• Author / Uploaded
• Kristine Ann Villanueva

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Problem 4.10-2 Geankoplis, Transport Processes and Unit Operations A loaf of bread having a surface temperature of 373 K is being baked in an oven whose walls and air are at 477.4 K. The bread moves continuously through the large oven on an open chain belt conveyor. The emissivity of the bread is estimated as 0.85 and the loaf can be assumed a rectangular solid 114.3 mm high x 114.3 mm wide x 330 mm long. Calculate the radiation heat-transfer rate to the bread, assuming that it is small compared to the oven and neglecting natural convection heat transfer.

Given: Tb=373 K Ts=477.4 K e=0.85 h=114.3 mm w=114.3 mm l=330 mm

Required: Q Solution: Q= e σ A [Ts4 – Tb4] Where e, emissivity σ, Stefan-Boltzmann constant A, surface area Ts, temperature of surrounding Tb, temperature of body considered

A in this problem is the surface area of a rectangular solid. A= ]}mm2( A={ [ A=0.17700498 m2

Q = (0.85) (5.67x10-8 Q = 277.9869 W

) (0.17700498 m2) [(477.4K)4 – (373K)4]

)2

Problem 4.10-3 Geankoplis, Transport Processes and Unit Operations A horizontal oxidized steel pipe carrying steam and having an OD of 0.1683 m has a surface temperature of 374.9 K and is exposed to air at 297.1 K in a large enclosure. Calculate the heat loss for 0.305 m of pipe from natural convection plus radiation. For the steel pipe use an ε of 0.79. Given: OD = 0.1683 m Tb = 374.9 K Ts = 297.1 K L = 0.305 m e = 0.79 Required: Heat loss due to natural convection and radiation Solution: Qt = Qnc + Qr Where Qt, total heat Qnc, heat in convection Qr, heat in radiation

Qnc = hA(Tb-Ts) h=6.12 W/m-K obtained from natural convection correlation for h Qnc = 6.12 Qnc = 76.78295945 W Qr = e σ A [Ts4 – Tb4] Qr = 0.79 (5.67x10-8 Qr = 86.41394471 W

(374.9-297.1)K

)

Qt = 76.78295945 W + 86.41394471 W Qt = 163.1969042 W

[(374.9K)4 – (297.1K)4]