Homework _2 Solutions - Chapter 2

Homework _2 Solutions - Chapter 2 - PROBLEM 2.2 KNOWN Hot...

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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 2.4 KNOWN: Symmetric shape with prescribed variation in cross-sectional area, temperature distribution and heat rate. FIND: Expression for the thermal conductivity, k. SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in x-direction, (3) No internal heat generation. ANALYSIS: Applying the energy balance, Eq. 1.11a, to the system, it follows that, since ± ± , E E in out = ( ) x q Constant f x . = Using Fourier’s law, Eq. 2.1, with appropriate expressions for A x and T, yields ( ) ( ) x x 2 3 dT q k A dx d K 6000W=-k 1-x m 300 1 2x-x . dx m = − Solving for k and recognizing its units are W/m K, ( ) ( ) ( ) ( ) 2 2 -6000 20 k= . 1 x 2 3x 1-x 300 2 3x = + < COMMENTS: (1) At x = 0, k = 10W/m K and k as x 1. (2) Recognize that the 1-D assumption is an approximation which becomes more inappropriate as the area change with x, and hence two-dimensional effects, become more pronounced.
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PROBLEM 2.20 KNOWN: Temperature distribution, T(x,y,z), within an infinite, homogeneous body at a given instant of time. FIND: Regions where the temperature changes with time. SCHEMATIC: ASSUMPTIONS: (1) Constant properties of infinite medium and (2) No internal heat generation. ANALYSIS: The temperature distribution throughout the medium, at any instant of time, must satisfy the heat equation. For the three-dimensional cartesian coordinate system, with constant properties and no internal heat generation, the heat equation, Eq. 2.15, has the form α 2 2 2 1 T x T y T z T t 2 2 2 + + = . (1) If T(x,y,z) satisfies this relation, conservation of energy is satisfied at every point in the medium.
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