Kreith - cap 03

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Chapter 3
PROBLEM 3.1 Show that in the limit Δx → 0, the difference equation for one-dimensional steady conduction with heat generation, Equation (3.1), is equivalent to the differential equation, Equation (2.27). GIVEN • One dimensional steady conduction with heat generation SHOW (a) In the limit of small Δx, the difference equation is equivalent to the differential equation SOLUTION FromEquation (3.1) Ti + 1 – 2Ti + Ti – 1 = − By definition Ti – 1 = T (x – Δx) Ti = T (x) Ti + 1 = T (x + Δx) so we can rewrite Equation (3.1) as follows
T ( x + Δx ) − 2T ( x ) + T ( x − Δx ) Δx

Δx 2  qG ,i k

= −

 qG ( x ) k d 2T so we dx 2

Now, in the limit Δx → 0, from calculus, the left hand side of the above equation becomes have
d 2T  = − qG ( x ) dx 2 which is equivalent toEquation (2.27).


PROBLEM 3.2 ‘What is the physical significance of the statement that the temperature of each node is just the average of its neighbors if there is no heat generation’ [with reference to Equation (3.2)]? SOLUTION The significance is that in regions without heat generation, the temperature profile must be linear. Compare the subject equation with the solution of the differentialequation
d 2T =0 dx 2 which is T(x) = a + bx, which is also linear.
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PROBLEM 3.3 Give an example of a practical problem in which the variation of thermal conductivity with temperature is significant and for which a numerical solution istherefore the only viable solution method. SOLUTION From Figure 1.6, the thermal conductivity of stainless steel (either 304 or 316) is a fairly strong function of temperature. For example kss 316 (100°C) = 14.2 (W/m K) kss 316 (500°C) = 19.6 (W/m K) which is about a 38% difference. Suppose a stainless steel sheet is to receive a heat treatment that involves heating the sheet to 500°C and thenplunging it into a water bath. The water near the sheet would probably boil producing a sheet surface temperature near 100°C while the interior of the sheet would be at 500°C, at least for a short time. One would expect the large variation in thermal conductivity to be important in this type of problem. PROBLEM 3.4 Discuss advantages and disadvantages of using a large control volume. SOLUTION Theadvantages of a large control volume are (1) the numerical solution can be carried out quickly (2) manual calculation for all control volumes are feasible for the purpose of verifying the numerical calculation (3) energy will be conserved Disadvantages are (1) large temperature gradients cannot be accurately represented with large control volumes (2) it is difficult to accommodate all but rectangulargeometries. PROBLEM 3.5 For one-dimensional conduction, why are the boundary control volumes half the size of interior control volumes? GIVEN • One-dimensional conduction EXPLAIN (a) Why the boundary control volume is half the size of internal control volumes SOLUTION There is a node on the boundary as well as one a distance Δx to the interior of the boundary. Since the interior nodes arecentered within a control volume of width Δx, the control volume associated with the first non-boundary node comes within Δx/2 of the boundary. So, there is a volume of only Δx/2 left over for the boundary node. PROBLEM 3.6 Discuss advantages and disadvantages of two methods for solving one-dimensional steady conduction problems.

© 2011 Cengage Learning. All Rights Reserved. May not be scanned,copied or duplicated, or posted to a publicly accessible website, in whole or in part.

SOLUTION The two methods for solving one-dimensional steady conduction problems are matrix inversion and iteration. Matrix inversion requires that we have some method (usually software) for inverting the matrix or for solving a tridiagonal system of equations. The method is difficult to apply to problems...
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