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daniel-ramirez
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%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