Week 4/Lesson 2 – Hydraulic motors Fluid Power Engineering Week 4/Lesson 2 – Hydraulic motors

Hydraulic motors In this lesson we shall Compare hydraulic motors with hydraulic pumps Look at power and efficiency calculations for hydraulic motors Introduce rotary actuators Discuss controlling speed vs position Work an example that shows calculations for hydraulic motors

Hydraulic motors Hydraulic motors are like backwards hydraulic pumps Hydraulic power is input, mechanical power comes out The mechanical power out is Phyd-out = T·w The hydraulic power in is Phyd-in = p·Q Thus, the overall efficiency is hovrall = T·w/p·Q

Hydraulic motors vs pumps There are similar relationships for motors that we had with pumps Mechanical efficiency does not take leakage into account 𝜂 𝑚𝑒𝑐ℎ = 𝑇∙𝜔 𝑝∙ 𝑄 𝑇 And like pumps, Q T = VD·w In the case of hydraulic motors, because of internal leakage, the actual flow rate to produce the hydraulic power out is > Q T Q T and Q A are related by hV , the volumetric efficiency 𝜂 𝑉 = 𝑄 𝑇 𝑄 𝐴 Here, Q A = Q T + Q leak

Hydraulic motors vs pumps So the relationships are a little different because they are backwards But they make sense if you keep in mind the function of a hydraulic motor vs that of a hydraulic pump

Hydraulic motors vs pumps Many of the details of pumps apply also to motors There are three basic types of hydraulic motors Gear Vane Piston But in general you cannot drive a pump as a motor The backwards flow of power through the motor—from hydraulic to mechanical—requires internally different mechanics for the two devices

E.g., vane motors vs pumps For example, a vane pump is driven mechanically The vanes are flung outward by centrifugal force So it is unnecessary to provide any mechanical means (a spring) to push the springs outward When the pump starts to turn, the vanes move outward, and the pump chambers are sealed But a vane motor might be hard to start from a dead stop if the vanes are not held outward to seal the chambers

Suggestion for understanding motors To understand the operation of a motor First understand the operation of the corresponding pump Then pose the question, what must be different to make this device operate as a motor

Rotary actuator Besides a hydraulic motor, there is also a rotary actuator This device is just like a motor, but it has a limited range of angular motion It does not turn through 360° Such a device would be used for a backhoe, for example Because of the limited range of motion, a rotary actuator is like a circular hydraulic cylinder The symbol for a rotary actuator shows this characteristic

Rotary actuator A rotary actuator comes to the end of its stroke and stops there, just like a cylinder At the end of its stroke, up against a stop, the pressure builds as it does in a cylinder So one needs to guard against over-pressure, like with a cylinder

Unbalance motors Watch out for unbalance motors The radial and thrust load on motors can be high if the internal geometry is not arrange to compensate for this This can shorten the life of the motor This can limit the amount of torque that the motor can deliver There are balanced motors, where the internal geometry is so laid out that unbalanced pressure forces are designed out of the motor

Variable flow rate Vane and piston motors can be constructed so that the flow rate is variable In fact, the internal volume can be adjusted so that the motor runs either forward or backward This arrow indicates adjustable flow rate Dual arrows show forward/backward rotation The flow-rate adjustment is effected by changing the position of a cam ring (vane), or the angle of a bent-axis motor or of a swash plate

Causality of speed control At this juncture, it is useful to give a brief look forward to closed-loop speed control and how the causality works in such a control loop Let’s say that the system at right is part of a speed control loop Its job is to hold the speed steady even if the load changes If the load were to increase, T would increase This would cause the motor to slow down An automatic controller would increase the motor’s output by changing the eccentricity of the cam ring (vane), or q in a piston pump Having to push harder against T to increase the speed with a greater flow rate, the pressure drop across the motor would also increase

If the load decreases If the load on the motor (T) suddenly decreased The motor would make the load spin faster The control for the cam ring or angle for a piston motor would actuate to reduce the eccentricity or angle This would reduce flow This would cause the motor to slow down The lighter torque against which the motor was pushing would cause a decrease in the pressure drop across the motor We shall explore this more when we cover closed-loop, feedback control later

Power/Efficiency relationships The power and efficiency relationships for a hydraulic motor are similar but different than those for a pump. Let’s look first at volumetric efficiency. QT , the theoretical flow rate, is the product of swept volume and speed. 𝑄 𝑇 = 𝑉 𝐷 ∙𝜔 But there is leakage, so the actual volume to drive the motor is greater than what’s theoretically required. 𝜂 𝑉 = 𝑄 𝑇 𝑄 𝐴

Power/Efficiency relationships For a pump, the actual flow rate produced is less than what it could theoretically produce, so QA < QT and 𝜂 𝑉−𝑝𝑚𝑝 = 𝑄 𝑇 𝑄 𝐴 Thus: Mechanically, for the power delivered by the motor, the theoretical torque can be calculated. But, like with a pump, there is internal friction, so the torque actually delivered by the motor, TA , is less than the theoretical torque, TT . 𝜂 𝑚 = 𝑇 𝐴 𝑇 𝑇 vs: 𝜂 𝑚−𝑝𝑚𝑝 = 𝑇 𝑇 𝑇 𝐴

Motor torques From the drawing below we see that Phyd is opposed by the frictional torque in the motor So only a lesser amount of the theoretical torque arrives at the output shaft of the motor The input Phyd rotates the motor counterclockwise… TA = TT - Tf …but that motion creates a counter torque (clockwise)

This is the same relationship we encountered with pumps Ideal case to real case If there were no leakage and no friction, the motor would be 100% efficient: 𝑃 𝑖𝑛 = 𝑃 ℎ𝑦𝑑 = 𝑃 𝑜𝑢𝑡 = 𝑃 𝑠ℎ𝑎𝑓𝑡 𝑝∙ 𝑄 𝑇 = 𝑇 𝑇 ∙𝜔 With the efficiency relationships, 𝑄 𝑇 = 𝜂 𝑉 ∙ 𝑄 𝐴 and 𝑇 𝑇 = 𝑇 𝐴 𝜂 𝑚 , so 𝑝∙ 𝜂 𝑉 ∙ 𝑄 𝐴 = 𝑇 𝐴 𝜂 𝑉 ∙𝜔 This is the same relationship we encountered with pumps 𝜂 𝑉 ∙ 𝜂 𝑚 = 𝑇 𝐴 ∙𝜔 𝑝∙ 𝑄 𝐴 = 𝑃 𝑜𝑢𝑡 𝑃 𝑖𝑛 = 𝜂 𝑜 and

TT and VD Also, if we go back to the theoretical case 𝑝∙ 𝑄 𝑇 = 𝑇 𝑇 ∙𝜔 and employ the relationship for the theoretical flow rate 𝑄 𝑇 = 𝑉 𝐷 ∙𝜔 we get another useful relationship: 𝑝∙ 𝑉 𝐷 ∙𝜔= 𝑇 𝑇 ∙𝜔 𝑇 𝑇 =𝑝∙ 𝑉 𝐷

Motor vs pump Pump Motor 𝑄 𝐴 < 𝑄 𝑇 (supplied) 𝑄 𝐴 > 𝑄 𝑇 (required) 𝜂 𝑉 = 𝑄 𝐴 𝑄 𝑇 𝜂 𝑉 = 𝑄 𝑇 𝑄 𝐴 𝑇 𝐴 > 𝑇 𝑇 (required) 𝑇 𝐴 < 𝑇 𝑇 (supplied) 𝜂 𝑚 = 𝑇 𝑇 𝑇 𝐴 𝜂 𝑚 = 𝑇 𝐴 𝑇 𝑇 Thus, some relationships are the same, but some are different, yet they all make sense

Example A hydraulic motor has a volumetric displacement of 80 cm3/rev. It operates at 70 bar. It receives oil from a pump at 35 lpm. Assume no leakage and no internal friction. Find: wmtr TT PT-out Solution: 𝑇 𝑇 = 𝑉 𝐷 ∙𝑝= 80 𝑐𝑚 3 𝑟𝑒𝑣 ∙70 𝑏𝑎𝑟∙ 1.0𝐸5 𝑁 𝑏𝑎𝑟∙𝑚 2 ∙ 𝑚 100 𝑐𝑚 3 ∙𝑚∙ 𝑟𝑒𝑣 2∙𝜋 𝑇 𝑇 =89.1 𝑁∙𝑚

Example Of course the motor will have leakage, but we will make a first-pass calculation to see the best limits of motor performance. No leakage means Q T = Q A 𝜔= 𝑄 𝑇 𝑉 𝐷 = 𝑄 𝐴 𝑉 𝐷 = 35 𝑙∙𝑟𝑒𝑣 80 𝑐𝑚 3 ∙𝑚𝑖𝑛 1000 𝑐𝑚 3 𝑙 =437.5 𝑟𝑝𝑚 𝑃 𝑇 = 𝑇 𝑇 ∙𝜔=89.1 𝑁∙𝑚∙437.5 𝑟𝑒𝑣 𝑚𝑖𝑛 ∙ 𝑚𝑖𝑛 60 𝑠𝑒𝑐 ∙ 2∙𝜋 𝑟𝑒𝑣 =4082 𝑁∙𝑚 𝑠𝑒𝑐 𝑃 𝑇 =4082 W=4.08 kW

Example Note also that the hydraulic power delivered to the pump is 𝑃 ℎ𝑦𝑑−𝑖𝑛 =𝑝∙ 𝑄 𝐴 =70 𝑏𝑎𝑟 ∙35 𝑙 𝑚𝑖𝑛 ∙ 1.0𝐸5𝑁 𝑏𝑎𝑟∙ 𝑚 2 ∙ 𝑚𝑖𝑛 60 𝑠𝑒𝑐 ∙ 𝑚 3 1000 𝑙 𝑃 ℎ𝑦𝑑−𝑖𝑛 =4083 𝑊=4.083 𝑘𝑊

Example A hydraulic motor has a volumetric displacement of 160 cm3/rev. It operates at 70 bar. The motor’s speed is 2000 rpm. The flow rate through the motor is 0.006 m3/sec. The actual torque delivered by the motor is 170 N·m. Find the motor efficiencies and the mechanical power delivered by the motor. Solution: The theoretical torque is TT = p·VD , so 𝑇 𝑇 =70 𝑏𝑎𝑟∙160 𝑐𝑚 3 𝑟𝑒𝑣 ∙ 1𝐸5 𝑁 𝑏𝑎𝑟∙ 𝑚 2 ∙ 𝑚 100 𝑐𝑚 3 ∙𝑚∙ 𝑟𝑒𝑣 2∙𝜋 =178 𝑁∙𝑚

Example Now we know the theoretical torque; the actual torque was given, so 𝜂 𝑚 = 𝑇 𝐴 𝑇 𝑇 = 170 𝑁∙𝑚 178 𝑁∙𝑚 =0.954=95.4% The volumetric efficiency, hV , is 𝜂 𝑉 = 𝑄 𝑇 𝑄 𝐴 = 𝑉 𝐷 ∙𝜔 𝑄 𝐴 𝜂 𝑉 =160 𝑐𝑚 3 𝑟𝑒𝑣 ∙2000 𝑟𝑒𝑣 𝑚𝑖𝑛 ∙ 𝑠𝑒𝑐 0.006 𝑚 3 ∙ 𝑚 100 𝑐𝑚 3 𝑚𝑖𝑛 60 𝑠𝑒𝑐 𝜂 𝑉 =0.889=88.9%

Example With hV and hm , we can calculate ho 𝜂 𝑜 = 𝜂 𝑉 ∙ 𝜂 𝑚 =0.889∙0.954=0.848=84.8% The power output of the motor is 𝑃 𝑚𝑡𝑟 =𝑇 𝐴 ∙𝜔=170 𝑁∙𝑚∙2000 𝑟𝑒𝑣 𝑚𝑖𝑛 ∙ 2∙𝜋 𝑟𝑒𝑣 ∙ 𝑚𝑖𝑛 60 𝑠𝑒𝑐 𝑃 𝑚𝑡𝑟 =35,605 𝑊=35.6 𝑘𝑊

Outside learning To better understand this subject matter, view the following Don’t forget to turn the closed-captioning on to be able to understand better the details of the lectures Watch: Fluid motor vs electric motor Hydraulic motor in lawn tractor drive

End of Week 4/Lesson 2