Question

A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 900 K, T2 = 300 K and emissivities ε1 = 0.3 and ε2 =0.7, respectively. Thin aluminum sheet has an ε3 of 0.1 facing on the sides of plate 1 and ε3 of 0.2 facing on the sides of plate 2. Determine (a) the net radiation heat transfer between the two plates per unit surface area of the plates (b) compare the result to that without the shield. (c) temperature of the radiation shield in steady operation

Answer 1

The net radiation heat transfer between the two plates per unit surface area of the plates with shield and without shied are respectively; 2282.76 W/m² and 9766.75 W/m²

How to find the net radiation heat transfer?

We are given;

Temperature 1; T₁

Temperature 2; T₂

Temperature 3; T₃

Emissivity 1; ε₁ = 0.3

Emissivity 2; ε₂ = 0.7

Emissivity 3; ε₃ = 0.2

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with shield is;

Q'₁₂ = σ(T₁⁴ - T₂⁴)]/[((1/ε₁) + (1/ε₂) - 1) + ((1/ε₃,₁) + (1/ε₃,₂) - 1)]

Q'₁₂ = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[((1/0.3) + (1/0.7) - 1) + ((1/0.15) + (1/0.15) - 1)]

Q'₁₂,shield = 2282.76 W/m²

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with no shield is;

Q'₁₂,no shield = σ(T₁⁴ - T₂⁴)]/((1/ε₁) + (1/ε₂) - 1))

Q'₁₂,no shield = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[(1/0.3) + (1/0.7) - 1)]

Q'₁₂,no shield = 9766.75 W/m²

Then the ratio of radiation heat transfer for the two cases becomes;

Q'₁₂,shield/Q'₁₂,no shield = 2282.76/9766.75 = 0.2337 or 4/17

Read more about Net Radiation Heat Transfer at; brainly.com/question/14148915

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