MNIRH_TP6
-
Upload
robert-aguedo-tahua -
Category
Documents
-
view
48 -
download
4
description
Transcript of MNIRH_TP6
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