Gas viscosity The viscosity of a fluid is a measure of the internal fluid friction (resistance) to flow. If the friction between layers of the fluid is small, i.e., low viscosity, an applied shearing force will result in a large velocity gradient. As the viscosity increases, each fluid layer exerts a larger frictional drag on the adjacent layers and velocity gradient decreases. The viscosity of a fluid is generally defined as the ratio of the shear force per unit area to the local velocity gradient. Viscosities are expressed in terms of poises, centipoise, or micropoises. One poise equals a viscosity of 1 dynesec/cm 2. The gas viscosity is not commonly measured in the laboratory because it can be estimated precisely from empirical correlations. Like all intensive properties, viscosity of a natural gas is completely described by the following function: The above relationship simply states that the viscosity is a function of pressure, temperature, and composition. Two popular methods that are commonly used in the petroleum industry are the: Carr-Kobayashi-Burrows Correlation Method Lee-Gonzalez-Eakin Method Carr-Kobayashi-Burrows Correlation Method Carr, Kobayashi, and Burrows (1954) developed graphical correlations for estimating the viscosity of natural gas as a function of temperature, pressure, and gas gravity. In this method the following two figures should be used. 1
2
Example: A gas well is producing at a rate of 15,000 ft 3 /day from a gas reservoir at an average pressure of 2,000 psia and a temperature of 140 F. The specific gravity is 0.72. Calculate the gas viscosity. Solution: M a = 0.72 * 28.96 = 20.85 Determine µ1 from the first figure which equal to 0.0113 pt r = 600/395.5 = 1.52 pp r = 2000/668.4 = 2.99 determine viscosity rate (µ g /µ1) from the second figure which equal to 1.5. The Lee-Gonzalez-Eakin Method. Lee, Gonzalez, and Eakin (1966) presented a semi-empirical relationship for calculating the viscosity of natural gases. The authors expressed the gas viscosity in terms of the reservoir temperature, gas density, and the molecular weight of the gas. Their proposed equation is given by: * + 3
ρ g = gas density at reservoir pressure and temperature, lb/ft 3 T = reservoir temperature, R M a =apparent molecular weight of the gas mixture Example: Rework the previous example and calculate the gas viscosity by using the Lee-Gonzalez-Eakin method. Solution: Determine z from the chart which equal to 0.78 * + * + 4
Initial Gas In Place. Initial gas in place (G) can be determined through two methods: a. Volumetric Method. b. Material balance. Volumetric Method The standard cubic feet of gas in a reservoir with a gas pore volume of V g cu.ft. is simply (V g / B g ), where B g is expressed in units of cubic feet per standard cubic foot. As B g changes with pressure, the gas in place also changes as pressure declines. The V g may also be changing, owing to water influx into the reservoir. The V g is related to the bulk reservoir volume by the average porosity and the average connate water saturation. Where: is the initial gas in place, SCF is the bulk volume, acre-ft connate water saturation, fraction gas formation volume factor, ft 3 /SCF is the porosity, fraction Example: Calculate initial gas in place for a reservoir at 1500 psi and T = 140 ⁰C. given: Thickness = 40 ft. Reservoir area = 2178 * 10 4 ft 2. Average porosity = 22%. Connate water saturation = 23%. 5
assuming the following composition: Component Mole fraction (Y i ) CH 4 0.6 C 2 H 6 0.3 C 3 H 8 0.1 Solution: First of all we have to find Z i for each component at given P and T from charts which: Z i CH 4 = 0.99 Z i C 2 H 6 = 0.75 Z i C 3 H 8 = 0.435 Component Mole fraction (Y i ) Z i (from charts) Y i Z i CH 4 0.6 0.99 0.594 C 2 H 6 0.3 0.75 0.225 C 3 H 8 0.1 0.435 0.0435 Z m = 0.8625 T = 140 ⁰C = 5/9 (⁰F 32) T = 284 ⁰F 6
Material Balance Method If enough production-pressure history is available for a gas reservoir, the initial gas in place G, the initial reservoir pressure P i, and the gas reserves can be calculated without knowing A, h, or S w. This is accomplished by applying law of conservation of mass to the reservoir and associated production. The general material balance equation for a gas reservoir is: For most gas reservoirs, the formation and water compressibilites are too small compared to the gas compressibility, therefore, the second term on the left-hand side of the equation can be neglected. In a volumetric reservoir where neither water production nor water encroachment present, the equation reduced to: Where: G is the initial gas in place Gp is gas produced B gi is initial gas formation factor Substituting Because the production is isothermal process. 7
Rearranging: 8
Example: A dry gas reservoir contains gas of the following composition: Mole fraction Methane 0.75 Ethane 0.20 N - Hexane 0.05 The initial reservoir pressure was 4200 psia, and temperature of 180 F. The reservoir has been producing for some time, two pressure surveys have been made at different times: P/Z (psia) Gp (MMMSCF) 4600 0 3700 1 2800 2 a) What will be the cumulative gas produced when the average reservoir pressure has been dropped to 2000 psia? b) Assuming the reservoir rock has a porosity of 12%, the water saturation is 30% and the reservoir thickness is 15ft, how many acres does the reservoir cover? Solution: Comp. y i Pc i Tc i y i Pc i y i Tc i Methane 0.75 673.1 343.2 504.8 257.4 Ethane 0.20 708.3 504.8 141.7 110.0 n- Hexane 0.05 440.1 914.2 22.0 45.7 Total= 668.5 413.1 9
a) To get Gp at 2000 psia, calculate Z and then P/Z, use the pseudo critical properties: Z=0.8 Plotting P/Z vs. Gp will give: 10
Substituting the value of P/Z=2500 in this equation yields: b)substituting P/Z=0 into the straight line equation, Gp=IGIP Also: 11