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Purdue ME 606: Enclosure Characteristics (Lecture 9)

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Black Enclosures

For an enclosure with N black surfaces, the energy leaving a surface i is qi(leaving) = EbiAi. The energy arriving from another surface (noted as j) equals qi = EbjAjFji. For all N surfaces the net heat incident on surface i is,

q_i(incident)=\sum_{j=1}^N E_{bj}A_jF_{ji}.

Thus, the net energy flow to surface i in an enclosure is:

q_i(net)=A_iE_bi\sum_{j=1}^NF_{ij}~-~\sum_{j=1}^N E_{bj}A_jF_{ji},

or due to reciprocity, AjFji = AiFij,

simplifies to: q_i(net)=\sum_{j=1}^N(E_{bi}-E_{bj})A_iF_{ij} (eq. 10.1)

For an enclosure with N surfaces, there are N equations and 2N of unknown values.

To solve for the 2N unknowns you need either temperature or q at each i. If not enough information is known then a technique like least squared can be used to determine values. Finally, if just the heat flux is known for each surface then the resulting matrix would be singular.

Resistance Network Analogy

An analogy can be drawn between resistance diagrams and the heat tranfer between surfaces in an enclosure.

For simplicity the approach will be shown for an enclosure with three surfaces, but the technique can be readily extended to N surfaces.

The net heat flux between the different surfaces in this enclosure is as follows:

q_{1~to~2}=(E_{b1}-E_{b2})A_1F_{12},

q_{1~to~3}=(E_{b1}-E_{b3})A_1F_{13},

q_{2~to~3}=(E_{b2}-E_{b3})A_1F_{23}.

This is assuming that surface 1 does not emit any radiation on itself.

Recall for in a circuit, I_{1 to 2} =\frac{V_1-V_2}{R_{12}}. Extending this analogy to heat transfer, R_{12} =\frac{1}{A_iF_{ij}}.

Special Cases

These are special cases which may affect the heat flux or temperature used in circuts. Be mindful of these situations.

1) Re-radiating (refractory,adiabatic) well. Everything that is absorbed is emitted (q=0).

2) Current source provides a known heat flux in

3) Outside heating/cooling can effect the heat flux into a node on a circuit diagram, q=hA(T_inf-Ti).

4) If you have a window or opening of an enclosure the radiation leaving through the opening will be blackbody radiation. Also, eventually the temperature inside the cavity will come to the same temperature as the temperature of the surrounding air (which the opening is exposed to) assuming that there is no heating or cooling of the cavity.

Enclosures with Gray Surfaces

If the surfaces in an enclosure are gray, now we need to account for reflection from the surfaces.

In the following derivation several assumptions are made: 1) Emission and reflection diffuse 2) Intensity leaving is uniform over the surface 3) Energy arriving over the surface is uniform 4) All gases are non-participating 5) All surfaces are gray (i.e. the spectral emissivity is constant and equals the absorptivity)

Because th emitted and reflected energy is diffuse, a concept called radiosity (J) can be used:

Ji = εiEbi + ρiGi = εiEbi + (1 − εi)Gi.

This is the total rate (per area) at which radiation is leaving surface (i). This flux leaves in all directions. The net energy reaching a surface (j) from surface (i) is then JiAiFij.

Knowing this and summing over all the surfaces in the enclosure the radiosity (flux) becomes,

J_i = \epsilon_iE_{bi} + (1-\epsilon_i)\sum_{j=1}^N J_jF_{ij}.

Now assume that there is some heat flux leaving or entering the surface (e.g. not adiabatic). From a first law balance,

qi = (Ji − Gi)Ai.

Combining our knowledge of Gi relative to the radiation coming from other surfaces the heat flux to surface (i) becomes:

\frac{q_i}{A_i} =\sum_{j=1}^{N}(J_i-J_j)F_{ij},

alternatively this can be expressed as

\frac{q_i}{A_i} =J_i-\frac{J_i-\epsilon_iE_{bi}}{1-\epsilon_i}~=~ \frac{\epsilon_i}{1-\epsilon_i}(E_{bi}-J_i).

This can be related to a circuit diagram as follows:

Solving for an enclosure problem:

For N gray surfaces we have 4N unknowns and 3N independent equations. We need to know N boundary conditions.

The 3N independent equations include:

G_i=\sum_{j=1}^{N}J_jF_{ij}

J_i=\epsilon_iE_{bi}+(1-\epsilon_i)\sum_{j=1}^{N}J_jF_{ij}

\frac{q_i}{A_i} =\sum_{j=1}^{N}(J_i-J_j)F_{ij}

Another useful equation relating the heat flux and radiosity is:

\frac{q_i}{\epsilon_i}=E_{bi}-\sum_{j=1}^{N}J_jF_{i-j} ~ H_{0i} from i=1,2…N . If you eliminate J then the equation becomes:

\frac{q_i}{\epsilon_i}-\sum_{j=1}^{N}(\frac{1}{\epsilon_j}-1)F_{i-j}q_j + H_{0i} =E_{bi}-\sum_{j=1}^{N}E_{bj}F_{i-j} , where H0i is an external flux.

Last modified on 15 Oct, 2008