Explanation Given: The mass flow through the tube is m ˙ = 0.8 ρ − 1 / 2 slug/s . Radius of the tank ( r ) is 1 ft. Length of the tank ( h ) is 4 ft. The mass of nitrogen in the tank ( m ) is 0.5 slugs. Calculation: Draw the fixed control volume of the tank as in Figure (1). The density of the nitrogen within the control volume changes with time and therefore contributes to local changes. Calculate the constant volume of the tank V . V = π r 2 h V = π × ( 1 ft ) 2 × 4 ft = 12.566 ft 3 Calculate the initial amount of nitrogen in the tank ρ 0 . ρ 0 = m V ρ 0 = 0.5 slug 12.566 ft 3 = 0.0397 slug / ft 3 Calculate the density of the nitrogen within the tank ρ . [ ∂ ∂ t ∫ c v ρ d V + ∫ c s ρ V ⋅ d A ] = 0 Here, differential volume is d V and the differential area is d A Substitute ∫ c s ρ V ⋅ d A = − m ˙ in the above equation. ∂ ∂ t ∫ c v ρ d V − m ˙ = 0 ∂ ρ ∂ t V − m ˙ = 0 ∂ ρ ∂ t ( 12.566 ft 3 ) − ( 0.8 ρ − 1 2 ) slug / s = 0 ∂ ρ ∂ t = 0

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 4, Problem 90P

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The density of the nitrogen within the tank.

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Chapter 4, Problem 89P

Chapter 4, Problem 91P

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Fluid Mechanics (2nd Edition)

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