Thermodynamics HW Solutions 381

Thermodynamics HW Solutions 381 - Chapter 4 Transient Heat...

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Unformatted text preview: Chapter 4 Transient Heat Conduction 4-115E A plate, a long cylinder, and a sphere are exposed to cool air. The center temperature of each geometry is to be determined. Assumptions 1 Heat conduction in each geometry is one-dimensional. 2 The thermal properties of the bodies 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 approximate solutions (or the transient temperature charts) are applicable (this assumption will be verified). Properties The properties of bronze are given to be k = 15 Btu/h.ft.°F and α = 0.333 ft2/h. Analysis After 5 minutes Plate: First the Biot number is calculated to be Bi = hL (7 Btu/h.ft 2 .°F)(0.5 / 12 ft ) = 0.01944 = (15 Btu/h.ft.°F) k 2 ro The constants λ 1 and A1 corresponding to this Biot number are, from Table 4-1, 2 ro λ1 = 0.1410 and A1 = 1.0033 The Fourier number is τ= αt L2 = (0.333 ft 2 /h)(5 min/60 min/h) (0.5 / 12 ft) 2 = 15.98 > 0.2 2L Then the center temperature of the plate becomes θ 0, wall = 2 2 T − 75 T0 − T∞ = A1 e −λ1 τ ⎯ ⎯→ 0 = (1.0033)e − (0.1410) (15.98) = 0.730 ⎯ T0 = 312°F ⎯→ Ti − T∞ 400 − 75 Cylinder: Bi = 0.02 ⎯Table 9⎯→ λ1 = 0.1995 and A1 = 1.0050 ⎯ ⎯ −1 θ 0,cyl = 2 2 T − 75 T 0 − T∞ = A1 e − λ1 τ ⎯ ⎯→ 0 = (1.0050)e − (0.1995) (15.98) = 0.532 ⎯ T0 = 248°F ⎯→ Ti − T∞ 400 − 75 Sphere: Bi = 0.02 ⎯Table 9⎯→ λ1 = 0.2445 and A1 = 1.0060 ⎯ ⎯ −1 θ 0, sph = 2 2 T − 75 T 0 − T∞ = A1 e − λ1 τ ⎯ ⎯→ 0 = (1.0060)e − (0.2445) (15.98) = 0.387 ⎯ T0 = 201°F ⎯→ Ti − T∞ 400 − 75 After 10 minutes τ= αt L2 = (0.333 ft 2 /h)(10 min/60 min/h) (0.5 / 12 ft) 2 = 31.97 > 0.2 Plate: θ 0, wall = 2 2 T − 75 T0 − T∞ = A1 e −λ1 τ ⎯ ⎯→ 0 = (1.0033)e − (0.1410) (31.97 ) = 0.531 ⎯ T0 = 248°F ⎯→ Ti − T∞ 400 − 75 4-92 ...
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This note was uploaded on 01/19/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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