Contributor
When I had that problem it was an unloader valve sticking shut. Unloader valves release air in the cylinders so the motor isn't starting against pressure. Does the compressor, when stopping release a small amount of air? Had it previously? Otherwise a marginal circuit or circuit breaker would be a likely candidate. Could you run a heavy duty extension cord to the compressor from another circuit?
Contributor
A 2 HP motor will do poorly on a 20A 120 volt circuit, a 30A or larger circuit is required which is doable but it is much simpler to run it at 240V, 1 1/2 HP is the limit for common 120 volt circuits.
[OP]
Contributor
Thanks Rollie. FWIW, the compressor has a NEMA 5-15 plug on it.Originally Posted by Rollie Meyers
Edit: This compressor is also rated at 1500 Watts, it's in the model name P1.5IU-A9, so thats 12.5 Amps. Perhaps the 2HP figure is from unscrupulous marketing.
Last edited by mark mcfarlane; 11-14-2019 at 1:29 PM.
Mark McFarlane
Member
I don't want to interrupt the troubleshooting flow here but there are a couple of minor points I want to bring up.
Not many clamp meters are capable of capturing inrush current. If yours is not (and it should be in the manual if it is) and you motor has a proper nameplate you'll find it there. There will be a code letter for the Locked Rotor Current (LRC). It's usually labelled "CODE". Look up the code letter in the table:
LR Codes.JPG
Then make the appropriate calculation. For example, if your motor has a code G and a rated voltage of 115V then you start current is 2HP x 5.9kVA/HP x 1000VA/kVA / 115V = 103A.
Both HP and Watts are units of power. Watt is metric and HP is Imperial. Motor ratings generally use a rounded number of 750 Watt per Horsepower. So 2HP x 750W/HP = 1500W. That is the rated LOAD not the input.
Input power is output load divided by efficiency and is also measured in Watts. Apparent input power is input power divided by power factor and is measured in Volt-Amps. Input current is then apparent input power divided by input voltage. For example, a 2HP (1.5kW) motor with 85% efficiency and 90% power factor would have an input current at rated load of 1500/.85/.9/120 = 16.3A.
Beranek's Law:
It has been remarked that if one selects his own components, builds his own enclosure, and is convinced he has made a wise choice of design, then his own loudspeaker sounds better to him than does anyone else's loudspeaker. In this case, the frequency response of the loudspeaker seems to play only a minor part in forming a person's opinion.
L.L. Beranek, Acoustics (McGraw-Hill, New York, 1954), p.208.
[OP]
Contributor
Thanks David, I can't see the language you have referenced on my compressor motor.
IMG_4493.jpg
Mark McFarlane
Member
No, I don't see it either. Those are IEC specs not NEMA on that nameplate, but it's not complete.
The laboratory method for measuring locked rotor current is to literally lock the rotor and then measure the current. You usually don't have a lot of time before the breaker opens but most ammeters are fast enough to get a measurement. It's not difficult, you just need a good way to keep the rotor from spinning.
Beranek's Law:
It has been remarked that if one selects his own components, builds his own enclosure, and is convinced he has made a wise choice of design, then his own loudspeaker sounds better to him than does anyone else's loudspeaker. In this case, the frequency response of the loudspeaker seems to play only a minor part in forming a person's opinion.
L.L. Beranek, Acoustics (McGraw-Hill, New York, 1954), p.208.
Moderator
I doubt that the apparent leak in the network is related to the breaker thing. The circuit itself is the first place to look. Even though this machine will run on 120v, it's still wanting it's own circuit so if it shares, it could merely be a coincidence of timing. It could also be a bad breaker. It could also be something like Curt describes.
--
The most expensive tool is the one you buy "cheaply" and often...
Contributor
The unloader is your 1st place to look. They get "sticky" (and a bit later just stop working altogether). They are not difficult to replace.
[OP]
Contributor
Thanks Lisa Is this the unloader? It does actuate when the compressor shuts off.
unloader.jpg
Does one typically just replace the little brass fitting with the valve, or the actuating arm assembly which is part of the pressure switch? Unfortunately my manual and IR's web site don't even show this valve as one of the parts.
Mark McFarlane
Contributor
The first thing I would try is to lube the bearings/bushings on the motor, whether they are designed to be lubed, or not. I have a motor on a wet tile saw, that was doing the same thing, that has kept working for several years, so far.
[OP]
Contributor
Yea, the leak is somewhere in the many joints of RapidAir tubing.
It is a dedicated circuit. I just exercised the two relief valves and will watch it for a week. If the breakers blow again I'll try an extension cord to another circuit, I have an unused dedicated circuit under my workbench and a 20 amp extension.
Mark McFarlane
Contributor
Clip on ampmeters are cheap and help with troubleshooting. Will tell you actual current draw.
MK
Contributor
You need to convert to 240 volts or pull new 120 volt wire to carry 30 amps at 120 V. The breaker and wire is fine you are just at the limits.
How far from the service entrance does the wire run? Over fifty feet can cause a noticeable volt drop under load and a corresponding amp increase.
Bill D
[OP]
Contributor
Thanks Bill. The service panel is about 7 feet away from the outlet where the compressor is plugged in. Add 5 foot to the ceiling and 5 foot back down to the outlet you’re looking at about 18’ for the wire.
Contributor
36' is nothing to the source of unlimited amps so voltage drop is not a problem due to long wire runs.
Bill D
Posting Permissions
Reply With Quote