1. Week 3/Lesson 2 – Power & efficiency
Fluid Power Engineering
Week 3/Lesson 2 – Power & efficiency

2. Power & efficiency In this lesson we shall
Discuss what power and efficiency are in the realm of fluid power
Learn how to compute the efficiency of an existing hydraulic system
Examine a design application

3. Power takes many forms Power is the rate at which work is done
Thus its units are N·m/sec = W (watt)
Power comes in many forms
Mechanical power
Electrical power
Hydraulic power
Power can be converted from one form to another, but always at some cost (i.e. some is lost)

4. Efficiency (h) Efficiency is the ratio of power out/power in
Since some power is lost, power out < power in
So h always < 1
𝜂= 𝑃 𝑜𝑢𝑡 𝑃 𝑖𝑛 <1

5. Power crosses boundaries
Power can be measured at a domain boundary
That is, we can draw a control volume around a device and see where power enters and leaves the volume
Hydraulic power out
The pump raises the energy state of the fluid passing through the pump
Shaft power in
It does this at a certain rate
Hydraulic power in
So we can characterize a pump simply as a device that converts mechanical power (the shaft) into hydraulic power (the fluid flow)

6. Power in general
Power, in general, involves two types of physical quantities
Power is their product: power = effort · flow
Domain
Effort
Flow
Units
Electric V i
volt·amp = watt
Mechanical – translation F v
N·m/sec = Joule/sec = watt
Mechanical – rotational T w
N·m·(rad/sec) = Joule/sec = watt
Hydraulic/Pneumatic p Q
Pa·m3/sec = (N/m2) ·m3/sec = N·m/sec = Joule/sec = watt
But we do not have time here to cover this field.
Note that, no matter the domain, power has units watt
There is an entire methodology in the field of system dynamics called bond graphs that tracks power flow in multi-domain systems

7. Pump power conversion
The drawing below shows how power in a pump is converted from electrical power to mechanical shaft power (rotational) into hydraulic power
This is really Dp·Q, but the inlet p is close to atmospheric, and since we use gage pressures here, pin = 0
But we do not have time here to cover this field.
If the system were 100% efficient, i.e. no losses, then V·i = T·w = p·Q
But, of course, there will be losses, so V·i > T·w > p·Q

8. Efficiencies stack up
The power conversions are one after the other, i.e. serially configured
𝜂 𝑚𝑡𝑟 = 𝑇∙𝜔 𝑉∙𝑖
𝜂 𝑝𝑚𝑝 = 𝑝∙𝑄 𝑇∙𝜔
𝜂 𝑜𝑣𝑟𝑎𝑙𝑙 = 𝑝∙𝑄 𝑉∙𝑖 = 𝜂 𝑚𝑡𝑟 ∙ 𝜂 𝑝𝑚𝑝
But we do not have time here to cover this field.
So if both motor and pump efficiencies were 85%
𝜂 𝑜𝑣𝑟𝑎𝑙𝑙 =0.85∙0.85=0.7225
In a string of power conversions, overall efficiency can be low

9. If we add a cylinder…
The power conversions are one after the other, i.e. serially configured
The power delivered to the load is mechanical translation, so
𝑃 𝑐𝑦𝑙 =𝐹∙𝑣
But we do not have time here to cover this field.
and
𝜂 𝑜𝑣𝑟𝑎𝑙𝑙 = 𝐹∙𝑣 𝑉∙𝑖 = 𝜂 𝑚𝑡𝑟 ∙ 𝜂 𝑝𝑚𝑝 ∙ 𝜂 𝑐𝑦𝑙

10. The overall power-conversion view
So overall, the power conversion chain is
Electric motor: electricity to rotational mechanical
Pump: rotational mechanical to flow
But we do not have time here to cover this field.
Cylinder: flow to linear mechanical

11. Why need hydraulic system at all?
Q: But if had rotational mechanical power output from electric motor, why not just convert it directly to translational mechanical power directly without going through hydraulic system?
Could use a rack and pinion system for that
But we do not have time here to cover this field.
A: May need Pmech-trans in a confined space
Like in an aircraft wing and tail
Mechanical linkage to get from motor (big) out there may be too complicated
…and heavy
…and too subject to failure

12. Hydraulic system as mechanical linkage
So a hydraulic linkage can be thought of as a sort of flexible linkage that can deliver a lot of power through tortuous, flexible geometry exactly to the spot where it is needed
But we do not have time here to cover this field.

13. Two types of efficiency
With pumps and motors we break the efficiency down into two parts
Volumetric efficiency (hV)
Mechanical efficiency (hm)
Volumetric efficiency (hV)
The geometry of the pump is such that with every revolution of the pump, a distinct volume of fluid is pushed out, VD , the displaced volume
This leads to the concept of the theoretical flow rate, Q T
𝑄 𝑇 = 𝑉 𝐷 ∙𝜔

14. Volumetric efficiency
But of course there is leakage from high-pressure regions into low-pressure regions, so the actual flow rate, Q A , is less than Q T
Actually 𝑄 𝐴 = 𝑄 𝑇 − 𝑄 𝑙𝑒𝑎𝑘
The volumetric efficiency is the ratio of these two flow rates
𝜂 𝑉 = 𝑄 𝐴 𝑄 𝑇
The actual flow rate of a pump is given by the manufacturer from test results
This is usually given as a function of pressure too
The greater the pressure rise, the greater the leakage
So hV goes down with greater pressures

15. Mechanical efficiency
Mechanical efficiency (hm ) has to do simply with the internal friction in the pump
It does not take leakage into account
It is quoted by the manufacturer, who says…
…if you have a certain flow, Q T, going through the pump…
…at a certain pressure, p , then it will take this much input power to generate that flow
𝜂 𝑚 = 𝑝∙𝑄 𝑇 𝑇 𝐴 ∙𝜔
The actual power out is, of course 𝑝∙𝑄 𝐴 = 𝑝∙ 𝜂 𝑉 ∙𝑄 𝑇
Thus, the overall efficiency is 𝜂 𝑜 = 𝑝∙ 𝜂 𝑉 ∙𝑄 𝑇 𝑇 𝐴 ∙𝜔 = 𝜂 𝑉 ∙ 𝜂 𝑚

16. Overall efficiency This can be reworked into
𝜂 𝑜 = 𝑝∙ 𝜂 𝑉 ∙𝑉 𝐷 ∙𝜔 𝑇 𝐴 ∙𝜔 = 𝑝∙ 𝜂 𝑉 ∙𝑉 𝐷 𝑇 𝐴
And since 𝜂 𝑜 = 𝜂 𝑉 ∙ 𝜂 𝑚
𝜂 𝑚 = 𝑝∙𝑉 𝐷 𝑇 𝐴

17. A second look at this
We can calculate the relationships between power into and out of the pump by drawing a control volume around it:
Hydraulic power out:
The overall efficiency is the ratio of these two quantities:
Shaft power in:
𝑝∙𝑄 𝐴
𝜂 𝑜 = 𝑝∙𝑄 𝐴 𝑇 𝐴 ∙𝜔
𝑇 𝐴 ∙𝜔

18. Volumetric and mechanical efficiencies
But we also have two other efficiency quantities:
Volumetric efficiency:
The actual flow rate is less than the theoretical flow rate because of leakage
𝜂 𝑉 = 𝑄 𝐴 𝑄 𝑇
Mechanical efficiency:
The theoretical shaft torque needed by the pump is less than the actual torque due to internal friction in the pump
𝜂 𝑚 = 𝑇 𝑇 𝑇 𝐴

19. Volumetric and mechanical efficiencies
𝑇 𝐴 = 𝑇 𝑇 𝜂 𝑚
So:
𝑄 𝐴 = 𝜂 𝑉 ∙ 𝑄 𝑇
and:
𝜂 𝑜 = 𝑝∙𝑄 𝐴 𝑇 𝐴 ∙𝜔 = 𝑝∙ 𝜂 𝑉 ∙ 𝑄 𝑇 ∙ 𝜂 𝑚 𝑇 𝑇 ∙𝜔
Thus:

20. Ideal case: no leakage, no friction
If there were no leakage and no internal friction, there would be no losses, and the mechanical power into the pump would equal the hydraulic power out of the pump
𝑇 𝑇 ∙𝜔
𝑝∙𝑄 𝑇
𝑝∙𝑄 𝑇 = 𝑇 𝑇 ∙𝜔
𝑝∙ 𝑄 𝐴 𝜂 𝑉 = 𝑇 𝐴 ∙ 𝜂 𝑚 ∙𝜔

21. Ideal case: no leakage, no friction
But we can rearrange this equation:
𝑝∙ 𝑄 𝐴 𝜂 𝑉 = 𝑇 𝐴 ∙ 𝜂 𝑚 ∙𝜔
𝜂 𝑉 ∙ 𝜂 𝑚 = 𝑝∙ 𝑄 𝐴 𝑇 𝐴 ∙𝜔 = 𝑃 𝑝𝑚𝑝−𝑜𝑢𝑡 𝑃 𝑝𝑚𝑝−𝑖𝑛 = 𝜂 𝑜
Also, returning to the ideal case:
𝑝∙𝑄 𝑇 = 𝑇 𝑇 ∙𝜔
, since
𝑄 𝑇 = 𝑉 𝐷 ∙𝜔
𝑝∙ 𝑉 𝐷 ∙𝜔= 𝑇 𝑇 ∙𝜔
means
𝑝∙ 𝑉 𝐷 = 𝑇 𝑇

22. Summary
Thus, needed to calculate input power, output power, overall efficiency, and actual flowrate of a pump are
hV , hm , VD , w , p
Then
𝜂 𝑜 = 𝜂 𝑉 ∙𝜂 𝑚
𝑄 𝑇 = 𝑉 𝐷 ∙𝜔
𝑄 𝐴 = 𝜂 𝑉 ∙ 𝑄 𝑇
𝑃 𝑝𝑚𝑝−𝑖𝑛 = 𝜂 𝑚 ∙𝑝∙ 𝑄 𝑇
𝑃 𝑝𝑚𝑝−𝑜𝑢𝑡 = 𝜂 𝑜 ∙ 𝑃 𝑝𝑚𝑝−𝑖𝑛

23. Example
A pump has a volumetric efficiency of 92%, a mechanical efficiency of 95%, a displaced volume of 7 in3/rev, turns at 1800 rpm against a pressure of 3000 psi. Find the overall efficiency (mechanical with leakage), the actual flow rate, the pump’s input mechanical power, and the pump’s output hydraulic power.
𝜂 𝑜 = 𝜂 𝑉 ∙𝜂 𝑚 =0.92∙0.95=0.874=87.4%
𝑄 𝑇 = 𝑉 𝐷 ∙𝜔=7 in 3 rev ∙1800 rev min ∙ gal 231 in 3 =54.5 gpm

24. Example, continued 𝑄 𝐴 = 𝜂 𝑉 ∙ 𝑄 𝑇 =0.92∙54.5 gpm=50.2 gpm
𝑃 𝑝𝑚𝑝−𝑜𝑢𝑡 =𝑝 ∙𝑄 𝐴 =3000 𝑙𝑏 𝑖𝑛 2 ∙50.2 gal min ∙ 231 in 3 gal ∙ ft 12 in
𝑃 𝑝𝑚𝑝−𝑜𝑢𝑡 = ft∙lb min ∙ min∙hp 33,000 ft∙lb =90.6 hp
𝑃 𝑝𝑚𝑝−𝑖𝑛 = 𝑃 𝑝𝑚𝑝−𝑜𝑢𝑡 𝜂 𝑜 = 90.6 hp =104 hp

25. Power loss in fluid conduits
Note that there will also be a pressure drop in pipes/tubes/hoses between the hydraulic power that in generated by the pump and the hydraulic power with which the cylinder does its work
This conduit run between the pump discharge and the cylinder could be a twisting and turning geometry

26. The engine and hydraulic pump are located on the main chassis here
For example, a backhoe
Think of a backhoe, for example
But the fluid arrives at the bucket cylinder through a long, complicated conduit run
The engine and hydraulic pump are located on the main chassis here
There is a pressure loss between the pump discharge and the bucket cylinder because of the friction in the conduit, fittings, and valves

27. Pressure losses in fluid conduits
When fluid flows through conduits, fittings, valves, it generates friction, and that represents a power loss as heat to the environment
We’ll cover this in detail later
Friction loss depends on length of the conduit line
Friction loss depends on roughness of internal surface of conduit
Friction loss depends on number and types of fittings
Friction loss depends on resistance in valves
Friction loss depends on thickness of fluid

28. A quick look at a design problem
Shown at right is a pump driving a cylinder that raises a mass
For part of the motion cycle, we want to accelerate the mass upward at a steady acceleration rate, a
Then after we reach a certain speed, v , we continue at that speed for a certain part of the motion cycle
What we need to know is what pump, cylinder, and conduit is needed to perform these motion specifications
We have a coupled problem
We would like to calculate the needed flow rate by using the given v , but we also need the A of the cylinder to calculate Q
And A of the cylinder depends on the load
Besides the gravitational load (m·g), the upward load depends on the acceleration too

29. A quick look at a design problem
Below is shown a free-body diagram and mass-acceleration diagram for the mass
From the force balance, we see that
F = m·(g+a) = p·A
Now we have a decision to make. What pressure do we want the system to operate at?
There are standard pressures: 70 bar, 210 bar, 350 bar
The higher the pressure, the smaller the cylinder
The smaller the cylinder, the lower the flow rate (Q = A·v)

30. A quick look at a design problem
Also the higher the pressure, the more expensive the conduit, fittings, and valves
But with a high-pressure system, since everything is smaller, the system is lighter
As you can see, often a fluid-power design problem can be complicated, even in a simple situation like this
Often with decisions you come to a fork in a road that has no clear indicator which road needs to be taken
This is the challenge of fluid-power design
Effective decision making comes with experience

31. Still another complication
Besides the options for designing a pump and cylinder to perform a certain task, there is another device that can be added to a system for short-term power needs
Accumulators are storage devices for fluid and pressure that can be called on to deliver that power during limited amounts of time
We shall look at accumulators in the next lesson to see how they can be designed and deployed in our circuits to provide benefits with smaller costs

32. End of Week 3/Lesson 2