Abaste. Redes de Distribuc

ACUIFEROS

Problema Nº 01 Problema Nº 02 Problema Nº 03 Problema Nº 04

Problema Nº 01 Una Una hoya hoya con con una una exten extensi sion on de de 2500 2500 Ha se se encue encuent ntra ra un acui acuife fero ro libre libre que que tiene tiene las sigui sigui Largo Promedio= 5.8 Km Ancho= 3.3 Km Esesor= 18 m Area= 2500 Ha

= = =

5800 3300

m m

25000000 m2

!e la in"estigacion reali#ada se determina$ α= poro!"a" %& ' = 0(0)* Ppm= = 0.3000 m#a$o +00 mm,a-o Per"!"a por e%apora&!o' "el or"e' "e(

)3 *

Se pre+,'-a( 1 /ol,me' -o-al "e a+,a ,e p,e"e alma&e'are e' el a&,!ero /alm= Ae4=

5858)00.0

m3

2/ol,me' "e a+,a ,e re&!b!ra a',alme'-e el a&,!ero Se e%apora= /=

)3 *

)275000.00

6,e"a =

57 * 57 *

m3a$o

3Ca,"al "e bombeo perm!!ble !' pel!+ro "e a+o-arlo bombea'"o( Hora= 12 9ora 12 9ora 6=

6= 6=

/ol."e A+,a 9ora per"!"a35"!a 0.271 m3e+ 271 l-e+

)N!%el "e la a+,a ,b-erra'ea "ep,e "e 7 mee "e e,!a e:-ra;e'"o el +a-o a'- Cal&,lamo pr!mero &,a'-o "e %ol,me' %amo a e:-raer e'( /= mee"!a9ora6e:-rae300e+ /= 2)58512 m3 1 m3
0.0) m3 2)58512 m3

5017371) m3 11)0000 m2

7 mee 7 mee

e= 2.21 Problema Nº 02

.i se conoce el coeficiente de ermeabilidad de un material es de$ K=

e=

0.003 &me+ 1.2
Hallar( 1>C,?l e el &oe!&!e'-e "e per&ola&!o' "el ma-er!al "el ma-er!al@ Kp=
&me+ ; , poro!"a" e

5).5)

K'

K= Kp' K= 0.003 &me+

Problema Nº 03

En un terreno formado or + estratos de diferentes materiales y diferentes esesores se de ermeabilidad "ertical /"/ y hor#ontal /h/ ara cada estrato1 como se muestra en la coeficiente de ermeabilidad del conunto( "%= h%=

H= 750.00

0(00*&2 cm6seg 0(00%72 cm6seg

H1= 2.00 m

200

"2= 0(0000%5* cm6seg h2= 0(00002)2 cm6seg

H2= 3.00 m

300

"+= 0(000025+ cm6seg h+= 0(0000+27 cm6seg

H3= 2.50 m

250

K%p=

0.0000207

cm6seg

K9p=

0.000)5255

cm6seg

K VP 

H 1 K V 1 1

Kp=

0.000108

cm6seg

K P K VP *K HP 

Problema Nº 04



H H 2 K V 2



H 3 K V 3

K HP  * H 1 K H 1  H 2 K H 2  H 3K H 3 H

e= 2.21 Problema Nº 02

.i se conoce el coeficiente de ermeabilidad de un material es de$ K=

e=

0.003 &me+ 1.2
Hallar( 1>C,?l e el &oe!&!e'-e "e per&ola&!o' "el ma-er!al "el ma-er!al@ Kp=
&me+ ; , poro!"a" e

5).5)

K'

K= Kp' K= 0.003 &me+

Problema Nº 03

En un terreno formado or + estratos de diferentes materiales y diferentes esesores se de ermeabilidad "ertical /"/ y hor#ontal /h/ ara cada estrato1 como se muestra en la coeficiente de ermeabilidad del conunto( "%= h%=

H= 750.00

0(00*&2 cm6seg 0(00%72 cm6seg

H1= 2.00 m

200

"2= 0(0000%5* cm6seg h2= 0(00002)2 cm6seg

H2= 3.00 m

300

"+= 0(000025+ cm6seg h+= 0(0000+27 cm6seg

H3= 2.50 m

250

K%p=

0.0000207

cm6seg

K9p=

0.000)5255

cm6seg

K VP 

H 1 K V 1 1

Kp=

0.000108

cm6seg

K P K VP *K HP 

Problema Nº 04



H H 2 K V 2



H 3 K V 3

K HP  * H 1 K H 1  H 2 K H 2  H 3K H 3 H

15 .e hi#o un bombeo de rueba en una cuenca que tenia una rofundidad de$ imermeable imermeable encontrado encontrado a esa rofundid rofundidad ad ( El ni"el del agua agua freactico freactico o ni"el de acuifer acuifer 5 a ras del terreno natural osea en la suerficie( A una distancia$ m y se hicieron 2 o#os de obser"acion como se muestra en la figura( El estado ermanente 13 lt6min el abaquiento del ni"el de agua en los 2 o#os de rueba la descarga era$ 1.52 m e' el ma &er&a'o al poo "e bombeo 0.335 m e' el o-ro 1>Cal&,lar el &oe!&!e'-e "e permeab!l!"a" e' la are'a @

%(52

= 13 = 0.013

0(++5

15 m 15 m h% %+()8

K=

h2 %)(7&

q K

r1=

0.

5 m r2= 8 m





entes dimensiones

er!or

*

eterminaron los coeficientes figura3u4l seria el

&m

&m

&m



m y descansaba sobre un estrato inicialmente se encontraba 8 m del o#o de rueba e establecio en los o#os cuando era$

l-m!' m3m!'

0000580 *

h

Ln 2 2



mm!'

r 2 r 1 h 12



Problema Nº 01 12 0.015 Una tuberia de desague de$ ulg de arcilla "itrificada n= 0.03 con una .= en la direccion norte una tuberia de concreto de$ S= 0.035 8 otra tuberia de $ ulg de arcilla "etrifi S= 0.007) Las 2 tuberias de desague de 8/ estan fluyendo a tubo l Hallar e' la -,ber!a "e 12D( 1p,l+= a El &a,"al ,e "!&,rre

N 62 = '= S= R=

0.2032

A=

0.032)

0.013 0.035

61

0.0508



= 0.30)8 '= 0.015 S= 0.030 R= 0.072 A= 0.0730

E

63 = '= S= R= A=

0.2032

Q

S



R

2 / 3

n

0.015 0.007)

62= 63=

0.0508 0.032)

61= 6263 61= 0.00

0.05) 0.0255

m3

bG!ra'-e "el l,o 6<12D= 6<12D=

y d 0

0.15 15.0

y

0.53



ba %elo&!"a" "e "eplaam!e'-o

V r V t



0.12



V TUBOLLENO

V REAL



Q12" m

%vel . *V TUBOLLENO

1.03



Qr

m3 l-e+



V REAL

V TUBOLLENO

Problema Nº 02

S 1 / 2 A



QTUBOLLENO ATUBOLLENO

2.27)

me+

2.3)2

me+



&5

E' ,'a &!,"a" e -!e'e ,' &ole&-or "e pe'"!e'-e S 1 * ; '= rela&!o'(

y d 0

m "e lo'+!-," ;

3onsiderando que el colector de

0.010

1p,l+=

0.5



S= Pob.A&-,al= r= -= 2= Bo-a&!o'= Per"!"a "e a+,a= = '= lo'+.= C=

3 10

9ab * a$o

Pob(9ut(=

1.8

l-9ab"!a * )0 0.355 m 220

Qmax 



0)7.2)

Pob. Fut . * Dot 86400

6ma:&=

0.010 75

0.025)

Pob. Fut .  P 0 1

0.01 )5000

1)

221.75

m

0.8

R= 0.088 m A= 0.0)57 m2

Q

R

2 /3



S 1 / 2 A n

6= 6=

aBe-erm!'ar ! el &ole&-or -!e'e &apa&!"a" para &o'",&!r la "ema'"a ,-,ra

0.081 8.13

221.75 J 8.1 No -!e'e la &apa&!"a"

bS! 'o -!e'e la &apa&!"a" &,al "eber!a er la pe'"!e'-e m!' 'e&ear!a ; &,al e la % ! e ma'-!e'e el "!ame-ro ; la rela&!o' ;"

1/ 2

S



Q*n A * R

S= S=

2/3

0.0503 5.03

*

&S! e !'&reme'-o la pe'"!e'-e "el &ole&-or &,a'-o "ebo pro,'"!are el b,o' e: abao "el -ramo Hpro,'"!ar =

3.77

m

1* 5 *

"S! e ma'-!e'e la pe'"!e'-e "el &ole&-or &,al era el "!ame-ro &omer&!al a"e&,a"o ;" =0L50 1p,l+=

 8 / 3

Q*n

 1/ 2

S

*

1 8

1

* 4

2/3





 

0.7)

m

2.11

p,l+

 

73.38

mm

eS! e ma'-!e'e la pe'"!e'-e "el &ole&-or &,al era el "!ame-ro &omer&!al a"e&,a"o ;" =0L75 y 0.75

d 0



M= 8/3



Q * 7257 .15 * n * ( 2 ) 2 / 3  1/ 2 S * ( 2  360 !en  ) 5 / 3

2)0

º

  2 * co

 

0.507

 

1.1

 

20.000

corre rumbo de este 8 ulg ingresa cada entra desde el sur leno( 0.025)

m3 m3

0.5)7

m

p,l+ &o' ,'a

be trabaar con una m

r 

t

9ab

* K * C

l-e+

m3e+ l-e+

elo&!"a"

!-e'-e a+,a

3.77 m

! e ma'-!e'e 0.025)

30.00

m

p,l+

! e &amb!a

1

m p,l+ p,l+

y   1  2  d  

DATOS pa pf dotacion

2136 3166 120

1030

densidad por tramo area actual densidad por tramo area futura SOLUCO! "m %&' %U!T

99 77

#$397222 6$(9()33 0$0020)3 T*A&O

A$ ACTUAL

A-C CD D, ,+ A+ - C' D , AD

CA!TDAD D, 'A-TA!T,S ACTUAL +UTU*A 277 0 17) 92 2)7 0 3#7 10# 193 1(( 1() 13( 12# 0 13# 196 119 20# 12# 0 19( 1## 3166 #196

A$ +UTU*A

2$) 1$) 2$9 3$( 1$9( 1$6 1$2( 1$3( 1$2 1$2( 1$97 3#$9(

1$2 1$3( 2$01 1$7( 2$(( 2$6( 1$)7 21#9$3)

!. TOTAL D, 'A-$

%&A/'O* T*A&O

277 270 2)7 #(1 3#) 293 12# 330 323 12# 339 3166

0$(770) 0$(62(0 0$(9792 0$939() 0$72(00 0$610#2 0$2()33 0$6)7(0 0$67292 0$2()33 0$7062( 6$60

LS

*,S,*4O*O



6$60 0$)3(

5

-

1$2

110& A

1$2

0$36#6

0$72

0$)69 +

0$3313

<<<+*,DD=

3$12

C

0$))(#

D

0$6(2

,

1$2(0 '

0$9)3 

2$917 

CO&*O-AC

0$)17( 1

Circuito 

Tramo A-C CD AD

Lonitud 107 77 107 77



Diametro 1$(00 1$000 1$000 1$(00 :%;

% 1$2 0$36#6 0$))(# 0$72

3$(9)3 2$)((7 1#$7#3) 1$39)6

8 3$(9) 2$)(6 1#$7## 1$399 9$6))

8% 2$999 7$)33 16$6(2 1$9#2 29$#26

:% 0$17)0 0$17)0 0$17)0 0$2(97 :%;

0$17)0 1

Circuito



AD D, ,+

Tramo 

Lonitud 112 77

Diametro 1$(00 1$000 1$000

% 0$72 0$6(21 0$3313

1$39)6 )$3723 2$391(

8 1$399 )$372 2$392

8% 1$9#2 12$)39 7$220

:% 0$2(97 0$0)1) 0$0)1)



6$60 0$)3(

5

-

1$2

110&

A

1$2

0$36#6

0$72

0$)69 +

0$3313

<<<+*,DD=

3$12

C

0$))(#

D

0$6(2

,

1$2(0 '

0$9)3 

2$917 

CO&*O-AC

0$)17( 1

Circuito 

Tramo A-C CD AD

Lonitud 107 77 107 77



Diametro 1$(00 1$000 1$000 1$(00 :%;

% 1$2 0$36#6 0$))(# 0$72

3$(9)3 2$)((7 1#$7#3) 1$39)6

8 3$(9) 2$)(6 1#$7## 1$399 9$6))

8% 2$999 7$)33 16$6(2 1$9#2 29$#26

:% 0$17)0 0$17)0 0$17)0 0$2(97 :%;

0$17)0 1

Circuito



AD D, ,+

A+

 

Tramo 

Lonitud 112 77

Diametro 1$(00 1$000 1$000

112

1$(00

*,SU&,! D,L CALCULO 'D*AULCO D, LA *,D D, DST*-UCO! 1 2 3 # T*A&O ASTO LO! D T*A&O DS,?O *,SA 6$60 110 2$000 A1$37# 107 1$(00 -C 0$(3) 77 1$000 DC 0$712 107 1$000 AD 0$#76 77 1$(00 D, 0$()1 112 1$000 +, 0$#02 77 1$000 A+ 1$271 112 1$(00

% 0$72 0$6(21 0$3313

1$39)6 )$3723 2$391(

1$2

3$(9)3

8 1$399 )$372 2$392

8% 1$9#2 12$)39 7$220

3$(9) 3$7)1

2$999 2($000

:% 0$2(97 0$0)1) 0$0)1)

0$0)1)

:%;

0$0)1)

:%;

(

6

7 ) 9 10 ,*DDA D, CA*A COTA ,>O&,T*CA U!T T*A&O !CAL +!AL 20$7670 22$)#37 3(20$0000 3#97$1(63 #$6209 #$9### 3#97$1(63 3#92$2119 ($)721 #$(21( 3#92$2119 3#)7$690# 9$)#30 10$(320 3#96$6(63 3#)6$12#3 0$6#93 0$(000 3#97$1(63 3#96$6(63 6$7706 7$()30 3#96$6(63 3#)9$0733 3$#20) 2$63#0 3#92$67(9 3#90$0#19 #$000# #$#)0# 3#97$1(63 3#92$67(9

4 3$2(#2 1$20#9 1$0623 1$#0#( 0$#171 1$1#73 0$7933 1$11#6

A+

112

1$(00 :%;

*,SU&,! D,L CALCULO 'D*AULCO D, LA *,D D, DST*-UCO! 1 2 3 # T*A&O ASTO LO! D T*A&O DS,?O *,SA 6$60 110 2$000 A1$37# 107 1$(00 -C 0$(3) 77 1$000 DC 0$712 107 1$000 AD 0$#76 77 1$(00 D, 0$()1 112 1$000 +, 0$#02 77 1$000 A+ 1$271 112 1$(00

 

#$6#7# ($9()1 9$73#6 0$6112

8 #$6#7 ($9() 9$73( 0$611 0$260

8% 3$373 10$9)2 13$760 1$32) 29$##2

:% 0$00#) 0$00#) 0$00#) 0$0161

%3 1$37320#3) 0$(37) 0$7122 0$#763#(20#

:%;

8 0$611 6$(3( 3$(97

2$999 2($000

0$0)1)

0$0)1)

:%;

6

7 ) 9 10 ,*DDA D, CA*A COTA ,>O&,T*CA U!T T*A&O !CAL +!AL 20$7670 22$)#37 3(20$0000 3#97$1(63 #$6209 #$9### 3#97$1(63 3#92$2119 ($)721 #$(21( 3#92$2119 3#)7$690# 9$)#30 10$(320 3#96$6(63 3#)6$12#3 0$6#93 0$(000 3#97$1(63 3#96$6(63 6$7706 7$()30 3#96$6(63 3#)9$0733 3$#20) 2$63#0 3#92$67(9 3#90$0#19 #$000# #$#)0# 3#97$1(63 3#92$67(9

4 3$2(#2 1$20#9 1$0623 1$#0#( 0$#171 1$1#73 0$7933 1$11#6

#$6177 ($)616 9$)(63 0$6(13

8 #$61) ($)62 9$)(6 0$6(1 0$02)

8% 3$363 10$)99 13$)39 1$367 29$#6)

:% %# 0$000( 1$37372333 0$000( 0$(3)3 0$000( 0$7117 0$000) 0$#7((6#(29 :%;

0$000(

2 0$6112 6$(3#6 3$(966

3$(9) 3$7)1

3

0$00#)

% 0$#6027(26 0$(703 0$#130

3$(9)3

(

2 % 1$377972(3 0$(#26 0$707# 0$#6027(262

1$2

0$0000

3 8% 1$32) 11$#() )$709

:% 0$0161 0$0113 0$0113

% 0$#763#(2 0$()16 0$#017

0$6(13 6$7762 3$#167

8 0$6(1 6$776 3$#17

8% 1$367 11$6(0 )$(06

:% 0$0007)067( 0$0003 0$0003

%# 0$#7(6 0$()1# 0$#020

2 % 1$377972(3 0$(#26 0$707# 0$#6027(262

#$6#7# ($9()1 9$73#6 0$6112

8 #$6#7 ($9() 9$73( 0$611 0$260

3 8% 3$373 10$9)2 13$760 1$32) 29$##2

:% 0$00#) 0$00#) 0$00#) 0$0161

%3 1$37320#3) 0$(37) 0$7122 0$#763#(20#

:%;

0$00#)

#$6177 ($)616 9$)(63 0$6(13

0$6112 6$(3#6 3$(966

1$2)17(2212

#$06#9

:% %# 0$000( 1$37372333 0$000( 0$(3)3 0$000( 0$7117 0$000) 0$#7((6#(29 :%;

0$0000

3

8 0$611 6$(3( 3$(97

8% 1$32) 11$#() )$709

#$06( 0$(16

3$171 2#$66(

:% 0$0161 0$0113 0$0113

% 0$#763#(2 0$()16 0$#017

0$6(13 6$7762 3$#167

0$0113

1$270#(0#21

3$99)9

:%;

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1.758

60.00

1.50

1.542

72.914

4.375

2462.000

2457.625

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2443.00

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0.191

1.758

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1.50

1.542

72.914

9.114

2457.625

2448.511

2443.00

2432.00

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0.252

0.747

97.00

1.50

0.655

14.967

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2430.00

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0.250

0.250

67.00

1.00

0.493

14.209

0.952

2447.059

2446.107

2430.00

2424.00

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0.245

0.245

105.00

1.00

0.484

13.720

1.441

2447.059

2445.618

2430.00

2434.00

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0.156

0.820

109.00

1.50

0.720

17.802

1.940

2448.511

2446.570

2432.00

2412.00

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0.664

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0.582

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2392.00

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86.00

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0.367

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0.317

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0.626

22.071

2.207

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0.302

5.737

0.688

2408.613

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72.914

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0.493

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105.00

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0.484

13.720

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2434.00

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17.802

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0.338

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0.0576

0.393

0.4221

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0.8330

3.744

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0.2955

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Problema Nº 01 8 0.015 Una tuberia de desague de$ ulg de arcilla "itrificada n= 0.15285 con una .= en la direccion norte una tuberia de concreto de$ S= 0.035 8 otra tuberia de $ ulg de arcilla "etrifi S= 0.01210 Las 2 tuberias de desague de 8/ estan fluyendo a tubo l Hallar e' la -,ber!a "e 12D( 1p,l+= a El &a,"al ,e "!&,rre

N 62

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0.2032

A=

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0.015 0.00

#17#11

0.0508



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E

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0.2032

Q

S

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0.015 0.0121

62= 63=

0.0508 0.032)

0.021 0.032

61= 6263 61= 0.017 m3

bG!ra'-e "el l,o 6<12D= 6<12D=

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0.108

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V REAL

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0.03180 31.80)

33.)8 J 31.80 No -!e'e la &apa&!"a"

bS! 'o -!e'e la &apa&!"a" &,al "eber!a er la pe'"!e'-e m!' 'e&ear!a ; &,al e la % ! e ma'-!e'e el "!ame-ro ; la rela&!o' ;"

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2.3

m

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1 8

1

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d 0

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8/3

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