physics

A 5.4 cm diameter horizontal pipe gradually narrows to 3.7 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 35.0 kPa and 21.0 kPa, respectively. What is the volume rate of flow?

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asked by Cara
  1. (1/2) rho v^2 + p = constant (Bernoulli)
    Q = area* speed = constant

    area of big pipe = pi d^2/4 = pi(.054)^2/4 = .00229 m^2

    area of small pipe = pi (.037)^2/4 = .00108 m^2

    Vsmall = Vbig *.00229/.00108
    so
    Vsmall = 2.12 Vbig

    Pbig = 35*10^3
    Psmall = 21*10^3

    rho = 1000 kg/m^3

    so
    35*10^3 + (1/2) (10)^3 Vbig^2 = 21*10^3 + (1/2)(10)^3 (2.12 Vbig)^2

    35 - 21 = (1/2)(4.5 - 1) Vbig^2

    28 = 3.5 Vbig^2
    Vbig = .894 m/s
    Q = Area big* Vbig = .00229 m^2 * .894 m/s
    = .00205 m/s = 2.05 cm/s

    check my arithmetic !

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    posted by Damon
  2. when i was doing it ... everything was alright until the last part...


    (28 = 3.5 Vbig^2
    --> Vbig = 2.828 m/s <--
    Q = Area big* Vbig = .00229 m^2 * 2.828 m/s
    = .00647 m/s = 6.047 cm/s

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  3. Stephanie's answer for m/s will also work for m^3/s.

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    posted by Wendi

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