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More than 40 million students have trusted Schaum’s Outlines for their expert knowledge and helpful solved problems. Written by renowned experts in their respective fields, Schaum’s Outlines cover ...
ESASFull description
More than 40 million students have trusted Schaum’s Outlines for their expert knowledge and helpful solved problems. Written by renowned experts in their respective fields, Schaum’s Outlines…Full description
More than 40 million students have trusted Schaum’s Outlines for their expert knowledge and helpful solved problems. Written by renowned experts in their respective fields, Schaum’s Outlines…Descrição completa
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More than 40 million students have trusted Schaum’s Outlines for their expert knowledge and helpful solved problems. Written by renowned experts in their respective fields, Schaum’s Outlines…Full description
solved problems in heat transferFull description
More than 40 million students have trusted Schaum’s Outlines for their expert knowledge and helpful solved problems. Written by renowned experts in their respective fields, Schaum’s Outlines…Full description
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Problem 4.10-2
G e an an k o p l i s , Tr Tr a n s p o r t P r o c e s s e s a n d U n i t O p e r a t i o n s
A loaf of bread having a surface temperature 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 -Boltzmann constant A, surface area area Ts, temperature of surrounding surrounding Tb, temperature of body considered
A in this problem problem is the surface area area of a rectangular rectangular solid. A=
2 4 4 Q = (0.85) (5.67x10 -8 ) (0.17700498 m ) [(477.4K) – (373K) ] Q = 277.9869 W
2
Problem 4.10-3
G e an k o p l i s , Tr a n s p o r t P r o c e s s e s a n d U n i t O p e r a t i o n s
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: Q t = 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
(374.9-297.1)K
Qnc = 76.78295945 W
Qr = e σ A [Ts4 – Tb4] ) Qr = 0.79 (5.67x10 -8
[(374.9K) – (297.1K) ]
Qr = 86.41394471 W
Qt = 76.78295945 W + 86.41394471 W Qt = 163.1969042 W
