By Je-Chin Han

Entrance conceal; Contents; Preface; bankruptcy 1. warmth Conduction Equations; bankruptcy 2. 1-D Steady-State warmth Conduction; bankruptcy three. 2-D Steady-State warmth Conduction; bankruptcy four. temporary warmth Conduction; bankruptcy five. Numerical research in warmth Conduction; bankruptcy 6. warmth Convection Equations; bankruptcy 7. exterior pressured Convection; bankruptcy eight. inner pressured Convection; bankruptcy nine. traditional Convection; bankruptcy 10. Turbulent

Chapter 14. Radiation move via GasesAppendix A: Mathematical family and features; again Cover. Read more...

summary: entrance hide; Contents; Preface; bankruptcy 1. warmth Conduction Equations; bankruptcy 2. 1-D Steady-State warmth Conduction; bankruptcy three. 2-D Steady-State warmth Conduction; bankruptcy four. brief warmth Conduction; bankruptcy five. Numerical research in warmth Conduction; bankruptcy 6. warmth Convection Equations; bankruptcy 7. exterior compelled Convection; bankruptcy eight. inner compelled Convection; bankruptcy nine. typical Convection; bankruptcy 10. Turbulent stream warmth move; bankruptcy eleven. basic Radiation; bankruptcy 12. View issue; bankruptcy thirteen. Radiation alternate in a Nonparticipating Medium.

Chapter 14. Radiation move via GasesAppendix A: Mathematical kinfolk and features; again disguise

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**Extra resources for Analytical Heat Transfer**

**Example text**

14) has a cylindrical cap of radius ro and height L, and is attached to a base plate at temperature Tb . Show that the heat dissipated is qf = 2πkro t Tb − T∞ m I0 (mro ) sinh mL + I1 (mro ) cosh mL I0 (mro ) cosh mL + I1 (mro ) sinh mL where the metal thickness is t , m = (h/kt )1/2 , and the heat transfer coefficient on the sides and top is assumed to be the same, h, and outside temperature, T∞ . 14 A hollow transistor modeled as a fin. 37 1-D Steady-State Heat Conduction SOLUTION The cap is split into two fins, (1) a straight fin of width 2πro and length L, and (2) a disk fin of radius ro .

Performing an energy balance for the heater, E˙in = E˙out , it follows that qh (2πr2 ) = qi + qo = + Th − T∞,i (hi 2πr1 )−1 + (ln(r2 /r1 )/2πkB ) + Rtc,B Th − T∞,o (ho 2πr3 )−1 + (ln(r3 /r2 )/2πkA ) + Rtc,A c. 2. 5a. The side surfaces lose heat by convection to a liquid at temperature T∞ . Obtain the steady-state temperature distributions for the following cases: a. q˙ = q˙ o 1 − (x/L)2 , with x measured from the centerplane. b. q˙ = a + b(T − T∞ ) SOLUTION a. q˙ = q˙ o 1 − (x/L)2 x 2 q˙ o d2 T =− 1− 2 k L dx q˙ o 1 x3 dT =− x− dx k 3 L2 T =− q˙ o k + C1 1 2 1 x4 x − 2 12 L2 + C1 x + C2 33 1-D Steady-State Heat Conduction dT = 0 (symmetry); x = L, dx Boundary conditions: x = 0, −k dT = h (T − T∞ ) dx Substituting the boundary conditions into the above equation to replace C1 and C2 , T − T∞ = q˙ o L2 2k 5 1 x 4 x 2 4 + − + 6 6 L L 3Bi b.

34 by setting θ = T − T∞ − qr h 4 ), the above energy balance However, if we consider qr = −εσ(T 4 − Tsur equation can be rewritten as qx − qx + dqx 4 )=0 dx − hAs (T − T∞ ) − εσAs (T 4 − Tsur dx 27 1-D Steady-State Heat Conduction 2 )(T + T ), the above equation can be If we let T∞ = Tsur , hr = εσ(T 2 + T∞ ∞ written as d2 (T − T∞ ) (h + hr )P − (T − T∞ ) = 0 kAc dx2 The solution of the above equation can be obtained by numerical integration. 4 Conduction through Fins with Variable Cross-Sectional Area: Bessel Function Solutions Most often, we would like to have a fin with decreasing fin cross-sectional area in the heat conduction direction in order to save material costs.