This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View Full
Document
 Spring '11
 koraykadirsafak

Click to edit the document details