Amount of water condensed per day helps determine the water content of a hydrocarbon gas.
Water condensed = natural gas water content at 380C x change in water content
= (2.83 x 106 x 698) = 1975 x 106 mg/day = 1975 kg/day
Methanol inhibitor concentration is necessary for developing the quantity XX methanol XXXXXX in relation to the operating conditions.
d = XX – X = XX0X
X (XXXOH)= (12 + 3 + 16 + X) = 32
X1 = dMWX ÷ (XH + dMW1)
= XX x 32 x X.098 ÷ (XXXX + XX x XX x X.098)
= 0.XXX
Mass rate of inhibitor XXXXXXXX in XXXXX phase XXXXX XXXXXXXXX the XXXX XX inhibitor solution XXXXXX in XXX water phase.
mX= (XR • m XXO)/ (XX − XR)
= (X.XXX x XXXX)/ (X – X.XXX)
= 749 kg/XXX
XXXXXXXXXXXX losses represent a vital XXXX of operation XXX XXX lead to issues in XXXXXXXXXX process units.
Daily XXXXXX = (1.XX x 10X x 2.X x 106 x27.X) = 1310 kg/XXX
The XXXXXXXX XX hydrate XXXXXXXXX XX XX included in processing XX gas and in the transmission XXXXXX XX prevent formation XX hydrate XXXX XX XXXXXXXXXX XX prevent XXXXXXXX XX the XXXXXXXXX in XXX XXXXX phase and also to XXXXXX for the XXXXXXXXXXX vapor XXXXX content of the XXXXXXXXX XXX XXX XXXX XX the XXXXXXXXX in XXX XXXXXX hydrocarbon.
(2.83 x XX6 SmX/day x 56 m3 x XX-6 XX-X x XXX kg m-3 = 881 kg XXX/XXX
kg XXX XXXXXXXX = (881 x X.002) = 1.XX kg XXXX/XXX
kg methanol = (1.76 x 32) = XX XX/day.
XXXXX methanol injection XXXX = (XXX + XXXX + XX) = 2115 XX/day.
X1 = XXXX ÷ (KX + XXX1)
= XX x 62 x 0.5811 ÷ (XXXX + XX x XX x 0.5811)
= X.XX
mX= (XX • m XXO)/ (XL − XR)
= (0.XX x 1975)/ (0.X – 0.28)
= 106
3 kg/day
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