p0560 - = ⇒ 16 gh = 2 p b − p a ρ Thus the pressure...

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5.60. CHAPTER 5, PROBLEM 60 519 5.60 Chapter 5, Problem 60 Figure 5.36: Pressurized tank with a jet. 5.60(a): From Bernoulli’s equation, we have p b + 1 2 ρ · 0 2 + ρ gh ± 1 Free surface = p a + 1 2 ρ U 2 + ρ g · 0 ± 1 Jet = U = ³ 2 gh +2 p b p a ρ Clearly, for p b = p a , the jet velocity would be 2 gh . Thus, to have triple this value, i.e., to have U =3 2 gh , 3 ´ 2 gh = ³ 2 gh +2 p b p a ρ Squaring both sides of this equation yields 18 gh =2 gh +2 p b p a ρ
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Unformatted text preview: = ⇒ 16 gh = 2 p b − p a ρ Thus, the pressure required to yield U = 3 √ 2 gh is p b = p a + 8 ρ gh 5.60(b): For the given values, we find p b = 2116 . 8 lb ft 2 + 8 X 1 . 99 slug ft 3 ~ X 32 . 174 ft sec 2 ~ (18 ft) = 11337 lb ft 2 = 5 . 36 atm U = 3 5 2 p 32 . 174 ft / sec 2 Q (18 ft) = 102 . 1 ft sec...
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