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A 3-ft-diameter duct is used to carry ventilating air into a vehicular tunnel at a rate of 9000 ft3/min. Tests show that the pressure drop is 1.5 in. of water per 1500 ft of duct. What is the value of the friction factor for this duct and the approximate size of the equivalent roughness of the surface of the duct?

Answer :

davidlcastiblanco

Answer:

The roughness surface

ε = 0.0132

Explanation:

Given:

D = 3 ft , υ = 1.58 x 10 ⁻⁴ ft² / s, ρ = 2.38 x 10 ⁻³ slug / ft ³, L = 1500 ft

rH₂O = 62.4 lb / ft , Q = 9000 ft³ / min ,

Now to determine using the equation

ΔP = rH₂O * h = ( 62.4 lb / ft ) (* 1.5 / 1.2 ft )  = 7.80 lb / ft²

V = Q / A = [ 9000 ft³ / min * 1 min / 60 seg ] / [ π / 4 * ( 3ft )² ]

V = 21.2 ft / s

f = 2 * ΔP / ρ * L * V²

f = [ 2 * 7.80 lb / ft² ] / [ 2.38 slug / ft³ * 1500 ft * (21.2 ft /s )²

f = 0.0292  The kinetic density

Now determine the Reynolds number

Re = V*D / υ

Re = (21.2 ft / s * 3 ft ) / (1.58 x 10⁻⁴ ft² / s)

Re = 4.05 x 10 ⁵

Finally  surface roughness of the duct as

ε = D * 0.0044

ε = 3 ft * 0.0044 = 0.0132

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