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Fluid Mechanics Question - Pipes and Pumps

  1. Apr 4, 2012 #1
    Fluid Mechanics Question -- Pipes and Pumps

    1. The problem statement, all variables and given/known data

    Design a system to deliver [itex]1 m^{3}/s[/itex] of water from reservoir A at elevation 300m to reservoir B at 500m. The distance from A to B is 1000km (1000,000m) and the elevation between the two reservoirs can be approximated as parabolic with a maximum elevation of 660m reached 600km from A.

    Negative pressures shouldn’t be allowed to exceed about 70% of atmospheric pressure, use this figure to prevent cavitation.

    The steel pipe with a roughness of e=0.5cm and diameter of 1.1m and a pressure rating of 150m head is available. The pumps can deliver [itex]1m^3/s[/itex] at 100m head.

    *Since the horizontal distance is significantly greater than the vertical distance. I approximated the parabola with straight lines from A to the maximum point of 600,000m from A and elevation 660m and then from that point to B.
    I am then doing calculations separately for the 2 parts

    *I am trying to calculate amount of pumps that would make this system work and prevent cavitation

    *Any help would be appreciated, thanks!

    2. Relevant equations

    Extended bernoulli’s equation:
    [itex]\frac{V_{1}^{2}}{2g}[/itex] + [itex]\frac{p_{1}}{\gamma}[/itex] + [itex]Z_{1}[/itex] + [itex]h_{p}[/itex] = [itex]\frac{V_{2}^{2}}{2g}[/itex] + [itex]\frac{p_{2}}{\gamma}[/itex] + [itex]Z_{2}[/itex] + [itex]h_{L}[/itex]
    V: Velocity
    P: Pressure
    [itex]\gamma[/itex]: Specific Weight
    Z: Height/ Altitude
    [itex]h_{p}[/itex]: Pump Head
    [itex]h_{L}[/itex]: Head Loss

    Head Loss:
    [itex]
    h_{L} = f \frac{L}{D} \frac{V^{2}}{2g}
    [/itex]
    f: friction factor
    L: Pipe Length
    D: Pipe Diameter

    Volumetric flow rate to Velocity:

    V = Q/A

    Q: Volumetric Flow Rate[itex] (m^{3}/s)[/itex]
    A: Area
    V = Velocity



    3. The attempt at a solution

    For part 1 [0km to 600km] only:
    D = 1.1m
    A = 0.95 m^{2}
    Q = 1[itex]m^{3}/s[/itex]
    V = [itex]\frac{Q}{A}[/itex] = 1.053 m/s

    f = 0.0171 (from moody diagram)
    R[itex]_{e}[/itex] = 1157490

    P[itex]_{minimum}[/itex] (absolute) = 101315Pa*0.7 = 70927.5Pa
    P[itex]_{minimum}[/itex] (gage) = P_{min,absolute} – P_{atmospheric} = -30397.5Pa

    H[itex]_{p}[/itex] = 100m
    L ~= 600km = 600,000m
    Using Bernoulli’s formula solve for maximum Length before cavitation:

    [itex]\frac{V_{1}^{2}}{2g}[/itex] + [itex]\frac{p_{1}}{\gamma}[/itex] + Z[itex]_{1}[/itex] + h[itex]_{p}[/itex] = [itex]\frac{V_{2}^{2}}{2g}[/itex] + [itex]\frac{p_{2}}{\gamma}[/itex] + Z[itex]_{2}[/itex] + [itex]h_{L}[/itex]

    P[itex]_{1}[/itex] = 0
    P[itex]_{2}[/itex] = -30397.5
    V[itex]_{1}[/itex] = V[itex]_{2}[/itex]
    Z[itex]_{1}[/itex] = 300m
    Z[itex]_{2}[/itex] = 660m
    [itex]\gamma[/itex] = 9800

    Sub in Head Loss:

    [itex]\frac{V_{1}^{2}}{2g}[/itex] + [itex]\frac{p_{1}}{\gamma}[/itex] + Z[itex]_{1}[/itex] + h[itex]_{p}[/itex] = [itex]\frac{V_{2}^{2}}{2g}[/itex] + [itex]\frac{p_{2}}{\gamma}[/itex] + Z[itex]_{2}[/itex] + [itex]f*\frac{L}{D}*\frac{V^{2}}{2g}
    [/itex]
    Rearrange:

    L = [itex](\frac{V_{1}^{2}}{2g}[/itex] + [itex]\frac{p_{1}}{\gamma}[/itex] + Z[itex]_{1}[/itex] – Z[itex]_{2}[/itex] + h[itex]_{p}[/itex] - [itex]\frac{V_{2}^{2}}{2g}[/itex] - [itex]\frac{p_{2}}{\gamma}[/itex] ) [itex]*\frac{2D*g}{fV^{2}}[/itex]
    Cancelling terms:

    L = (Z[itex]_{1}[/itex] – Z[itex]_{2}[/itex] +h[itex]_{p}[/itex] - [itex]\frac{p_{2}}{\gamma}[/itex]) [itex]*\frac{2D*g}{fV^{2}}[/itex]

    L = (300m – 660m + 100m +3.0986) *1144.77
    L = -293076m

    I don’t understand why this is the answer, what am I doing wrong?
     
  2. jcsd
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