Hardy Cross
8
ESTUDIO DE REDES DE TUBERIA (METODO DE HARDY CROSS) SOLUCION PROBLEMA N° 01 Determinar por el método de Hardy Cross, los caudales que circulan por cada tubería, coeficiente de Hazen y Williams es 100 para todos los tubos. SOLUCION: 1ra ITERACION CIRCUITO I Tubo D(pulg ) L(km ) C H Qo hfo (m) hfo/Qo ▲Q Q1 = Qo + ▲Q 1-2 2-3 1-3 8” 6” 6” 0.5 0.5 0.6 100 100 100 +35 +10 -15 +5.109 +2.037 -5.175 0.146 0.204 0.345 -1.533 -1.533- 1.6546 -1.533 33.467 10.113 -16.533 = ∑ +1.970 = ∑ 0.695
Transcript of Hardy Cross
ESTUDIO DE REDES DE TUBERIA (METODO DE HARDY CROSS)
SOLUCIONPROBLEMA N° 01Determinar por el método de Hardy Cross, los caudales que circulan por cada tubería, coeficiente de Hazen y Williams es 100 para todos los tubos.
SOLUCION:
1ra ITERACIONCIRCUITO I
Tubo D(pulg) L(km) CH Qo hfo (m) hfo/Qo ▲Q Q1 = Qo + ▲Q1-22-31-3
8”6”6”
0.50.50.6
100100100
+35+10-15
+5.109+2.037-5.175
0.1460.2040.345
-1.533-1.533-1.6546-1.533
33.46710.113-16.533
∑ = +1.970 ∑ = 0.695CIRCUITO II
Tubo D(pulg) L(km) CH Qo hfo (m) hfo/Qo ▲Q Q1 = Qo + ▲Q2-42-33-4
8”6”8”
0.70.50.6
100100100
+25-10-25
+3.838-2.037-3.29
0.1540.2040.132
+1.6546+1.6546+1.533+1.6546
26.646-9.887-23.354
∑ = -1.489 ∑ = 0.489
2da ITERACION
CIRCUITO ITubo D(pulg) L(km) CH Q1 hf1 (m) hf1/Q1 ▲Q Q2 = Q1 + ▲Q1-22-31-3
8”6”6”
0.50.50.6
100100100
+33.65+7.00-16.35
+4.702+2.08-6.196
0.1410.2060.375
+0.17+0.17+0.35+0.17
33.827.52-16.18
∑ = -0.586 ∑ = 0.721
CIRCUITO IITubo D(pulg) L(km) CH Q1 hf1 (m) hf1/Q1 ▲Q Q2 = Q1 + ▲Q2-42-33-4
8”6”8”
0.70.50.6
100100100
+26.65-7.0-23.35
+4.2-1.05-2.87
0.160.150.12
-0.35-0.35-0.17-0.35
26.30-7.52-23.70
∑ = +0.28 ∑ = 0.43
3ra ITERACIONCIRCUITO I
Tubo D(pulg) L(km) CH Q2 hf2 (m) hf2/Q2 ▲Q Q3 = Q2 + ▲Q1-22-31-3
8”6”6”
0.50.50.6
100100100
+33.2+7.52-16.18
+4.54+1.20-5.75
0.130.160.35
+0.008+0.008-0.046 +0.008
33.217.48-16.18
∑ = -0.01 ∑ = 0.64
CIRCUITO IITubo D(pulg) L(km) CH Q2 hf2 (m) hf2/Q2 ▲Q Q3 = Q2 + ▲Q2-42-33-4
8”6”8”
0.70.50.6
100100100
+26.30-7.52-23.7
+4.1-1.20-2.92
0.150.160.12
+0.046+0.046-0.008+0.046
26.35-7.48-23.65
∑ = -0.02 ∑ = 0.43Error = ▲Q*100/Q = 0.008*100/16.17= 0.00005% < 1% ¡OKEY¡
Error = ▲Q*100/Q = 0.046*100/23.65 = 0.002% < 1% ¡OKEY¡
Q1-2 = 33.21 lt/s Q2-3 = 7.48 lt/s Q1-3 = 16.17 lt/s Q2-4 = 26.35lt/s Q3-4 = 23.65 lt/s
50 lt/seg35lt/s
10lt/s
25lt/s
25lt/s
15lt/s
50 lt/seg1
2
4
3
PROBLEMA N° 02Determinar por el método de Hardy Cross los caudales que circulan por cada uno de los ramales de la red de tuberías mostradas.
SOLUCION:
15 lt/seg 45lt/s 55lt/s
20 lt/seg
20 lt/seg
25 lt/seg
25 lt/seg
17 lt/seg
25 lt
/seg
170
lt/se
g
30lt
/s
30lt
/s
20lt/s
25lt
/s
35lt
/s
70lt
/s
25lt/s
8lt/s15lt/s
23 lt
/seg
I II
III
2 1 3
6 5 4
789
1ra ITERACION
CIRCUITO ITubo D(pulg) L(km) CH Qo hfo (m) hfo/Qo ▲Q Q1 = Qo + ▲Q1-22-66-51-5
12”10”10”16”
1.21.51.21.5
100100120100
-45-30+25+70
- 2.64-3.45+1.56+1.27
0.05860.11500.06230.0182
+ 6.95+ 6.95+6.95-0.525+6.95+5.64
-38.05-23.05+31.425+82.59
∑ = -3.26 ∑ = 0.2541
CIRCUITO IITubo D(pulg) L(km) CH Qo hfo (m) hfo/Qo ▲Q Q1 = Qo + ▲Q1-33-44-51-5
14”10”10”16”
2.01.52.01.5
100100120100
+55+30-20-70
+3.00+3.45-1.66-1.88
0.05450.11500.08300.0268
-5.64-5.64-5.64-0.525-5.64-6.95
+49.366+24.36- 26.165- 82.59
∑ = +2.91 ∑ = 0.2973
CIRCUITO IIITubo D(pulg) L(km) CH Qo hfo (m) hfo/Qo ▲Q Q1 = Qo + ▲Q6-56-99-88-77-44-5
10”12”8”8”10”10”
1.21.51.22.01.52.0
120100100100100120
-25-35-15+8+25+20
-1.56-2.1-2.51+1.3+2.55+1.7
0.06230.060.16760.16240.10200.0850
0.525-6.950.5250.5250.5250.5250.525+5.64
-31.425-34.475-14.4758.525+25.525+26.165
∑ = +0.62 ∑ = 0.6393
2da ITERACION
CIRCUITO ITubo D(pulg) L(km) CH Q1 hf1 (m) hf1/Q1 ▲Q Q2 = Q1 + ▲Q1-22-66-51-5
12”10”10”16”
1.21.51.21.5
100100120100
-38.05-23.05+31.42+82.59
-1.92-2.25+2.28+2.40
0.05050.09780.07350.0291
-1.1-1.1-1.1+0.156-1.1-0.76
-39.15-24.15+30.48+80.73
∑ = +0.51 ∑ = 0.2509
CIRCUITO IITubo D(pulg) L(km) CH Q1 hf1 (m) hf1/Q1 ▲Q Q2 = Q1 + ▲Q1-33-44-51-5
14”10”10”16”
2.01.52.01.5
100100120100
+49.36+24.36-26.16-82.59
+2.4+2.4-2.8-2.4
0.05450.11500.08300.0268
+0.76+0.76+0.76+0.156+0.76+1.1
+50.12+25.12-25.249-80.76
∑ = -0.4 ∑ = 0.2835
CIRCUITO IIITubo D(pulg) L(km) CH Q1 hf1 (m) Hf1/Q1 ▲Q Q2 = Q1 + ▲Q6-56-99-88-77-44-5
10”12”8”8”10”10”
1.21.51.22.01.52.0
120100100100100120
-31.42-34.47-14.47+8.52+25.52+26.16
-2.28-2.025-2.4+1.4+2.7+2.8
0.07250.05890.16560.16400.10600.1073
-0.156+1.1-0.156-0.156-0.156-0.156-0.156-0.76
-30.481-34.631-14.631+8.369+25.369+25.249
∑ = +0.195 ∑ = 0.6743
3ra ITERACION
CIRCUITO ITubo D(pulg) L(km) CH Q2 hf2 (m) hf2/Q2 ▲Q Q3 = Q2 + ▲Q1-22-66-51-5
12”10”10”16”
1.21.51.21.5
100100120100
-39.15-24.15+31.48+80.73
-2.1-2.40+2.10+2.33
0.05380.09970.06900.0290
+0.15+0.15+0.15+0.025+0.15+0.425
-39.00-24.00+30.656+81.305
∑ = -0.07 ∑ = 0.2515
CIRCUITO IITubo D(pulg) L(km) CH Q2 hf2 (m) hf2/Q2 ▲Q Q3 = Q2 + ▲Q1-33-44-51-5
14”10”10”16”
2.01.52.01.5
100100120100
+50.12+25.12-25.249-80.73
+2.5+2.55-2.5-2.33
0.0500.10100.09900.0290
-0.425-0.425-0.425+0.025-0.425-0.15
+49.695+24.695-25.649-81.305
∑ = -0.22 ∑ = 0.2790
CIRCUITO IIITubo D(pulg) L(km) CH Q2 hf2 (m) hf2/Q2 ▲Q Q3 = Q2 + ▲Q6-56-99-88-77-44-5
10”12”8”8”10”10”
1.21.51.22.01.52.0
120100100100100120
-30.481-34.631-14.631+8.369+25.369+25.249
-2.10-2.10-2.22+1.4+2.55+2.5
0.07250.05890.16560.16400.10600.1073
-0.025-0.15-0.025-0.025-0.025-0.025-0.025+0.42
-30.656-34.656-14.656+8.344+25.344+25.649
∑ = +0.03 ∑ = 0.6482
Como: ▲Q < 1% Q3 los caudales Q3 en cada tramo son:
Q1-2 = 39l/s Q2-6 = 24l/s Q6-5 = 30.656l/s Q1-5 = 81.305l/s
Q1-3 = 49.695l/s Q3-4 = 24.695l/s Q4-5 = 25.649l/s Q6-9 = 34.656l/s
Q9-8 = 14.655l/s Q8-7 = 8.344l/s Q7-4 = 25.344l/s
