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22 May, 14:17

Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0-cm diameter pipe in the basement under a pressure of 3.03x10^5 Pa, what will be the velocity and pressure in a 2.6-cm diameter pipe on the second floor 5.0m above?

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  1. 22 May, 14:19
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    velocity and pressure in a 2.6-cm:

    P2 = 2.53x10^5Pa, v2 = 1.18m/s

    Explanation:

    Pressure = P, Velocity = v, Height = h, Diameter = d, Radius = r, Area = A

    Area = πr^2

    From the question:

    v1 = 0.5m/s

    d1 = 4cm = 0.04m

    r1 = d1/2 = 0.04/2 = 0.02m

    Since water was pumped from basement, h1 = 0m

    P1 = 3.03x10^5 Pa

    A1 = π*0.02*0.02

    A1 = 0.0004πm^2

    v2 = unknown

    d2 = 2.6cm = 0.026m

    r2 = d2/2 = 0.026/2 = 0.013m

    h2 = 5m

    P2 = unknown

    A2 = π*0.013*0.013

    A2 = 0.000169πm^2

    Using continuity equation:

    A1v1 = A2v2

    0.0004π * 0.5 = 0.000169π * v2

    v2 = (0.0004π * 0.5) / (0.000169π)

    v2 = 1.18m/s

    Applying a Bernoulli principle

    P + 1/2*density*v^2 + density*g*h = C

    C = constant

    P1 + 1/2*density*v1^2 + density*g*h1

    = P2 + 1/2*density*v2^2 + density*g*h2

    Let g = 9.81m/s

    density of water = 1000kg/m^3

    (P1-P2) = 1/2 * density (v2^2 - v1^2) + (density*g*h2) - (density*g*h1)

    (P1-P2) = 1/2 * density (v2^2 - v1^2) +

    density * g (h2-h1)

    (3.03x10^5 - P2) = 1/2*1000 (1.18^2-0.5^2) + 1000 (9.81 (5-0))

    (3.03x10^5 - P2) = 500 (1.3924-0.25) + 49050

    3.03x10^5 - P2 = 571.2 + 49050

    3.03x10^5 - P2 = 49621.2

    3.03x10^5 - 49621.2 = P2

    P2 = 253378.8

    P2 = 2.53x10^5Pa

    P2 = 2.53x10^5Pa, v2 = 1.18m/s
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