Linearized Model of Hydraulic Motor

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Linearized Model of a Hydraulic Motor

Jack Johnson, P.E.

Tue, 2014-02-18 14:36

Most motion-control applications are of a critical nature they must meet accuracy, bandwidth, or

some other performance demand. The most sensible and expedient way to design such systems is to

use performance requirements as the design goals at the very outset of the design process. The

techniques are analytical in nature, so they require mathematical descriptions of all elements of the

system. Only then can synthesis and simulation methods be applied to direct the design processtoward the end result without undue trial-and-error techniques. This is how motion-control and

mathematical models complement and enhance one another.

The nature of the mathematical model is dictated as much by the intended use as it is by the nature

of the device being modeled. Individual modelers beliefs and biases have been known to influence

models, too. However, most designers would agree that models fall into two broad categories: steady

state and dynamic. A hydraulic motor will be modeled in steady state and analyzed through some

examples of how the models can be used.

Hydraulic motor models

The analytical schematic of the hydraulic motor has three internal leakage paths, and one internal

friction-windage resistance. However, note in Figure 1, that output is mechanical power in the form

of speed and torque, whereas the input is hydraulic in the form of pressure and flow. Well begin by

visualizing the real physical processes that the three leakage resistances represent in, say, a piston

motor.

First, a direct path exists between

the rotating barrel and the port

plate, characterized by the laminar

leakage resistance,RIpp. Second,there is a leakage from the high-

pressure side, past the pistons and

their bores, that ends up in the motor case. Another leakage component feeds the slipper faces

through the piston centers and also leads to the motor case. Its leakage resistance is symbolized by

R1. Lastly, the same effects exist on the low-pressure side, leading to a low-pressure leakage

component to case drain. It is characterized by R2.

In addition, friction and windage account for a torque loss that depends on speed. It is symbolized

withRfwin Figure 1. This completes the steady-state, high-speed, linearized mathematical model of a

hydraulic motor. It can be used on any motor type, provided sufficient data exists to evaluate theleakage resistance and the friction and windage resistance.

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An application scenario

Imagine a hydraulic motor has been tested at a load torque of 823 lb-in. at 2,400 rpm. The inlet

supply pressure was 3,000 psi while the motor outlet and case drain were essentially at 0 psig. The

case-drain flow was measured at 3.39 in.3/sec and the motor inlet flow was 82.9 in.3/sec. If the

motor has a displacement of 1.88 in.3/rev, determine the values forR1andRfw.

With outlet and case-drain ports at zero pressure, the full 3,000 psi is impressed acrossR1,Rpp, and

the ideal displacement element of the motor. First, we need to find the ideal flow, QI, using the well-

known relationship:

We can calculate theR1coefficient directly from given data by

assuming that the leakage flow is laminar and, therefore, directly proportional to pressure and

inversely proportional to the resistance coefficient:

Input flow continuity requires that:

Now the port-to-port leakage resistance can be found:

To find the friction resistance, we must first calculate the ideal torque using the well-known

relationship between inlet differential pressure and output torque in the ideal motor:

The measured torque was given as 823 lb-in. Therefore, the total friction torque loss is:




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Because we assume that this is all viscous friction loss, and that the loss is directly proportional to

speed, then:

Now the coefficients for the motor model have been evaluated. Formulas for calculating leakage

resistance directly from motor efficiencies exist, but space prevents their inclusion here. Most

technical data sheets on motors lack a specific value of case-drain leakage, which is necessary to

evaluate port-to-case-drain resistance. The motor manufacturer must be consulted for that

information.

Adding proportional control

Now consider that the same motor is being used in a circuit controlled by a proportional valve. A

low-pressure shaft seal in the motor allows case drain flow to return to tank through separateplumbing. Due to valve pressure drops, pressure is 2,160 psig at the motor inlet port, 915 psig at the

motor outlet port, and the motor shaft spins at 1,722 rpm. Assuming that R1=R2, we will calculate

case-drain flow, motor-inlet flow, motor-outlet flow, and load torque.

Well start with Figure 2, an

illustration of an analytical

schematic that lists all the known

values. Notice that the entire supply

pressure is impressed across the

leakage resistance (R1). Therefore:

Similarly, the outlet pressure,Pb, is impressed across R2, therefore:

The motor differential pressure is impressed across the port-to-port leakage resistance:

The operating speed is given as 1,722 rpm, therefore, the ideal flow can be found:




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The total inlet flow is found using the summation of flows at the A-port node:

The case-drain flow is the sum of two components:

The outlet flow is comprised of three components:

The load torque can be found by first calculating the ideal torque:

The load torque is the ideal torque less the loss due to viscous friction:

Now:

Summarizing, then, case-drain flow (Qcd) is 3.475 in.3/sec; motor-inlet flow (QaMi) is 58.19 in.

3/sec;

motor-outlet flow (QaMo) is 54.72 in.3/sec; and load torque (ToM) is 144.45 lb-in.

Synchronizing the speed of two motors

Hydraulic-system designers often connect two motors in series in an attempt to synchronize their

speeds. In principle, this is a sound idea. In actuality, however, the degree of synchronizing is




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imperfect because of finite internal leakage resistances. The accompanying box illustrates a practical

use of a mathematical model to quantify the degree of this nonequality of the two motor speeds.

Connecting two hydraulic motors in series in an attempt to synchronize their speeds is a sound idea.

In reality, though, the synchronization is imperfect because of internal leakage resistances. Well now

examine a scenario using a mathematical model to quantify the inequality of the two motor speeds.

Assume two hydraulic motors each

identical to that described previously

are to be connected in series and

powered by a 60-in.3/sec constant-flow

source. As shown in Figure 3, the outlet

port of the low-pressure motor is

connected directly to tank, as are both

case-drain ports. The high-pressure

motor is connected to a 650-lb-in. load,

but the shaft of the low-pressure motor

is completely free. Both motors have a

displacement of 1.88 in.3/rev; leakage resistance from each motor port to case of 885 psi/(in.3/sec);

port-to-port leakage resistance of 696 psi/(in.3/sec); and torque loss from friction and windage of

0.031 lb-in./rpm.

There are four unknowns:P1, P2, N1, andN2, so four equations will be written and solved

simultaneously. Note from the illustration thatP4andP3equal 0. Two node equations represent the

summation of flows (P1andP2nodes) and two torque summation equations (N1andN2).

Flow summation atP1:

Flow summation atP2:

Torque summation atN1:

Torque summation atN1:

Substitution and linear algebra matrix are two common methods of solving four equations with four

unknowns. However, the most practical method is by computer, and all popular spreadsheet




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programs have a simultaneous equation-solving capability. I solved these equations using the

eQsolver capability in IDAS Engineering software. The results are:

P1= 2,537 psig,

P2= 186.6 psig,

N1= 1,716 rpm, and

N2= 1,802 rpm.

The solution to this problem demonstrates that there is nearly a 100-rpm difference between the two

motor speeds. If we now solve the problem with the loads reversed (the upper motor is unloaded and

the lower motor is loaded) we find that:

P1= 2,521 psig,

P2= 2,333 psig,

N1= 1,815 rpm, and

N2= 1,549 rpm.

This solution shows that there is nearly a 300-rpm change in the speed of the lower motor a

condition that certainly is less than ideal for the application, but without more specific information,

judgments cannot be passed.

The point of this analysis is not to provide a means for achieving perfect motor speed

synchronization. Rather, there is a more limited goal. First, using reasonable models of hydraulicmachinery, it is possible to evaluate the consequences of implementing a given circuit concept before

any hardware is even assembled. Second, circuit developers and designers can explore the endless

what ifs that always occur at circuit design time.

The broader issue of perfect motor-speed synchronization requires closed-loop speed-control

systems and will have to wait for some later discussion. Additionally, closed-loop control modeling

must expand to include dynamic response because of the possibility of hunting and sustained

oscillations.

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Linearized Model of Hydraulic Motor

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