 K cp = 4178; //[J/kg.K] specific heat //Table A.4 Air Properties T = 353 K k = .670; //[W/m] Thermal Conductivity u = 352*10^-6; //[N.s/m^2] Viscosity Pr = 2.2; //Prandtl Number cp = 4178; //[J/kg.K] specific heat L = m*cp*(To-Ti)/(%pi*D*q); //Using equation 8.6 Re = m*4/(%pi*D*u); printf("\n (a) Length of tube for required heating = %.2f m\n\n (b)As Reynolds Number is %i. The flow is laminar.",L,Re); Nu = 4.364; //Nusselt Number h = Nu*k/D; //[W/m^2.K] Heat convection Coefficient Ts = q/h+To; //[K] printf("\n Surface Temperature at tube outlet = %i degC",Ts-273); //END
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//Example 6.17 clc clear x = [0 1 4]; y = [2 1 4]; n = length(x); del = %nan*ones(n,3); del(:,1) = y'; for j = 2:3 for i = 1:n-j+1 del(i,j) = (del(i+1,j-1) - del(i,j-1)) / (x(i+j-1) - x(i)); end end del(:,1) = []; Y = 0; X = 2; for i = 1:n t = x; t(i) = []; p = 1; for j = 1:length(t) p = p * (X-t(j))/(x(i)-t(j)); end Y = Y + p*y(i); end disp(Y,"y(2) = ")
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//Example 2.4 clc; clear; close; //Given data : A=0.2;//m^2 dy=0.02/100;//m du=20/100;//cm/s mu=0.001;//Ns/m^2 tau=mu*du/dy;//in N/m^2 F=tau*A;//N disp(F,"Force required in N : "); Power=F*du;//Watts disp(Power,"Power required in W : ");
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clc disp("Example 7.23") printf("\n") //Sketch voltage 'v' t=-.001:0.00005:0 t1=0:0.00005:.001 T=1*10^-3; V0=10; v=V0*exp(t/T) v1=V0*exp(-t1/T) figure a= gca (); plot(t,v) plot(t1,v1) xtitle ('v vs t','t (ms)','v '); a. thickness = 2; //Sketch current 'i' t=-.001:0.00005:0 t1=0:0.00005:.001 T=1*10^-3; I0=10*10^-3; i=I0*exp(t/T) i1=-I0*exp(-t1/T) figure a= gca (); plot(t,i) plot(t1,i1) xtitle ('i vs wt','t (ms)','i (mA)'); a. thickness = 2;
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//**************************** Speech ********************************** if (blk_name.entries(bl) =='speech') then addvmm = %t; mputl("# speech",fd_w); for ss=1:scs_m.objs(bl).model.ipar(1) speech_str= '.subckt speech in[0]=net' + string(blk(blk_objs(bl),2))+"_1 in[1]=net"+string(blk(blk_objs(bl),3))+"_1 in[2]=vcc out[0]=net'+ string(blk(blk_objs(bl),2+numofip))+"_" + string(ss)+" out[1]=net'+ string(blk(blk_objs(bl),3+numofip))+"_" + string(ss)+" #c4_ota_bias[0] =" +string(sprintf('%1.3e',scs_m.objs(blk_objs(bl)).model.rpar(2*ss-1)))+"&c4_ota_bias[1] =" +string(sprintf('%1.3e',scs_m.objs(blk_objs(bl)).model.rpar(2*ss)))+"&speech_fg[0] =0&c4_ota_p_bias[0] =105e-9&c4_ota_n_bias[0] =105e-9&c4_ota_p_bias[1] =105e-9&c4_ota_n_bias[1] =105e-9&speech_peakotabias[0] =100e-9&speech_pfetbias[0] =2e-11&speech_peakotabias[1] =9e-10"; mputl(speech_str,fd_w); mputl(" ",fd_w); select board_num case 2 then plcloc=[plcloc;'net'+string(blk(blk_objs(bl),2+numofip))+'_'+ string(ss),'6 '+string(ss)+' 0']; case 3 then plcloc=[plcloc;'net'+string(blk(blk_objs(bl),2+numofip))+'_'+ string(ss),'1 '+string(ss)+' 0']; end end end
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clc // //Bar with Square Cross Section //Variable declaration tALL=40 // Stress(MPa) //Calculation // Bar with square cross section a=0.040 // Length(m) b=0.040 // Length(m) temp=(a/b) c1=0.208 // Coefficient tmax=tALL // Maximum stress(MPa) T1=(40)*((10**6))*(0.208)*((0.040**3)) // Torque(N.m) // Bar with Rectangular Cross Section. a=0.064 // Length(m) b=0.025 // Length(m) temp2=(a/b) T2=(40)*((10**6))*(0.259)*(0.064)*((0.025**2)) // Torque(N.m) //Square Tube A=(0.034)*(0.034) // Area bounded by the center line of the cross section(m**2) T3=((40)*((10**6))*(2)*(0.006)*(1.156)*((10**-3))**0) // Torque(N.m) // Result printf("\n Largest torque on bar with square cross section = %1f N.m' ,T1) printf("\n Largest torque on bar with rectangular cross section = %1f N.m' ,T2) printf("\n Largest torque on square tube = %1f N.m' ,T3)
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clc clear disp("Example 8.49a") printf("\n") disp("Prove the following boolean identities") disp("A+BC=(A+B)(A+C)") A=[0 0 0 0 1 1 1 1] B=[0 0 1 1 0 0 1 1] C=[0 1 0 1 0 1 0 1] for i=1:length(A) Y(i)=A(i)+(B(i)*C(i)) if(Y(i)==2) Y(i)=1 end end for i=1:length(A) Z(i)=(A(i)+B(i))*(A(i)+C(i)) if(Z(i)==2) Z(i)=1 end if(Z(i)==3) Z(i)=1 end if(Z(i)==4) Z(i)=1 end end for i=1:length(A) if(Z(i)==Y(i)) printf("_") else printf("NOT") abort end end printf("proved")
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//EXAMPLE 2-74 PG NO-112 ZA=10+%i*7.226; //IMPEDANCE ZB=5+%i*10.99; //IMPEDANCE V=200+%i*0; //VOLTAGE IA=V/ZA; //CURRENT disp('i) CURRENT (IA) is in polar form = '+string (IA) +' A '); IB=V/ZB; disp('ii) CURRENT (IB) is in polar form = '+string (IB) +' A '); I=IA+IB; disp('iii) CURRENT (I) is in polar form = '+string (I) +' A '); S=V*I; disp('i) Apparent Power (S) is = '+string (S) +' VA '); P=V*I*0.63; disp('i) Active Power (P) is = '+string (P) +' W '); Q=V*I*0.775; disp('i) Reactive Power (Q) is = '+string (Q) +' Var ');
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function count=mtlb_fwrite(fid,a,prec) prectbl=['c' 'char' 'char' '?' 'schar' 'signed char' 's' 'short' 'short' 'l' 'int' 'int' 'l' 'long' 'long' 'f' 'float' 'float' 'd' 'double' 'double' 'uc' 'uchar' 'unsigned char' 'us' 'ushort' 'unsigned short' 'ul' 'uint' 'unsigned int' 'ul' 'ulong' 'unsigned long' 'c' 'char' 'char*1' 'f' 'float32' 'real*4' 'd' 'float64' 'real*8' 'c' 'int8' 'integer*1' 's' 'int16' 'integer*2' 'l' 'int32' 'integer*4' '?' '' 'integer*8' '?' 'intN' '' '?' 'uintN' '' ] [lhs,rhs]=argn(0) if rhs<3 then prec='uchar';end [l,k]=find(prec==prectbl) if l==[] then error('The format:'+prec+' is unknown') end Prec=prectbl(l,1) if Prec=='?' then error('The format:'+prec+' is not yet handled') end mput(size(a,'*'),Prec,fid) count=size(a,'*')
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clc //initialisation of variables m=27*10^-3//N.s/m^2 sg=0.90 m1=27//cp v1=5.6*10^-4//lbf.sec/ft^2 v2=2.5*10^-2//m y=9802//N/m^3 g=9.8//m/s^2 Nr=4000 Nr1=2000 //CALCULATIONS P=(y*sg)/g//N.s^2/m^4 V1=(Nr*m)/(v2*P)//m/s V2=(Nr1*m)/(v2*P)//m/s //RESULTS printf('The critical velocity range is=% f m/s',V2)
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clc;funcprot(0); //Example 3.3 //Initializing the variables D = 1.8; //Depth of tank h = 1.2; //Depth of water l = 3; //Length of wall of tank p = 35000; //Air pressure rho = 10^3; //Density of water g = 9.81; //Acceleration due to gravity //Calculations Ra = p*D*l; //Force due to air Rw = .5*(rho*g*h)*h*l; //Force due to water R = Ra + Rw; // Resultant force x = (Ra*0.9+Rw*0.4)/R; // Height of center of pressure from base disp(x,"Height of the centre of pressure above the base(m) :",R/1000,"Resultant force on the wall(kN)");
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clear; clc; //Example - 1.8 //Page number - 24 printf("Example - 1.8 and Page number - 24\n\n"); // Given V = 10;// [m^(3)] - volume of vessel P_1 = 1;// [bar] - initial pressure V_liq_sat = 0.05;// [m^(3)] - saturated liquid volume V_gas_sat = 9.95;// [m^(3)] - saturated vapour volume //At 1 bar pressure V_liq_1 = 0.001043;// [m^(3/kg)] - specific saturated liquid volume U_liq_1 = 417.33;// [kJ/kg] - specific internal energy V_gas_1 = 1.69400;// [m^(3/kg)] - specific saturated vapour volume U_gas_1 = 2506.06;// [kJ/kg] M_liq_1 = V_liq_sat/V_liq_1;// [kg] - mass of saturated liqid M_gas_1 = V_gas_sat/V_gas_1;// [kg] - mass of saturated vapour M = (M_liq_1+M_gas_1);// [kg] - total mass U_1t = (M_liq_1*U_liq_1)+(M_gas_1*U_gas_1);// [kJ] - initial internal energy V_gas_2 = (V/M);//[m^(3/kg)] //from steam table at 10 bar pressure as reported in the book V_vap_2 = 0.19444;// [m^(3/kg)] U_vap_2 = 2583.64;// [kJ/kg] //from steam table at 11 bar pressure as reported in the book V_vap_3 = 0.17753;//[m^(3/kg)] U_vap_3 = 2586.40;//[kJ/kg] //Now computing pressure when molar volume of saturated vapour=Vg_2 //By interpolation (P2-10)/(11-10)=(Vg_2-Vv_2)/(Vv_3-Vv_2) P_2 = (((V_gas_2 - V_vap_2)/(V_vap_3 - V_vap_2)*1)+10);// [bar] - final pressure //By interpolation calculating internal energy at state 2 //(P2-10)/(11-10)=(U2-Uv_2)/(Uv_3-Uv_2) U_2 = (((P_2-10)/(11-10))*(U_vap_3 - U_vap_2))+U_vap_2;//[kJ/kg] U_2t = U_2*M;//[kJ] H = U_2t - U_1t;//[kJ] - Heat supplied H = H/1000;//[MJ] printf(" Total heat supplied is %f MJ',H); // since volume is constant,no work is done by the system and heat supplied is used in increasing the internal energy of the system.
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clc //to calculate radius of curvature lambda=5900*10^-10 //wavelength in m n=10 Dn=5*10^-3 // diameter of 10th dark ring in m R=Dn^2/(4*n*lambda) disp("the radius of curvature of the lens is R="+string(R)+"m") //to calculate thichness t=n*lambda/2 disp("the thickness of the air film is t="+string(t)+"m")
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clc clear //Input data CC=(80*10^6)//Capital cost in Rs L=30//Useful life in years S=5//Salvage value of the capital cost in percent i=0.06//Yearly rate of compound interest //Calculations A=((100-S)/100)*CC//Difference of capital cost and salvage value in Rs P=((A*i)/((1+i)^L-1))//The amount of money to be saved annually in Rs //Output printf('The amount of money to be saved annually is Rs.%3.0f/-',P)
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clear; clc; close; disp("Example 4.17") M0=0.7 //Mach no. T0=228 // in K p0=16 //kPa eprop=0.85 // prop efficiency m=10 //Kg/s pd=0.98 //diffuser pressure ratio pc=30 //compressor pressurer ratio ec=0.92 //thermal efficiency of compressor Tt4=1600 //in K Qr=42000000 //in kJ/kg eb=0.99 //thermal efficiency of burner pb=0.96 //burner pressure ratio etHPT=0.82 emHPT=0.99 alfa=0.85 emLPT=0.99 eLPT=0.88 egb=0.995 en=0.95 gmc=1.4 //gamma of compressor Cpc=1004 // in J/kg.K gmt=1.33 //gamma of turbine Cpt=1152 // in J/kg.K Tt0=T0*(1+((gmc-1)*(M0)^2)/2) pt0=p0*(1+((gmc-1)*(M0)^2)/2)^(gmc/(gmc-1)) a0=((gmc-1)*Cpc*T0)^(1/2); V0=a0*M0 pt2=pt0*pd Tt2=Tt0 //Adiabatic pt3=pt2*pc tc=pc^((gmc-1)/(ec*gmc)) Tt3=Tt2*tc f=(Cpt*Tt4-Cpc*Tt3)/(Qr*eb-Cpt*Tt4) pt4=pt3*pb ht45=Cpt*Tt4-(Cpc*Tt3-Cpc*Tt2)/((1+f)*emHPT) Tt45=ht45/Cpt pt45=pt4*(Tt45/Tt4)^(gmt/((gmt-1)*etHPT)) m9=(1+f)*m sp=(1+f)*m*eLPT*alfa*ht45*(1-(p0/pt45)^((gmt-1)/gmt))/10^6 Tt5=(ht45-sp*10^6/((1+f)*m))/Cpt tt=Tt5/Tt45 et=log(tt)/(log(1-((1-tt)/eLPT))) pt=tt^(gmt/(et*(gmt-1))) pt5=pt45*pt p9=p0 //assumption pi=p9/pt5 ti=pi^((gmt-1)/gmt) T9i=Tt5*ti T9=Tt5-en*(Tt5-T9i) V9=(2*Cpt*(Tt5-T9))^(1/2) Fprop=eprop*egb*emLPT*sp*10^3/V0 a9=((gmt-1)*Cpt*T9)^(1/2) M9=V9/a9 pt9=p9*(1+((gmt-1)*M9^2)/2)^(gmt/(gmt-1)) pn=pt9/pt5 Fncore=m*((1+f)*V9-V0)/1000 spp=egb*emLPT*sp Ft=Fprop+Fncore mp=((m9*V9^2)/2-m*(V0^2)/2)/10^3 mf=m9-m PSFC=mf*10^6/((spp*10^3)+mp) TSFC=mf*10^3/(Ft) eth=(spp*10^3+mp)*10^3/(mf*Qr) ep=(Ft*V0)/(spp*10^3+mp) eo=eth*ep disp("a(1)Total pressures throughout the engine in kPa:") disp(pt0,"Total pressure of flight:") disp(pt2,"Total pressure at engine face:") //disp(p19,"Static pressure at nozzle exit:") disp(pt3,"Total pressure at compressor exit:") disp(pt4,"Total pressure at burner exit:") disp(pt45,"Total pressure across HPT :") disp(pt5,"Total pressure at turbine exit:") disp(pt9,"Total pressure at nozzle exit:") disp("a(2)Total temperatures across the engine in K:") disp(Tt0,"Total temperature of flight:") disp(Tt2,"Total temperature at engine face:") //Tt0=Tt2, since adiabatic! disp(Tt3,"Total temperature at compressor exit:") disp(Tt4,"Total temperature at burner exit:") disp(Tt45,"Total temperature across HPT :") disp(Tt5,"Total temperature at turbine exit:") disp(f,"a(3)fuel-to-air ratio in burner :") disp(Fncore,"(b)Engine core thrust in kN :") disp(Fprop,"(c)Propeller thrust in kN :") disp(PSFC,"(d)Power-specific fuel consumption in mg/s/kW :") disp(TSFC,"(e)TSFC in mg/s/N :") disp(ep,"f(1)Propulsive efficiency :") disp(eth,"f(2)Thermal efficiency :") disp(eo,"(g)Overall efficiency :")
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//In this example flow over the wing is both turbulent and laminar.so to find drag we need to find drag on both laminar and turbulent layer and add them. b=12.202;//wing span in meter S=23.69;//wing area in m^2 c=S/b //wing width Ret=6.5*10^5;//transition reynolds number or critical reynolds number D=1.225;//density at standard sea level,Kg/m^3 u=1.79*10^-5;//Viscosity in at standard sea level in kg/(m)(s) V=48.3*5/18 //velocity of flyer q=D*V^2/2 //dynamic pressure Re=D*V*c/u //reynolds no. at trailing edge Xcr=(Ret*u)/(D*V) //distance from leading edge where transition occur A=Xcr*b //area over which laminar flow occur in m^2 B=(c-Xcr)*b //area over which turbulent flow occur in m^2
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//Example 4.13 clc printf("Enter value for Item, partno, cost : "); printf("\n [Enter values in single line seperated by spaces]) "); [a,Item, partno,cost]=mscanf("%11s %*d %f");; //due to uncertainity of values assigned, omitted use of printf here disp( " = cost",cost," = Partno ",partno, " = Item ",Item);
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// Theory and Problems of Thermodynamics // Chapter 9 // Air_water Vapor Mixtures // Example 13 clear ;clc; //Given data mw3 = 1000 // cooling tower supply rate in kg/min T1 = 303.15 // Temp of air entering cooling tower in K RH1 = 0.3 // relative humidity of air entering cooler T2 = 308.15 // Temp of air leaving cooling tower in K RH2 = 0.8 // relative humidity of air leaving cooler T3 = 318.15 // Temp of water entering cooling tower in K T4 = 300.15 // Temp of water leaving cooling tower in K // subscript 1 and 2 denotes the state of air entering and leaving the cooling tower respectively // subscript 3 and 4 denotes the state of water entering and leaving the cooling tower respectively // data from psychometric chart for T = 30 degree C and RH = 0.3 SH1 = 0.0078 // in kg H2O/kg air h1 = 51 // in kJ/kg air // data from psychometric chart for T = 35 degree C and RH = 0.8 SH2 = 0.029 // in kg H2O/kg air h2 = 110 // in kJ/kg air hw3 = 188.45 // in kJ/kg hw4 = 113.25 // in kJ/kg // mass balance for H2O: //mw3-mw4 = ma*(SH2-SH1) // energy balance gives: // mw3*hw3 - mw4*hw4 = ma*(h2-h1) // x(1) = ma; x(2)= mw4; function[f] =F(x) f(1) = mw3-x(2)-x(1)*(SH2-SH1); f(2) = mw3*hw3 - x(2)*hw4 -x(1)*(h2-h1); endfunction x = [10 10]; y = fsolve(x,F) ma = y(1); // air flow rate in kg/min mw4 = y(2); wat_mak = mw3-mw4; // make up water required // Output Results mprintf('Make up water required = %4.2f kg/min' , wat_mak); mprintf('\n Air flow rate = %4.1f kg/min' , ma);
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x=0.1 y=0.2 MAcid=98.1 MS=32 MSalt=142 MBase=40 MWater=18 MNa=46 basis=1000 //g T2=35 T1=25 T3=40
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clc // Given that l_0 = 1// let initial length of rod in m l = 0.99 // Observed length in m c = 3e8 // speed of light in m/s // Sample Problem 10 on page no. 27 printf("\n # PROBLEM 10 # \n") printf(" Standard formula used \n") printf(" l = l_0/((1-v^2/c^2)^1/2) \n") v = c* sqrt(1-(l/l_0)^2)// speed of rocket in m/s printf("\n Speed of rocket is %e m/s.",v)
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/* * By: Jatin Kumar Mandav * * Frequency Modulation * * Inputs: Amplitude of Carrier Wave, Frequency of Carrier Wave * Amplitude of Message Wave, Frequency of Message Wave * Sampling Frequency or Sampling/sec * */ function [] = fm_mod(amp_carrier, freq_carrier, amp_message, freq_message, freq_sampling) t = 0:1/freq_sampling:1 message_signal = amp_message*sin(2*%pi*freq_message*t) carrier_signal = amp_carrier*sin(2*%pi*freq_carrier*t) k = 1.5 modulated_signal = amp_carrier*sin(2*%pi*freq_carrier*t + (k*amp_message/freq_message)*cos(2*%pi*freq_message*t)) subplot(3, 1, 1) plot(message_signal) title("Message Signal") subplot(3, 1, 2) plot(carrier_signal) title("Carrier Signal") subplot(3, 1, 3) plot(modulated_signal) title("Modulated Signal") endfunction
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//3.4 clc; tq=50*10^-6; Vin=40; Vo=230; IL=2; IL_ref=2*Vo/Vin; // C/L=(IL-ref/Vin)^2; (i) // Assume that circuit is reverse biased for one-fourth period of resonant circuit. thus //%pi/3*(L*C)^0.5=50*10^-6; (ii) // on solving (i) and (ii) C=13.73*10^-6; L=C/(IL_ref/Vin)^2*10^6; C=13.73*10^-6*10^6; printf("C=%.3f uF",C) printf("\nL=%.2f uH",L)
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Ex4_2.sce
//example-4.2 //page no-115 //given //height of zinc unit cell h=4.935*10^-10 //m //side of the lattice a=2.66*10^-10 //m //as we know that zinc has HCP unit cell. //the number of effective atoms Ne=6 //as we know //tan(%pi/3)=x/(a/2) //so x=a/2*tan (%pi/3) //m //area of basal plane Ar=6*a*x/2 //m^2 //volume of the unit cell V=Ar*h //m^3 //atomic weight of zinc Aw=65.37 //avogadro's number NA=6.023*10^23 //density of zinc rho=Aw*Ne/(NA*V)/1000 //kg/m^3 printf ("the no of effective atoms, the volume of unit cell and density of zinc are 6, 9.07*10^-29 and 7180 kg/m^3 resp")
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Ex9_14.sce
clc //initialisation of variables P1=14.7//lbf/in^2 T1=520 //R P2=70 //lbf/in^2 T2=814 //R Wc=70.6 //Btu/lbm P3=70 //lbf/in^2 T3=2060 //R Cp=0.24 //R Wt=70.6 //lbf/in^2 T=294//R T5=1318//R P=1.167 //Btu/lbm p=1.715 //Btu/lbm g=32.17//lbf/in^2 t=778//F //CALCULATIONS T4=T3-T//R P4=P3/p//lbf/in^2 V=sqrt (2*g*t)*Cp*(T4-T5)//ft/sec //RESULTS printf('The velocity of the air leaving the nozzle=% f ft/sec',V)
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// Chapter 11 example 2 //------------------------------------------------------------------------------ clc; clear; // Given data Ap_Pe_diff = 30000; // difference between apogee and perigee in Km a = 16000; // semi major axis of orbit // Calculations e = Ap_Pe_diff/(2*a); // Eccentricity // Output mprintf('Eccentricity = %3.2f',e); //------------------------------------------------------------------------------
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clc //Chapter8 //Example8.19, page no 350 //Given Zo=100// Characteristic Impedance P=100e-3//Load power Zr=140//Load Resistance f=100e3// Operating freq //a K=(Zr-Zo)/(Zo+Zr)//Vtg Reflection coeff //b S=(1+K)/(1-K)//VSWR //c+d Emax=sqrt(Zr*P)//Max line vltg Imin=Emax/Zr//Min line current Emin=Emax/S// Min line vltg Imax=S*Imin//Max line current //e R=14000/40 Zr=(Zo^2)/R// mprintf('\nThe voltage reflection coeff is %f\nThe VSWR is %f\n\n\nThe Max and min voltage and crresponding crrent is\n Emax= %fV Imin= %fmA\n Emin= %fV Imax= %fmA\n\n The Termination resistance should be %f ohm',K,S,Emax,Imin*1e3,Emin,Imax*1e3,Zr)
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EX4_3.sce
//Clearing console clc clear //Intializing Variables b = 40 h = 40 I = (b*h^3)/12 L1=300 L2=300 L3=200 E1 = 207*10^3 E3 = 69*10^3 A1= 1600 A3 = 78.54 //Calculating elemental stiffness matrices K1 = ((E1*I)/(L1^3))*[12 6*L1 -12 6*L1;6*L1 4*(L1)^2 -6*L1 2*(L1)^2; -12 -6*L1 12 -6*L1; 6*L1 2*(L1)^2 -6*L1 4*(L1)^2] K2 = K1 //as L1 = L2 and both are of same material (E1 = E2) K3 = (A3*E3/L3)*[1 -1;-1 1] //Constructing Global Stiffness matrix K(1,[1:7])= [K1(1,[1:4]) 0 0 0] K(2,[1:7])= [K1(2,[1:4]) 0 0 0] K(3,[1:7])= [K1(3,[1:2]) K1(3,3)+K2(1,1)+K3(1,1) K1(3,4)+K2(1,2) K2(1,[3:4]) K3(1,2)] K(4,[1:7])= [K1(4,[1:2]) K1(4,3)+K2(2,1) K1(4,4)+K2(2,2) K2(2,[3:4]) 0] K(5,[1:7])= [0 0 K2(3,[1:4]) 0 ] K(6,[1:7])= [0 0 K2(4,[1:4]) 0 ] K(7,[1:7])= [0 0 K3(2,1) 0 0 0 K3(2,2)] //Constructing Force matrix (required values) F([2:6],1) = [0; 0; 0; -10000; 0] //Solving for displacements U(2:6,1)=linsolve(K(2:6,2:6),-F(2:6,1)) //K*U=F (equlibrium equation) //Solving Axial stress of BD element stress_BD = E3*[-1/L3 1/L3]*[0 1 0 0;0 0 0 1]*[0;U(3,1);0;0] U(1,1)=0 U(7,1)=0 //Calculating Reaction forces F = [K]*[U] //Printing Results printf('\nResults\n') printf('\nNode-C Displacement Components \nU=%fmm \nTheta=%frad',U(3,1),U(4,1)) printf('\nReaction Forces \nR1=%fN \nR4=%fN',F(1,1),F(7,1))
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// Example 7.1: Transfer curve clc, clear IDSS=12; // in mili-amperes VP=-5; // in volts // Plotting transfer curve VGS=[0:-0.01:VP]; // Gate source voltage in volts // Using Shockley's equation ID=IDSS*(1-VGS/VP)^2; // Drain current in mili-amperes plot(VGS,ID); xtitle("Transfer Curve","VGS (V)","ID (mA)");
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FLAnova1WayUdt-NZ-UM-01.tst
-- Fuzzy Logix, LLC: Functional Testing Script for DB Lytix functions on Netezza -- -- Copyright (c): 2016 Fuzzy Logix, LLC -- -- NOTICE: All information contained herein is, and remains the property of Fuzzy Logix, LLC. -- The intellectual and technical concepts contained herein are proprietary to Fuzzy Logix, LLC. -- and may be covered by U.S. and Foreign Patents, patents in process, and are protected by trade -- secret or copyright law. Dissemination of this information or reproduction of this material is -- strictly forbidden unless prior written permission is obtained from Fuzzy Logix, LLC. -- -- -- Functional Test Specifications: -- -- Test Category: Hypothesis Testing Functions -- -- Last Updated: 05-29-2017 -- -- Author: <deept.mahendiratta@fuzzylogix.com> -- -- BEGIN: TEST SCRIPT --timing on -- BEGIN: TEST(s) -----******************************************************************************************************************************* ---FLAnova1WayUdt -----******************************************************************************************************************************* DROP VIEW UM_view_ANOVA1Way; CREATE VIEW UM_view_ANOVA1Way AS SELECT s.serialval AS GroupID, t.City, t.SalesPerVisit FROM tblCustData t, fzzlserial s WHERE City <> 'Boston' AND serialval <= 1; --Created VIEW SELECT * FROM UM_view_ANOVA1Way LIMIT 20; --Output Table SELECT a.* FROM(SELECT a.GroupID, a.City, a.SalesPerVisit, NVL(LAG(0) OVER (PARTITION BY a.GroupID ORDER BY a.City), 1) AS begin_flag, NVL(LEAD(0) OVER (PARTITION BY a.GroupID ORDER BY a.City), 1) AS end_flag FROM UM_view_ANOVA1Way a) AS z, TABLE(FLANOVA1WAYUdt(z.GroupID, z.City, z.SalesPerVisit, z.begin_flag, z.end_flag)) AS a; DROP VIEW UM_view_ANOVA1Way; -- END: TEST(s) -- END: TEST SCRIPT --timing off
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// This file is part of www.nand2tetris.org // and the book "The Elements of Computing Systems" // by Nisan and Schocken, MIT Press. // File name: projects/02/Add4.tst load Mult4.hdl, output-file Mult4.out, compare-to Mult4.cmp, output-list a%B1.4.1 b%B1.4.1 c%B1.1.1 out%B1.8.1; set a %B0000, set b %B0000, set c 0, eval, output; set a %B0000, set b %B1111, set c 0, eval, output; set a %B1111, set b %B0001, set c 0, eval, output; set a %B0100, set b %B0010, set c 0, eval, output; set a %B0011, set b %B0000, set c 0, eval, output; set a %B0001, set b %B1001, set c 0, eval, output;
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clc(); clear; //To determine the ratio of vacancies k=8.625*10^-5; //Boltzmann constant ineV/K //n1000/n500=ln[n1000/n500]=Ev/1000k Ev=1.08; //average energy required to create a vacancy in eV N=exp(Ev/(1000*k)) //n1000/n500 printf("The ratio of vacancies is %f",N);
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// Example 6.3;// feedback output clc; clear; close; A= 600;// open voltage gain Af=50;// Beta=( (A/Af)-1)/A;// feedback ratio fop= (Beta*100);//percentage of output voltage which is fedback to the input is disp(fop,"percentage of output voltage which is fedback to the input is ")
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// Test # 6 : Range test for Input Argument #4 or Input Argument #5 exec('./zpklp2mb.sci',-1); [z,p,k,n,d]=zpklp2mb(0.2,1,4,4,0.7); //!--error 10000 //Wo must lie between 0 and 1 //at line 48 of function zpklp2mb called by : //[z,p,k,n,d]=zpklp2mb(0.2,1,4,4,0.7);
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clc k = 1.33 // die opening factor l = 1200 // bend length in mm sigma = 455 // ultimate tensile strength in N/mm^2 t = 1.6 // blank thickness in mm w = 8*t // width of die opening in mm F = k*l*sigma*t^2/w // bending force in N printf("\n bending force = %0.2f kN", F/1000)
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path= get_absolute_file_path('loader.sce'); exec(path+"/loader_inc.sce"); functions1 = ["dmumpsc"]; functions2 = ["zmumpsc"]; entrypoint1 = "scidmumps"; entrypoint2 = "scizmumps"; addinter(objects,entrypoint1,functions1) num_interface = floor(funptr("dmumpsc")/100); intppty(num_interface) addinter(objects,entrypoint2,functions2) num_interface = floor(funptr("zmumpsc")/100); intppty(num_interface) [units,typs,nams]=file(); clear units typs for k=size(nams,'*'):-1:1 l=strindex(nams(k),'loader.sce'); if l<>[] then DIR_SCIMUMPS = part(nams(k),1:l($)-1); break end end DIR_SCIMUMPS_DEM=DIR_SCIMUMPS+ "examples/"; getf(DIR_SCIMUMPS+"initmumps.sci") getf(DIR_SCIMUMPS+"dmumps.sci") getf(DIR_SCIMUMPS+"zmumps.sci") add_help_chapter("Interface to the MUMPS package",path+"Help");
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//pathname=get_absolute_file_path('11.11.sce') //filename=pathname+filesep()+'11.11-data.sci' //exec(filename) //Time of trial(in hrs): t=24 //Pressure at which steam is generated(in bar): p=16 //Coal consumed(in kg): c=10000 //Rate of steam generation(in kg/hr): r=2500 //Feed water temperature(in C): Tf=27 //Total heating surface area(in m^2): hsa=3000 //Total grate area(in m^2): ga=4 //Calorific value of coal(in kJ/kg): C=28000 //From steam tables: hg=2794 //kJ/kg //Latent heat at 100 C: L=2257 //Coal burnt per hour(in kg/hr): m=c/t //Coal burnt per m^2 of grate per hour: mg=m/ga //Rate of steam generated per kg of coal(in kg steam/kg coal): r1=r/m //Heat added to steam per kg of coal(in kJ): Q=r1*(hg-4.18*Tf) //Equivalent evaporation from and at 100 C per kg of coal(in kg): Ee=Q/L //Equivalent evaporation from and at 100 C per m^2 of total surface per hour(in kg): Eepm=Ee*m/hsa //Boiler efficiency: n=Ee*L/C*100 printf("\n RESULT \n") printf("\nMass of coal burnt per m^2 of grate per hour = %f kg",mg) printf("\nEquivalent evaporation from and at 100 C per kg of coal = %f kg",Ee) printf("\nEquivalent evaporation from and at 100 C per m^2 of total surface per hour = %f",Eepm) printf("\nBoiler efficiency = %f percent",n)
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//Scilab Code for Example 1.7 of Signals and systems by //P.Ramakrishna Rao clear; clc; n=1; for t=-10:0.1:10; //Function for Even signal y1(n)=0.5*(exp(-t)*u(t)+exp(t)*u(-t)); n=n+1; end a=gca(); a.x_location="origin"; a.y_location="origin"; t=-10:0.1:10; //Plot of Even Signal plot(t,y1); title('y1(t)'); xlabel('Time in seconds'); n=1; for t=-1:0.01:1; //Function for Odd signal y2(n)=0.5*(exp(-t)*u(t)-exp(t)*u(-t)); n=n+1; end figure(1); a=gca(); a.x_location="origin"; a.y_location="origin"; t=-1:0.01:1; //Plot of Odd Signal plot(t,y2) disp('plotted the signal both in even and odd forms'); title('y2(t)'); xlabel('Time in seconds');
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ex2_51.sce
errcatch(-1,"stop");mode(2);//Caption:Calculate the speed and BHP of the motor //Exam:2.51 ; ; V=250;//applied emf(in V) R_sh=0.05;//field resistance (in Ohm) R_a=0.1;//armature resistance(in Ohm) I=80;//motor current(in Amp) A_s=240;//armature slots C_s=4;//number of conductor per slot Z=A_s*C_s;//total number of conductor E_b=V-I*(R_a+R_sh);//Back emf(in V) A=2;//number of parallel paths for wave wound P=6;//poles F=1.75;//flux per pole (in megalines) F_1=1.75*10^-2;//flux per pole (in Wb) N=E_b*60*A/(F_1*Z*P);//speed of the motor (in rpm) disp(N,'speed of the motor (in rpm)='); I_p=V*I;//input to the motor(in watts) L_c=(I^2)*(R_a+R_sh);//copper losses(in watts) L_i=900;//iron and friction losses(in watts) L_t=L_c+L_i;//total losses(in watts) O_p=I_p-L_t;//output(in watts) B.H.P=O_p/746;//B.H.P. of the motor disp(B.H.P,'B.H.P. of the motor(in H.P)='); exit();
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//Part A Chapter 7 Example 6 clc; clear; close; p=2;//MPa T=500+273.15;//K dh_by_ds=T;//for constant pressure disp("Slope of an isobar is "+string(dh_by_ds));
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composeRT.sci~
function [RotVec3 TransVec3] = composeRT(RotVec1,TransVec1,RotVec2,TransVec2) [RotVec3 TransVec3] = opencv_composeRT(RotVec1,TransVec1,RotVec2,TransVec2) endfunction
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clc; pathname=get_absolute_file_path('3_17_soln.sce') filename=pathname+filesep()+'3_17_data.sci' exec(filename) // Solution: // Acceleration due to gravity, g=9.81; //m/s^2 // Energy Equation between Station 1 and Station 2 is given by, // (Z+P1+K1+Hp-Hm-Hl)=(P2+K2) // since, There is no Hydraulic motor between Station 1 and 2, // Therefore Motor Head, Hm=0; //m // also, cross section of oil tank is very large, as a result oil is at rest, v1=0; //m/s // Kinetic Energy Head at inlet, K1=(v1^2)/(2*g); //m // Pressure Head at inlet, P1=p1/SG; //m // specific weight of oil, gamma1=round(SG*9797); //N/m^3 // Pump Power, W=HHP*1000; //W // Pump Head, Hp=(W/(Q*gamma1)); //m // Area of pipe, A=((%pi)*(D^2))/4; //m^2 // Therefore, velocity in pipe, v2=Q/A; //m/s // Kinetic Energy head at Station 2, K2=(v2^2)/(2*g); //m // Therefore, Pressure Head at outlet, P2=Z+P1+K1+Hp-Hm-Hl-K2; //m // Pressure available at inlet of hydraulic motor at station 2, p2=floor((P2*gamma1)/1000); // kPa gage // Results: printf("\n Results: ") printf("\n The Pressure available at inlet of hydraulic motor at Station 2 is %.0f kPa gage.",p2)
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load Prog1.vm, output-file Prog1.out, compare-to Prog1.cmp, output-list RAM[16]%D2.6.2, set RAM[0] 256, set RAM[16] 8, set RAM[17] 5, repeat 10 { vmstep; } output; load Prog1.vm, set RAM[0] 256, set RAM[16] -1, set RAM[17] 5, repeat 10 { vmstep; } output; load Prog1.vm, set RAM[0] 256, set RAM[16] 0, set RAM[17] 0, repeat 10 { vmstep; } output; load Prog1.vm, set RAM[0] 256, set RAM[16] 10, set RAM[17] 20, repeat 10 { vmstep; } output;
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//Example 7.2// power required clc; clear; close; format('v',6) l=0.2;//length in meter w=0.1;//width in meter th=25;//thickness in mm vw=l*w*th*10^-3;//volume in m^3 ww=600;//weight of wood in kg/m^3 ww1=vw*ww;//weight of wood kg shw=1500;//specific heat of wood in J/kg/degree celsius t=200;//temperature in degree celsius rg=t*shw*ww1;//energy in joules h=(rg/(3.6*10^3));//Wh t=15;//time in minutes pr=h*(60/t);//power required in Watt disp(pr,"power required is,(W)=")
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errcatch(-1,"stop");mode(2);; ; //Example 14.8 Is1=10^-14; Is2=1.05*10^-14; Vt=0.026; Vos=Vt*log(Is2/Is1); printf('\nthe offset voltage =%fV\n',Vos) exit();
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// // Autor: Jonas Vieira de Souza // // TODO: Função para ajuste de curvas por regressão linear // Exemplo de chamada: // $$ exec( caminho + 'regressao_linear.sci', -1 ); // $$ x [1 2 5 7 9 21]; // $$ y [4 5 6 7 9 20]; // $$ [ vi ] = interpolador_newton( x, y ); // // Retornos: // $$ vi $$ variavel independente do interpolado // // Argumentos: // $$ _x $$ vetor de valores da Variavel Dependente // $$ _y $$ vetor de valores da Variavel Independente // $$ _pa $$ ponto de avalição do Polinônio Interpolador // clc function v_indep = interpolador_newton( _x, _y, _pa ) [ mx, nx ] = size(_x); [ my, ny ] = size(_y); if nx ~= ny then disp("Dados incompatíveis - Tamanho dos dados desiguais"); error("x e y devem ter a mesma dimensão"); end b = _y; //Encontrando os termos do polinomio de Newtom // f(n-1) = b1+b2(x-x1)+...+bn(x-x1)(x-x2)...(x-xn) for i = 2:nx for j = nx:-1:i b(j) = (b(j)-b(j-1))/(_x(j)-_x(j-(i-1))); // end end disp(b,"Termos de b -->"); //Avaliando o ponto no polinomio interpolador v_indep = 0; for i = nx:-1:1 jota = 1; for j = 1:i-1 jota = jota*(_pa-_x(j)); end v_indep = v_indep + jota*b(i); end endfunction
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//clear// //Caption:Matched Filter output of rectangular pulse //Figure7.41 //Matched Filter Output clear; clc; T =4; a =2; t = 0:T; g = 2*ones(1,T+1); h =abs(convol(g,g)); for i = 1:length(h) if(h(i)<0.01) h(i) =0; end end h = h-T; t1 = 0:length(h)-1; figure a =gca(); a.data_bounds = [0,0;6,4]; plot2d(t,g,5) xlabel('t--->') ylabel('g(t)---->') title('Rectangular pulse duration T = 4, a =2') figure plot2d(t1,h,6) xlabel('t--->') ylabel('Matched Filter output') title('Output of filter matched to rectangular pulse g(t)')
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//Example 24.4 P=1*10^3;//Power (W) A=0.30*0.40;//Area (m^2) I_ave=P/A;//Intensity (W/m^2) printf('a.Intensity = %0.2e W/m^2',I_ave) I_0=2*I_ave;//Peak intensity (W/m^2) printf('\n Peak Intensity = %0.2e W/m^2',I_0) c=3*10^8;//Speed of light (m/s) eps_0=8.85*10^-12;//Permittivity of free space (C^2/N.m^2) E_0=sqrt(2*I_ave/(c*eps_0));//Peak electric field strength (V/m) printf('\nb.Peak electric field strength = %0.2e V/m',E_0) B_0=E_0/c;//Peak magnetic field strength (T) printf('\nc.Peak magnetic field strength = %0.2e T',B_0) //Openstax - College Physics //Download for free at http://cnx.org/content/col11406/latest
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// Electric Machinery and Transformers // Irving L kosow // Prentice Hall of India // 2nd editiom // Chapter 4: DC Dynamo Torque Relations-DC Motors // Example 4-6 clear; clc; close; // Clear the work space and console. // Given data T_old = 150 ; // Torque developed by a motor in N-m. disp("Example 4-6") disp("Given data : ") printf("\n \t\t\t phi \t I_a \t T "); printf("\n \t\t\t ________________________"); printf("\n Original condition \t 1 \t 1 \t 150 N-m "); printf("\n New condition \t\t 0.9 \t 1.5 \t ? "); // Calculation T_new = T_old * ( 0.9 / 1 ) * ( 1.5 / 1 ) ; // New torque produced in N-m // Display the result printf("\n\n Solution : ") printf("\n Using the ratio method, the new torque is the product "); printf("\n of two new ratio changes : "); printf("\n T = %.1f N-m ", T_new );
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errcatch(-1,"stop");mode(2);//Example 1.3 ; ; R1=10*10^3; R2=100*10^3; Ri=R1;//Input Resistance Ro=0;//Output Resistance A=-(R2/R1);// Ideal Overall Gain printf("Ri=%.2f kohms",(Ri/1000)); printf("\nRo=%.f ohms",Ro); printf("\nA=%.2f V/V",A); exit();
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// Scilab Code Ex3a.a.5: Page-135 (2008) clc; clear; phi1 = 0; // Phase of the first SHM, degree phi2 = 60; // Phase of the second SHM, degree phi3 = 90; // Phase of the third SHM, degree a1 = 1.0; // Amplitude of the first SHM, cm a2 = 1.5; // Amplitude of the second SHM, cm a3 = 2.0; // Amplitude of the third SHM, cm A = sqrt((a1 + a2*cosd(phi2)+a3*cosd(phi3))^2 + (a2*sind(phi2)+a3*sind(phi3))^2); // Resultant amplitude relative to the first SHM, cm phi = atand((a2*sind(phi2)+a3*sind(phi3))/(a1 + a2*cosd(phi2)+a3*cosd(phi3))); // Resultant phase angle relative to the first SHM, degree printf("\nThe resultant amplitude and phase angle relative to the first SHM = %4.2f cm and %2d degrees respectively", A, phi); // Result // The resultant amplitude and phase angle relative to the first SHM are 3.73 cm and 62 degrees respectively
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function [stk,txt,top]=sci_pause() // Copyright INRIA txt=[] if rhs<1 then stk=list('halt()','0','0','0','1') else if stk(top)(5)=='10' then write(logfile,'Warning: pause('+stk(top)(1)+') ignored') txt='//! pause('+stk(top)(1)+') ignored' write(logfile,txt) stk=list(' ',-2,'0','0','1') else stk=list('xpause(1000*('+stk(top)(1)+'))','0','0','0','1') end end
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clc;funcprot(0);//Example 5.13 //Initilisation of Variables L=3;......//Length of rectangular box in m W=2;...//Width of rectangular box in m H=7;.......//height of rectangular box in m Ts=290;....//Surface Temparature of box in K Ta=330;....//Temparature of air in K U=16.67;....//Velocity of air in m/s //Properties of air at 310 K rho=1.147;......//Density in kg/m^3 mu=16.48*10^-6;......//Viscocity in m^2/s K=0.0271;........//Thermal conductivity in W/mK Cp=1.005;......//Specific heat in kJ/kg K //calculation Re=(U*H)/mu;....//reynolds number Pr=(Cp*rho*mu)/(K*10^-3);.....//Prandtl number Nua=0.036*Re^(0.8)*Pr^(1/3);....//Nusselt number ha=(Nua*K)/H;...//Heat transfer coefficient in W/m^2 K Q=ha*H*2*(L+W)*(Ta-Ts)*10^-3;....//Heat transfer rate from plate in W disp(Q,"Heat transfer rate from plate in W:")
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//Solving four linear system of equations with Gauss-Seidel and SOR method //the convergence is much faster in SOR method clear; close(); clc; format('v',7); x1=[0,0]; x2=[0,0]; x3=[0,0]; x4=[0,0]; x1(1,2)=-0.33333*(1-x2(1,1)-3*x4(1,1)); x2(1,2)=0.16667*(1-x1(1,2)-x3(1,1)); x3(1,2)=0.16667*(1-x2(1,2)-x4(1,1)); x4(1,2)=-0.33333*(1-3*x1(1,2)-x3(1,2)); i=1; while (abs(x1(1,1)-x1(1,2))>0.5*10^-2 | abs(x2(1,1)-x2(1,2))>0.5*10^-2 | abs(x3(1,1)-x3(1,2))>0.5*10^-2 | abs(x4(1,1)-x4(1,2))>0.5*10^-2) x1(1,1)=x1(1,2); x2(1,1)=x2(1,2); x3(1,1)=x3(1,2); x4(1,1)=x4(1,2); x1(1,2)=-0.33333*(1-x2(1,1)-3*x4(1,1)); x2(1,2)=0.16667*(1-x1(1,2)-x3(1,1)); x3(1,2)=0.16667*(1-x2(1,2)-x4(1,1)); x4(1,2)=-0.33333*(1-3*x1(1,2)-x3(1,2)); i=i+1; end disp([x1; x2; x3; x4],'Answers are:') disp(i,'Number of Iterations :') w=1.6; x1=[0,0]; x2=[0,0]; x3=[0,0]; x4=[0,0]; x1(1,2)=x1(1,1)-0.33333*w*(1+3*x1(1,1)-x2(1,1)-3*x4(1,1)); x2(1,2)=x2(1,1)+0.16667*w*(1-x1(1,2)-6*x2(1,2)-x3(1,1)); x3(1,2)=x3(1,1)+0.16667*w*(1-x2(1,2)-6*x3(1,2)-x4(1,1)); x4(1,2)=x4(1,1)-0.33333*w*(1-3*x1(1,2)-x3(1,2)+3*x4(1,1)); i=1; while (abs(x1(1,1)-x1(1,2))>0.5*10^-2 | abs(x2(1,1)-x2(1,2))>0.5*10^-2 | abs(x3(1,1)-x3(1,2))>0.5*10^-2 | abs(x4(1,1)-x4(1,2))>0.5*10^-2) x1(1,1)=x1(1,2); x2(1,1)=x2(1,2); x3(1,1)=x3(1,2); x4(1,1)=x4(1,2); x1(1,2)=x1(1,1)-0.33333*w*(1+3*x1(1,1)-x2(1,1)-3*x4(1,1)); x2(1,2)=x2(1,1)+0.16667*w*(1-x1(1,2)-6*x2(1,2)-x3(1,1)); x3(1,2)=x3(1,1)+0.16667*w*(1-x2(1,2)-6*x3(1,2)-x4(1,1)); x4(1,2)=x4(1,1)-0.33333*w*(1-3*x1(1,2)-x3(1,2)+3*x4(1,1)); i=i+1; end disp([x1; x2; x3; x4],'Answers are :') disp(i,'Number of Iterations :')
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clear; clc; //after calculating //t=w_m/6000-%pi/360 N=1000; w_m=2*%pi*N/60; t=w_m/6000-%pi/360; printf("time reqd=%.5f s",t); //printing mistake in the answer in book
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###grammar %define token1 token2 %ifdef token1 %ifdef token2 S -> in : out1 %else S -> in : out2 %endif %else %ifdef token2 S -> in : out3 %else S -> in : out4 %endif %endif S -> in : out ###input in ###enum out1 out ###grammar %define token1 %ifdef token1 %ifdef token2 S -> in : out1 %else S -> in : out2 %endif %else %ifdef token2 S -> in : out3 %else S -> in : out4 %endif %endif S -> in : out ###input in ###enum out2 out ###grammar %define token2 %ifdef token1 %ifdef token2 S -> in : out1 %else S -> in : out2 %endif %else %ifdef token2 S -> in : out3 %else S -> in : out4 %endif %endif S -> in : out ###input in ###enum out3 out ###grammar %ifdef token1 %ifdef token2 S -> in : out1 %else S -> in : out2 %endif %else %ifdef token2 S -> in : out3 %else S -> in : out4 %endif %endif S -> in : out ###input in ###enum out4 out
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//Calculations //Part a // 200*1.141 4 //v1(t)=-------------(1- - cos628t) // 3.14 3 // 200*1.141 800*1.141 //v2(t)=----------- - ------------ cos(628t+<(V2/V1)) // 3.14 3*3.14 // //V2/V1|w=0 =0.8;V2/V1|w=628 =6.43*10^-4 <V2/V1|w=628 =180 //v2(t)=72.02+0.0538 cos628t //Part b vrms=0.0538 vdc=sqrt(2)*72.02 r=vrms/vdc //Results printf ("ripple factor is %.2e",r)
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M=10; p=0.7; nb_lancers = 5000; nb_iterations = 15; X = grand(1,nb_lancers,"bin",M,p); function y=F(x,K,U) y=x; if (K>x) &(U<=p) then y=x+1; elseif (K<=x)&(U>p) then y= x-1 else y=x end endfunction for i =1 : nb_iterations for j=1 : nb_lancers K = grand(1,1,"uin",1,M); U = grand(1,1,"unf",0,1); X(j)= F(X(j),K,U); end end histplot(0.5:10.5,X,style=2)
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clc m=1; //kg p=20; //bar T_sup=400; //0C x=0.9; c_ps=2.3; //kJ/kg.K disp("(i) Internal energy of 1 kg of superheated steam") // At 20 bar: From steam tables T_s=212.4; //0C h_f=908.6; //kJ/kg h_fg=1888.6; //kJ/kg v_g=0.0995; //m^3/kg h_sup = h_f+h_fg+c_ps*(T_sup-T_s); v_sup=v_g*(T_sup+273)/(T_s+273); u=h_sup-p*v_sup*10^2; disp("Internal energy=") disp(u) disp("kJ/kg") disp("(ii) Internal energy of 1 kg of wet steam") h=h_f+x*h_fg; u=h-p*x*v_g*10^2; disp("Internal energy=") disp(u) disp("kJ/kg")
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clc //Example 1.4 //Convert 327 miles/hr into ft/s V=327//miles/hr //1 mile = 5280 ft //1 hour = 3600 sec V1=V*(5280/3600)//ft/s printf("327 miles/hr = %f ft/s",V1);
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clc V=0.04; //m^3 p=120*10^5; //Pa T=293; //K R0=8314; disp("(i) kg of nitrogen the flask can hold") M=28; //molecular weight of Nitrogen R=R0/M; m=p*V/R/T; disp("kg of nitrogen=") disp(m) disp("kg") disp("(ii) Temperature at which fusible plug should melt") p=150*10^5; //Pa T=p*V/R/m; //K t=T-273; //0C disp("Temperature =") disp(t) disp("°C")
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2127135785=0 978402736="Jingyuan" 1180449268="Omar" 27322186="The Harrowing of Hell" 319044975="The Desperate Man" 105019561=true 1866459795="Jingyuan" 988370360="Omar" 1076862205="Potsdamer Platz" 29737522="Fitz" 570352297=2 353461707="Jingyuan" 840018027=true 1300594462="Jingyuan" 948989387="Bobby"
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//Example 1_25 clc(); clear; //To find the value of the slit width lemda=6500 //units in angstroam theta=30 //units in degrees a=lemda/sin(theta*%pi/180) printf("The value of the slit is %.0f angstroam",a)
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function [x1,crit]=karmarkar(a,b,c,x0,eps,gamma) maxiter=200; epsdef=1.d-5 // [lhs,rhs]=argn(0) select rhs case 4 then gamma=1/4,eps=epsdef case 5 then gamma=1/4 case 6, else error('[x1,crit]=karmarkar(a,b,c,x0 [,eps [,gamma]])') end //verification des donnees [n,p]=size(a) w=size(b) if w(1)<>n then error('invalid B dimension'),end w=size(c) if w(1)<>p then error('invalid A dimension'),end w=size(x0) if w(1)<>p then error('invalid x0 dimension'),end if mini(x0)<0|norm(a*x0-b)>eps then error('x0 is not feasible'), end // x1=x0;tc=c'; crit=tc*x1; test=eps+1 count=0 while test>eps&count<=maxiter count=count+1 ax=a*diag(x1);xc=x1.*c y=(ax*ax')\(ax*xc) d=-xc+ax'*y dk=x1.*d if mini(dk)>0 then error('Unbounded problem!'),end alpha=-gamma/mini(d) test=alpha*(norm(d)**2)/maxi(1,abs(crit)) x1=x1+alpha*dk crit=tc*x1 write(%io(2),[count,crit,test],'(f3.0,3x,e10.3,3x,e10.3)') end
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clear;lines(0); a=[1,2,3;4,5,6]; a=1,b=1;c=2
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printf("\t example 13.5 \n"); // for a Basis of one Hour printf("\t approximate values are mentioned in the book \n"); c(1)=1544; // Flow rate of CO2, Lb/hr h(1)=4500; // Flow rate of H20, Lb/hr c(2)=35; //Flow rate of CO2, Mol/hr h(2)=250;//Flow rate of H20, Mol/hr t(1)=c(1)+h(1); //Total flow rate , Lb/hr t(2)=c(2)+h(2); //Total flow rate, Mol/hr Pt = (30+14.7)/(14.7); //Total Pressure in atm printf("\t Pt is %.2f\n",Pt); Pw = ( h(2)/t(2) )*Pt; //Partial pressure of Water in atm printf("\t Partial Pressure of Water: %.2f atm \n",Pw); Tw = 267; // from table 7 at 2.68atm Mm = (t(1)/t(2)); printf("\t mean molecular weight : %.1f \n",Mm); // weighted temperature difference // overall balance //for Inlet Pv=2.68; // water vapour pressure, atm Pg=Pt-Pv; // Inert pressure //for Exit Pw1 = 0.1152 // Partial pressure of water at 120 F Pv1 = 0.115; // Water vapor pressure Pg1 = 2.935; // Inert pressure w1 = 250; //Pound mols steam inlet w2 = c(2)*(Pv1/Pg1); printf("\tPound mols steam exit:%.2f\n",w2); w3 = w1 - w2; printf("\tPound mols steam condessed:%.2f\n",w3); //Assume points at 267, 262, 255,225,150,120 deg F //For the interval from 267 to 262 F Pv2 = 2.49; // From table 7 at 262 F Pg2 = Pt - Pv2; //Inert pressure printf("\tPg is %.2f",Pg2); w4 = c(2) * (Pv2/Pg2); //Mol steam remaining w5 = h(2) - w4; //Mol steam condensed printf("\tMol steam remaining:%.0f\n",w4); printf("\tMol steam condensed:%.0f\n",w5); h1 = (w5*18*937.3) + (0.46*(267-262) * w5 * 18); //Heat of condensation h2 = (w4 * 18 * 0.46*(267-262)); //Heat from uncondensed steam h3 = c(1)*0.22*5.0; //Heat from noncondensable printf("\tHeat of condensation:%.2e\n",h1); printf("\tHeat from uncondensed steam:%.2e\n",h2); printf("\tHeat from noncondensable:%.1e\n",h3); ht = h1+h2+h3;//Total heat printf("\tTotal heat:%.0f\n",ht); //Similarily calculating the Heat balance for other intervals printf("\tInterval,F\tTotal Heat\n\t267-262\t1,598,000\n\t262-255\t1,104,000\n\t255-225\t1,172,000\n\t225-150\t751,000\n\t150-120\t177,000\n\tTotal\t4,802,000\n"); w=4802000/(115-80); //Total water printf("\tTotal water: %.2e\n",w); //Water coefficient Nt = 246; at1 = 0.302; n = 4; at = Nt * (at1/(144*n)); // From eq 7.48 printf("\tat is %.3f ft^2\n",at); Gt = w/at; printf("\tGt is %.2e lb/(hr)(ft^2)\n",Gt); ro = 62.5; V = Gt/(3600*ro); printf("\tV is %.2f fps\n",V); hi = 1120; // From fig. 25 ID = 0.62; OD = 0.75; hi0= hi *(ID/OD); //From eq 6.5 printf("\thi0 is %.0f\n",hi0); //Mean properties at 267 F c = ((c(1)*0.22)+(h(1)*0.46))/t(1); // Calculation mistake in Book printf("\tMean c:%.3f Btu/(lb)(F)\n",c); k = ((c(1)*0.0128)+(h(1)*0.015))/t(1); // Calculation mistake in Book printf("\tMean k:%.4f Btu/(hr)(ft^2)(F/ft)\n",k); mu = (((c(1)*0.019)+(h(1)*0.0136))/t(1))* 2.42; // Calculation mistake in Book printf("\tMean mu:%.4f lb/(hr)(ft)\n",mu); ID1 = 21.25; C = 0.25; B = 12; PT = 1.0; as = ID1 * C * (B/(144*PT)); //From eq 7.1 printf("\tas is %.3f ft^2\n",as); Gs = t(1)/as //From eq 7.2 printf("\tGs is %.3e lb/(hr)(ft^2)\n",Gs); Ds = 0.0792; // From Fig 28 Res = Ds * (Gs/0.0363); // From eq 7.3 printf("\tRes is %.2e\n",Res); jH = 102; // From Fig 28 x = ((c*mu)/k)^(1/3); printf("\t(c.mu/k)^1/3 is %.0f\n",x); h0 = jH * 0.0146 * (x/Ds); //From eq 6.15b printf("\th0 is %.0f\n",h0); y = 0.62 // y = (mu/ro * kd)^(2/3) z = 1.01; // z = ((c*mu)/k)^(2/3) K = (h0*z)/(0.407*Mm*y); //KG = K/p0f printf("\tK is %.2f\n",K); //at point 1 Tg = 244; // F tW = 115; delt=(Tg-tW); printf("\t delt is %.0f F \n",delt);
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2_9.sce
//2.9 clc; Vm=150*2^0.5; Vdc=(Vm/(%pi))*(1+cosd(45)); R=30; Load_current_average=Vdc/R; printf("\nAverage Load current = %.2f A", Load_current_average) Vrms=Vm*(((%pi-(%pi/4))/(2*%pi))+(sind(90)/(4*%pi)))^0.5; printf("\nRMS voltage = %.1f V", Vrms) RMS_current=Vrms/R; printf("\nRMS current = %.3f A", RMS_current)
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7_9.sce
clc //initialisation of variables G1= -7800 //cal G2= -24600 //cal G3= -39700 //cal R= 1.987 //cal/mol K T= 25 //C //CALCULATIONS G= G1+G2-G3 Ksp= 10^(-G/(2.303*R*(273.2+T))) //RESULTS printf (' solubility product constant = % 1e ',Ksp)
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Ex6_30.sce
clear all; clc; disp("Scilab Code Ex 6.30 : ") //Given: sigma_y = 250; //MPa t = 12.5; //mm w = 200; //mm h = 225; //mm c = (h/2)+t; I = 82.44*10^6;//mm^4 Mp = 188; //kN //Calculations: sigma_allow = (Mp*10^6*c)/(I); y = (sigma_y*c)/(sigma_allow); //Display: printf("\n\nThe point of zero normal stress = %1.2f mm',y); printf("\nThe Residual Stress distribution is shown in the text book."); //------------------------------------------------------------------------END---------------------------------------------------------------------------------------
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imm equ $55 label equ $3456 port equ $102 rd equ $78 rs equ $34 adc b,a adc rs,a adc rs,b adc rs,rd adc #imm,a adc #imm,b adc #imm,rd add b,a add rs,a add rs,b add rs,rd add #imm,a add #imm,b add #imm,rd and b,a and rs,a and rs,b and rs,rd and #imm,a and #imm,b and #imm,rd andp a,port andp b,port andp #imm,port br label br label(b) br [rs] btjo b,a,jmpdst btjo rs,a,jmpdst btjo rs,b,jmpdst btjo rs,rd,jmpdst btjo #imm,a,jmpdst btjo #imm,b,jmpdst btjo #imm,rd,jmpdst btjop a,port,jmpdst btjop b,port,jmpdst btjop #imm,port,jmpdst btjz b,a,jmpdst btjz rs,a,jmpdst btjz rs,b,jmpdst btjz rs,rd,jmpdst btjz #imm,a,jmpdst btjz #imm,b,jmpdst btjz #imm,rd,jmpdst btjzp a,port,jmpdst btjzp b,port,jmpdst btjzp #imm,port,jmpdst call label call label(b) call [rs] clr a clr b clr rd clrc cmp b,a cmp rs,a cmp rs,b cmp rs,rd cmp #imm,a cmp #imm,b cmp #imm,rd cmpa label cmpa label(b) cmpa [rs] dac b,a dac rs,a dac rs,b jmpdst dac rs,rd dac #imm,a dac #imm,b dac #imm,rd dec a dec b dec rd decd a decd b decd rd dint djnz a,jmpdst djnz b,jmpdst djnz rd,jmpdst dsb b,a dsb rs,a dsb rs,b dsb rs,rd dsb #imm,a dsb #imm,b dsb #imm,rd eint idle inc a inc b inc rd inv a inv b inv rd jc jmpdst jeq jmpdst jge jmpdst jgt jmpdst jhs jmpdst jl jmpdst jlt jmpdst jmp jmpdst jn jmpdst jnc jmpdst jne jmpdst jnz jmpdst jp jmpdst jpz jmpdst jz jmpdst lda label lda label(b) lda [rs] ldsp mov a,b mov a,rd mov b,a mov b,rd mov rs,a mov rs,b mov rs,rd mov #imm,a mov #imm,b mov #imm,rd movd rs,rd movd #imm(b),rd movd #imm,rd movp a,port movp b,port movp port,a movp port,b movp #imm,port mpy b,a mpy rs,a mpy rs,b mpy rs,rd mpy #imm,a mpy #imm,b mpy #imm,rd nop or b,a or rs,a or rs,b or rs,rd or #imm,a or #imm,b or #imm,rd orp a,port orp b,port orp #imm,port pop a pop b pop rd pop st push a push b push rd push st reti rets rl a rl b rl rd rlc a rlc b rlc rd rr a rr b rr rd rrc a rrc b rrc rd sbb b,a sbb rs,a sbb rs,b sbb rs,rd sbb #imm,a sbb #imm,b sbb #imm,rd setc sta label sta label(b) sta [rs] stsp sub b,a sub rs,a sub rs,b sub rs,rd sub #imm,a sub #imm,b sub #imm,rd swap a swap b swap rd trap 00 trap 01 trap 02 trap 03 trap 04 trap 05 trap 06 trap 07 trap 08 trap 09 trap 10 trap 11 trap 12 trap 13 trap 14 trap 15 trap 16 trap 17 trap 18 trap 19 trap 20 trap 21 trap 22 trap 23 tsta tstb xchb a xchb b xchb rd xor b,a xor rs,a xor rs,b xor rs,rd xor #imm,a xor #imm,b xor #imm,rd xorp a,port xorp b,port xorp #imm,port
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file while in test branch
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Ex10_14.sce
clc clear printf("Example 10.14 | Page number 363 \n\n"); //Find the flow rate of feed water into the heater. //Given data m1 = 0.2 //kg/s p = 4 //bar //Solution //From superheated steam table h1 = 2752.8 //kJ/kg h2 = 209.31 //kJ/kg h3 = 604.73 //kJ/kg m2 = (m1*h1-m1*h3)/(h3-h2) //kg/s printf("The flow rate of feed water into the heater = %.3f kg/s",m2)
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39_03.sce
//Problem 39.03: A 2000 pF capacitor has an alternating voltage of 20 V connected across it at a frequency of 10 kHz. If the power dissipated in the dielectric is 500 μW, determine (a) the loss angle, (b) the equivalent series loss resistance, and (c) the equivalent parallel loss resistance. //initializing the variables: P = 500E-6; // in Watt C = 2000E-12; // in Farads V = 20; // in Volts f = 10000; // in Hz //calculation: //power loss = w*C*V^2*tan(del) //loss angle del = atan(P/(2*%pi*f*V*V*C)) //for an equivalent series circuit, //tan(del) = (Rs*w*Cs) Cs = C Cp = C Rs = (tan(del))/(2*%pi*f*Cp) //for an equivalent parallel circuit //tan(del) = 1/(Rp*w*Cp) Rp = 1/(2*%pi*f*Cp*tan(del)) printf("\n\n Result \n\n") printf("\n (a)loss angle %.6f rad.",del) printf("\n (b)series resistance %.2f ohm.",Rs) printf("\n (c)parallel resistance %.2E ohm.",Rp)
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clc //initialisation of variables SR= 0.5 //V/us Vcon= 12 //V //CALCULATIONS f= SR*1000/(2*%pi*Vcon) //RESULTS printf ('full power = %.2f kHz ',f)
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//(x^2)/(a^2) + (y^2)/(b^2)=1 a=1; b=2; x=-1:.0001:1; y=(b/a)*sqrt((a^2)-(x^2)); plot(x,y) plot(x,-y) legend('(x^2)/(a^2) + (y^2)/(b^2)=1') title('ELLIPSE')
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syms G1 G2 G3 G4 G5 H2 H3; T1=G1*G3*G2; T2=G1*G3*G5; T3=G4*G2*G3; T4=-G4*G2*G5*G3*H2 L1=-G2*H3; L2=-G3*H3; L3=-G5*H2*H3*G3; L4=-G1*G2*G3; L5=-G1*G3*G5; L6=-G2*G3*G4; L7=G2*G3*G4*G5*H2; delta=1-(L1+L2+L3+L4+L5+L6+L7) del1=1; del2=1; del3=1 del4=1; TF=(T1*del1 + T2*del2 + T3*del3 + T4*del4)/delta ; disp(TF,"C/R = ")
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//Caption:Calculate (i)-input power in watts,(ii)-output power in watts, (iii)-efficiency //Exa:8.3 clc; clear; close; n=2; V_o=300;//in volts I_o=20*10^-3;//in A V_i=40;//in volts J=1.25;//J(X') P_dc=V_o*I_o; P_ac=2*V_o*I_o*J/(2*n*%pi-%pi/2); eff=(P_ac/P_dc)*100; disp(P_dc,'Input power (in watts) ='); disp(P_ac,'Output power (in watts) ='); disp(eff,'Efficiency (in percent) =');
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//Detemine the load and pf of the other machine clc; clear; Pa=3000*(10^3);// Lighting load Pma=5000*(10^3); // Aggregate Motor load pfm=0.71; // power factor of motor load P1a=5000*(10^3); // One Machine load pf1=0.8;// Power factor machine 1 (lagging) Pta=Pa+Pma; // Total load active power requirement // Reactive power Pr=0; // Lighting Pmr=Pma*tand(acosd(pfm)); // Motor P1r=P1a*tand(acosd(pf1)); // Machine 1 P2a=Pta-P1a; // Active power by other machine P2r=Pr+Pmr-P1r; // Reactive power by other machine pf2=cosd(atand(P2r/P2a)); // Power factor of other machine printf('The other machine supplies:\n') printf(' A load of %g kw at a p.f of %g\n',P2a/1000,pf2)
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//Example 19_5 clc(); clear; //To find the time constant of the circuit and the final energy stored l=0.5 //Units in H r1=2 //Units in Ohms r2=4 //Units in Ohms r=r1+r2 //Units in Ohms l_r=l/r //Units in sec i=2 //Units in A ene=0.5*l*i^2 printf("The time constant is L/R=%.4f Sec\n The energy stored is=%d J",l_r,ene)
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--> A=[5 3 -11; 4 -5 4; 3 -13 19] A = 5. 3. -11. 4. -5. 4. 3. -13. 19. --> [L,U,P]=lu(A) L = 1. 0. 0. 0.6 1. 0. 0.8 0.5 1. U = 5. 3. -11. 0. -14.8 25.6 0. 0. 1.421D-15 P = 1. 0. 0. 0. 0. 1. 0. 1. 0.
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function[y]=cmd_shift_in_xcos(dataPin,clockPin,latchPin,ledPin,clockLed) dataPinout=O; cmd_digital_out(1,latchPin,1)// parallel load mode for j=1:10 //to perform 10 iterations of parallel loading disp('Give input, Parallel load mode') sleep(2000) cmd_digital_out(1,clockPin,1) //positive edge of clock pulse disp('Inputs stored, Serial shift mode:') sleep(500) cmd_digital_out(1,clockPin,0)//negative edge of clock pulse cmd_digital_out(1,latchPin,0)// serial out mode value=zeros(1,8);//matrix representing 8 bit number value2=zeros(8,8); for i=1:8 dataPinout=cmd_digital_in(1,dataPin);//reads input at the dataPin given by the register as its LSB disp(dataPinout); if(dataPinout==1) cmd_digital_out(1,ledPin,1); //Led glows sleep(100); else if(dataPinout==0) cmd_digital_out(1,ledPin,0); sleep(100); end end value2(i,i)=cmd_digital_in(1,dataPin); //sets the (i,i)th element of 8x8 zeros matrix as the logic level of the serial data out of register value = value | value2((i),:); cmd_digital_out(1,clockPin,1); cmd_digital_out(1,clockLed,1); sleep(100); cmd_digital_out(1,clockPin,0);cmd_digital_out(1,clockLed,0); end cmd_digital_out(1,latchPin,1); y=value; disp(y); end endfunction
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clc //Intitalisation of variables clear R= 1.987 //cal T= 25 //C p= 23.76 //mm //CALCULATIONS dF= R*(273.2+T)*log(760/p) //RESULTS printf ('Free energy change = %.f cal mole^-1',dF+1)
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clc pathname=get_absolute_file_path('9_3_1.sce') filename=pathname+filesep()+'931.sci' exec(filename) printf(" All the values in the textbook are Approximated hence the values in this code differ from those of Textbook") Hr=5*HCO2+6*HH2O-HC5H12 printf(" \n Heat of the rxn= %f KJ/mol",Hr)
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TABLE_SIZE=256 ANNOTATIONS=1 QIF=$(cat<<'EOQ' dude nude dude nude dude where is my car? ## t 0 one fish two fish red fish blue fish EOQ )
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//Optoelectronics - An Introduction, 2nd Edition by J. Wilson and J.F.B. Hawkes //Example 5.5 //OS=Windows XP sp3 //Scilab version 5.5.2 clc; clear; //given lambda=0.84e-6;//wavelength in m DeltaNu=1.45e13;//Transition linewidth in Hz Gamma=3.5e3;//Loss coefficient in m^(-1) n=3.6;//Refractive index of GaAs medium n1=1;//Refractive index of air medium l=300e-6;//Length in m d=2e-6;//Diameter in m etai=1;//Internal quantum efficiency e=1.6e-19;//Electronic charge in C R=((n-n1)/(n+n1))^2;//Reflectance at GaAs/air interface by Fresnel equation mprintf("\n R = %.2f",R); Kth=Gamma+1/(2*l)*log(1/R^2);//Threshold gain in m^(-1) mprintf("\n Kth = %.1f m^(-1)",Kth);//The answers vary due to round off error Jth=8*%pi*e*d*DeltaNu*(n^2)/(etai*(lambda^2))*Kth;//Threshold current density in A m^(-2) mprintf("\n Jth = %.1f A mm^-2",Jth/1e6);//Dividing by 10^6 to convert into A mm^(-2) //The answers vary due to round off error
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//ex12.1 A_ol=100000; //open loop voltage gain A_cm=0.2; //common mode gain CMRR=A_ol/A_cm; CMRR_dB=20*log10(CMRR); disp(CMRR,'CMRR') disp(CMRR_dB,'CMRR in decibels')
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clear //Given l = 400 //mm - Length b = 300 //mm - breath F = 20 //KN _ the force applied on the beam F_d = 0.75 //KN-m - The force distribution d = 2 //mt - the point of interest from the free end //calculations //From moment diagram M = F*d - F_d*d*1 I = b*(l**3)/12 //mm4 - Bending moment diagram c = l/2 // the stress max at this C S = I/c //The maximum shear stress shear_max = M*(10**6)/S //MPa - the maximum stress printf("\n The maximum stress at 2 mt is %0.2f MPa",shear_max)
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//Page Number: 9.15 //Example 9.8 clc; //Given p=0.99; u=1; q=1-p; //As exp(-Ac^2/4*n*B)=1-p //AndAC^2/2*n*B=S/N //Therefore exp(-(1/2)*(S/N))=1-p SN=2*(log(1/q)); SN1=(round(SN)+1); //Upper limit disp('db',SN1,'S/N:'); //Hence proved
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clc //Initialization of variables D = 1.94 // slugs/ft^3 A1 = 0.06 //ft^2 V1 = 10 // ft/s // Calculations Fax = -D*A1*(V1)^2 Faz = D*A1*(V1)^2 // Results printf (" the force required in x direction is %.2f(1-cosT) lb",Fax) printf ("\n the force required in z direction is %.2f(sinT) lb",Faz)
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format(12) function dy = f(t,y) dy = y - t^2 + 1 return dy endfunction function yt = y(t) yt = (t+1)^2 -0.5*%e^t return yt endfunction clc disp("> La aproximación a la solución de la ecuación diferencial") disp(" usando el método de Euler es: ") E = UN_ecua_dif_Euler(0,0.5,1,10) disp(E) disp("> La aproximación a la solución de la ecuación diferencial") disp(" usando el método de Runge-Kutta es: ") RK4 = UN_ecua_dif_RK4(0,0.5,1,10) disp(RK4) disp("> El valor de la solución exacta de la ecuación diferencia es: ") disp(y(1))
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errcatch(-1,"stop");mode(2); // Example 1.4.b :correction , // given : vm=42; // pressure in bar vt=41.4; // pressure in bar Es=vm-vt; Cs=-Es; disp(Cs,"static corrction,Cs = (bar)") exit();
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// PG (334) // dy/dt=-y function ydot=f(y,t),ydot=-y, endfunction y0=0;t0=0;t=0:1:%pi; y=ode(y0,t0,t,f) plot(t,y)