Marty, you¡¯re confusing instantaneous energy vs energy over time. If you turned your 1500w electric heater on for the same amount of time as a cut on the table saw ¨C 5 seconds or so - you would be equally disappointed in the amount of heat
generated. If you want to come over and feel the heat out of my CNC vacuum hold down table blower or the output of a large dust collector that runs for tens of minutes or several hours at a time, you will find it much easier to compare that to how we normally
use electric heaters.
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BTW, I live in Minnesota, so I¡¯m familiar with heaters and cold temperatures.
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To take more of a devils advocate position on this topic. If all the energy was being converted to heat then we could substitute our saws for heaters. ?
Any of you who live in cold weather know that you can work in your shop all night long when it¡¯s freezing outside and your shop would never warm to a comfortable level. ?Your body temperature may warm up from the work you¡¯re doing.
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On May 23, 2020, at 7:36 AM, Brian Lamb <blamb11@...> wrote:
?You all are missing the point that a lot of the hp/watts is actually consumed doing the ¡°work¡±, whether it¡¯s sawing the wood, jointing, planing, sanding, lighting or even compressing air. It takes force to do most of these things (not
lighting of course), and the force of the blade cutting through the wood is a large percentage of the watts consumed. Yes, there is always a heat load, but it¡¯s not anywhere close to 100% of consumed power¡. even here in AZ where it will be 111? later this
week.
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I think perhaps the issue then is the usual workload of a typical shop. Although in an it system things are, I imagine, a bit more stable in terms of draw and power use, that is not the
case for most shops not in a factory environment. We tend to turn on and off equipment and not be steadily throwing plank after plank in rapid succession thorough a saw at max force leading to an actual nameplate type draw. Whereas computer fans are typically
more blunt force objects running at max 100% of the time to account for a potentially catastrophic consequence of failing to control heat in a limited space and the consequences of those failures. And so, I may allow for a total heat release into your shop
for the sake of argument, but even then, the draw is variable and subsequently the heat release. If the OP is concerned about utilizing a motor and dealing with the heat for 100% utilization, he is probably 1) not using the right equipment as Felder doesn¡¯t
spec the duty as 100%, and 2) in the wrong forum and needs to be talking to factory owners.
But in more real world situations, we don¡¯t use 100% draw even when we are using machines most of the time, turn them off between runs, have leaks under a doors, windows to radiate out,
poor insulation to not contain heat or cold, and a myriad of other issues. Sure, running motors will increase heat but it¡¯s not a 1:1 in any of our shops, and especially not at nameplate levels.
I think you just said mostly what I did ¨C that the energy turns into heat, just not entirely in the motor itself.?
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8KW of electricity into a room turns into 27,000 of BTU in that room, either directly or indirectly, unless that motor is driving a shaft through a wall where some of that energy is converted into heat somewhere else. That 8KW of heat may
not stay in the room due to diffusion through the room walls and ceiling, but it¡¯s put into the room, no if¡¯s, and¡¯s, or but¡¯s, just as if you had an 8KW electric resistance heater.
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Energy is not returned to the environment on the return leg ¨C the neutral wire in a single phase 120v world, or the other hot legs in a 240v or 3-phase world. If the energy is not needed for work output or system inefficiency losses it
is not drawn in the first place; e.g. a 5HP/3KW motor has a free-running power consumption value of maybe 1KW, and that power consumption increases when the motor is asked to do work. Only when the motor is fully loaded does it draw 3KW of power. Yes there
is voltage drop on the electrical service wiring and yes there is heat generated from that lost energy, but that¡¯s a different problem in a different room. Power distribution is sized to deliver nominal voltage at the end point, factoring in losses in distribution.
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RE designing HVAC based on electrical load ¨C yes, this is exactly how it¡¯s done. My day job is in IT, part of which includes managing a datacenter and it¡¯s power and cooling. Cooling load is absolutely sized based on power draw of computing
equipment as well as the expected environmental factors. ??
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Small spaces like workshops ¨C small closed systems ¨C will show the temperature rise of power consumption more quickly than a larger system which has a lot more thermal sinking capability.
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I agree it¡¯s complicated, and I¡¯m glad nothing is simple on the Felder forum, which I¡¯m new to. There are few things in life that can truly be expressed simply.? Learning stuff ¨C the reason I joined the group ¨C happens when the complexity
is welcomed.??
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No. While you may be correct that the differences between potential energy and kinetic reach equilibrium through heat transfer, it¡¯s not so clear cut as the statement that an 8kw input to a motor yields 8kw of heat within a workshop. Within
systems there are many components, heat sinks, and losses. So for example, the spinning motor creates heat in bearings through friction, the blade creates heat through friction in the wood and drag through air, some energy is passed through entirely in the
electrical supply and leaves the system and recovers its potential in a ground, some energy is absorbed through wood fibers/saw dust and heat sinked. Heat is lost through inefficient insulation, air drafts, radiation through windows etc. Some energy is released
slowly and muddies the results like the specific heat of cast iron and sawdust. All causes, yes, 8kw input leads to 8kw of heat. But the closed system needs to be very large which simply is out of line with a real world workshop. I don¡¯t know much about sizing
hvac but I don¡¯t think this is the way to do it.
Nothing is simple on a Felder forum :) And I¡¯m waiting for my finish to dry.
All forms of energy ultimately end in heat, so yes, 8KW of energy coming in results in 8KW of heat in your shop.
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A 3KW 5HP motor produces 3KW worth of heat ¨C electrical resistance heat in the power cord and motor windings, sliding friction heat in the bearings and air friction in parts rotating in air. Even the work output of the motor ¨C the cutting,
sanding, blowing, etc., ultimately ends up as heat ¨C if you stick your hand in a pile of just cut sawdust, it will be quite warm from the cutting tool friction and the forced deformation of the wood.
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It is accurate to say that a 3KW motor itself doesn¡¯t itself give off 10,000 BTU of heat, but if you factor all of the losses in the system and especially the work output into whatever the motor is doing, you end up with 10,000 BTU of heat
in your room as a result of the motor running. It¡¯s counter-intuitive, but it¡¯s true.
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I am not an expert but I am pretty sure this is not correct.
¡°?That's about 8KW of electric coming in that all turns to heat, either motor heat, or friction heat from cutting etc.¡±
Only a small portion of power being consumed is generating heat.
?Mark, I understand the thermal mass.? I often run the saw or shaper for an hour or two straight.? Sucking 110F air into my shop would definitely be a problem.? When running I have a 5HP dust collector and a 5HP saw, shaper, or sander running.
That's about 8KW of electric coming in that all turns to heat, either motor heat, or friction heat from cutting etc.? My lighting is another 2.3kw.? 1kw of electric is 3412btu so 10kw of electric in is about 34K btu.? Over 3 tons of AC.? If I ran machines
all the time and wanted to keep it cool when it's over 110F I would have had to have 10 tons of AC per the mechanical engineer.? That's without dumping exhaust outside.? Now if I were heating the machine heat would work for me and not against me.
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