Kreith - cap 03

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 From Equation (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
2

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 to Equation (2.27).

k

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 differential equation 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 is therefore 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 then