Sizing of Accumulators

 Fig.9.13 shows a simple weight-loaded type accumulator.

SIZING OF ACCUMULATORS

1. Analysis of Weight-Loaded Type Accumulators

Fig.9.13 shows a simple weight-loaded type accumulator.

Capacity of accumulator: The maximum amount of energy that the accumulator can store is known as the capacity of the accumulator.

Derivation:

Let

A = Area of the sliding ram = π / 4 D2,

D = Diameter of the ram,

L = Stroke or lift of the ram,

P = Intensity of pressure of hydraulic fluid supplied by the pump, and

W = Total weight of the ram including the weight of the dead-load on the ram.

We know that,

W = P × A

Work done in lifting the ram = W × Lift of ram

Example 9.1 

An accumulator has a ram of 200 mm diameter and a lift of 6 m. If the hydraulic fluid is supplied at 60 bar, find: (i) the necessary load on the ram, and (ii) capacity of the accumulator.

Given Data: 

D = 200 mm; L= 6m; P = 60 bar.

Solution:

(i) Load on the ram (W):

(ii) Capacity of the accumulator :

Example 9.2

An accumulator is loaded with 400 kN weight. The ram has a diameter of 300 mm and stroke of 6 m. Its friction may be taken as 5 percent. It takes two minutes to fall through its full stroke. Find the total work supplies and power delivered to the hydraulic appliance by the accumulator, when 0.0075 m3/s of liquid is being delivered by a pump, while the accumulator descends with the stated velocity. Take the density of oil as 1000 N/m3. 

Given Data: 

W=400 kN ; D = 300 mm; L = 6m; Friction = 5%.

Solution:

1. Total work supplied to the hydraulic machine :

Net load on accumulator, when it descends = Actual load × (1 - % of friction)

= 400 × 0.95 = 380 kN

Time taken by ram to fall through full stroke, t = 2 min or 120 s

Liquid supplied by pump = 0.0075 m3/s       …(Given)

We know that, P = ρ g h

2. Power delivered to the hydraulic machine:

Power delivered Work = supplied per second

2. Analysis of Gas-Loaded Accumulators

• As we have discussed in Section 9.4, gas-loaded accumulators are governed by the Boyle's gas law. That means, the compression and expansion of the gas in gas-loaded accumulators are governed by the Boyle's law of gas.

• The rate of discharge determines whether the process is isothermal or adiabatic. If the rate of discharge is quick, the expansion process can be assumed to be adiabatic (P1V1 = P2V2). Isothermal (P1V1 = P2V2) relations can be used for compression if the process is slow.

• The precharge pressure should be selected so that use is made of all liquid in the accumulator.

• The compressed gas volume is a function of the charge and discharge time. The required liquid volume can be determined from the performance of the accumulator.

The following solved numerical problems explain how the sizing of the gas-loaded accumulator can be determined in hydraulic circuits.

Example 9.3 

A gas-loaded accumulator has the following data:

Precharge pressure = 125 bar

Charge accumulator (or maximum) pressure = 200 bar

Pressure falls from 200 bar to 175 bar when 20 litres of oil is discharged.

What is the size (i.e., volume) of the accumulator used? Take the charging and discharging processes as isothermal.

Given Data:

P1 = 125 bar; P2 = 200 bar; P3 = 175 bar;

Volume of oil discharged = 20 litres =20 × 10-3 m3.

Solution: Fig.9.14 shows the three important states of the accumulator operation.

For isothermal process,

P1 V1 = P2 V2 = P3 V3

From Figs.9.14(b) and (c), we can write

P2 V2 = P3 V3

Example 9.4 

The accumulator of Fig.9.15 is to supply 8000 cm3 of oil with a maximum pressure of 225 bars and a minimum pressure of 140 bars. If the nitrogen precharge pressure is 95 bars, find the size of the accumulator. The hydraulic cylinder piston diameter is 200 mm.

Given Data: Discharge of oil = 8000 cm3 = 8000 × 106 m3; Pmax = P2 = 225 bars;

Pmin = P3 = 140 bars; Pprecharge = P1 = 95 bars; D =200 mm.*

Solution: The solution to this problem is quiet similar to the previous solved problem. So proceeding in the same way, we get the solution as below:

Using Boyle's gas law, we write-

Example 9.5

For the accumulator of Example 9.4, find the load force Fload that the cylinder can carry over its entire stroke. What would be the total stroke of the cylinder if the entire output of the accumulator were used?

Given Data: Refer Example 9.4.

Solution: Since the pressure of the oil supplied by the accumulator to the cylinder varies from maximum (225 bars) to minimum (140 bars), we can consider the average pressure ( 225 + 440 / 2 = 182.5 bars ) for our calculations.

• This data is a superfluous data. That means, this data may be an important data, but not required to solve this particular problem.

1. Load force (Fload) : 

2. Total stroke of the cylinder (L) :

Given that the entire output of the accumulator (i.e., V1 = 0.0312 m3 of volume) were used.

Example 9.6 

A hydraulic circuit has been designed to crush a car body into bale size using a 175 mm diameter hydraulic cylinder. The hydraulic cylinder is to extend 2.4 m during a period of 10 s. The time between crushing strokes is 6 min. The accumulator has the following other details :

Gas precharge pressure = 90 bars

Gas charge pressure when pump is turned on = 225 bars

Minimum pressure required to actuate load = 125 bars

(i) Calculate the required size of the accumulator.

(ii) What are the pump flow requirements and the pump hydraulic kW power with and without an accumulator ?

Given Data : D = 175 mm; Extension = 2.4 m; Pprecharge = P1 = 90 bars;

Pcharge = Pmax = P2 = 225 bars; Pmin = P3 = 125 bars.

Solution: Fig.9.16 shows the three important states of the accumulator operation.

The hydraulic cylinder diameter and stroke (extension) are given. Therefore the volume of the hydraulic cylinder is calculated as

Also, it should be noted from Figs.9.16(b) and (c) that the volume of oil displaced (V3 - V2) from the accumulator is nothing but the volume of oil in the hydraulic cylinder.

(i) Size of the accumulator (V1):

We know that, P1 VI = P2 V2 = P3 V3

For Figs.9.16(b) and (c), we can write

On solving equations (i) and (ii), we get

Now for Figs.9.16(a) and (b), we can write

(ii) Pump flow rate (Qpump) and pump kW power (kWpump) :

(a) With accumulator: It is given that the time between crushing strokes is 6 min (360 sec). It means that within 6 min time the pump must recharge the accumulator only and to the extent of the volume displaced in the cylinder during extension and retraction. This volume is equal to 2 (V3 - V2).

We know that the kW power,

(b) Without accumulator: When there is no accumulator, then the pump flow rate is given by

Comment: From the above results, it may be noted that by employing the accumulator, the pump size and the pump kW power requirement can be drastically reduced.

Example 9.7 

Design a car crushing system. The crushing force required is such that a 20 cm diameter cylinder is required at a working pressure of 130 kg/cm2. Time for crushing is about 10 sec and the stroke required to flatten the car is 30 cm. Compare the power requirements of a circuit without and with accumulator.

Accumulator details: It is a gas loaded accumulator.

Time taken for charging = 6 min

Initial pressure of charging (precharged), P1 = 100 kg/cm2

Charged pressure of accumulator, P2 = 225 kg/cm2

Minimum pressure for crushing, P3 = 130 kg/cm2

Given Data: 

Solution:

(i) Power requirement of the circuit without accumulator:

The hydraulic cylinder diameter and stroke (extension) are given. Therefore the volume of the hydraulic cylinder is calculated as

(ii) Power requirement of the circuit with accumulator:

Fig.9.17 shows the three important states of the accumulator operation.

Also, it should be noted from Fig.9.17(b) and (c) that the volume of oil displaced (V3 - V2) from the accumulator is nothing but the volume of oil in the hydraulic cylinder.

Vhydraulic cylinder = V3 - V2 = 0.00942 m3

To find the size of the accumulator (V1) 

We know that, P1 V1 = P2 V2 = P3 V3

For Figs.9.17 (b) and (c), we can write

On solving equations (i) and (ii), we get

V2 = 0.129 m3 and V3 = 0.223 m3

Now for Figs.9.17(a) and (b), we can write

P1 V1 = P2 V2

To find pump kW power (kWpump) :

It is given that the time between crushing strokes (i.e., charging) is 6 min (360 sec). It means that within 6 min time the pump must recharge the accumulator only and to the extent of the volume displaced in the cylinder during extension and retraction. This volume is equal. to 2 (V3 - V2).