Heat Conduction through a Wall 2

[WARose]

I have a thermo problem here....for a mechanical guy this is probably pretty easy, but for a structural guy like me....I am struggling (first time I've opened a thermo book in about 20 years). The problem is this: how thick does a concrete wall have to be (on a per square foot basis) to facilitate a 104 degree (Fahrenheit) drop?

One way I tried to do it is to relate specific heat for material to the conduction equation. Specifically I set it up this way:


Q=M*Cp*Delta(T)=Delta(T)/(X/Ck*A)

Where:
M=mass (lbm); Concrete is about 150 lbs/cubic foot. I got mass by dividing by 32.2. I included X (see below) in this calculation.
Cp=Specific Heat (I used 0.2 BTU/lbm*F)
Delta(T)=104 F
X=Thickness of the wall (in feet)
Ck=1.0 BTU/h*ft*F
A= 1 ft^2

Plugging these values in....I get:

4.65X*(0.201)*104=104/X; solving for X I get 1.03' of thickness which doesn't seem right.

Looking at the units, the Ck variable includes a time element which I do not consider here (I was looking at it as a steady-state problem).

Could someone backcheck me here? Does what I am doing make sense? Thanks in advance.

 

[chicopee]

Before anyone is to commit to an answer, for those who have some knowledge in this area, there are a few questions to ask:
1) are you asking, how long would it take to drop the internal energy within the block 104 dF, or
2) are you asking how thick should the block be so that the temperature gradient from one side of the block to the other side is 104 df
3) is the temperature drop to be under transient or steady state conditions
4) is there a medium surrounding the concrete block
To note, your equation is not properly used unless you have either a heat sink or a heat generation within the block and a time element provided.
You should revised your OP by providing more information about the problem.

 

[WARose]

Thanks for the reply chicopee.....to answer your questions:

1) No, I don't care about time. This will be a steady-state (day to day) state of affairs.

2) Yes.

3) Steady-state.

4) If I understand the question correctly: air on both sides.

 

[racookpe1978]

Please confirm:

Solid concrete NOT concrete blocks or masonry, right?
Static air on both sides, but free to heat up (and to cool down) naturally: No wind, no fans, no insulation covering either side of the solid wall? No roof or barrier in the local area of the heat transfer?
Wall implies a vertical structure - is that correct?
Static conditions, right? A steady-state air temperature on one side of the wall is established by a heater, a furnace, a pipe, a steam leak, a diesel exhaust pipe or something; and (somehow) a steady-state temperature on the other side is maintained by something else that constantly must remove enough heat energy to keep that 104 deg F difference, right?
We are to ignore heat losses or gains everywhere else: roof, floor, other walls, etc, right?

 

[WARose]

racookpe1978,

--Yes, solid (cast in place) reinforced concrete wall. No hollow parts.

--Yes, treat it like a static air problem. No insulation on the walls. Assume no losses other than the conduction through the walls.

--Yes, wall is vertical. But I figured it would be easier to treat it on per square foot basis.

--Yes, on the other side there is a HVAC system that will account for maintaining that difference. (Although that brings up an interesting question: Part of what I have to figure is the temperature load in the wall. If that HVAC system did malfunction would that mean the wall would have to be twice as thick for the temperature in the middle of the wall to have lost that 104 degree increase?)

--Yes, ignore all other losses. I'd treat it like a pure conduction problem.

Thanks for your time.

 

[gruntguru]

You seem to have two unknowns. The thickness of the wall and the heat flux through the wall. I suspect you are asking for thickness of the wall given that each side is in contact with air with delta T (air) = 104*F. If so, you also have a convection problem to solve and that will require knowledge of geometry, orintation and forcing.

je suis charlie

 

[IRstuff]

Your problem statement is still incomplete. How much heat is there, or, how hot is the air on the other side?

It's basically a straight thermal conductivity problem, for the most part:

Q = thermal conductivity * area * delta temperature / thickness

Since the equation ends up with watts, you need to give either the amount of heat flow, or air temperature, so that natural convection can be applied to get the heat flow.

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[racookpe1978]

We are being picky - and, in part, the equations themselves are very straightforward - ONCE ALL of the picky little details are figured out.

See, the hot side and cold side wall each have a film layer of air-being-heated by natural convection, right? Well, as a wall gets taller, the motive force for the movement of air next to the wall increases. A tall wall will conduct more heat (have less resistance) than a short wall bounded by a floor and a ceiling. So "height of wall" is an important variable even though the final average film coefficient is defined by an average heat rate/sq foot of wall area.
Floor-to-room air, wall-to-room air, and ceiling-to-room air are very, very different problems if natural convection is the only thing moving the room air. Also important in each of those cases, which side is hotter, which side is higher or lower. If you have forced convection with a lot of velocity everywhere, it almost doesn't matter what side of the room you are looking at.

Your temperatures are modest, so radiation is not a big factor.
No condensation or boiling/evaporation = no latent heat issues. Thank you. 8<)

A very hot room air against a room wall that is much colder will create more natural convection than a mild room room air and a mild wall side temperature = more convection forces = less film resistance. All of the different individual properties of air change as room air temp - wall
side temp change too.

Concrete as a solid is nice and forgiving. Thickness (what you are looking for) is about the only variable. But you are looking for that variable,

The film resistance depends on the temperature of the mid-point of the film: this temperature is halfway between wall temp and room air temp (more or less - but good enough for approximations like yours.) So, what is the hot room air temp, the hot side wall edge temp, cold side wall temp, and cold side room air temp? You approximate or define those, then you run the equation through everything from hot room air to cold room air. But you have to start with either an amount of energy that is going to be transferred, or a difference in temperature that will be maintained.

Get results_1, put those back in the problem, re-run to get results_2. Usually only 3-4 runs give adequate results. Film coefficients and film resistances change a little in each iteration, so the net heat transfer changes each time.

Rinse, wash, repeat.

All of the above require steady state assumptions: Changing temps over time add thermal mass and thermal delays to the problem.

Heat (amount of energy transferred = your A/C load to cool the cold room) transferred can either be the start of the problem (it is simply defined in most textbook problems, because it is the hardest to figure out in real life and the easiest to just blindly state as an assumption in a textbook. )

In your case: Define what you want temperatures you want your two rooms to be held at. See what the maximum amount of heat energy is. Then see if that is a reasonable solution, or if you have to add insulation to reduce heat transfer. A 1/2 of foam behind fireproof cement board or a 2x4 wall of stuffed fiberglass will be cheaper than pouring more concrete. Probably.

IR Stuff. Anything I've forgotten to ask or address?

 

[WARose]

racookpe1978,

Concrete is all I can use. (No insulation.) The "hot wall" temp will be a maximum (hopefully) of 174 F. The "cold wall" temp I'm aiming for is 70 F. Wall height is about 17'.

 

[IRstuff]

@racooke, looks reasonable

@WARose, you still haven't described what the air temps are on the two sides of the wall, unless they are 174 and 70. If so, then we can say with nearly absolute certainty that the outer surface can never achieve 70, since that would require extremely small thermal conductivity and convection in the air. Likewise, the inner surface should not get to 174, since the wall will suck heat out of the hot side, resulting a thermal gradient.

The general approach for solving this is to set up two simultaneous equations with the two air temps and the two unknown surface temps.

Eq1. h*Area*(Tin-Tin_wall) = k*Area*(Tin_wall-Tout_wall)/thk
Eq2. sigma*Area*(Tin_wall-Tout_wall)/thk = h*Area*(Tout_wall-Tout)

where h = convection coefficient, which you could start by assuming 10 W/m^2-K
and k = thermal conductivity, which is on the order of 1 W/m-K

With that, 1-ft thick wall results in outer surface temp of 90.6F, while a 2-ft wall would get to 82.8F. That's because concrete, while being a relatively good insulator when compared to, say, aluminum, Styrofoam is 0.033W/m-K, which is 30 times more effective. So, 2 inches of styrofoam ONLY, would result in an outer surface temperature of 70.6F. So, some combination of concrete and styrofoam. or similar, would be the best approach to minimizing wall thickness and getting very close to your desired operational conditions.



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[WARose]

@WARose, you still haven't described what the air temps are on the two sides of the wall, unless they are 174 and 70.

That's what they are. It's something just adjacent to the wall on one side that's causing that (according to the vendor). Obviously the whole room won't get to that.

With that, 1-ft thick wall results in outer surface temp of 90.6F, while a 2-ft wall would get to 82.8F. That's because concrete, while being a relatively good insulator when compared to, say, aluminum, Styrofoam is 0.033W/m-K, which is 30 times more effective. So, 2 inches of styrofoam ONLY, would result in an outer surface temperature of 70.6F. So, some combination of concrete and styrofoam. or similar, would be the best approach to minimizing wall thickness and getting very close to your desired operational conditions.

Thanks.

 

[gruntguru]

Aerated concrete is a much better insulator than concrete. If you calculate the required conrete wall to be ridiculously thick, you could use aerated concrete for the entire wall or just an inner layer if the exposed faces need to be full strength concrete.

je suis charlie

 

[racookpe1978]

Well, you're going to need form works for the concrete wall and its rebar. So make the wall structurally strong, then add a single layer of foam inside the form on the cold side of the wall, pour the concrete. If fireproofing is required over the foam, then add a layer of drywall over the foam.

 

[Tunalover]

@IRStuff-
What is the term "sigma" in your second equation?



Tunalover

 

[IRstuff]

oops, that should have been k

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