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At 2:00 PM, a plant operator notices a drop in pressure in a pipeline transporting a...

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 4 atmg. At 2:30 PM, a 1.9 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. Note: SGhydrocarbon is 0.816

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Answer #1

Area of hole = (3.14/4)*D²

= (3.14/4)*(1.9in x 1ft/12in)²

= 0.01968 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 = 4 atmg x 14.7psig/atmg

= 57.6 psig

Mass flow rate from hole

= ACo (2 x density x gc x P)^0.5

= 0.01968 ft2 x 0.61 x [2 x (50.9184 lbm/ft3) x (32.174 lbm-ft/lbf-s2) x (57.6 lbf/in2) x (144in2/ft2)]^0.5

= 0.0120048 x [27174044.37]^0.5

= 62.58 lbm/s

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

Mass flow rate = (62.58 lbm/s) x (30 min) x (60s/min)

= 112643.095 lbm

Volume spilled= mass/density

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

= 2212.23 ft3 x (7.481 gal/ft3)

= 16550 gal

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