P15_089 - A 2 = A 1 / 2 treated in the Sample Problem. Note...

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89. (a) This is similar to the situation treated in Sample Problem 15-8, and we refer to some of its steps (and notation). Combining Eq. 15-35 and Eq. 15-36 in a manner very similar to that shown in the textbook, we Fnd R = A 1 A 2 s 2∆ p ρ ( A 2 1 A 2 2 ) . for the flow rate expressed in terms of the pressure di±erence and the cross-sectional areas. Note that this reduces to Eq. 15-38 for the case
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Unformatted text preview: A 2 = A 1 / 2 treated in the Sample Problem. Note that ∆ p = p 1 − p 2 = − 7 . 2 × 10 3 Pa and A 2 1 − A 2 2 = − 8 . 66 × 10 − 3 m 4 , so that the square root is well deFned. Therefore, we obtain R = 0 . 0776 m 3 /s. (b) The mass rate of flow is ρR = 68 . 9 kg/s....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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