Ejemplos de cálculo de tuberías en paralelo y ramificadas

1. Tuberías en paralelo
Balance del punto 1 al 2:
∆𝑃
𝜌
+
∆𝑣
2
+ 𝑔∆𝑧 + 𝑊 + 𝑊 = 0
( 𝑊 + 𝑊 ) = ( 𝑊 + 𝑊 ) = ( 𝑊 + 𝑊 )
1 2
( 𝑊 + 𝑊 ) = −
∆𝑃
𝜌
+
∆𝑣
2
+ 𝑔∆𝑧
Depende del
camino elegido
NO depende del
camino

2. Ejemplo 25. a) Calcular el caudal
1
2 Datos
Tubería A
(𝐿 + 𝛴𝐿𝑒) = 200 𝑚
𝐷 = 0.2 𝑚
𝜀
𝐷 = 0.005
𝑃1 = 2 𝑏𝑎𝑟 𝑧 = 5𝑚 𝜌 = 800 𝑘𝑔/𝑚
𝑃2 = 1 𝑏𝑎𝑟 𝑧 = 1 𝑚 𝜇 = 0.4 𝑐𝑃
∆𝑃
𝜌
+
∆𝑣
2
+ 𝑔∆𝑧 + 𝑊 + 𝑊 = 0
∆𝑃
𝜌
+ 𝑔∆𝑧 + 𝑊 =
(100 000 − 200 000) 𝑃𝑎
800 𝑘𝑔/𝑚
+ 9.81
𝑚
𝑠
(1 − 5)𝑚 + 𝑊 = 0
𝑊 = 164.24
𝑊 = 𝑓
( 𝐿 + 𝛴𝐿𝑒) ∗ 𝑣2
2 𝐷
= 164.24
Suponiendo 𝑓 = 0.03
0.03 ∗
(200 𝑚) ∗ 𝑣
2 ∗ (0.2 𝑚)
= 164.24 𝑣 = 3.3089 𝑚/𝑠
𝑄 = 𝑣 ∗ 𝐴 = 3.3089
𝑚
𝑠
∗
𝜋
4
∗ (0.2 𝑚) = 0.1039 𝑚 /𝑠
𝑄 = 0.1039 𝑚 /𝑠
Leyendo 𝑓 comprobar que 𝑓 = 0.03

3. b) Suponiendo que el caudal A= ctte y que agregamos entre los puntos 1 y 2 una tubería B,
con (𝐿 + 𝛴𝐿𝑒) = 230 𝑚, con 𝐷 = .15 𝑚 y 𝜀
𝐷 = 0.006. Determine que % aumenta el caudal total.
1
Datos
Tubería B
(𝐿 + 𝛴𝐿𝑒) = 230 𝑚
𝐷 = 0.15 𝑚
𝜀
𝐷 = 0.006
( 𝑊 + 𝑊 ) = ( 𝑊 + 𝑊 )
0.03 ∗ (200 𝑚) ∗ (3.3089)
2 ∗ (0.2 𝑚)
= 𝑓
(230 𝑚) ∗ 𝑣2
2 (0.15 𝑚)
= 164.24
Suponiendo 𝑓 = 0.035 , 𝑣 = 2.474 𝑚/𝑠
𝑅𝑒 =
(0.15 𝑚) ∗ 2.474
𝑚
𝑠
∗ 800
𝑘𝑔
𝑚
( 0.0004 𝑃𝑎 ⋅ 𝑠)
= 7.42 ∗ 10

4. c) Que caudal pasará por las tuberías A y B conservando las longitudes, 𝜀
𝐷 y 𝐷 , calcular
también P1. Si el 𝑄 = 0.2 𝑚 /𝑠
𝑓 = 0.032 𝑣 = 2.5874 𝑚/𝑠 𝑅𝑒 = 7.76 ∗ 10
𝑄 = 𝑣 ∗ 𝐴 = 2.5874
𝑚
𝑠
∗
𝜋
4
∗ (0.15 𝑚) = 0.04572 𝑚 /𝑠
𝑄 = 0.04572 𝑚 /𝑠
𝑄 = 𝑄 + 𝑄 = 0.1 + 0.046 = 0.146 Aumenta el 43%

5. 𝑄 = 𝑄 + 𝑄 = 0.2
𝑓 𝐿 𝑄
2 ∗
𝜋
4
∗ 𝐷
=
𝑓 𝐿 𝑄
2 ∗
𝜋
4
∗ 𝐷
0.03 ∗ (200 𝑚) ∗ 𝑄
2 ∗
𝜋
4
∗ (0.2 𝑚)
=
0.032 ∗ (230 𝑚) ∗ 𝑄
2 ∗
𝜋
4
∗ (0.15 𝑚)
𝑄 = 2.3 𝑄
𝑄 + 𝑄 = 0.2
𝑄 = 0.14 𝑚 /𝑠 𝑄 = 0.06 𝑚 /𝑠
Calcular P1
(100 000 − 𝑃 ) 𝑃𝑎
800 𝑘𝑔/𝑚
+ 9.81
𝑚
𝑠
(1 − 5)𝑚 +
(0.03) ∗ (200 𝑚) ∗ (0.14)
2 ∗
𝜋
4
∗ (0.2 𝑚)
= 0
𝑃 = 3.23 ∗ 10 𝑃𝑎

6. Ejemplo 26. Una bomba es de 1 HP con 𝜂 = 75%. Determine 𝑄 y 𝑄
Tubería 𝐿 + 𝛴𝐿𝑒 (m) 𝐷 (m) 𝜀
𝐷
A 200 0.15 0.001
B 300 0.10 0.002
1 2
𝑄 = 0.12 𝑚 /𝑠 Agua a 20°C 𝜇 = 1 ∗ 10 𝑃𝑎 ⋅ 𝑠
A
B
𝑓 𝐿 𝑄
2 ∗
𝜋
4
∗ 𝐷
= −
(𝑃 +) 𝜂
𝑚̇
+
𝑓 𝐿 𝑄
2 ∗
𝜋
4
∗ 𝐷
𝑚̇ = 𝑄 ∗ 𝜌
Si 𝑓 = 0.022 ; 𝑓 = 0.025
(0.022) (200) 𝑄
2
𝜋
4
(0.15)
= −
746 (0.75)
𝑄 (1000)
+
0.025 (300) 𝑄
2
𝜋
4
(0.10)
𝑄 = 0.12 = 𝑄 + 𝑄
(4.7 ∗ 10 ) ∗ (0.12 − 𝑄 ) = −
0.5595
𝑄
+ (6.08 ∗ 10 ) ∗ 𝑄
(5.61 ∗ 10 ) 𝑄 + (1.128 ∗ 10 ) 𝑄 − 676.8 𝑄 − 0.5595 = 0
𝑄 = −0.046 𝑚 /𝑠 X
𝑄 = 0.0266 𝑚 /𝑠
𝑄 = −8.16 ∗ 10 𝑚 /𝑠 X
𝑄 = 0.0266 𝑚 /𝑠 𝑄 = 0.0934 𝑚 /𝑠

7. Ejemplo 27. Determine la potencia de la bomba para transportar 80 L/s de agua a 20°C y 1 a
2 (𝜂 = 0.8) QF= 0.08 𝑚3
/𝑠
Tubería 𝐿 (m) 𝐷 (m) 𝑓
A 100 0.10 0.025
B 50 0.12 "
C 100 0.08 "
D 150 0.06 "
E 200 0.08 "
F 100 0.10 "
Punto 𝑍 (m) 𝑃 (𝑏𝑎𝑟)
1 3 (Atmosférica)
2 12 "
B

8. Sistema BC 𝑊 = 𝑊
0.025 (50) 𝑄
2 ∗
𝜋
4
∗ (0.12)
=
0.025 (100) 𝑄
2 ∗
𝜋
4
∗ (0.08)
40718.71 𝑄 = 618415.427𝑄
𝑄 = 3.897 𝑄
𝑄 + 𝑄 = 𝑄
(3.897 𝑄 ) + 𝑄 = 𝑄 𝑄 =
1
4.897
𝑄 ‫؞‬ 𝑄 = 0.204 𝑄
𝑄 +
1
3.897
𝑄 = 𝑄 𝑄 =
1
1.256
𝑄 ‫؞‬ 𝑄 = 0.796 𝑄
Sistema C E D 𝑊 + 𝑊 = 𝑊
0.025 (100) 𝑄
2 ∗
𝜋
4
∗ (0.08)
+
0.025 (200) 𝑄
2 ∗
𝜋
4
∗ (0.08)
=
0.025 (150) 𝑄
2 ∗
𝜋
4
∗ (0.06)
(100) ∗ (0.204 𝑄 )
(0.08)
+
200 ∗ 𝑄
(0.08)
=
150 ∗ 𝑄
(0.06)
(6.1 ∗ 10 ) 𝑄 = (1.93 ∗ 10 ) 𝑄
𝑄 = 1.77 𝑄 𝑄 + 𝑄 = 𝑄 = 0.08
1.77 𝑄 + 𝑄 = 0.08
𝑄 = 0.029 𝑚 /𝑠
𝑄 = 0.051 𝑚 /𝑠
𝑄 = 0.041 𝑚 /𝑠
Balance de masa
𝑄 = 𝑄 + 𝑄 + 𝑄
𝑄 + 𝑄 = 𝑄
𝑄 + 𝑄 = 𝑄
𝑄 = 𝑄 = 0.08 𝑚 /𝑠
Ecuación de Bernoulli
∆𝑃
𝜌
+
∆𝑣
2
+ 𝑔∆𝑧 + 𝑊 + 𝑊 + +𝑊 + +𝑊 = 0

9. Tubería ramificada
9.81 (12 − 3) −
( 𝑃 𝑂 +)(0.8)
0.08(1000)
+
0.025 (100)0.082
2 ∗
𝜋
4
2
∗ (0.10)5
𝐴
+
0.025 (150)0.0292
2 ∗
𝜋
4
2
∗ (0.06)5
𝐷
+
0.025 (100)0.082
2 ∗
𝜋
4
2
∗ (0.10)5
𝐹
= 0
𝑃 = 5.97 ∗ 10 𝑊 = 800.5 𝐻𝑃
A B
C
D
∆𝑃
𝜌
+
∆𝑣
2
+ 𝑔∆𝑧 + 𝑊 + 𝑊 = 0 ∗
1
𝑔
∆𝑃
𝜌 𝑔
+
∆𝑣
2 𝑔
+ ∆𝑧 + ℎ + ℎ = 0
∆𝑃
𝜌 𝑔
+
∆𝑣
2 𝑔
+ ∆𝑧 ≡ ℎ
∆ℎ + ∆ℎ + ∆ℎ = 0
Si no hay máquinas
∆ℎ + ∆ℎ = 0
∆ℎ +
𝑓 𝐿 𝑄
2 𝑔
𝜋
4 𝐷
= 0
Altura piezométrica
Pérdida de carga

10. Ejemplo 28. Determine caudal y dirección de cada una de las tuberías.
Punto Z (m) P bar Tub, m 𝐿 + 𝛴𝐿𝑒
(m)
𝐷 (m) 𝑓
1 20 0 A 50 0.08 0.025
2 15 −0.10 B 40 0.06 "
3 10 0.5 C 100 0.08 "
4 5 1.0 D 80 0.05 "
X 8 - Escriba aquí la ecuación.
Fluido; 𝜌 = 900 ; 𝜇 = 8 ∗ 10 𝑘𝑔/𝑚𝑠
Solución:
Calculando 
Punto 
1 0.00563
2 0.00307
3 0.00398
4 0.00138
𝑄 =
2 𝑔
𝜋
4 𝐷
𝑓 𝐿
(−∆ℎ)
/
𝑄 = 𝛼√−∆ℎ
𝛼 =
2 𝑔
𝜋
4 𝐷
𝑓 𝐿
∆𝑃
𝜌 𝑔
+
∆𝑣
2 𝑔
+ ∆𝑧 ≡ ℎ
∆𝑃
𝜌 𝑔
+ ∆𝑧 ≡ ℎ
Punto Z (m) P bar
1 20 0
ℎ =
∆𝑃
𝜌 𝑔
+ ∆𝑧 = 0 + 0 + 20 = 20 𝑚

11. Los datos obtenidos anteriormente se presentan en la siguiente tabla.
Punto Z (m)  P/  g ∆h
1 20 0.00563 0.000 20.00
2 15 0.00307 -1.133 13.87
3 10 0.00398 5.663 15.66
4 5 0.00138 11.326 16.33
x 8
Punto Z (m) P bar
2 15 −0.1
ℎ =
∆𝑃
𝜌 𝑔
+ ∆𝑧 =
(−0.1 𝑏𝑎𝑟)
100 ∗ 10 𝑃𝑎
𝑏𝑎𝑟
900
𝑘𝑔
𝑚
∗ 9.81
𝑚
𝑠
+ 15
ℎ = −1.1326 + 15 = 13.8674
Punto Z (m) P bar
3 10 0.5
ℎ =
∆𝑃
𝜌 𝑔
+ ∆𝑧 =
(0.5 𝑏𝑎𝑟)
100 ∗ 10 𝑃𝑎
𝑏𝑎𝑟
900
𝑘𝑔
𝑚
∗ 9.81
𝑚
𝑠
+ 10
ℎ = 5.6632 + 10 = 15.6632
Punto Z (m) P bar
4 5 1
ℎ =
∆𝑃
𝜌 𝑔
+ ∆𝑧 =
(1 𝑏𝑎𝑟)
100 ∗ 10 𝑃𝑎
𝑏𝑎𝑟
900
𝑘𝑔
𝑚
∗ 9.81
𝑚
𝑠
+ 5
ℎ = 11.3263 + 5 = 16.3263

12. Los siguientes pasos del problema son a prueba y error.
 Suponer hx
 Identificar direcciones de las tuberías
 Calcular Q
 Balance de materia
Prueba y error
A
B
C
D
Suponer hx=18 m
𝑄 = 𝑄 + 𝑄 + 𝑄
𝑄 = 𝛼√−∆ℎ = 0.0056 ∗ √20 − 18 𝑄 = 0.00797
1 − 𝑥
𝑄 = 𝛼√−∆ℎ = 0.00307 ∗ √18 − 13.87 𝑄 = 0.00624
𝑥 − 2
𝑄 = 𝛼√−∆ℎ = 0.00398 ∗ √18 − 15.66 𝑄 = 0.00609
𝑥 − 3
𝑄 = 𝛼√−∆ℎ = 0.00138 ∗ √18 − 16.33 𝑄 = 0.00178
𝑥 − 4
𝑄 = 𝑄 + 𝑄 + 𝑄
𝑄 = 0.00797 ≠ 0.00624 + 0.00609 + 0.00178 = 0.01410
20 m
15 m
10 m
5 m

13. 𝑄 + 𝑄 + 𝑄 = 0.01410
0.01410 > 0.00797 ∴ Sale más de lo que entra, bajar la altura.
Suponer hx=14 m
Es conveniente igualar hx a alguno de los h en los tanques.
Suponer hx = h3
𝑄 = 𝛼√−∆ℎ = 0.0056 ∗ √20 − 14 𝑄 = 0.01380
𝑄 = 𝛼√−∆ℎ = 0.00307 ∗ √14 − 13.87 𝑄 = 0.00112
𝑄 = 𝛼√−∆ℎ = 0.00398 ∗ √14 − 15.66 𝑄 = 0.00514
𝑄 = 𝛼√−∆ℎ = 0.00138 ∗ √14 − 16.33 𝑄 = 0.00210
𝑄 = 𝑄 + 𝑄 + 𝑄
𝑄 = 0.0138 ≠ 0.00835
𝑄 = 𝛼√−∆ℎ = 0.0056 ∗ √20 − 16.33 𝑄 = 0.01079
𝑄 = 𝛼√−∆ℎ = 0.00307 ∗ √16.33 − 13.87 𝑄 = 0.00481
𝑄 = 𝛼√−∆ℎ = 0.00398 ∗ √16.33 − 15.66 𝑄 = 0.00325
𝑄 = 𝛼√−∆ℎ = 0.00138 ∗ √16.33 − 16.33 𝑄 = 0.00008
𝑄 = 𝑄 + 𝑄 + 𝑄
𝑄 = 0.01079 ≠ 0.00815
ℎ = 16.73139
Punto Q
1 0.010183
2 0.005192
3 0.004116
4 0.000875