Explanation Given: Total heat transfer, Q ˙ t o t a l = 1000 W Air Temperatures T A = 2 0 ° C Surface area A s = 0.02 m 2 Emissivity ε = 0.6 Heat transfer coefficient h = 35 W/m 2 .K Concept used: The rate of heat transfer is calculated as, Q ˙ t o t a l = Q ˙ c o n v + Q ˙ r a d Here, Q ˙ t o t a l is the total heat transfer, Q ˙ c o n v is the rate of heat convection and Q ˙ r a d is the rate of heat radiation. Calculation: The total heat supplied to the iron will be dissipated to the surroundings by convention and radiation heat transfer. The rate of heat loss by convection is, Q ˙ c o n v = h A s Δ T Here, h is the heat transfer coefficient, A s is the surface area and ?T is temperature difference. Q ˙ c o n v = ( 35 ) ( 0.02 ) [ T s − ( 273 + 20 ) ] Q ˙ c o n v = 0.7 ( T s − 293 ) And, the rate of heat loss by radiation is, Q ˙ r a d = ε σ A s ( T s 4 − T R 4 ) Here, s is the Stefan Boltzmann constant and its value is σ = 5.67 × 10 − 8 W / m 2

6th Edition

ISBN: 9781260440027

Author: CENGEL

Publisher: MCG

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Chapter 1, Problem 96P

To determine

The temperature of the base of the iron.

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Chapter 1, Problem 95EP

Chapter 1, Problem 97P

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HEAT+MASS TRANSFER:FUND.+APPL.(LOOSE)

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