A heater rod (10 mm diameter, 100 mm length) of emissivity 0.75 is enclosed within a hollow cylindrical vacuum chamber (50 mm diameter, 100 mm length) of emissivity 0.25. The entire setup is insulated at the top and bottom ends by a low emissivity material, preventing any conductive heat dissipation from the ends. The heater rod is known to have a surface temperature of 1000 K, while the vacuum chamber is at a surface temperature of 300 K. How much heat is dissipated from the heater rod to the vacuum chamber (W)

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shallomisaiah19 shallomisaiah19 · Answer 1 · 2020-06-11T02:40:30+02:00

Answer:

Explanation:

Given that:

Heater temperature ,T₁ = 1000K

Vaccum Chamber ,T₂ = 300K

emissivity of heater E₁ = 0.75

emissivity vaccum E₂ = 0.25

Heater diameter d₁ = 10 * 10⁻³mm

vaccum chamber d₂ = 50 * 10⁻³mm

When there is vaccum, then no air resistance will be there,

F₁₂ = 1

F₁₁ = 0

[tex]R_1= \frac{1-E_1}{E_1A_1} \\\\=\frac{1-0.75}{0.75*\pi * 10^-^2*L}[/tex]

[tex]R_2=\frac{1}{F_1_2 * A_1} \\\\=\frac{1}{1* \pi *10^-^2*L}[/tex]

[tex]R_3=\frac{1-0.25}{F_1_2 * A_1} \\\\=\frac{1}{0.25* \pi *5*10^-^2*L}[/tex]

Heat leaving from heater surface 1 to vaccum

[tex]Q_1_2 = \frac{L \pi \sigma (T_1^4- T_2^4)}{R_1+R_2+R_3}[/tex]

[tex]Q_1_2 = \frac{1000*10^-^3*\pi * 5.67*10^-^8(1000^4-300^4)}{\frac{0.25}{0.75*10^-2}+\frac{1}{10^-2} +\frac{0.75}{0.25*10^-^2*5} }[/tex]

[tex]Q_1_2 = \frac{1000*10^-^3*\pi * 5.67*10^-^8(1000^4-300^4)} {0.3333+1+0.6}\\\\Q_1_2= 91.39 \text {watt}[/tex]