461_PartUniversity Physics Solution

461_PartUniversity Physics Solution - Fluid Mechanics...

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Fluid Mechanics 14-25 EXECUTE: (a) 22 3 33 1 3 3 2 ( ) 2(9.80 m/s )(8.00 m)(0.0160 m ) 0.200 m /s. vA g y y A =− = = (b) Since 3 p is atmospheric, the gauge pressure at point 2 is 2 2 3 23 1 3 2 11 8 () 1 , 9 A pv vv ρ gy y A ρρ ⎛⎞ ⎜⎟ = = ⎝⎠ using the expression for 3 v found above. Substitution of numerical values gives 4 2 6.97 10 p Pa. EVALUATE: We could also calculate 2 p by applying Bernoulli&s equation to points 1 and 2. 14.90. IDENTIFY: Apply Bernoulli&s equation to the air in the hurricane. SET UP: For a particle a distance r from the axis, the angular momentum is L mvr = . EXECUTE: (a) Using the constancy of angular momentum, the product of the radius and speed is constant, so the speed at the rim is about 30 (200 km/h) 17 km/h. 350 = (b) The pressure is lower at the eye, by an amount 2 32 2 3 m / s (1.2 kg/m )((200 km/h) (17 km/h) ) 1.8 10 Pa. . 6 k m / h p Δ= = × (c) 2 160 m 2 v g = . (d) The pressure difference at higher altitudes is even greater. EVALUATE: According to Bernoulli&s equation, the pressure decreases when the fluid velocity increases. 14.91. IDENTIFY: Apply Bernoulli&s equation and the equation of continuity. SET UP: Example 14.8 shows that the speed of efflux at point D is 1 2 gh . EXECUTE: Applying the equation of continuity to points at C and D gives that the fluid speed is 1 8 gh at C . Applying Bernoulli&s equation to points A and C gives that the gauge pressure at C is 1 , 43 ρ gh ρ gh ρ gh −= and this is the gauge pressure at the surface of the fluid at E . The height of the fluid in the column is 21 3. hh = EVALUATE: The gauge pressure at C is less than the gauge pressure 1 gh ρ at the bottom of tank A because of the speed of the fluid at C . 14.92. IDENTIFY: Apply Bernoulli&s equation to points 1 and 2. Apply 0 p pg h =+ to both arms of the U-shaped tube in order to calculate h . SET UP: The discharge rate is 2 2 vA vA = . The density of mercury is m 13.6 10 kg/m and the density of water is w 1.00 10 kg/m . Let point 1 be where 42 1 40.0 10 m A and point 2 is where 2 10.0 10 m A . 12 yy = . EXECUTE: (a) 1 6.00 10 kg/m 1.50 m/s v × == × . 2 6.00 m/s v × × (b) 111 2 2 2 y v y v ++ = . 3 2 2 4 ( ) (1000 kg/m )([6.00 m/s] [1.50 m/s] ) 1.69 10 Pa pp − = (c) 1w 2m p gh p gh += + and 4 2 mw 1.69 10 Pa 0.137 m 13.7 cm. ( ) (13.6 10 kg/m 1.00 10 kg/m )(9.80 m/s ) h g −× = = × EVALUATE: The pressure in the fluid decreases when the speed of the fluid increases. 14.93. (a) IDENTIFY: Apply constant acceleration equations to the falling liquid to find its speed as a function of the distance below the outlet. Then apply Eq.(14.10) to relate the speed to the radius of the stream. SET UP: Let point 1 be at the end of the pipe and let point 2 be in the stream of liquid at a distance 2 y below the end of the tube, as shown in Figure 14.93. Figure 14.93 Consider the free-fall of the liquid. Take + y to be downward. Free-fall implies . y ag = y v is positive, so replace it by the speed v .
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14-26 Chapter 14 EXECUTE: 22 21 0 2( ) vv a yy =+ gives 2 2 vv g y and 2 2 2. vvg y Equation of continuity says 11 2 2 vA vA = And since 2 Ar π = this becomes vr vr = and 2 1 2 (/). vv r r = Use this in the above to eliminate 2 : v 2 11 2 1 2 (/) 2 vrr v g y / 4 2111 2 /( 2 ) rrvv g y To correspond to the notation in the problem, let 10 = and , rr =
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461_PartUniversity Physics Solution - Fluid Mechanics...

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