p0542

# p0542 - Then from Bernoulli’s equation p B 1 2 ρ U I B 2...

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500 CHAPTER 5. CONSERVATION OF MASS AND MOMENTUM 5.42 Chapter 5, Problem 42 We must make a Galilean transformation in order to make the flow steady, wherefore we can use Bernoulli’s equation. The velocities transform as indicated in Figure 5.20. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..................... . . . . . . . . . . . . ..................................... . . . . . . ............. . . . . . . . ....... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... .... (Unsteady) U A =15 q U B =4 q U C =2 q U =0 q (Steady) U I A =0 q U I B =11 q U I C =13 q U I =15 q Figure 5.20: Galilean transformation details.
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Unformatted text preview: Then, from Bernoulli’s equation, p B + 1 2 ρ ( U I B ) 2 = p C + 1 2 ρ ( U I C ) 2 = ⇒ p B − p C = 1 2 ρ ± ( U I C ) 2 − ( U I B ) 2 = Hence, since the density of water is 1000 kg/m 3 , the pressure difference is p B − p C = 1 2 X 1000 kg m 3 ~ X 169 m 2 sec 2 − 121 m 2 sec 2 ~ = 24 · 10 3 kg m · sec 2 = 24 kPa...
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