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Engineering Chemical Engineering Fundamentals of Momentum, Heat, and Mass Transfer

The problem is to determine the pressure drop for a pipe of 50 ft The pressure drop for a pipe of 50 ft is 13.37 lb/ft 2 . The problem was solved by the following procedures Determination of the velocity from the volumetric flow rate Reynold’s number, for the flow pattern determination The pressure drop in the pipe due to its length. Given Information: Volumetric flow rate = 10 gal/h Kinematic viscosity = 0.08 ×10 -3 ft 2 /s Density = 57 lbm/ft 3 Length of the tube = 50 ft Explanation: Oil is flowing through a horizontal pipe, which is having a kinetic viscosity of 0.08 ×10 -3 ft 2 /s and a density of 57 lbm/ft 3 . The volumetric flow rate is 10 gal/h. As per the continuity equation, Q = A V ⇒ 10 26930 = π 4 × ( 0.24 12 ) 2 × V ⇒ V = 10 × 4 26930 × π × 0.02 2 ⇒ V = 1.18 f t / s From the relation of Reynold’s Number, Re = V D ν ⇒ Re = 1.18 × 0.24 12 0.08 × 10 − 3 ⇒ Re = 295 Hence, the flow is laminar from the relation of kinematic viscosity and dynamic viscosity we can write. ⇒ ν = μ ρ ⇒ μ = ν ρ ⇒ μ = ( 0.08 × 10 − 3 ) × 57 ⇒ μ = 4.56 × 10 − 3 l b m f t 2 / s It is also known that ⇒ Δ P = 32 × μ × V × L D 2 ⇒ Δ P = 32 × ( 4.56 × 10 − 3 ) × 1.18 × 50 ( 0.24 12 ) 2 ⇒ Δ P = 21523.2 l b m / f t 2 The pressure drop for a pipe of 50 ft is 21523.2lbm/ft 2

The problem is to determine the pressure drop for a pipe of 50 ft The pressure drop for a pipe of 50 ft is 13.37 lb/ft 2 . The problem was solved by the following procedures Determination of the velocity from the volumetric flow rate Reynold’s number, for the flow pattern determination The pressure drop in the pipe due to its length. Given Information: Volumetric flow rate = 10 gal/h Kinematic viscosity = 0.08 ×10 -3 ft 2 /s Density = 57 lbm/ft 3 Length of the tube = 50 ft Explanation: Oil is flowing through a horizontal pipe, which is having a kinetic viscosity of 0.08 ×10 -3 ft 2 /s and a density of 57 lbm/ft 3 . The volumetric flow rate is 10 gal/h. As per the continuity equation, Q = A V ⇒ 10 26930 = π 4 × ( 0.24 12 ) 2 × V ⇒ V = 10 × 4 26930 × π × 0.02 2 ⇒ V = 1.18 f t / s From the relation of Reynold’s Number, Re = V D ν ⇒ Re = 1.18 × 0.24 12 0.08 × 10 − 3 ⇒ Re = 295 Hence, the flow is laminar from the relation of kinematic viscosity and dynamic viscosity we can write. ⇒ ν = μ ρ ⇒ μ = ν ρ ⇒ μ = ( 0.08 × 10 − 3 ) × 57 ⇒ μ = 4.56 × 10 − 3 l b m f t 2 / s It is also known that ⇒ Δ P = 32 × μ × V × L D 2 ⇒ Δ P = 32 × ( 4.56 × 10 − 3 ) × 1.18 × 50 ( 0.24 12 ) 2 ⇒ Δ P = 21523.2 l b m / f t 2 The pressure drop for a pipe of 50 ft is 21523.2lbm/ft 2

Solution Summary: The author explains that the problem is to determine the pressure drop for a pipe of 50 ft. The flow is laminar from the relation of kinematic and dynamic viscosity

Fundamentals of Momentum, Heat, and Mass Transfer

Fundamentals of Momentum, Heat, and Mass Transfer

6th Edition

ISBN: 9781118947463

Author: James Welty, Gregory L. Rorrer, David G. Foster

Publisher: WILEY

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