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4-53

Transient Heat Conduction in Multidimensional Systems 4-74C The product solution enables us to determine the dimensionless temperature of two- or threedimensional heat transfer problems as the product of dimensionless temperatures of one-dimensional heat transfer problems. The dimensionless temperature for a two-dimensional problem is determined by determining the dimensionlesstemperatures in both directions, and taking their product. 4-75C The dimensionless temperature for a three-dimensional heat transfer is determined by determining the dimensionless temperatures of one-dimensional geometries whose intersection is the three dimensional geometry, and taking their product. 4-76C This short cylinder is physically formed by the intersection of a long cylinder and a plane wall. Thedimensionless temperatures at the center of plane wall and at the center of the cylinder are determined first. Their product yields the dimensionless temperature at the center of the short cylinder. 4-77C The heat transfer in this short cylinder is one-dimensional since there is no heat transfer in the axial direction. The temperature will vary in the radial direction only.

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4-54

4-78 A short cylinder is allowed to cool in atmospheric air. The temperatures at the centers of the cylinder and the top surface as well as the total heat transfer from the cylinderfor 15 min of cooling are to be determined. Assumptions 1 Heat conduction in the short cylinder is two-dimensional, and thus the temperature varies in both the axial x- and the radial r- directions. 2 The thermal properties of the cylinder are constant. 3 The heat transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is τ > 0.2 so that the one-term approximatesolutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The thermal properties of brass are given to be ρ = 8530 kg/m 3 , c p = 0.389 kJ/kg ⋅ °C ,
k = 110 W/m ⋅ °C , and α = 3.39 × 10 −5 m 2 /s .

Analysis This short cylinder can physically be formed by the intersection of a long cylinder of radius D/2 = 4 cm and a plane wall of thickness 2L= 15 cm. We measure x from the midplane. (a) The Biot number is calculated for the plane wall to be

Bi =

hL (40 W/m 2 .°C)(0.075 m) = = 0.02727 k (110 W/m.°C)
Air T∞ = 20°C

D0 = 8 cm
z

The constants λ1 and A1 corresponding to this Biot number are, from Table 4-2,

L = 15 cm
r

λ1 = 0.1620 and A1 = 1.0045
The Fourier number is

Brass cylinder Ti = 150°C
= 5.424 > 0.2

τ=αt
L2

=

(3.39 × 10 −5 m 2 /s)(15 min × 60 s/min) (0.075 m) 2

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable. Then the dimensionless temperature at the center of the plane wall is determined from

θ 0, wall =

2 2 T0 − T∞ = A1 e − λ1 τ = (1.0045)e −( 0.1620) (5.424) = 0.871 Ti − T∞

We repeat the same calculations for the longcylinder,

Bi =

hro (40 W/m 2 .°C)(0.04 m) = = 0.01455 k (110 W/m.°C)

λ1 = 0.1677 and A1 = 1.0036 τ= αt
ro2 = (3.39 × 10 −5 m 2 /s)(15 × 60 s) (0.04 m) 2 = 19.069 > 0.2

θ o,cyl =

2 2 T o − T∞ = A1e − λ1 τ = (1.0036)e −( 0.1677 ) (19.069) = 0.587 Ti − T∞

Then the center temperature of the short cylinder becomes

⎡ T (0,0, t ) − T∞ ⎤ = θ o, wall × θ o,cyl = 0.871× 0.587 = 0.511 ⎢ ⎥short ⎣ Ti − T∞ ⎦ cylinder T (0,0, t ) − 20 = 0.511 ⎯ T (0,0, t ) = 86.4°C ⎯→ 150 − 20 (b) The center of the top surface of the cylinder is still at the center of the long cylinder (r = 0), but at the outer surface of the plane wall (x = L). Therefore, we first need to determine the dimensionless temperature at the surface of the wall.
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc....