열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera 업로드
열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera 업로드
열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera
열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera
열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed IncroperaPROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal
conductivity k and inner temperature, T1. FIND: The outer temperature of the wall, T2. SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law,
q cond = q x = q ′′ ? A = -k x
Solving for T2 gives
T ?T dT ? A = kA 1 2 . dx L
T2 = T1 ?
q cond L . kA
Substituting numerical values, find
T2 = 415 C -
3000W × 0.025m 0.2W / m ? K × 10m2
T2 = 415 C - 37.5 C T2 = 378 C.
COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.
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PROBLEM 1.2
KNOWN: Inner surface temperature and thermal conductivity of a concrete wall. FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from -15 to 38°C. SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional condu
자료출처 : http://www.ALLReport.co.kr/search/Detail.asp?pk=11020346&sid=sanghyun7776&key=
[문서정보]
문서분량 : 500 Page
파일종류 : PDF 파일
자료제목 : 열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera
파일이름 : 열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera.pdf
키워드 : 열전달,5판,솔루션,Introduction,to,the,Heat,Transfer,5th,ed
자료No(pk) : 11020346
열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera
열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera
열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed IncroperaPROBLEM 1.1 KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal
conductivity k and inner temperature, T1. FIND: The outer temperature of the wall, T2. SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) Constant properties. ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law,
q cond = q x = q ′′ ? A = -k x
Solving for T2 gives
T ?T dT ? A = kA 1 2 . dx L
T2 = T1 ?
q cond L . kA
Substituting numerical values, find
T2 = 415 C -
3000W × 0.025m 0.2W / m ? K × 10m2
T2 = 415 C - 37.5 C T2 = 378 C.
COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.
<
PROBLEM 1.2
KNOWN: Inner surface temperature and thermal conductivity of a concrete wall. FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from -15 to 38°C. SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional condu
자료출처 : http://www.ALLReport.co.kr/search/Detail.asp?pk=11020346&sid=sanghyun7776&key=
[문서정보]
문서분량 : 500 Page
파일종류 : PDF 파일
자료제목 : 열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera
파일이름 : 열전달 5판 솔루션 Introduction to the Heat Transfer 5th ed Incropera.pdf
키워드 : 열전달,5판,솔루션,Introduction,to,the,Heat,Transfer,5th,ed
자료No(pk) : 11020346
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