A 1.5 cm diameter ball bearing at a temperature of 100 C is cooled by passing water at 27 C and a velocity of 0.3 m/s over it. Calculate heat dissapation by the bearing. The thermo-physical properties of the water are µ = 18.46 x 10“ N.s/m2, µs = 20.82 x 10 N.s/m2 p= 1.1614 kg/m', Pr 0.707 and k= 0.0263 W/m.K. Determine:

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
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Author:Kreith, Frank; Manglik, Raj M.
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Chapter5: Analysis Of Convection Heat Transfer
Section: Chapter Questions
Problem 5.65P
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A 1.5 cm diameter ball bearing at a temperature of 100°C is cooled by passing water at 27 C
and a velocity of 0.3 m/s over it. Calculate heat dissapation by the bearing. The thermo-physical
properties of the water are u = 18.46 x 10* N.s/m2, µs = 20.82 x 10 N.s/m? p= 1.1614 kg/m',
Pr 0.707 and k= 0.0263 W/m.K. Determine:
Transcribed Image Text:A 1.5 cm diameter ball bearing at a temperature of 100°C is cooled by passing water at 27 C and a velocity of 0.3 m/s over it. Calculate heat dissapation by the bearing. The thermo-physical properties of the water are u = 18.46 x 10* N.s/m2, µs = 20.82 x 10 N.s/m? p= 1.1614 kg/m', Pr 0.707 and k= 0.0263 W/m.K. Determine:
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