Hydraulic Fluid Power - Andrea Vacca

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      The following chapters of the book focus on the analysis of hydraulic systems operating in steady‐state conditions. Hence, after the presentation of the basic equations for hydraulic resistance and conservation of mass, it is now appropriate to provide the reader with the general approach that can be used to model a flow network.

      A flow network can be defined as any collection of elements (valves, cylinders) and sources (pumps). The network interconnections are the fluid conveyance elements.

      According to the approach also presented by Merritt [32], the flow and the pressure distribution within a network must satisfy three constraints:

      1 Flow–pressure relationshipEvery element of the circuit is characterized by a flow–pressure relationship. The simplest example is the case of the hydraulic resistance that can be used to describe pipes, fittings, and certain hydraulic valves, previously shown in Eqs. (3.38) and (3.39). The next chapters will present also relations for other elements, such as pumps, motors, and linear actuators.

      2 Flow lawThe flow law applies at any junction of pipes, and it was already presented as a direct consequence of the conservation of mass principle:(3.41) Equation (3.41) can be seen as the equivalent of the Kirchhoff's current law in the electric domain. Essentially, this equation states that the sum of the flow rates entering a junction has to be equal to the sum of the flow rates exiting it (Figure 3.16a).

      3 Pressure lawThe pressure law can be seen as the equivalent of the Kirchhoff's voltage law in the electric domain. It states that the overall pressure drop around any closed circuit has to be null:(3.42)

      Example 3.1 Series and parallel hydraulic connections

      The pressure drop–flow rate relation for three different pipes is known to be linear, as shown in the figure below. Find the pressure drop–flow rate relation for different configurations of the three pipes: (a) series; (b) parallel; and (c) series–parallel.

Schematic illustration of the pressure drop-flow rate relation for different configurations of the three pipes: (a) series, (b) parallel, and (c) series-parallel.

       Given:

      The linear characteristic of three pipe sections:

      ΔpA = RA · QA; ΔpB = RB · QB; ΔpC = RC · QC

       Find:

      The equivalent hydraulic resistance for the three cases:

      Δpseries = Rseries · Q; Δpparallel = Rparallel · Q; Δpseries/parallel = RC · Q

      Solution:

       Case (a) series

      This case can be solved by considering the quantities shown in the figure below:

Schematic illustration of considering the quantities of series pipe. p 1 minus p 2 equals upper Delta p Subscript upper A Baseline equals upper R Subscript upper A Baseline dot upper Q Subscript upper A Baseline semicolon p 2 minus p 3 equals upper Delta p Subscript upper B Baseline equals upper R Subscript upper B Baseline dot upper Q Subscript upper B Baseline semicolon p 3 minus p 4 equals upper Delta p Subscript upper C Baseline equals upper R Subscript upper C Baseline dot upper Q Subscript upper C Baseline p 1 minus p 4 equals left-parenthesis upper R Subscript upper A Baseline plus upper R Subscript upper B Baseline plus upper R Subscript upper C Baseline right-parenthesis dot upper Q

      which means

upper R Subscript series Baseline equals upper R Subscript upper A Baseline plus upper R Subscript upper B Baseline plus upper R Subscript upper C

       Case (b) parallel

      The approach is similar to case (a). With reference to the figure below,

Schematic illustration of considering the quantities of parallel pipe. p 1 minus p 2 equals normal upper Delta p Subscript upper A Baseline equals upper R Subscript upper A Baseline dot upper Q Subscript upper A Baseline equals normal upper Delta p Subscript upper B Baseline equals upper R Subscript upper B Baseline dot upper Q Subscript upper B Baseline equals normal upper Delta p Subscript upper C Baseline equals upper R Subscript upper C Baseline dot upper Q Subscript upper C

      Considering the flow law,

upper Q equals upper Q Subscript upper A Baseline plus upper Q Subscript upper B Baseline plus upper Q Subscript upper C

      which gives

upper Q equals StartFraction p 1 minus p 2 Over upper R Subscript upper A Baseline EndFraction plus StartFraction p 1 minus p 2 Over upper R Subscript upper B Baseline EndFraction plus StartFraction p 1 minus p 2 Over upper R Subscript upper C Baseline EndFraction

      or also

upper Q equals left-parenthesis p 1 minus p 2 right-parenthesis dot left-parenthesis StartFraction 1 Over upper R Subscript upper A Baseline EndFraction plus StartFraction 1 Over upper R Subscript upper B Baseline EndFraction plus StartFraction 1 Over upper R Subscript upper C Baseline EndFraction right-parenthesis

      which means

upper R Subscript parallel Baseline equals left-parenthesis StartFraction 1 Over upper R Subscript upper A Baseline EndFraction plus StartFraction 1 Over upper R Subscript upper B Baseline EndFraction plus StartFraction 1 Over upper R Subscript upper C Baseline EndFraction right-parenthesis Superscript negative 1

       Case (c) series–parallel

      In this case, the reference schematic is shown below:

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