%PROGRAMA P121%FECHA 30/05/2015%RAMIREZ PEREZ DANIEL%TALLER DE MECANICA DE FLUIDOS%PERDIDA DE CARGA EN TUBERIAS % CASO SECCION CONSTANTE%CARGA ESTATICA (P1/RHOG)=CE1%CARGA ESTATICA (P2/RHOG)=CE2%CARGA GOEDESICA EN 1 Z1=CG1%CARGA GEODESICA EN 2 Z2=CG2%CARGA DINAMICA EN 1 CD1=V^2/2GR=(8*Q1^2)/(PI^2*D1^4*G)%CARGA DINAMICA EN 1 CD2=V^2/2GR=(8*Q2^2)/(PI^2*D2^4*G)%ENERGIA TOTAL EN 1; E1=CE1+CD1+CG1%ENERGIA TOTAL EN 2; E2=CE2+CD1+CG2%ENERGIA PERDIDA ENTRE 1 Y 2 EP12=E1=E2%Q=CAUDAL%COEFICIENTE DE RESISTENCIA R=EP12/(Q^2)%FACTOR DE FRICCION F=((G*(PI^2)*(D^5))/(8*L12))*R%DATOS%******************************************%INTERPOLAH1=87;Q1=0.75;H2=88;Q2=0.78;HX=87.2;7TX=21.8QX=Q1+((HX-H1))*(Q2-Q1)/(H2-H1)QINT=QX*(10^-3)Q=QINTQ1=Q;Q2=Q;PI=3.1416;D1=0.075;D2=0.075;G=9.81;L12=2;CE1=226.5/1000;CE2=226.35/1000;CG1=0;CG2=0;D=D1;CD1=(8*Q1^2)/(PI^2*D1^4*G)CD2=(8*Q2^2)/(PI^2*D2^4*G)E1=CE1+CD1+CG1E2=CE2+CD1+CG2EP12=E1-E2R=EP12/(Q^2)F=((G*(PI^2)*(D^5))/(8*L12))*RNU=1.007*(10^-6)-((0.203)*(TX-20)*(10^-6)/10)RE=(4*Q)/(PI*D*NU)syms xif RE> P121

ans =

7

TX =

21.8000

QX =

0.7560

QINT =

7.5600e-04

Q =

7.5600e-04

CD1 =

0.0015

CD2 =

0.0015

E1 =

0.2280

E2 =

0.2278

EP12 =

1.5000e-04

R =

262.4507

F =

0.0038

NU =

9.7046e-07

RE =

1.3225e+04

FU = 1/x^(1/2) - (2*log((1817614736827489*x^(1/2))/137438953472))/log(10) + 4/5 >> fsolve('1/x^(1/2) - (2*log((1817614736827489*x^(1/2))/137438953472))/log(10) + 4/5',0.1)

Equation solved.

fsolve completed because the vector of function values is near zeroas measured by the default value of the function tolerance, andthe problem appears regular as measured by the gradient.

ans =

0.0287