electric bandmills

[dean herring]

Does w WM or TK offer an electric driven mill.

[Dave Shepard]

I don't know if TK does, but there is a link to their website on the left. I bet they do. WM offers an electric option on LT15 and the LT40 Super and larger at the very least. As mentioned, you need the support equipment to get the potential of a higher production mill.

[PC-Urban-Sawyer]

...  It takes three people to start to use the production potential of the mill.

Which means that since Bibbyman has ProSawyer Mary on payroll they've got plenty of capability to use the full potential of their mill!

8) 8) 8) 8) 8) 8) 8) 8)

Herb

[SPD748]

I hope this isn't a hijack however I do have a question about electric motors and generators. How does one go about sizing a generator for an electric motor (mill)? For instance, say I have a 30 hp, 3 phase motor on my mill and I want to run it with a diesel generator. What size generator would it take to run this load? Or... would a more detailed list of motor specs be required to properly size the generator?

-lee

[schmism]

would a more detailed list of motor specs be required to properly size the generator?

AMPS!

or total watts or volt-amps if you perfer, either way you need to know amps.    google tells me the avg 30hp electric motor would pull something like 35 amps at 460v   which is 16,100w  or 16kw.   you might squeeze that out on a 15Kw genset as the 35 amps listed may be locked rotor condition... operating amp draw is much less (half or more)   

but if your sizeing your genset, id leave "room" to power other items.   electric chain saws, lights, blowers, etc.    sum up the total number of watts needed for all your intended use, and find a genset that outputs that amount.  20...25...30kW are all easly attainable portable genset numbers.    There really is no upper end,   Catapillar genesets currently run from 450kW up to 7MW.

couple of other notes.... technically there is a conversion between hp and w  as 1hp = 745w  but thats purely a theoretical conversion.  It doesnt take into account any losses etc.   so to say 30*745 = 22350w isnt quite accurate.   One might also say that because the above electric motor only uses 16kw its only a 21hp motor... again  there is marketing, electric hp vs gas hp vs diesel hp....   much of this portion of the discussion devolves quickly which is why brigs and stratton no longer lists hp and only lists displacement.  (due to the lawsuit)

The point is, it boils down to total watts.... or volt-amps (kVA) if you want to be technical (as your not considering power factor)

[SPD748]

Ah...

This particular motor says 74 full load amps @ 230 volts with 94.1% full load efficiency and a 1.25 service factor. I'm told there is a "locked rotor amps" rating however it's not on the name plate.

Oh... the plate says 230/460 volts. Forgive my electrical ignorance here: If wired for 460, will the required generator size/rating be lower?

-lee

[Ianab]

The issue with the generator size is that it has to be able to handle the start current surge of whatever the motor draws for the first few seconds while it's spinning up. This will be considerably higher than it's running load.  Maybe 50% more?

It also varies with with the motor design, the type of starter (A VFD / soft start will usually start an motor with less current) and the genset itself will have different continuous and peak loads as well.

But 50% more is going to be in the right ball park.

This also gives you some "extra" to run accessories as schism mentions, although you may have to start the largest motor first.

[schmism]

the most common rateings you can sometimes find for electric motors are as follows

-full load amps, -  max current that the motor can draw and still maintain listed rpm
-startup or surge amps - current that the motor draws durring startup.  generally this value and the full load amps are very close,  enough so that you can assume to the 2 to be the same unless otherwise stated (eg for special motor designs)
- locked rotor amps - max current the the motor can draw period.  this occurs when the motor is stalled or the rotor (the part that spins) is locked up unable to move.   can also be called stall-current.
- no load current - this is the current the motor draws when operating under no load.  (free spinning on the bench with nothing attached) 

You can assume a fairly linear line between the no load current and full load current based on percent load.   eg if no load current is 10 amp and full load is 110 amp.  then 50% load will draw ~60 amp.   The inverse of this is also true, which is how load meters generally work,  they read the current draw and display that as % load.

IN the case of a band mill its unlikely you'll ever see a locked rotor condition as the system is belt driven and the belt drives on saw mills will slip long before a locked rotor condition especially on a 30hp motor.   (you may be able to achieve that much torque on a small 5hp though)

[schmism]

The issue with the generator size is that it has to be able to handle the start current surge of whatever the motor draws for the first few seconds while it's spinning up. This will be considerably higher than it's running load.  Maybe 50% more?

So gensets include a "surge" rateing also.   Generally to handle startup currents for motors.   In the above example 74amp*230v = 17020 VA  or roughly 17kVA.     What would be the smallest genset that could run that motor?  Id say 15kVA genset.   A 15kW Kohler genset lists a 19kW surg rateing.  Enough to start the motor.   Again, if you assume you ran it at 50% load its likely only going to draw roughly 10kW which is well under the 15kVA nominal output of the genset.

[schmism]

A few other notes.....

If your looking at purchasing a factory built 30hp electric machine,  its as simple as reading the numbers off the face plate to size the generator.

If your thinking of upgrading (or building from scratch) an existing gas powered mill to electric, and think hey, 20hp gas might as well step it up to 30hp electric, be warned,  your likely not likely to see addtional saw rate increase unless you "re-gear" the mill for the higher hp.

here is a quote from Cooks

Here are some guidelines: 8 hp max speed 3,500 fpm, 12 hp max speed 4,000 fpm, 16 to 18 hp max speed 4,500 fpm, 25 hp max speed 5,000 fpm, 30 hp and above max speed 5,500 fpm.

You "use" your hp in feed rate.    high feed rates = high hp.   but there are physical limits to feed rate that have to do with how fast you can remove the sawdust for the lenght of cut your makeing.  You will physically fill the gullet of the tooth with sawdust,  Once its full, if you try to pack more material in it you get problems,  most notably, diving or climbing.     So high hp, high feed rate production mills use very large gullet blades = few teeth per inch and the larger the gullet the wider the blade has to be. 

OR the other option is to increase the band speed so that you get more teeth through the cut in a given time thus allowing you to increase your feed rate.  which is why you see cooks recommend higher band speeds for higher hp mills (with the same 1.25" blades "everyone" uses)

So moving from 20hp gas to a 30hp electric, without increasing your "gearing" allowing for more bandspeed, you will not likely see the advantage of that extra hp you just bolted on.

[SPD748]

Ok so...

An electric motor which has a full load amp requirement of 74 amps @ 230 volts would require a minimum of a 15 kw genset to start and run the motor?

-lee

[Ianab]

I wouldn't guarantee that under all situations. It might work, or it might not.

The 15 kva is the max draw while running, so a 15 kva generator would KEEP it running. The start current will actually be something higher. Maybe higher than surge capacity of the generator.

Most Generator suppliers would have you with at least 20kva, and that depends on the motor / starter design. A DOL starter would need more like 38 kva to get a 30 hp motor spinning. (Which is why those are not usually used)

[sandsawmill14]

1 hp= .745 kw      30 hp=22.35 kw    NEC allows for 800% fla of conductors with acceptable thermal overload protection to cover start loads. this allows up to an 800 amp breaker to be used on #2 thhn wire but remember the THERMAL OVERLOAD thing or you will burn the shed down !!!!  I installed a reduced voltage starter, auto transformer type, on a 75 hp 480 volt dust blower at a sawmill, start load with rvs was 563 amps without the starter i could not check start load because it went higher than my amp probe (1000 amp). this only happens for 2-3 seconds depending on load, we also have a 5' chipper on the same type starter that is set for 100 second start to get start load down to about 950 amps. we had to install cap. banks because we were dimming the lights at the bank and post office but everyones happy now

but on this motor if it says 74 amps @240 volts it will draw 37 amps on 480 volts  far as sizing genset take highest single amp load X 2, then take total combined load  and use which ever that is bigger and that will be MINIMUM size kw genset. the biggest i have installed is 1200 kw running complete sawmill down to 10kw residential units.  Im not trying to argue but it cost to much to do it twice.  SPD748  if you have any doubts call the company and they will tell you

[WoodenHead]

If you have an electric motor with a soft-start or VFD (Variable Frequency Drive), the minimum recommended genset size (kW) would be roughly 2 x rated HP of the electric motor (assuming this is the only load you wish to start).  Theoretically the genset can be smaller, but I have read that any smaller and your genset will have trouble trying to regulate voltage resulting in potential damage to the genset.

Without a soft-start or VFD, the situation changes.  You need to have a genset that can handle the initial start-up current (Amps) which can be 6 x the rated full load amps (FLA) if you are under full load depending on motor type.  Normally in a sawmill scenario you are not in a full load situation, but start with no-load to lightly loaded.  This makes sizing a little trickier.  If I remember correctly, 2 or 3 times FLA is what you are looking for (or 2 to 3 times the rated HP of the electric motor for easier calculation).

So the absolute minimum genset size for a 30HP motor is 2 x 0.745 kW/HP x 30 HP = 44.7 kW.  For quick sizing many take the motor HP and multiply by two or three (i.e. 60kW or 90kW).

[sandsawmill14]

well stated  woodenhead
If you size a genset by these guide lines you will be happy ;D

dont for get to calculate the total load if more than 1 motor or other load is being used. never load genset past 80% for continuous load.

[4x4American]

say_what  Say what?  Explain.

The drive side bandwheel you can spin it around easily.  Be nice for cleaning the wheel and checking tracking.  Idk how your mill is but I can't spin the drive side bandwheel around unless it is disconnected from the engina.

[Magicman]

OK, I understand. 

Different sawmill manufacturers/models are set up differently.  My drive side band wheel  is secured by a brake when the clutch is disengaged.  To manually turn the band wheel, you relieve the break tension.