heat gain question

Eugene Silberstein 3

I've never had to deal with a situation like this, but let's get all of our ducks in line.

From what I gather we have an approximate r-value between the basement space and the steam pipe of about R-20. Let's check the math:

12" concrete around steam pipe @ R = 0.08/inch = 1.0

12" concrete walls of the vault @ R = 0.08/inch = 1.0

Air space between the insulate steam pipe and the wall of the vault = Approximately 0.8

30" of soil between the vault and the foundation wall @ R = 0.5/inch = 15

24" of fieldstone foundation wall @0.09/inch = 2.16

Total approximate R value = 20

Now, let's take a look at the layout of the walls.

You have 1728 square feet of total wall space in the basement and, of that, we have 126 square feet of vault wall space that lies in-line with the basement wall.

Does this sound right so far, William?

William Faust

heat gain question

A steam company has a manhole/vault approx. 2' from one of my basement foundation walls. Pursuant to installing central air conditioning in my basement, I need to figure out how much heat comes through my foundation and then add that figure to Manual J Calculations.

I can get a tool that reads the temperature of both the subject foundation wall and compare it to my "normal," unaffected foundation walls, but I don't where to go from there. How do I translate this into Btu's per hour?

Please help. Thanks.

Eugene Silberstein 3

A Couple of Questions

Do you know how big the vault is?

What is the area of the walll where the vault is?

What is the total wall area in the basement (including the wall in question)?

William Faust

heat gain question

Our home's footprint is approx. 43' x 53'. Our basement floor is wood over concrete.

The subject foundation wall is 24" thick and is comprised of Pennsylvania schist field stone. The wall is 9' deep x 43' long x 2' wide. The area of the wall is thus 387 s.f.

The steam company's (expansion joint) vault's exterior dimensions are 16' deep x 14' long x 10' wide. Therefore, the area of the vault that faces the wall is 224 s.f. (16x14), of which 98 s.f. (7x14) extends below the wall's lowest level.

The center of a 36" steam-carrying pipe is suspended in the vault at 11' below grade. This pipe is covered or wrapped by 12" layer of concrete.

The vault's concrete walls are 12" thick.

There is 2' 5" of dirt fill between the vault and the wall.

William Faust

You've never had to...

Yes, you have the details correctly stated.

I would again point out that 98 s.f. of vault wall space lies at a depth below the lowest point of the basement wall (98 below +126 inline = 224 total). I have presumed that that this 98 s.f. radiates heat toward and into the soil that lies UNDER our concrete basement floor. Use of an infra-red temperature sensing tool indicated that the floor indeed picks up heat across a horizontal distance (moving away from the basement wall and vault) of approx. 20'.

I cannot drop a thermometer into the vault as I lack the tool to open a manhole cover. If it helps, however, I can add the following information:

Thermometer A - set on basement floor, standing vertically and leaning against subject basement wall, near vault wall. This thermometer reading was 65.

Thermometer B - set instead against other basement wall far away from vault wall. This thermometer reading was 59.

In the summertime, use of an infra-red temperature tool indicated that the two basement WALLS' temperature difference was in the range of 12 to 14 degrees.

This is very much appreciated.

Bill

William Faust

Prof. Silberstein - I failed to add the further detail:

Attached to the outside face of the basement wall is 1.5 to 2" of cement parging.

Attached to the inside face of the wall is 1/2" thick terra-cotta tile and, directly thereon, 1" of plaster (not sheet rock).

Bill

Eugene Silberstein 3

Here we go

> Prof. Silberstein - I failed to add the further

> detail:

>

> Attached to the outside face of the

> basement wall is 1.5 to 2" of cement

> parging.

>

> Attached to the inside face of the

> wall is 1/2" thick terra-cotta tile and, directly

> thereon, 1" of plaster (not sheet rock).

>

> Bill

Eugene Silberstein 3

Here we go

Given the additional thicknes and wall material, we will alter our R value to R-22. This gives us a conduction factor of 0.05. Now, let's average our heat source over the affected wall and floor area and we get about 40 btu/square foot, giving us a total heat gain rate of 1.8 btu/square foot.

For the affect area (wall and and floor) of 1,247 square feet, we have a heat gain of 2,246 btu/hour.

So, it looks like you are dealing with an additional 80 cfm or so for that area.

I will double check my math to be sure, but I am pretty certain this is what you have.