Fluid Power Journal

System Integrators Energy Challenge: Follow Up

There are some prerequisites for reading this article: you need to read last month’s article entitled “System Integrators: An Energy Challenge,” you need to have your calculator handy, and you probably ought to have your Lightning Reference Handbook at your fingertips (available at www.ifps.org) because this article is going to examine the thought process used when trying to provide the most energy-efficient fluid power system.

A quick recap: two fictitious fluid power applications were described: one hydraulic and one pneumatic. The challenge is to see who can come up with the most energy-efficient system to do the work. The hydraulic system will be rated by the kW used and the pneumatic system by the standard cubic feet of air consumed.

This article will cover what I believe should be considered for the two systems. I will not give away my answer, but I will tell you what I think the targets should be.

With the hydraulic system, the first thing we need to do is determine the power requirements for the system. This is a challenge because of the changing power requirements throughout the cycle. As you recall (or maybe just read), a cylinder rod has to extend horizontally for two seconds with almost no load at a rate of 12″/second. The rod then comes in contact with a heavy object that it has to move 96″ in 8 seconds. It then has to retract under no load for the full 120″ and cover the distance in 10 seconds. The system now rests for a full 2 minutes waiting to start again. We were given the fact that the cylinder has a 6″ bore, a 4″ rod, and a 120″ stroke. We are to assume a return line pressure of 100 psi and an overall efficiency of 85%. The weight of the object to move is 300,000 pounds, but it is slid sideways, not lifted, and there is a given coefficient of friction of 0.25.

Now, at first look, it seems pretty straightforward; the speed is always 12″/second, so we only have to calculate the flow rate for that speed and then add the pressure required. But there is a wrinkle here. The heavy object is not moving when the rod comes in contact with it, so it will have to be accelerated at a rate that will allow it to cover the distance in 8 seconds. It seems reasonable that there will be a hesitation of about 1 second as the rod contacts the object, builds pressure, and satisfies the capacitance of the plumbing. This means I have to cover the 96″ in just 7 seconds.

Now pull out your Lightning Reference Handbook and turn to page 101 where the formulas are listed. Take a pencil and write these acceleration formulas in the margin:

F = ma, d = 0.5at2, and V = at where F equals force in pounds, m equals mass, a equals imperial acceleration, d equals distance traveled in feet, t equals time in seconds, and V equals velocity in feet/second. You’ll be glad you have these the next time you take an IFPS certification exam.

We first have to solve for a. We know the distance the object has to move and we know how much time it will take to move it. It travels 96″ or 8′ in 7 seconds. So warm up your calculator and what do you get? 8 = 0.5a72. So, 16 = 49a. 16/49 = a.  a = 0.3265.

Who remembers how to find the mass of the object? Okay, you with the pocket protector full of pens, how do we do it? That’s right! We divide the weight by the acceleration of gravity, which in imperial speak is 32.2. So 300,000 / 32.2 = 9317ish so m = 9317.

Force and velocity will now fall into place. F = ma. So F = 9317 x 0.3265 giving us F = 3042 pounds of force.  V = at.  So V = 0.3265 x 7 giving us V = 2.29 feet/second maximum speed.

I like to use Q = 2.45 Vd2 to find the flow rate where Q equals flow in gpm, V equals velocity in feet/second, and d equals the inside diameter through which the fluid moves. When we apply this to our 6″ bore cylinder, we find we need a pretty high flow rate by the time the rod reaches the end of its stroke. Q = 2.45 x 2.29 x 36 so Q = 202 gpm.

We have the maximum flow rate, but who can tell me how to find what pressure we need to move the load? Come on. You know the answer. Just shout it out! Right! P = F/A where P equals pressure, F equals force, and A equals the area acted on by the pressure. We know the area (62 x 0.7854) so A = 28.27 in2, but what is the resistive force? Careful now! We have the 300,000-pound object with a coefficient of friction of 0.25 providing a resistive load of 75,000 pounds. But that’s not all. We also have an acceleration force of 3,042 pounds and a pressure of 100 psi in the return line acting on the annuls area of the piston…Quick! Who has that figured already? 1571 pounds? Everybody agree? There is a total resistive load of 75,000 + 3,042 + 1571 pounds for a total of 79,613 pounds.  P = 79,613 / 28.27 so P = 2816 psi.

We know that hp = gpm x psi / 1714. This means that, as the cylinder approaches the end of its stroke, we have to provide 202 x 2816 / 1714 or 332 hp in order to do the work. Throw in the 85% efficiency, and we are looking at 391 hp (292 kW).

A 400-hp motor with a 205-gpm pump would do the job, but remember, the contest is to find the most energy-efficient system. That means we need to use the smallest electric motor/pump combination that will do the job and then push it to its nameplate power as long as it is running. We need to take a look, not at the maximum power consumption, but at the average power consumption.

To find the average power consumption, let’s take a second-by-second snapshot at what is going on in the system. I know, I know. It sounds like a lot of work, but we are the professionals, right? It really will not be that bad. The total cycle time is 140 seconds, but the working portion is only 20 seconds. So we need 20 snapshots of the energy usage; 2 seconds no-load extending, 8 seconds full-load extending, and 10 seconds no-load retracting. When we calculate the flow of pressure at each second, we can find the kW usage at each second. We then add them all up and divide by the total cycle time to get our average power usage.

I see some of you are getting glassy eyed already, so I will give you the answer. The average power consumption for this system is 7.51 kW (10.07 hp). Given the 85% overall efficiency, we have a target of 8.84 kW (11.85 hp). The winning system will be the one that comes closest to this number.

Okay, let’s take a ten-minute break and then we will come back and take a look at the pneumatic challenge….

dan challenge followup

The pneumatic system poses a somewhat different set of challenges. We discussed in an earlier article that hydraulics and pneumatics differ in the way they use energy. Hydraulics is a “pay as you go” system where we put in the energy and then immediately either use it by doing useful work or release it by turning it into heat. Pneumatics is more of a “store it now and use it later” system where we add energy to a volume of air, store it in a receiver, and then use it later as needed. The amount of air we use is described in terms of Standard Cubic Feet (scf). In order to determine the amount of scf to be consumed, we need to find two things: the maximum pressure required and the volume to be filled.

This is what we know. A 4″ bore cylinder having a 1″ rod and with a 24″ stroke is mounted horizontally and is fully extended. It is attached to an 85-pound spacer that is wedged between two parts. It will take 1000 pounds of force to break the spacer free, and then the cylinder retracts in 1 second, dragging the weight of the spacer with a coefficient of friction of 0.15. After a 30-second dwell time, the cylinder extends in 1 second against the inertia of the spacer and the 0.15 coefficient of friction. The parts are then placed against the spacer and 10 seconds after the extension, the process repeats.

We will again use the formula P = F/A. The F will be the given 1000 pounds plus the acceleration force required of the 85-pound spacer plus the frictional resistance. To be on the safe side, let’s add in a 15-psi resistive load as the air is exhausted from the cylinder. We find the F to be 1000 + 10.56 + 12.75 + 188.55 or 1211.86 pounds. Remember that, in this case, the A is the area of the rod end of the cylinder, which is 11.78 in2.  So P = 1211.86 / 11.78 or 102.87 psig. We now have the maximum pressure.

We next have to calculate the volume. This is easy. We already know the annulus area and the stroke, so the volume is 11.78 x 24 or 282.72 in3. We will add in the volume to fill when we extend, which is 12.57 x 24 or 301.68 for a total volume of 584.40 in3.

The conventional way of calculating the required scf would be to take the maximum pressure in terms of psia and divide it by the atmospheric pressure and multiply that result times the total volumes of the cylinder. We would use (102.87 + 14.7) / 14.7 giving us about 8. So 8 x 584.40 in3 = 4675.2 in3. Now we want scf not in3. So we divide 4675.2 in3 by 1728 in3 and get 2.71 scf for each complete cycle of the system. We would use meter-out flow controls to control the cylinder speed.

The fact is we never need both the maximum pressure and the full volume at the same time. Once we break the spacer free, we only have 211.86 pounds of resistive force requiring only 18 psig. When we are extending, we have the same relatively small resistive load. Your job is to find a way to use the least scf and still get the work done. I am suggesting a target of 0.4 scf per cycle.

Give it your best shot.

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