Re: Building a closet for dust collector
The law of conservation of energy says energy can¡¯t be destroyed, or created, only have its form changed.
The actual kw usage, the ¡°draw¡± you actually pay for - that energy comes into the shop and ¡°stays¡±... in many of the forms discussed here and more (the sound you can here outside is energy dissipating, transferring heat via vibration, the universe seeking equilibrium.
(If you don¡¯t think sound waves are ¡°heat¡± or energy, look up vibratory or sonic welding for extreme vibration demonstrating its heat)
So while you¡¯ll likely never constantly draw the nameplate rating, what you DO draw, well its energy in the shop, and a lot of it is ¡°pure¡± heat you can feel. ? Go hold the Motor on my 5hp clearvue after 30 minutes of running... if it was a cold day you¡¯d be glad to have it near you. ?You wouldn¡¯t think of it as an efficient heater - it¡¯s not meant to be - it is supposed to try to use energy to move air.. but even In that endeavor it ends of shedding a lot of heat converting from electricity to moving wind...
Note - ¡°more efficient¡± motors do what? ?Either run cooler, or on less power. ?But the ¡°total energy¡± in the system isn¡¯t changed with a more efficient motor, it¡¯s simply wasting less energy to heat as it turns electricity into whatever it¡¯s supposed to do mechanically.
So a more reasonable way to calculate average hvac load with be a delta off of that measured average use and some building thermal and environmental considerations to boot. ?My $.02
Thanks, Brian Sent from a device with less than stellar autocorrect
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On May 23, 2020, at 9:43 AM, imranindiana via groups.io <imranindiana@...> wrote:
? Brian,
That is where I started. At least 80% of the energy is consumed in doing work otherwise it is a pretty inefficient setup. But these guys have convinced me that the energy used to do the work is also released as thermal energy in the environment. I have never thought of this before. It makes sense to me.
Imran?
On May 23, 2020, at 10:36 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.
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.
Get?
I think you just said mostly what I did ¨C that the energy turns into heat, just not entirely in the motor itself.? ? 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. ? 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. ? 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. ?? ? 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. ? 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.?? ? 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. ? 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. ? 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. ? 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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Re: Building a closet for dust collector
I¡¯m not sure I agree, show me the heat involved in this:
I agree that there is heat created by electrical input, but I don¡¯t think all the electricity consumed is dedicated to heat. For example, incandescent lighting takes 100W to produce roughly 1500 lumens, while a CFL takes roughly 25W and an LED takes 15W. The LED is much more efficient at producing light per watt than incandescent which mainly produces heat per watt with light as the secondary result. So not all electrical input is converted to heat.
You speak of metal working, which I am intimately acquainted with, so I can throw out another example. Take a punch press, we had one that we used for punching out parts, it ran a 5hp motor, spinning a huge pulley/flywheel, storing kinetic energy, to be triggered to punch the die through the metal. That part does not get physically hot, oh, you might get a couple degrees of temperature ?rise, but nothing to speak of. It would be completely different to mill or grind the part out, much more heat generated for that process.
Yes, there are methods of work that are less efficient than others, and produce a lot of heat as the by product, but it¡¯s not always the case, and it¡¯s rarely all converted to heat.
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On May 23, 2020, at 8:17 AM, TJ Cornish < tj@...> wrote:
Brian, yes, all electrical input is converted to heat. The force to cut something turns into heat eventually as well. The power is not all consumed by the motor, but all of that work ultimately ends up as heat. Have you ever stuck your hand in a bucket of just cut sawdust? It¡¯s hot. How about hand sanded something with a piece of sandpaper? It¡¯s hot. Work ends up as heat. A 3KW table saw produces exactly as much heat as a 3KW electric heater (assuming they¡¯re on for the same amount of time ¨C again, see the difference between KW and KWH). The only difference is with a saw you get a board cut in the process of making heat, while the heater takes a shorter path to making heat. ? Another obvious example is drilling a hole in metal with a drill press. The chips come of hot ¨C really hot. Hot enough to change the temper of the metal (chips turning straw colored or blue). That energy to tear and deform the metal turns into heat. It¡¯s not all friction ¨C deforming a material takes work, too. ? Unless you¡¯re storing the energy somewhere ¨C in a battery, in a flywheel, in lifting something, that energy ends up as heat. That stored energy changes to heat too ¨C just maybe not right away. ? ? ? No, not all energy to do the work is converted to heat. For example, your saw idles at say 1hp, a heavy cut is consuming 3hp, that¡¯s a 2hp or 1500W increase. Picture a 1500W heater, which efficiently converts all it¡¯s watts to heat, do you really think you pumped out 1500W of heat into the shop taking a cut with the saw? No, you might have gotten 100W of heat in the cut, but everything else was consumed doing the actual work. ? ? That is where I started. At least 80% of the energy is consumed in doing work otherwise it is a pretty inefficient setup. But these guys have convinced me that the energy used to do the work is also released as thermal energy in the environment. I have never thought of this before. It makes sense to me.
On May 23, 2020, at 10:36 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. ? ? 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.? ? 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. ? 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. ? 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. ?? ? 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. ? 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.?? ? 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. ? 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. ? 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. ? 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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Was there a reference earlier to using a loading dock light arm to suspend the vac hose and power cord?? Perhaps none were long enough and or could not support the load of the hose?? There is one at Amazon with a 60" boom that might work if the work table/bench was against the wall.
Just asking, as I'd like to get a vacuum sanding table and it would have to put it against the shop wall.
Bob
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Re: Building a closet for dust collector
Electric motors (like pumps) only draw the energy from the power source according?to the amount of WORK being done. A motor idling will only draw the energy from the source through mechanical losses (friction, transmission inductance losses) and when it encounters resistance will draw more energy to meet the work required of it. E.g. a 1kW motor may idle full speed at .2A, encounter?a normal load at 10A, and be forced into temporary overloads at 20A.? If one cuts the wood, after the cut the amps return?to the lower idle rate. It's not sitting there burning 1kWh of work constantly.
A pump "deadheading" with no water flowing at maximum pressure build is essentially doing no work and as such is a much lower power draw compared to a pump running at maximum volume flow rate and low pressure. Throttling?a pump by increasing head pressure will actually lower the amps in use as the pump is doing less work. A dust collector will pull more amps with a clean filter and fully open duct system than it will with all ports but 1 closed, despite the perceived velocity and pressure changes at that one open port.
Enthalpy, Entropy, and the 0th, 1st, 2nd, and 3rd laws of thermodynamics can be counterintuitive, but essentially the entire universe is on the way to converting energy in whatever form now into.....sadly, waste heat that won't return to a higher form until the next big bang forces conversion.?
What's really trippy is thinking the waste heat you generated today was the conversion of mechanical energy from the chemical bonds in the wood from the suns photosynthesis of radiant energy transmitted from solar nuclear fusion. The path of electricity down the wires to run the saw is much less fascinating in my opinion, but still ultimately points to energy conversion from nuclear fusion in the sun.
What's this about dust collecter?closets we were discussing, Joe? hahaha!
On Sat, May 23, 2020 at 10:17 AM TJ Cornish < tj@...> wrote:
Brian, yes, all electrical input is converted to heat. The force to cut something turns into heat eventually as well. The power is not all consumed by the motor, but all of that work ultimately ends up as heat. Have you ever stuck your
hand in a bucket of just cut sawdust? It¡¯s hot. How about hand sanded something with a piece of sandpaper? It¡¯s hot. Work ends up as heat. A 3KW table saw produces exactly as much heat as a 3KW electric heater (assuming they¡¯re on for the same amount of time
¨C again, see the difference between KW and KWH). The only difference is with a saw you get a board cut in the process of making heat, while the heater takes a shorter path to making heat.
?
Another obvious example is drilling a hole in metal with a drill press. The chips come of hot ¨C really hot. Hot enough to change the temper of the metal (chips turning straw colored or blue). That energy to tear and deform the metal turns
into heat. It¡¯s not all friction ¨C deforming a material takes work, too.
?
Unless you¡¯re storing the energy somewhere ¨C in a battery, in a flywheel, in lifting something, that energy ends up as heat. That stored energy changes to heat too ¨C just maybe not right away.
?
?
?
No, not all energy to do the work is converted to heat. For example, your saw idles at say 1hp, a heavy cut is consuming 3hp, that¡¯s a 2hp or 1500W increase. Picture a 1500W heater, which efficiently converts all it¡¯s watts to heat, do
you really think you pumped out 1500W of heat into the shop taking a cut with the saw? No, you might have gotten 100W of heat in the cut, but everything else was consumed doing the actual work.
?
?
That is where I started. At least 80% of the energy is consumed in doing work otherwise it is a pretty inefficient setup. But these guys have convinced me that the energy used to do the work is also released as thermal energy in the environment.
I have never thought of this before. It makes sense to me.
On May 23, 2020, at 10:36 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.
?
?
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.?
?
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.
?
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.
?
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. ??
?
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.
?
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.??
?
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.
?
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.
?
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.
?
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.
?
?
-- Brett Wissel Saint Louis Restoration 1831 S Kingshighway Blvd (at Shaw Blvd) St Louis, MO 63110 314.772.2167 brett@...
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Re: Building a closet for dust collector
You are splitting hairs. My point is simply whatever the consumed electrical power is for a given day as measured by your electric meter, ALL of that power turns into heat in your room. Obviously if you run your tool for less time or run it at less than 100% capacity there will be less electrical power used than a 100% duty cycle situation, however, whatever power is ?consumed will turn into exactly the same amount of heat as an equivalently sized electric heater.?
For example, if my total electrical consumption for the day was 10 kilowatt hours - doesn¡¯t matter if I used that power in a huge one minute burst or sporadically split between multiple machines, I have generated 34,100 BTU-hours worth of thermal energy into my space. Maybe it¡¯s winter and that heat is desirable and leaks out to the environment. Maybe it¡¯s 110F in the shade and the outside environment is putting even more in my space. The point is that my 10KWH worth of tool/lighting use generates a very simple calculatable amount of heat based on the formula that 1 watt = 3.41 BTU of thermal energy. This formula holds both for power tools that produce other temporary forms of energy before they turn to heat and for simple heaters that just turn power directly to heat.?
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On May 23, 2020, at 10:24 AM, habacomike via groups.io <habacomike@...> wrote:
? TJ, you are presuming that a motor always runs at a constant load. ?That is not true. ?And the power consumption of a motor, and the current, depends on the load on the motor. On May 23, 2020, at 9:17 AM, TJ Cornish < tj@...> wrote:
Brian, yes, all electrical input is converted to heat. The force to cut something turns into heat eventually as well. The power is not all consumed by the motor, but all of that work ultimately ends up as heat. Have you ever stuck your hand in a bucket of just cut sawdust? It¡¯s hot. How about hand sanded something with a piece of sandpaper? It¡¯s hot. Work ends up as heat. A 3KW table saw produces exactly as much heat as a 3KW electric heater (assuming they¡¯re on for the same amount of time ¨C again, see the difference between KW and KWH). The only difference is with a saw you get a board cut in the process of making heat, while the heater takes a shorter path to making heat. ? Another obvious example is drilling a hole in metal with a drill press. The chips come of hot ¨C really hot. Hot enough to change the temper of the metal (chips turning straw colored or blue). That energy to tear and deform the metal turns into heat. It¡¯s not all friction ¨C deforming a material takes work, too. ? Unless you¡¯re storing the energy somewhere ¨C in a battery, in a flywheel, in lifting something, that energy ends up as heat. That stored energy changes to heat too ¨C just maybe not right away. ? ? ? No, not all energy to do the work is converted to heat. For example, your saw idles at say 1hp, a heavy cut is consuming 3hp, that¡¯s a 2hp or 1500W increase. Picture a 1500W heater, which efficiently converts all it¡¯s watts to heat, do you really think you pumped out 1500W of heat into the shop taking a cut with the saw? No, you might have gotten 100W of heat in the cut, but everything else was consumed doing the actual work. ? ? That is where I started. At least 80% of the energy is consumed in doing work otherwise it is a pretty inefficient setup. But these guys have convinced me that the energy used to do the work is also released as thermal energy in the environment. I have never thought of this before. It makes sense to me.
On May 23, 2020, at 10:36 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. ? ? 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.? ? 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. ? 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. ? 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. ?? ? 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. ? 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.?? ? 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. ? 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. ? 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. ? 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.
?
?
|
Re: Building a closet for dust collector
TJ, you are presuming that a motor always runs at a constant load. ?That is not true. ?And the power consumption of a motor, and the current, depends on the load on the motor.
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Show quoted text
On May 23, 2020, at 9:17 AM, TJ Cornish < tj@...> wrote:
Brian, yes, all electrical input is converted to heat. The force to cut something turns into heat eventually as well. The power is not all consumed by the motor, but all of that work ultimately ends up as heat. Have you ever stuck your hand in a bucket of just cut sawdust? It¡¯s hot. How about hand sanded something with a piece of sandpaper? It¡¯s hot. Work ends up as heat. A 3KW table saw produces exactly as much heat as a 3KW electric heater (assuming they¡¯re on for the same amount of time ¨C again, see the difference between KW and KWH). The only difference is with a saw you get a board cut in the process of making heat, while the heater takes a shorter path to making heat. ? Another obvious example is drilling a hole in metal with a drill press. The chips come of hot ¨C really hot. Hot enough to change the temper of the metal (chips turning straw colored or blue). That energy to tear and deform the metal turns into heat. It¡¯s not all friction ¨C deforming a material takes work, too. ? Unless you¡¯re storing the energy somewhere ¨C in a battery, in a flywheel, in lifting something, that energy ends up as heat. That stored energy changes to heat too ¨C just maybe not right away. ? ? ? No, not all energy to do the work is converted to heat. For example, your saw idles at say 1hp, a heavy cut is consuming 3hp, that¡¯s a 2hp or 1500W increase. Picture a 1500W heater, which efficiently converts all it¡¯s watts to heat, do you really think you pumped out 1500W of heat into the shop taking a cut with the saw? No, you might have gotten 100W of heat in the cut, but everything else was consumed doing the actual work. ? ? That is where I started. At least 80% of the energy is consumed in doing work otherwise it is a pretty inefficient setup. But these guys have convinced me that the energy used to do the work is also released as thermal energy in the environment. I have never thought of this before. It makes sense to me.
On May 23, 2020, at 10:36 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. ? ? 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.? ? 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. ? 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. ? 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. ?? ? 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. ? 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.?? ? 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. ? 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. ? 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. ? 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.
?
?
|
Re: Building a closet for dust collector
Brian, yes, all electrical input is converted to heat. The force to cut something turns into heat eventually as well. The power is not all consumed by the motor, but all of that work ultimately ends up as heat. Have you ever stuck your
hand in a bucket of just cut sawdust? It¡¯s hot. How about hand sanded something with a piece of sandpaper? It¡¯s hot. Work ends up as heat. A 3KW table saw produces exactly as much heat as a 3KW electric heater (assuming they¡¯re on for the same amount of time
¨C again, see the difference between KW and KWH). The only difference is with a saw you get a board cut in the process of making heat, while the heater takes a shorter path to making heat.
?
Another obvious example is drilling a hole in metal with a drill press. The chips come of hot ¨C really hot. Hot enough to change the temper of the metal (chips turning straw colored or blue). That energy to tear and deform the metal turns
into heat. It¡¯s not all friction ¨C deforming a material takes work, too.
?
Unless you¡¯re storing the energy somewhere ¨C in a battery, in a flywheel, in lifting something, that energy ends up as heat. That stored energy changes to heat too ¨C just maybe not right away.
?
?
?
No, not all energy to do the work is converted to heat. For example, your saw idles at say 1hp, a heavy cut is consuming 3hp, that¡¯s a 2hp or 1500W increase. Picture a 1500W heater, which efficiently converts all it¡¯s watts to heat, do
you really think you pumped out 1500W of heat into the shop taking a cut with the saw? No, you might have gotten 100W of heat in the cut, but everything else was consumed doing the actual work.
?
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Show quoted text
?
That is where I started. At least 80% of the energy is consumed in doing work otherwise it is a pretty inefficient setup. But these guys have convinced me that the energy used to do the work is also released as thermal energy in the environment.
I have never thought of this before. It makes sense to me.
On May 23, 2020, at 10:36 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.
?
?
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.?
?
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.
?
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.
?
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. ??
?
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.
?
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.??
?
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.
?
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.
?
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.
?
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.
?
?
|
Try waxing the table surface first.
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On May 22, 2020, at 6:59 PM, johnjgram via < johnjgram@...> wrote:
When are you use the thickness planer the rowers seem to be turning but when I put the word in there I can Them it doesn¡¯t feed the wood I have to push like hell and then pull it out penny Any thoughts
|
Re: Building a closet for dust collector
No, not all energy to do the work is converted to heat. For example, your saw idles at say 1hp, a heavy cut is consuming 3hp, that¡¯s a 2hp or 1500W increase. Picture a 1500W heater, which efficiently converts all it¡¯s watts to heat, do you really think you pumped out 1500W of heat into the shop taking a cut with the saw? No, you might have gotten 100W of heat in the cut, but everything else was consumed doing the actual work.
toggle quoted message
Show quoted text
Brian,
That is where I started. At least 80% of the energy is consumed in doing work otherwise it is a pretty inefficient setup. But these guys have convinced me that the energy used to do the work is also released as thermal energy in the environment. I have never thought of this before. It makes sense to me.
Imran? On May 23, 2020, at 10:36 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.
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.
Get?
I think you just said mostly what I did ¨C that the energy turns into heat, just not entirely in the motor itself.? ? 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. ? 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. ? 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. ?? ? 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. ? 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.?? ? 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. ? 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. ? 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. ? 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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The holes for the bearings are through holes, correct?
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The holes I reamed, I reamed to a slight hair below 26mm..the ID hole on the bearing is 10mm..I used a 10mm bolt to secure everything together
On Saturday, May 23, 2020, 06:38:41 AM PDT, imranindiana via < imranindiana@...> wrote:
Randy,
I guess the nut at bottom is smaller than the ID of the bearing. I have only seen smaller reamers and the nut or the threaded rod is close to the size of the reamer OD. Just was curious how you achieved flat bottom.
Imran ?
Thanks Imran.. the reamer I used was an adjustable reamer.. works exactly like you described.. I had to adjust it 4 or 5 times till the desired size was achieved..
On Saturday, May 23, 2020, 05:49:53 AM PDT, imranindiana via < imranindiana@...> wrote:
Randy,
Well done. Looks great. What kind of reamer did you use? Ones I have seen work for thru holes with adjusting nut on the bottom.
Imran ?
I will say, the seating of the bearing in the aluminum took a little time.? I used 4" wide x 6" long x 5/16" thick 6061 aluminum plates.? I put some green masking tape on both plates and used CA glue to glue the 2 plates together so they would not move.? I used a 15/16" drill bit ( bearings are 26mm in dia.) and drilled the holes on both ends of the plates,? Took a chisel and separated the plates after all the holes were drilled and then used an adjustable reamer to ream open the holes drilled for the bearings until the holes were just right and pressed the bearings in..The are seated very tightly and will not budge.? This is what prevents the sag of the arms when assembled?
On Friday, May 22, 2020, 09:53:15 PM PDT, joelgelman via < joelgelman@...> wrote:
For the mid-section of what I hope will be the super-easy-and-inexpensive-to-make-in-your-shop version, I made a quick demo of the plan for the pivot mechanism (gear will be used but omitted at this time). ?In picture 1, you see a 2 inch Forstner bit being used to drill partially through one of the pieces. ?The center indentation can be a guide for the drilling of the smaller hole in the middle for the bolt. ?Another way to insure a dead center hole is to clamp the workpiece, and change bits without moving the wood. ?Figure 2 shows a nice fit of the 2 inch OD angled thrust bearing. ?After the hole is bored (figure 3), a hole is bored in the other piece of wood, and with washers, and a nut and bolt, the units are assembled. ?For those who do not want to deal with seating bearings in aluminum, this bearing method may work well. ?I just need to make sure there will be adequate strength in the wood.?
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Re: Building a closet for dust collector
Mark,
Thanks for the explanation. You are right, I just chose simple words. Law of conservation of energy is something learned early but this last bit I never realized or if I was taught/told did not comprehend or forgot. I am happy to be aware of this now.
Imran
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On May 23, 2020, at 10:44 AM, mark thomas <murkyd@...> wrote: ?Imran, I think perhaps it's more intuitive when you realize that it's not actually "thermal loss".? It's not quite right to say that it's "loss", and likewise not right to say that this is "not true if some of the energy to do work is stored temporarily".? Energy is not neither lost nor stored, it's just moved around.??
Loss is just a concept relative to a some chosen boundary (the "closed system") and relative to your objectives for the energy.? ? If the boundary is the motor, and the goal is cutting wood, one can say that all the the energy pumped into the motor/cutting that doesn't result in cut wood is "lost", such as the friction that heats up the bearings.? ?But suppose that in addition to a saw blade, there's also a generator attached to the motor and it's charging a battery. The energy going into battery is just as "lost" from the motor/cutting system as is the bearing friction heat.? But neither is a loss, it's just movement, resulting in the energy being "stored" somewhere else.? ?Some of the energy is moved to (stored in) the battery, and some of the energy is moved to (stored in) the motor housing and air.?
With respect to physics,? it's not that one is "lost" and the other is "stored".
When you expand the system boundary -- eg, to the shop -- it's the same.? ?As Glen noted, when running his equipment, some of the energy pumped into his shop resulted in cut wood and some of it resulted in (desirable) warming of the shop air.? The energy that heated his shop wasn't lost, it was moved into the air, and is still inside the system boundary (his shop).? But if he had a heat pump hooked up to his equipment and the motor heat was pumped outside his shop, he'd say the energy was "lost" because it's now moved outside his boundary of concern.? But maybe the heat pump dumps the energy outside where it warms someone's tush.? That person won't say the energy is lost, they'll say they're getting a free tush warming!
Here's a silly non-energy metaphor of system boundary and loss and storage.? If I drop a $20 bill on the street, I might say I lost $20.? It's indeed "lost" with respect to the system boundary of my finances.? ?But my neighbor who finds it and "stores" it in their pocket doesn't see it was a loss.? And at the boundary of "neighborhood finances", there is no net change, just money moving around.
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Re: Building a closet for dust collector
Imran, I think perhaps it's more intuitive when you realize that it's not actually "thermal loss".? It's not quite right to say that it's "loss", and likewise not right to say that this is "not true if some of the energy to do work is stored temporarily".? Energy is not neither lost nor stored, it's just moved around.??
Loss is just a concept relative to a some chosen boundary (the "closed system") and relative to your objectives for the energy.? ? If the boundary is the motor, and the goal is cutting wood, one can say that all the the energy pumped into the motor/cutting that doesn't result in cut wood is "lost", such as the friction that heats up the bearings.? ?But suppose that in addition to a saw blade, there's also a generator attached to the motor and it's charging a battery. The energy going into battery is just as "lost" from the motor/cutting system as is the bearing friction heat.? But neither is a loss, it's just movement, resulting in the energy being "stored" somewhere else.? ?Some of the energy is moved to (stored in) the battery, and some of the energy is moved to (stored in) the motor housing and air.?
With respect to physics,? it's not that one is "lost" and the other is "stored".
When you expand the system boundary -- eg, to the shop -- it's the same.? ?As Glen noted, when running his equipment, some of the energy pumped into his shop resulted in cut wood and some of it resulted in (desirable) warming of the shop air.? The energy that heated his shop wasn't lost, it was moved into the air, and is still inside the system boundary (his shop).? But if he had a heat pump hooked up to his equipment and the motor heat was pumped outside his shop, he'd say the energy was "lost" because it's now moved outside his boundary of concern.? But maybe the heat pump dumps the energy outside where it warms someone's tush.? That person won't say the energy is lost, they'll say they're getting a free tush warming!
Here's a silly non-energy metaphor of system boundary and loss and storage.? If I drop a $20 bill on the street, I might say I lost $20.? It's indeed "lost" with respect to the system boundary of my finances.? ?But my neighbor who finds it and "stores" it in their pocket doesn't see it was a loss.? And at the boundary of "neighborhood finances", there is no net change, just money moving around.
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Re: Building a closet for dust collector
Brian,
That is where I started. At least 80% of the energy is consumed in doing work otherwise it is a pretty inefficient setup. But these guys have convinced me that the energy used to do the work is also released as thermal energy in the environment. I have never thought of this before. It makes sense to me.
Imran?
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On May 23, 2020, at 10:36 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.
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.
Get?
I think you just said mostly what I did ¨C that the energy turns into heat, just not entirely in the motor itself.? ? 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. ? 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. ? 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. ?? ? 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. ? 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.?? ? 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. ? 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. ? 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. ? 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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Joel,
Is there going to be that much space between the two pieces or is thus where the gears go?
Randy¡¯s version has gears on one side only, if your version is similar than is there enough depth in wood to get two pieces of wood closer?
I am guessing that if the two pieces of wood are abutted, with UHMW tape between, that should increase rigidity.?
Imran?
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On May 23, 2020, at 12:53 AM, joelgelman via groups.io <joelgelman@...> wrote: ?For the mid-section of what I hope will be the super-easy-and-inexpensive-to-make-in-your-shop version, I made a quick demo of the plan for the pivot mechanism (gear will be used but omitted at this time). ?In picture 1, you see a 2 inch Forstner bit being used to drill partially through one of the pieces. ?The center indentation can be a guide for the drilling of the smaller hole in the middle for the bolt. ?Another way to insure a dead center hole is to clamp the workpiece, and change bits without moving the wood. ?Figure 2 shows a nice fit of the 2 inch OD angled thrust bearing. ?After the hole is bored (figure 3), a hole is bored in the other piece of wood, and with washers, and a nut and bolt, the units are assembled. ?For those who do not want to deal with seating bearings in aluminum, this bearing method may work well. ?I just need to make sure there will be adequate strength in the wood.?
<1.JPG> <2.JPG> <3.jpg> <4.jpg>
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Re: Building a closet for dust collector
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.
toggle quoted message
Show quoted text
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.
Get?
I think you just said mostly what I did ¨C that the energy turns into heat, just not entirely in the motor itself.? ? 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. ? 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. ? 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. ?? ? 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. ? 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.?? ? 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. ? 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. ? 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. ? 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.
|
The holes I reamed, I reamed to a slight hair below 26mm..the ID hole on the bearing is 10mm..I used a 10mm bolt to secure everything together
On Saturday, May 23, 2020, 06:38:41 AM PDT, imranindiana via groups.io <imranindiana@...> wrote:
Randy,
I guess the nut at bottom is smaller than the ID of the bearing. I have only seen smaller reamers and the nut or the threaded rod is close to the size of the reamer OD. Just was curious how you achieved flat bottom.
Imran
On May 23, 2020, at 9:22 AM, Randy Child via groups.io <strongman_one@...> wrote: ?
Thanks Imran.. the reamer I used was an adjustable reamer.. works exactly like you described.. I had to adjust it 4 or 5 times till the desired size was achieved..
On Saturday, May 23, 2020, 05:49:53 AM PDT, imranindiana via groups.io <imranindiana@...> wrote:
Randy,
Well done. Looks great. What kind of reamer did you use? Ones I have seen work for thru holes with adjusting nut on the bottom.
Imran
On May 23, 2020, at 8:34 AM, Randy Child via groups.io <strongman_one@...> wrote:
?
I will say, the seating of the bearing in the aluminum took a little time.? I used 4" wide x 6" long x 5/16" thick 6061 aluminum plates.? I put some green masking tape on both plates and used CA glue to glue the 2 plates together so they would not move.? I used a 15/16" drill bit ( bearings are 26mm in dia.) and drilled the holes on both ends of the plates,? Took a chisel and separated the plates after all the holes were drilled and then used an adjustable reamer to ream open the holes drilled for the bearings until the holes were just right and pressed the bearings in..The are seated very tightly and will not budge.? This is what prevents the sag of the arms when assembled?
On Friday, May 22, 2020, 09:53:15 PM PDT, joelgelman via groups.io <joelgelman@...> wrote:
For the mid-section of what I hope will be the super-easy-and-inexpensive-to-make-in-your-shop version, I made a quick demo of the plan for the pivot mechanism (gear will be used but omitted at this time). ?In picture 1, you see a 2 inch Forstner bit being used to drill partially through one of the pieces. ?The center indentation can be a guide for the drilling of the smaller hole in the middle for the bolt. ?Another way to insure a dead center hole is to clamp the workpiece, and change bits without moving the wood. ?Figure 2 shows a nice fit of the 2 inch OD angled thrust bearing. ?After the hole is bored (figure 3), a hole is bored in the other piece of wood, and with washers, and a nut and bolt, the units are assembled. ?For those who do not want to deal with seating bearings in aluminum, this bearing method may work well. ?I just need to make sure there will be adequate strength in the wood.?
|
Randy,
I guess the nut at bottom is smaller than the ID of the bearing. I have only seen smaller reamers and the nut or the threaded rod is close to the size of the reamer OD. Just was curious how you achieved flat bottom.
Imran
toggle quoted message
Show quoted text
On May 23, 2020, at 9:22 AM, Randy Child via groups.io <strongman_one@...> wrote: ?
Thanks Imran.. the reamer I used was an adjustable reamer.. works exactly like you described.. I had to adjust it 4 or 5 times till the desired size was achieved..
On Saturday, May 23, 2020, 05:49:53 AM PDT, imranindiana via groups.io <imranindiana@...> wrote:
Randy,
Well done. Looks great. What kind of reamer did you use? Ones I have seen work for thru holes with adjusting nut on the bottom.
Imran
On May 23, 2020, at 8:34 AM, Randy Child via groups.io <strongman_one@...> wrote:
?
I will say, the seating of the bearing in the aluminum took a little time.? I used 4" wide x 6" long x 5/16" thick 6061 aluminum plates.? I put some green masking tape on both plates and used CA glue to glue the 2 plates together so they would not move.? I used a 15/16" drill bit ( bearings are 26mm in dia.) and drilled the holes on both ends of the plates,? Took a chisel and separated the plates after all the holes were drilled and then used an adjustable reamer to ream open the holes drilled for the bearings until the holes were just right and pressed the bearings in..The are seated very tightly and will not budge.? This is what prevents the sag of the arms when assembled?
On Friday, May 22, 2020, 09:53:15 PM PDT, joelgelman via groups.io <joelgelman@...> wrote:
For the mid-section of what I hope will be the super-easy-and-inexpensive-to-make-in-your-shop version, I made a quick demo of the plan for the pivot mechanism (gear will be used but omitted at this time). ?In picture 1, you see a 2 inch Forstner bit being used to drill partially through one of the pieces. ?The center indentation can be a guide for the drilling of the smaller hole in the middle for the bolt. ?Another way to insure a dead center hole is to clamp the workpiece, and change bits without moving the wood. ?Figure 2 shows a nice fit of the 2 inch OD angled thrust bearing. ?After the hole is bored (figure 3), a hole is bored in the other piece of wood, and with washers, and a nut and bolt, the units are assembled. ?For those who do not want to deal with seating bearings in aluminum, this bearing method may work well. ?I just need to make sure there will be adequate strength in the wood.?
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Thanks Imran.. the reamer I used was an adjustable reamer.. works exactly like you described.. I had to adjust it 4 or 5 times till the desired size was achieved..
On Saturday, May 23, 2020, 05:49:53 AM PDT, imranindiana via groups.io <imranindiana@...> wrote:
Randy,
Well done. Looks great. What kind of reamer did you use? Ones I have seen work for thru holes with adjusting nut on the bottom.
Imran
On May 23, 2020, at 8:34 AM, Randy Child via groups.io <strongman_one@...> wrote: ?
I will say, the seating of the bearing in the aluminum took a little time.? I used 4" wide x 6" long x 5/16" thick 6061 aluminum plates.? I put some green masking tape on both plates and used CA glue to glue the 2 plates together so they would not move.? I used a 15/16" drill bit ( bearings are 26mm in dia.) and drilled the holes on both ends of the plates,? Took a chisel and separated the plates after all the holes were drilled and then used an adjustable reamer to ream open the holes drilled for the bearings until the holes were just right and pressed the bearings in..The are seated very tightly and will not budge.? This is what prevents the sag of the arms when assembled?
On Friday, May 22, 2020, 09:53:15 PM PDT, joelgelman via groups.io <joelgelman@...> wrote:
For the mid-section of what I hope will be the super-easy-and-inexpensive-to-make-in-your-shop version, I made a quick demo of the plan for the pivot mechanism (gear will be used but omitted at this time). ?In picture 1, you see a 2 inch Forstner bit being used to drill partially through one of the pieces. ?The center indentation can be a guide for the drilling of the smaller hole in the middle for the bolt. ?Another way to insure a dead center hole is to clamp the workpiece, and change bits without moving the wood. ?Figure 2 shows a nice fit of the 2 inch OD angled thrust bearing. ?After the hole is bored (figure 3), a hole is bored in the other piece of wood, and with washers, and a nut and bolt, the units are assembled. ?For those who do not want to deal with seating bearings in aluminum, this bearing method may work well. ?I just need to make sure there will be adequate strength in the wood.?
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Re: Building a closet for dust collector
Michael, Joe, Mark and TJ,
Thanks for the education. I have really not thought about this before and it is not intuitive to equate 100% of the energy, consumed to do mechanical work, to thermal loss in environment. This as we discussed is not true if some of the energy to do work is stored temporarily.
Good stuff.
Imran?
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On May 22, 2020, at 11:19 PM, Michael Tagge <mike.j.tagge@...> wrote: ? Imran ¨C ? That¡¯s more akin to a heat sink- It messes with the temporality of the heat into the system, ie stores it and decays. A wood shop equivalent is the sawdust and the specific heat of cast iron / steel that dissipate heat over an extended period of time. Great for IT heat management strategies to allow for short term overclocking and increased heat generation. ? In a pump, the conversion is into potential energy actually, and there is the analogue. Its not converted to heat but to potential which then can be utilized at a different time in its path to heat equilibration. ? So the real crux is how to actually measure heat generation. On a theoretical level, all energy into a system is the energy out of the system. But the systems are leaky, time delayed, and fluctuant in real world situations is my point. So yes, thermodynamics hold true but you have to measure further than anyone wants to really go, or maybe they do, and in which case, that¡¯s cool too ¨C I¡¯m really fond of getting into arguments over details. ? ? ? I am intrigued by this conversation. In a data center I follow that energy is turning into heat. But in case of a motor I am not convinced. Imagine a well pump. There is energy stored in the water that has been lifted. So per the law of conservation of energy not all energy got converted into heat.
On May 22, 2020, at 10:34 PM, TJ Cornish <tj@...> wrote:
? I think you just said mostly what I did ¨C that the energy turns into heat, just not entirely in the motor itself.? ? 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. ? 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. ? 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. ?? ? 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. ? 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. ? ? 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. ? 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. ? 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. ? 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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Randy,
Well done. Looks great. What kind of reamer did you use? Ones I have seen work for thru holes with adjusting nut on the bottom.
Imran
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On May 23, 2020, at 8:34 AM, Randy Child via groups.io <strongman_one@...> wrote: ?
I will say, the seating of the bearing in the aluminum took a little time.? I used 4" wide x 6" long x 5/16" thick 6061 aluminum plates.? I put some green masking tape on both plates and used CA glue to glue the 2 plates together so they would not move.? I used a 15/16" drill bit ( bearings are 26mm in dia.) and drilled the holes on both ends of the plates,? Took a chisel and separated the plates after all the holes were drilled and then used an adjustable reamer to ream open the holes drilled for the bearings until the holes were just right and pressed the bearings in..The are seated very tightly and will not budge.? This is what prevents the sag of the arms when assembled?
On Friday, May 22, 2020, 09:53:15 PM PDT, joelgelman via groups.io <joelgelman@...> wrote:
For the mid-section of what I hope will be the super-easy-and-inexpensive-to-make-in-your-shop version, I made a quick demo of the plan for the pivot mechanism (gear will be used but omitted at this time). ?In picture 1, you see a 2 inch Forstner bit being used to drill partially through one of the pieces. ?The center indentation can be a guide for the drilling of the smaller hole in the middle for the bolt. ?Another way to insure a dead center hole is to clamp the workpiece, and change bits without moving the wood. ?Figure 2 shows a nice fit of the 2 inch OD angled thrust bearing. ?After the hole is bored (figure 3), a hole is bored in the other piece of wood, and with washers, and a nut and bolt, the units are assembled. ?For those who do not want to deal with seating bearings in aluminum, this bearing method may work well. ?I just need to make sure there will be adequate strength in the wood.?
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imran