cálculo toma lateral
TRANSCRIPT
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CÁLCULO HIDRAULICO DE UN ALIVIADERO - ALCANTA
DATOS DEL PROBLEMA:UNIDAD
Q = 300 Lt/SegElevac A = 94.60 !Elevac B = 94.20 !
0.40 !
"# = 0.80
!"$ = 0.30 !% = 0.023L = 3.00 !
DESARROLLO
& = ' (' m
A = ' #)#
E*t+*ce,:
dh
L = 3.00 m
Elev.A d 1
Elev.B
Δh=0.40m
h = d 1
− d 2
+ Δh − H t
h = 0 . 80 − 0 . 30 + 0 . 40 − 0 . 10
Q=0.5 A√ 2gh, despej
A=0.300 .5√ 19.62(0.80)
A= π x D2
4,remplazando :
√ Ax 4π = D , quedando :
dh
L = 3.00m
Elev. A d 1
Elev.B d 2
h +HtΔh= 0.40m
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D = ' . ! 0 1
A= π x D
2
4,remplazando :
√ Ax 4π = D , quedando :
V 2
2 g=
1.9872
19.62= 0.201
he= 0 . 5 V 2
2 g=0 . 5 (0.201 )=0.101
hf =f L D(V
2
2g)=0.023 3.000.439(0.201)=0.032h s= 0 . 2 he= 0 . 2 (0.101 )=0.0202 H t = he+hf +h s= 0.101 +0.032 +0.0202 H t = 0.153 m
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& = ' 22 !
he=0 . 5 V 2
2 g=0 . 5 (0.173 )=0.087 m
hf =f L D (V 22g)=0.023 3.000.456 (0.173)=0.02
h s= 0. 2 he= 0 . 2 (0.087 )= 0 . 0174 m H t = h e+hf +h s= 0.087 +0.026 +0.017 H t = 0 .13 m
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RILLA
!$
d 2
h + H t
5 A√ 2gh, despejando A:305√ 19.62(0.80)
=0.151m2
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&e = 0.085 m
& % = 0.025 m
& , = 0.016985 m
H t = ' #. !
Q = ' .$ !./,
)=0.087 m
056
(0.173)=0.026m
)= 0 . 0174 m+0.026 +0.0174
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34lg
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