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Unformatted text preview: 5.58. CHAPTER 5, PROBLEM 58 655 5.58 Chapter 5, Problem 58 Problem: The figure depicts incompressible flow through a pipe and nozzle that emits a vertical jet. The flow is steady, irrotational, has density ρ and the only body force is gravity. What is the velocity of the jet at the nozzle exit, Uj ? To what height, zmax , will the jet of fluid rise? Express your answers in terms of ρ, g , h, Ui and pi − pa . Determine the numerical values of Uj and zmax if pi − pa = 60 kPa, h = 1 m, Ui = 4 m/sec and ρ = 1000 kg/m3 . . .. .. .............. . ................... . .......... . . ........... . . ........ . ... ...... . .. .. ........... .. .. .......... .. . .. . ...... . . ......... . ..... . . ......... . ... .. . ...... ......... .. . .. .. .......... ..... .. ... ... ........................ .. .. . ......... .. . .. . .......... ...... .. ............ . .. .. . .............. . ............. . .. . .. . max .. ............... ............... ... . .. ............. ................ ................ . .. . ... .. . . .. . ... . . .. .. . ................ . .................. . .. . . ........................... .......................... . . . . .. . .... . . . . ......... .................. . . . . .......... . . . .. .. . .................. .. .. . . .. .. . . .•. . . .. .................. ................... . .. ..... .................... ..................... .... ... ....................... ......................... . .... .... . .................... ....... .. .......................................................................................................................................................................................................................... ........................................................................................................................................................................................................................... .............................................................................................................................................................................................................................................................................................. . .... . .............. .. . ........................................................................................................................................................................................... .. . .. ... .. .......................................................................................................................................................................................... ............................................................................................................................................... .................. .. . . ........... ................ .... .. . .. .... . g . . .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. . . . z h Uj , pa Ui , pi Solution: Using Bernoulli’s equation, we know that throughout the flow 1 1 p + ρ u · u + ρgz = pi + ρ Ui2 2 2 At the nozzle exit, p = pa , and z = h. So, 1 1 pa + ρ Uj2 + ρgh = pi + ρ Ui2 2 2 Therefore, the initial jet velocity is Uj = 1 1 ρ Uj2 = pi − pa + ρ Ui2 − ρgh 2 2 =⇒ 2 pi − pa + Ui2 − 2gh ρ In the jet, p = pa , and at z = zmax we have u = 0. Hence, 1 pa + ρgzmax = pi + ρ Ui2 2 =⇒ zmax = pi − pa Ui2 + ρg 2g For the given values, Uj = 60000 N/m2 2 1000 kg/m3 +4 m sec 2 − 2 9.807 m2 m (1 m) = 10.79 sec sec and, zmax = 60000 N/m2 1000 kg/m3 9.807 m/sec2 + (4 m/sec)2 2 9.807 m/sec2 = 6.93 m ...
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