Sistemas ramificados P1

Por: publicado el Oct 13, 2022

Gráfico del sistema

Sistema ramificado mecanica de fluidos

Datos del sistema

TuberıˊaL[m]D[m]ϵ/D1200010,00015223000,600,001325001,200,002\begin{array}{|l|l|l|l|} \hline Tubería & L {[}m{]} & D{[}m{]} & ϵ/D \\ \hline 1 & 2000 & 1 & 0,00015 \\ \hline 2 & 2300 & 0,60 & 0,001 \\ \hline 3 & 2500 & 1,20 & 0,002 \\ \hline \end{array}

Resolución por el primer método

Realizando un análisis de las tuberías tendríamos

zA−hL1=hJ→zA−f1L1D1v122g=hJ→120−f120001v122⋅9,81=80  (1)zB−hL2=hJ→zB−f2L2D2v222g=hJ→100−f223000,6v222⋅9,81=80  (2)hJ−hL3=zC→hJ−f3L3D3v322g=zC→80−f325001,20v322⋅9,81=28  (3)\begin{array}{llr} z_{A} - h_{L1} = h_{J} \rightarrow & z_{A} - f_{1}\frac{L_{1}}{D_{1}}\frac{v_{1}^{2}}{2g} = h_{J} \rightarrow & 120 - f_{1}\frac{2000}{1}\frac{v_{1}^{2}}{2 \cdot 9,81} = 80\space\space\textbf{(1)}\\ \\ z_{B} - h_{L2} = h_{J} \rightarrow & z_{B} - f_{2}\frac{L_{2}}{D_{2}}\frac{v_{2}^{2}}{2g} = h_{J} \rightarrow & 100 - f_{2}\frac{2300}{0,6}\frac{v_{2}^{2}}{2 \cdot 9,81} = 80 \space\space\textbf{(2)}\\ \\ h_{J} - h_{L3} = z_{C}\rightarrow & h_{J} - f_{3}\frac{L_{3}}{D_{3}}\frac{v_{3}^{2}}{2g} = z_{C}\rightarrow & 80 - f_{3}\frac{2500}{1,20}\frac{v_{3}^{2}}{2 \cdot 9,81} = 28 \space\space\textbf{(3)} \end{array}

Utilizando la ecuación de Von Karman:

1f=−2log⁡(ϵ/D3,7)\frac{1}{\sqrt{f}} = - 2\log\left( \frac{\epsilon/D}{3,7} \right)

f1=0,013; f2=0,020;f3=0,023\begin{array}{ccc} f_{1} = 0,013;& \ f_{2} = 0,020 ; & f_{3} = 0,023 \end{array}

Utilizando las ecuaciones (1), (2) y (3):

v1=5,494[m/s]→Q1=4,315[m3/s]v2=2,262[m/s]→Q2=0,640[m3/s]v3=4,614[m/s]→Q3=5,218[m3/s]\begin{array}{ccc} v_{1} = 5,494 [ m/s] & \rightarrow & Q_{1} = 4,315 [ m^{3}/s]\\ v_{2} = 2,262 [ m/s] & \rightarrow & Q_{2} = 0,640 [ m^{3}/s]\\ v_{3} = 4,614 [ m/s] & \rightarrow & Q_{3} = 5,218 [ m^{3}/s]\\ \end{array}

Calculamos ΔQ\mathrm{\Delta}Q:

ΔQ=(Q1+Q2)−Q3=(4,315+0,640)−5,218\mathrm{\Delta}Q = \left( Q_{1} + Q_{2} \right) - Q_{3} = \left( 4,315 + 0,640 \right) - 5,218

ΔQ=−0,263 [m3/s]\mathrm{\Delta}Q = - 0,263\ [ m^{3}/s]

Asumiendo un valor de hJ=76 [m]h_{J} = 76\ [ m]

Realizando un análisis para cada tubería tendríamos:

zA−hL1=hJ→zA−f1L1D1v122g= hJ→120−f120001v122⋅9,81=76 (1)zB−hL2=hJ→zB−f2L2D2v222g= hJ→100−f223000,6v222⋅9,81=76 (2)hJ−hL3=zC→ hJ−f3L3D3v322g=zC→76−f325001,20v322⋅9,81=28 (3)\begin{array}{lrlr} z_{A} - h_{L1} = h_{J} \rightarrow & z_{A} - f_{1}\frac{L_{1}}{D_{1}}\frac{v_{1}^{2}}{2g} = \ h_{J} \rightarrow & 120 - f_{1}\frac{2000}{1}\frac{v_{1}^{2}}{2 \cdot 9,81} = 76 \ &\textbf{(1)}\\ \\ z_{B} - h_{L2} = h_{J} \rightarrow & z_{B} - f_{2}\frac{L_{2}}{D_{2}}\frac{v_{2}^{2}}{2g} = \ h_{J}\rightarrow & 100 - f_{2}\frac{2300}{0,6}\frac{v_{2}^{2}}{2 \cdot 9,81} = 76\ &\textbf{(2)}\\ \\ h_{J} - h_{L3} = z_{C}\rightarrow & \ h_{J} - f_{3}\frac{L_{3}}{D_{3}}\frac{v_{3}^{2}}{2g} = z_{C}\rightarrow & 76 - f_{3}\frac{2500}{1,20}\frac{v_{3}^{2}}{2 \cdot 9,81} = 28\ &\textbf{(3)} \end{array}

Utilizando la ecuación de Von Karman:

f1=0,013; f2=0,020;f3=0,023\begin{array}{ccc} f_{1} = 0,013; & \ f_{2} = 0,020; & f_{3} = 0,023\\ \end{array}

Utilizando las ecuaciones (1), (2) y (3):

v1=5,762 [m/s] →Q1=4,525 [m3/s]v2=2,478 [m/s]  →Q2=0,701 [m3/s]v3=4,433 [m/s]  →Q3=5,014 [m3/s]\begin{array}{lcr} v_{1} = 5,762\ [ m/s] &\ \rightarrow & Q_{1} = 4,525\ [ m^{3}/s]\\ v_{2} = 2,478\ [ m/s] &\ \ \rightarrow & Q_{2} = 0,701\ [ m^{3}/s]\\ v_{3} = 4,433\ [ m/s] &\ \ \rightarrow & Q_{3} = 5,014\ [ m^{3}/s]\\ \end{array}

Calculando ΔQ\mathrm{\Delta}Q:

ΔQ=(Q1+Q2)−Q3=(4,525+0,701)−5,014\mathrm{\Delta}Q = \left( Q_{1} + Q_{2} \right) - Q_{3} = \left( 4,525 + 0,701 \right) - 5,014

ΔQ=0,212 [m3/s]\mathrm{\Delta}Q = 0,212\ [ m^{3}/s]

Trabajando ahora con el valor de hJ=77,785 [m]h_{J} = 77,785\ [ m]

Realizando un análisis para cada tubería tendríamos:

zA−hL1=hJ→  zA−f1L1D1v122g=hJ→120−f120001v122⋅9,81=77,785  (1)zB−hL2=hJ→zB−f2L2D2v222g=  hJ→100−f223000,6v222⋅9,81=77,785 (2)hJ−hL3=zC→  hJ−f3L3D3v322g=zC→77,785−f325001,20v322⋅9,81=28 (3)z_{A} - h_{L1} = h_{J}\rightarrow \ \ z_{A} - f_{1}\frac{L_{1}}{D_{1}}\frac{v_{1}^{2}}{2g} = h_{J}\rightarrow 120 - f_{1}\frac{2000}{1}\frac{v_{1}^{2}}{2 \cdot 9,81} = 77,785 \ \textbf{ (1)}\\ z_{B} - h_{L2} = h_{J}\rightarrow z_{B} - f_{2}\frac{L_{2}}{D_{2}}\frac{v_{2}^{2}}{2g} = \ \ h_{J}\rightarrow 100 - f_{2}\frac{2300}{0,6}\frac{v_{2}^{2}}{2 \cdot 9,81} = 77,785 \textbf{ (2)}\\ h_{J} - h_{L3} = z_{C}\rightarrow \ \ h_{J} - f_{3}\frac{L_{3}}{D_{3}}\frac{v_{3}^{2}}{2g} = z_{C}\rightarrow 77,785 - f_{3}\frac{2500}{1,20}\frac{v_{3}^{2}}{2 \cdot 9,81} = 28 \textbf{ (3)}\\

Utilizando la ecuación de Von Karman:

f1=0,013  ; f2=0,020  ; f3=0,023f_{1} = 0,013 \ \ ; \ f_{2} = 0,020 \ \ ; \ f_{3} = 0,023

Utilizando las ecuaciones (1), (2) y (3):

v1=5,644  [m/s] →Q1=4,433 [m3/s]v2=2,384 [m/s]  →Q2=0,674 [m3/s]v3=4,515 [m/s]  →Q3=5,106 [m3/s]v_{1} = 5,644\ \ [ m/s] \ \rightarrow Q_{1} = 4,433\ [ m^{3}/s]\\ v_{2} = 2,384\ [ m/s] \ \ \rightarrow Q_{2} = 0,674\ [ m^{3}/s]\\ v_{3} = 4,515\ [ m/s] \ \ \rightarrow Q_{3} = 5,106\ [ m^{3}/s]\\

Calculando ΔQ\mathrm{\Delta}Q:

ΔQ=(Q1+Q2)−Q3=(4,433+0,674)−5,106ΔQ=0,001 [m3/s]\mathrm{\Delta}Q = \left( Q_{1} + Q_{2} \right) - Q_{3} = \left( 4,433 + 0,674 \right) - 5,106\\ \mathrm{\Delta}Q = 0,001\ [ m^{3}/s]\\

Entonces los caudales buscados son:

Q1=4,433 [m3/s]Q2=0,674 [m3/s]Q3=5,106 [m3/s]Q_{1} = 4,433\ [ m^{3}/s]\\ Q_{2} = 0,674\ [ m^{3}/s]\\ Q_{3} = 5,106\ [ m^{3}/s]\\

Segundo Método

Asumiendo un valor de hJ=80 [m]h_{J} = 80\ [ m]

Realizando un análisis para cada tubería tendríamos:

zA−hL1=hJ  →zA−f1L1D1v122g=hJ   →120−f120001v122⋅9,81=80   (1)zB−hL2=hJ  →  zB−f2L2D2v222g=  hJ   →100−f223000,6v222⋅9,81=80 (2)hJ−hL3=zC  →hJ−f3L3D3v322g=zC   →80−f325001,20v322⋅9,81=28 (3)z_{A} - h_{L1} = h_{J}\ \ \rightarrow z_{A} - f_{1}\frac{L_{1}}{D_{1}}\frac{v_{1}^{2}}{2g} = h_{J} \ \ \ \rightarrow 120 - f_{1}\frac{2000}{1}\frac{v_{1}^{2}}{2 \cdot 9,81} = 80 \ \ \ (1)\\ z_{B} - h_{L2} = h_{J} \ \ \rightarrow \ \ z_{B} - f_{2}\frac{L_{2}}{D_{2}}\frac{v_{2}^{2}}{2g} = \ \ h_{J} \ \ \ \rightarrow 100 - f_{2}\frac{2300}{0,6}\frac{v_{2}^{2}}{2 \cdot 9,81} = 80 \ (2)\\ h_{J} - h_{L3} = z_{C} \ \ \rightarrow h_{J} - f_{3}\frac{L_{3}}{D_{3}}\frac{v_{3}^{2}}{2g} = z_{C} \ \ \ \rightarrow 80 - f_{3}\frac{2500}{1,20}\frac{v_{3}^{2}}{2 \cdot 9,81} = 28 \ (3)\\

Utilizando la ecuación de Von Karman, para empezar la suposición de los factores de fricción:

f1=0,013;   f2=0,020;   f3=0,023f_{1} = 0,013 ; \ \ \ f_{2} = 0,020 ; \ \ \ f_{3} = 0,023\\

Reynolds:  NRe=vDν  ;Colebrook:1f=−0,86Ln(ϵ/D3,7+2,51NRef)   Reynolds: \ \ N_{\text{Re}} = \frac{\text{vD}}{\nu} \ \ ; Colebrook: \frac{1}{\sqrt{f}} = - 0,86Ln\left( \frac{\epsilon/D}{3,7} + \frac{2,51}{N_{\text{Re}}\sqrt{f}} \right)\text{\ \ \ }

TABLA DE ITERACION

fsup1v1 [m/s]Q1 [m3/s]NRe1f10,0135,4944,3154777391,300,0140,0145,2944,1584603478,260,014\begin{array}{ccccc} f_{sup1} & v_{1}\ [m/s] & Q_{1}\ [ m^{3}/s] & N_{Re1} & f_{1}\\ \hline 0,013 & 5,494 & 4,315 & 4777391,30 & 0,014\\ 0,014 & 5,294 & 4,158 & 4603478,26 & 0,014\\ \end{array}

fsup2v2 [m/s]Q2 [m3/s]NRe2f20,0202,2620,6401180173,910,020\begin{array}{ccccc} f_{sup2} & v_{2}\ [ m/s] & Q_{2}\ [ m^{3}/s] & N_{Re2} & f_{2} \\ \hline 0,020 & 2,262 & 0,640 & 1180173,91 & 0,020\\ \end{array}

fsup2v2 [m/s]Q2 [m3/s]NRe2f20,0234,6145,2184814608,700,0240,0244,5175,1094713391,300,024\begin{array} {ccccc} f_{sup2} & v_{2}\ [ m/s] & Q_{2}\ [ m^{3}/s] & N_{Re2} & f_{2}\\ \hline 0,023 & 4,614 & 5,218 & 4814608,70 & 0,024\\ 0,024 & 4,517 & 5,109 & 4713391,30 & 0,024\\ \end{array}

Entonces, se tiene:

Utilizando las ecuaciones (1), (2) y (3):

v1=5,294  [m/s] →  Q1=4,158 [m3/s]v2=2,262 [m/s]  →  Q2=0,640 [m3/s]v3=4,517 [m/s]  →  Q3=5,109 [m3/s]v_{1} = 5,294\ \ [ m/s] \ \rightarrow \ \ Q_{1} = 4,158\ [ m^{3}/s]\\ v_{2} = 2,262\ [ m/s] \ \ \rightarrow \ \ Q_{2} = 0,640\ [ m^{3}/s]\\ v_{3} = 4,517\ [ m/s] \ \ \rightarrow \ \ Q_{3} = 5,109\ [ m^{3}/s]\\

Calculando ΔQ\mathrm{\Delta}Q:

ΔQ=(Q1+Q2)−Q3=(4,158+0,640)−5,109ΔQ=−0,311 [m3/s]\mathrm{\Delta}Q = \left( Q_{1} + Q_{2} \right) - Q_{3} = \left( 4,158 + 0,640 \right) - 5,109\\ \mathrm{\Delta}Q = - 0,311\ [ m^{3}/s]

Asumiendo un valor de hJ=76 [m]h_{J} = 76\ [ m]

Realizando un análisis para cada tubería tendríamos:

zA−hL1=hJ  →zA−f1L1D1v122g=hJ   →120−f120001v122⋅9,81=76   (1)zB−hL2=hJ  →  zB−f2L2D2v222g=  hJ   →100−f223000,6v222⋅9,81=76 (2)hJ−hL3=zC  →hJ−f3L3D3v322g=zC   →76−f325001,20v322⋅9,81=28 (3)z_{A} - h_{L1} = h_{J} \ \ \rightarrow z_{A} - f_{1}\frac{L_{1}}{D_{1}}\frac{v_{1}^{2}}{2g} = h_{J} \ \ \ \rightarrow 120 - f_{1}\frac{2000}{1}\frac{v_{1}^{2}}{2 \cdot 9,81} = 76 \ \ \ \textbf{(1)}\\ z_{B} - h_{L2} = h_{J} \ \ \rightarrow \ \ z_{B} - f_{2}\frac{L_{2}}{D_{2}}\frac{v_{2}^{2}}{2g} = \ \ h_{J} \ \ \ \rightarrow 100 - f_{2}\frac{2300}{0,6}\frac{v_{2}^{2}}{2 \cdot 9,81} = 76 \ \textbf{(2)}\\ h_{J} - h_{L3} = z_{C} \ \ \rightarrow h_{J} - f_{3}\frac{L_{3}}{D_{3}}\frac{v_{3}^{2}}{2g} = z_{C} \ \ \ \rightarrow 76 - f_{3}\frac{2500}{1,20}\frac{v_{3}^{2}}{2 \cdot 9,81} = 28 \ \textbf{(3)}\\

Utilizando la ecuación de Von Karman, para empezar la suposición de los factores de fricción:

f1=0,013;   f2=0,020;   f3=0,023f_{1} = 0,013 ; \ \ \ f_{2} = 0,020 ; \ \ \ f_{3} = 0,023

Reynolds:  NRe=vDν  ;Reynolds: \ \ N_{\text{Re}} = \frac{\text{vD}}{\nu} \ \ ;

Colebrook:1f=−0,86Ln(ϵ/D3,7+2,51NRef)Colebrook: \frac{1}{\sqrt{f}} = - 0,86Ln \left( \frac{\epsilon/D}{3,7} + \frac{2,51}{N_{\text{Re}}\sqrt{f}} \right)

TABLA DE ITERACION

fsup1v1 [m/s]Q1 [m3/s]NRe1f10,0135,7624,5255010434,780,013\begin{array} {ccccc} f_{sup1} & v_{1}\ [ m/s] & Q_{1}\ [ m^{3}/s] & N_{Re1} & f_{1}\\ \hline 0,013 & 5,762 & 4,525 & 5010434,78 & 0,013\\ \end{array}

fsup2v2 [m/s]Q2 [m3/s]NRe2f20,0202,4780,7011292869,560,020\begin{array} {ccccc} f_{sup2} & v_{2}\ [ m/s] & Q_{2}\ [ m^{3}/s] & N_{Re2} & f_{2}\\ \hline 0,020 & 2,478 & 0,701 & 1292869,56 & 0,020\\ \end{array}

fsup2v2 [m/s]Q2 [m3/s]NRe2f20,0234,4335,0144625739,130,0240,0244,5175,1094713391,300,024\begin{array} {ccccc} f_{sup2} & v_{2}\ [ m/s] & Q_{2}\ [ m^{3}/s] & N_{Re2} & f_{2}\\ \hline 0,023 & 4,433 & 5,014 & 4625739,13 & 0,024\\ 0,024 & 4,517 & 5,109 & 4713391,30 & 0,024\\ \end{array}

Utilizando las ecuaciones (1), (2) y (3):

v1=5,762  [m/s] →  Q1=4,525 [m3/s]v2=2,478 [m/s]  →  Q2=0,701 [m3/s]v3=4,517 [m/s]  →  Q3=5,109 [m3/s]v_{1} = 5,762\ \ [ m/s] \ \rightarrow \ \ Q_{1} = 4,525\ [ m^{3}/s]\\ v_{2} = 2,478\ [ m/s] \ \ \rightarrow \ \ Q_{2} = 0,701\ [ m^{3}/s]\\ v_{3} = 4,517\ [ m/s] \ \ \rightarrow \ \ Q_{3} = 5,109\ [ m^{3}/s]\\

Calculamos ΔQ\mathrm{\Delta}Q:

ΔQ=(Q1+Q2)−Q3=(4,525+0,701)−5,109\mathrm{\Delta}Q = \left( Q_{1} + Q_{2} \right) - Q_{3} = \left( 4,525 + 0,701 \right) - 5,109

ΔQ=0,117 [m3/s]\mathrm{\Delta}Q = 0,117\ [ m^{3}/s]

Trabajando ahora con el valor de hJ=77,093 [m]h_{J} = 77,093\ [ m]

Realizando un análisis para cada tubería tendríamos:

zA−hL1=hJ  →zA−f1L1D1v122g=hJ   →120−f120001v122⋅9,81=77,093   (1)zB−hL2=hJ  →  zB−f2L2D2v222g=  hJ   →100−f223000,6v222⋅9,81=77,093 (2)hJ−hL3=zC  →hJ−f3L3D3v322g=zC   →77,093−f325001,20v322⋅9,81=28 (3)z_{A} - h_{L1} = h_{J} \ \ \rightarrow z_{A} - f_{1}\frac{L_{1}}{D_{1}}\frac{v_{1}^{2}}{2g} = h_{J} \ \ \ \rightarrow 120 - f_{1}\frac{2000}{1}\frac{v_{1}^{2}}{2 \cdot 9,81} = 77,093 \ \ \ \textbf{(1)}\\ z_{B} - h_{L2} = h_{J} \ \ \rightarrow \ \ z_{B} - f_{2}\frac{L_{2}}{D_{2}}\frac{v_{2}^{2}}{2g} = \ \ h_{J} \ \ \ \rightarrow 100 - f_{2}\frac{2300}{0,6}\frac{v_{2}^{2}}{2 \cdot 9,81} = 77,093 \ \textbf{(2)}\\ h_{J} - h_{L3} = z_{C} \ \ \rightarrow h_{J} - f_{3}\frac{L_{3}}{D_{3}}\frac{v_{3}^{2}}{2g} = z_{C} \ \ \ \rightarrow 77,093 - f_{3}\frac{2500}{1,20}\frac{v_{3}^{2}}{2 \cdot 9,81} = 28 \ \textbf{(3)}

Utilizando la ecuación de Von Karman, para empezar la suposición de los factores de fricción:

f1=0,013;   f2=0,020;   f3=0,023f_{1} = 0,013 ; \ \ \ f_{2} = 0,020 ; \ \ \ f_{3} = 0,023

Reynolds:  NRe=vDν  ;Reynolds: \ \ N_{\text{Re}} = \frac{\text{vD}}{\nu} \ \ ;

Colebrook:1f=−0,86Ln(ϵ/D3,7+2,51NRef)   Colebrook: \frac{1}{\sqrt{f}} = - 0,86Ln\left( \frac{\epsilon /D}{3,7} + \frac{2,51}{N_{\text{Re}}\sqrt{f}} \right)\text{\ \ \ }

TABLA DE ITERACION

fsup1v1 [m/s]Q1 [m3/s]NRe1f10,0135,6904,4694947826,090,013\begin{array} {ccccc} f_{sup1} & v_{1}\ [ m/s] & Q_{1}\ [ m^{3}/s] & N_{Re1} & f_{1}\\ \hline 0,013 & 5,690 & 4,469 & 4947826,09 & 0,013\\ \end{array}

fsup2v2 [m/s]Q2 [m3/s]NRe2f20,0202,4210,6851263130,430,020\begin{array} {ccccc} f_{sup2} & v_{2}\ [ m/s] & Q_{2}\ [ m^{3}/s] & N_{Re2} & f_{2}\\ \hline 0,020 & 2,421 & 0,685 & 1263130,43 & 0,020\\ \end{array}

fsup2v2 [m/s]Q2 [m3/s]NRe2f20,0234,4835,0704677913,040,0240,0244,5175,1094713391,300,024\begin{array} {ccccc} f_{sup2} & v_{2}\ [ m/s] & Q_{2}\ [ m^{3}/s] & N_{Re2} & f_{2}\\ \hline 0,023 & 4,483 & 5,070 & 4677913,04 & 0,024\\ 0,024 & 4,517 & 5,109 & 4713391,30 & 0,024\\ \end{array}

Calculamos ΔQ\mathrm{\Delta}Q:

ΔQ=(Q1+Q2)−Q3=(4,469+0,685)−5,109ΔQ=0,045 [m3/s]\mathrm{\Delta}Q = \left( Q_{1} + Q_{2} \right) - Q_{3} = \left( 4,469 + 0,685 \right) - 5,109\\ \mathrm{\Delta}Q = 0,045\ [ m^{3}/s]

Entonces los caudales bus cados son:

Q1=4,469 [m3/s]Q2=0,685 [m3/s]Q3=5,109 [m3/s]Q_{1} = 4,469\ [ m^{3}/s]\\ Q_{2} = 0,685\ [ m^{3}/s]\\ Q_{3} = 5,109\ [ m^{3}/s]\\