p0544 - 5.44. CHAPTER 5, PROBLEM 44 639 5.44 Chapter 5,...

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Unformatted text preview: 5.44. CHAPTER 5, PROBLEM 44 639 5.44 Chapter 5, Problem 44 Problem: Body A travels through kerosene at a constant speed of UA = 20 ft/sec. Velocities at points B and C are induced by the moving body and have magnitudes of UB = 7 ft/sec and UC = 3 ft/sec. If the density of kerosene is 1.58 slug/ft3 and effects of gravity can be ignored, what is pB − pC in psi? UB U ... .................. .. . ......... .. . A ............................... ... .................................... . ..................................... ................................... ...... ......................................... ......................................... .. .......................................... . ................................................................................. .. ........................................... . ... ... .. . . . .................................................................. . . ................................. ..................... ........................................... .............. .. .... .......................................... . ....................................... ........................................ ... .................... . ........................................ .................................. .... .................................. ................................ .. ........................ ... .............. ...... ... .... UC . .. .. .. .. .. .. .. .. . . ............ ........... .. .. .. .. .. .. .. .. .. ...... ...... . • B • A • C Solution: 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 the figure below. ........ . ........... ............... .... .... ... ................. .................................. . ................................... ... ............ .... ....................................... .......................................... A . ............................................... .. . ............................................ . . ............................................ .............................................................. . . ... . ... . . .. .. . . .............................................. ............... . . ............................................... ............................................. .. . ............................................ . . . . .. . . ............................................ ... .......................................... ..................................... ..................................... ... ........................................................ .... .......................... .. .......... ... .... ... . U = 20 UB = 7 . . •.................. • ... .... ..... .. ........ ..... ..... .......................... ............................ . .................................... .. ... ...................................... ......................................... .. ........................................... ............................................ .............................................. . . ... .. . .............................................. . .............................................. . .............................................. . . .............................................. .............................................. . ............................................ ........................................... .. . ......................................... ...................................... .................................... . .................................... ............................... .................... .......... . ...... . UC = 3 . . •......... U∞ = 0 • (Unsteady) UA = 0 UB = 13 .................• . ................... . . • ........ UC = 17 • .. ........................ ....................... U∞ = 20 • . . ............................ ............................ (Steady) Then, from Bernoulli’s equation, 1 1 2 2 pB + ρ (UB ) = pC + ρ (UC ) 2 2 =⇒ 1 2 2 pB − pC = ρ (UC ) − (UB ) 2 Hence, since the density of kerosene is 1.58 slug/ft3 , the pressure difference is pB − pC = 1 slug 1.58 3 2 ft 289 ft2 ft2 − 169 sec2 sec2 = 94.8 slug = 0.66 psi ft · sec2 ...
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