A very long tunnel has the shape of an isosceles triangle as shown in Fig. 10.7. The tunnel has a thin partition wall down the middle as shown. Determine all the relevant angle factors for diffuse radiant interchange among the surfaces that make up the tunnel. Also, bring out the effect of shadowin

Question

A very long tunnel has the shape of an isosceles triangle as shown in Fig. 10.7. The tunnel has a thin partition wall down the middle as shown. Determine all the relevant angle factors for diffuse radiant interchange among the surfaces that make up the tunnel. Also, bring out the effect of shadowing by comparing [latex]F_{AB−DC}[/latex] in the presence and absence of the partition.

Accepted Answer

In all, there are 8 surfaces in this two-dimensional enclosure, considering the two sides of the partition (represented by [latex]DE_{l}[/latex] facing ABD and [latex]DE_{r}[/latex] facing ACD) and the gap (represented by [latex]AE_{l}[/latex] facing ABD and [latex]AE_{r}[/latex] facing ACD) as individual surfaces.

Step 1 The The lengths of all the sides are first calculated. Since the triangle ABC is an isosceles triangle, AB = AC. Also [latex]BD = DC = \frac{BC}{2} =\frac{2}{2} =1 m[/latex] since the partition runs down the middle of the tunnel. Then [latex]AB = AC = \sqrt{AD^{2}+BD^{2} } =\sqrt{2^{2}+1^{2}} = 2.236 m[/latex] . In the calculations, we also require BE and CE that are equal to each other and given by [latex]BE = CE = \sqrt{DE^{2}+BD^{2} } =\sqrt{1^{2}+1^{2}} = 1.414 m[/latex]. Now we are ready to calculate the angle factors.

Step 2 Since all surfaces are flat, all self-angle factors are zero. Because of symmetry, we note the following:

[latex]F_{BD−AB} = F_{DC−AC} F_{BD−DE_{l}} = F_{DC−DE_{r}} [/latex]

[latex]F_{BD−AC} = F_{DC−AB} F_{AB−DE_{l}}= F_{AC−DE_{r}} [/latex]

[latex]F_{AB−BD} = F_{AC−DC} F_{AB−AC}= F_{AC−AB}[/latex]

[latex]F_{AB−DC} = F_{AC−BD}[/latex]

Reciprocity will indicate other similar equalities. The reader may draw up the complete list based on these. We look at individual angle factors now.

Step 3 Angle factor [latex]F_{AB−BD}[/latex] :
This may be obtained by applying the triangle formula to triangle ABD as

[latex]F_{AB-BD} =\frac{AB+BD-AD}{2AB} =\frac{2.236+1-2}{2 imes 2.236} =0.276[/latex]

Step 4 Angle factor [latex]F_{AB−DE_{l}}[/latex] :
This requires the application of the decomposition rule.We determine first the angle factors [latex]F_{AB−AD_{l}}[/latex] and [latex]F_{AB−AE_{l}}[/latex] . The former is obtained by the use of the sum rule applied to surface AB.

[latex]F_{AB−AD_{l}} = 1 − F_{AB−BD} = 1 − 0.276 = 0.724[/latex]

The latter is obtained by the use of the triangle formula applied to triangle ABE.

[latex]F_{AB-AE_{l} } =\frac{AB+AE_{l}-BE}{2AB} =\frac{2.236+1-1.141}{2 imes 2.236} =0.407[/latex]

Using the decomposition rule, we get

[latex]F_{AB-DE_{l} } = F_{AB−AD_{l}} − F_{AB−AE_{l}} = 0.724 − 0.407 = 0.317[/latex]

Step 5 Angle factor [latex]F_{BD-AC }= F_{DC−AB}[/latex] :

From the figure, it is clear that this angle factor is the same as [latex]F_{BD-AE_{l} } = F_{DC−AE_{r}}[/latex] which may be obtained again by the use of the decomposition rule as done below. Applying the triangle formula to triangle ABD, we have

[latex]F_{BD-AD_{l} } =\frac{BD+AD_{l}-AB}{2 imes BD} =\frac{1+2-2.236}{2 imes 1} =0.382[/latex]

Applying the triangle formula to triangle BED, we have

[latex]F_{BD-ED_{l} } =\frac{BD+ED_{l}-BE}{2 imes BD} =\frac{1+1-1.141}{2 imes 1} =0.293[/latex]

By the decomposition rule, we then have

[latex]F_{BD-AE_{l} } = F_{BD-AD_{l} } − F_{BD-ED_{l} } = 0.382 − 0.293 = 0.089[/latex]

We thus have [latex]F_{BD-AC } = F_{DC-AB } = 0.089[/latex]. Using the reciprocity rule, we then have

[latex]F_{AB-DC }=F_{AC-BD } =\frac{DC}{AB}F_{DC-AB }=\frac{1}{2.236} imes 0.089 =0.040[/latex]

Step 6 Angle factor [latex]F_{AB-AC } = F_{AC-AB}[/latex] :
By inspection of the figure, it is clear that [latex]F_{AB-AE_{l} } = F_{AB−(AC+DC) } = 0.407[/latex] as obtained earlier. By decomposition rule, we then have

[latex]F_{AB-AC } = F_{AB−(AC+DC) } − F_{AB−DC} = 0.407 − 0.040 = 0.367[/latex]

Alternately, this angle factor may be determined by using the crossed string method. Assume that a string is tightly stretched between B and C. Obviously, the string will be represented by BEC. AB − BEC − CA may be treated as a triangle that has three convex sides. The triangle rule may be applied to this to get

[latex]F_{AB-AC }=\frac{AB+AC-(BE+EC)}{2AB}[/latex]

[latex]=\frac{2AB-2BE}{2AB}=\frac{2.236-1.141}{2.236}=0.367[/latex]

Table 10.1 summarizes the results for this example.

Table 10.1 Angle factors in Example 10.2

[latex]F_{AB-BD }[/latex]	0.276	[latex]F_{BD-AB}[/latex]	0.618	[latex]F_{DE_{l}-AB}[/latex]	0.707
[latex]F_{AC-CD }[/latex]		[latex]F_{DC-AC }[/latex]		[latex]F_{DE_{r} -AC }[/latex]
[latex]F_{AB-DE_{l} }[/latex]	0.317	[latex]F_{BD-DE_{l} }[/latex]	0.293	[latex]F_{DE_{l} -BD }[/latex]	0.293
[latex]F_{AC-DE_{r} }[/latex]		[latex]F_{DC-DE_{r}}[/latex]		[latex]F_{DE_{r} -DC }[/latex]
[latex]F_{AB-AC}[/latex]	0.367	[latex]F_{BD-AC}[/latex]	0.089
[latex]F_{AC-AB }[/latex]		[latex]F_{DC-AB}[/latex]
[latex]F_{AB-DC }[/latex]	0.04
[latex]F_{AC-BD }[/latex]

Step 7 Angle factor [latex]F_{AB-DC }[/latex] intheabsenceofthepartition :

We need to use the decomposition rule to obtain the desired angle factor. By applying the triangle formula to ABC, we get

[latex]F_{AB-BC } =\frac{AB+BC-AC}{2AB}=\frac{2.236+2-2.236}{2 imes 2.236}=0.447[/latex]

Applying the triangle formula to triangle ABD, we get

[latex]F_{AB-BD } =\frac{AB+BD-AD}{2AB}=\frac{2.236+1-2}{2 imes 2.236}=0.276[/latex]

By the decomposition rule we then get, in the absence of the partition,

[latex]F_{AB-DC } = F_{AB-BC } − F_{AB-BD } = 0.447 − 0.276 = 0.171[/latex]

Step 8 Observation: On comparison, it is seen that the the effect of the partition (shadowing body) is to reduce the angle factor between AB and DC from
0.171 to a mere 0.04!

Question 10.2: A very long tunnel has the shape of an isosceles triangle as...

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