Thermodynamics HW Solutions 383

Thermodynamics HW Solutions 383 - Chapter 4 Transient Heat...

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Unformatted text preview: Chapter 4 Transient Heat Conduction 4-116E 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 geometries 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 cast iron are given to be k = 29 Btu/h.ft.°F and α = 0.61 ft2/h. Analysis After 5 minutes Plate: First the Biot number is calculated to be 2 ro hL (7 Btu/h.ft 2 .°F)(0.5 / 12 ft ) = 0.01006 Bi = = (29 Btu/h.ft.°F) k 2 ro The constants λ 1 and A1 corresponding to this Biot number are, from Table 4-1, λ1 = 0.0998 and A1 = 1.0017 The Fourier number is τ= αt L2 = (0.61 ft 2 /h)(5 min/60 min/h) (0.5 / 12 ft) 2 = 29.28 > 0.2 2L Then the center temperature of the plate becomes θ 0, wall = 2 2 T − 75 T0 − T ∞ = A1e − λ1 τ ⎯ ⎯→ 0 = (1.0017)e − (0.0998) ( 29.28) = 0.748 ⎯ T0 = 318°F ⎯→ Ti − T∞ 400 − 75 Cylinder: Bi = 0.01 ⎯Table 4⎯→ λ1 = 0.1412 and A1 = 1.0025 ⎯ ⎯ −1 θ 0,cyl = 2 2 T − 75 T0 − T∞ = A1 e −λ1 τ ⎯ ⎯→ 0 = (1.0025)e −( 0.1412) ( 29.28) = 0.559 ⎯ T0 = 257°F ⎯→ Ti − T∞ 400 − 75 Sphere: Bi = 0.01 ⎯Table 4⎯→ λ1 = 0.1730 and A1 = 1.0030 ⎯ ⎯ −1 θ 0, sph = 2 2 T − 75 T0 − T∞ = A1 e −λ1 τ ⎯ ⎯→ 0 = (1.0030)e −( 0.1730) ( 29.28) = 0.418 ⎯ T0 = 211°F ⎯→ Ti − T∞ 400 − 75 After 10 minutes τ= αt L2 = (0.61 ft 2 /h)(10 min/60 min/h) (0.5 / 12 ft) 2 = 58.56 > 0.2 Plate: θ 0, wall = 2 2 T − 75 T0 − T ∞ = A1e − λ1 τ ⎯ ⎯→ 0 = (1.0017)e − (0.0998) (58.56) = 0.559 ⎯ T0 = 257°F ⎯→ Ti − T∞ 400 − 75 4-94 ...
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