Replies

1. 
   John_D | Dec 14, 2005 01:58am | #1

   I just had to do the same figuring, and found the following information at McMaster-Carr:

   Heat Output Required— A heater should be large enough to replace the heat lost through the floor, walls, and ceiling of the space you are trying to heat. The amount of heat lost depends on how well an area is sealed and insulated. Follow these steps to estimate your heat requirement:

   Step 1: Determine the surface area of your floor, walls, and ceiling in square feet:

     (2 x length x width)+(2 x length x height)+(2 x width x height)

   Step 2: Estimate your heat loss factor by choosing the description that best fits your building:

   Very well sealed and insulated = .25

   Well sealed, but not insulated = .75

   Not well sealed or insulated = 1.25

   Step 3: Decide how much you want the temperature to rise in ° F:

   If you don't currently have interior heat this would be the difference between the outside temperature and your desired temperature. If adding to existing heat this would be the difference between your current temperature and your desired temperature.

   Step 4: Multiply the results from steps 1-3 for your estimated Btu/hr. requirement. You may need to consider more than one heater to meet your total requirement.

   Step 5: If you are sizing an electric heater by watts, multiply the result in Step 4 by .293.

   For Example— Your 20-ft. x 20-ft. area has a 12-ft. ceiling and is well sealed and insulated. Your current heat source can only maintain 50° F, yet you want your area to be 65° F.

   Step 1: Your surface area is: (2 x 20 x 20)+(2 x 20 x 12)+(2 x 20 x 12) = 1760 sq. ft.

   Step 2: Your heat loss factor is .25 since your room is well sealed and insulated.

   Step 3: You need a 15° F temperature rise (the difference between current 50° F temperature and desired 65° F temperature).

   Step 4: Multiplying the results of steps 1-3 determines your heat required: 1760 x .25 x 15 = 6600 Btu/Hr.

   Note: Your result in step 4 is the minimum requirement to reach your desired temperature. If this result is between two heater sizes, you should select the larger size.

   Hope this helps!

   My goal is for my work to outlast me.  Expect my joinery to get simpler as time goes by.
1. 
   kywood50 | Dec 14, 2005 03:48am | #2

   John,

   Thanks, how did this calulation work for you?  I just did it and was surprised by the amount of heat btu's needed.  I have R-49 ceiling and R-19 walls.  The only question is does the .25 heat lose factor represent the insulation level I used.

   Thanks again,

   Les Otte

1. 
   John_D | Dec 14, 2005 11:09am | #4

   The jury's not in yet.  I've yet to order the heater!My goal is for my work to outlast me.  Expect my joinery to get simpler as time goes by.

2. 
   BG | Dec 14, 2005 04:24am | #3

   Kywood50,
   There may be a fast and relatively accurate way, but doing it correctly is more important. A call to your supplier (or the power company) will probably get you more accurate results.
   With all due respect to John's example it does leave out a few very important factors. You have a very large area to heat, some would say it's equal to a small house. I have no idea what climate you live in but I'd suspect 15 degree differential is insufficient. Walls on a super insullated building loose about 5-6 btu's per hour/sq. ft. while floor and ceiling loose about 7-9 btu's per hour/sq. ft. Glass looses 80 btu's per hour/sq. ft. Also, generally you count on two(2) air changes per hour..in a room with one door. With a vac system, however the air changes can go up drmatically..ten (10) changes per hour or more. As you can see McMasters numbers did not provide for glass or air changes...and they are, or can be in your situation, significant.

1. 
   John_D | Dec 14, 2005 11:11am | #5

   Hey, I was just passing on a link -- the example is McMaster's, not mine.  Actual expertise is welcome, I'm sure. :)My goal is for my work to outlast me.  Expect my joinery to get simpler as time goes by.

1. 
   BG | Dec 14, 2005 03:34pm | #6

   John_D,
   I struggled to complement you for all the work you did on the post and not offend you for McMasters methodology...sorry if that didn't come across.
   When I was a kid(late 60's) my job was determining the heat/air needs and layout from the blueprints. It would take several hours to crunch the numbers and my boss went over my work very carefully. When all was said and done we doubled the estimated heat loss numbers in sizing the furnace (so, why all the accuracy?). The fact that he is going electric with a heat exchanger down the road implies he lives in a warmer temp area of the country...dunno. My ballpark guess would be he will need about 20,000 to 30,000 btu's available to support a building that size.

1. 
   kywood50 | Dec 14, 2005 05:38pm | #7

   BG,  John D, others

   Thanks for all the help/information, I live in central Kentucky.  Their are no windows, and I have sealed all the ceiling, wall, floor seams with foam insulation.  The celing is two layers of insulation, top being R-30 as a continous layer wo/structural members. 

    The floor is a slab concrete, with one edge exposed w/R-6 insulation, the other edges are inside a 75' by 72' insulated wo/heat building.  I need to do more checking on the slab temperature, I am guessing that the ground temp is somewhere around 55 degrees F.  I have measured that the main building is warmer then the outside temperature.  Also the slab had a sealer applied while it was finished for strength and reduced moisture infiltration.

   For the walls I have the option to add between R-6 to R-9 insulation on the outside of the wall studs, do you think this is a good idea?  My thinking is to do it for the one wall that is exposed to the outside as a minimum, other three walls are interior to the main building.

   Just a note, I am looking into an open geothermal HP system, if I can do it for minimal dollars.  Have lots of groundwater near the surface ( low cost to drill wells/materials), pick-up a used HP and blower ( have friend in the HP service business).   We would do it as a kind of experiment. 

   Again thanks for all your help!

   Les Otte

    

1. 
   BG | Dec 14, 2005 09:44pm | #8

   kywood50,
   Boy, it sounds like John's numbers may be a lot closer to reality than mine...(my face red)...lol. I would think you may have a problem getting a unit that is small(output wise) enough for your building...and still be cost justified with all the setup costs....they can provide air too? It may be that after an hour or two the electric could be turned off. Usually each person gives off about 300 btu's per hour..and then there is some heat from the equipment. How many will be working in the space?

2. 
   John_D | Dec 14, 2005 09:47pm | #9

   I didn't take offense, sorry if it seemed so.  Heck, you're right, this McMaster formula has a LOT of obvious slop.  The multiplier in step 2 ranges from .25 to 1.25, a 5x factor!  So after you measure your shop the square inch and do that math to three significant figures, then you just come up with your own multiplier.  Heh.

   Still, it gives you a decent range of possibilities.  I just ordered a portable heater that produces 19K btu/hr and draws 240v/23A.  Perfect for the 30A sockets around my shop -- it even has the 6-30 plug.  The 19K btu/hr is in the range the formula points me to.

   I can afford to be loosey-goosey with the calculations.  If the shop's too cold, I'll just supplement with a radiant heater or something, or buy a second unit.  If it's too much heater, well, it has a thermostat and it just won't run that much. :)My goal is for my work to outlast me.  Expect my joinery to get simpler as time goes by.

3. 
   WillGeorge | Dec 16, 2005 10:48pm | #10

   and add electric or geothermal heat pump later?? You live in ICELAND?

   Wish I had that!

1. 
   PeterFarquharson | Dec 21, 2005 09:43am | #11

   I also built a new shop lately and used a radiant floor to heat the space (700'). Without running equipment, I can only get to about 60 degrees (when it's 40 below outside), but even small amounts of tool usage raise the temp to a very warm 70 degrees. Morning's might be crisp, but you'll be dripping by noon...

   Edited 12/21/2005 1:44 am ET by PeterFarquharson

2. 
   jpattillo | Dec 21, 2005 09:14pm | #12

   I think he means one of those geothermal heat pumps that use the earth as a heat source/sink by burying a network of coolant filled pipes a few feet underground (or vertically in a well bore). They are extremely efficient. Unfortunately they are pretty expensive to install.

1. 
   kywood50 | Dec 24, 2005 06:10pm | #13

   You are correct on the geothermal heat pump.  Their are two basic types of systems open and closed.  The closed is as you described a closed loop of pipe in the ground. The open type uses two wells.  One well is used to pumps ground water at 55 to 60 degrees up out of the ground to a heat pump that extracts heat out of the water.  The second well is a discharge well that receives the water after heat has been recovered.  My plan is to see if I can use very shallow wells because of our high GW table and very fractured bedrock.  This along w/ a used heat pump could make my cost much lower. 

   Les

1. 
   jpattillo | Jan 03, 2006 11:54pm | #14

   The open system sounds like it would be less expensive to install, with the big cost being boring two wells. Are you planning to convert a conventional air heat pump to use with the geothermal system? Speaking of heat pumps, I see a lot of discussion about heating shops, but very little about air conditioning. I'm in the deep south, so when I build my in the next year or so I think I'll be using a heat pump for all my climate control in the shop. Air conditioning will be a big concern for me, as I'll be doing most of my woodworking in the summer, and I don't want my tools to rust from the high humidity.

1. 
   tinkerer2 | Jan 04, 2006 07:30am | #15

   Can you explain basically the difference between open and closed system.  I have an idea but am not sure about the open system.

2. 
   WillGeorge | Jan 05, 2006 09:10pm | #16

   but am not sure about the open system.With OLD steam heat... the steam/water gushed out a pipe stickin' out the roof as I recall!EDIT:: Closed uses a expansion tank I thinks'

   Edited 1/5/2006 1:11 pm by WillGeorge

3. 
   jpattillo | Jan 06, 2006 12:21am | #17

   I'm not an expert on this by any stretch, but basically the difference is in how the heat energy is extracted from or moved into the ground. In a closed system, a loop of tubing is buried in trenches or vertical shafts. A coolant (water or water/alcohol mix) is pumped through this tubing, to pump heat in or out of the system. The coolant never comes in direct contact with the ground. In an open system, as described in the earlier post, water pumped from a well is used as the source/sink for the heat pump and then the water is pump back down another well. In this case, the groundwater itself is the coolant.I think this is correct, but I'm happy to be corrected

4. 
   tinkerer2 | Jan 06, 2006 02:24am | #18

        This makes a lot of sense.  I had the closed system down but was a little unsure of the open system.  From your explanation it would seem that a loop that ran out into a lake would be a closed system but if you pumped the lake water to the unit it would be an open system.  Am I right?

5. 
   jpattillo | Jan 06, 2006 03:00am | #19

   Right, You would pump the "used" water back into the lake and the heat of the earth would keep the lake at pretty much the same temperature.