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3 February, 12:36

At 2:00 PM, a plant operator notices a drop in pressure in a pipeline transporting a volatile hydrocarbon. The pressure is immediately restored to 7 atmg. At 2:30 PM, a 1.2 inch diameter hole is found in the pipeline and immediately repaired. Provide a worst-case estimate for the total amount of hydrocarbon spilled in kilograms.

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  1. 3 February, 13:03
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    ujhljyjg

    Explanation:

    Area of hole = (3.14/4) * D²

    = (3.14/4) x (1.2in x 1ft/12in) ² = 0.785*0.1

    Area of hole, A = 0.0785 ft²

    Density of hydrocarbon = specific gravity x density of water

    = 0.816 x 62.4 lbm/ft³

    = 50.9184 lbm/ft³

    Discharge coefficient from orifice, Co = 0.61

    Gauge pressure P = 7 atmg x 14.7psig/atmg

    = 102.9 psig

    Mass flow rate from hole

    = A x Co (2 x density x gc x P) ^0.5

    = 0.0785 ft2 x 0.61 x [2 x (50.9184 lbm/ft³) x (32.174 lbm-ft/lbf-s²) x (102.9 lbf/in²) x (144in2/ft²) ]^0.5

    = 0.047885 x [48549824.96]^0.5

    = 0.047885 x6967.70444

    = 333.6485 lbm/s

    From 2:00 pm to 2:30 pm (30 min)

    Mass flow rate = (333.6485 lbm/s) x (30 min) x (60s/min)

    = 600,567.3488lbm

    Volume spilled = mass/density

    = (112643.095 lbm) / (50.9184 lbm/ft³)

    = 11794.70189 ft³ x (7.481 gal/ft³)

    Volume spilled = 88,236 gal.

    1 gallon (gal) = 3.785411784 kilogram (kg).

    Therefore, 88,236 gal = 3.785411784 x 88,236

    = 334009.5942kg

    Volume spilled = 334009.59kg
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