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Emissive and Absorptive Power

The top of an...

Question

The top of an insulated cylindrical container is covered by a disc having emissivity 0.6, conductivity $0.167 W K^{- 1} m^{- 1}$ and thickness 1 cm. The temperature is maintained by circulating oil.
(a) Find the radiation loss to the surroundings in $J m^{- 2} s^{- 1}$ if temperature of the upper surface of the disc is $127^{o} C$ and temperature of surroundings is $27^{o} C$ .
(b) Also find the temperature of the circulating oil. Neglect the heat loss due to convection.

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Solution

By Stephan Boltzman law ,

$\frac{P}{A} = e σ (T^{4} - T_{S}^{4})$

Hence, $\frac{P}{A} = (0.6) (5.67 X 10^{- 8}) (400^{4} - 300^{4}) = 595 J / m^{2} s$

Now, let the temperature of coil be T. Since the temperature is constant and steady state is achieved, heat coming in must equal heat radiated out. Hence,

$595 A = \frac{0.167 A}{0.01} (T - 127)$

$T = {162.3}^{0} C$

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Similar questions

0.6

0.167

W K^{- 1} m^{- 1}

J m^{- 2} s^{- 1}

127^{\circ}

27^{\circ}

127^{\circ}

27^{\circ}

σ = \frac{17}{3} \times 10^{- 8} W / m^{2} K^{4}

127^{\circ}

27^{\circ}

σ = \frac{17}{3} \times 10^{- 8} W / m^{2} K^{4}

K = 0.149

/ (m -^{o} C

t = 5

= 0.6

T_{l} = 127^{o} C

T a = 27^{o}

J s^{- 1} m^{- 2}

5 \times 10^{- 3} m

T

127^{\circ} C

27^{\circ} C

J s^{- 1} m^{- 2})

T

^{\circ} C)

0.149 W m^{- 1} C^{- 1}

0.6

σ = \frac{17}{3} \times 10^{- 8} W m^{- 2} K^{- 4}

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