1. Week 5/Lesson 1 – Cylinders
Fluid Power Engineering
Week 5/Lesson 1 – Cylinders

2. Cylinders In this lesson we shall
First look at an important example with a hydraulic motor, a hydrostatic transmission
Investigate the important construction parameters of cylinders
Look at what calculations are needed regarding loads on cylinders
Explore power input and output of a cylinder
Look at pressure distribution in a cylinder circuit with a DCV
Investigate cylinder mounting possibilities
Investigate cylinder configurations for lifting weights
Explain cylinder cushioning

3. The reservoir is simply to replace fluid lost from the system
But before we start…
A hydro-static transmission
This is a variable-speed pump coupled with a bi-directional motor
The reservoir is simply to replace fluid lost from the system
Note that the pump and motor are directly connected, i.e. not through a reservoir
The variable flow rate of the pump allows the motor (and load) to turn at infinitely variable speeds
The pump turning in one direction can drive the motor at infinitely variable speeds in forward or reverse
On a mobile platform, the pump is driven by a gasoline or diesel engine

4. Hydrostatic transmission
If this is on a tractor or a piece of earth-moving equipment
Could set up engine for speed control
If load starts to increase and motor speed starts to decrease…
…more fuel is added to engine to pick speed back up
Thus the system acts like a cruise control on a car, except the resistance encountered is lifting a weight or encountering some obstacle in the soil
But the engine picks up speed, just like it does with a car going up a hill

5. Hydrostatic transmission
Operator can control system with one lever
If the pump is, for example, a swash-plate pump, the lever controls the swash-plate angle, q
With the lever pushed forward, q > 0 and thus Q > 0, and the motor turns forward
With the lever pushed backward, q < 0 and thus Q < 0, and the motor turns backward
With the lever upright, q = 0 and thus Q = 0
The greater q, the faster the motor turns forward
The greater q is set back, the faster the motor turns backward

6. But with swash-plate pump…
If the pump is a swash-plate pump, reverse flow is accomplished by changing the swash-plate angle
The engine and pump always turn in the same direction
But the swash plate allows the pump to pump backwards, even though its direction of rotation stays the same
Not many engines will run in two directions
But if the pump is driven by an electric motor, there are electric motors that do run in reverse
With such an electric motor drive, perhaps it would be possible to use a simpler pump, one that must not itself pump backward

7. Only part of the flow from the pump is used to turn the motor
Example: hydrostatic transmission
The following parameters for a hydrostatic transmission are given
Parameter
Pump
Motor p
70 bar VD
80 cm3/rev ? hV
82%
92%
83%
90% w
500 rpm
400 rpm TA
N/A
Calculate the missing parameters for the motor
Solution:
𝑄 𝑇−𝑝𝑚𝑝 = 𝑉 𝐷−𝑝𝑚𝑝 ∙ 𝜔 𝑝𝑚𝑝 =80 𝑐𝑚 3 𝑟𝑒𝑣 ∙500 𝑟𝑒𝑣 𝑚𝑖𝑛
𝑄 𝑇−𝑝𝑚𝑝 =40,000 𝑐𝑚 3 𝑚𝑖𝑛
𝑄 𝐴−𝑝𝑚𝑝 = 𝜂 𝑉−𝑝𝑚𝑝 ∙ 𝑄 𝑇−𝑝𝑚𝑝 =0.82∙40,000 𝑐𝑚 3 𝑚𝑖𝑛 =32,800 𝑐𝑚 3 𝑚𝑖𝑛
Only part of the flow from the pump is used to turn the motor
𝑄 𝑇−𝑚𝑡𝑟 = 𝜂 𝑉−𝑚𝑡𝑟 ∙ 𝑄 𝐴−𝑚𝑡𝑟 =0.92∙32,800 𝑐𝑚 3 𝑚𝑖𝑛 =30,176 𝑐𝑚 3 𝑚𝑖𝑛
The rest of the flow leaks internally

8. Example: hydrostatic transmission
Parameter
Pump
Motor p
70 bar VD
80 cm3/rev ? hV
82%
92% hm
83%
90% w
500 rpm
400 rpm TA
N/A
But
𝑄 𝑇−𝑚𝑡𝑟 = 𝑉 𝐷−𝑚𝑡𝑟 ∙ 𝜔 𝑚𝑡𝑟 So
𝑉 𝐷−𝑚𝑡𝑟 = 𝑄 𝑇−𝑚𝑡𝑟 𝜔 𝑚𝑡𝑟 = 30, 𝑐𝑚 3 𝑚𝑖𝑛 𝑚𝑖𝑛 𝑟𝑒𝑣 =75.4 𝑐𝑚 3 𝑟𝑒𝑣
As we have learned, the pressure actually is caused by the load that the system drives. But if the load is maximum, so that the pressure is the maximum that the pump can produce, what torque could the motor output?
𝑇 𝑇 =𝑝∙𝑉 𝐷−𝑚𝑡𝑟 =70 𝑏𝑎𝑟∙75.4 𝑐𝑚 3 𝑟𝑒𝑣 ∙ 1𝐸5 𝑁 𝑏𝑎𝑟∙ 𝑚 𝑚 100 𝑐𝑚 3 ∙𝑚∙ 𝑟𝑒𝑣 2∙𝜋 =84.0 𝑁∙𝑚
𝑇 𝐴 = 𝜂 𝑚−𝑚𝑡𝑟 ∙ 𝑇 𝑇 =0.90∙84.0 𝑁∙𝑚=75.6 𝑁∙𝑚

9. Cylinder details Cylinders produce linear motion from hydraulic flow
The cylinder at left has a single rod on the right
The rod-less end is called the cap end
Cap end
The rod end is called the rod end
Rod end
The cylinder has two ports, A and B
When the cylinder extends, this pushes fluid in the rod end out of port B
Flow into port A causes the cylinder to extend
Flow into port B causes the cylinder to retract
When the cylinder retracts, this pushes fluid in the cap end out of port A

10. Cylinder details S is the stroke of the cylinder
It is shorter than the cylinder length because of the thickness of the piston
v , the velocity of extension/retraction is important
This is the rate at which volume is swept as the cylinder moves
During cylinder extension, the flow rate into the cylinder in port A is Q A = A cap·v
During cylinder extension, the flow rate out the cylinder at port B is Q B = A rod·v
Thus, for a given v, the flow rate in ≠ the flow rate out: Q A > Q B

11. Cylinder details Because of the unequal areas on the cap and rod ends
If an equal pressure were applied to both ends, the cylinder would extend
Note the seals in the cylinder
There are actually double-rod cylinders with rods on both ends. These are used for precise motion control, because the unequal areas on each side cause a non-linearity in the system that is more difficult to control
These seals around the piston prevent internal leakage around the piston from high pressure to low pressure
These seals that grab the rod prevent leakage from the cylinder into the surrounding environment

12. Cylinder details Because of the unequal areas on the cap and rod ends
A given flow Q into port A will cause the extension of the cylinder to be slower than the same flow rate into port B
𝑣 𝑒𝑥𝑡 = 𝑄 𝐴 𝑐𝑎𝑝
𝑣 𝑟𝑒𝑡 = 𝑄 𝐴 𝑟𝑜𝑑

13. Force analysis on cylinder
Let’s look at the force analysis on a cylinder working against a load
If the cylinder is moving at a constant speed, the p·A cap = Fload
If the cylinder is moving at a constant speed, the p·A rod = Fload
Comparing the two cases, if Fload is the same in both cases, then p for the case where the flow enters the rod end must be > p with flow into the cap end
The rod end here is connected directly to the reservoir, so prod-end = 0

14. Power analysis on cylinder
If, in both cases, the force and velocity of the load are the same, then the power driving the load is the same, namely Fload·v
Then for the extension case, assuming no losses, Fload·v = pcap·Q cap
…and for the retraction case Fload·v = prod·Q rod
For the first case, pcap is lower but Q cap is greater…
…and in the second case, prod is greater but Q rod is lower…

15. Pressure depends on load
Notice that if the load is light, then the pressure will be low
This is because the pump does not have anything to push against
The load can be increased to pmax·A, where pmax is the pressure setting of a PRV (not shown); past that the pressure is not great enough to push the load
So, again we see that the load determines the system pressure; the pump determines the flow

16. There’s friction too
Not considered so far is that there is friction in the system
The movement of a mass driven by the cylinder may have a great deal of friction associated with it
The seals on the piston and the rod exit from the cylinder have some friction associated with them
So undergoing a constant velocity, pA = Fload + Ffriction

17. But this is a simple case
But in a normal case, the pump flow will be routed through a DCV
There will be pressure drops in a DCV
So for a real case with flow conduits, fittings, and valves, we must perform a thorough analysis of the pressure drops in the system from pump discharge to tank

18. Force analysis on cylinder
Let’s look at the simple case of a double-rod cylinder pushing a load
What’s different here is that pA and pB are not pP and pT , as in the previous example
There is a pressure drop from pP to pA and another from pB to pT
If the load is being driven at a constant velocity, Fload = (pA – pB)·A
Thus, the greater the load, the greater the difference between pA and pB
Most DCVs are symmetrical, in that Dp across the pressure side = Dp across the tank side, i.e. pP – pA = pB - pT
Thus, if the imposed load is very light, then pA ≈ pB , and there are large pressure drops across the DCV

19. Force analysis on cylinder
Thus, if the imposed load is very light, then pA ≈ pB , and there are large pressure drops across the DCV
This means that pA ≈ pP/2
Thus pB ≈ pP/2 too
On the other hand, if the imposed load is very high, then pA → pP , and pB → pT
The pressure drop from pP to pT occurs across the piston in order to drive the load
Thus, again the load determines how the pressure is distributed in the system

20. Design context for cylinder
In general, the design of a system begins with the load
How much of a load is there (p and A)?
How fast do we need to move it (Q)?
How far do we have to move it (S)?
With 3, we have the stroke of the cylinder
But with 1, there is still a choice to make
Do we use a small (A), high-pressure cylinder (and lines, and pump) with a low flow rate?
Or do we use a large (A), low-pressure cylinder (and lines, and pump) with a large flow rate?
The answers to these questions depend very much on the context of the problem

21. Design context for cylinder
A good way to start would be to select a standard operating pressure for our system: 70 bar, 210 bar, 350 bar
Once that is done, then the A of the cylinder can be selected
With A and the specified velocity, we can determine Q
With Q and p, we can select a pump
With the pump selected, we can determine the power needed to drive the system
This is an overview of how design decisions are made in the design of a hydraulic system

22. Cylinder construction details
The figure below shows the construction details of a hydraulic cylinder
Notice too the internal seal on the piston to prevent leakage from the high-pressure side to the low-pressure side, i.e. internal leakage
Besides the external seals where the rod leaves the cylinder, there is a rod wiper to scrape off foreign material and prevent it from entering the cylinder it retracts
In general, the cylinder consists of a pipe-like barrel bound between two end plates, held together on the barrel by tie rods

23. Cylinder mounting options
Cylinders come with a variety of mounting options, shown in the figure below
The trunnion- and clevis-mounted cylinders allow for a flexible geometry, so that the cylinder can rotate during extension/retraction

24. Options for lifting weights
For lifting weights, the geometries shown below are standard
First class
Second class
Third class
For each a kinematic analysis of the movement is required to determine where the forces and cylinder speeds will be highest

25. Cylinder cushions
When pistons come to the end of their travel, we want to prevent them from slamming into the cylinder ends, damaging the cylinder
When the cushion spear enters the hole at the cap end of the cylinder…
This needle can be adjusted into or out of the orifice to change the resistance of flow through the orifice
The same arrangement is set up on the rod end for cylinder retraction
The fluid is forced to exit through this small orifice

26. Outside learning
To better understand this subject matter, view the following videos on the “Big Bad Tech Channel” and elsewhere
Don’t forget to turn the closed-captioning on, if it’s available, to be able to understand better the details of the lectures
Watch:
Hydraulic Cylinders
Cylinder Selection: The Key to Better Hydraulic Systems

27. End of Week 5/Lesson 1