17 From solution of equation for Unsteady state conduction of heat:yT - Ti = (Ts - T₁) erfc(-2√atT = Ti at y> 0 and t = 0Ts at y = 0 and t > 0T =T →>>>Ti at y → ∞o and t > 0Prove that the heat flux at the surface of the solid (y=0):k(Ts - Ti)Vπατqy= 0 =

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From solution of equation for Unsteady state conduction of heat: y 2√at T = T₁ at y> 0 and t = 0 T = Ts at y = 0 and t > 0 T →>> Ti at y → ∞ and t > 0 T T₁ = (Ts T₁) erfc(- - Prove that the heat flux at the surface of the solid (y = 0): qy = 0 k(Ts - Ti) (πατ

From solution of equation for Unsteady state conduction of heat: y 2√at T = T₁ at y> 0 and t = 0 T = Ts at y = 0 and t > 0 T →>> Ti at y → ∞ and t > 0 T T₁ = (Ts T₁) erfc(- - Prove that the heat flux at the surface of the solid (y = 0): qy = 0 k(Ts - Ti) (πατ

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Conduction and Convection

When energy travels from one object to another it is said to have undergone heat transfer. Heat transfer is nothing but the transfer of thermal energy or internal energy from one thing to another, like e.g., when a vessel with water, kept on top of a burner, heats up because of the flame underneath it.

Question

From solution of equation for Unsteady state conduction of heat:
y
T - Ti = (Ts - T₁) erfc(-
2√at
T = Ti at y> 0 and t = 0
Ts at y = 0 and t > 0
T =
T →>>>
Ti at y → ∞o and t > 0
Prove that the heat flux at the surface of the solid (y=0):
k(Ts - Ti)
Vπατ
qy
= 0 =

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