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Fluid Reservoir Properties PEEG 216 PEEG 216 RESERVOIR FLUID PROPERTIES Dr. John Williams Class 14 Reservoir Fluid Properties PEEG 216 Outline Summary of Last Class ( & complete) Notice about planning Second presentation on Compositional Analysis Hand out Support Document Homework Reservoir Fluid Properties PEEG 216 Planning notice Mid-term exam Next quiz I will be teaching an industry course on 7th and 8th April. Mr. mimoune may take these classes. Reservoir Fluid Properties PEEG 216 Summary of Last Class Compositional analysis of petroleum fluids is complicated. Pressurised samples are usually separated into subsamples to facilitate analysis. Gas chromatography (GC) is used extensively, but distillation and other analyses can also be involved. Analyses are simplified by grouping hydrocarbons between the normal (straight chain) components. Some particular components may be used as genetic indicators of the origin of the fluid. Heavy fractions are characterised by cryoscopy and density. Reservoir Fluid Properties PEEG 216 Composition Conversions Compositions have units and can be interconverted. Pure = 100 mol % = 1 000 000 ppm = mole fraction of 1 (all are based on molecules) 100 ppm = ? However, 50 mol% 50 wt% due to differences in molar masses. Reservoir Fluid Properties PEEG 216 Conversion from %wt to %mol Comp. C1 C2 C3 C4 C5 C6 C7+ Total Wt% (g/100g) 32.73 14.41 10.51 8.38 6.90 4.10 23.02 100.00 MM (g/mol) 16.0 30.1 44.1 58.1 72.2 86.2 120.2 Col2/Col3 (mol/100g) 2.0456 0.4787 0.2383 0.1442 0.0956 0.0476 0.1915 3.2416 Mol% (mol/100mol) 63.11 14.77 7.35 4.45 2.95 1.47 5.91 100.01 The inverse of 3.2416 mol/100g is 30.848 g/mol, the average molar mass Reservoir Fluid Properties PEEG 216 GOR Conversion As seen earlier, most laboratories flash samples to give atmospheric gases and liquids for analysis. This gives a gas-oil ratio which must be converted to a molar ratio. Reservoir Fluid Properties PEEG 216 Calculation of mole ratio The lab. flash gives 95 cm3 of liquid and 11 769 cm3 of gas (this is a GOR of 123.9 m3/m3) Amount of liquid nl = Vl x / M = 95 cm3 x 0.785 g/cm3 / 124.1 g/mol = 0.6009 mol Amount of gas ng = P x Vg /(Z x R x T) = 101325 Pa x 11 769 cm3 /(1 x 8.3145 JK-1mol-1 x 298.15 K) = 0.4810 mol Mole fraction of gas = 0.4810 mol/(0.4810 mol + 0.6009 mol) = 0.4446 what is the mole fraction of liquid? Reservoir Fluid Properties PEEG 216 Sep. Note: Liquid Further Analysis recomb. from With PVT Sep. G Report seen earlier Reservoir Fluid Properties PEEG 216 The Plus Fraction Remember that the plus fraction (e.g. C7+) is a grouping of heavy components. The plus fraction is not separated as an individual sub-sample. It is similar to the stock tank liquid, but also usually includes the heavier components from the stock tank gas. Reservoir Fluid Properties PEEG 216 On-Site Analyses Some petroleum fluid components, notably H2S and other sulphur-containing compounds, are reactive, and their concentration may decrease or become zero before analyses can be performed in a laboratory. Use of length of stain detector is tubes one way of determining concentrations in the field Detector tubes are available for numerous gas components, such as H2S, CO2 and the mercaptans, represented as RSH (where R is an alkyl group such as methyl) Reservoir Fluid Properties PEEG 216 Natural Gas with little hydrocarbons Although rare, extremely unusual gas concentrations can occur. The only real limit is that all components must total 100%! Beware: The compositions of unusual fluids are frequently erroneous because analysis systems have not been properly calibrated for their concentrations. Reservoir Fluid Properties PEEG 216 Compositional Analysis Reference Properties of Oils and Natural Gases K.S. Pedersen, Aa. Fredenslund, and P. Thomassen Chapter 2: Compositional Determinations Reservoir Fluid Properties PEEG 216 What have we learnt? Compositional analysis of natural gases and crude oils is extremely complicated. Compositional units include mol%, wt%, ppm Sub-sample analyses are re-combined to get the original fluid. The plus fraction does not exist as a physically separate part of the fluid (it is not the same as the residue) Reactive components need to be measured in the field Reservoir Fluid Properties PEEG 216 For the gas analysis in Figure 2.1: Homework 1 due after the break Skim-read the handout on compositional analysis How many of the components shown in Table 2.1 would be grouped into the C8 (octanes) fraction based on their retention time (detection order)? What do you notice about some of these components? How many different components have been detected (they may not be given names) in the gas chromatogram for the condensate sample shown in Figure 2.3? A more detailed composition in Table 2-18, is provided with PNA distribution. What does PNA stand for? Reservoir Fluid Properties PEEG 216 Homework 2 (Individual) Due after the break. Calculate new gas and liquid mole fractions from the volumes of flash gas and flash liquid in the following table. (Use the same properties as in the example during the class) Then recombine the flash gas and liquid according to the new ratio to get the separator liquid composition The GOR must be calculated by hand. You can use Excel for the composition, but only if you make the spreadsheet yourself. Reservoir Fluid Properties PEEG 216 Class 01 Student 920010738 920010784 920011482 920011503 920011566 920011821 920012228 920012269 920012363 920012495 920012565 Select all cells and use Oil volume cm3 95 93 91 89 87 83 81 79 77 75 73 Paste special & Unformat Gas volume cm3 9800 9700 9600 9500 9400 9300 9200 9100 9000 8900 8800 Ed text Select all cells and use Student 920012647 920012665 920012769 920012885 920013024 920013035 920013038 920013046 920013374 920013450 Class 01 Oil volume cm3 95 93 91 89 87 83 81 79 77 75 Gas volume cm3 7800 7700 7600 7500 7400 7300 7200 7100 7000 7900 Paste special & Unformat Ed text Reservoir Fluid Properties PEEG 216 Class 50 Student 920012211 920012596 920012912 920013119 920013723 920013732 Oil volume cm3 95 93 91 89 87 83 Gas volume cm3 11800 11700 11600 11500 11400 11300 ... View Full Document

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