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Thus the rate of energy transfer by conduction through the rod is (20.15) Substances that are good thermal conductors have large thermal conductivity val- ues, whereas good thermal insulators have low thermal conductivity values. Table 20.3 For a compound slab containing several materials of thicknesses L 1 , L 2 , . . . and thermal conductivities k 1 , k 2 , . . . , the rate of energy transfer through the slab at steady state is (20.16) where T c and T h are the temperatures of the outer surfaces (which are held constant) and the summation is over all slabs. Example 20.9 shows how this equation results from ! ! A(T h & T c ) ' i (L i /k i ) ! ! k
A $ T h & T c L % S E C T I O N 2 0 . 7 • Energy Transfer Mechanisms 625 Example 20.9 Energy Transfer Through Two Slabs Two slabs of thickness L 1 and L 2 and thermal conductivities k 1 and k 2 are in thermal contact with each other, as shown in Figure 20.12. The temperatures of their outer surfaces c and T h , respectively, and T h ( T c . Determine the tem- perature at the interface and the rate of energy transfer by Solution To conceptualize this problem, notice the phrase when the system is in steady state. We categorize this as a The rate at which energy is transferred through slab 2 is When a steady state is reached, these two rates must be Solving for T gives (3) T ! Substituting Equation (3) into either Equation (1) or Equa- (4) ! ! To finalize this problem, note that extension of this proce- A(T h & T c ) (L 1 /k 1 ) " (L 2 /k 2 ) k 1 L 2 T c " k 2 L 1 T h k 1 L 2 " k 2 L 1 k 1 A $ T & T c L 1 % ! k 2 A $ T h & T L 2 % (2)
! 2 ! k 2 A $ T h & T L 2 % (1)
! 1 ! k 1 A $ T & T c L 1 % L
2 L
1 T
h k
2 k
1 T
c T Figure 20.12 (Example 20.9) Energy transfer by conduction through two slabs in thermal contact with each other. At steady state, the rate of energy transfer through slab 1 equals the rate of energy transfer through slab 2. |