Help Calculating time to recharge air tanks [Archive]

nuclearnerd

29-01-2015, 10:06

This one is making me feel dumb:

I'm trying to optimize the performance of our (rookie team) pneumatics system. One basic parameter is how long it takes the KOP compressor to charge the tanks from the working pressure (60 psig) to the storage pressure (120 psig). I've tried estimating this a couple of different ways, but they all go pretty much like this:

1) Assuming temperature stays constant (not a fair assumption maybe):

P1 * V1 = P2 * V1 (from the ideal gas law)

So ending with a storage volume of 35 in^3 at 70 psig (84psia), to get to start with a larger volume at 60 psig (74 psia):

V1 = P2/P1 * V1
V1 = 84 psia / 74 psi * 35 in^3
V1 = 39.73 in^3

2) and since we started with 35 in^3 of 60 psi air, we need to add:
deltaV = 39.73 - 35 = 4.73 in^3

3) Looking up the performance values for the ViaAir 090c (http://www.viaircorp.com/90C.html#tabs-2), we get a flow rate at 60 psig of
Vdot60 = 0.53 cfm = 15.26 in^3/s

4) So the time to bring the tank from 60 to 70 psig is:
t60to70 = deltaV/Vdot60
t60to70 = 4.73 in^3 / 15.26 in^3/s = 0.3s

5) Repeating this for each 10 psig interval between 60 and 120, I get a total recharge time of about 1.7s per 35 in^3 tank.

Here's my problem - we're seeing much longer recharge times on our robot. With three tanks at above 60 psig, the compressor runs for over 20 s before turning off :confused:

Have I made a mistake in my arithmetic or assumptions? Should I have paid more attention in fluids class? Does ~2s per tank to recharge from 60 to 120 psi sound reasonable in other teams' experience?

(note -I get pretty much the same answer if I just use the viaair flowrate at 60 psig and do the calculation in one interval between 60 and 120 psig)

eddie12390

29-01-2015, 10:15

I'm not entirely sure of how to do this myself, but I don't think your solution accounts for the fact that the amount of gas provided will decrease as the pressure in the chamber increases. The Ideal Gas Law also makes a lot of assumptions that generally don't hold true in the real world so there will be some deviance between the actual time to fill and the expected time to fill. Honestly, I would plot time and pressure in the tank and try to do a regression on it.

MrRoboSteve

29-01-2015, 10:17

I don't think you're accounting for the lower flow rate at higher pressures. If you plot how long it takes to make each 10psi increment you'll find that the higher ones take a lot longer.

Your experimental value of 20 seconds to run from 60 to 120 psi sounds close to what our experience has been in the past. Unfortunately we're not using pneumatics this year so I can't easily get you a comparison value.

Hint for your compressor testing: put a fan on the heat sink. The approved compressors have a very low duty cycle and it's easy to overrun it during tests.

pfreivald

29-01-2015, 10:17

MrBasse

29-01-2015, 10:20

I would go through and check all the math, but I learned a long time ago that with air storage it is faster and easier to just plumb it up and try it with a stopwatch to get an accurate measurement of time. I want to say that when we run our systems they charge from empty to full pressure in around 30-40 seconds. That is with reasonable hose runs and 4 stainless clippard tanks. From 60-90 would probably be in the 5-10 second range and 90-120 is at least 10 seconds.

With that said:

2 seconds sounds fast to me, but when it is per tank it isn't as unreasonable.

as the pressure increases the compressor has to work harder to fill the tank, thus working slower.

compressed air is a tricky thing to measure with math, and fluid dynamics was never my friend

you'll likely never see 120 PSI in FRC

Pratik Kunapuli

29-01-2015, 10:24

We see much longer recharge times than 0.3 or 2s as well.

I haven't done the calculus, but this is what the curve looks like if you want to play around with it. (It's not an excellent fit, but it's close. Ish.)

Using the function that Excel generated, the integral from 0-120 yields 88.332 seconds for a full charge from 0-120 psi.

jee7s

29-01-2015, 10:25

pfreivald

29-01-2015, 10:31

Using the function that Excel generated, the integral from 0-120 yields 88.332 seconds for a full charge from 0-120 psi.

I'm tired and don't feel well, which is why I'm not doing the math myself, but something seems fishy to me about just integrating that curve and getting an answer because it's independent of the size of your storage tanks. The integration also doesn't yield time.

You've got volume/minute on the Y axis, and pressure on the X axis, so the integration will yield units of volume*pressure/time, or power (in the charming non-SI unit of inch-lbs/minute)--not time.

pfreivald

29-01-2015, 10:35

JamesCH95

29-01-2015, 10:42

Okay, I think I've got it. Integrate the curve between the pressures whose charge time you want, set that equal to the difference in energy of the gas (delta-P*V) over time, and solve for time.

Yes?

(Man, I need a nap.)

This is what I would do. Even this is slightly optimistic as the air will cool off as it flows into, and sits in, the storage tanks. Accurately modeling the compressors performance will be a bit more than trivial.

A decent sanity check would be to take the compressors electrical power input, multiply by a conservative assumption of efficiency, and divide the difference in stored energy by this power to get time.

OP - your initial approach makes too many optimistic assumptions such as no temperature change, constant flow rate vs pressure, etc.

MrForbes

29-01-2015, 10:43

I don't think you're accounting for the lower flow rate at higher pressures.

this.

http://www.viaircorp.com/90C.html#tabs-2

lower right corner, "Performance Data" shows 0.34 cfm at 120 psi

It's been my experience with FRC robotics, that if you need to worry about the refill rate of the storage tanks with this much detail, then you're probably using pneumatics for something that would be better handled with a more powerful motor, like a CIM.

nuclearnerd

29-01-2015, 10:45

Thanks for the responses so far.

Using the function that Excel generated, the integral from 0-120 yields 88.332 seconds for a full charge from 0-120 psi.

...I'm not sure you can just take the integral of that curve. The integral of flow(P) dP would give you something with units of volume/time*pressure (not just time). We would need a function for pressure(volume) for the units to come out right, which is what I was grabbing at (probably wrongly) with the ideal gas law

I don't think you're accounting for the lower flow rate at higher pressures. If you plot how long it takes to make each 10psi increment you'll find that the higher ones take a lot longer.
I only showed the step from 60 to 70 psig. You would need to repeat the next step with the viaair flowrate at 70 psig and so forth for every increment up to 120 psig (where the flowrate is .34 cfm). In fact, the time to fill each increment of pressure goes *down* in my calculation because the pressure increases faster (for a given volume) than the flowrate decreases.

I would go through and check all the math, but I learned a long time ago that with air storage it is faster and easier to just plumb it up and try it with a stopwatch to get an accurate measurement of time.
Yeah, when it comes down to it, that's what we'll have to do. I would prefer to have a theoretical grasp of it though, so we could make intelligent choices during the design stage about how many tanks to buy, or how often we can actuate our cylinders.

It's been my experience with FRC robotics, that if you need to worry about the refill rate of the storage tanks with this much detail, then you're probably using pneumatics for something that would be better handled with a more powerful motor, like a CIM.

But it should work! (sigh).

pfreivald

29-01-2015, 10:45

This is what I would do.

Hooray!

Pro-tip: if your students don't yet know calculus, approximate the curve with a trapezoid (or several) to teach them the *idea* of calculus, and of using unit analysis on axes to determine the physical quantities yielded by integration.

Mike Bortfeldt

29-01-2015, 11:07

AdamHeard

29-01-2015, 11:12

Hooray!

Pro-tip: if your students don't yet know calculus, approximate the curve with a trapezoid (or several) to teach them the *idea* of calculus, and of using unit analysis on axes to determine the physical quantities yielded by integration.

Yup.

This method is actually how a great deal of engineering calcs are done. Before I knew what I was doing, I brute forced a lot of calculations this way after being inspired by JVN using the method in his drive design calc. Later one I learned better ways to do some calcs, but a lot are still easier to do numerically.

With excel's easy method of repeating formulas, it can be done with plenty precision using small timesteps.

Euler's Method (http://en.wikipedia.org/wiki/Euler_method) is the simplest (could be wrong...), but there are other methods that can get a lot better, especially if the curve changes direction a lot.

JamesCH95

29-01-2015, 11:12

I think the error you are making is that the CFM compressor performance values you are using are at the inlet to the compressor, not the discharge. The compressor does not put out 0.53 cfm @ 60 psi, it compresses 0.53 cfm of atmospheric air (14 psia) when the discharge is @60 psi.

Mike

Using this tidbit (which makes sense, since it rates the compressor independent of ambient conditions and output temperature) methinks one can do a mass-conservation sort of calculation to take air at STP and a given flow rate and bring it to ST & raised pressure. Integrating the aforementioned flow-rate function (combined with another function or coefficient to account for mass conservation) is still the final way to solve this problem. I think.

GeeTwo

29-01-2015, 11:19

1) Assuming temperature stays constant (not a fair assumption maybe): P1 * V1 = P2 * V1 (from the ideal gas law)

This is a bad assumption. The ideal gas law assumes that the energy in the system in constant. The compressor of course puts energy into the system, so that violates PV=PV.

The ideas gas law does not require that the energy in the system remain constant. However, on the other side of that equation, in addition to R (constant) and T (assumed constant) is n - which is definitely changing. You need to add the "new" air, which began at a gauge pressure of zero. This serves as a reminder that you need to add about 15psi to every gauge pressure that you're measuring to use it.

If you did not account for the slowing, the time to fill a 35 in^3 tank from zero pressure would be based on the volume of this gas after it were returned to ambient. Since 120 = 15 * 8, we started with 8 x 35 in^3 = 280 in^3 = .162 ft^3 outside the tank, in the atmosphere. For a 1 cfm pump, this would take about 9.7 seconds to compress. I haven't measured it, but it seems that the sound of an air compressor drops about two octaves as it goes from zero to 120 psi. Assuming the same amount of air is compressed in each cycle, this means that at the end, it is pumping about 1/4 cfm, which would take 39 seconds to fill. Your time of 20 seconds to fill from 60 to 120 sounds about right.

MrBasse

29-01-2015, 11:21

Yeah, when it comes down to it, that's what we'll have to do. I would prefer to have a theoretical grasp of it though, so we could make intelligent choices during the design stage about how many tanks to buy, or how often we can actuate our cylinders.

The thing that everyone always seems to overlook is that adding more tanks won't make your compressor run less. It will just run for a longer period of time less times.

Kids always stress over the need to have ten tanks on the robot and then they complain that the compressor runs for the whole match. When we reduce it to three or four tanks, the compressor may run twice a match, but for only 15-20 seconds each time. Unless you need to dump 80 cubic inches of air at a time, you rarely need 80 cubic inches of storage.

JamesCH95

29-01-2015, 11:36

The thing that everyone always seems to overlook is that adding more tanks won't make your compressor run less. It will just run for a longer period of time less times.

Kids always stress over the need to have ten tanks on the robot and then they complain that the compressor runs for the whole match. When we reduce it to three or four tanks, the compressor may run twice a match, but for only 15-20 seconds each time. Unless you need to dump 80 cubic inches of air at a time, you rarely need 80 cubic inches of storage.

To add to this (quite correct) post: the only upside (that I can see) to lots of storage is that one can start the match with a lot of stored energy. This effectively increases the total power that the pneumatic system can expend over the duration of the match.

Ether

29-01-2015, 11:41

the only upside (that I can see) to lots of storage

You need enough storage so that you can disable the compressor during periods of high current draw and still have enough pressure to operate pneumatics during those (hopefully not too prolonged) blackout periods.

clicato100

29-01-2015, 11:55

Originally Posted by JamesCH95 View Post
the only upside (that I can see) to lots of storage

Bit of an aside, I think for many pneumatic shooters last year, the closer your reservoir of air approached "infinite" the higher sustained pressure you maintained at the cylinder throughout the shot.

I believe some students on my team have some interesting graphs we made of "real-time" pressure at the cylinder inlet throughout the shot varying the amount of reservoir tanks both HP and LP side, but besides the point...::rtm::

JamesCH95

29-01-2015, 11:59

You need enough storage so that you can disable the compressor during periods of high current draw and still have enough pressure to operate pneumatics during those (hopefully not too prolonged) blackout periods.

I suppose saving that little bit of current could help.

Bit of an aside, I think for many pneumatic shooters last year, the closer your reservoir of air approached "infinite" the higher sustained pressure you maintained at the cylinder throughout the shot.

I believe some students on my team have some interesting graphs we made of "real-time" pressure at the cylinder inlet throughout the shot varying the amount of reservoir tanks both HP and LP side, but besides the point...::rtm::

Those plots would be most interesting to see.

nuclearnerd

29-01-2015, 12:07

I think the error you are making is that the CFM compressor performance values you are using are at the inlet to the compressor, not the discharge. The compressor does not put out 0.53 cfm @ 60 psi, it compresses 0.53 cfm of atmospheric air (14 psia) when the discharge is @60 psi.
Oh crap, so it's SCFM then, not cfm? Ugh, that changes things significantly. Now I get closer to 13 seconds to recharge one tank from 60 to 120 psi. I'll compare that with the robot performance tonight, but it sounds more like what we were seeing. I wish that were clearer on the compressor spec.

- that moment when you realize you need to make a big change, and it's almost week 5 :(

Mr V

29-01-2015, 12:12

If you go to the link Mr Forbes provided you'll find another tab labeled refill rates. Of course those are only valid with the listed sizes and pressure ranges. Of course the listed pressure ranges are not exactly what we see in FRC and it is unlikely that you are using a .5 gallon tank. However you could probably use that data along with the CFM at the different pressures to derive how it would work with your tank size and pressure range.

The thing to keep in mind is that it is a piston type compressor. That piston has a set volume however you don't get to use all of that volume with each stroke. At zero psi the entire volume of the cylinder is discharged. As the pressure rises the initial portion of the stroke is used to reach the current tank pressure and no air flows out of the cylinder until the current pressure is met. The other thing is that at zero psi there is minimal load on the compressor's motor. As the pressure rises the load on the motor increases so it's rpm decreases. Not that the chart lists different current consumption at different pressures to reflect that change in load and rpm of the motor.

The listed CFM in the chart is what flows out of the compressor when working against a given pressure but the air does cool and contract when it reaches the tank further complicating the calculations.

Jon Stratis

29-01-2015, 12:12

Kids always stress over the need to have ten tanks on the robot and then they complain that the compressor runs for the whole match. When we reduce it to three or four tanks, the compressor may run twice a match, but for only 15-20 seconds each time. Unless you need to dump 80 cubic inches of air at a time, you rarely need 80 cubic inches of storage.

This.

When designing our pneumatic system this year, our first focus was on actual air usage - how often would our "ideal" system be able to actuate. From there, we worked backwards to determine how much air was required at working pressure. This helped to directly inform us of the required storage volume - how many times do you want to be able to activate your mechanism before you run out of storage? How much airflow do you need to support the system over the entire match? Can the compressor handle it?

In the end, our math showed that we could handle the load at our theoretical top actuation speed (a speed we'll probably never actually hit, but it's good to have a bit of a cushion in these calculations). It also showed us the affect we would have if we had 2 tanks for storage, or 4, or 8. Doing the math to figure out how much the storage pressure drops with each activation is critical!

MrBasse

29-01-2015, 12:59

Bit of an aside, I think for many pneumatic shooters last year, the closer your reservoir of air approached "infinite" the higher sustained pressure you maintained at the cylinder throughout the shot.

This is where smart design takes over and eliminates the need for "infinite" air storage. Our shooter used a 2" bore 12" stroke cylinder and we were never wanting for air. We precharged our cylinder and let springs throw the ball.

jee7s

29-01-2015, 13:29

The ideas gas law does not require that the energy in the system remain constant. However, on the other side of that equation, in addition to R (constant) and T (assumed constant) is n - which is definitely changing.

I stand corrected. Boyle's Law (PV=PV) assumes that mass (n) and temperature (T) -- hence energy -- are constant.

GeeTwo

30-01-2015, 07:58

I stand corrected. Boyle's Law (PV=PV) assumes that mass (n) and temperature (T) -- hence energy -- are constant.

Even in isothermal processes (T remains constant), the gas does work as it expands and work must be done on it to compress it.

Michael Hill

30-01-2015, 12:42

GeeTwo

01-02-2015, 00:07

I'm not sure I trust Viair's CFM data...has anyone actually measured it. It doesn't jive mathematically if you look at the 0.5 gallon and 1.0 gallon performance...(assuming my calculation is right). If there's a math mistake please late me know.

I'm not sure exactly what you're doing in your spreadsheet, but I certainly do find it suspicious that you wind up with a temperature of over 1000R (550F). I believe that this temperature would cause plastic tubing (and maybe tanks) to soften and rupture. This may be correct for an adiabatic compression of air, but there is certainly some temperature equalization taking place.
I did the calculations based on constant temperature (not right, but probably closer), and came up with 2 minutes and 9 seconds for 0-120 on a 1 gallon tank. This is still shorter than Viair's 3 minutes, but a good bit closer. There does seem to be a disagreement between the "performance" and "fill rate" data.

Michael Hill

01-02-2015, 08:26

I'm not sure exactly what you're doing in your spreadsheet, but I certainly do find it suspicious that you wind up with a temperature of over 1000R (550F).

Yup there's a problem there. Didn't even notice it. The compressor gets hot, but not that hot.

My assumption is isentropic compression, but I obviously have a mistake in my math somewhere. That's what happens when you take someone who was once pretty good at aerodynamics and thermo and give them a job dealing with spacecraft...

Ether

01-02-2015, 08:50

Is the compressor motor startup current at 100psi specified anywhere? Has anyone measured it with a current probe and a scope?

GeeTwo

01-02-2015, 23:28

Is the compressor motor startup current at 100psi specified anywhere? Has anyone measured it with a current probe and a scope?

By start up, do you mean a transient? If not, it's given on the VIAIR web page referenced in the spreadsheet: http://www.viaircorp.com/90C.html#tabs-2 as being 9A. I have not measured this myself. As my team is not using pneumatics this year, it's unlikely that I'll have an opportunity until after build season is over.

GeeTwo

01-02-2015, 23:37

My assumption is isentropic compression..
I don't believe that the air compressor would be well modeled as an isentropic (reversible) process. Even if the mechanical compression were isentropic, the presence of a heat sink at the compressor output radiating/convecting heat from the compressed air into the environment would make the process irreversible.

Michael Hill

02-02-2015, 00:07

I don't believe that the air compressor would be well modeled as an isentropic (reversible) process. Even if the mechanical compression were isentropic, the presence of a heat sink at the compressor output radiating/convecting heat from the compressed air into the environment would make the process irreversible.

Don't forget that any inefficiencies are going to be released as heat. That would include inefficiencies in the mechanics of the compressor as well as the motor inside of it. That being said, tanks aren't perfect insulators either, so some heat will escape through the walls (plastic is better than metal in that respect), so even if you model the compressor perfectly, you still have to model the whole system to get a decent model. For general FRC use, who really cares? Just use the manufactures' specs, but as an academic exercise, there are just far too many unknowns to take into account. There are things like back pressures that get involved with fittings, pressure loss in pneumatic lines, as well as heat loss in some of the larger metallic things like the pressure switch and any manifolds that are used. All that can add up. At any rate, I think isentropic is generally a good model for a compressor you don't know much about in terms of construction as long as you take efficiency into account. That being said, I do acknowledge that there probably is a math mistake in my spreadsheet somewhere because that temperature is extremely high (though the temperature does get extremely hot). We have melted high temp pressure tube coming off the compressor before switching to copper.

Richard Wallace

02-02-2015, 07:43

Is the compressor motor startup current at 100psi specified anywhere? Has anyone measured it with a current probe and a scope?

Good idea, Russ. I will try that tonight and report results. My team has three types of compressors on hand so I'll attempt to measure starting surge current for each.

-------edit after testing--------

We tested three compressors, using an oscilloscope and current probe to measure (1) initial peak surge current, (2) average current draw during charging, and (3) time to full charge. The test system capacity was 48 cubic inch (3x Clippard AVT 32-16). In each case the duration of the initial surge current was ~0.05 second, rising to its peak in about 0.005 second and decaying exponentially to steady value after that. I have attached a typical oscilloscope plot of the compressor current, 50 seconds full horizontal scale and +/- 40 Ampere full vertical scale. This plot is for the Thomas compressor; other plots are similar in form but differ in their initial peak surge, average current draw, and time to full charge as summarized below.

Viair 90C: initial peak surge 18 Ampere, average current draw during charging 12 Ampere, time to full charge 40 seconds.

Thomas 405ADC38/12: initial peak surge 32 Ampere, average current draw during charging 10 Ampere, time to full charge 36 seconds.

Viair 250C-IG: initial peak surge 40 Ampere, average current draw during charging 9 Ampere, time to full charge 33 seconds.

GeeTwo

03-02-2015, 08:07

..In each case the duration of the initial surge current was ~0.05 second, rising to its peak in about 0.005 second and decaying exponentially to steady value after that...

Viair 90C: initial peak surge 18 Ampere, average current draw during charging 12 Ampere, time to full charge 40 seconds.

Thomas 405ADC38/12: initial peak surge 32 Ampere, average current draw during charging 10 Ampere, time to full charge 36 seconds.

Viair 250C-IG: initial peak surge 40 Ampere, average current draw during charging 9 Ampere, time to full charge 33 seconds.

Interesting that the efficiency correlates so well with initial peak surge. Assuming the voltage was constant at 12.5V throughout, the 90C needed 6.0 kJ, the Thomas 4.5 kJ, and the 250C-IG only needed 3.7 kJ from the battery to do the same job.

Mark McLeod

03-02-2015, 08:15

We've run similar testing and found a wide variation sometimes between compressors of identical models.
For instance, a ViAir 90c that pulled ~6amps and a Thomas that pulled under 10amps while another Thomas pulled ~12 amps.
I'd put the Thomas variations down to age and abuse, but I was surprised to see your ViAir 90C pulling so many more amps than some of ours.

Jon Stratis

03-02-2015, 09:02

Richard Wallace

03-02-2015, 11:11

How was the test system controlled? I'm curious what difference (if any) we would see controlling the compressor via a spike versus the new PCM. I'm particularly interested in the surge current for the PCM... Can we measure it both between the PCM and the compressor, and between the PCM and the PDP? The results of such a test may impact whether teams hook their PCM up through the dedicated port with a fuse, or go through a WAGO and a circuit breaker.I am interested in that, also.

My test results above were made using the highly illegal method of series connecting a pressure switch to the compressor -- the opposite of soft-starting!

Can someone try the tests Jon suggests using the PCM? We have already returned our beta hardware, and it may be a couple of days before we are ready to test this on our 2015 robot.

Mr V

03-02-2015, 11:15

Good idea, Russ. I will try that tonight and report results. My team has three types of compressors on hand so I'll attempt to measure starting surge current for each.

-------edit after testing--------

We tested three compressors, using an oscilloscope and current probe to measure (1) initial peak surge current, (2) average current draw during charging, and (3) time to full charge. The test system capacity was 48 cubic inch (3x Clippard AVT 32-16). In each case the duration of the initial surge current was ~0.05 second, rising to its peak in about 0.005 second and decaying exponentially to steady value after that. I have attached a typical oscilloscope plot of the compressor current, 50 seconds full horizontal scale and +/- 40 Ampere full vertical scale. This plot is for the Thomas compressor; other plots are similar in form but differ in their initial peak surge, average current draw, and time to full charge as summarized below.

Viair 90C: initial peak surge 18 Ampere, average current draw during charging 12 Ampere, time to full charge 40 seconds.

Thomas 405ADC38/12: initial peak surge 32 Ampere, average current draw during charging 10 Ampere, time to full charge 36 seconds.

Viair 250C-IG: initial peak surge 40 Ampere, average current draw during charging 9 Ampere, time to full charge 33 seconds.

By time to full charge do you mean starting at 0 psi? The original question on start up current was asking what it was with 100psi already in the tank. Personallly I'd like to see the start up current and time to fill using the on-off range of the legal pressure switch. Time to fill from 0psi and the start up current at that pressure is less critical since you can fill before the match starts and should only occur once per match while the 85-115psi event can potentially happen multiple times in a match. If the fuse were to blow before a match that could easily be corrected before going on the field.

Richard Wallace

03-02-2015, 11:22

By time to full charge do you mean starting at 0 psi? Yes, we started with the tanks empty. I was looking mostly at peak surge current with the 'scope. A stopwatch would suffice to measure refill time, and that should be much shorter than the 0-120 fill times reported above.

Also, I would expect bigger differences in the compressor comparisons for 100-120 refill times, based on their published flow curves.

BTW, the peak surge transient (50 msec) we observed would not trip a breaker or blow a fuse.

Jon Stratis

03-02-2015, 11:34

BTW, the peak surge transient (50 msec) we observed would not trip a breaker or blow a fuse.

I'm no expert, but I've been repeating the "conventional wisdom" for years as an inspector and making teams replace their spike fuse with a breaker for the compressor. I had heard that the fuse will wear out over time due to the start up surge until it eventually pops...

Can anyone with a better understanding correlate the readings posted in this thread with the compressor breaker rule we've all been following for years?

Richard Wallace

04-02-2015, 12:15

I'm no expert, but I've been repeating the "conventional wisdom" for years as an inspector and making teams replace their spike fuse with a breaker for the compressor. I had heard that the fuse will wear out over time due to the start up surge until it eventually pops...

Can anyone with a better understanding correlate the readings posted in this thread with the compressor breaker rule we've all been following for years?
I am not an expert, but my team has made some tests which may shed some light. See attached summary of our set up and results for recharging a 48 cubic inch system using three FRC-legal compressors.

When recharging from ~100 to ~120 PSI, the compressor must start against load. As Mr. V pointed out above, this case is much more pertinent to actual FRC operation than the tests I reported earlier.

The initial peak surge current observed using Viair compressors is relatively short duration (<0.1 sec), but the Thomas compressor's initial peak surge current is much longer duration (~0.5 sec). Comparing the Thomas surge against typical automotive fuse curves suggests an explanation for the "conventional wisdom" Jon mentioned above. The Thomas compressor has much less margin against blowing a 20A fuse during surge, and historically that is the compressor on which our FRC "conventional wisdom" is based.

Mr V

05-02-2015, 00:57

I am not an expert, but my team has made some tests which may shed some light. See attached summary of our set up and results for recharging a 48 cubic inch system using three FRC-legal compressors.

When recharging from ~100 to ~120 PSI, the compressor must start against load. As Mr. V pointed out above, this case is much more pertinent to actual FRC operation than the tests I reported earlier.

The initial peak surge current observed using Viair compressors is relatively short duration (<0.1 sec), but the Thomas compressor's initial peak surge current is much longer duration (~0.5 sec). Comparing the Thomas surge against typical automotive fuse curves suggests an explanation for the "conventional wisdom" Jon mentioned above. The Thomas compressor has much less margin against blowing a 20A fuse during surge, and historically that is the compressor on which our FRC "conventional wisdom" is based.

Excelent and very useful data, thanks very much.

What people need to keep in mind is that a fuse and the type of circuit breakers that we use are thermal devices. When they reach a certain temp they will open. With a large overload it will open almost instantaneously. With a slight overload it will open but in a longer period. With intermittent medium overloads it can eventual heat to the point where it opens.

So the old Thomas compressor if it starts multiple times during a match could certainly eventually heat up the fuse to the point it opens. I'm not sure why they choose the ATM form factor instead of the ATO/ATC form factor as used for the 20/30 amp circuits on the PDP, PDB and Spike. The ATM is not that much smaller than the ATO/ATC.

yara92

20-02-2015, 05:33

Richard Wallace

20-02-2015, 05:58

Richard HI
I try the test. noting move may be because of the spike, I use defrent spike , nothing move.I saw your questions in the other thread. Have you been able to determine if your Thomas compressor is still functioning normally? Do you have another compressor to try?

yara92

20-02-2015, 13:14

I saw your questions in the other thread. Have you been able to determine if your Thomas compressor is still functioning normally? Do you have another compressor to try?

YesI have ,
for testing Iconnect dirict itis working.
My quation i you knowhow to use the spike as relay. I use Jumber, the output 0 [V]

Richard Wallace

20-02-2015, 14:40

YesI have ,
for testing Iconnect dirict itis working.
My quation i you knowhow to use the spike as relay. I use Jumber, the output 0 [V]
Ok, now I understand your question. I caused confusion by showing the control connection between the pressure switch and the Spike incorrectly in the diagram attached in my earlier post. The connection used on my team's bench test board is shown in the attached (revA) version of that diagram.