ch02 - PROBLEM 2.1 KNOWN Steady-state one-dimensional heat...

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PROBLEM 2.1 KNOWN: Steady-state, one-dimensional heat conduction through an axisymmetric shape. FIND: Sketch temperature distribution and explain shape of curve. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, one-dimensional conduction, (2) Constant properties, (3) No internal heat generation. ANALYSIS: Performing an energy balance on the object according to Eq. 1.11a, ± ± , E E in out = 0 it follows that ± ± E E q in out x = and that q q x x x ± $ . That is, the heat rate within the object is everywhere constant. From Fourier’s law, q kA dT dx x x = − , and since q x and k are both constants, it follows that A dT dx Constant. x = That is, the product of the cross-sectional area normal to the heat rate and temperature gradient remains a constant and independent of distance x. It follows that since A x increases with x, then dT/dx must decrease with increasing x. Hence, the temperature distribution appears as shown above. COMMENTS: (1) Be sure to recognize that dT/dx is the slope of the temperature distribution. (2) What would the distribution be when T 2 > T 1 ? (3) How does the heat flux, ′′ q x , vary with distance?
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PROBLEM 2.2 KNOWN: Hot water pipe covered with thick layer of insulation. FIND: Sketch temperature distribution and give brief explanation to justify shape. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional (radial) conduction, (3) No internal heat generation, (4) Insulation has uniform properties independent of temperature and position. ANALYSIS: Fourier’s law, Eq. 2.1, for this one-dimensional (cylindrical) radial system has the form q kA dT dr k 2 r dT dr r r = − = − π ± 1 6 where A r and r = 2 π ± ± is the axial length of the pipe-insulation system. Recognize that for steady- state conditions with no internal heat generation, an energy balance on the system requires ² ² ² ² . E E since E E in out g st = = = 0 Hence q r = Constant. That is, q r is independent of radius (r). Since the thermal conductivity is also constant, it follows that r dT dr Constant. ± ! " $ # = This relation requires that the product of the radial temperature gradient, dT/dr, and the radius, r, remains constant throughout the insulation. For our situation, the temperature distribution must appear as shown in the sketch. COMMENTS: (1) Note that, while q r is a constant and independent of r, ′′ q r is not a constant. How does ′′ q r r 1 6 vary with r? (2) Recognize that the radial temperature gradient, dT/dr, decreases with increasing radius.
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PROBLEM 2.3 KNOWN: A spherical shell with prescribed geometry and surface temperatures. FIND: Sketch temperature distribution and explain shape of the curve. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in radial (spherical coordinates) direction, (3) No internal generation, (4) Constant properties.
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