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15 June, 11:23

H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation f s m o o t h ≈ 0.3164 Re - 1 / 4 fsmooth≈0.3164Re-1/4 is reasonably accurate for Reynolds numbers between 4000 and 105. Use this information for the following scenario. Water at 20 °C is to flow through a 3-cm I. D. plastic pipe at the rate of 0.001 m3/s. Find the incline angle of the pipe needed to make the static pressure constant along the pipe

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  1. 15 June, 11:40
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    Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

    Explanation:

    The first step to take is to calculate the the velocity of flow through a pipe

    Q = Av

    Where Q = is the discharge through pipe

    A = Area of the pipe

    v = the flow of velocity

    We substitute 0.001 m^3/s for Q and 0.03 m for D

    Q = Av

    0.001=Av

    Substitute π/4 D² for A

    0.001 = π/4 D² (v)

    v = 0.004/πD²

    D = he diameter of the pipe

    substitute 3 cm for D

    v = 0.004/π * [3 cm * 1 m/100 cm]²

    v = 1.414 m/s

    Obtain fluid properties from the table Kinematic viscosity and Dynamic of water

    p = 1000 kg / m³

    μ = 1.002 * 10^ ⁻³ N. s/m³

    Thus,

    we write the expression to determine the Reynolds number of flow

    Re = pvD/μ

    Re = is the Reynolds number

    p = density

    μ = dynamic viscosity at 20⁰C

    We then substitute 1000 kg / m³ in place of p, 1.002 * 10^ ⁻³ N. s/m³ for μ,

    1.414 m/s for v and 0.03 m for D

    Thus,

    Re = 1000 * 1.414 * 0.03 / 1.002 * 10^ ⁻³ = 42335

    The next step is to calculate the friction factor form the Blasius equation

    f = 0.3164 (Re) ^1/4

    f = friction factor

    We substitute 42335 for Re

    f = 0.3164 (42335) 1/4

    =0.022

    The next step is to write the expression to determine the friction head loss

    hl = flv²/2gD

    hl = head loss

    l = length of pipe

    g = acceleration due to gravity

    We then again substitute 0.022 for f, 1.414 m/s for v, 0.03 m for D, and 9.8 m/s² for g.

    so,

    hl = flv²/2gD

    hl/L = 0.022 * 1.414²/2 * 9.81 * 0.03

    sinθ = 0.07473

    θ = 4° 16'

    Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'
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