Solar recharged, battery powered air compressor system

[tr1ntx]

This may be a trivial question for you electrical types, but it's stuff I just don't know . . .

We have a customer, sort of an "All ya gotta do" type and not as clever as he thinks he is, who knows how to pull the strings of my boss (the dolt). He wants us to design / piece together a system with solar recharged batteries powering an air compressor (or set of compressors).

The equivalent single AC-powered compressor would be 10HP, 230V 3-phase, 24 Amp. The supplied air would be 100 psi, so the tank pressure would be 175 I suppose. 20 cfm rate required for 20-30 seconds. The good news is that it would only run intermittently, 4 times per day, maybe 5. We can install a lot of storage capacity and pre-charge the system, so it generally wouldn't need to fill from zero using battery power. This is a piece of mobile equipment, on site for maybe 2 weeks.

The scheme is to have 3 or 4 separate compressors. There can be manual valves accessing various fully-charged storage vessels.

My thought is that 3 or 4 DC motors "summed" to get the equivalent output of the above AC-powered unit would require much more current. Is that right, AC 3-phase delivers much more power for a given Voltage and Current?

Secondly, and actually over-riding the above issue, my thought is that compressing air to 175 psi into 60 gallon tanks requires much too large a current draw for a "transportable" battery bank. It's simply too much work being done in a short period of time, even if you're only recharging from 125 psi kick-in pressure. The battery bank would be quite large to move around, right? Getting that amount of work from solar-generated DC electricity 4 times a day would be a significant feat, wouldn't it? Not to mention 24 hrs/day and handling 3 rainy days in a row or something. That's out of the question I believe.

We currently have one 12V 8D battery on the unit, with a 12W solar charger. He's thinking the compressors can be powered with several larger panels and 3-4 more batteries. I'm thinking he's greatly underestimating the power from 230V 3-phase electricity and the current draw required to compress air.

Any insight?

 

[waross]

Your 10 HP motor will require about 7500 Watts plus losses. Say about 9000 Watts at full load.
To get one hour running time from 8 hours of optimum sunlight, it's simple division.
9000 / 8 = 1125 Watts charge rate. Yes the motor may not always be fully loaded but the sun may not always be bright.
Now for a charge rate of 1125 Watts you will need 1125 / 12 = 94 twelve Watt panels.
The batteries;- Anyone up for batteries?? I dropped out at the solar panels. Don't forget the three day carry over.


Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[itsmoked]

One word:

Absurd.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[DRWeig]

Considering 12V batteries, your load seems to be only about 5 or 6 amp-hours per 30-second cycle. The size of a fully-charged battery stack to hold 30 AH per day for two weeks wouldn't be very outrageous. That's if it truly does only run 30 seconds, 5 times per day.

I'd be inclined to forget about solar and all its complexity (plus rainy-day performance), charge a bunch of batteries before leaving home and take them along. OR, a small portable generator. If I have to carry energy with me, gasoline is hard to beat. Finally, has anyone considered a gasoline-driven compressor? They're mighty cheap.

Good on ya,

Goober Dave

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[waross]

There are lots of off the shelf gas engine compressors.

https://www.acklandsgrainger.com/AG...balProductDetailDisplay.do?item_code=MTM2YJ61

If he may need electric power also or welding capabilioty these are interesting.

http://www.e-compressor.com/Air-Com...nit-s/87.htm?gclid=CIvwo7Sg868CFQZ_hwoduSjeWw

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[DRWeig]

Also, re-reading the OP made me question the compressor size. On the surface, it seems that your tool demand is 20 cfm at 100 PSI. The compressor doesn't need to meet this demand, that's why we use storage tanks.

Figure out how much storage you need for a 30-second run if the tank starts at 175 PSI. Should be close to 100 gallons, maybe 120. Then determine how fast you need to replace the 10 cf you spent on the 30-second use and size the charging compressor to meet your time limit. If you can wait 10 minutes between 30-second tool uses, the compressor only has to spit out 1 cfm. That'll only take 1 horsepower or so.

Or am I missing something?

Good on ya,

Goober Dave

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[tr1ntx]

I forgot to mention . . . no internal combustion engines near this equipment.

I realize there are ways to do it, just not his "All ya gotta do is ..." argument.

I guess I spent too much time explaining/rambling and forgot the key thing I wanted to ask. It's about the 240V 3-phase 24 Amp compressor draw. I can't find a "3-phase Electricity for Dummies" manual or a "Rules of Thumb" guide. To determine the power supplied by 3-ph electricity, you need to know 2 other things besides the Voltage and Current, right? Phase Angle and is it Power Factor?

Or is Power Factor derived from Phase Angle and there's some other variable you need to know? If I could find someone with an equal size compressor, I don't have a way to measure those values anyway. Does anyone have a Rule of Thumb or good average value for estimating purposes?

Well, I guess Bill did it for me, ~9000W, but I need to be able to defend that number.

Thanks everyone for the input.

Dave, I'm wondering the same thing, "Am I missing something?"

Yes, the compressor doesn't have to meet the 20 cfm @ 100 psi demand, but the system does and it could be back-to-back-to-back-to-back. They want compressor & storage capacity to handle "continuous" load, not an hour straight or anything, but 30 second bursts with only slight pauses, maybe 15 in a row. So even if you have enormous storage, the compressor(s) have to run a long time to recharge it. It simply struck me to first consider the power required to do that much work, then the whole concept might be ruled unworkable.

 

[waross]

240 Volts times 24 Amps times 1.73 (three phase factor) = 9976 VA
The power factor is the cosine of phase angle. Typically 80% to 90% for a fully loaded motor.
9976 VA times 0.9 PF = 8979 Watts.
An 8D will supply about 225 Amp hours.
225 Amps times 12 Volts = 2700 Watts for one hour.
Or, 9000 Watts for about 18 minutes. 225 Hours of 100% output from your 12 Watt panel to recharge.
These estimates are approximate. There will be inverter losses and the capacity of the battery varies with the discharge rate.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[badservo]

"I forgot to mention . . . no internal combustion engines near this equipment."

So just get a gas engine compressor and a hose (or pipe) that is longer than what you are calling "near", it will be much more reliable and easier to transport.

 

[itsmoked]

So the compressed air is to breath?

Keith Cress
kcress -

http://www.flaminsystems.com

 

[LionelHutz]

Power is power. The basic thing you need to use is efficiency = power out divided by power input. Your power required at the electrical input to the motor is the power required at the compressor shaft divided by the motor efficiency. The power required at the batteries is the motor input power divided by the inverter efficiency.

You still haven't defined the run time of the compressor. 7.5 minutes of 20cfm air usage would require the compressor to pump 150cfm. What average cfm can the pump produce?

 

[tr1ntx]

Thanks, Bill.

I did manage to find a similar example. It was a 10HP 3-PH motor. It divided the 7460W by an "nominal overall motor efficiency" of 87.5% (= 8525W input).

After doing some rectal-cranial inversion reduction exercises, I think I was confusing myself recalling a past packaged unit I dealt with. I was combining motor efficiency and compressor efficiency concerns, I believe, in my musings of this concept.

I was tasked with determining how many more solar panels and batteries we would have to install to power a compressor supplying air to some items the customer (Mr. All Ya Gotta Do Is . . .) wants added. I knew off the bat it wasn't a little cigarette lighter plug-in tire inflator we would need. The equipment manufacturer didn't have any idea about a solar DC system, but they did spec a "normal" compressor, the 10 HP, 230V, 3-PH unit.

When I saw the 230V 3-PH, I thought "Uh oh, a DC motor will have to draw one heck of a current to power that same compressor if it needs 3-PH now." That was my mistake. You wouldn't want to power that same 2-stage compressor with DC. So along this line of argument, the cost of the compressor will be high. I wasn't thinking of an inverter to run the same AC motor.

I believe Bill provides the ammunition to shoot this down. I think I'm right in saying that you only want to draw about 10% from a battery, so one would give 270W. 9000 / 270 = 34 batteries. Call it 40 for to account for losses. 5000 lb and $9000 of batteries. We can probably add the weight, and they would pay the price, but there is not room for 40 freakin' batteries. This line of thinking leads to more advanced (expensive) batteries. And they better be very advanced to hold that power in about 1/7 the space.

Then there is the 94 solar panels to deal with. We can easily go up to the 18W size, so that requires 63. And that's bare minimum. I guess I could check with NASA to see if they have any surplus fold-up satellite solar arrays.

 

[DRWeig]

Park a Chevy Volt next your machine and couple a drive wheel to the compressor shaft?

Good on ya,

Goober Dave

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[LionelHutz]

I keep finding your comments about how difficult it would be to power the compressor with a DC motor compared to an AC motor rather odd. 10hp is 10hp, no matter what kind of motor you use. And you can still buy 10hp DC motors.

Ideally, you would want to find the system that is the most efficient to go between the batteries and the pipe leading to the storage tank. These days, a brushless DC motor might be the most efficient way to go between the batteries and the compressor shaft.

You are not correct about the power draw from the batteries and the calculations so far are not complete.

A 225Ah rating is typically measured at a 20 hour rate. However, 20 hours is often the rate used which would leave the battery completely dead and you don't want to do that, so you might discharge until 40% capacity is left or for say 12 hours. This really depends on the battery in question because the final capacity used for this test can vary a lot between brands. Better makes will get you closer to 20 hours of practical run time at the specified discharge rate.

Typically, you could draw more current with the capacity suffering. For example, the 225Ah battery @ 20hr might allow you to draw 10A for 20 hours (in practice it's a 200Ah @ 20hr battery). At 100A you might get 1.5 hours or 150Ah. The good batteries are typically 6V so you can draw 100A or approximately 600W for 1.5 hours of run time. 15 of these batteries can power a 9kW load for 1.5 hours.

You now need to figure out how many hours you need the batteries to power the compressor. Obviously, if the requirement is to run for 6 hours off batteries without recharging then you need 60 batteries. However, if the requirement is to run for 15 minutes off batteries without recharging then 15 batteries would be more than enough. This required running time off the batteries is what you have failed to define and all calculations are incomplete without defining this.