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Unformatted text preview: W/m 2 K. The surface emissivity of the sphere is 0.65. Calculate the total heat loss due to both convection and radiation. q = hA(T s – T air ) + σɛA(T s 4 – T sur 4 ) = (9.8)(4π × 0.02 2 )(150 – 20) + (5.67×108 )(0.65)(4π × 0.02 2 ) × (423 4 – 293 4 ) = 6.40 + 4.56 = 10.96 W 3. A wall 2.0 cm thick is to be constructed from material which has an average thermal conductivity of 1.3 W/m K. The wall is to be insulated with material having an average thermal conductivity of 0.035 W/m K, so that the heat loss will not exceed 1830 W/m 2 . Assuming that the temperature difference across the combined wall is 1270ºC, calculate the thickness of insulation required. ins w o i k x k x T T q " 035 . 3 . 1 02 . 1270 1830 x 0238 . x m 4....
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 Spring '11
 koraykadirsafak
 Heat, Heat Transfer, average thermal conductivity, maximum allowable chip, mm Coolant chip

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