Question

The overall heat transfer coefficient due to convection and radiation for a steam maintained at 200°C running in a large room at 30°C is 17.95 W/m²K. If the emissivity of the pipe surface is 0.8; the value of σ = 5.67 × 10 ^{- 8} W/m²K⁴; the heat transfer coefficient due to radiation will be nearly 
1. 17 W/m²K
2. 14 W/m²K
3. 11 W/m²K
4. 8 W/m²K

Answer 1

Correct Answer - Option 3 : 11 W/m²K

Concept:

Heat transfer by Radiation = Heat transfer by Convection

\(Q = \sigma {{\rm{A}}_{{\rm{pipe}}}}{ \in _{{\rm{pipe}}}}\left( {{\rm{T}}_{{\rm{pipe}}}^4 - {\rm{\;T}}_{{\rm{o}}}^4} \right) = {\rm{\;}}{{\rm{h}}_{{\rm{rad}}}}{{\rm{A}}_{{\rm{pipe}}}}\left( {{{\rm{T}}_{{\rm{pipe}}}} - {\rm{\;}}{{\rm{T}}_{{\rm{o}}}}} \right)\)

Calculation:

Given:

 \({ \in _{pipe}}\) = 0.8, To = 273 + 30 = 303 K, Tpipe = 273 + 200 = 473 K

Heat transfer by radiation:

∴ 5.67 ×10 ^{- 8}× 0.8 × (473⁴ – 303⁴) = h_rad × (473 – 303)

= 11.1 W/m²K