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Effect of radiation shield between plates

Short answer

Net radiant heat exchange without shield: 8747 W/m². With a radiation shield (ε=0.1 on both sides) placed between them, heat exchange reduces to 926 W/m², a reduction of 89.4%. The shield temperature is 701.5 K.

Two large parallel grey plates have emissivities 0.8 and 0.5 and are at 800 K and 500 K respectively. Calculate (a) the net radiant heat exchange per square metre, (b) the heat exchange if a thin radiation shield of emissivity 0.1 on both sides is placed between them, (c) the percentage reduction in heat transfer, and (d) the temperature of the shield. Use sigma = 5.67 x 10^-8 W/m^2 K^4.

For two parallel grey plates q=σ(T14−T24)1ε1+1ε2−1q = \dfrac{\sigma(T_1^4 - T_2^4)}{\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_2}-1}.

σ(T14−T24)=5.67×10−8 (8004−5004)=5.67×10−8(4.096×1011−6.25×1010)=19 681 W/m2\sigma(T_1^4-T_2^4) = 5.67\times10^{-8}\,(800^4 - 500^4) = 5.67\times10^{-8}(4.096\times10^{11}-6.25\times10^{10}) = 19\,681\ \text{W/m}^2

(a) Without shield

q12=19 68110.8+10.5−1=19 6812.25=8747 W/m2q_{12} = \frac{19\,681}{\dfrac{1}{0.8}+\dfrac{1}{0.5}-1} = \frac{19\,681}{2.25} = 8747\ \text{W/m}^2

(b) With shield (emissivity ε3=0.1\varepsilon_3 = 0.1 on both sides)

The shield splits the space into two gaps in series: plate 1 to shield, and shield to plate 2. The total resistance is the sum of the resistances of the two gaps:

(1ε1+1ε3−1)+(1ε3+1ε2−1)=(1.25+10−1)+(10+2−1)=10.25+11=21.25\left(\frac{1}{\varepsilon_1}+\frac{1}{\varepsilon_3}-1\right)+\left(\frac{1}{\varepsilon_3}+\frac{1}{\varepsilon_2}-1\right) = (1.25+10-1)+(10+2-1) = 10.25 + 11 = 21.25 
 Plate 1    Shield    Plate 2
 800 K   |  T3  |    500 K
 e=0.8   | e=0.1|    e=0.5
qwith=19 68121.25=926 W/m2q_{with} = \frac{19\,681}{21.25} = 926\ \text{W/m}^2

(c) Reduction

8747−9268747×100=89.4%\frac{8747 - 926}{8747}\times100 = 89.4\%

(d) Shield temperature

Heat from plate 1 to shield equals qwithq_{with}:

qwith=σ(T14−T34)10.25  ⇒  T34=4.096×1011−926×10.255.67×10−8=2.422×1011q_{with} = \frac{\sigma(T_1^4-T_3^4)}{10.25}\;\Rightarrow\;T_3^4 = 4.096\times10^{11}-\frac{926\times10.25}{5.67\times10^{-8}} = 2.422\times10^{11} T3=701.5 KT_3 = 701.5\ \text{K}

Check with the other side: T34−T24=2.422×1011−6.25×1010=1.797×1011T_3^4 - T_2^4 = 2.422\times10^{11}-6.25\times10^{10} = 1.797\times10^{11}; σ×1.797×1011/11=926\sigma\times1.797\times10^{11}/11 = 926 W/m2^2. Correct.

Answer: (a) 8747 W/m2^2; (b) 926 W/m2^2; (c) 89.4 % reduction; (d) 701.5 K.

✓ Answer checked by Claude Sonnet 5.5

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