my cyclone is makin me hot....

[Michael W. Clark]

Do the calcs on the heat rise and room volume and figure out where the heat goes for yourself. The temperature rise in the shop will not equal the electrical input. Size the A/C for your electrical input, its your money, enjoy. Mike

[Carl Beckett]

Do the calcs on the heat rise and room volume and figure out where the heat goes for yourself. The temperature rise in the shop will not equal the electrical input. Size the A/C for your electrical input, its your money, enjoy. Mike

Good morning Michael,

First, I want to say that for sure you have a lot more practical, direct experience with DC systems than I do. I do appreciate your comments, and your willingness to share your insights and experience with the forum.

At the same time, having some pretty significant experience in thermal systems myself, there are comments that you make that I like to try to understand better. Because some of these comments do not support my understanding and experience, and by listening to others like yourself, I am able to learn something.

Your suggestion to collect empirical data is shifting the problem definition. In fact, I HAVE done building heat loss calculations and there is considerable heat loss to the outside. The prior statements were qualified as a 'perfectly insulated' building - similar to comments such as 'ignoring friction'. These are common assumptions when doing mathematical modeling. And they dont relate to the real work practical results - except that they ARE useful to gain a fundamental understanding of the system energy sources (and sinks).

You sound frustrated that your statements and understanding of the system is being challenged in some way. Im sorry for that, it wasnt my intention to upset anyone, and already its past the point where I feel the original posters question was lost. This has happened in prior threads on the topic, where I simply gave up.

But I firmly believe in the power of these type of public forums - where multiple minds can discuss a problem for the benefit of better understanding, from which we all gain knowledge. In no way would I presume to know everything, nor have 'all the answers' on a particular topic (and even in areas where I am considered a deep technical expert, I spend most of my time thinking about how little we actually understand of the area). And without rattling off qualifications, it turns out I do have some relevant credentials.

I for one appreciate your comments and insights and hope that you do not get discouraged, nor close yourself off to looking at these problems from a different perspective. As a scientist, my philosophy is that at the end of the day nature/physics win out, and that by understanding this we are able to better understand the practical implementation/behavior of these systems.

This is a great site. I very much appreciate being a part of it. If the moderators want to take this post down as off topic then I respect that. If there is ever a feeling of disrespect then I am open to discussing it, since it is most certainly not the intent.

[Matt Meiser]

Well, one thing is for sure--your question sure generated a lot of hot air!

[Michael W. Clark]

Matt, that's classic!

Carl, Thank you for the civility. Yes, these debates can get "heated". I'm an engineer and figure out ways to solve problems/answer questions. I would suggest to anyone that takes an interest in this to collect some data so we can evaluate it. Get the shop volume, temperature before starting the cyclone and other motors, shop temperature after running the motors, and the time that the motors were running. With this information, we can quickly calculate the BTU input into heating the shop air. This BTU would be converted to HP to compare to the running HP.

This topic came up some time back and most that had a larger temperature gradient also had high ceilings. The DC is going to circulate the air so some of the hot air from the ceiling is going to be mixed in. I know one person had 26' ceilings and had a significant temperature stratification.

At the end of the day, everyone is going to do as they will in their own shop, its how it should be. I'm here because I like woodworking, not installing extra A/C. I have a small shop, 11x22 with 8.5ft celings. I run a single stage 1.5 HP collector and a 5HP tablesaw (I know its not enough DC, but an upgrade is planned and thats another topic). I often run this setup continuously when ripping and sizing sheet goods. I haven't noticed any unusual temperature rise, especially not on the order of the electrical input energy.

If anyone wants to collect the data above and post it, I would be glad to convert it and I'll show an example calc so you can repeat it and know that I am not cheating.
Mike

P.S. I understand your theory on the DC air slowing and imparting its energy into heat, but what about the energy from the TS motor? Where does all that energy go?

[Dan Friedrichs]

The temperature rise in the shop will not equal the electrical input.

From a practical perspective, you may be entirely right. I'm idealizing everything, which is not very practical

You bring electrical energy into the shop - some is lost through the boundary (perhaps some moving air gets blown out a door, taking its kinetic energy with it, some heat is transferred through the not-perfectly-insulated walls, etc), but whatever energy remains in the shop eventually turns into heat. To suggest otherwise violates conservation of energy

[mreza Salav]

[QUOTE=Michael W. Clark;1941092]

You cannot "create" energy.QUOTE]

That is exactly what you are doing when you account for the kinetic energy in the air moving and the heat generated if you assume that all of the electrical energy is heat. You can't get both at the same time. Don't misunderstand me, I know there will be some temperature rise due to inefficiencies, fan friction etc. However, the law you are referring to says energy cannot be created or destroyed, it can only change forms. The electrical energy is converted into kinetic/potential energy in the air system. When you apply the electrical power, the air moves. When you take away the power, the air stops. If you don't think it takes energy to move the air, then why do we need a motor if all we are doing is heating the shop?

Take John's post as an example and assume 10' ceilings. We have a room volume of 6250ft3 raised 5F in one hour. The Btu requirement is 562.5 Btu/hr = 0.22 HP. Note, he had 6HP running according to the post. Granted, the drum sander was probably not pulling FLA the entire time, but I think the point is clear. If all 6HP is converted to heat (15,267 Btu/hr), then you have a temperatuer rise of 136F, assuming no leaking to the surroundings, the dimensions stated above, and ignoring motor inefficiencies. Feel free to check my math.

Work is a scaler (does not matter the direction). Work = Force x Distance in mechanical systems. Without a force or distance, no work is done. Without friction, no force is required, so no work is done. The more friction, the more force required, the more work done, and the faster you want the work done, the more power requried. Work is different from Heat, which is different from Energy. The change in energy of a system is proportional to the change in work + the change in heat (1st Law of Thermo). Yes, watt=watt=watt, but total energy watts = work watts + heat watts, thats the First Law of Thermo. You guys are ignoring the work watts and assuming that all of the energy is going into frictional heat.

Revision: should be 27 degree temp rise instead of 136. I thought that looked extreme.

Respectfully,
Mike

If you disagree with the statement of Dan then you are disagreeing with one of the fundamental basis of physics.
Aside from nuclear reactions, you simply CAN NOT "create" energy, period, you are only transforming one form to another. Ryan said it well.

If you consider the boundary of shop and you assume the loss of heat is constant in all times then all that electricity that goes into the shop needs to be transformed to something.
Heat (at the heart of it) is movement, btw. So if there is no significant force that is being moved then that energy will be transformed to heat at the end.
Even if you move air, eventually that energy is transformed into heat. And to overcome friction you generate heat.

[Michael W. Clark]

[QUOTE=mreza Salav;1941487]

If you disagree with the statement of Dan then you are disagreeing with one of the fundamental basis of physics.
Aside from nuclear reactions, you simply CAN NOT "create" energy, period, you are only transforming one form to another. Ryan said it well.

If you consider the boundary of shop and you assume the loss of heat is constant in all times then all that electricity that goes into the shop needs to be transformed to something.
Heat (at the heart of it) is movement, btw. So if there is no significant force that is being moved then that energy will be transformed to heat at the end.
Even if you move air, eventually that energy is transformed into heat. And to overcome friction you generate heat.

I am agreeing that energy is not created, that is the basis of my argument. If you spend energy to heat the shop, that costs money. If you spend the same energy to cut parts at the tablesaw and say that same energy also heated the shop just as the heater did, then cutting the project parts were free, from an energy standpoint. That, to me, is free work, which is not possible. Like I said, collect the data, do the calcs, and I will stand corrected if I'm wrong. We can debate theory, but what matters is how hot the shop gets and if this equals the electrical input.

Mike

[Ryan Brucks]

The example of measuring the electrical draw of a whole shop and then comparing to the heat rise would have so many pitfalls it would be basically pointless, aside from the small scale 3 box test I already mentioned.

Why? The wires themselves would be dissipating a good mount of wattage into the walls. If the electric panel was in another room, you'd have another huge source of dissipation elsewhere. Depending on the type of lighting fixtures, most of the lighting heat could be rising through installation holes into the space above the shop or between floors. Just to name a few.

I don't quite see where you are coming from when you say some energy gets just turned into "work"... As previously stated, it gets turned into heat eventually or stored as potential energy somewhere (a good example would be the static charge that gets left behind on the DC pipes).


When you mention the cutting being "free"... it's not!! Try cutting acrylic and if you aren't careful you will see it melt before you eyes. The very cutting itself causes exaclty as much heat as the energy it took to make the cut! Even the little melted chips and spirals all heated up as they were cut, some melting onto the blade.

[mreza Salav]

[QUOTE=Michael W. Clark;1941497]

I am agreeing that energy is not created, that is the basis of my argument. If you spend energy to heat the shop, that costs money. If you spend the same energy to cut parts at the tablesaw and say that same energy also heated the shop just as the heater did, then cutting the project parts were free, from an energy standpoint. That, to me, is free work, which is not possible. Like I said, collect the data, do the calcs, and I will stand corrected if I'm wrong. We can debate theory, but what matters is how hot the shop gets and if this equals the electrical input.

Mike

The problem is your view of "work". At any given point of time if you look at the state of everything in the shop there is certain amount of energy stored in them.
If you move an object from point A to a higher ground part of the energy spent is stored in the object.
If you move an object from point A to point B and the amount of energy stored in that object is the same (say horizontal movement) then the energy you spent to
move that object has gone somewhere else, but where? It's all heat now.
If you cut a piece of wood the energy spent to cut it is all transformed to "heat". There is no "stored" energy in the pieces you get. As simple as that.
If you move a lot of air in the shop, at the end (when everything settles down) the energy spent has transformed to heat (it cannot go anywhere else).

[Michael W. Clark]

The example of measuring the electrical draw of a whole shop and then comparing to the heat rise would have so many pitfalls it would be basically pointless, aside from the small scale 3 box test I already mentioned.

Why? The wires themselves would be dissipating a good mount of wattage into the walls. If the electric panel was in another room, you'd have another huge source of dissipation elsewhere. Depending on the type of lighting fixtures, most of the lighting heat could be rising through installation holes into the space above the shop or between floors. Just to name a few.

I don't quite see where you are coming from when you say some energy gets just turned into "work"... As previously stated, it gets turned into heat eventually or stored as potential energy somewhere (a good example would be the static charge that gets left behind on the DC pipes).


When you mention the cutting being "free"... it's not!! Try cutting acrylic and if you aren't careful you will see it melt before you eyes. The very cutting itself causes exaclty as much heat as the energy it took to make the cut! Even the little melted chips and spirals all heated up as they were cut, some melting onto the blade.

My wires are all located in my shop (as are all the wires for the basement and many of the ones for upstairs), so is the shop sub-panel, and so is the main panel for the house. The shop walls are insulated with 1/2" OSB and I have 10" thick concrete basement walls behing those, mostly below grade. The ceiling is insulated and has 5/8" sheetrock If you are still adament about telling the OP that they need to add 1 ton of A/C to compensate for the 5HP DC, then that's fine. If you want to figrure out if its true or not, get the measurements mentioned above and we can go through them and calculate the BTU added to the shop and compare that to the motor HP. I'll give you a pass, and we won't consider the heat load from the lighting or ther ancillaries. We can compare it to the DC hp only and assume the tool motor is partially loaded, or we may not have to consider it at all. Let's let the data drive the commentary, not our understandings of theory. I'll gladly admit that I'm wrong, and would like to know the answer as I intend to build a shop in the next 5 years and would like to have space conditioning.

Edit: Add shop info
I just ran the numbers for my shop at dimensions mentioned previously and, for a 10 degree F temp rise, it would take 0.15 HP. Please verify my math, but this is a factor of 10 off of my DC nameplate motor HP, it doesn't even account for the extra electrical supplied to overcome the motor inefficiencies, or any other equipment running. I got 370 Btu/hr and converted that to horsepower using a cp of 0.24 Btu/lb-F and an air density of .075 lb/ft3 for 2057 ft3. I'm going off of memory on the cp, so please let me know if I made a mistake.
Mike

[Alan Lightstone]

...,snip.... I also had the industrial ceiling fan running. The air temp jumped somewhere between 5 and 10 degrees even though the outside temp fell a good 15 during that time.

You turned on the fan???????? That's why the room got hotter.




Sorry, couldn't resist.

[Steve Peterson]

I don't get the logic.

my air conditioner uses about 3000 watts.
2 space heaters?

Your air conditioner sends its waste heat out the window. If it pulls 1000 watts of heat out of the room, then a total of 4000 watts of heat is sent outside.

The case with the cyclone is that all of the heat is held inside the shop.

Steve

[Van Huskey]

If your cyclone is making you hot I suggest you make it wear more and/or less revealing clothing...

Sorry...

[Ryan Brucks]

To be perfectly honest, I am not much capable of debating this topic at the level required to prove one way or another. I just have my layman's knowledge and some basic examples but I will admit also could be mistaken here.

I think there are still pitfalls when you expect the 0.15hp heater to also heat the room the same as a 0.15hp blower. The heater is very efficient at transferring the energy to the air quickly... so it is able to rise the air temp quickly. A general purpose motor itself heats up like a heatsink, and uses a fan to remove it, but still the motor itself will be holding quite a bit of the heat. So even if a 0.15hp motor can produce the energy to heat the room (on paper), it does no good if that energy warms the air as slowly as the room gives off heat to the outside world.

Same with the electric panel losses. even with the panel in the room, those wires are insulated and won't heat the air much as they will conduct heat in the wires themselves. Same with lighting, even if the light is all inside the room, the visible light striking the walls will be turned to heat, but then leaving the room as quickly as it re enters (ie, drywall will radiate heat in and out). so some types of energy exit the system MUCH faster than others.

again, that wasn't really a technical rebuttal, more of a "maybe this explains the discrepancy?" answer. granted its a huge discrepancy and I've neve run those calcs myself to check them.

[Jason Roehl]

One thing I haven't seen in the calculations offered is anything beyond the rise in temp of the air. To raise the temperature 10ºF in Michael's shop may only take 0.15HP, but then that air is also warming all the equipment, walls, ceiling, floor, plus the air will stratify...

I'm with Dan on this one. If you're using a 5HP motor at FLA (5 HP), then there's 5 HP worth of heat going into the room.

Here's another way of looking at it. Temperature of a system is, by definition, based on the average speed of the molecules in the system. Turn on a DC and you're speeding up the air molecules, raising the average speed, which means you're raising the temperature.