# Accumulators not as simple as they appear.

Accumulators are to a hydraulic system what starting batteries are to engine systems; they both store energy for future use and both can become discharged and must be recharged. However, accumulators have broader functions in a variety of hydraulic systems. In order to carry out these functions, accumulators are designed to accept a given volume of fluid under pressure from a hydraulic system, store it and return it to the system on command.The broad range of equipment on which accumulators are used translates ham a need for knowledge of what they are, how they work, where they are likely to be located, their potential failure modes and their maintenance procedures. The dangers associated with the application of accumulators in hydraulic systems are many and usually associated with Lack of knowledge or carelessness.

The first rule is to know your equipment.

Fig. 1 shows several types of accumulators, all of which are, or have been, used in practice. Bladder, piston and spring-loaded accumulators use compressible storage media. When fluid is pumped into the hydraulic side of the barrier element, the gas or spring is compressed. This action increases pressure (or force, in the spring type) in the storage media and the pressure in the storage volume below the barrier element.

[FIGURE 1 OMITTED]

The fluid storage pressure in a gravimetric accumulator is constant and proportional to the weight and the area of the piston. Only this type of accumulator stores a volume of hydraulic fluid at constant pressure.

Accumulators appear to be deceptively simple until one gets into the details of applying them in practical hydraulic circuits. For instance, air could be used in a gas accumulator, but nitrogen is usually selected because it will not burn or support combustion, is readily available and will not contaminate the environment should it escape confinement.

Accumulator size selection

Gas-loaded piston, diaphragm and bladder-type accumulators operate on the principle of Boyle's Law of Gases. Therefore, the basic relationship between the pressure and the volume of a gas is expressed by the equation: [P.sub.1][V.sub.1.sup.n.] = [P.sub.2][V.sub.sup.n], where [P.sub.1] and [P.sub.2] are the initial and final gas pressures and [V.sub.1] and [V.sub.2] are the corresponding gas volumes.

In an isothermal condition, where gas Compression and expansion occur slowly, providing sufficient time for heat to be dissipated, n = 1.

[P.sub.1][V.sub.1] = [P.sub.2][V.sub.2] ... Isothermal condition

In an adiabatic condition, however, where there is no transfer (i.e., loss or gain) of heat, n - the ratio between the specific heat of a gas at constant volume and its specific heat at constant pressure. For nitrogen gas in an adiabatic condition, for example, n = 1.4.

[P.sub.1][V.sub.1.sup.1.4] = [P.sub.2][V.sub.2.sup. 1.4] ... Adiabatic condition for nitrogen

Generally, an adiabatic condition call be said to exist if the compression or expansion of a gas occurs in less than one minute. It is also the condition typically developed in gas-loaded diaphragm or bladder-type accumulators where the separator is fabricated of synthetic rubber materials possessing high insulating characteristics.

These materials have the effect of reducing the amount of heat transferred into or out of the gas during the expansion or compression cycle so as to most closely approximate an adiabatic, rather than an isothermal, condition.

To fully utilize the potential energy of the practically incompressible fluid and compressed gas stored in the accumulator, the size of the accumulator must be properly calculated.

Auxiliary power source

For sizing the accumulator as an auxiliary power source, the amount of fluid required from the accumulator to meet the need of the system must be known. The following formulas and their explanations present a simplified method of determining the capacity of the accumulator to be used.

Isothermal condition: (when system temperature remains constant)

[V.sub.1] = [V.sub.x]([P.sub.3]/[P.sub.1])/1-([P.sub.3]/[P.sub.1])

Adiabatic condition: (when no heat is transferred from the system) n = 1.4 for nitrogen gas

[V.sub.1] = [V.sub.x] [([P.sub.3]/[P.sub.1]).sup.1/n]/1-[([P.sub.3]/[P.sub.2]).sup.1/n]

Where (ref. Fig. 2):

[V.sub.1] = Size of accumulator necessary ([in.sup.3]). This is the maximum volume occupied by the gas at precharge pressure.

[V.sub.x] = The volume of fluid discharged from accumulator. It is the additional volume of fluid demanded by the system ([in.sup.3])

[P.sub.1] = Gas precharge of accumulator (psia). Note: This pressure must be less than or equal to minimum system pressure (P.sub.3]).

[P.sub.2] = the maximum system design operating pressure (psia).

[V.sub.2] = The compressed volume of gas at maximum system pressure ([in.sup.3])

[P.sub.3] = The minimum system pressure (psia) at which the additional volume of fluid is needed.

[V.sub.3] = The expanded volume ([in.sup.3]) of gas at minimum system pressure.

(The above mentioned formula are based on the principles stated in Boyle's Law of Gases: [P.sub.1][V.sub.1] = [P.sub.2][V.sub.2]=[P.sub.3][V.sub.3].)

Isothermal condition:

(1) [V.sub.x] = [V. sub.3] - [V.sub.2]

[FIGURE 2 OMITTED]

Equation (1) thus shows that [V.sub.x] is actually the volume of gas expanded during the pressure drop from [P.sub.2] to [P.sub.3], thereby displacing the equal volume of stored fluid from the accumulator.

(2) [V.sub.3] = [V.sub.x] + [V.sub.2]

[P.sub.2][V.sub.2] = [P.sub.3][V.sub.3] From Boyle's Law

(3) [P.sub.2][V.sub.2] = [P.sub.3]([V.sub.x] + [V.sub.2])

(4) [P.sub.2][V.sub.2] = [P.sub.3][V.sub.x] + [P.sub.3][V.sub.2]

(5) [V.sub.2]([P.sub.2]-[P.sub.3]) = [P.sub.3][V.sub.x]

(6) [V.sub.2] = [P.sub.3][V.sub.x]/[P.sub.2]-[P.sub.3]

[P.sub.1][V.sub.1] = [P.sub.2][V.sub.2] From Boyle's Law

(7) [V.sub.1] = [P.sub.2][V.sub.2]/[P.sub.1]

(8) [V.sub.1] = [P.sub.2]([P.sub.3][V.sub.x])/[P.sub.1]([P.sub.2])-[P.sub.3])

(9) [V.sub.1] = [V.sub.x]([P.sub.3]/[P.sub.1])/1-([P.sub.3]/[P.sub.1]

Adiabatic condition:

(10) [P.sub.1][V.sub.1.sup.n] = [P.sub.2][V.sub.2.sup.n] = [P.sub.3][V.sub.3.sup.n]

Or (11) [V.sub.1][([P.sub.1]).sup.1/n] = [V.sub.2][([P.sub.2]).sup.1/n] = [V.sub.3][([P.sub.3]).sup.1/n]

(1) [V.sub.x] = [V.sub.3] - [V.sub.2]

(2) [V.sub.3] = [V.sub.x] + [V.sub.2]

(11) [V.sub.2][([P.sub.2]).sup.1/n] = [V.sub.3] [([P.sub.3]).sup.1/n]

From equation

(12) [V.sub.2][([P.sub.2]).sup.1/n] = [([P.sub.3]).sup.1/n]([V.sub.x] + [V.sub.2]

(13) [V.sub.2][([P.sub.2]).sup.1/n] = [([P.sub.3]).sup.1/n] [V.sub.x] + [([P.sub.3]).sup.1/n] [V.sub.2]

(14) [V.sub.2][[([P.sub.2]).sup.1/n] - [([P.sub.3]).sup.1/n]] = [V.sub.x] [([P.sub.3]).sup.1/n]

(15) [V.sub.2] = ([V.sub.x])[([P.sub.3]).sup.1/n]/[([P.sub.2]).sup.1/n]-[([P.sub 3]).sup.1/n]

[V.sub.1][([P.sub.1]).sup.1/n] = [V.sub.2][([P.sub.2]).sup.1/n]

From Equation

(16) [V.sub.1] = [([P.sub.2]/[P.sub.1]).sup.1/n] ([V.sub.2])

(17) [V.sub.1] = [([P.sub.2]/[P.sub.1]).sup.1/n][([P.sub.3]).sup.1/n] ([V.sub.x])/([P.sub.2]).sup.1/n] - ([P.sub.3]).sup.1/n]

(18) [V.sub.1] = [V.sub.x][([P.sub.3]/[P.sub.1]).sup.1/n]/1-[([P.sub.3]/[P.sub.2]).sup.1/n]

Example:

What size of accumulator is necessary to supply 300 [in.sup.3] of fluid in a hydraulic system of a maximum operating pressure of 3000 psia, which drops to minimum 1500 psia? Assuming nitrogen gas precharge of accumulator is 1000 psia.

[V.sub.1] = ? (size of accumulator) [in.sup.3] [P.sub.1] = 1000 psia [P.sub.2] = 3000 psia [P.sub.3] = 1500 psia [V.sub.x] = 300 [in.sup.3]

Isothermal Solution:

[V.sub.1] = [V.sub.x] ([P.sub.3]/[P.sub.1])/ 1 - ([P.sub.3]/[P.sub.1)

Adiabatic Solution:

[V.sub.1] = [V.sub.x] [(P.sub.3]/[P.sub.1]).sup.1/n/ 1 - ([P.sub.3]/[P.sub.2]).sup.1/n]

1/n = 1/1.4 = 0.714 for [N.sub.2]

[V.sub.1] = 300(1500/1000)/1 - (1500/3000)

[V.sub.1] = 300[(1500/1000).sup..714]/1 - [(1500/3000).sup..714]

[V.sub.1] = (300)(3/2)(2) [V.sub.1] = (300)(3.69)

[V.sub.1] = 900 [in.sup.3] [V.sub.1] = 1105 [in.sup.3]

[V.sub.1] = 900/231 = 3.9 U.S. gal. * [V.sub.1] = 1105/231 = 4.78 U.S. gal. **

* Necessary accumulator size = 3,9 U.S. gallons

** Necessary accumulator size = 5 U.S. gallons

In addition to utilizing accumulators to store energy for release during the work cycle of a machine, other applications for these devices include: compensating for thermal expansion, isolating two different fluids in a system, acting as a fluid 'spring' in suspension systems, shock or surge suppression, and pulsation damping.

Thermal expansion compensation

In a closed hydraulic circuit, expanded fluid volume due to thermal expansion can readily increase system pressure beyond safety limits, ultimately causing damage to system components as well as line breakage. To prevent this, a properly sized accumulator is located in the system to absorb the increased volume of fluid mad then return it into the lines as the system temperature decreases.

The factors that must be considered for calculation of accumulator sizing--under the conditions described--are shown in the following formula:

[V.sub.1] = [V.sub.a]([t.sub.2]--[t.sub.1] ([Beta]--[3.sub.[Alpha]] [([P.sub.2]/[P.sub.1]).sup.1/n]/1--[([P.sub.2]/[P.sub.3]).sup.1/n]

where:

[V.sub.1 = Size of accumulator required ([in.sup.3]). This is maximum volume occupied by the gas at precharge pressure.

[P.sub.1] = Gas precharge pressure of accumulator (psia). Note: This pressure must be less than or equal to minimum system pressure (P2).

Va = Total volume of fluid in the pipe line [area of pipe ([in.sup.2]) x pipe length (in.)]

[t.sub.1] = Initial temperature of the system, ([degrees]F).

[t.sub.2] = Final temperature of the system ([degrees]F).

[P.sub.2] = System pressure (psia) at temperature ([t.sub.1]) - minimum system pressure.

[P.sub.3] = System pressure (psia) at temperature ([t.sub.2) - maximum system pressure.

[Alpha] = Coefficient of linear expansion of pipe material per [degrees]F

[Beta] = Coefficient of cubical expansion of the fluid per [degrees]F.

n = 1.4 for nitrogen.

This formula is based on the principles Of thermal expansion of liquids and metals and Boyle's Law of Gases.

Shock suppression using accumulators

One of the more interesting applications is the use of accumulators to suppress shock in hydraulic systems--'water hammer, load impact feedback, etc. If this is to be successful, the accumulator must be properly located in the system. It has concluded that most accumulators put into systems end up being very expensive Tee's--not effective shock suppressors. A quick look at hydraulic transmission line dynamics may provide insights as to wig this may be true.

Consideration of transmission line dynamics includes concepts of-spring rate and natural frequency. In addition, the designer must be concerned with transmission line effects on system response. Of special interest are pressure transients--or shock or water hammer--as these transients are sometimes called.

Celerity--Celerity is the speed at which a pressure wave travels along a fired transmission line.

c = [square root of [Beta]f/[Rho]],

where:

c - speed of sound in a hydraulic fluid.

[Beta]f - fluid bulk modulus.

[Rho] - fluid density. The associated wave length in a robe is:

[Lambda] = c/f - 2[pi]c/[omega]

where:

[Lambda] - wave length

[omega] - 2[pi]f

f - frequency, Hz.

Examples of transmission lines--Several cases must be considered:

1. Open end line. Resonance occurs when input impedance is zero as when: L = 0, [Lambda]/2,[Lambda], 3[Lambda]/2.

2. Closed end line. Resonance occurs when: [omega] L/c = [pi]/2, 3[pi]/2, 5[pi]/2 ...

Fundamental frequency is [f.sub.f] = c/4L Hz.

Pressure ratio [p.sub.4]/[p.sub.1] = 1/coso[omega]L/c,

where:

[p.sub.4] is load pressure; [p.sub.1] is supply pressure;

[p.sub.4]/[p.sub.1] = 1 when [omega] L/c = 0, [Lambda], 2[Lambda] ... [infinity]

Given a seven-piston pump rotating at 2400 rpm in a system filled with MIL H-5606 fluid. A valve is closed at the end of the hue. What transmission lengths should be avoided to prevent resonance?

a) Celerity, c = [square root of [Beta]f/[Rho]], = 4480 fps x 12 = 53,800 ips.

b) [omega] = 7(2400/6002[pi] = 1759. Rad/sec.

c) Since resonance occurs at odd multiples of [pi]/2, [L.sub.res] = [([pi]/2 53,800]/ 1759.3 = 48.04 in. 4.01 ft.

This length then, or odd multiples of it, should be avoided to prevent resonance.

3. Line with an end chamber.

a) Limits: If the volume of the end chamber [V.sub.c] = 0, the condition is that of a closed tube. If [V.sub.c] = [infinity], the condition is that of an open tube. If the cross-sectional area of the cylinder is greater than 10 times its length, the effect of [L.sub.f] is small; that of [R.sub.H] is negligible and the chamber only adds capacitance to the system.

b) Excluding the above limits, resonance occurs at [f.sub.LC] = (n[pi] - [phi]) (c/2[pi]L).

c) Solutions to specific problems involve procedures beyond the scope of tiffs article.

Pulses in transmissions

Two types of pressure pulses are encountered in hydraulic systems: 1) Periodic pulses such as those resulting from pump ripple; and 2) transient pulses, as typified by shock or water hammer.

Periodic pulses--Designers may consider three techniques for suppressing pressure ripple; 1) A closed-end side branch, 2) a Quincke robe, or 3) an accumulator.

Case 1: The requisite condition to suppress pressure ripple with a closed-end side branch is that its input impedance be zero. Then all of the ripple will pass into the branch and none into the system.

To be an effective pulse damper, an accumulator must have zero input impedance. This occurs when: [omega] Llc = [pi]/2, 3[pi]/2, 5[pi]/2, etc., or if fL = c/4L.

Accumulator circuits

As mentioned earlier, accumulators are fluid power components that store potential energy and return it to the circuit on demand. Accumulators are used in fluid power circuits for two purposes: 1) To store energy and provide pressurized fluid to the circuit, and 2) to act as surge suppressors and reduce pressure shocks in a system.

The circuit shown in Fig. 3 illustrates how an accumulator is used as a primary source of pressurized fluid. The fixed displacement pump supplies pressure fluid to the gas-charged accumulator during nonworking or dwell periods in the work cycle. An unloading valve opens when the desired maximum pressure is reached, bypassing pump output flow to tank at low pressure. When the cam-operated, spring-returned directional control valve is shifted, the accumulator delivers pressurized fluid into the circuit.

[FIGURE 3 OMITTED]

Fig. 4 shows a circuit in which the accumulator maintains a constant pressure in the system. This circuit appears to be identical to that of Fig. 3, except that a relief valve is used instead of an unloading valve. This type of circuit would be used where it is necessary to maintain a certain pressure, yet it may also be desirable at times to stop the pump.

[FIGURE 4 OMITTED]

Fig. 5 is a modification of the basic pressure-holding circuit: the weighted accumulator is connected to the cylinder line downstream of the directional control valve. When a control device is in placed, it controls only that particular line. Thus, the accumulator in Fig. 5 only pressurizes the cap end of the cylinder; the accumulator in Fig. 4 holds pressure in both lines to the rotary actuator.

[FIGURE 5 OMITTED]

Fig. 6 shows an accumulator used in a failsafe-retract mode. Should the pump fail, the accumulator would provide the pressurized fluid needed to retract the cylinder. There are many applications in which this safety feature is desirable.

[FIGURE 6 OMITTED]

Accumulators may also be used where electric power failure may create a hazardous situation (Fig. 7). Here, loss of electric power would de-energize the solenoid operators of main directional control valve A, allowing it to center. Should this happen, secondary directional control valve B could then be shifted manually to control the cylinder, using pressurized fluid stored in the accumulator. Auxillary directional control valve C provides the necessary, flow paths during emergency operation.

[FIGURE 7 OMITTED]

Fig. 8 illustrates an application where the accumulator reduces pump pressure surges. All positive-displacement pumps exhibit a pulsating output characteristic due to the cyclic mechanical input to the pumping mechanism. In some applications, these pulsations must be damped out with an accumulator.

[FIGURE 8 OMITTED]

Fig. 9 is an application similar to that illustrated in Fig. 8. Here, the accumulator reduces any shock that might be generated in the circuit by impact loading on the cylinder piston rod. A resistive type impact on the cylinder rod could momentarily convert the cylinder into a pump and send a pressure shock back into the system. An overrunning type load impact would cause cavitation on the cap end of the cylinder, and fluid rushing in to the fill the void could create a pressure shock. The accumulator damps out the pressure shock in either case.

[FIGURE 9 OMITTED]

These examples are but a few of the unlimited varieties of circuits that can be used to solve fluid power design problems.

Some information for this article is from "Fluid Power Systems & Circuits" by Russell W. Henke, published by Penton.

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Title Annotation: | hydraulic systems trends |
---|---|

Author: | Henke, Russ |

Publication: | Diesel Progress North American Edition |

Geographic Code: | 1USA |

Date: | Sep 1, 2003 |

Words: | 3122 |

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