= 200 M = 50 M = 400 M = 80 M = 4.0 M = 0.022 = 3.2 M = 0.018 = 1100m/s = 900 M/s

Para el siguiente sistema reservorio, tubería o canal de aducción, tubería forzada y válvula en el extremo final, se pide determinar las alturas de presión y velocidades en los puntos señalados con letras (a, b, c) y números arábigos (0, 1, 2, 3, 4, 5, 6) para el caso de flujo permanente t=0 y no permanente en los dos primeros intervalos de tiempo mayor que cero. (Debe escribir todo el procedimiento de cálculo obligatorio).

L1=200 m

alfa 0=25

∆ S1 =50 m

alfa 1=60 °

L2=400 m

alfa 2=10 °

∆ S2 =80 m

alfa 3=30°

D1=4.0 m

alfa 4=0°

f 1=0.022

alfa 5=50°

D2=3.2 m

Tc=10TR

f 2=0.018

Bn=2.5

Hr=20.0 m

C 1=1100 m/s C 2=900 m/s Calculo de H 0

H 0  H r  L1Sen(alfa 0)  S 2 ( Sen(alfa1)  Sen(alfa 2)  Sen(alfa3)  Sen(alfa5)) H 0  20  200  Sen(25º )  80  ( Sen(60º )  Sen(10º )  Sen(30º )  Sen(50º )) H 0  261.1974m H  H 0  H f Hf 0 Vamos a calcular la velocidad en el conducto de diámetro D 2 ,V 2 La fórmula:

V2 

2 g H 1  K mf  K pm

Donde:

K mf 

f1 L1 D2 4 f 2 L2 ( )   K pm D1 D1 D2 2

2

 D   3.2  K pm  1  ( 2 ) 2   1  ( ) 2   0.1296 D1   4   0.022  200 3.2 4 0.018  400 K mf  ( )   0.1296 4 4 3.2 K mf  2.8302 2  9.81 261.1974 1  2.8302  0.1296 V2  35.9749m / s V2 

Por continuidad

V1  (

A2 )V2 A1

V1  (

D2 2 ) V2 D1

V1  (

D2 2 ) V2 D1

V1  (

3.2 2 )  35.9747  23.0239m / s 4

Calculo del intervalo de tiempo en cada tramo Escogemos el menor

(t1 , t2 )

t1  t2

t  t1  0.045 Determinando los valores de

H i en cada punto.

En la entrada; punto "0"

H 0t0  H 0 

(0) f1S1V12 (0)(0.022)(50)(23.0239)  261.1974  2 gD1 2(9.81)(4)

H 0t0  261.1974m En el punto " a "

H at0  H 0 

(1) f1S1V12 (1)(0.022)(50)(23.0239)  261.1974  2 gD1 2(9.81)(4)

H at0  253.7673m En el punto " b "

H bt0  H 0 

(2) f1S1V12 (2)(0.022)(50)(23.0239)  261.1974  2 gD1 2(9.81)(4)

H bt0  246.3373m

En el punto " c "

H ct0  H 0 

(3) f1S1V12 (3)(0.022)(50)(23.0239)  261.1974  2 gD1 2(9.81)(4)

H ct0  238.9072m En el punto "1E "

H1t0E  H 0 

(4) f1S1V12 (4)(0.022)(50)(23.0239)  261.1974  2 gD1 2(9.81)(4)

H1t0E  231.4771m Para el punto "1S " ,consideramos efectos de pérdidas en el mismo punto.

V2   D  H1t0S  H1t0E  2 1   2  2 g   D1  

2

   

2

35.9749   3.2  H  231.4771  1    2(9.81)   4  t0 1S

2

  

2

H1t0S  222.9283m Para el punto "2"

(1) f 2 S2V22 (1)(0.018)(80)(35.9749) H H   261.1974  2 gD2 2(9.81)(3.2) t0 2

t0 1S

H 2t0  193.2450m Para el punto "3"

H 3t0  H1t0S 

(2) f 2 S 2V22 (2)(0.018)(80)(35.9749)  261.1974  2 gD2 2(9.81)(3.2)

H 3t0  163.5617 m Para el punto "4"

H 4t0  H1t0S 

(3) f 2 S 2V22 (3)(0.018)(80)(35.9749)  261.1974  2 gD2 2(9.81)(3.2)

H 4t0  133.8784m Para el punto "5"

(4) f 2 S 2V22 (4)(0.018)(80)(35.9749) H H   261.1974  2 gD2 2(9.81)(3.2) t0 5

t0 1S

H 5t0  104.1950m Para el punto "6"

H 6t0  H1t0S 

(5) f 2 S2V22 (5)(0.018)(80)(35.9749)  261.1974  2 gD2 2(9.81)(3.2)

H 6t0  74.5117m PARA LAS ECUACIONES DE LAS CARACTERÍSTICAS:

CR   HR  Z  VR    VR  G  R  VR  VR CS   HS  Z  VS    VS  G  R  VS  VS a1 1100   112.1305 g 9.81 a tf (1100)(0.0455)(0.022) R1  1 1  =0.0140 2 gD1 2(9.81)(4)

Z1 

A1 

 D12 =12.5664m 2 4

;

a2 900   91.7431 g 9.81 a tf (900)(0.0455)(0.018) R2  2 2  =0.0117 2 gD2 2(9.81)(3.2)

Z2 

; ;

A2 

 D22  8.0425m 2 4

G1   (0.0455) sen(25)  0.0192 G2   (0.0455) sen(60)  0.0394 G3  (0.0455) sen(10)  0.0079 G4   (0.0455) sen(30)  0.0227 G5   (0.0455) sen(0)  0 G6   (0.0455) sen(0)  0.0348 PARA EL TIEMPO t=0.045 (Flujo no permanente) Punto 0:

H S  H 0to  a1

t H 0to  H a to   S1

H S  261.1974  1100 

0.0455   261.1974  253.7673 50

H S  253.7673m VS  V0to  a1

t V0to  Va to   S1

VS  23.0239  1100 

0.0455   23.0239  23.0239  50

VS  23.0239m / s Cálculo de Cs:

Cs   Hs  Z1Vs  G1Vs  R1Vs Vs Cs  253.7673  112.1305  23.0239  0.0192  23.0239  0.0140  23.02392 Cs  2320.9 Entonces:

H 0t1  H 0t 0  261.1974m Cs  H 0t1 2320.9  261.1974  Z 112.1305 t1 V0  23.0279m / s V0t1 

Punto a:

H R  H a to  a1

t  H ato  H 0to  S1

H R  253.7673  1100 

0.0455   253.7673  261.1974  50

H R  261.2048m VR  Va to  a1

t Va to  V0to   S1

VR  23.0239  1100 

0.0455   23.0239  23.0239  50

VR  23.0239m / s Cálculo de CR:

CR   H R  Z1VR  G1VR  R1VR VR CR  261.2048  112.1305  23.0239  0.0192  23.0239  0.0140  23.02392 CR  2835.009 H S  H a to  a1

t H a to  H bto   S1

H S  253.7673  1100 

0.0455   253.7673  246.3373 50

H S  246.3373m VS  Va to  a1

t  Vato  Vbto  S1

VS  23.0239  1100 

0.0455   23.0239  23.0239  50

VS  23.0239m / s Cálculo de Cs:

Cs   Hs  Z1Vs  G1Vs  R1Vs Vs Cs  246.3373  112.1305  23.0239  0.0192  23.0239  0.0140  23.02392 Cs  2328.4 Entonces:

 C  CR H a t1    S 2  t1 H a  253.3250

  2328.4  2835.00      2   

 C  CR   2328.4  2835.00  Va t1    S      2  112.1305   2 Z1  Va t1  23.0239 Punto b:

H R  H b to  a1

t H bto  H a to   S1

H R  246.3373  1100 

0.0455   246.3373  253.7673  50

H R  253.7673m VR  Vbto  a1

t Vb to  Va to   S1

VR  23.0239  1100 

0.0455   23.0239  23.0239  50

VR  23.0239m / s Cálculo de CR:

CR   H R  Z1VR  G1VR  R1VR VR CR  253.7673  112.1305  23.0239  0.0192  23.0239  0.0140  23.02392 CR  2827.6 H S  H b to  a1

t  H bto  H cto  S1

H S  246.3373  1100 

0.0455   246.3373  238.9072  50

H S  238.9072m VS  Vbto  a1

t Vb to  Vc to   S1

VS  23.0239  1100  VS  23.0239m / s Cálculo de Cs:

0.0455   23.0239  23.0239  50

Cs   Hs  Z1Vs  G1Vs  R1Vs Vs Cs  238.3373  112.1305  23.0239  0.0192  23.0239  0.0140  23.02392 Cs  2335.8 Entonces:

 C  CR H b t1    S 2  H b t1  245.8950  C  CR Vb t1    S  2 Z1 Vb t1  23.039

  2335.8  2827.6      2   

  2335.8  2827.6       2  112.1305  

Punto C:

H R  H c to  a1

t H c to  H b to   S1

H R  238.4649  1100 

0.0455   238.9072  246.3373 50

H R  246.3373m VR  Vc to  a1

t  Vcto  Vbto  S1

VR  23.0239  1100 

0.0455   23.0239  23.0239  50

VR  23.0239m / s Cálculo de CR:

CR   H R  Z1VR  G1VR  R1VR VR CR  246.4649  112.1305  23.0239  0.0192  23.0239  0.0140  23.02392 CR  2820.1 H S  H c to  a1

t H c to  H1to   S1

H S  238.9072  1100  H S  231.4771m

0.0455   238.9072  231.4771 50

VS  Vc to  a1

t  Vcto  V1to  S1

VS  23.0239  1100 

0.0455   23.0239  23.0239  50

VS  23.0239m / s Cálculo de Cs:

Cs   Hs  Z1Vs  G1Vs  R1Vs Vs Cs  238.4771  112.1305  23.0239  0.0192  23.0239  0.0140  23.0239 2 Cs  2343.2 Entonces:

 C  CR H c t1    S 2  t1 H c  238.4649  C  CR Vc t1    S  2 Z1 Vc t1  23.039

  2343.2  2820.1      2   

  2343.2  2820.1       2 112.1305  

Punto 1:

H R  H1E to  a1

t H1E to  H c to   S1

H R  231.4771  1100 

0.0455   231.4771  238.9072  50

H R  238.9072m VR  V1E to  a1

t V1E to  Vc to   S1

VR  23.0239  1100 

0.0455   23.0239  23.0239  50

VR  23.0239m / s Cálculo de CR:

CR   H R  Z1VR  G1VR  R1VR VR CR  238.9072  112.1305  23.0239  0.0192  23.0239  0.0140  23.02392 CR  2812.7

H S  H1S to  a1

t  H1S to  H 2to  S1

H S  222.9283  1100 

0.0455   222.9283  193.2450  50

H S  207.7494m VS  V1S to  a1

t V1S to  V2to   S1

VS  35.9749  900 

0.0455   35.9749  35.9749  80

VS  35.9749m / s Cálculo de Cs:

Cs   Hs  Z 2Vs  G2Vs  R2Vs Vs Cs  207.7494  91.7431 35.9749  0.0394  35.9749  0.0394  35.97492 Cs  3079.5 Hallando:

fL K  A1  a 1 1    2 gD1 2 g  A2 

2

0.022  200 0.1296  12.5664     2  9.81 4 2  9.81  8.0425  a  0.0722 a

b

a2  12.5664     143.3486 g  8.0425 

c  (Cs  Ho) c  (3079.5  261.1974) c  3340.13 De forma general:

b  b 2  4ac 2a t1 V1E  23.0336 Vp 

H1t1E    CR  Z1V1tE1  H1t1E  (2812.7  112.1305  23.0336) H1t1E  229.9522

Aplicando la ecuación de continuidad:

V1tE1 A1  V1tS1 A2 A  V1tS1  V1tE1  1   A2   12.5664  V1tS1  23.0336    8.0425  V1tS1  35.9900 H1tS1  Cs  Z 2V1tS1 H1tS1  3079.5  91.7431 35.9900 H1tS1  222.8962 Punto 2:

H R  H 2to  a2

t  H 2to  H1to  S2

H R  193.2450  900 

0.0455   193.2450  222.9283 80

H R  208.4240m VR  V2to  a2

t V2to  V1to   S 2

VR  35.9749  900 

0.0455   35.9749  35.9749  80

VR  35.9749m / s Cálculo de CR:

CR   H R  Z 2VR  G2VR  R2VR VR CR  208.4240  91.7431 35.9749  0.0394  35.9749  0.0394  35.9749 2 CR  3492.3 H S  H 2to  a2

t H 2to  H 3to   S 2

H S  193.2450  900  H S  178.0660m

0.0455   193.2450  163.5617  80

VS  V2to  a2

t  V2to  V3to  S 2

VS  35.9749  900 

0.0455   35.9749  35.9749  80

VS  35.9749m / s Cálculo de Cs:

Cs   Hs  Z 2Vs  G3Vs  R2Vs Vs Cs  178.066  91.7431 35.9749  0.0079  35.9749  0.0117  35.9749 2 Cs  3106.9 Entonces:

 C  CR H 2t1    S 2  t1 H 2  192.6789

  3106.9  3492.3      2   

 C  CR   3106.9  3492.3  V2t1    S      2  91.7431   2Z1  V2t1  35.9656 Punto 3:

H R  H 3to  a2

t H 3to  H 2to   S 2

H R  163.5617  900 

0.0455   163.5617  193.2450  80

H R  177.7900m VR  V3to  a2

t V3to  V2 to S2



VR  35.9749  900 



0.0455   35.9749  35.9749  80

VR  35.9749m / s Cálculo de CR:

CR   H R  Z 2VR  G3VR  R2VR VR CR  177.7900  91.7431 35.9749  0.0394  35.9749  0.0394  35.9749 2 CR  3464.3

H S  H 3to  a2

t  H 3to  H 4to  S 2

H S  163.5617  900 

0.0455   163.5617  133.8784  80

H S  148.3827 m VS  V3to  a2

t V3to  V4to   S2

VS  35.9749  900 

0.0455   35.9749  35.9749  80

VS  35.9749m / s Cálculo de Cs:

Cs   Hs  Z 2Vs  G4Vs  R2Vs Vs Cs  148.3827  91.7431 35.9749  0.0079  35.9749  0.0117  35.97492 Cs  3137.7 Entonces:

 C  CR H 3t1    S 2  H 3t1  163.2949

  3137.7  3464.3      2   

 C  CR   3106.9  3492.3  V3t1    S      2  91.7431   2 Z1  V3t1  35.9809

Punto 4:

H R  H 4to  a2

t  H 4to  H 3to  S 2

H R  133.8784  900  H R  168.8020m

0.0455   133.8784  163.5617  80

VR  V4to  a2

t V4to  V3 to S 2



VR  35.9749  900 



0.0455   35.9749  35.9749  80

VR  35.9749m / s Cálculo de CR:

CR   H R  Z 2VR  G4VR  R2VR VR CR  168.8020  91.7431 35.9749  0.0394  35.9749  0.0394  35.9749 2 CR  3433.5 H S  H 4to  a2

t H 4to  H 5to   S 2

H S  133.8784  900 

0.0455   133.8784  104.1950  80

H S  118.6994m VS  V4to  a2

t V4to  V5to   S 2

VS  35.9749  900 

0.0455   35.9749  35.9749  80

VS  35.9749m / s Cálculo de Cs:

Cs   Hs  Z 2Vs  G4Vs  R2Vs Vs Cs  118.6994  91.7431 35.9749  0  35.9749  0.0117  35.97492 Cs  3166.6 Entonces:

 C  CR H 4t1    S 2  t1 H 4  133.4696

  3166.6  3433.5      2   

 C  CR   3166.6  3433.5  V4t1    S      2  91.7431   2Z1  V4t1  35.9704 Punto 5:

H R  H 5to  a2

t  H 5to  H 4to  S 2

H R  104.1950  900 

0.0455   104.1950  133.8784  80

H R  202.3423m VR  V5to  a2

t V5to  V4 to S 2



VR  35.9749  900 



0.0455   35.9749  35.9749  80

VR  35.9749m / s Cálculo de CR:

CR   H R  Z 2VR  G5VR  R2VR VR CR  202.3423  91.7431 35.9749  0  35.9749  0.0394  35.97492 CR  3404.6 H S  H 5to  a2

t H 5to  H 6to   S 2

H S  104.1950  900 

0.0455   104.1950  74.5117  80

H S  89.0161m VS  V5to  a2

t  V5to  V6to  S2

VS  35.9749  900 

0.0455   35.9749  35.9749  80

VS  35.9749m / s Cálculo de Cs:

Cs   Hs  Z 2Vs  G5Vs  R2Vs Vs Cs  89.0161  91.7431 35.9749  0.0348  35.9749  0.0117  35.97492 Cs  3197.5 Entonces:

 C  CR H 5t 1    S 2  t1 H 5  103.5687

  3197.5  3404.6      2   

 C  CR   3197.5  3404.6  V5t1    S      2  91.7431   2 Z1  V5t1  35.9817 Punto 6 (válvula):

H R  H 6to  a2

t  H 6to  H 5to  S 2

H R  74.5117  900 

0.0455   74.5117  104.1950  80

H R  238.8131m VR  V6to  a2

t V6to  V5 to S 2



VR  35.9749  900 



0.0455   35.9749  35.9749  80

VR  35.9749m / s Cálculo de CR:

CR   H R  Z 2VR  G5VR  R2VR VR CR  238.8131  91.7431 35.9749  0.0348  35.9749  0.0394  35.97492 CR  3373.7 Por dato nos da un valor de B=2.5

  (1  TR 

t 2.5 ) tC

10.L a

Para nuestro caso.

TR 

2  L1 2  L2 2  200 2  400     1.2525 a1 a2 1100 900

TC  TR  10  12.5253"

  (1 

t 0.0455 )  (1  )  0.9910 tC 12.5253

 nV0  Kn 

2

H

 0.9910  35.9749   253.7673

2

 5.0085

Bkn  Kn.Z  Kn.Z 2 Bkn  446.3844  91.7431  446.3844 Ckn  Kn  CR  hf  Ckn  5.0085   3373.7  0   16415.08 Y de la fórmula:

 Bkn  Bkn 2  4Ckn Vp  2 446.3844  446.38442  4  ( 16415.08) V  2 t1 V6  34.1594m / s t1 6

En forma general:

Hp    CR  ZVP  H 6t1    3373.7  91.7431 34.1594  H 6t1  239.8195m

Para t=0.0909 Para el punto 0

Hs  H ot1 

a1 * t H 0t1  H1t1   s1

Hs  261,1974 

1100*0.0455  261.1974  253.3250  50

Hs  253.3250 a * t t 1 Vs  Vot1  1  Vo  V1t1  S1 Vs  23.0387  Vs  23.0239

1100*0.0455 *(23.0387  23.0239) 50

Cs   Hs  Z1 *Vs  G1 *Vs  R1 *Vs * Vs Cs  253.3250  112.1305*23.0239  0.0192* 23.0239  0.0140* 23.0239* 23.0239 Cs  2322,5843 H 0t 2  H 0t 1  261.1974 Vot 2 

Cs  H 0t2 2322,5843  261.1974   23.0426 Z1 112.1305

punto a

H R  H At1 

a1t ( H At1  H 0t1 ) S 1

H R  253.3250  VR  VAt1 

1100*0.0455 (253.3250  261.1974)  261.1974 50

a1t t1 (VA  V0t 1 ) S 1

VR  23.0239 

1100*0.0455 (23.0239  23.0387)  23.0387 50

CR   H R  Z1 *VR  G1 *VR  R1 *VR VR CR  261.1974  1112.1305* 23.0387  0.0192* 23.0387  0.0.140* 23.0387 2 CR  -2836,6546 H S  H At1 

a1t ( H At1  H Bt1 ) S 1

H S  253.3250  VS  VAt1 

1100 *0.0455*(253.3250  245.8950)  245.8950 50

a1t t1 (VA  VBt1 ) S 1

VS  23.0239 

1100*0.0455 (23.0239  23.0239)  23.0239 50

CS   H S  Z1 *VS  G1 *VS  R1 *VS VS CS  245.8950  112.1305* 23.0239  0.0192* 23.0239  0.0140* 23.0239 23.0239 Cs  2328,8016

(Cs  CR )   2328,8016  2836,6546)   253.9265 2 2 C  CR 2328,8016  2836,6546 VAt2  s   23.0332 2* Z1 2*112.1305 H At2 

punto b

H R  H Bt1 

a1t ( H Bt1  H At1 ) S 1

H R  245.8950  VR  VBt1 

1100*0.0455 (245.8950  253.3250)  253.3250 50

a1t t1 (VB  VAt 1 ) S 1

VR  23.0239 

1100*0.0455 (23.0239  23.0239)  23.0239 50

CR   H R  Z1 *VR  G1 *VR  R1 *VR VR CR  253.3250  1112.1305* 23.0239  0.0192* 23.0239  0.0140* 23.02392 CR  -2827,1370 H S  H Bt1 

a1t ( H Bt1  H Ct1 ) S 1

H S  245.8950  VS  VBt1 

1100 *0.0455*(245.8950  238.4649)  238.4649 50

a1t t1 (VB  VCt1 ) S 1

VS  23.0239 

1100*0.0455 (23.0239  23.0239)  23.0239 50

CS   H S  Z1 *VS  G1 *VS  R1 *VS VS CS  238.4649  112.1305* 23.0239  0.0192* 23.0239  0.0140* 23.0239 23.0239 Cs  2336,2316 (Cs  CR )   2336.2316  2827.1370)   245.4527 2 2 C  CR 2336.2316  2827.1370 VBt2  s   23.0239 2* Z1 2*112.1305 H Bt2 

punto C

punto 1

H R  H1tE1 

a1t ( H1tE1  H Ct1 ) S 1

H R  229.9522  VR  V1tE1 

1100*0.0455 (229.9522  238.4649)  238.4649 50

a1t t1 (V1E  VCt 1 ) S 1

VR  23.0336 

1100*0.0455 (23.0336  23.0239)  23.0239 50

CR   H R  Z1 *VR  G1 *VR  R1 *VR VR CR  238.4649  1112.1305* 23.0239  0.0192* 23.0239  0.0140* 23.02392 CR  -2812,2769 H S  H1tS1 

a2 t ( H1tS1  H 2t1 ) S 2

H S  222.8962  VS  V1tS1 

900 *0.0455*(222.8962  192.6789)  207.4442 80

a2 t t1 (V1S  V2t1 ) S 2

VS  35.9900 

900*0.0455 (35.9900  35.9656)  35.9775 80

CS   H S  Z 2 *VS  G2 *VS  R2 *VS VS CS  207.4442  91.7432*35.9775  0.0394*35.9775  0.0117 *35.9775 35.9775 Cs  3079,4812   D 2  k  1   2     D1    

2

2

  3.20 2  k  1    0.1296   4.00     f *L k  A1  a 1 1   2 gD1 2* g  A2 

2

2

0.022* 200 0.1296  4 2  a     0.0722 2*9.81* 4 2*9.81  3.2 2 

b

a2  A1  *  g  A2 

b

900  42  *   143.3486 9.81  3.22 

c  (Cs  Ho)  (3079.4812  261.1974) c  3340,6786 V1tE2 

b  b 2  4ac 2a

V1tE2 

143.3486  143.34862  4*(0.0722) *(3340.6786) 2*(0.0722)

V1tE2  23.0373 H1t2E  (C R  Z1 *V1tE2 ) H1t2E  (2812.2769  112.1305* 23.0373)  229.0935 V2tS2 

A1 t2 V2 E A2

V2tS2 

42 * 23.0373  35.9958 3.24

H 2t2s  C s  Z 2 *V2ts2 H 2t 2s  3079.4812  91.7432*35.9958 H 2t2s  222.8839 punto 2

H R  H 2t1 

a2 t ( H 2t1  H1tS1 ) S 2

H R  192.6789  VR  V2t1 

900*0.0455 (192.6789  222.8962)  208.1309 80

a2 t t1 (V2  V1tS1 ) S2

VR  35.9656 

900 *0.0455 (35.9656  35.9900)  35.9781 80

CR   H R  Z 2 *VR  G2 *VR  R2 *VR VR CR  208.1309  91.7432*35.9781  0.0394*35.9781  0.0117 *35.97812 CR  -3492,2741

H S  H 2t1 

a2 t t 1 ( H 2  H 3t1 ) S 2

H S  192.6789  VS  V2t1 

900 *0.0455*(192.6789  163.2949)  177.6530 80

a2 t t1 (V2  V3t 1 ) S 2

VS  35.9656 

900*0.0455 (35.9656  35.9809)  35.9734 80

CS   H S  Z 2 *VS  G3 *VS  R2 *VS VS CS  117.6530  91.7432*35.9734  0.0079*35.9734  0.0117 *35.9734 35.9734 Cs  3107,2003 (Cs  CR )   3107.2003  3492.2741)   192.5369 2 2 C  CR 3107.2003  3492.2741 V2t2  s   35.9671 2* Z 2 2*91.7432 H 2t2 

punto 3

H R  H 3t1 

a2 t ( H 3t 1  H 2t1 ) S 2

H R  163.2949  VR  V3t1 

900*0.0455 (163.2949  192.6789)  178.3208 80

a2 t t1 (V3  V2t1 ) S 2

VR  35.9809 

900*0.0455 (35.9809  35.9656)  35.9731 80

CR   H R  Z 2 *VR  G3 *VR  R2 *VR VR CR  178.3208  91.7432*35.9731  0.0079*35.9731  0.0117 *35.97312 CR  3463,7104

H S  H 3t1 

a2 t t 1 ( H 3  H 4t1 ) S 2

H S  163.2949  VS  V3t1 

900 *0.0455*(163.2949  133.4696)  148.0433 80

a2 t t1 (V3  V4t 1 ) S 2

VS  35.9809 

900*0.0455 (35.9809  35.9704)  35.9755 80

CS   H S  Z 2 *VS  G4 *VS  R2 *VS VS CS  148.0433  91.7432*35.9755  0.0227 *35.9755  0.0117 *35.9755 35.9755 Cs  3138,1036 (Cs  C R )   3138.1036  3463.7104)   162.8034 2 2 C  C R 3138.1036  3463.7104 V2t2  s   35.9799 2* Z 2 2*91.7432 H 2t2 

punto 4

H R  H 4t1 

a2 t ( H 4t1  H 3t1 ) S 2

H R  133.4696  VR  V4t1 

900*0.0455 (133.4696  163.2949)  148.7211 80

a2 t t1 (V4  V3t1 ) S 2

VR  35.9704 

900 *0.0455 (35.9704  35.9809)  35.9758 80

CR   H R  Z 2 *VR  G4 *VR  R2 *VR VR CR  148.7211  91.7432*35.9758  0.0227 *35.9758  0.0117 *35.97582 CR  -3433,2544

H S  H 4t1 

a2 t t 1 ( H 4  H 5t1 ) S 2

H S  133.4696  VS  V4t1 

900 *0.0455*(133.4696  103.5687)  118.1794 80

a2 t t1 (V4  V5t 1 ) S 2

VS  35.9704 

900*0.0455 (35.9704  35.9817)  35.9762 80

CS   H S  Z 2 *VS  G5 *VS  R2 *VS VS CS  118.1794  91.7432*35.9762  0*35.9762  0.0117 *35.9762 35.9762 Cs  3167,2098 (Cs  CR )   3167.2098  3433.2544)   133.0223 2 2 C  CR 3167.2098  3433.2544 V2t2  s   35.9725 2* Z 2 2*91.7432 H 2t2 

punto 5

H R  H 5t1 

a2 t ( H 5t 1  H 4t1 ) S 2

H R  103.5687  VR  V5t1 

900*0.0455 (103.5687  133.4696)  118.8589 80

a2 t t1 (V5  V4t1 ) S 2

VR  35.9817 

900*0.0455 (35.9817  35.9704)  35.9759 80

CR   H R  Z 2 *VR  G5 *VR  R2 *VR VR CR  118.8589  91.7432*35.9759  0*35.9759  0.0117*35.97592 CR  3404,2247

H S  H 5t1 

a2 t t 1 ( H 5  H 6t1 ) S 2

H S  103.5687  VS  V5t1 

900 *0.0455*(103.5687  239.8195)  173.2424 80

a2 t t1 (V5  V6t1 ) S 2

VS  35.9817 

900*0.0455 (35.9817  34.1594)  35.0498 80

CS   H S  Z 2 *VS  G6 *VS  R2 *VS VS CS  173.2424  91.7432*35.0498  0.0348*35.0498  0.0117 *35.0498 35.0498 Cs  3029,1514 (Cs  C R )   3029.1514  3404.2247)   187.5367 2 2 C  C R 3029.1514  3404.2247 V2t2  s   35.0619 2* Z 2 2*91.7432 H 2t2 

punto 6 (Válvula)

H R  H 6t1 

a2 t ( H 6t1  H 5t1 ) S 2

H R  239.8195  VR  V6t1 

900*0.0455 (239.8195  103.5687)  170.1458 80

a2 t t1 (V6  V5t1 ) S2

VR  34.1594 

900 *0.0455 (34.1594  35.9817)  35.0913 80

CR   H R  Z 2 *VR  G6 *VR  R2 *VR VR CR  170.1458  91.7432*35.0913  0.0348*35.0913  0.0117 *35.09132 CR  3373.8629

 t * i  n  1   tc  

2.5

 2*0.0455  n  1   12.5253  

 n *V0  kn 

2

2.5

 0.9820

 0.9820*35.9749  

2

 4.7776 H 261.1974 Bkn  kn * Z  kn * Z 2  4.7776*917432

Bkn  438.3141 Ckn  kn(CR  hf )  4.7776*( 3373,8629  0)  -16119,0490 V6t2 

 Bkn  Bkn 2  4Ckn 2

438.3141  438.31412  4 *(16119.0490)  34.1192 2 H 6t2  (Cr  Z 2 *V6t2 )

V6t2 

H 6t2  (3373.8629  91.7432 *34.1192)  243.6609