I have attached example 8.2 along with the answer for your reference.

8.6 In Example 8.2, it was assumed that no condensate is produced in the line. Using the same operating conditions, and the dosages given in the example, if 0.02 GPM (2.83 m liquid/106 Nm gas) of condensate (molar mass = 80, SpGr=0.65) forms in the line, estimate the addi- tional amount (lb per MMscf, g per 109 Nm) of methanol required to account for loss in condensate. Example 8.2 A sweet gas with a specific gravity of 0.73 leaves a gas-liquid separator at 100F (37.8C) and 600 psia (41.4 bara) saturated with water. The gas drops to 35F (1.7C) before reaching the next booster station at 500 psia (34.5 bara). Assume that no hydrocarbon condensate has formed in the line. Calculate how much methanol must be added per MMSC (10 Nm) to prevent hydrate for- mation between the separator and the booster station. Repeat the calculation for ethylene glycol, which is added in an 80 wt% mixture with water. Solution Use of Equation 8.1 or Figure 8.13 shows that the hydrate formation temperature at 600 psia (41.4 bara) is 59F (15C). This value means a 59F - 35F = 24F (13.3C) subcooling into the hydrate region, and hydrate formation is probable without inhibition. Determination of the inhibitor rate requires: 1. Determination of the amount of liquid water formed 2. Calculation of the required amount of inhibitor in the water phase 3. Calculation of the required amount of inhibitor in the gas phase 1. The water content of the gas leaving the separator is 95 lb/MMscf and 13 lb /MMscf at 35F and 600 psia (see Figure 11.1 to obtain these data). Therefore, 95 lb, - 13 lb, = 82 lb. of water per MMscf (1,390 kg/10 Nm?) drops out in the line, assuming maximum water condensation. METHANOL REQUIREMENT 2. To estimate the aqueous phase concentration of methanol, rearrange Equation 8.4 to give In xw -Af(F) -24 = -0.185 129.6 129.6 xw = 0.831 or Xcoh = 1.0 - 0.831 = 0.169 mole fraction. Because the molar mass of MeOH is 32, the weight fraction is (0.169 x 320.831 x 18 + x 0.169 X 32) = 0.266 or 0.362 lb, methanol/lb. water. Thus, the methanol needed in the water phase is 0.362 x 82 = 29.7 lb/MMSCf (500 g/10 Nm). 3. To estimate the methanol in the vapor phase use Figure 8.14, which, at 35F and 500 psia, gives a distribution coefficient of approximately 1.7. Then the vapor content is 1.7 x 16.9 mol % MeOH in aqueous phase = 29 lb/MMSC (470 g/10 Nm). Thus, the vapor phase holds as much methanol as the aqueous phase. The total amount of methanol required is (30 + 29) = 59 lb. MMSc (990 g/10 Nm). If a condensate phase was present as well, the losses would have to be added for that phase. The condensate amount of methanol will be much less than what goes into the aqueous and vapor phases. ETHYLENE GLYCOL REQUIREMENT 4. Use Equation 8.3 with the constant 2,335 (1,297) and a molar mass of ethylene glycol of 62 to obtain the mass fraction of pure glycol required 24 x 62 X = 24x62 +2,335 However, the glycol is diluted to 80 wt%. To obtain the mass of inhibitor solution added per unit mass of free water present to obtain the desired concentration, use X/X - X), where X, is the weight fraction of inhibitor in the solution to be added. The amount of inhibitor solution added per pound of free water initially present is then 0.8/0.8 -0.39) = 1.95, and the total amount of ethylene glycol solution added is 1.95 x 82 = 160 lb/MMSC (2.70 g/10 Nm). At these conditions, glycol loss into the vapor phase is negligible, so the total amount of solution required is 160 lb/MMscf (2.78 g/10 Nm). =0.39