p0168 - -500 500 1000 1500 2000 T o F 4 8 12 10 7 μ Figure...

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72 CHAPTER 1. INTRODUCTION 1.68 Chapter 1, Problem 68 1.68(a): From Sutherland’s law, the viscosity at T =68 o F = 527.67 o Ris μ = 2 . 27 · 10 8 (527 . 67) 3 / 2 527 . 67 + 198 . 6 =3 . 789 · 10 7 slug ft · sec Hence, to determine the coefficient A in the power-law viscosity, we have A (527 . 67 o R) 0 . 7 =3 . 789 · 10 7 slug ft · sec = A =4 . 708 · 10 9 slug ft · sec ( o R) 0 . 7 1.68(b): First, note that all computations must be done in terms of absolute temperature, which means that 250 o F T 2000 o F corresponds to 209 . 67 o R T 2459 . 67 o R . Figure 1.10 compares Sutherland-law and power-law viscosities, with μ in slug/ft · sec and T in degrees Fahrenheit.
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Unformatted text preview: -500 500 1000 1500 2000 T ( o F) 4 8 12 10 7 μ ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ........................................................................................................ .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. .. ...... .. .. .. .. .. .. .. .. Figure 1.10: Comparison of Sutherland-law (—–) and power-law (- - -) viscosities....
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