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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
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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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
Transcribed Image Text: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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