Example: The Resistance Network Method Applied to Unsteady Flow

branch contains a compensating flow regulator valve. The valve, which is area orifice at the top. The valve is free to move in its bore due to the forces that develop from the oil pressure and spring forces. The top orifice is configured as a rectangular flow slot. Therefore, the flow area of the top orifice will vary as the valve moves in its bore. The flow slot has zero overlap and its area is a maximum when the system is not operating.

The goal of the analysis that follows is to determine suitable operating characteristics for the flow regulator valve. Therefore, the upper flow branch has an orifice that is fixed for analysis, but could be varied in size by the user to simulate a circuit workload.

Because the hydraulic resistance method will be used to study the sys-tem, an appropriate resistance term R must be established for each circuit element. Resistance values may be combined in a flow branch, or in an Figure 13.7shows a fluid power system with two flow branches. The lower shown inFigure 13.8, has a fixed area orifice at the left end and a variable

Figure 13.7

: System with two flow branches.

entire system, in the same manner that is used in electric circuits [1, 7-9].

Values for flow, can then be determined with the use of the circuit equations The last part of the analysis incorporates the dynamic aspects of some of the circuit com-ponents. These components will be analyzed in the conventional fashion by applying continuity, (dp/dt = (β/V )dV /dt) and Newton’s Second Law, (F = ma). The solution to the problem will require that the differential equations with displacement and pressure as the state variables be solved iteratively. As the solution moves forward in time, the values of resistance and flow will be updated automatically. The variables that are used to We shall start the analysis by developing the resistance terms for the flow regulator valve and the actuator. Expressions must be written for the resistance terms R_o, R_x, and R_L. The first two are of the orifice flow type and may be written as:

Ro=r ρ 2

1 CdAo

describe the operation of the system are presented inTable 13.4.

[1, 9] that have been reviewed inSections 13.4 and 13.5.

Chapter 13 317

Figure 13.8

: (A) Compensating flow regulator valve and (B) Forces on the valve spool.

and:

Rx=r ρ 2

1 Cdw(`0− x) as:

RL=

√pL

A_py˙

may be combined into a single resistance, RoxL: R_oxL=

q

R²_o+ R²_x+ R²_L

The resistance in the upper branch, Rz, is simply an orifice so:

Rz =r ρ 2

1 C_dA_z

Total parallel resistance RT, the total system resistance seen by the pump, may be expressed as:

RT = R_zR_oxl Rz+ Roxl

The series resistance terms in the lower flow branch, as shown onFigure 13.7, As explained inSection 13.4, the resistance for a actuator can be written

Table 13.4

: Characteristics for flow division analysis

Characteristic Description Units

ps, pv, pL System pressure values N/m² Qp, Qi, Qo, Qz System flow values m³/s

Az Variable load orifice area m²

Av Valve end area m²

m_V valve mass kg

Ao Fixed valve orifice m²

w Valve flow gradient m²/m

k Valve spring rate N/m

F Valve spring preload N

c Damping coefficient N.s/m

Cd Flow coefficient

`₀ Length of flow slot m

Vc Oil volume inside valve m³

ρ Oil mass density kg/m³

g Acceleration of gravity m/s²

βe oil bulk modulus N/m²

A_p Cylinder piston area m²

d Cylinder piston diameter m

`_sl Piston seal width m

µst Coefficient of static friction (seal)

VL Cylinder oil volume m³

m_L Cylinder load mass kg

Ro, Rz, Rx, RL Element flow resistance Pa^1/2s/m³ R_T, R_oxl Combined resistance Pa^1/2s/m³

x Valve motion m

y Cylinder piston motion m

Most of the pressures in the circuit will be state variables derived from dp/dt = (βe/V )dV /dt expressions. An exception is the system pressure ps. In this problem, the driving function is the pump flow so the system

Chapter 13 319 pressure will need to be upgraded at each step of the the iterative solution.

The hydraulic ohm method will be used. Values for pressure at the pump outlet p_s, are given by:

p_s= R²_TQ²_p(t)

The flow in the upper system branch Qz may then be determined as:

Qz= R_T Rz

Qp(t) (13.27)

There would appear to be two methods of determining Qo. Both approaches involve using a resistance method:

Qo= RT Because the major part of the solution involves solving for state variables using a differential equation solver, it is usually desirable to employ the state variables as much as possible. It will be shown shortly that pv is a state variable, so Equation 13.28 is the preferred equation. Another reason to consider in choosing Equation 13.28 is that this formulation only involves one resistance, R_o, and not the three, R_o, R_x, and R_L, that are needed to calculate R_T. The remaining flow that must be determined before the dynamic equations can be formulated is the flow from the valve into the load, Q_i. This flow is determined from:

Qi=

√pv− pL

R_x (13.29)

Now that the flows in and out of the various components have been established using the hydraulic ohm method, the dynamic aspects of the system can be evaluated. Values for the cylinder pressure pL, and the pressure p_v inside the control valve, can be established with use of the continuity of flow principle. Flow continuity, as it applies to the cylinder and the flow control valve, includes flow into and out of the confined volumes V_L and V_c. Also, the influence of fluid compressibility and the effect of moving parts must be included. Flow continuity applied to volume V_c may be written as:

Qo= Qi− Avx +˙ Vc

β_ep˙v

Where flows Qoand Qihave been determined in Equations 13.28 and 13.29.

The volume Vc is essentially constant since motion x for the valve is small.

The equation for continuity may be rearranged as follows:

d

dtp_v= ˙p_v=β_e Vc

(Q_o− Qi+ A_vx)˙ (13.30) The notation dpv/dt has been used to show that pv is a state variable, i.e., one variable in a set of ordinary differential equations that are being developed to complete solution of the problem.

Flow continuity applied to the volume V_L may be written as:

Q_i = A_py +˙ VL

β_ep˙_L

The cylinder oil volume V_L is a function of the cylinder piston motion y and may be expressed as:

VL= Apy

This equation may be rearranged as follows to provide another equation in the set.

d

dtp_L= ˙p_L= β_e

Apy(Q_i− Apy)˙ (13.31)

A). The flow control element, afterwards just called a valve, must be free to move. The equation of motion for the valve can be established from the forces shown on the valve free body diagram shown in Figure 13.8 (panel B). From Newton’s Second Law:

Avps− Avpv− kx − F − c ˙x = mvx¨

This equation can be rearranged so that it can be integrated to find the valve velocity and valve position x:

d

dtx = ¨˙ x = (Avps− Avpv− kx − F − c ˙x)/mv (13.32) Incidentally a further differential equation must be written:

d

dtx = ˙x (13.33)

because most differential equation solvers only work with sets of first order equations.

A reasonable value for the damping coefficient c has been established from laboratory test results. This value applies to spring loaded valves A schematic for the flow regulator valve is shown in Figure 13.8(panel

Chapter 13 321 operating in close fitting bores. For this purpose the damping coefficient can be expressed as a fraction of the theoretical critical damping as:

c = 0.5(2p

km_v) (13.34)

For most valves of this type, the proper fraction shown here as 0.5, will vary from 0.5 to 0.7.

The equation of motion for the cylinder piston and load can be estab-lished from the free body diagram shown in Figure 13.9:

Figure 13.9

: Forces on cylinder piston and load controlled by a flow regulator valve.

AppL− mLg − πd`slpLµst

˙ y

| ˙y| = mLy¨

To determine load piston velocity ˙y and position y the equation may be arranged as:

d

dty = ¨˙ y =



AppL− mLg − πd`slpLµst

˙ y

| ˙y|



/mL (13.35)

As before, add the following equation to the set:

d

dty = ˙y (13.36)

A value for the damping coefficient is difficult to determine for a cylinder piston. It is well known from laboratory work, however, that a friction force

exists at the piston and rod seals. In the above equation the friction has been modeled as a single value that incorporates piston diameter d, seal width `_sl, load pressure p_L, and seal Coulomb friction µ_st. The friction force always opposes motion of the piston. Therefore, the direction of the friction force must be corrected with the ratio of piston velocity over absolute velocity ˙y/| ˙y|.

The example that we have examined requires a set of steps for its so-lution. There are many packages that can be used for solution and each may have different methods of interaction between the user and the pack-age. All, however, must follow the same internal sequence. Sufficient initial values of flow and pressure must be submitted in addition to the system, i.e., component, characteristics. The solution formulation is dynamic, that is, a set of simultaneous, ordinary differential equations with pressures, dis-placements, and velocities as state variables is solved numerically. The user must ensure that the coefficients in these state variable equations that de-pend on varying resistance values are updated during the iteration process.

sequence that is necessary.

In document Hydraulic Power System Analysis (Page 93-100)