Problem 2.37

# Problem 2.37 - = ν 1.33 10 5 − × m 2 s ⋅ = ν 4.129...

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Problem 2.37 [Difficulty: 2] Given: Sutherland equation Find: Corresponding equation for kinematic viscosity Solution: Governing equation: μ bT 1 2 1 S T + = Sutherland equation p ρ R T = Ideal gas equation Assumptions: Sutherland equation is valid; air is an ideal gas The given data is b 1.458 10 6 × kg ms K 1 2 = S 110.4 K = R 286.9 J kg K = p 101.3 kPa = The kinematic viscosity is ν μ ρ = μ R T p = RT p bT 1 2 1 S T + = Rb p T 3 2 1 S T + = b' T 3 2 1 S T + = where b' Rb p = b' 4.129 10 9 × m 2 K 1.5 s = b' 286.9 Nm kg K 1.458 × 10 6 × kg ms K 1 2 m 2 101.3 10 3 × N × 4.129 10 9 × m 2 sK 3 2 = = S 110.4 K = Hence ν b' T 3 2 1 S T + = with b' 4.129 10 9 × m 2 sK 3 2 =

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Check with Appendix A, Table A.10. At T0 ° C = we find T
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Unformatted text preview: = ν 1.33 10 5 − × m 2 s ⋅ = ν 4.129 10 9 − × m 2 s K 3 2 ⋅ 273.1 K ⋅ ( ) 3 2 × 1 110.4 273.1 + = ν 1.33 10 5 − × m 2 s ⋅ = Check! At T 100 °C = we find T 373.1 K = ν 2.29 10 5 − × m 2 s ⋅ = ν 4.129 10 9 − × m 2 s K 3 2 ⋅ 373.1 K ⋅ ( ) 3 2 × 1 110.4 373.1 + = ν 2.30 10 5 − × m 2 s ⋅ = Check! 20 40 60 80 100 1.5 10 5 − × 2 10 5 − × 2.5 10 5 − × Calculated Table A.10 Viscosity as a Function of Temperature Temperature (C) Kinematic Viscosity (m2/s)...
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## This note was uploaded on 10/19/2011 for the course EGN 3353C taught by Professor Lear during the Fall '07 term at University of Florida.

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Problem 2.37 - = ν 1.33 10 5 − × m 2 s ⋅ = ν 4.129...

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