Hidraulica

CÁLCULOS BOMBA CENTRÍGUFA

Determinación de carga de la bomba desarrollada:
HB=Pd-PBɣ
HB1=344750+45717.759810=39.80m
HB2=344750+ 45717.759810=39.80m
HB3=344750+ 474119810=39.98m
HB4=344750+49104.259810=40.14m
HB5=344750+ 49104.259810=40.14m
HB6=344750+ 49104.259810=40.14m
HB7=330960+ 49104.259810=38.74m
HB8=317170+ 52490.759810=37.68m
HB9=310275+ 541849810=37.15m

Determinacióndel caudal que circula por el sistema:
Q=0.60.01(Hvertedero-30.3)5/2
Q1=0.60.01(0-30.3)5/2=0
Q2=0.60.01(37.1-30.3)5/2=0.72 lts/s = 0.00072m3/s
Q3=0.60.01(38.2-30.3)5/2=1.05 lts/s = 0.00105m3/sQ4=0.60.01(39.9-30.3)5/2=1.71 lts/s = 0.00171m3/s
Q5=0.60.01(40.3-30.3)5/2=1.89 lts/s = 0.00189m3/s
Q6=0.60.01(41.8-30.3)5/2=2.69 lts/s = 0.00269m3/s
Q7=0.60.01(42.4-30.3)5/2=3.06 lts/s = 0.00306m3/sQ8=0.60.01(42.5-30.3)5/2=3.12 lts/s = 0.00312m3/s
Q9=0.60.01(42.6-30.3)5/2=3.18 lts/s = 0.00318m3/s

Calculo de la potencia hidráulica desarrollada por la bomba:
PHID=ɣQHB
PHID1=9810039.80=0PHID2=98100.0007239.80=281.11watts
PHID3=98100.0010539.98=411.81watts
PHID4=98100.0017140.14=673.35watts
PHID5=98100.0018940.14=744.23watts
PHID6=98100.0026940.14=1059.25wattsPHID7=98100.0030638.74=1162.92watts
PHID8=98100.0031237.68=1153.28watts
PHID9=98100.0031837.15=1158.92watts

Cálculo de la eficiencia de la bomba:
ɳB=PBombaPEléctrica*100%
ɳB1=02400*100%=0ɳB2=281.112400*100%=11.71
ɳB3=411.812400*100%=17.16
ɳB4=673.352400*100%=28.06
ɳB5=744.232400*100%=31.01
ɳB6=1059.252400*100%=44.14
ɳB7=1162.922400*100%=48.46
ɳB8=1153.282400*100%=48.05
ɳB9=1158.922400*100%=48.29Calculo de Q2
Q21=(0)2=0
Q22=(0.72)2=0.5184
Q23=(1.05)2=1.1025
Q24=(1.71)2=2.9241
Q25=(1.89)2=3.5721
Q26=(2.69)2=7.2361
Q27=(3.06)2=9.3636
Q28=(3.12)2=9.7344
Q29=(3.18)2=10.1124

GRAFICASVemos en la grafica mostrada que:
H0= 39.415

Vemos en la grafica que:
k = m = -2.7794

Ecuación General de la carga:
HB = H0 – kQ2

Ecuación de la carga encontrada:
HB= 39.415-2.7794Q2...