# Basic hydraulic circuit design: considerations of circuit dynamics.

In our last several discussions, we have seen how cycle profiling
offers a systematic approach to fluid power design (see September 2007,
Diesel Progress) and how load analysis can be considered the key to
cycle profiling and parametric definition. Now we'll continue by
looking at dynamic conditions in circuits, such as overrunning and
inertial loads.

A cycle profile for the transfer application of Fig. 1 is shown in Fig. 2. It is based on a quantitative, steady-state analysis of the example problem we have been discussing and would be analogous to the profile for one linear actuator in the excavator problem we discussed earlier.

We have so far considered open-loop hydraulic circuit design and performance based on steady-state conditions. Steady-state design and analysis presume no change in the state of a variable with time. Fig. 3 shows the velocity profile of Fig. 2 with simple harmonic motion added as Method 4. When we earlier discussed the choice of velocity profile, Method 1 was presented as a case of steady-state velocity, which is true during time interval [t.sub.3] - [t.sub.1]. We accommodated acceleration and deceleration periods, [t.sub.i] - 0 and [t.sub.5] - [t.sub.4] by assuming uniform (steady-state) acceleration:

[[bar.a].sub.a] = ([v.sub.1] - [v.sub.0])/([t.sub.1] - [t.sub.0])

and

[[bar.a] = ([v.sub.5] - [v.sub.3])/([t.sub.5] - [t.sub.4]).

We thus assumed the slope of the velocity curve in those time intervals to be constant. For most designs, this approach might be accurate enough, especially since the system designer may be compelled to guess at many machine parameters. However, when variations in a parameter such as velocity must be considered, the more accurate expression for velocity and acceleration would be:

v = [??] = dx/dt

and

a = [??] = [??] = [d.sup.2]x/[dt.sup.2] = dv/dt.

[FIGURE 1 OMITTED]

In the case of Method 4's velocity profile, the variable must be expressed as a function of time since they vary constantly and steady-state, as previously discussed, has no meaning.

For simple harmonic motion,

S = N sin [omega]t,

[??] = v = [omega]N cos [omega]t,

and

[??] = v = a = [[omega].sup.2]N sin [[omega]t.

Overrunning Loads

Fig. 4 is a simple illustration of an overrunning load. The important part of the work cycle is the "climb milling" operation, during which the cutter produces a load reaction in the same direction as the motion of the piston rod.

[FIGURE 2 OMITTED]

[FIGURE 3 OMITTED]

In the first step, the acceleration and rapid transverse phases of the cycle are analyzed in the same way as in the previous example. Next, at the point where the work piece contacts the cutter, the velocity must be changed to match the desired fee rate. This is shown in the velocity plot in Fig. 4.

The conventional method of achieving this is to shift a direction control valve so as to bring flow control valve in a meter-out position into the circuit. With a [micro]P/[micro]C controlled system, the change to meter-out might be programmed or, with a servo system, the reduction in transfer velocity would be programmed and a servo valve used to control flow.

The feed rate and the piston velocity are determined from the machining characteristics of the material. So far as the fluid power circuit design is concerned, it is an independent variable.

In the third step, the magnitude of the overrunning load can be calculated from the size of the cutter and the horsepower of the motor that drives it. The total load during the feed part of the cycle consists of the cutter force minus the friction force resisting motion. Note that this is a negative force, that is, opposite in the sense that normal pressure must be developed on the rod end of the piston to hold it back.

In the final step, the direction of motion must be reversed during the rapid-return phase of the cycle in order to return the table to its initial position. The method of analysis is the same as that used in the previous example. Figure 7 indicates how the velocity, flow, force and pressure plots would look.

[FIGURE 4 OMITTED]

[FIGURE 5 OMITTED]

[FIGURE 6 OMITTED]

Inertial Loads

Fig. 5 shows a hydraulic motor driving a flywheel. The two fundamental problems with an application such as this, or a wheel drive on a mobile machine, are starting up the load from a zero velocity condition and changes in rotational velocity after the load is running.

First, consider the problem of starting up the load. The relation between applied torque and angular acceleration is T = [J.sub.m][alpha] where [J.sub.m] is the mass moment in the inertia and [alpha] is the angular acceleration. The torque necessary to bring a rotating load up to speed depends on how rapidly the speed is to be achieved.

This is a matter for the designer to decide on the basis of the application parameters. The usual procedure has been to match the motor to the operating cycle and let acceleration take care of itself. If a specified acceleration must be achieved, the motor must be matched to acceleration not steady-state rotational velocity.

[FIGURE 7 OMITTED]

When specifying the motor drive for this application, the designer must take care to choose one with a sufficient torque rating. The horsepower alone does not reflect the ability of a motor to furnish the required torque at low speeds.

Second, if a fixed-displacement hydraulic pump is used for this application, part of the pump output will be dumped over the relief valve during acceleration. This is shown in the cycle plot of Fig. 6.

The pump must be sized to supply enough oil to run the process at design speed. During acceleration, the pump will supply more oil than the motor can absorb, since the motor does not reach design speed until the end of the acceleration period. This type of application is a good place for a pressure-compensated, variable-volume pump. Such a pump will adjust its output to that required by the system as a function of system pressure.

Next, consider how the situation in which the process is running at design speed and a change in speed is called for. The speed change may be either an increase in velocity or a decrease in velocity. If an increase in rotational velocity is required, then an increase in torque, [DELTA]T, will be needed, [DELTA]T = [J.sub.m] [DELTA]T.

There are two possible approaches to the problem of increasing the velocity. One is to determine the fastest possible response with available maximum torque. Since the running torque will be less than the torque required for acceleration, there will be an excess capability for the speed change. The second is to determine the torque required to achieve the change in a given time interval. Suppose it is necessary to make the speed change in [DELTA]T sec to accommodate some process requirement. Then the torque required for the change can be calculated based on [DELTA]T.

Consider the second case, where the speed change is a reduction in rotational velocity rather than an increase. In this case, the system will have to be capable of absorbing energy rather than supplying it. Frictional losses assist in slowing down the load. Thus, the deceleration occurs 20% faster than the acceleration with identical parameters.

It should be recognized that the circuit must be designed to absorb the excess energy during a period of deceleration.

The illustration in Fig. 7 puts the situation in perspective. It shows a simple pump-valve-motor drive circuit represented in several ways: in a block diagram, in ISO symbols, in a functional diagram where component characteristics replace hardware symbols and a parametric diagram showing how each parameter varies using pressure as the independent variable. Note that the axes for pump and motor are inverted, representing that there is a pressure rise across a pump while there is a pressure drop across a motor. The load analyses and cycle profile we have discussed must reflect these kinds of parametric relationships.

BY RUSS HENKE, PE, CFPE

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

A cycle profile for the transfer application of Fig. 1 is shown in Fig. 2. It is based on a quantitative, steady-state analysis of the example problem we have been discussing and would be analogous to the profile for one linear actuator in the excavator problem we discussed earlier.

We have so far considered open-loop hydraulic circuit design and performance based on steady-state conditions. Steady-state design and analysis presume no change in the state of a variable with time. Fig. 3 shows the velocity profile of Fig. 2 with simple harmonic motion added as Method 4. When we earlier discussed the choice of velocity profile, Method 1 was presented as a case of steady-state velocity, which is true during time interval [t.sub.3] - [t.sub.1]. We accommodated acceleration and deceleration periods, [t.sub.i] - 0 and [t.sub.5] - [t.sub.4] by assuming uniform (steady-state) acceleration:

[[bar.a].sub.a] = ([v.sub.1] - [v.sub.0])/([t.sub.1] - [t.sub.0])

and

[[bar.a] = ([v.sub.5] - [v.sub.3])/([t.sub.5] - [t.sub.4]).

We thus assumed the slope of the velocity curve in those time intervals to be constant. For most designs, this approach might be accurate enough, especially since the system designer may be compelled to guess at many machine parameters. However, when variations in a parameter such as velocity must be considered, the more accurate expression for velocity and acceleration would be:

v = [??] = dx/dt

and

a = [??] = [??] = [d.sup.2]x/[dt.sup.2] = dv/dt.

[FIGURE 1 OMITTED]

In the case of Method 4's velocity profile, the variable must be expressed as a function of time since they vary constantly and steady-state, as previously discussed, has no meaning.

For simple harmonic motion,

S = N sin [omega]t,

[??] = v = [omega]N cos [omega]t,

and

[??] = v = a = [[omega].sup.2]N sin [[omega]t.

Overrunning Loads

Fig. 4 is a simple illustration of an overrunning load. The important part of the work cycle is the "climb milling" operation, during which the cutter produces a load reaction in the same direction as the motion of the piston rod.

[FIGURE 2 OMITTED]

[FIGURE 3 OMITTED]

In the first step, the acceleration and rapid transverse phases of the cycle are analyzed in the same way as in the previous example. Next, at the point where the work piece contacts the cutter, the velocity must be changed to match the desired fee rate. This is shown in the velocity plot in Fig. 4.

The conventional method of achieving this is to shift a direction control valve so as to bring flow control valve in a meter-out position into the circuit. With a [micro]P/[micro]C controlled system, the change to meter-out might be programmed or, with a servo system, the reduction in transfer velocity would be programmed and a servo valve used to control flow.

The feed rate and the piston velocity are determined from the machining characteristics of the material. So far as the fluid power circuit design is concerned, it is an independent variable.

In the third step, the magnitude of the overrunning load can be calculated from the size of the cutter and the horsepower of the motor that drives it. The total load during the feed part of the cycle consists of the cutter force minus the friction force resisting motion. Note that this is a negative force, that is, opposite in the sense that normal pressure must be developed on the rod end of the piston to hold it back.

In the final step, the direction of motion must be reversed during the rapid-return phase of the cycle in order to return the table to its initial position. The method of analysis is the same as that used in the previous example. Figure 7 indicates how the velocity, flow, force and pressure plots would look.

[FIGURE 4 OMITTED]

[FIGURE 5 OMITTED]

[FIGURE 6 OMITTED]

Inertial Loads

Fig. 5 shows a hydraulic motor driving a flywheel. The two fundamental problems with an application such as this, or a wheel drive on a mobile machine, are starting up the load from a zero velocity condition and changes in rotational velocity after the load is running.

First, consider the problem of starting up the load. The relation between applied torque and angular acceleration is T = [J.sub.m][alpha] where [J.sub.m] is the mass moment in the inertia and [alpha] is the angular acceleration. The torque necessary to bring a rotating load up to speed depends on how rapidly the speed is to be achieved.

This is a matter for the designer to decide on the basis of the application parameters. The usual procedure has been to match the motor to the operating cycle and let acceleration take care of itself. If a specified acceleration must be achieved, the motor must be matched to acceleration not steady-state rotational velocity.

[FIGURE 7 OMITTED]

When specifying the motor drive for this application, the designer must take care to choose one with a sufficient torque rating. The horsepower alone does not reflect the ability of a motor to furnish the required torque at low speeds.

Second, if a fixed-displacement hydraulic pump is used for this application, part of the pump output will be dumped over the relief valve during acceleration. This is shown in the cycle plot of Fig. 6.

The pump must be sized to supply enough oil to run the process at design speed. During acceleration, the pump will supply more oil than the motor can absorb, since the motor does not reach design speed until the end of the acceleration period. This type of application is a good place for a pressure-compensated, variable-volume pump. Such a pump will adjust its output to that required by the system as a function of system pressure.

Next, consider how the situation in which the process is running at design speed and a change in speed is called for. The speed change may be either an increase in velocity or a decrease in velocity. If an increase in rotational velocity is required, then an increase in torque, [DELTA]T, will be needed, [DELTA]T = [J.sub.m] [DELTA]T.

There are two possible approaches to the problem of increasing the velocity. One is to determine the fastest possible response with available maximum torque. Since the running torque will be less than the torque required for acceleration, there will be an excess capability for the speed change. The second is to determine the torque required to achieve the change in a given time interval. Suppose it is necessary to make the speed change in [DELTA]T sec to accommodate some process requirement. Then the torque required for the change can be calculated based on [DELTA]T.

Consider the second case, where the speed change is a reduction in rotational velocity rather than an increase. In this case, the system will have to be capable of absorbing energy rather than supplying it. Frictional losses assist in slowing down the load. Thus, the deceleration occurs 20% faster than the acceleration with identical parameters.

It should be recognized that the circuit must be designed to absorb the excess energy during a period of deceleration.

The illustration in Fig. 7 puts the situation in perspective. It shows a simple pump-valve-motor drive circuit represented in several ways: in a block diagram, in ISO symbols, in a functional diagram where component characteristics replace hardware symbols and a parametric diagram showing how each parameter varies using pressure as the independent variable. Note that the axes for pump and motor are inverted, representing that there is a pressure rise across a pump while there is a pressure drop across a motor. The load analyses and cycle profile we have discussed must reflect these kinds of parametric relationships.

BY RUSS HENKE, PE, CFPE

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

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Title Annotation: | HYDRAULIC SYSTEMS TRENDS |
---|---|

Comment: | Basic hydraulic circuit design: considerations of circuit dynamics.(HYDRAULIC SYSTEMS TRENDS) |

Author: | Henke, Russ |

Publication: | Diesel Progress North American Edition |

Geographic Code: | 1USA |

Date: | Dec 1, 2007 |

Words: | 1384 |

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