Derive the expression for the temperature profile and heat transfer for the four Conditions cosh mL-1sinh mLHeat Transfer at the base of the fincosh mL-1,sinhmLHeat Transfer from the fin9x=0= √hPKA9₁ =√√hPKA9x=L=9x=0-9f=√√hPkA00₂-0₂(0₁ + 0₂)-% cosh mLsinh mL-0₂sinh[m(L−x)]+/%_sinh mxsinh mL

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cosh mL-1 sinhmL Heat Transfer at the base of the fin 9x=0=√√hPKA -cosh mL-1 (₁+0) sinh mL Heat Transfer from the fin 9=√hPkA cosh mL-1 00 9x=L=9x=0-9f=√√hPkA - cosh mL sinhmL -0₂ sinh[m/L_x)]+0 sinh mà

cosh mL-1 sinhmL Heat Transfer at the base of the fin 9x=0=√√hPKA -cosh mL-1 (₁+0) sinh mL Heat Transfer from the fin 9=√hPkA cosh mL-1 00 9x=L=9x=0-9f=√√hPkA - cosh mL sinhmL -0₂ sinh[m/L_x)]+0 sinh mà

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

Free undamped vibrations

Whenever an object/system displaces in either side of its equilibrium position with the help of an external/internal means, then the object/system moves regularly or randomly on both sides of its equilibrium position. This motion is called vibration. Generally, there are three types of vibration: longitudinal vibration, transverse vibration, and torsional vibration.

Question

Derive the expression for the temperature profile and heat transfer for the four Conditions

cosh mL-1
sinh mL
Heat Transfer at the base of the fin
cosh mL-1,
sinhmL
Heat Transfer from the fin
9x=0= √hPKA
9₁ =√√hPKA
9x=L=9x=0-9f=√√hPkA
0
0₂
-0₂
(0₁ + 0₂)
-% cosh mL
sinh mL
-0₂
sinh[m(L−x)]+/%_sinh mx
sinh mL

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