Physics Solution Manual for 1100 and 2101

# The lift force then is the pressure difference times

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Unformatted text preview: ing the equation of continuity, as given by Equation 11.9: A1v1 = A2v2 or π r12v1 = π r22v2 or v2 = r12v1 r22 Chapter 11 Problems 601 Here, we have used that fact that the area of a circle is A = π r . Substituting this result for v2 into Bernoulli’s equation, we find that 2 r14 2 P = P2 + ρ 4 − 1 v1 1 r 2 1 2 Taking the density of water to be ρ = 1.00 × 103 kg/m3 (see Table 11.1), we find that the absolute pressure of the water in the pipe is r14 2 P = P2 + ρ 4 − 1 v1 1 r 2 1 2 = 1.01 × 105 Pa + 1 2 ( ) ( 1.9 × 10−2 m 3 3 1.00 × 10 kg/m −3 4.8 × 10 m ( ) − 1 ( 0.62 m/s)2 = 4 ) 4 1.48 × 105 Pa 65. REASONING We assume that region 1 contains the constriction and region 2 is the normal region. The difference in blood pressures between the two points in the horizontal artery is 2 2 given by Bernoulli’s equation (Equation 11.12) as P2 − P = 1 ρ v1 − 1 ρ v2 , where v1 and 1 2 2 v2 are the speeds at the two points. Since the volume flow rate is the same at the two points, the speed at 1 is related to the speed at 2 by Equation 11.9, the equation of continuity: A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the artery. By combining these two relations, we will be able to determine the pressure difference. SOLUTION Solving the equation of continuity for the blood speed in region 1 gives v1 = v2A2/A1. Substituting this result into Bernoulli’s equation yields P2 − P = 1 Since A1 = 1 4 1 ρ v2 1 2 − 1 ρv2 2 2 v A = 1 ρ 2 2 2A 1 2 − 1 2 2 ρ v2 A2 , the pressure difference is 2 P2 − P = 1 1 2 v A ρ 12 2 − A 4 2 = 1 2 (1060 kg/m3 ) ( 0.11 m/s )2 (15) = 96 Pa 1 2 2 ρ v2 = 1 2 We have taken the density ρ of blood from Table 11.1. 2 ρ v2 (16 − 1) 602 FLUIDS 66. REASONING The volume of water per second leaking into the hold is the volume flow rate Q. The volume flow rate is the product of the effective area A of the hole and the speed v1 of the water entering the hold, Q = Av1 (Equation 11.10). We can find the speed v1 with the aid of Bernoulli’s equation. SOLUTION According to Bernoulli’s equation, which relates the pressure P, water speed v, and elevation y of two points in the water: 2 2 P + 1 ρ v1 + ρ gy1 = P2 + 1 ρ v2 + ρ gy2 12 2 (11.11) In this equation, the subscript “1” refers to the point below the surface where the water enters the hold, and the subscript “2” refers to a point on the surface of the lake. Since the amount of water in the lake is large, the water level at the surface drops very, very slowly as water enters the hold of the ship. Thus, to a very good approximation, the speed of the water at the surface is zero, so v2 = 0 m/s. Setting P1 = P2 (since the empty hold is open to the atmosphere) and v2 = 0 m/s, and then solving for v1, we obtain v1 = 2 g ( y2 − y1 ) . Substituting v1 = 2 g ( y2 − y1 ) into Equation 11.10, we find that the volume flow rate Q of the water entering the hold is ( Q = Av1 = A 2 g ( y2 − y1 ) = 8.0 × 10−3 m 2 ) 2 (9.80 m/s2 ) ( 2.0 m ) = 5.0 ×10−2 m3/s 67. REASONING The pressure P, the fluid speed v, and the elevation y at any two points in an ideal fluid of density ρ are related by Bernoulli’s equation: 2 2 P + ρ v1 + ρ gy1 = P2 + ρ v2 + ρ gy2 (Equation 11.11), where 1 and 2 denote, respectively, 1 1 2 1 2 the first and second floors. With the given data and a density of ρ = 1.00 × 103 kg/m3 for water (see Table 11.1), we can solve Bernoulli’s equation for the desired pressure P2. SOLUTION Solving Bernoulli’s equation for P2 and taking the elevation at the first floor to be y1 = 0 m , we have ( ) 2 2 P2 = P + ρ v1 − v2 + ρ g ( y1 − y2 ) 1 1 2 = 3.4 × 105 Pa + 1 2 (1.00 ×103 kg/m3 ) ( 2.1 m/s )2 − (3.7 m/s )2 ( )( ) + 1.00 ×103 kg/m3 9.80 m/s 2 ( 0 m − 4.0 m ) = 3.0 × 105 Pa Chapter 11 Problems 68. REASONING a. The drawing shows two 1 points, labeled 1 and 2, in the fluid. Point 1 is at the top of the water, and point 2 is where it flows out of the dam at the bottom. Bernoulli’s y equation, Equation 11.11, 1 can be used to determine the speed v2 of the water exiting the dam. 603 = 1 atmosphere v1 = 0 m/s P1 2 P2 = 1 atmosphere v2 y2 b. The number of cubic meters per second of water that leaves the dam is the volume flow rate Q. According to Equation 11.10, the volume flow rate is the product of the cross-sectional area A2 of the crack and the speed v2 of the water; Q = A2v2. SOLUTION a. According to Bernoulli’s equation, as given in Equation 11.11, we have 2 2 P + 1 ρ v1 + ρ gy1 = P2 + 1 ρ v2 + ρ gy2 12 2 Setting P1 = P2, v1 = 0 m/s, and solving for v2, we obtain ( ) v2 = 2 g ( y1 − y2 ) = 2 9.80 m/s 2 (15.0 m ) = 17.1 m/s b. The volume flow rate of the water leaving the dam is ( ) Q = A2v2 = 1.30 ×10−3 m 2 (17.1 m/s ) = 2.22 ×10−2 m3 /s (11.10) 69. SSM REASONING Since the pressure difference is known, Bernoulli's equation can be used to find the speed v 2 of the gas in the pipe. Bernoulli's equation also contains the unknown speed v 1 of the g...
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