Explanation Given: The temperature of the outer surface of the grill, T 1 = 47 ° C Surrounding temperature, T 2 = 27 ° C Stefan Boltzmann’s constant σ = 5.67 × 10 − 11 kWm -2 K -4 The emissivity of the body ∈ o = 0.93 Heat transfer coefficient for the convection h = 10 Wm -2 Heat transfer is the process of heat energy transfer due to temperature differences. Heat can be transferred by conduction, convection, and radiation. In convection, the transfer of heat takes place without contact between two bodies through an external medium such as air or water. The radiation heat transfer can take place without any medium. Convection heat transfer per unit area can be given as: Q A = h Δ T Q A is the rate of heat transfer per unit area in Wm-2 h is the heat transfer coefficient for the convection in Wm-2 Δ T is the temperature difference in Kelvin(K) Radiation heat transfer per unit area can be given as: P A = [ ∈ o σ ( T 1 4 − T 2 4 ) ] 1 − A 1 A 2 ( 1 − ∈ o ) 2 P A is the rate of radiation per unit area in Wm-2 ∈ o is the emissivity of the body σ is the Stefan Boltzmann’s constant in Wm- 2 K-4 A is the unit surface area of the body in m 2 Calculation: To convert Celsius degrees to Kelvin add 273 to the Celsius temperature. T = 47 + 273 T = 320 K T ' = 27 + 273 T ' = 300 K To convert the unit of convection heat transfer coefficient from Wm-2 to kWm-2 divide it by 1000 h = 0

Fundamentals Of Engineering Thermodynamics, 9th Edition Epub Reg Card Loose-leaf Print Companion Set

9th Edition

ISBN: 9781119456285

Author: Michael J. Moran

Publisher: Wiley (WileyPLUS Products)

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Chapter 2, Problem 2.41P

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Fundamentals Of Engineering Thermodynamics, 9th Edition Epub Reg Card Loose-leaf Print Companion Set

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The net rate of heat transfer between the grill hood and the surroundings by convection and radiation.

Solution Summary: The author explains that the net rate of heat transfer between the grill hood and the surroundings by convection and radiation is 0.325 kWm-2.

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