Physics Solution Manual for 1100 and 2101

# The siphon will stop working when v2 0 ms or y 0 m ie

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Unformatted text preview: as in the Venturi meter; therefore, we must first express v 1 in terms of v 2 . This can be done by using Equation 11.9, the equation of continuity. 604 FLUIDS SOLUTION a. From the equation of continuity (Equation 11.9) it follows that v1 = A2 / A1 v 2 . Therefore, 0.0700 m 2 v1 = v = ( 1.40 ) v 2 0.0500 m 2 2 c h Substituting this expression into Bernoulli's equation (Equation 11.12), we have 2 1 1 P1 + 2 ρ (1.40 v 2 ) 2 = P2 + 2 ρ v 2 Solving for v 2 , we obtain v2 = 2 ( P2 − P1 ) ρ (1.40 ) − 1 2 = 2( 120 Pa) (1.30 kg / m 3 ) (1.40) 2 − 1 = 14 m / s b. According to Equation 11.10, the volume flow rate is Q = A2 v 2 = (0.0700 m 2 )( 14 m / s) = 0.98 m 3 / s 70. REASONING The gauge pressure in the reservoir is the pressure difference P2 − P 1 between the reservoir (P2) and the atmosphere (P1). The muzzle is open to the atmosphere, so the pressure there is atmospheric pressure. Because we are ignoring the height difference between the reservoir and the muzzle, this is an example of fluid flow in a horizontal pipe, for which Bernoulli’s equation is 2 2 P + 1 ρ v1 = P2 + 1 ρ v2 12 2 (11.12) Solving Equation 11.12 for the gauge pressure P2 − P yields 1 2 2 2 P2 − P = 1 ρ v1 − 1 ρ v2 = 1 ρ v1 1 2 2 2 (1) where we have used the fact that the speed v2 of the water in the reservoir is zero. We will determine the speed v1 of the water stream at the muzzle by considering its subsequent projectile motion. The horizontal displacement of the water stream after leaving the muzzle is given by x = v0 xt + 1 axt 2 (Equation 3.5a). With air resistance neglected, the water stream 2 undergoes no horizontal acceleration (ax = 0 m/s2). The horizontal component v0x of the initial velocity of the water stream is identical with the velocity v1 in Equation (1), so Equation 3.5a becomes Chapter 11 Problems ( ) x = v1t + 1 0 m/s 2 t 2 2 or v1 = x t The vertical displacement y of the water stream is given by 605 (2) y = v0 yt + 1 ayt 2 2 (Equation 3.5b), where ay = −9.8 m/s2 is the acceleration due to gravity, with upward taken as the positive direction. The velocity of the water at the instant it leaves the muzzle is horizontal, so the vertical component of its velocity is zero. This means that we have v0y = 0 m/s in Equation 3.5b. Solving Equation 3.5b for the elapsed time t, we obtain y = ( 0 m/s ) t + 1 a y t 2 2 or t= 2y ay (3) SOLUTION Substituting Equation (3) into Equation (2) gives v1 = ay x =x 2y 2y ay (4) Substituting Equation (4) for v1 into Equation (1), we obtain the gauge pressure P2 − P : 1 2 2 ay ay ρ x ay 1 ρx 1 ρ x2 =2 P2 − P = 2 = 1 2y 4y 2y (1.000 × 103 kg/m3 ) ( 7.3 m )2 ( −9.80 m/s2 ) = 1.7 × 105 Pa = 4 ( −0.75 m ) 71. REASONING AND SOLUTION a. Taking the nozzle as position 1 and the top of the tank as position 2 we have, using Bernoulli's equation, P1 + (1/2)ρv12 = P2 + ρgh, we can solve for v1 so that 2 v1 = 2[(P2 − P1)/ρ + gh] v1 = 2[(5.00 × 105 Pa)/(1.00 × 103 kg/m3 ) + (9.80 m/s 2 )(4.00 m)] = 32.8 m/s 606 FLUIDS 2 b. To find the height of the water use (1/2)ρv1 = ρgh so that 2 2 2 h = (1/2)v1 /g = (1/2)(32.8 m/s) /(9.80 m/s ) = 54.9 m 72. REASONING In level flight the lift force must balance the plane’s weight W, so its magnitude is also W. The lift force arises because the pressure PB beneath the wings is greater than the pressure PT on top of the wings. The lift force, then, is the pressure difference times the effective wing surface area A, so that W = (PB – PT)A. The area is given, and we can determine the pressure difference by using Bernoulli’s equation. SOLUTION According to Bernoulli’s equation, we have 2 2 PB + 1 ρ vB + ρ gyB = PT + 1 ρ vT + ρ gyT 2 2 Since the flight is level, the height is constant and yB = yT, where we assume that the wing thickness may be ignored. Then, Bernoulli’s equation simplifies and may be rearranged as follows: 2 2 2 2 PB + 1 ρ vB = PT + 1 ρ vT or PB − PT = 1 ρ vT − 1 ρ vB 2 2 2 2 Recognizing that W = (PB – PT)A, we can substitute for the pressure difference from Bernoulli’s equation to show that ( ) 2 2 W = ( PB − PT ) A = 1 ρ vT − vB A 2 = 1 2 (1.29 kg/m3 ) ( 62.0 m/s )2 − (54.0 m/s )2 (16 m2 ) = 9600 N We have used a value of 1.29 kg/m3 from Table 11.1 for the density of air. This is an approximation, since the density of air decreases with increasing altitude above sea level. 73. REASONING The top and bottom surfaces of the roof are at the same height, so we can 2 2 use Bernoulli’s equation in the form of Equation 11.12, P + 1 ρ v1 = P2 + 1 ρ v2 , to 12 2 determine the wind speed. We take point 1 to be inside the roof and point 2 to be outside the roof. Since the air inside the roof is not moving, v1 = 0 m/s. The net outward force ΣF acting on the roof is the difference in pressure P1 − P2 times the area A of the roof, so ΣF = (P1 − P2)A. Chapter 11 Problems 607 SOLUTION Setting v1 = 0 m/s in Bernoulli’s equation and solving it for the speed v2 of the wind, we obtain 2 ( P − P2 ) 1 v2 = ρ Since the pressure difference is eq...
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