Put a diverted on the exhaust pipe. In the warmer months switch it to exhaust outside.
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Put a diverted on the exhaust pipe. In the warmer months switch it to exhaust outside.
[OP]
Member
Originally Posted by Ole Anderson
I really just wanted to know if others experienced this much heat gain from running a cyclone.....sorry to have not responded till now Ole Anderson, but to answer your question in a rather ambiguous way: it depends......if I have many operations to perform on the machines separated by simple measurements and such I leave it running. If I'm making a quick couple cuts I shut it off. Not surprisingly it heats up more when left on for long periods of time. And by long periods I mean like anything over an hour and half or so.
Now for a new question: Has anyone who's experienced this heat gain done anything proactive to keep things cooler?
John L: thanks for that info, I was wondering what size AC unit I'd need for my 20 x 25 shop with two (very un-insulated) garage doors......and that's exactly the type of info that will help! Funny you mention running a sander, that's exactly what I was doing that heated things up! I love my drum sander but it's so slowwww!
And David: if your DC outlet is in the attic where does your make-up air come from?
I'm thinking about venting outside in the summer and then using air from the rest of the house as the make-up air so the AC from the house would keep the shop cool. Of course I'd have to put some sort of filter medium on the return so that I could be sure to keep the dust out of the rest of the house. And I'd probably like to have some sort of a cover so that when finishing anything I could be really sure that the air spaces were separate.
And Carl: no need to apologize for anything.....thermodynamics is the root of my troubles...I just need to find a way to keep the heat part of the equation outta my shop! Thanks guys!
[OP]
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We must have been typing at the same time....this is exactly what I'm thinking about!
Member
I think you kind of answered your own question. Venting outside is the only way to deal with the slightly warmer air that will be exhausting from the blower.
The other thing is putting the motor in some other space, like an attic or space between floors. Even in those situations it might still add to the load of cooling the rest of your house slightly. I like the diverter idea because if gives you options for when the weather is nice outside.
Contributor
Jeff, blower is in the attic above the cyclone and the outlet pipe drops back down to the ceiling plenum into four filter cartridges. since I heat in the winter and A/C in the summer I keep the air in and monitor with a Dylos. Dave
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[QUOTE=Dan Friedrichs;1941024]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
Last edited by Michael W. Clark; 06-12-2012 at 7:23 PM. Reason: Dumb math error
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+1. I too apoligize if I distracted from the OP getting the information requested. If I understand correctly, that is the primary purpose of this site. I stand behind my comments and don't want someone going off to fix a problem with mis-information. I intend for all of my comments to be in a respectful tone, but sometimes I type quickly between doing other tasks, so they may not come across that way.must........... be ............. strong........... and .......... resist ................ urge............ to .... reply
Maybe we should start a new thread titled DC physics?
The goal would be to get more of us on the same page and talking the same language, so that we dont propagate confusion and ill advise to the community. Not to mention OVERWHELMING a relatively innocent question.
Im sure I can learn a lot from everyone here, and always enjoy a good free body diagram.....
Mike
Contributor
Mike,
The energy that goes into moving the air certainly isn't IMMEDIATELY turned into heat. As you say, it takes some energy to get it moving. However, that energy is then transformed into kinetic energy in the air. Once the air hits something and stops moving, that energy must have gone somewhere. Of course, it was turned into heat.
Let's assume that OP's shop is solid (no open windows) and well-insulated. At the start of the day, he turns on the blower and runs it for 1 hour. The only energy "entrance" to the shop is via the power line. He drew 5kW*1hr=5kWhr of energy into the shop. He shuts the blower down and leaves for the day. The moving air stops moving. There is now 5kWh more energy inside the shop than was there at the beginning of the day. Where did the energy go? It cannot be created or destroyed; it was ALL converted to heat.
Member
I don't get the logic.
my air conditioner uses about 3000 watts.
2 space heaters?
Member
Dan,
A % of the energy was converted to work. Delta Energy = Delta Heat + Delta Work. he work being performed stops when the energy supply is taken away. All these units convert, but it doesn't mean all these properties are the same or that everything goes to heat. Work and Heat are defined as interactions between the system boundary, they are not properties of the system. Drawing a boundry around the shop does not provide enough information for evaluation. The only comment that can be made about the shop as a system is that the energy increased in the shop becuase there is nothing going out, but energy is going in.
As Carl stated previously and in another thread, you have to draw the boundry where the system can be analyzed. If you draw the boundary around the motor or the air in the shop/system, it becomes more clear the breakdown of the energy that was added to the shop.
Mike
Contributor
My view of the system is close to Dans
I think we get tripped up here when making a statement that applies to one boundary definition, and associating it to a different boundary definition
J
And also in how we estimate ' work ' component.
I believe most all of the electrical energy in, eventually contributes to shop heating. I would also guess that in this case, the warm air out of the closet is due to dissipating the electrical and blower inefficiencies, with relatively low air flow ( which means it gets hot to transfer the heat). These two statements are based on different boundary definitions, and can be mutually inclusive
Member
Carl/Dan,
If you haven't done this, I would encourage it:
Measure the delta T per unit time, room volume, and prove that all the electrical energy going in the shop heated the air. The heat gain in the air should equal the total electrical input if you are correct. You will have to include lights, shop refrigerators, TVs, etc. All this electrical usage turns to heat according to your theory.
Measure the temperature of the air entering the duct and the air leaving the DC exhaust. Estimate the flow based on the system rating and compare the heat added to the electrical input.
Let us know what you find out.
If I pour a bucket of water throgh a sealed opening in the roof of a shop (and the shop is sealed), how much heat did it add to the shop? Where did the KE go? The energy was put into the shop and did not come out. The shop, as a system, gained energy while the water was poured.
I've performed system balancing and design on countless exhaust systems in a wide variety of industries. I also have a thourough understanding of the physics, energy principles we are discussing throuhg my education and I design high efficiency industrial cyclones for my day job. Having said all this, I have never measured a noticeable increase in the air temp as it goes across a fan (very large fans included). If anything, the air is usually cooler and more directly correlates to the inlet air temp. To my knowledge, there is no accomodation in the ACGIH vent manual to account for air heating due to friction caused by the air moving through the duct or across the fan. I encourage you to prove me wrong through your gathering of emperical data just as my emperical data suggests otherwise.
Mike
Contributor
To the OP.....often when I am sanding a project like I was yesterday, I will have my 3 HP DC running and hooked to my ROS and both of my Jet Air Filtration systems running too. It seems to keep the dust down pretty well.
And I don't care where the energy comes from or goes ...or what form it takes.......the 3 motors running....DC and 2 AFs creates a noticable rise in temperature in my well insulated shop.
Ken
So much to learn, so little time.....
Member
I worked a few hours in my shop last Friday evening. Cyclone running most of that time and either the TS or the jointer/planer. 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. Should have opened a window.
The heat is noticeable near my filter at any time.
Contributor
Of course it will. Let me ask you this - if the motor is 100% efficient, and we say that 100% of the electrical energy is converted to "work", where does that energy eventually go? The air stops moving (gives up its kinetic energy), energy can't be destroyed - it must have gone somewhere, right? Where are you suggesting it goes?
Energy cannot be "turned" into work. Doing work on an object increases its potential energy or kinetic energy or something - but 100% of the applied energy gets turned into some other form of energy. Ultimately, it all goes to heat.
(FYI - I apologize to the OP for our little mini-hijack. It's always interesting to discuss these practical implications of physics with learned individuals)
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