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fluid_mechanics

claudia.marcela.becerra.rativa
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212 Chapter 5 ■ Finite Control Volume Analysis When the gate is closed, the horizontal forces acting on the contents of the control volume are identified in Fig. E5.16c. Application of Eq. 5.22 to the contents of this control volume yields 0 1no flow2 (1) Note that the hydrostatic pressure force, gH 2 b2, is used. From Eq. 1, the force exerted on the water by the gate 1which is equal to the force necessary to hold the gate stationary2 is R x 1 2 gH 2 b which is equal in magnitude to the hydrostatic force exerted on the gate by the water. When the gate is open, the horizontal forces acting on the contents of the control volume are shown in Fig. E5.16d. Application of Eq. 5.22 to the contents of this control volume leads to cs urV nˆ dA 1 2 gH 2 b R x 1 2 gh 2 b F f Note that because the water at sections (1) and (2) is flowing along straight, horizontal streamlines, the pressure distribution at those locations is hydrostatic, varying from zero at the free surface to g times the water depth at the bottom of the channel (see Chapter 3, Section 3.4). Thus, the pressure forces at sections (1) and (2) (given by the pressure at the centroid times the area) are gH 2 b2 and gh 2 b2, respectively. Also, the frictional force between the channel bottom and the water is specified as F f . The surface integral in Eq. 3 is nonzero only where there is flow across the control surface. With the assumption of uniform velocity distributions, cs urV nˆ dA 1u 1 2r1u 1 2Hb 1u 2 2r1u 2 2hb Thus, Eqs. 3 and 4 combine to form ru 2 1Hb ru 2 2hb 1 2 gH 2 b R x 1 2 gh 2 b F f (2) (3) (4) (5) cs urV nˆ dA 1 2 gH 2 b R x If the upstream velocity, is much less than so that the Closed sluice gate Control volume H Control volume H Open sluice gate h z x u (a) Control volume Water only F I G U R E E5.16 Control volume H h, u 1 , contribution of the incoming momentum flow to the control surface integral can be neglected and from Eq. 5 we obtain R x 1 2 gH 2 b 1 2 gh 2 b F f ru 2 2hb By using the continuity equation, m # bHu 1 bhu 2 , Eq. (6) can be rewritten as R x 1 2 H 2 b 1 2 h 2 b F f m # 1u 2 u 1 2 Hence, since u 2 7 u 1 , by comparing the expressions for R x (Eqs. 2 and 7) we conclude that the reaction force between the gate and the water (and therefore the anchoring force required to hold the gate in place) is smaller when the gate is open than when it is closed. (Ans) (b) Water only u 1 R x 1_ ( 2 γ H ) Hb Section (2) u 2 1_ ( 2 γ H ) Hb 1_ Section F f ( ) (c) R x (1) (d) 2 γ h hb u 2 (6) (7) The linear momentum equation can be written for a moving control volume. V5.9 Jelly fish All of the linear momentum examples considered thus far have involved stationary and nondeforming control volumes which are thus inertial because there is no acceleration. A nondeforming control volume translating in a straight line at constant speed is also inertial because there is no acceleration. For a system and an inertial, moving, nondeforming control volume that are both coincident at an instant of time, the Reynolds transport theorem 1Eq. 4.232 leads to D Dt sys Vr dV 0 0t cv Vr dV cs VrW nˆ dA When we combine Eq. 5.23 with Eqs. 5.19 and 5.20, we get 0 0t cv Vr dV cs VrW nˆ dA a F contents of the control volume (5.23) (5.24) When the equation relating absolute, relative, and control volume velocities 1Eq. 5.142 is used with Eq. 5.24, the result is 0 0t cv 1W V cv 2r dV cs 1W V cv 2rW nˆ dA a F contents of the control volume (5.25)

5.2 Newton’s Second Law—The Linear Momentum and Moment-of-Momentum Equations 213 For a constant control volume velocity, V cv , and steady flow in the control volume reference frame, 0 0t cv 1W V cv 2r dV0 Also, for this inertial, nondeforming control volume (5.26) cs 1W V cv 2rW nˆ dA cs WrW nˆ dA V cv cs rW nˆ dA (5.27) For steady flow 1on an instantaneous or time-average basis2, Eq. 5.15 gives cs rW nˆ dA 0 (5.28) Combining Eqs. 5.25, 5.26, 5.27, and 5.28, we conclude that the linear momentum equation for an inertial, moving, nondeforming control volume that involves steady 1instantaneous or timeaverage2 flow is The linear momentum equation for a moving control volume involves the relative velocity. Example 5.17 illustrates the use of Eq. 5.29. cs WrW nˆ dA a F contents of the control volume (5.29) E XAMPLE 5.17 Linear Momentum—Moving Control Volume GIVEN A vane on wheels moves with constant velocity V 0 when a stream of water having a nozzle exit velocity of V 1 is turned 45° by the vane as indicated in Fig. E5.17a. Note that this is the same moving vane considered in Section 4.4.6 earlier. The speed of the water jet leaving the nozzle is 100 fts, and the vane is moving to the right with a constant speed of 20 fts. FIND Determine the magnitude and direction of the force, F, exerted by the stream of water on the vane surface. Nozzle A 1 = 0.006 ft 2 45° Moving vane V 1 V0 F I G U R E E5.17 (a) V CV = V 0 Moving R control z volume z 1 ft Moving vane Nozzle x R x V 1 V0 (1) w (2) (c) (b)

5.2 Newton’s Second Law—The Linear Momentum and Moment-of-Momentum Equations 213<br />

For a constant control volume velocity, V cv , and steady flow in the control volume reference frame,<br />

0<br />

0t cv<br />

1W V cv 2r dV0<br />

Also, for this inertial, nondeforming control volume<br />

(5.26)<br />

cs<br />

1W V cv 2rW nˆ dA cs<br />

WrW nˆ dA V cv cs<br />

rW nˆ dA<br />

(5.27)<br />

For steady flow 1on an instantaneous or time-average basis2, Eq. 5.15 gives<br />

cs<br />

rW nˆ dA 0<br />

(5.28)<br />

Combining Eqs. 5.25, 5.26, 5.27, and 5.28, we conclude that the linear momentum equation for<br />

an inertial, moving, nondeforming control volume that involves steady 1instantaneous or timeaverage2<br />

flow is<br />

The linear momentum<br />

equation for a<br />

moving control volume<br />

involves the<br />

relative velocity.<br />

Example 5.17 illustrates the use of Eq. 5.29.<br />

cs<br />

WrW nˆ dA a F contents of the<br />

control volume<br />

(5.29)<br />

E XAMPLE 5.17<br />

Linear Momentum—Moving Control Volume<br />

GIVEN A vane on wheels moves with constant velocity V 0<br />

when a stream of water having a nozzle exit velocity of V 1 is<br />

turned 45° by the vane as indicated in Fig. E5.17a. Note that<br />

this is the same moving vane considered in Section 4.4.6<br />

earlier. The speed of the water jet leaving the nozzle is 100 fts,<br />

and the vane is moving to the right with a constant speed of<br />

20 fts.<br />

FIND Determine the magnitude and direction of the force, F,<br />

exerted by the stream of water on the vane surface.<br />

Nozzle<br />

A 1 = 0.006 ft 2 45°<br />

Moving<br />

vane<br />

V 1<br />

V0<br />

F I G U R E E5.17<br />

(a)<br />

V CV = V 0<br />

Moving<br />

R<br />

control<br />

z<br />

volume<br />

z<br />

1 ft Moving<br />

vane<br />

Nozzle<br />

x<br />

R x<br />

V 1 V0 (1)<br />

w<br />

(2)<br />

(c)<br />

(b)

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