Thermodynamics HW Solutions 593

# Thermodynamics HW Solutions 593 - 10 2 2 conv 2 ° = ⎯→...

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Chapter 7 External Forced Convection 7-61 A cylindrical bottle containing cold water is exposed to windy air. The average wind velocity is to be estimated. Assumptions 1 Steady operating conditions exist. 2 Radiation effects are negligible. 3 Air is an ideal gas with constant properties. 4 Heat transfer at the top and bottom surfaces is negligible. Properties The properties of water at the average temperature of ( T 1 + T 2 )/2 = (3+11)/2 = 7 ° C are (Table A- 9) C J/kg. 4200 kg/m 8 . 999 3 ° = = p C ρ The properties of air at 1 atm and the film temperature of ( T s + T )/2 = (7+27)/2 = 17 ° C are (Table A-15) 7317 . 0 Pr /s m 10 489 . 1 C W/m. 02491 . 0 2 5 - = × = ° = υ k Analysis The mass of water in the bottle is kg 2.356 m)/4 (0.30 m) (0.10 ) kg/m 8 . 999 ( 4 2 3 2 = π = ρπ = ρ = L D V m Then the amount of heat transfer to the water is J 79,162 = C 3) - C)(11 J/kg. kg)(4200 356 . 2 ( ) ( 1 2 ° ° = = T T mC Q p Air V T = 27 ° C Bottle D =10 cm L = 30 cm The average rate of heat transfer is W 32 . 29 s 60 45 J 162 , 79 = × = Δ = t Q Q The heat transfer coefficient is C . W/m 55 . 15 C 7) - )(27 m (0.09425 W 32 . 29 ) ( m 0.09425 = m) m)(0.30
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Unformatted text preview: 10 . ( 2 2 conv 2 ° = ⎯→ ⎯ ° = ⎯→ ⎯ − = = = ∞ h h T T hA Q DL A s s s & ππ The Nusselt number is 42 . 62 C W/m. 0.02491 m) C)(0.10 . W/m 55 . 15 ( 2 = ° ° = = k hD Nu Reynolds number can be obtained from the Nusselt number relation for a flow over the cylinder ( ) [ ] ( ) [ ] 856 , 12 Re 000 , 282 Re 1 7317 . / 4 . 1 ) 7317 . ( Re 62 . 3 . 42 . 62 000 , 282 Re 1 Pr / 4 . 1 Pr Re 62 . 3 . 5 / 4 8 / 5 4 / 1 3 / 2 3 / 1 5 . 5 / 4 8 / 5 4 / 1 3 / 2 3 / 1 5 . = ⎯→ ⎯ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + = ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎣ ⎡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + + = Nu Then using the Reynolds number relation we determine the wind velocity m/s 1.91 V V V = ⎯→ ⎯ × = ⎯→ ⎯ υ = ∞ − ∞ ∞ /s m 10 489 . 1 ) m 10 . ( 856 , 12 Re 2 5 D 7-48...
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