समान उत्सर्जकता 0.5 के दो लंबे समानांतर प्लेटों को अलग-अलग तापमान पर बनाए रखा जाता है और उनके बीच विकिरण ऊष्मा विनिमय होता है। बीच में रखा उत्सर्जकता 0.25 का विकिरण कवच विकिरण ऊष्मा विनिमय को _______ तक कम करेगा।

This question was previously asked in
BPSC AE Paper 5 (Mechanical) 2019 Official Paper
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  1. 1/2
  2. 1/4
  3. 3/10
  4. 3/5

Answer (Detailed Solution Below)

Option 3 : 3/10
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Detailed Solution

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स्पष्टीकरण: -

हम जानते हैं कि,

समानांतर प्लेटों के बीच विकिरण ऊष्मा विनिमय

\(Q = \frac{{\sigma \times \left( {\;T_1^4 - T_2^4\;} \right)}}{{\left( {\;\frac{1}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} - 1\;} \right)}}\)

विकिरण ऊष्मा विनिमय जब प्लेटों के बीच विकिरण कवच डाला जाता है

\(Q = \frac{{\sigma \times \left( {\;T_1^4 - T_2^4\;} \right)}}{{\left( {\;\frac{1}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} + \frac{2}{{{\varepsilon _3}}} - 2\;} \right)}}\)

गणना:-

दिया हुआ:-

\({\varepsilon _1} = \;{\varepsilon _2} = \varepsilon = 0.5\;\left( {{\rm{\;emissivity\;of\;plates\;}}} \right)\)

\({\varepsilon _3} = 0.25\left( {{\rm{\;emissivity\;of\;radiation\;shield\;}}} \right)\;\)

\({Q_1} = \frac{{\sigma ~ \times \left( {\;T_1^4 - T_2^4\;} \right)}}{{\left( {\;\frac{1}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} - 1\;} \right)}}\)

\({Q_1} = \frac{{\sigma ~ \times \left( {\;T_1^4 - T_2^4\;} \right)}}{{\left( {\;\frac{1}{{0.5}} + \frac{1}{{0.5}} - 1\;} \right)}}\)

\({Q_1} = \frac{{\sigma \times \left( {\;T_1^4 - T_2^4\;} \right)}}{3}\)

अभी,

यदि प्लेटों के बीच विकिरण कवच डाला जाता है

\({Q_2} = \frac{{\sigma \times \left( {\;T_1^4 - T_2^4\;} \right)}}{{\left( {\;\frac{1}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}} + \frac{2}{{{\varepsilon _3}}} - 2\;} \right)}}\)

\({Q_2} = \;\frac{{\sigma \; \times \;\left( {\;T_1^4 - T_2^4\;} \right)}}{{\left( {\;\frac{1}{{0.5}} + \frac{1}{{0.5}} + \frac{2}{{0.25}} - 2\;} \right)}}\)

\({Q_2} = \frac{{\sigma \times \left( {\;T_1^4 - T_2^4\;} \right)}}{{10}}\)

\(\frac{{{Q_2}}}{{{Q_1}}} = \frac{3}{{10}}\)

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