Storyboard

When hydraulic elements are connected in series, the flow remains constant, but each hydraulic element experiences a pressure drop. The sum of these pressure drops equals the total drop, and therefore, the total hydraulic resistance is equal to the sum of all individual hydraulic resistances. On the other hand, the inverse of the total hydraulic conductivity is equal to the sum of the inverses of the hydraulic conductivities.

>Model

ID:(2109, 0)

Iframe

>Top

ID:(15955, 0)

Concept

>Top

In the case of a sum where the elements are connected in series, the total hydraulic resistance of the system is calculated by summing the individual resistances of each element.

One way to model a tube with varying cross-section is to divide it into sections with constant radius and then sum the hydraulic resistances in series. Suppose we have a series of the hydraulic resistance in a network ($R_{hk}$), which depends on the viscosity ($\eta$), the cylinder k radio ($R_k$), and the tube k length ($\Delta L_k$) via the following equation:

In each segment, there will be a pressure difference in a network ($\Delta p_k$) with the hydraulic resistance in a network ($R_{hk}$) and the volume flow ($J_V$) to which Darcy's Law is applied:

the total pressure difference ($\Delta p_t$) will be equal to the sum of the individual pressure difference in a network ($\Delta p_k$):

therefore,

$\Delta p_t=\displaystyle\sum_k \Delta p_k=\displaystyle\sum_k (R_{hk}J_V)=\left(\displaystyle\sum_k R_{hk}\right)J_V\equiv R_{st}J_V$

Thus, the system can be modeled as a single conduit with the hydraulic resistance calculated as the sum of the individual components:

ID:(15957, 0)

Top

>Top

Parameters

$R_{h1}$

R_h1

Hydraulic Resistance 1

kg/m^4s

$R_{h2}$

R_h2

Hydraulic Resistance 2

kg/m^4s

$R_{h3}$

R_h3

Hydraulic Resistance 3

kg/m^4s

$R_{pt}$

R_pt

Total hydraulic resistance in parallel

kg/m^4s

$R_{st}$

R_st

Total hydraulic resistance in series

kg/m^4s

Variables

$J_{Vt}$

J_Vt

Flujo de Volumen Total

m^3/s

$\Delta p_1$

Dp_1

Pressure Difference 1

Pa

$\Delta p_3$

Dp_3

Pressure Difference 3

Pa

$\Delta p_t$

Dp_t

Total pressure difference

Pa

$J_{V1}$

J_V1

Volume flow 1

m^3/s

$J_{V2}$

J_V2

Volume flow 2

m^3/s

Calculations

• Alternative method
• Traditional method
• Strategy
• 2D
• 3D

Equation

Number

First, select the equation: to , then, select the variable: to

---

Calculations

Symbol

Equation

Solved

Translated

Calculations

Symbol

Equation

Solved

Translated

Variable Given Calculate Target : Equation To be used

Equations

ID:(15956, 0)

Equation

>Top, >Model

The series combination of the hydraulic Resistance 1 ($R_{h1}$) and the hydraulic Resistance 2 ($R_{h2}$) results in a total sum of the total hydraulic resistance in series ($R_{st}$):

$R_{st} = R_{pt} + R_{h3}$

$R_{st} = R_{h1} + R_{h2}$

Conditions

Variables

$R_{h1}$

$R_{pt}$

Total hydraulic resistance in parallel

$kg/m^4s$

5429

$R_{h2}$

$R_{h3}$

Hydraulic Resistance 3

$kg/m^4s$

5427

$R_{st}$

Total hydraulic resistance in series

$kg/m^4s$

5428

Relation

ID:(3854, 0)

Equation

>Top, >Model

The parallel combination of the hydraulic Resistance 1 ($R_{h1}$) and the hydraulic Resistance 2 ($R_{h2}$) results in a total equivalent of the total hydraulic resistance in series ($R_{st}$):

$\displaystyle\frac{1}{ R_{pt} }=\displaystyle\frac{1}{ R_{h1} }+\displaystyle\frac{1}{ R_{h2} }$

Conditions

Variables

$R_{h1}$

Hydraulic Resistance 1

$kg/m^4s$

5425

$R_{h2}$

Hydraulic Resistance 2

$kg/m^4s$

5426

$R_{pt}$

Total hydraulic resistance in parallel

$kg/m^4s$

5429

Relation

ID:(3858, 0)

Equation

>Top, >Model

In the case of hydraulic resistances in series, the pressure drops across each of them, and the sum of these pressure drops is equal to the total pressure difference across the entire series.

For two resistances in series, the hydraulic Resistance 1 ($R_{h1}$) and the hydraulic Resistance 2 ($R_{h2}$), with their respective pressure drops the pressure Difference 1 ($\Delta p_1$) and the pressure Difference 2 ($\Delta p_2$), the sum of these drops equals the total pressure difference the total pressure difference ($\Delta p_t$):

$\Delta p_t = \Delta p_1 + \Delta p_3$

$\Delta p_t = \Delta p_1 + \Delta p_2$

Conditions

Variables

$\Delta p_1$

Pressure Difference 1

$Pa$

9841

$\Delta p_2$

$\Delta p_3$

Pressure Difference 3

$Pa$

5447

$\Delta p_t$

Total pressure difference

$Pa$

9842

Relation

ID:(9943, 0)

Equation

>Top, >Model

The flujo de Volumen Total ($J_{Vt}$) represents the total sum of the individual contributions from the volume flow 1 ($J_{V1}$) and the volume flow 2 ($J_{V2}$), from the elements connected in parallel:

$J_{Vt} = J_{V1} + J_{V2}$

Conditions

Variables

$J_{Vt}$

Flujo de Volumen Total

$m^3/s$

6611

$J_{V1}$

Volume flow 1

$m^3/s$

8478

$J_{V2}$

Volume flow 2

$m^3/s$

8479

Relation

ID:(12800, 0)

Equation

>Top, >Model

Darcy rewrites the Hagen Poiseuille equation so that the pressure difference ($\Delta p$) is equal to the hydraulic resistance ($R_h$) times the volume flow ($J_V$):

$\Delta p_t = R_{st} J_{Vt}$

$\Delta p = R_h J_V$

Conditions

Variables

$R_h$

$R_{st}$

Total hydraulic resistance in series

$kg/m^4s$

5428

$\Delta p$

$\Delta p_t$

Total pressure difference

$Pa$

9842

$J_V$

$J_{Vt}$

Flujo de Volumen Total

$m^3/s$

6611

Relation

Development

The volume flow ($J_V$) can be calculated from the hydraulic conductance ($G_h$) and the pressure difference ($\Delta p$) using the following equation:

$J_V = G_h \Delta p$

Furthermore, using the relationship for the hydraulic resistance ($R_h$):

$R_h = \displaystyle\frac{1}{ G_h }$

results in:

$\Delta p = R_h J_V$

ID:(3179, 1)

Equation

>Top, >Model

Darcy rewrites the Hagen Poiseuille equation so that the pressure difference ($\Delta p$) is equal to the hydraulic resistance ($R_h$) times the volume flow ($J_V$):

$\Delta p_1 = R_{pt} J_{Vt}$

$\Delta p = R_h J_V$

Conditions

Variables

$R_h$

$R_{pt}$

Total hydraulic resistance in parallel

$kg/m^4s$

5429

$\Delta p$

$\Delta p_1$

Pressure Difference 1

$Pa$

9841

$J_V$

$J_{Vt}$

Flujo de Volumen Total

$m^3/s$

6611

Relation

Development

The volume flow ($J_V$) can be calculated from the hydraulic conductance ($G_h$) and the pressure difference ($\Delta p$) using the following equation:

$J_V = G_h \Delta p$

Furthermore, using the relationship for the hydraulic resistance ($R_h$):

$R_h = \displaystyle\frac{1}{ G_h }$

results in:

$\Delta p = R_h J_V$

ID:(3179, 2)

Equation

>Top, >Model

Darcy rewrites the Hagen Poiseuille equation so that the pressure difference ($\Delta p$) is equal to the hydraulic resistance ($R_h$) times the volume flow ($J_V$):

$\Delta p_3 = R_{h3} J_{Vt}$

$\Delta p = R_h J_V$

Conditions

Variables

$R_h$

$R_{h3}$

Hydraulic Resistance 3

$kg/m^4s$

5427

$\Delta p$

$\Delta p_3$

Pressure Difference 3

$Pa$

5447

$J_V$

$J_{Vt}$

Flujo de Volumen Total

$m^3/s$

6611

Relation

Development

The volume flow ($J_V$) can be calculated from the hydraulic conductance ($G_h$) and the pressure difference ($\Delta p$) using the following equation:

$J_V = G_h \Delta p$

Furthermore, using the relationship for the hydraulic resistance ($R_h$):

$R_h = \displaystyle\frac{1}{ G_h }$

results in:

$\Delta p = R_h J_V$

ID:(3179, 3)

Equation

>Top, >Model

Darcy rewrites the Hagen Poiseuille equation so that the pressure difference ($\Delta p$) is equal to the hydraulic resistance ($R_h$) times the volume flow ($J_V$):

$\Delta p_1 = R_{h1} J_{V1}$

$\Delta p = R_h J_V$

Conditions

Variables

$R_h$

$R_{h1}$

Hydraulic Resistance 1

$kg/m^4s$

5425

$\Delta p$

$\Delta p_1$

Pressure Difference 1

$Pa$

9841

$J_V$

$J_{V1}$

Volume flow 1

$m^3/s$

8478

Relation

Development

The volume flow ($J_V$) can be calculated from the hydraulic conductance ($G_h$) and the pressure difference ($\Delta p$) using the following equation:

$J_V = G_h \Delta p$

Furthermore, using the relationship for the hydraulic resistance ($R_h$):

$R_h = \displaystyle\frac{1}{ G_h }$

results in:

$\Delta p = R_h J_V$

ID:(3179, 4)

Equation

>Top, >Model

Darcy rewrites the Hagen Poiseuille equation so that the pressure difference ($\Delta p$) is equal to the hydraulic resistance ($R_h$) times the volume flow ($J_V$):

$\Delta p_1 = R_{h2} J_{V2}$

$\Delta p = R_h J_V$

Conditions

Variables

$R_h$

$R_{h2}$

Hydraulic Resistance 2

$kg/m^4s$

5426

$\Delta p$

$\Delta p_1$

Pressure Difference 1

$Pa$

9841

$J_V$

$J_{V2}$

Volume flow 2

$m^3/s$

8479

Relation

Development

The volume flow ($J_V$) can be calculated from the hydraulic conductance ($G_h$) and the pressure difference ($\Delta p$) using the following equation:

$J_V = G_h \Delta p$

Furthermore, using the relationship for the hydraulic resistance ($R_h$):

$R_h = \displaystyle\frac{1}{ G_h }$

results in:

$\Delta p = R_h J_V$

ID:(3179, 5)