A flat plate is experiencing a parallel air flow over the top surface with an average friction coefficient (CF) of 1.842×104. All properties of air are listed in the figure. What is the average convection coefficient (h) on the top surface? Please show the complete analysis procedure. u = 40 m/s Pr = 0.7 Ts > Too p=1.018 kg/m³ C = 1.842x10-4 Cp = 1009 J/kgK Modified Reynolds Analogy f Cƒ = 2St Pr²/3 (0.6 < Pr < 60) (6.70)
A flat plate is experiencing a parallel air flow over the top surface with an average friction coefficient (CF) of 1.842×104. All properties of air are listed in the figure. What is the average convection coefficient (h) on the top surface? Please show the complete analysis procedure. u = 40 m/s Pr = 0.7 Ts > Too p=1.018 kg/m³ C = 1.842x10-4 Cp = 1009 J/kgK Modified Reynolds Analogy f Cƒ = 2St Pr²/3 (0.6 < Pr < 60) (6.70)
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter5: Analysis Of Convection Heat Transfer
Section: Chapter Questions
Problem 5.25P
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
Transcribed Image Text:A flat plate is experiencing a parallel air flow over the top surface with an average friction
coefficient (CF) of 1.842×104. All properties of air are listed in the figure. What is the
average convection coefficient (h) on the top surface? Please show the complete analysis
procedure.
u = 40 m/s
Pr = 0.7
Ts > Too
p=1.018 kg/m³
C = 1.842x10-4
Cp
= 1009 J/kgK
Modified Reynolds Analogy
f
Cƒ = 2St Pr²/3 (0.6 < Pr < 60)
(6.70)
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