Example14-1(2).pdf

# Example14-1(2).pdf - EXAMPLE 4.4 Choice of Control Volume...

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Unformatted text preview: EXAMPLE 4.4 Choice of Control Volume tor Momentum Analysis Water from a stationary nozzle strikes a ﬂat plate as shown. The water leaves the nozzle at 15 mls; the nozzle area is 0.01 m2. Assuming the water is directed normal to the plate, and ﬂows along the plate, determine the horizontal force on the Nozzle support. / Plate —o EXAMPLE PROBLEM 4.4 GIVEN: Water from a stationary nozzle is directed normal to the plate; subsequent ﬂow is parallel to plate. - I v Jet velocity, V = 15? mls 3—. Nozzle area, A. = 0.01 m2 FIND: Horizontal force on the support. SOLUTION: We chose a coordinate system in deﬁning the problem above. We must now choose a suitable control volume. Two possible choices are shown by the dashed lines below. both cases. water from the nozzle crosses the control surface through area A. (assumed equal to the 1,9. . ,1 ) and is assumed to leave the control volume tangent to the plate surface in the + y or — y direction. j_.trying to decide which is the “best” control volume to use. let us write the basic equations. _==Fs+ F, = 31 VpdV+I VpV-di and 1! 9431+] pidir‘eo 3' cv cs 3' cv cs ' y s: (1) Steady ﬂow (2) Incompressible ﬂow (3) Uniform ﬂow at each section where ﬂuid crosses the CV boundaries " Iessofourchoiee of control volume. the ﬂow is steady and the basic equations become F-F‘S+F,=J we; and IpV-dft'=0 cs CS r. 'ng the momentum ﬂux term will lead to the same result for both control volumes. We should control volume that allows the most straightforward evaluation of the forces. a bet in applying the momentum equation that the force, F. represents all forces acting on the true. us solve the problem using each of the control volumes. ' volume has been selected so that the area of the left surface is equal to the area of the right surface. . 'sarea by A. ' omtrol volurm cuts through the support. We denote the components of the reaction force of the "if 3 ﬁre control volume as R, and ll’y and assume both to be positive. (The force of the control volume . is equal and opposite to R, and R,.) M1 is the reaction moment (about the z axis) from the the control volume. pheric pressure acts on all surfaces of the control volume. (The distributed force due to atmo- f m sure has been shown on the vertical faces only.) find)! force on the control volume is denoted as W. Z , _‘ we are looking for the horizontal force. we write the x component of the steady ﬂow momentum F5, +F5. =J MPV'dK CS There are no body forces in the x direction, so F a. - 0, and F5, I I upV - (1; cs To evaluate F5" we must include all surface forces acting on the control volume FS. = PaA - PaA + Rx force due to atmospheric force due to atmospheric force of support on pressure acts to right pressure acts to left control volume (positive direction) on (negative direction) on (assumed positive) left surface right surface Consequently. F3, = R3, and _ ... -. _ -. - For mass crossing top and bottom Rx ‘ JCS “PV 4‘4 ' L “9V “"4 [surfaces, u = 0. ] At V-dK= - v dA.sin edirection =[ ul-lqudAI} [ 9" - "’ ' ' ° ] 4| Of V] anddAl are [80° apart. = -u1|pV|A|I {properties uniform over A1} 2 . 2 -_lSm 999§x159x0mm Ns {u;=15mls} s m3 3 kg - m R, = —2.25 kN {Rx acts opposite to positive direction assumed} The horizontal force on the support is K, = -R, = 2.25 W [force our support acts to the right} K. CV" with Horizontal Forces Shown The control volume has been selected so the areas of the left surface and of the right surface are equal to the area of the plate. Denote this area by A,. The control volume is in contact with the plate over the entire plate surface. We denote the horizontal reaction force from the plate on the control volume as B. (and assume it to be positive). Atmospheric pressure acts on the left surface of the control volume (and on the two horizontal surfaces). The body force on this control volume has no component in the x direction. Then the A: component of the momentum equation. F5, = J upV-dI-l. cs yields upV-di- I u{-lpV1dAl} = -2.25 kN Then . . .‘ thenetforoc ontheplate. wcneedaﬁBe-bOdydiam ofthe plate: 2F; =0: -Bx_p¢Ap+R; R; a p‘AP + B; . R, '- pail, + ("paAp - 2.25 W) I *225 ILN ﬁlﬁhﬂ‘izonlalforceonthewmort is K. =- -R, s 2.25 kN. Note that the choice of CW resulted in the need for an additional free-body diagram In general it is : 1 us to select the control volume so that the force sought acts explicitly on the control volume. problem illustrates the application of the momentum equation to an inertial control volume, with -, -. -: is on choosing a suitable control volume. ...
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