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p1132

# p1132 - 2 = ⇒ r min = Γ 2 π ± ρ 2 p ∞ The density...

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38 CHAPTER 11. POTENTIAL FLOW 11.32 Chapter 11, Problem 32 For a potential vortex, we know that u r = 0 , u θ = Γ 2 π r So, from Bernoulli’s equation, p ρ + 1 2 p u 2 r + u 2 θ Q = p ρ + 1 2 w Γ 2 π r W 2 = p ρ = p = p 1 2 ρ w Γ 2 π r W 2 Clearly, the solution is meaningless when p < 0 , which can occur as r approaches zero. Hence, the minimum value of r corresponds to the point where p = 0, viz., 0 = p 1 2 ρ w Γ 2 π r min W 2
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Unformatted text preview: 2 = ⇒ r min = Γ 2 π ± ρ 2 p ∞ The density of mercury at 20 o C is ρ = 13550 kg/m 3 . Also, the pressure as r → ∞ is p ∞ = 100 kPa = 10 5 kg/(m · sec 2 ). Thus, for the given values, r min = 10 m 2 / sec 2 π > ² ² : 13550 kg / m 3 2 [10 5 kg / (m · sec 2 )] = 0 . 41 m...
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