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How do you add UA values for a space if you have walls and a ceiling?

jasno999 (Aerospace) (OP)
6 May 08 23:06
How do you add UA values for a space if you have walls and a ceiling.  I have a space and I have walls built up of several materials and a ceiling and then an air gap and a metal outter surface (call it a roof).

I know how to find the U values for the wall or the ceiling and therefore I can determine the UA value for the wall or for the ceiling and outter surface.  However I am not sure how you find the overall UA value for the entire space.  Do you simply add the wall UA value and the celing UA value together or do you need to do some type of series / parallel type adding???  
 
MintJulep (Mechanical)
7 May 08 7:38
One method is to simply treat the space between the ceiling and the roof (let's call it an attic) as a space.

If you know UA of the attic floor (or room ceiling), and of the roof, then you can calculate the attic temperature.

Now you can use that temperature in the UA delta-T caclulation for the room ceiling.
jasno999 (Aerospace) (OP)
7 May 08 9:08
I don't think that answered the basic question.  How do you get the overall UA value for the room.  Is the process simply you add the UA value you get for the wall to the UA value for the ceiling?


How about this as an example.  Pretend you have a two storry building that is open and has nothing in it.  Attaced to one wall is a 1 story garage.  And between the ceilign and the roof you have a 3 inch air gap.

How do you go about finding the UA value for the buiding???

I know how to get a UA value across the wall to the outside and I knwo how to find the UA value through the ceiling, air gap and roof.  The U value between the buildign and the garage should be the same as for the otehr walls if the wall itself has the same layers to it.

Granted the Q (heat transfer) into the garage section will be lower because the temeprature in that area will be greater than what you have outside- I get that.

the question becomes how do I determine the overall UA value for the building????  I thought I would jsut add the UA values for each area together however I saw an example in a book that leads me to beleive that this might not be true.  However this example I saw is unclear and that is whay I need some help.
vpl (Nuclear)
7 May 08 10:59
I would take the inverse of each of the UA's (1/UA), add all those, then take the inverse of that.

So 1/[1/UA(wall)+ 1/US(ceiling) + 1/UA(whatever else)

Patricia Lougheed

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jasno999 (Aerospace) (OP)
8 May 08 7:40
vpl - I beleive what you stated is not correct.  When you are looking for the overall U value you use the equation:

U = 1 / (R1 + R2 + R3...)

However I don't think you add the invers of the UA values together to get an overall UA value????


Just htink abotu the math...  Say the Wall UA=45 and the ceiling UA = 23

You method says (1/45)+(1/23)= 1/UA = 1/0.0657 = 15.2

I beleive the process is a simple addition of 45 + 23 to give you an overall room UA value of 68 in this example.

 
vpl (Nuclear)
8 May 08 8:26
Jasno

I think we're both wrong.

I should have told you to remove the area from each UA value then inverse the U's, calculate the total U and the total area separately then multiply back together.  For example, assume your room is a perfect cube, where one wall is 8x8 and the roof is 8x8.  So each has an area of 64.  Divide 45/64 to give you your U value for your wall.  Do the same with the ceiling.  Invert the U values and add and then invert again.  Add your 2 areas.  Then multiply U*A again.

Doing this, I came up with a total UA of around 30.

Peace

Patricia Lougheed

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jasno999 (Aerospace) (OP)
8 May 08 9:37
Sorry I have to dissagree again.  I don't htink that is how it works cause the numbers jsut don't make sense.

When you think about it Heat transfer or Q is equal to UA(Delta T).

So if your wall UA value is 45 and lets assume a delta T = 40F then Q is 1800

That is just for one wall.  If you do the addition like you suggest you are telling me the Q for the entire room is less than what really goes thru jsut one wall....

Like I said I think I answered my own question-  it appears that you simply ad UA values to come up with the overall UA value.
IRstuff (Aerospace)
8 May 08 12:03
No, you do not set the deltaT.  You set the heat.  The deltaT is a consequence of the amount of heat available.

If you only have 500 W of heat, then the deltaT is correspondingly lower.

TTFN

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DRWeig (Electrical)
8 May 08 13:56
Answer to the original posted question is just add 'em up.  If you've calculated the UA correctly for the walls and ceiling, then UAroom = UAwalls + UAceiling.

In your case, though, it might not make sense to do this addition.  Your ceiling delta-T will probably be different from your wall delta-T, so you need to keep them separate to determine total heat load:

UAdeltaTwall + UAdeltaTceiling = UAdeltaTroom

Good on ya,

Goober Dave
jasno999 (Aerospace) (OP)
8 May 08 15:54
DRWeig - I agree with you totally.

IRstuff - I am nto sure what you are talkign about.  The deltaT is the difference in temperature between the inside area and the outside area or between the inside space and the garage space.

If you want to keep a house at 80F and the outdoor design temperature is 20F then the difference between inside and outside is 60F.  If yo uare looking at gettign the UA between the space and say a garage area then you need to look at all the UA values from the garage to the outside and solve for the garage temperature- then with that you can fidn out the UA between the garage and the space.

I have the answer to the original question which is UA, once you know them, are used to find the Q or heat tranfer out of each section (wall/ceiling/garage/etc.)  Then you can simply add all the Q values to find out what the overall heat transfer out of the room is.
IRstuff (Aerospace)
8 May 08 16:15
The deltaT is a manifestation, i.e., the RESULT of heat flow.  You cannot arbitrarily set the interior temperature without some means of making heat flows create the deltaT in question.  

In your 60ºF deltaT example, if you had diamond walls, the interior temperature would rapidly drop due to the fact that the resistance to heat flow is low and there isn't enough heat generated inside to compensate.  Likewise, if your walls are made from vacuum insulation packs, the interior temperature will actuall climb, since the walls cannot move the heat out.

The classic heat flow equation is about a HEAT FLOW, meaning that for the equation to remain with a static deltaT, you need a heat source that is constantly supplying the heat flow.  Without that heat source, the equation becomes transient, resulting in the eventual result of a 0ºF deltaT.  So, that's consistent with poor insulation, which is that heat leaks out, and the inside gets colder.

TTFN

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jasno999 (Aerospace) (OP)
8 May 08 17:28
IRstuff -  Here is how I see it.  I do understand what you are saying but my goal is to maintain a specific temperature in a given space.  Therfore the steps to take would be to look at the given temperature and the outdoor air desing temperatures (both hot and cold).  Solving for all the UA values and addign together all of the Q (heat transfers out of the space) you come up with an overall heat transfer in (hot day) or out (cold day).

Now that I know Q I can determine how much heat or coolign I need to put into the space via forced air through a heatign or cooling system...
vpl (Nuclear)
9 May 08 8:25
Jasno999

It sounds like you've made up your mind, so it doesn't matter what anyone else tells you.

However, using your method, in a hypothetical one-room house with 4 walls with a UA of 45 and a completely uninsulated roof, the total UA value would be 200.  This sounds like a really good UA value, but I sure wouldn't want to be the homeowner paying the heating bills.
 
It doesn't make real world sense.
 

Patricia Lougheed

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DRWeig (Electrical)
9 May 08 9:41
Jasno9999 --

If you get stuck, consider posting over in the HVAC/R Engineering forum.  When it comes to heating and cooling a house, there are lots of simple rules-of-thumb that can be applied, proven to work by hundreds of millions of installations.

I agree with each answer I've seen here, from IRStuff, MintJulep, and vpl (Patricia) -- none are wrong, they're just from different perspectives.  There are just easier ways to get to your end goal.

Good on all of y'all,

Goober Dave

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