Question

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.

Answer 1

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Step-by-step 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