Heat transfer Calculation

[KKolev]

Is it possible to calculate the cold face temperature of insulation panel knowing :
k (W/m.K) thermal conductivity
A (m2) Area
Th (K) Hot Face Temperature
x (m) wall thickness

I don't know:
Q (W) Heat Transfer
Tc (K) Cold Face Temperature

Guess I need to use Fourier's equation:
Q=k.A(Th-Tc)/x
BUT I NEED SOMETHING MORE...
another equation?

Thanks

 

[Twoballcane]

You will have to calculate q from the internal wall temp out to the external ambient, so your thermal resistances are both the conduction and natural convection (or forced) and maybe radiant if you want to go that far. Once you know q you can calculate the external wall temp.

Tobalcane
"If you avoid failure, you also avoid success."

 

[IRstuff]

Is this for school?

TTFN

FAQ731-376

 

[KKolev]

I'm just trying to understand how to crudely calculate the cold side temperature of a car exhaust insulation. Obviously I've never done this before and some of my questions might sound stupid.

Twoballcane, thanks for the reply.
So ignoring radiation I need to calculate the composite Heat transfer (cond+conv) & from this get the cold face temp.


CONDUCTION CONVECTION
k,W/mK h,W/m2K
x,m A,m2
A,m2
|/////////|
|/////////|
|/////////|
Thot,K |/////////| Tcold,K ========> Tamb,K
|/////////|
|/////////|

k.A.(Thot-Tcold)
Qcond=--------------- Qconv=A.h.(Tcold-Tamb)
x


REScond=x/(k.A) RESconv=1/(h.A)


Thot-Tamb Thot-Tcold
Qcomposite= ------------------- = ------------
REScond+RESconv REScond

A.k.Thot-x.Qcomposite
Tcold=----------------------
k.A

 

[vpl]

KKolev

Right now you've got two many unknowns to solve your problem. As a first approximation, just worry about the conduction through the insulation and set T_cold equal to T_ambient and calculate Q.

I'd also point out that in the equation at the bottom of your second post, the terms "A.k" and "k.A" would cancel.

However, assuming that this is a work problem, the easiest way to solve the problem would be by direct measure.

If it's not a work problem, and you're more a car enthusiast, I'd recommend that you think about why car manufacturers wouldn't add insulation to their car exhaust.


Patricia Lougheed

******

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[IRstuff]

You need to use the continuity assumption, i.e., the heat entering the insulation is the same as the heat leaving the insulation. This eliminates one of your excess unknowns. Given that, you now only have a single unknown, which is the cold side temperature, which you now can solve for.

Your algebra at the bottom seems wonky. It should be:

stuff1*(Thot-Tcold)=stuff2(Tcold-Tamb), solve for Tcold:

Tcold = (stuff1*Thot+stuff2*Tamb)/(stuff1+stuff2)

TTFN

FAQ731-376

 

[Twoballcane]

VPL, I have to disagree, the external wall temperature can be closer to the internal wall temperature than ambient temp. So you can not assume this scenario. You will have a smaller q which may be unrealistic (almost an adiabatic boundary condition).

KKolev, I think one of your Ak is suppose to be Ah, but in any case, you have to find q for the system. Thus this will equal the sum of the resistance (conduction plus convection) times the internal wall temp mines ambient temp. Once you find q, you can recalculate either the conduction or convection alone and get the external wall temp. Your biggest challenge will be to calculate h.


Tobalcane
"If you avoid failure, you also avoid success."

 

[IRstuff]

If you have all the external parameters, you don't need Q. The cold wall temperature is solely determined by the other two temperatures and the conductivity and convection coefficients.

TTFN

FAQ731-376

 

[Twoballcane]

Yup I agree, however, it is always (well to me) good practice to start quantifing the model so that we can get a better understanding of what is going on. Does the Q make sense, can we corelate that back to any data? From there you can get a feeling of what the temps should be.

Tobalcane
"If you avoid failure, you also avoid success."

 

[IRstuff]

I'm not suggesting not determing the Q.

But, once you have the temperatures, the Q can be calculated and compared against any other data or experience you might have.

TTFN

FAQ731-376

 

[FredRosse]

The simple answer to your question: NO, it is not possible to calculate the cold wall temperature with the data you have.

The cold wall temperature will depend on the data you have (except for the surface area, which is not relevant), plus the mechanism for removing heat from the cold wall.

For example, if the cold wall is connected to ordinary air, and temperatures are moderate, then the cold wall temp will have one solution, somewhat higher than the ambient air temperature. Qdot = Hcair * (Tcoldwall - Tair ambient), Hcair is generally small.


If the cold wall is connected to a highly turbulent flow of cooling water, then the cold wall temperature would usually be very close to the water temperature. Qdot = Hcwater * (Tcoldwall - Twater ambient), Hcwater is generally large.

If the cold wall is connected to ordinary air, and temperatures are high, then the cold wall temp will again have one solution, somewhat higher than the ambient air temperature, but with both convective and radiation heat transfer mechanisms. Qdot = Hcair * (Tcoldwall - Tair ambient) + Emissivity * (Tcoldwall^4 - Tambient^4)

For any of these cases, the fundamental equations dictate that the specific heat transfer via conduction through the insulation will be set equal to the heat removal from the cold surface via its various mechanisms. A unique Tcoldwall exists for each condition.