tarea de flujo
Post on 15-Feb-2016
214 Views
Preview:
DESCRIPTION
TRANSCRIPT
Q H0 21
0.0566 180.0725 16.50.0858 150.0978 13.50.1082 12
0.116 10.50.124 9
0.1298 7.50.1339 6
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.160
5
10
15
20
25
f(x) = − 796.529207873027 x² + 0.601147500975311 x + 20.865688460461R² = 0.995180776500963
H vs Q
HPower (H)Polynomial (H)
Q(m^3/s)
H(m
)
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.160
5
10
15
20
25
f(x) = − 796.529207873027 x² + 0.601147500975311 x + 20.865688460461R² = 0.995180776500963
H vs Q
HPower (H)Polynomial (H)
Q(m^3/s)
H(m
)
L: A-j L: j-B L: j-C D e Qa Hbomba H piezom. 0.0566
Q H0 21
0.0566 180.0725 16.50.0858 150.0978 13.50.1082 12
0.116 10.50.124 9
0.1298 7.50.1339 6
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.160
5
10
15
20
25
f(x) = − 796.529207873027 x² + 0.601147500975311 x + 20.865688460461R² = 0.995180776500963
H vs Q
HPower (H)Polynomial (H)
Q(m^3/s)
H(m
)
L: A-j Za L: j-B Zb L: j-C Zc D e30 5 0.25
Qa Hbomba Vj Re f hf A H. piezom. hfB0.0566 18.3482906 1.153043035 288260.759 0.02 0.16263098 #VALUE! H piezom. + vj - zb 0.0725 16.7228189
Re(raiz de f ) luego con R - f y se halla vB H piezom. + vj - zb
top related