p1144 - 11.44 CHAPTER 11 PROBLEM 44 53 11.44 Chapter 11...

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11.44. CHAPTER 11, PROBLEM 44 53 11.44 Chapter 11, Problem 44 U ρ ,p y x ........ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .............................. . . . . . . . . . . . . ........... . . . . . . . . . . . . . . . ... .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 11.14: Flow over a surface with a source. 11.44(a): Using superposition of a uniform stream of velocity U and a source of strength Q ,wehave ψ ( r, θ )= Ur sin θ + Q 2 π θ and φ ( θ cos θ + Q 2 π f nr To verify that y =0 is a streamline, consider the negative x axis. Since θ = π ψ ( π U | x | sin π + Q 2 π π = Q 2 On the positive x axis, θ , wherefore ψ ( 0) = Ux sin 0 + Q 2 π · 0=0 Therefore, we find ψ = l Q/ 2 ,x< 0 0 ,x > 0 so that ψ does not vary along y . Thus, the surface is a coincident with the streamlines for this flow. 11.44(b): Working with the velocity potential, the velocity components are u r = ∂φ r = U cos θ + Q 2 π r and u θ = 1 r ∂θ = U sin θ Now, we know from Bernoulli’s equation that p + 1 2 ρ p u 2 r + u 2 θ Q = p + 1 2 ρ U 2 For x< 0 , u r = U cos π + Q 2 π ( x ) = w U + Q 2 π x W ,u θ = U sin π
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54 CHAPTER 11. POTENTIAL FLOW So, the pressure on the negative x axis is given by p + 1 2 ρ w U + Q 2 π x W 2 = p + 1 2 ρ U 2
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p1144 - 11.44 CHAPTER 11 PROBLEM 44 53 11.44 Chapter 11...

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