“Free” Compressed air? Really? (Part 2)
(This is Part 2 of a 2-part article on the issues involved with using compressed air for load applications. Link to Part 1 of the article).
I had used an example of a hydraulic bottle jack in order to illustrate the basics of Pascal’s Law in Part 1. Now let us do the same exercise, except this time we assume that we have somehow converted the hydraulic bottle jack to function using compressed air (i.e. pneumatic) instead of hydraulic oil.
(I would like to mention, again, that this article has been written so that a reader with little or no technical knowledge can follow along and comprehend the logic. As a result, this may prove to be a bit too basic for those with more technical and engineering knowledge. )
An exercise to demonstrate improper use of compressed air: Pneumatic jack
Consider the converted bottle jack to be hooked up to a 100 psig compressed air pressure line. That pressure (see schematic below) is now acting on what used to be area A2 in the previous illustration of the hydraulic bottle jack.
Let us now calculate the cross-sectional area A2 required to lift 3,000 lbs of load (F2). Since force is equal to pressure time area, therefore, to lift 3,000 lbs we shall require 100 psig (pounds per sq in gage pressure) acting on 30 sq in of area.Now compare the 30 sq in
Now compare this 30 sq in area value in the pneumatic application to say 3 sq in of area required in a hydraulic bottle jack to do the same work. The cylinder diameter for the pneumatic application calculates out to be 6.2 inches vs approximately 1 inch in diameter for the hydraulic jack. That explains why we don’t see many pneumatic bottle jacks. They would be too heavy, bulky and unwieldy. Using compressed air in this application is unwise.
Incredibly, pneumatic jacks actually do exist. The photographs below should illustrate the point made above about the size difference between a hydraulic and a pneumatic jack.
From the example above one can intuitively understand why using compressed air for performing work through the application of load (i.e. pressure) is wasteful. Pressure has (theoretically) a one to one transfer through a hydraulic fluid because it does not compress while air gets compressed and does not have a 1-to-1 transfer of load. But how does that translate to cold hard cash?
Attempting to show how energy consumption is calculated from first principles was turning out to be too lengthy and complicated for this article. For that reason, I decided to use some empirical facts and data to provide some simple rules of thumb to aid in estimating the energy costs for load transfer using hydraulics as well as air (pneumatics).
This excerpt below, from Bud Trinkel’s comments, sums up, quite well, how expensive it can be to operate an air motor.
“….cost of an air circuit may be less than a hydraulic circuit but operating cost can be five to ten times higher. Compressing atmospheric air to a nominal working pressure requires a lot of horsepower.
….. It takes approximately one horsepower (hp) to compress 4 CFM (cubic feet per minute) of atmospheric air to 100 psi.
A 1-hp air motor can take up to 60 CFM to operate, so the 1-hp air motor requires (60/4) or 15 compressor horsepower when it runs. Fortunately, an air motor does not have to run continuously but can be cycled as often as needed.”
Therefore, when using compressed air in an air motor or diaphragm pump, it takes about 10 to 15 units of electrical energy at the compressor (compressor hp) to produce about one unit (motor or pump hp) of actual mechanical output to the work.
So what does it cost, in dollars and cents, to generate one cubic feet (CF) of compressed air at a given pressure? Let’s illustrate how to calculate that using an actual example.
We go back to the thought experiment illustrating the hypothetical pneumatic bottle jack. The size of the cylinder required for the pneumatic jack had to be six times larger to transfer air pressure at standard shop air (between 80 – 120 psi) to lift only 3,000 lbs.
Imagine how large a cylinder would have to be in order to move, say, 30,000 pounds (approx. 14 MT). For example, if we were, say, trying to punch steel sheets by simply using compressed air. A 62-inch diameter cylinder (theoretical without accounting for any losses),to be precise, going by the parameters used in the thought experiment.
We will use this example of a 62-inch diameter cylinder only for the purpose of demonstrating how the calculations work. (In actuality there are other methods to intensify air pressure and one would not really use a 62-inch diameter cylinder in a pneumatic press of 15 MT capacity). However, the energy consumption would be conceptually similar but calculated using different methods. Using this simple cylinder equivalent will help you get into the ball park of energy costs for that kind of tonnage.
We are going to first calculate the air consumption of a 62? bore cylinder with a 1-inch stroke under full load, operating 30 complete cycles (out and back) per minute at 100 psi inlet pressure. The calculations follow the method described in The Mead Pneumatic Handbook.
Step 1: Calculate the area of the piston by converting the bore diameter into square inches. (62 in. bore/2)2 x 3.1416 (?) = 3,019 sq. in.
Step 2: Determine air consumption per single stroke. (Volume of a cylinder) 3,019 sq. in. x 1 in. stroke = 3,019 cu. in.
Step 3: Determine consumption per complete cycle – to (Load application stroke) and fro (return stroke) (disregard the piston rod that will take up some of the volume because it is generally not significant). ? 3,019 cu. in. x 2 = 6,038 cu. in. of air at 100 psi, per cycle.
Step 4: Determine volume of 100 psi air that is consumed per minute. 6,038 cu. in. x 30 cycles/minute = 181,140cu. in./min. of 100 psi air
Step 5: Convert cu. in. to cu. ft.: 105 cu ft/min or CFM of 100 psi air
Step 6: Calculate compression ratio, which is how many times free air has to be compressed so as to get it to 100 psi. (This equation was discussed in Part 1) (100 psi + 14.7)/14.7 = 7.8
Step 7: Now determine cubic feet of free air used per minute (SCFM).? 105 cu. ft. x 7.8 compression ratio = 819 cu. ft. of free air used per minute
Now we go on to calculate energy costs to run this pneumatic press of approximately 14 MT capacity. In other words, we will calculate how much energy is expended to compress 819 cu ft of air per minute.
We learned (from Part 1) that it takes 1 hp to compress 4 SCFM of free air. So 819 SCFM would require 205 hp. Also, let’s assume that this pneumatic press runs for 2,000 hours per year.
Annual Electricity Costs = (horsepower required) x (0.746 kW/hp) x (Annual Hours of Operation) x (Electricity Cost in $/kWh)
= 205 x 0.746 x 2000 x 0.10
= $30,586 annually or $2,548 per month for a pneumatic press.
This cost can be compared to approximately $200 per month to operate a hydraulic press of the same capacity!!! (Reference)
I hope that this illustrates how much money is wasted in using compressed air for load applications.
Energy costs and sheer impracticality make compressed air an “unconscionable” choice for any applications that require the application of loads! Especially when there are less expensive alternatives.
Smarter Alternatives to Compressed Air
Having said all this about improper use of pneumatics for load applications, what then is the better alternative? Well, it depends….
In both parts of this article, I have been referring to hydraulic systems when comparing the ineffectiveness of compressed air systems. I don’t to make it appear that I am recommending hydraulic systems as a default alternative to pneumatic systems. Hydraulic systems are not without their own inefficiencies and energy waste. Certainly not as much as pneumatic systems, yet enough to be a factor of concern for a lean and green environment.
John A. Neun, explained the inefficiencies of hydraulic systems quite clearly in his online article.
“The worst offenders of low efficiency are hydraulic systems using fixed-displacement pumps sized to meet peak flow and pressure demands. Constantly running at full speed, pumps in these circuits continuously produce maximum flows at maximum system pressure. But except for a few moments in an operating cycle, most machines only require a fraction of a system’s total pressure and flow capabilities. During standby, positioning, pick-and-place, or holding operations, the system’s pump consumes maximum energy while the hydraulic system is performing little or no work. As a result, hydraulic system operating efficiency suffers dramatically.”
Apart from hydraulic presses, mechanical and servo presses are also among some of the better-known alternatives. There is an informative chart (reproduced below) that summarizes the difference in costs between mechanical, hydraulic and servo systems.
As you can see, these three systems are not uniformly better or worse than the other in the different aspects that were measured.
Note that the data in the chart above was normalized to the values of a servo press in each of the categories being measured. So everything is relative to a servo press. Also, the graph does not give actual values which makes it less useful to someone seeking to justify one system or another by doing a payback analysis or ROI. Another aspect is one of omission of data, namely, the capital cost for the presses. That would have probably shown the servo press as the more expensive option. Regardless, this schematic is useful for making some broad comparative choices in choosing one system over another.
There is another source of useful data comparing servo presses to hydraulic presses. This data is quantitative and has been replotted for easier viewing.
(Caution when viewing the above data: Each of the two charts shown above have been provided by companies that are selling Servo presses. Hence they are not from unbiased sources, yet considered useful enough to make a point, to be included here.)
There is another type of system that combines the benefits of air (speed) and hydraulics (power) to yield a more energy efficient system (compared to pneumatic or hydraulic systems) for load applications up to 60 or 100 MT. These are known as the hydropneumatic or air-over-oil systems.
For example, the air-over-hydraulic bottle jack (like the one shown in the photo) that can go up to 20 MT. This jack is operated using compressed air which acts on a hydraulic system (very similar to what was shown in the hydraulic bottle jack example in Part 1) The compressed air is not used to do the mechanical work. The compressed air works on the hydraulic fluid (through pressure intensification as was discussed in Part 1) and the hydraulic fluid, at the intensified pressure, is what does the ultimate work of load trans. I will have a separate article explaining, in details, how these systems work.
Air over hydraulic systems are by far more energy efficient than pneumatic and even hydraulic systems and depending on the application, much less expensive that servo systems.
More on that in the upcoming article.
(Full disclosure – Phoenix Manufacturing Systems sells air-over-hydraulic cylinders and presses.)
About the author:
Rahul Sarkar is a registered professional engineer with 30+ years of experience in manufacturing environments. Rahul’s passion is in improving profitability in manufacturing operations through the applications of common sense problem solving, as well as Lean Manufacturing and Quality Systems principles.
(Feel free to email me with any questions at email@example.com. I shall be delighted to help in whatever way I can.)