p0532 - z = h at x = 0 when the coffee is at the brink of...

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5.32. CHAPTER 5, PROBLEM 32 485 5.32 Chapter 5, Problem 32 Figure 5.8: Accelerating coffee cup. When you accelerate, the three components of the Euler equation are ρ a = p x 0= p y 0= p z ρ g Clearly, since p/ y =0 , the pressure is at most a function of x and z . Integrating first over x , p ( x, z )= ρ ax + f ( z ) where f ( z ) is a function of integration. Then, differentiating with respect to z and using the z component of Euler’s equation, we find p z = f I ( z )= ρ g = f ( z )= C ρ gz where C is a constant. Thus, the pressure throughout the coffee is p ( x, z )= C ρ ax ρ gz Onthefreesurface , p = p a . Thus, the equation of the free surface is C ρ ax ρ gz = p a Now, if we let x =0
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Unformatted text preview: z = h at x = 0 when the coffee is at the brink of spilling. Hence, C = p a + gh and ax + gz = gh 486 CHAPTER 5. CONSERVATION OF MASS AND MOMENTUM Then, at the front of the cup where x = d , gh min = gh ad = h min = h ad g Now, we are given the fact that h o = 1 2 ( h min + h ) Substituting for h min from above, h o = 1 2 X h ad g + h ~ = h o = h ad 2 g So, if the acceleration is a = g , we have h o = h 2 d But, we are given h o = 0 . 75 h , which means . 75 h = h 2 d = = 0 . 5 h d Finally, for the given values = 0 . 5 w 4 in 3 . 5 in W = 0 . 57...
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p0532 - z = h at x = 0 when the coffee is at the brink of...

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