8/19/2019 1 -E -JAVIER

http://slidepdf.com/reader/full/1-e-javier 1/5

1

  e)

ECUACIÓN DE CHEZY V = C   .√  RS   entonces C =

V 

√  RS

  Q = V . A ;  ø deltubo = 80 mm =0.08 m ; A = π . D

2

4   = π

82

4   = 0.0016π   m2

; longt!" "e c#"# t$#mo% 1&0cm =1.& m

 'E(DIDA DE CA(A '*( +(ICCI*N, hf 

hf  A→B= P A

γ   + PB

γ 

EN E- '(IE( CAUDA-

  V =1.66∗0.001

0.0016   = 0.//m

s   ;

hf 1→2

=

 P1

γ   +

 P2

γ   =221.5−220.9=0.6cm  = 0.006 m

=hf 

1→2

 L  =0.006

1.2  = 0.00

C =0.33

√ 0.08∗0.005  = 16. m1

2   s−1

8/19/2019 1 -E -JAVIER

http://slidepdf.com/reader/full/1-e-javier 2/5

V =1.66∗0.001

0.0016   = 0.//m

s   ;

hf 3→4

=

 P1

γ   +

 P2

γ   =221.5−220.9=20cm=0.2m

=hf 

3→4

 L  =0.2

1.2  = 0.16

C =0.33

√ 0.08∗0.16  = &.2& m1

2   s−1

EN E- EUND* CAUDA-

V =1.96∗0.001

0.0016   = 0./2m

s   ; hf 1→2

= P

1

γ   + P

2

γ  =0.9cm=0.09m

=hf 

1→2

 L  =0.09

1.2  = 0.03

C =0.39

√ 0.08∗0.075  = .0/ m1

2   s−1

V =1.96∗0.001

0.0016   = 0./2m

s   ; hf 3→4

= P

1

γ   + P

2

γ   =26.5 cm=0.27

=

hf 3→4

 L  =

0.27

1.2  = 0.&&

C =0.39

√ 0.08∗0.225  = &.20 m1

2   s−1

EN E- 4E(CE( CAUDA-

8/19/2019 1 -E -JAVIER

http://slidepdf.com/reader/full/1-e-javier 3/5

V =2.28∗0.001

0.0016   = 0.5m

s   ; hf 1→2

= P

1

γ   + P

2

γ  =1cm=0.01 m

=hf 

1→2

 L  =0.01

1.2  = 0.008

C =0.45

√ 0.08∗0.008  = m1

2   s−1

V =2.28∗0.001

0.0016   = 0.5m

s   ; hf 3→4

= P

1

γ   + P

2

γ   =35.3cm=0.35m

=hf 

3→4

 L  =0.35

1.2  = 0.&2

C =0.45

√ 0.08∗0.29  = &.2 m1

2   s−1

EN E- CUA(4* CAUDA-

V =2.58∗0.001

0.0016   = 0.1m

s   ; hf 1→2

= P

1

γ   + P

2

γ  =1.4cm=0.014m

=hf 

1→2

 L  =0.014

1.2  = 0.01&

C =1.61

√ 0.08∗0.012  = 1.26 m1

2   s−1

V =2.58∗0.001

0.0016   = 0.1

m

s   ; hf 3→4=

 P1

γ   +

 P2

γ   =44.7 cm=¿ 0.5 m

8/19/2019 1 -E -JAVIER

http://slidepdf.com/reader/full/1-e-javier 4/5

=hf 

3→4

 L  =0.45

1.2  = 0./8

C =

0.51

√ 0.08∗0.38  = &.2/ m

1

2

 s−1

EN E- QUI4* CAUDA-

V =3.64∗0.001

0.0016   = 0.3&m

s   ; hf 1→2

= P

1

γ   + P

2

γ  =2.4cm=0.024m

=hf 

1→2

 L  =0.024

1.2  = 0.0&

C =0.72

√ 0.08∗0.02  = 18 m1

2   s−1

V =

3.64∗0.001

0.0016   = 0.3&

m

s   ; hf 3→4=

 P1

γ   +

 P2

γ   =80.9

cm=0.9

m

=hf 

3→4

 L  =0.9

1.2  = 0.3

C =0.72

√ 0.08∗0.75  = &.25 m1

2   s−1

EN E- E4* CAUDA-

V =5.05∗0.001

0.0016   = 1m

s   ; hf 1→2

= P

1

γ   + P

2

γ   =4.1cm=0.041m

=hf 

1→2

 L  =0.041

1.2  = 0.0/5

8/19/2019 1 -E -JAVIER

http://slidepdf.com/reader/full/1-e-javier 5/5