Explanation Given information: Diameter of the duct is 20 c m Average velocity is 5 m / s The size of square duct is equal to a Pressure drop is same in both ducts The pressure drop Δ P is defined as below, Δ P = f L D ( ρ V 2 2 ) In above equation, f - Friction factor L - Length of duct D h - Hydraulic diameter V - Velocity of the flow ρ - Density of the fluid The Reynolds’s number is defined as, Re D = ρ V D μ μ - Dynamic viscosity The hydraulic diameter for a square duct is defined as, D h = 4 A Ρ In above equation, A - Area of the square duct Ρ - Perimeter of square duct Assume, air at 20 ° C will have, ρ = 1.2 k g / m 3 μ = 1.8 × 10 − 5 k g / m s Calculation: For the circular duct, Calculate the Reynolds’s number, Re D = ρ V D μ = ( 1.2 k g / m 3 ) ( 5 m / s ) ( 0.2 m ) ( 1.8 × 10 − 5 k g / m s ) = 66666.67 Calculate the friction factor for smooth circular duct, 1 f 1 2 = 2.0 log ( Re D f 1 2 ) − 0.8 Substitute for Reynolds’s number, 1 f 1 2 = 2.0 log ( ( 66666.67 ) f 1 2 ) − 0.8 f = 0.0196 \ Calculate the pressure drop, Δ P = f L D ( ρ V 2 2 ) = ( 0.0196 ) 1 m 0.2 m ( ( 1.2 k g / m 3 ) ( 5 m / s ) 2 2 ) = 1.47 P a / m For the square duct, Calculate the hydraulic diameter, D h = 4 A Ρ = 4 a 2 4 a = a Equate flow rates to find the velocity in square duct, The flow arte is given as, Q = V A Therefore, ( 5 m / s ) ( π ( 0

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ISBN: 9789385965494

Author: Frank White

Publisher: MCGRAW-HILL HIGHER EDUCATION

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Chapter 6, Problem 6.94P

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The square duct size a.

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