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전류와 전압의 관계를 파악한다. , 옴의 법칙공학기술레포트 , 옴의 법칙
PROBLEM 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.
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 conduction …(생략(略)) in the x-direction, (2) Steady-state conditions, (3) Constant properties, (4) Outside wall temperature is that of the ambient air. ANALYSIS: From Fourier’s law, it is evident that the gradient, dT dx 〓 ? q′′ k , is a constant, and x hence the temperature distribution is linear, if q′′ and k are each constant. The heat flux must be x constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 〓 -15°C are
25 C ? ?15 C dT T1 ? T2 q′′ 〓 ? k 〓k 〓 1W m ? K 〓 133.3W m 2 . x dx L 0.30 m q x 〓 q′′ × A 〓 133.3 W m 2 × 20 m 2 〓 2667 W . x
Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15 ≤ T2 ≤ 38°C, with different wall thermal conductivities, k.
2500 Heat loss, qx (W)
-1500 -20 -10 0 10 20 30 40
Ambient air temperature, T2 (C) Wall t
전류와 전압의 관계를 파악한다.