Jet Pump on Inverter sizing
Cheezy
I have a 1/2 hp Craftsman Convertible Jet Pump configured to 110V and
am wanting to run it off of an inverter. I did some research and
determined that 1/2 hp corresponds to about 375W operating power. So I
figured that 2x that would be a good starting place for an inverter,
but the startup on a 750W inverter blows it out. Assuming that this
inverter has a max 1500W, I'm amazed that a little 375W motor would
need more than 1500W startup. I briefly plugged it into a 1000/2000W
inverter (not a reliable unit unfortunately) with the same results.
Is my math/reasoning off here, or does someone with a similar 1/2 hp
pump setup see the same behavior with required wattage to get the thing
going? I'm thinking of buying next a 1200 W inverter but it seems like
math should determine what I should buy and not trial and error.
Thanks in advance.
John Beardmore
Cheezy <chico...@gmail.com> writes
Some pumps and compressors can have a start up surge current of 20 times
the normal load current.
Peak value and duration probably depend on the motor size (power and
weight and mass distribution of rotating parts), mechanical load, and
any 'soft start' electronics that may be incorporated.
Ask the manufacturer or use a current clamp and a scope to find out ?
Cheers, J/.
--
John Beardmore
BobG
mains cycles" (epanorama). The inverter is putting out a modified
square wave of 140 volts peak and 120 volts average. (We all remember
from EE class that the RMS is the same as the average for a square
wave). You have 0 gauge wires to the battery? Put a voltmeter on the
12volts. Its probably sagging big time. Try it with a charger on it.
Steve Spence
> hello all:
>
> I have a 1/2 hp Craftsman Convertible Jet Pump configured to 110V and
> am wanting to run it off of an inverter.
My 1/2hp Gould Jet Pump won't start off my 2500/5000 inverter unless the
battery pack is VERY charged. The start surge drops the pack below the
low voltage shutdown.
--
Steve Spence
Dir., Green Trust, http://www.green-trust.org
Contributing Editor, http://www.off-grid.net
http://www.rebelwolf.com/essn.html
nicks...@ece.villanova.edu
>...We all remember from EE class that the RMS
is the same as the average for a square wave...
The average would be zero.
Nick
Solar Flare
Average of the absolute voltage = RMS = peak
<nicks...@ece.villanova.edu> wrote in message
news:dg7jgq$t...@acadia.ece.villanova.edu...
Solar Flare
"Steve Spence" <ssp...@green-trust.org> wrote in message
news:xZHVe.290$zN6...@fe10.lga...
BobG
but Milk of Magnesia and Vodka is a Phillips ScrewDriver! Thanks! I'll
be here all week! Be sure to tip your waitresses!
Steve Spence
> What storage voltage and amperehour cap. of the bank please?
>
>
12vdc 675ah pack.
ac delco 2500 watt inverter.
Solar Flare
then and not the inverter mostly?
If this is the case then perhaps a larger battery bank voltage would be a
better choice for heavy starting loads, offering less "strain" on the bank.
"Steve Spence" <ssp...@green-trust.org> wrote in message
news:ffMVe.499$H24...@fe11.lga...
Steve Spence
> You feel that the battery (source) impedance would be the cause of the dips
> then and not the inverter mostly?
>
> If this is the case then perhaps a larger battery bank voltage would be a
> better choice for heavy starting loads, offering less "strain" on the bank.
>
>
> "Steve Spence" <ssp...@green-trust.org> wrote in message
> news:ffMVe.499$H24...@fe11.lga...
>
>
> Solar Flare wrote:
>
>>What storage voltage and amperehour cap. of the bank please?
>>
>
>
> 12vdc 675ah pack.
>
> ac delco 2500 watt inverter.
>
>
would be necessary, with a decent sized battery pack.
BobG
starts? Might spin up faster with no backpressure. Put a pulley on it
and spin it like a lawnmower to start it.
Andy
"Cheezy" <chico...@gmail.com> wrote in message
news:1126646312.0...@o13g2000cwo.googlegroups.com...
Your reasoning, and therefore your math, is way off. A 1/2 horse motor
draws about 943 watts (115V X 8.2A) when working at rated capacity, and the
run winding will draw all of that until the motor comes up to speed. The
starting circuit, depending on how heavy it is, draws anywhere from one to
two KW until the starting points disconnect just as the motor reaches
operating speed. You're looking at a 3KW inverter, minimum, to start and
run that 1/2 horse motor and still have some kind of head-room for
longevity/reliability of the unit and concurrent loading. Four or five KW
would be better.
By the way; take the cover off the inverter that 'blew'. You may find a
fuse tucked away on the circuit board....
Regards,
Andy
BobG
A 1/2 horse motor draws about 943 watts (115V X 8.2A) when working at
rated capacity
Now you've got me confused.... explain a little more for me......
Solar Flare
thread branches.
"BobG" <bobga...@aol.com> wrote in message
news:1126744826....@g49g2000cwa.googlegroups.com...
Jens Kr. Kirkebø
Yep, me too. AFAIK 1 hp = 746W or so. 0.5hp would be ~373W. Is an
electric motor only 40% efficient (approx. like a diesel) ? I'd think
it would be at least double that ?
Anthony Matonak
Electric motors are not all created equal. Some are more efficient
than others. I believe it's not uncommon for common electric motors
to be in the 70% efficiency range and it wouldn't surprise me to
learn that some are much less efficient than that.
Anthony
Unknown
>A 1/2 horse motor
>draws about 943 watts (115V X 8.2A) when working at rated capacity
My 2HP submersible well pump draws 10.4A at 240V (2500W, 3.25
"electrical horsepower") for about 60 percent efficiency. When
starting, it draws about 70 amps (16.8KW, 35HP, about 18X the pump
rating) peak. It does, however, start OK on my generator (9KW,
derated to 8.2KW due to NG operation).
Just another datapoint...
Steve Spence
> Your reasoning, and therefore your math, is way off. A 1/2 horse motor
> draws about 943 watts (115V X 8.2A) when working at rated capacity, and the
> run winding will draw all of that until the motor comes up to speed. The
> starting circuit, depending on how heavy it is, draws anywhere from one to
> two KW until the starting points disconnect just as the motor reaches
> operating speed. You're looking at a 3KW inverter, minimum, to start and
> run that 1/2 horse motor and still have some kind of head-room for
> longevity/reliability of the unit and concurrent loading. Four or five KW
> would be better.
>
> By the way; take the cover off the inverter that 'blew'. You may find a
> fuse tucked away on the circuit board....
>
> Regards,
>
> Andy
>
>
1/2hp = 373 watts in a perfect world. My 2500 watt inverter will start
up a 1/2 hp pump.
-
Using the Inverter 1750 W, with a 12V 600AH Pack, She does not even medium
or high bars on the Power Meter. I suspect it is the Low Voltage disconnect
that is the problem with the setup.
"Cheezy" <chico...@gmail.com> wrote in message
news:1126646312.0...@o13g2000cwo.googlegroups.com...
daestrom
"Jens Kr. Kirkebø" <j...@scm.no> wrote in message
news:4nthi11r18ip1qcur...@4ax.com...
Well, a good motor might be 85% efficient, that takes us up to 438 watts.
Then most induction motors run at about a 0.8 power factor or better (power
factor and efficiency are *not* the same thing). So that takes us up to
about 550 VA.
The inverter has to supply the VA, while the DC battery usually just has to
supply the watts.
And then there is starting current/power. That can be three to five times
higher. Of course, many inverters have a 'surge' rating to support this
sort of thing. But the surge has to come from the battery so the voltage
dip on the DC side can still be substantial.
The easiest check is to look at the nameplate current rating of the motor.
If it claims nameplate is 8.2A, then that's what the inverter needs to
supply during normal operation. Between three and five times more during
starting.
OBTW, for centrifugal pumps, you can lower the starting time by *shutting*
the discharge. Just be sure to open it again once the pump is up to speed
to avoid overheating the pump with no flow.
daestrom
Andy
Alrighty.... looking at the plate on my Sears pump motor... it reads:
1/2 HP
115 / 230 V
8.2 / 4.1 A
W = V x A, therefore;
115V x 8.2A = 943 Watts (Motor set to run on 115 V)
230V x 4.1A = 943 Watts (Motor set to run on 230 V)
That is the RUN WINDING draw at maximum rated loading, which is the max the
motor will put out without stalling or slowing down to where the start
winding will cut in again. The actual technical detail of what determines
the rated HP is rather more complex than my simple semantics-free
explanation, but then who cares about that esoteric crap when all we want
to do is to be able to hook the damn thing up and know it will work for
us:-)
Note that you still need to add the draw of the start windings which varies
with design and construction.
HTH,
Regards,
Andy
Solar Flare
943 VAs at 70% PF = 660 Watts = .88 horsepower = 56% efficient.
"Andy" <nom...@here.net> wrote in message news:4329...@news.kos.net...
"Jens Kr. Kirkebx" <j...@scm.no> wrote in message
Solar Flare
I may be into the same situation in a few years.
<William P. N. Smith> wrote in message
news:ovoii11npfib0tcns...@4ax.com...
Solar Flare
load.
Now, mind you, the starting current has to be averaged over time. Most
meters cannot measure this in a few cycles. We did it with a scope measuring
the current. It only lasts a few cycles.
"Steve Spence" <ssp...@green-trust.org> wrote in message
news:c0fWe.679$H24...@fe11.lga...
Unknown
>How deep is your well and how many gpm please?
The hole is 755 feet deep, the pump is at 600 feet, the water is
coming in around the 150-foot level, and the static level is around 30
feet. I get something like 2GPM, but I've got a 900 gallon storage
tank that was _really_ expensive... 8*)
Solar Flare
Does the 150' & 30' level mean "above the pump", "above the bottom" or "from
the top of well/gorund"?
I am worried about 145' well here....LOL
<William P. N. Smith> wrote in message
news:641ki1d1lkpdsflqi...@4ax.com...
Paul
Leeson and others make 1/2 horse power motors that should
be able to drive a jet pump.
http://www.leeson.com/products/stock_dcmotors.htm
Unknown
>OMG!
Yeah, I finally sent the drill crew packing. Could have gotten 50GPM
another 10 feet down, but that's _always_ true. 8*)
>Does the 150' & 30' level mean
Below ground.
meow...@care2.com
> I have seen 1/4 hp synchronous motors take 30 x nameplate starting with no
> load.
>
> Now, mind you, the starting current has to be averaged over time. Most
> meters cannot measure this in a few cycles. We did it with a scope measuring
> the current. It only lasts a few cycles.
There are ways to enable motor startup on an underrated invertor or
genny. One is to put a series resistance on the motor, and once its up
to half speed, short the series R. Another is to set the output to 110v
with a 240v motor, and once up to the reduced speed, switch to 240.
These only work when the motor's mechanical load is low at low shaft
speed. This is true of most pumps, though not piston pumps.
These can only take you so far of course, but they can cut the startup
surge a lot. If the OPs invertor will deliver... I dont remember but if
its >2x run current, and the mechanical load drops toward zero at near
zero speed, it might be possible to use this approach to start it.
Whether it will start reliably I cant say, will have to suck and see.
Another way, which can be done with this if desperate, is to spin the
motor up to some extent using a bit of string, then connect immediately
while its still turning.
I havent tried any of this with an induction motor, so cant comment on
starting those this way. But perhaps the OP can work round the problem
this way.
NT
nicolas....@gmail.com
I bought an inverter Sunyima pure wave 800w/1600w peak on aliexpress very cheap... and it works, it can start the pump. I'm not sure it is a very high quality inverter so not sure if it will last so much. A friend have a Victron 800w continuous and it can also start 1/2 hp pump.