(a) Explanation Given information: Density of liquid is 850 kg / m 3 , radius of the first horizontal tube is 1.0 cm and the radius of second horizontal tube is 0.50 cm . The cross section area of first pipe is, A 1 = π r 1 2 A 1 is the cross section area of first pipe. r 1 is the radius of the first pipe. The cross section area of the second pipe is, A 2 = π r 2 2 A 2 is the cross section area of second pipe. r 2 is the radius of the second pipe. From the equation of continuity, A 1 v 1 = A 2 v 2 v 1 is the speed of liquid in first pipe. v 2 is the speed of liquid in second pipe. Substitute π r 1 2 for A 1 and π r 2 2 for A 2 in the above equation. π r 1 2 v 1 = π r 2 2 v 2 v 2 = v 1 ( r 1 r 2 ) 2 Substitute 1.0 cm for r 1 and 0.50 cm for r 2 in the above equation. v 2 = v 1 ( 1.0 cm 0.50 cm ) 2 = 4 v 1 (I) Apply Bernoulli’s equation for two pipes, P 1 + 1 2 ρ v 1 2 + ρ g h 1 = P 2 + 1 2 ρ v 2 2 + ρ g h 2 Since both the pipes are at same elevation then, h 1 = h 2 = h Therefore, P 1 + 1 2 ρ v 1 2 + ρ g h = P 2 + 1 2 ρ v 2 2 + ρ g h P 1 + 1 2 ρ v 1 2 = P 2 + 1 2 ρ v 2 2 P 1 − P 2 = 1 2 ρ ( v 2 2 − v 1 2 ) Δ P = 1 2 ρ ( v 2 2 − v 1 2 ) Substitute 4 v 1 for v 2 in the above equation (b) To determine The volume flow rate for Δ P = 6.0 kPa . (c) To determine The volume flow rate for Δ P = 12.0 kPa .

Physics for Scientists and Engineers With Modern Physics

9th Edition

ISBN: 9781133953982

Author: SERWAY, Raymond A./

Publisher: Cengage Learning

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Chapter 14, Problem 48P

(a)

To determine

The volume flow rate as a function of ΔP .

(b)

To determine

The volume flow rate for ΔP=6.0 kPa .

(c)

To determine

The volume flow rate for ΔP=12.0 kPa .

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Chapter 14, Problem 47P

Chapter 14, Problem 49P

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