practica tuberias

8/19/2019 Practica Tuberias

1/26

Tablas de datos experimentales

Tubo recto

CorridaGv

l

min

Tramo C-D

CCl4

∆ H cm

Tramo I-J

CCl4

∆ H cm

Tramo M-NHg∆Hcm

1 9 1.7 1.5 0.72 12 3 2.5 1.23 16 5.1 4.5 1.94 20 7.6 5.5 2.7

Ramal de accesorios

CorridaGv

litro!mi"

Tramo #-$Codo %Hg&

'H (m

Tramo )-*Com+,rta%C

Cl4&'H (m

Tramo G-HGloo %Hg&

'H (m

Tramo /-(to%CCl4&'H (m

1 0.2 4.5 0. 52 10 0. 6.5 1.1 7.53 12 1 10.5 1.4 114 16 1.3 17.3 2.7 19.3

c

m p g

g H P )( ρ ρ −∆=∆

ara

∆ P

d (ada dato


2/26

• IN)# D) T$/ )CT/

Tramo C-D %CCl4&

2

2

2

33

2

1 945.981.9

81.9

)10001585(107.1 m

kgf

skgf

mkg

s

m

m

kg

m

kg

m x P =

−

−−=∆

−

2

2

2

33

2

2 55.17

81.9

81.9

)10001585(100.3m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

3 83.29

81.9

81.9

)10001585(101.5m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

4 4.44

81.9

81.9)10001585(10.7

m

kgf

skgf

mkg sm

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

1 77.8

81.9

81.9

)10001585(105.1m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

Tramo I-J %CCl4&


3/26

2

2

2

33

2

2 2.14

81.9

81.9

)10001585(105.2m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

3 32.2

81.9

81.9)10001585(105.4

m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

4 17.32

81.9

81.9

)10001585(105.5

m

kgf

skgf mkg

s

m

m

kg

m

kg m x P =

−−

−=∆ −

Tramo M-N %Hg&

2

2

2

33

2

1 2.88

81.9

81.9)10001300(107.0

m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

2 2.151

81.9

81.9

)10001300(102.1m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−


4/26

2

2

2

33

2

3 4.239

81.9

81.9

)10001300(109.1m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

4 2.340

81.9

81.9)10001300(107.2

m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

•

#M# D) #CC)I/I/.

2

2

2

33

2

1 2.25

81.9

81.9

)10001300(102.0m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

Tramo #-$ (odo %Hg&

2

2

2

33

2

2 8.100

81.9

81.9

)10001300(108.0m

kgf

skgf

mkg sm

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

3 12

81.9

81.9

)10001300(101m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−


5/26

2

2

2

33

2

4 8.13

81.9

81.9

)10001300(103.1m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

Tramo )-* Com+,rta %CCl4&

2

2

2

33

2

1 32.2

81.9

81.9

)10001585(105.4

m

kgf

skgf mkg

s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

2 02.38

81.9

81.9

)10001585(105.m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

3 42.1

81.9

81.9)10001585(105.10

m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

4 20.101

81.9

81.9

)10001585(103.17m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−


6/26

Trato G-H Gloo %Hg&

2

2

2

33

2

1 8.100

81.9

81.9

)10001300(108.0m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

2 .138

81.9

81.9)10001300(101.1

m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

3 4.17

81.9

81.9)10001300(104.1

m

kgf

skgf

mkg sm

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

4 2.340

81.9

81.9)10001300(107.2

m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

1 2.29

81.9

81.9

)10001585(105m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

Tramo /- %CCl4&

2

2

2

33

2

2 87.43

81.9

81.9

)10001585(105.7m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−


7/26

2

2

2

33

2

3 35.4

81.9

81.9

)10001585(1011m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

2

2

2

33

2

4 9.112

81.9

81.9)10001585(103.19

m

kgf

skgf

mkg s

m

m

kg

m

kg m x P =

−

−−=∆

−

Tubo recto

CorridaGv

l

min

Tramo C-D

CCl4

∆ P kgf

m2

Tramo I-J

CCl4

∆ P kgf

m2

Tramo M-N%Hg&

∆ P kgf

m2

1 9 9.9 .7 .22 12 17.5 14.6 151.23 16 29. 26.3 239.4

4 20 44.4 32.1 340.2

Ramal de accesorios

CorridaGv

litro!mi"

Tramo #-$Codo %Hg&

∆ P kgf

m2

Tramo )-*Com+,rta%C

Cl4&

∆ P kgf

m2

Tramo G-HGloo %Hg&

∆ P kgf

m2

Tramo /-(to%CCl4&

∆ P kgf

m2

1 25.2 26.3 100. 29.22 10 100. 3.0 13.6 43.3 12 126 61.4 176.4 64.34 16 163. 101.2 340.2 112.9


8/26

T$/ )CT/ D) #-$ 8 0.619m

T$/ )CT/ D) )-* 8 1.074 m

T$/ )CT/ D) G-H 8 0.194m

Cálculos

Tubo recto

Serie de cálculo para el tramo C-D CCl4

ρCCl4

=1585 Kg

m3

ρ H 2

O=1000 Kg

m3

g=9.81m

s2 gc=9.81

Kg∗m

Kgf ∗s2

L=1.5m di=0.0254m μ H 2

O=0.001 Kg

m3

εhierro fundido=0.26mm

G#T/ /M:TIC/ %m3!&

Gv1=9

l

min∗1

m

3

1000 l ∗1min

60 s =1.5∗10−4

m3

s

Gv2=12

l

min∗1m3

1000 l ∗1min

60s =2.0∗10−4

m3

s

Gv3=16

l

min∗1m3

1000 l ∗1min

60s =2.7∗10−4

m3

s


9/26

Gv4=20

l

min∗1m3

1000 l ∗1min

60s =3.3∗10−4

m3

s

((i;" tra"vral i"tr"a d la t,r


10/26

N>mro d ?"old

ℜ=di∗ ∗ ρ H

2O

μ H 2O

ℜ1=

0.0254m∗0.2960 ms ∗1000 Kg

m3

0.001 Kg

m3

=7519.13

ℜ2=

0.0254m∗0.3947m

s∗1000

Kg

m3

0.001 Kg

m3

=10025.8

ℜ3=

0.0254 m∗0.5262m

s∗1000

Kg

m3

0.001 Kg

m3

=13367.7

ℜ4=

0.0254m∗0.6578 m

s∗1000

Kg

m3

0.001 Kg

m3

=16709.6

,goidad lativa

!= ε

di=

0.26mm

25.4mm=0.010236

!bteniendo el "actor de darc#$ con a#uda de Re # R

f 1=0.045 f 2=0.04 5 f 3=0.045 f 4=0.045


11/26

Ca


12/26

∆ P4=

f ∗ L∗ 2

2∗di∗g ∗ ρ H

2O∗g

gc=

0.045∗1.5m∗(0.6578 ms )2

2∗0.0254m∗9.81 m

s2

∗1000 Kg

m3∗9.81

m

s2

9.81

Kg∗m Kgf ∗s2

∆ P4=54.93

Kgf

m2

Serie de cálculo para el tramo I-J CCl4

ρCCl4=1585 Kg

m3

ρ H 2

O=1000 Kgm

3 g=9.81 m

s2 gc=9.81 Kg∗m

Kgf ∗s2

L=1.5m di=0.0254m μ H 2

O=0.001 Kg

m3

εhierro g$lv$ni%$do=0.15mm

G#T/ /M:TIC/ %m3!&

Gv1=9

l

min∗1m3

1000 l ∗1min

60 s =1.5∗10−4

m3

s

Gv2=12

l

min∗1m3

1000 l ∗1min

60s =2.0∗10−4

m3

s

Gv3=16

l

min∗1m3

1000 l ∗1min

60s =2.7∗10−4

m3

s


13/26

Gv4=20

l

min∗1m3

1000 l ∗1min

60s =3.3∗10−4

m3

s

((i;" tra"vral i"tr"a d la t,rmro d ?"old

ℜ=di∗ ∗ ρ H 2O

μ H 2O

ℜ1=

0.0254m∗0.2960m

s ∗1000

Kg

m3

0.001 Kg

m3

=7519.13


14/26

ℜ2=

0.0254m∗0.3947m

s∗1000

Kg

m3

0.001 Kg

m3

=10025.8

ℜ3=

0.0254 m∗0.5262m

s∗1000

Kg

m3

0.001 Kg

m3

=13367.7

ℜ4=

0.0254m∗0.6578 m

s∗1000

Kg

m3

0.001 Kg

m3

=16709.6

,goidad lativa

!= ε

di=

0.15mm

25.4mm=0.005906


f 1=0.04 f 2=0.04 f 3=0.036 f 4=0.036

Ca


15/26

" =f ∗ L∗ 2

2∗di∗g ∆ P= " ∗# # =

ρ H 2

O∗g

gc

∆ P1=

f ∗ L∗ 2

2∗di∗g∗ ρ H

2O∗g

gc=

0.04∗1.5m∗(0.2960 ms )2

2∗0.0254 m∗9.81 m

s2

∗1000 Kg

m3∗9.81

m

s2

9.81 Kg∗m Kgf ∗s2

∆ P1=10.88

Kgf

m2

∆ P2=

f ∗ L∗ 2

2∗di∗g∗ ρ H

2O∗g

gc=

0.04∗1.5m∗(0.3947 ms )2

2∗0.0254 m∗9.81m

s2

∗1000 Kg

m3∗9.81

m

s2


∆ P2=18.49

Kgf

m

2

∆ P3=

f ∗ L∗ 2

2∗di∗g∗ ρ H

2O∗g

gc=

0.036∗1.5m∗(0.5262 ms )2

2∗0.0254m∗9.81m

s2

∗1000 Kg

m3 ∗9.81

m

s2


∆ P3=31.61

Kgf

m2


16/26

∆ P4=

f ∗ L∗ 2

2∗di∗g ∗ ρ H

2O∗g

gc=

0.036∗1.5m∗(0.6578 ms )2

2∗0.0254m∗9.81 m

s2

∗1000 Kg

m3∗9.81

m

s2

9.81

Kg∗m Kgf ∗s2

∆ P4=48.12

Kgf

m2

Serie de cálculo para el tramo M-N (Hg)

ρ Hg=13600 Kg

m3

ρ H 2

O=1000 Kg

m3

g=9.81 m

s2 gc=9.81

Kg∗m

Kgf ∗s2

L=1.5m di=0.0127m μ H 2

O=0.001 Kg

m3

εhierrog $lv$ni%$do=0.15mm

G#T/ /M:TIC/ %m3!&

Gv1=9

l

min∗1m3

1000 l ∗1min60 s

=1.5∗10−4 m3

s


17/26

Gv2=12

l

min∗1m3

1000 l ∗1min

60s =2.0∗10−4

m3

s

Gv3=16

lmin

∗1m3

1000 l ∗1min

60s =2.7∗10−4

m3

s

Gv4=20

l

min∗1m3

1000 l ∗1min

60s =3.3∗10−4

m3

s

((i;" tra"vral i"tr"a d la t,rmro d ?"old


18/26

ℜ=di∗ ∗ ρ H

2O

μ H 2O

ℜ1=

0.0127m∗1.184 m

s

∗1000 Kg

m

3

0.001 Kg

m3

=15038.26

ℜ2=

0.0127m∗1.578 m

s∗1000

Kg

m3

0.001 Kg

m3

=20051.02

ℜ3=

0.0127m∗2.105m

s ∗1000 Kg

m3

0.001 Kg

m3

=26734.69

ℜ4=

0.0127m∗2.631m

s ∗1000

Kg

m3

0.001 Kg

m3

=33418.36

,goidad lativa

!= ε

di=

0.15mm

12.7mm=0.01181


f 1=0.044 f 2=0.044 f 3=0.042 f 4=0.042


19/26

Ca


20/26

∆ P3=31.61

Kgf

m2

∆ P4=

f ∗ L∗ 2

2∗di∗g ∗ ρ H

2O∗g

gc=

0.042∗1.5m∗(2.631 ms )2

2∗0.0254 m∗9.81m

s2

∗1000 Kg

m3 ∗9.81

m

s2


∆ P4=48.12

Kgf

m2

Cálculo de las relaciones de caídas de presión

$=

∆ P P C − D

∆ P P & −'

$1=

9.45 Kgf

m2

8.77 Kgf

m2

=1.13$2=

17.55 Kgf

m2

14.62 Kgf

m2

=1.20

$3=

29.83 Kgf

m2

26.32 Kgf m

2

=1.13$4=

44.46 Kgf

m2

32.17 Kgf m

2

=1.38

(=∆ P P ) − *

∆ P P & −'


21/26

(1=

88.2 Kgf

m2

8.77 Kgf

m2

=10.05(2=

151.2 Kgf

m2

14.62 Kgf

m2

=10.33

(3=

239.4 Kgf

m2

26.32 Kgf

m2

=9.09(4=

340.2 Kgf

m2

32.17 Kgf

m2

=10.57

c1=

∆ P+ C − D

∆ P+ & − '

c1=

12.1 Kgf

m2

10.88 K gf

m2

=1.11c2=

20.76 Kgf

m2

18.49 Kgf

m2

=1.12

c3=

35.83 Kgf

m2

31.61 Kgf

m2

=1.13 c4=

54.93 Kgf

m2

48.12 Kgf

m2

=1.14

d=∆ P+ ) − * ∆ P+ & −'

d1=

373.25 Kgf

m2

10.88 Kgf

m2

=34.28d2=

650.21 Kgf

m2

18.49 Kgf

m2

=35.16

d3=

1137.20 Kgf

m

2

31.61 Kgf

m2

=35.97d4=

1758.53 Kgf

m

2

48.12 Kgf

m2

=36.53

%ccesorios

Tramo A-B con 2 codos


22/26

1. Cal(,lar la vlo(idad d =,@o " la t,r


23/26

Tabla de Resultados

- Tubo Recto

Tramo &'(

ε

d=0.010236

CorridaGv(m

3

s ) ∆ P P(kgf

m2 ) v ( ms )

Re f D∆ P+ (kgf m2 )

1 1.5x10'4 9.94 0.290 7519.13 0.045 12.10

2 2.0x10'4 17.55 0.3947 10025.80 0.045 20.7

3 2.x10'4 29.83 0.522 1337.73 0.045 35.83

4 3.33x10'4 44.4 0.578 1709.7 0.045 54.93

Tramo )'*

ε

d=0.005906

CorridaGv(m

3

s ) ∆ P P(kg f

m2 ) v (ms )


1 1.5x10'4 8.77 0.290 7519.13 0.04 10.88

2 2.0x10'4 14.2 0.3947 10025.80 0.04 18.49

3 2.x10'4 2.32 0.522 1337.73 0.03 31.1

4 3.33x10'4 32.17 0.578 1709.7 0.03 48.12

Tramo +',

ε

d=¿ 00!!"!


24/26

CorridaGv(m

3

s ) ∆ P P(kgf

m2 ) v ( ms )


1 1.5x10'4 88.2 1.18 15038.2 .044 373.25

2 2.0x10'4 151.2 1.57 20051.01 .044 50.21

3 2.x10

'4

239.4 2.10 2734.8 .042 1137.20

4 3.33x10'4 340.2 2.3 33418.3 .042 1758.53

Relaci-n de resiones

Corrida a b c d

1 1.133 10.05 1.111 34.28

2 1.2 10.33 1.122 35.1

3 1.133 9.09 1.13 35.97

4 1.381 10.57 1.141 3.53

Tuber/a # %ccesorios

Tramo %' con 2 codos

Corrida

Gv

(m

3

s

) v

(m

s

) ∆ P

(kgf

m2

) ∆ Pcodos

(kgf

m2

)#e (m) ¿

di

1 1.33310-4 0.46 25.2 16.147 1.1041 57.957

2 1.66710-4 0.55 100. 7.221 3.9759 20.707

3 2.00010-4 0.702 126 106.04 3.2971 173.075


25/26

4 2.66710-4 0.936 163. 12.56 2.226 119.19

Tramo ' con lula de &ompuerta

CorridaGv(m

3

s ) v (m

s ) ∆ P(kgf m2 ) ∆ Pcom0uer-$(kgf

m2 ) #e (m)

¿

di

1 1.33310-4 0.46 26.325 10.61 0.726 3.110

2 1.66710-4 0.55 3.025 14.464 0.659 34.611

3 2.00010-4 0.702 61.425 26.69 0.35 43.37

4 2.6671

0-4 0.936 101.205 40.575 0.719 37.729

Tramo 6' con lula de 6lobo

CorridaGv(m

3

s ) v (m

s ) ∆ P(kgf m2 ) ∆ P glo(o(kgf

m2 ) #e (m)

¿

di

1 1.33310-4 0.46 226. 223.963 15.314 03.6

2 1.6671

0-4 0.55 13.6 134.344 6.124 321.4673 2.0001

0-4 0.702 176.4 170.15 5.29 277.6124 2.6671

0-4 0.936 340.2 329.24 5.32 306.15

Tramo !'

CorridaGv

(m

3

s

) v

(

m

s

)∆ P P

(

kg f

m2

)1 1.33310-4 0.46 29.25

2 1.66710-4 0.55 43.75

3 2.00010-4 0.702 64.35


26/26

4 2.66710-4 0.936 112.905

practica tuberias

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