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10 November, 00:15

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

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  1. 10 November, 01:36
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    5984.67N

    Explanation:

    A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

    from continuity equation

    v1A1=v2A2

    equation of continuity

    v1=4ft / s=1.21m/s

    d1=14 inch=.35m

    d2=14-2=0.304m

    A1=pi*d^2/4

    0.096m^2

    a2=0.0706m^2

    from continuity once again

    1.21*0.096=v2 (0.07)

    v2=1.65

    force on the pipe

    (p1A1 - p2A2) + m (v2 - v1)

    from bernoulli

    p1 + ρv1^2/2 = p2 + ρv2^2/2

    difference in pressure or pressure drop

    p1-p2=2psi

    13.789N/m^2=rho (1.65^2-1.21^2) / 2

    rho=21.91kg/m^3

    since the pipe is cylindrical

    pressure is egh

    13.789=21.91*9.81*h

    length of the pipe is

    0.064m

    AH=volume of the pipe (area * h)

    the mass = rho*A*H

    0.064*0.07*21.91

    m=0.098kg

    (193053*0.096 - 179263.6 * 0.07) + 0.098 (1.65 - 1.21)

    force = 5984.67N
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