Hidroestatica

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,

∆x1 = v1∆t ∆x2 = v2 ∆t

∆m1 = ρ A1v1∆t
dV

∆m2 = ρ A2 v2 ∆t
dV

Por conservacio′n de la masa : ∆m1 = ∆m2A1v1 = A2 v2 ecuacio′n de la continuidad

Av = constante.
Q = Av donde Q es el gasto o caudal.

# ! . ! /

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$

De acuerdo al teorema del trabajo y la energia 1 1 2 P A1 (∆x1 ) − P2 A2(∆x2 ) − ∆mg ( y2 − y1 ) = ∆mv2 − ∆mv12 1 2 2

P A(∆x ) − P2 A(∆x ) − ∆mg ( y2 − y1 ) = 1

1 1 2 ∆mv2 − ∆mv12 2 2

∆m = ρ∆V = ρA∆x
1 1 2 P A(∆x ) − P2 A(∆x ) − ρA∆xg ( y2 − y1 ) = ρA∆xv2 − ∆mv12 1 22

P A(∆x ) − P2 A(∆x ) − ρA∆xg ( y2 − y1 ) = 1 P − P2 − ρg ( y2 − y1 ) = 1
0

1 1 2 ρA∆xv2 − ρA∆xv12 2 2 1 2 1 2 ρv2 − ρv1 2 2

1 2 1 2 P + ρv1 + ρgy1 = P2 + ρv2 + ρgy2 1 2 2 1 2 P + ρv +ρgy = constante 2

"

1 2 4 5 " 6"04 3 3

2

,

!

1 2 P + ρv = constante 2
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(P3 − P4 ) = ρ ′gh
P+ 1

(1)

ρ ρ

1 2 1 ρv1 = P2 + ρv 2 2 2 2

1 (P1 − P2 ) = ρ 2

A1A2

2

−1 v

2 1

(2)

(

P3 = P + ρgy1 1 P4 = P2 + ρgy2
0 7 )8 7 (8

(3) (4)
2 1 A12 − A2 2 ρ ′gh = ρ v1 + ρgh 2 A2 2

(P3 − P4 ) = (P1 − P2 ) + ρg ( y1 − y2 )

v1 = A2

2 gh(ρ′ − ρ ) ρ A12 − A22

(

)

Aplicando Bernoulli entre (1) y(2) : 1 P0 + ρv12 + ρgy1 = P0 + ρgy2 2

+

1 2 ρv1 = + ρgy2 − ρgy1 2 1 2 ρv1 = + ρgh 2

+

1 2 ρv1 = + ρg ( y2 − y1 ) 2

+v1 = 2 gh

"

1 2 1 2 Pa + ρva = Pb + ρvb 2 2 Pb = Pa + 1 2 ρva 2

=0

Pb = Pa + ρ ′gh
1 2 ρva = ρ ′gh 2

va =

2 ρ ′gh

ρ

P+ 1

1 2 1 2 ρv1 + ρgH = P2 + ρv2 + 0 2 2

P =3atm = 3.04 ×105 Pascal 1
P+ 1
9(
v2 =

1 2 1 2 ρv1 + ρgH = P2 + ρv2 2 2
2 P − 2 P2 + 2 gH 1

ρ

=

2 P − 2 ρgh + 2 ρgH 1

ρ

v2 =

2 P + 2 ρg (H − h ) 1

ρ

2 3.04 ×105 +2(1000)(9.8)(5 − 0.5) m/s = 1000

(

)

v2 = 25.5 m

s

Q = Av2 =

πd 2
4

v2 =

π (0.03m )2
4

m3 25.5 m = 0.018 s s

:

1

Av = av2
9; #<

A −

dy = a 2 gy dt

#

v2...