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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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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