UNIVERSIDAD NACIONAL AGRARIA LA MOLINA

SEMESTRE : 2015 – I

DOCENTE : Ing. Jose Arapa

ALUMNO : Aguedo Tahua Robert Henry.

CURSO : Métodos numéricos en ingeniería

de recursos hídricos.

TEMA : Hidráulica de tuberías / calculo y

análisis de redes.

CODIGO : 20150824

ESCUELA DE POSTGRADO MAESTRIA EN RECURSOS HIDRICOS

60

200 500 m 400 m 40250 mm 150 mm

300 m 200 m 200 m250 mm 100 mm 100 mm

400 m600 m 150 mm

200 mm 3040 30

SoluciónPresion en la red

92 80

10 1080

2040

7595 90

1 2 3 4 5 6100 92 80 75 90 95

Datos.

7.3. La red mostrada en la figura tiene una válvula en la tubería 2-3, la cual se encuentra parcialmente cerrada y produce una perdida menor locade , la presión de 10.0 (v23) ^2/2g, en el nodo 1 es de 100 mca. Analizar los caudales y presiones en la red. Los diámetros (en milímetros) y las longitudes (en metros) para cada una de las tuberías son los indicados en la figura. Los caudales de demanda en los nodos están dados en l/s., la viscosidad cinemática es y Rugosidad absoluta en todas la tuberías: Ks=0.00005.

100 msnm

100 msnm

Nodo Presion m

Se puede suponer que en todos los tubos, salvo en la tuberia 2- 3, las perdidas menores son despreciables. Disenar e implementar el metdo de newton - Raphson para el analisis y diseno de redes cerradas en hidraulica de tuberias, el programa debe permitir ingresar los datos desde excel y construir matricial de la red. finalmente el programa debe permitir mostrar las iteraciones y reportar el resultado final en otra hoja Ecel

12

12

1

6

5 4

32

1

6

5 4

32

viscosidad cinematica v = 1.14E-06 m2/sCoeficiente rugosidad Ks = 6.00E-05

Aceleracion de la gravedad g = 9.81 m2/scoeficiente local Km = 10

Ec. 7.36 Ec. 7.37 Ec. 7.38 Ec. 7.33

1 2 100 92 8 500 0.25 0.049087 0.109347 31.63098 0.00683422 3 92 80 12 400 0.15 0.017671 0.039525 2.549773 47.06302 0.0016469

9.4502272 400 0.15 0.017671 0.034889 1.98665 60.40319 0.001453710.01335 400 0.15 0.017671 0.035961 2.110652 56.85446 0.0014984

9.8893477 400 0.15 0.017671 0.035727 2.083327 57.60018 0.00148869.9166731 400 0.15 0.017671 0.035779 2.089347 57.4342 0.00149089.9106525 400 0.15 0.017671 0.035768 2.088021 57.47069 0.00149039.9119791 400 0.15 0.017671 0.03577 2.088313 57.46264 0.0014904

3 4 80 75 5 200 m 0.1 0.007854 0.0123 39.99922 0.001234 5 75 90 15 400 m 0.15 0.017671 0.044395 46.63017 0.00147982 5 92 90 2 200 m 0.1 0.007854 0.00756 42.35118 0.001895 6 90 95 5 600 m 0.2 0.031416 0.043123 52.06667 0.00431231 6 100 95 5 300 m 0.25 0.049087 0.111698 18.94613 0.0111698

Ec. 7.35 Ec. 7.29

Q(m3/s)Hi(m) Hj(m) hf(m) L(m) d(m) A(m2)Tuberia hm(m)∑km + f*L/d

df/dx

Nodo N Tuberia N Hi (m) Hj (m) hf(m) Q (m3/s) df/dx df/dH f=∑Q-Qd2 -0.06

2 1 92 100 8 0.109347 0.0068342 3 92 80 -12 -0.03577 0.0016472 5 92 90 -2 -0.00756 0.00189 -0.010371 0.006017

3 -0.043 2 80 92 12 0.03577 0.0016473 4 80 75 -5 -0.0123 0.00123 -0.002877 -0.01653

4 -0.034 3 75 80 5 0.0123 0.001234 5 75 90 15 0.044395 0.00148 -0.00271 0.026695

5 -0.035 2 90 92 2 0.00756 0.001895 4 90 75 -15 -0.04439 0.001485 6 90 95 5 0.043123 0.004312 -0.007682 -0.02371

6 -0.046 1 95 100 5 0.111698 0.011176 5 95 90 -5 -0.04312 0.004312 -0.015482 0.028575

En forma matricial [Df / Dx] [Dx] = [-F]

Nodos 2 3 4 5 62 -0.010371 0.001647 0 0.00189 0 dH2 -0.006023 0.0016469 -0.00288 0.00123 0 0 dH3 0.016534 0 0.00123 -0.00271 0.00148 0 dH4 = -0.026695 0.00189 0 0.0014798 -0.00768 0.004312 dH5 0.0237126 0 0 0 0.004312 -0.01548 dH6 -0.02858

Solucion de Dx = [Df / Dx ] - 1 * [F] por inversa

-122.2717 -102.155 -75.21994 -52.8325 -14.7156 0.222836[Df / Dx] - 1 = -102.1551 -531.968 -294.4437 -97.0224 -27.0239 = -1.84691

-75.21994 -294.444 -587.9722 -156.19 -43.5042 8.820828-52.83249 -97.0224 -156.1903 -205.368 -57.2019 -0.35162-14.7156 -27.0239 -43.50416 -57.2019 -80.5237 1.747761

Los resultados para el vector dHi son:

dH2 = 0.22284 H2 = 92.2228

dH3 = ‐1.8469 H3 = 78.1531

dH4 = 8.82083 H4 = 83.8208

dH5 = ‐0.3516 H5 = 89.6484

dH6 = 1.74776 H6 = 96.7478

60

200 500 m 400 m 40250 mm 150 mm

300 m 200 m 200 m250 mm 100 mm 100 mm

400 m600 m 150 mm

200 mm 3040 30

Presion en la red

92.22284 78.153086

10 1080

2040

83.82082896.74776 89.64838

1 2 3 4 5 6100 92.22284 78.15309 83.82083 89.64838 96.74776

Datos.viscosidad cinematica v = 1.14E-06 m2/sCoeficiente rugosidad Ks = 6.00E-05

Aceleracion de la gravedad g = 9.81 m2/scoeficiente local Km = 10

100 msnm

100 msnm

Nodo Presion m

Segunda iteración

12

12

1

6

5 4

32

1

6

5 4

32

Ec. 7.36 Ec. 7.37 Ec. 7.38 Ec. 7.33

1 2 100 92.22284 7.7771645 500 0.25 0.049087 0.107749 31.6689 0.00692732 3 92.222836 78.15309 14.06975 400 0.15 0.017671 0.042941 3.009534 46.7506 0.001526

11.060216 400 0.15 0.017671 0.037878 2.341748 60.08224 0.001346111.728001 400 0.15 0.017671 0.039055 2.489514 56.51606 0.001387911.580236 400 0.15 0.017671 0.038798 2.456795 57.26873 0.001378811.612955 400 0.15 0.017671 0.038855 2.464038 57.10037 0.001380811.605712 400 0.15 0.017671 0.038842 2.462435 57.13756 0.001380311.607315 400 0.15 0.017671 0.038845 2.46279 57.12932 0.0013804

3 4 78.153086 83.82083 5.667742 200 m 0.1 0.007854 0.013139 39.73221 0.00115914 5 83.820828 89.64838 5.8275564 400 m 0.15 0.017671 0.027067 48.73782 0.00232232 5 92.222836 89.64838 2.5744515 200 m 0.1 0.007854 0.008651 41.62824 0.00168025 6 89.648384 96.74776 7.0993766 600 m 0.2 0.031416 0.051871 51.09463 0.00365321 6 100 96.74776 3.2522394 300 m 0.25 0.049087 0.089223 19.3137 0.0137172

Ec. 7.35 Ec. 7.29Nodo N Tuberia N Hi (m) Hj (m) hf(m) Q (m3/s) df/dx df/dH f=∑Q-Qd

2 -0.062 1 92.22284 100 7.777164 0.107749 0.0069272 3 92.22284 78.153086 -14.0697 -0.03885 0.0015262 5 92.22284 89.648384 -2.57445 -0.00865 0.00168 -0.010134 0.000253

3 -0.043 2 78.15309 92.222836 14.06975 0.038845 0.0015263 4 78.15309 83.820828 5.667742 0.013139 0.001159 -0.002685 0.011984

4 -0.034 3 83.82083 78.153086 -5.66774 -0.01314 0.0011594 5 83.82083 89.648384 5.827556 0.027067 0.002322 -0.003481 -0.01607

5 -0.035 2 89.64838 92.222836 2.574451 0.008651 0.001685 4 89.64838 83.820828 -5.82756 -0.02707 0.0023225 6 89.64838 96.747761 7.099377 0.051871 0.003653 -0.007656 0.003456

6 -0.046 1 96.74776 100 3.252239 0.089223 0.0137176 5 96.74776 89.648384 -7.09938 -0.05187 0.003653 -0.01737 -0.00265

En forma matricial [Df / Dx] [Dx] = [-F]

Nodos 2 3 4 5 62 -0.010134 0.001526 0 0.00168 0 dH2 -0.000253 0.001526 -0.00269 0.0011591 0 0 dH3 -0.011984 0 0.001159 -0.003481 0.002322 0 dH4 = 0.0160735 0.0016802 0 0.0023223 -0.00766 0.003653 dH5 -0.003466 0 0 0 0.003653 -0.01737 dH6 0.002647

hm(m)∑km + f*L/d

df/dxTuberia Hi(m) Hj(m) hf(m) L(m) d(m) A(m2) Q(m3/s)

Solucion de Dx = [Df / Dx ] - 1 * [F] por inversa

-122.3494 -98.9758 -68.20436 -52.8453 -11.1139 0.274021[Df / Dx] - 1 = -98.9758 -537.27 -251.5724 -108.971 -22.9178 = 2.736252

-68.20436 -251.572 -492.9771 -182.86 -38.4575 -4.36129-52.84528 -108.971 -182.8603 -219.741 -46.214 -0.9828-11.11395 -22.9178 -38.45755 -46.214 -67.2886 -0.3591

Los resultados para el vector dHi son:

dH2 = 0.27402 H2 = 92.4969

dH3 = 2.73625 H3 = 80.8893

dH4 = ‐4.3613 H4 = 79.4595

dH5 = ‐0.9828 H5 = 88.6656

dH6 = ‐0.3591 H6 = 96.3887