p0396 - 3.96. CHAPTER 3, PROBLEM 96 457 3.96 Chapter 3,...

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Unformatted text preview: 3.96. CHAPTER 3, PROBLEM 96 457 3.96 Chapter 3, Problem 96 Problem: Identical rectangular gates of height 2H and width w out of the page separate two liquids of densities ρ and 3 ρ from a central tank of fluid of density 2ρ as shown. 2 (a) Determine the hydrostatic forces acting on each gate from the left and right sides. (b) Determine the centers of pressure on each gate from the left and right sides. (c) As the depth of the top of the gates in the central tank, h, increases, which gate will open first? HINT: A gate is on the brink of opening when the force at the stop is zero. Ar 2H 4H ρ ............ ............ ............ ....... ............ ............ ............ ............ ............ ............ ............ ............ ............ ........ ............ .............. ... . . .... ... ............ ............ ............ . .. ............ ............. ......... ........... ........... .. ........... Ar s h ........... ...... ............ ............ ......... ............ ... . ............ ............ ............ ............ ............ ............ ............ ............ ....... ............ ............ ............ ............ ............ ............ ............ ........ ............ ............ ............ 2ρ s Ar 3 ρ 3H ............................................................................................................................................................ .. .. .. ... . . . 2 ..................................................................................................... . .. .. .................................................................................................... ..... .......... .. ..................................................................................................... ........................................................................................................... ................................................................................................. ............................................................................................ .................................................................. .. . . . ............................................................................................ ............................................................................................ ............................................................................................ . .. .............. .. ................. ... ... ... ... ... ......... ... ............................... ........................................................................................................................................................................................................................................................................................................................................................ .......... . . .. . . ...... . ... ...... ............................... .. .................................................. .............................. ............. .................... ..... ............. ..... ................................................................................................................................................................................................................................................................................................................................................................................................................... ............................................................................................................................................................................................................................................................................................................................................................. . .. . . . .. . ............ ....... ....... ... .................. ........................................................ ... ............................................... ........................ .................. . .................................... ........................... ...................................... .. . S op Solution: To solve, we must first determine the hydrostatic forces acting on the gate in Part (a). Then, in Part (b), we compute the centers of pressure. Finally, in Part (c), we balance moments to determine which gate will open first. (a) Clearly, the center fluid exerts the same hydrostatic force, Fc , on both gates. There are two additional hydrostatic forces acting viz the force on the left gate from the left fluid F , and the force on the right gate from the right fluid, Fr . Since the gate is rectangular with height 2H , its centroid is at a distance H below the top of the gate. So, for the fluids to the left, center and right, respectively, the centroid locations relative to the free surface are z = 2H + H = 3H, ¯ zc = H + h, ¯ zr = H + H = 2H ¯ Also, the area and moment of inertia for the gate are A = 2Hw and I= 1 2 w(2H )3 = wH 3 12 3 Therefore, the hydrostatic forces on the gate from left, center and right fluids are F = ρg z A = ρg (3H )(2Hw) = 6ρgH 2 w ¯ 458 CHAPTER 3. EFFECTS OF GRAVITY ON PRESSURE ¯ Fc = 2ρg zc A = 2ρg (h + H )(2Hw) = 4ρgH (H + h)w 3 3 Fr = ρg zr A = ρg (2H )(2Hw) = 6ρgH 2 w ¯ 2 2 (b) The centers of pressure are 2 wH 3 I 28 3 = 3H + =H zA ¯ (3H )(2Hw) 9 2 3 wH I 1 H2 3 = zc + ¯ =H +h+ =H +h+ zc A ¯ (H + h)(2Hw) 3H +h 2 3 wH 13 I 3 = 2H + =H = zr + ¯ zr A ¯ (2H )(2Hw) 6 ( zcp) = z + ¯ (c zcp) (r zcp) (c) There are four forces acting on each gate. In addition to the hydrostatic forces from each side, there are reaction forces at the top of the gate, Ft , and at the stop, Fs . We seek the depth of the center fluid, h, at which each gate is on the brink of opening, i.e., the depth that yields Fs = 0. To find the depth, we must balance moments. It is simplest to take moments about the top of the gate (which obviates the need to determine Ft ). Left Gate: ( (c F zcp) − 2H + Fs (2H ) = Ft (0) + Fc zcp) − h Now, the centers of pressure are ( zcp) = 28 H 9 and (c zcp) = H + h + 1 H2 3H +h Thus, substituting into the moment-balance equation with Fs = 0 yields 6ρgH 2 w 1 H2 10 H = 4ρgH (H + h)w H + 9 3H +h or, 20 3H + 3h + H ρgH 3 w = 4ρgH 2 w 3 3 Solving for h, there follows 1 h= H 3 =⇒ 5H = 4H + 3h Right Gate: (c (r Fc zcp) − h + Fs (2H ) = Ft (0) + Fr zcp) − H 3.96. CHAPTER 3, PROBLEM 96 459 Now, the centers of pressure are (c zcp) = H + h + 1 H2 3H +h and (r zcp) = 13 H 6 Thus, substituting into the moment-balance equation with Fs = 0 yields 4ρgH (H + h)w H + 1 H2 3H +h = 6ρgH 2 w 7 H 6 or, 3H + 3h + H = 7ρgH 3 w =⇒ 16H + 12h = 21H 3 Solving for h, there follows 5 h= H 12 5 Hence, since 1 H < 12 H , the LEFT gate will open first. 3 4ρgH 2 w ...
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This note was uploaded on 02/09/2012 for the course AME 309 taught by Professor Phares during the Spring '06 term at USC.

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