Example: Calculate the density of methane at 50 psig and 32 ⁰F. H.W. In previous example calculate the density of methane in gm/m 3.

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1 Gas density Because the density of a substance is defined as mass per unit volume, the density of gas (ρ g ), at given temperature and pressure can be derived as follows: If P in psia, T in ⁰R and R = ρ in Ib/ft 3. If P in Pa., T in ⁰K and R = ρ in gm/m 3. Example: Calculate the density of methane at 50 psig and 32 ⁰F. Solution: H.W. In previous example calculate the density of methane in gm/m 3. 1 atm = 14.7 psia 1 atm = Pa. psia = psig ⁰C = 5/9 ( ⁰F 32) ⁰K = ⁰C +273 ⁰R = ⁰F

2 Specific gravity The specific gravity is defined as the ratio of the gas density to that ofthe air. Both densities are measured or expressed at the same pressure and temperature. Commonly, the standard pressure Psc and standard temperature Tsc are used in defining the gas specific gravity: Assuming that the behavior of both the gas mixture and the air is described by the ideal gas equation, the specific gravity can then be expressed as: where = gas specific gravity. M air =apparent molecular weight of the air = M a =apparent molecular weight of the gas. Example 1: A gas well is producing gas with a specific gravity of 0.65 at a rate of 1.1 MMscf/day. The average reservoir pressure and temperature are 1,500 psi and 150 F. Calculate: a. Gas density at reservoir conditions. b. Flow rate in lb/day. 2

3 Solution: a) a) Because 1 lb-mol of any gas occupies scf at standard conditions, then the daily number of moles that the gas well is producing can be calculated from: Example 2: A gas well is producing a natural gas with the following composition: Component Yi C1 0.7 C2 0.2 C CO Calculate apparent molecular weight (assuming ideal gas). 3

4 Solution: Component Y i M i Y i M i C C C CO = 21.6 = M a M a = Y i M i = (16*0.7) + (30*0.2) + (44*0.06) + (44*0.04) = 21.6 Real Gas Law. In dealing with gases at a very low pressure, the ideal gas relationship is a convenient and generally satisfactory tool. At higher pressures, the use of the ideal gas equation-of-state may lead to errors as great as 500%, as compared to errors of 2 3% at atmospheric pressure. Basically, the magnitude of deviations of real gases from the conditions of the ideal gas law increases with increasing pressure and temperature and varies widely with the composition of the gas. Real gases behave differently than ideal gases. The reason for this is that the perfect gas law was derived under the assumption that the volume of molecules is insignificant and that no molecular attraction or repulsion exists between them. This is not the case for real gases. Numerous equations-of-state have been developed in the attempt to correlate the pressure-volume-temperature variables for real gases with experimental data. In order to express a more exact relationship between the variables p, V, 4

5 and T, a correction factor called the gas compressibility factor, gas deviation factor, or simply the z-factor, must be introduced to account for the departure of gases from ideality. The gas compressibility factor z is a dimensionless quantity and is defined as the ratio of the actual volume of n-moles of gas at T and p to the ideal volume of the same number of moles at the same T and p: ( ) z = 1 for ideal gas. All gases behave as an ideal gases near the standard conditions. Z depends on P, T and composition of the gas. The figure below shows the gas deviation factors of two gases, one of 0.90 sp.gr. and the other of sp.gr. These curves show that the gas deviation factors drop from unity at low pressures to a minimum value near 2500 psia. They rise again to unity near 5000 psia and the value greater than unity at still higher pressures. In the range of 0 to 5000 psia, the z factor at the same temperature will be lower for the heavier gas, and for the same gas they will be lower at the lower temperature. 5

6 Example 1: for the mixture below determine z m at 1000 psi and 104 ⁰F. Component y i z i y i z i CH C 2 H C 3 H Z m = Note: if z i is not given in the question you can determine them using charts. Example 2 Calculate the volume of 1000 SCF (standard cubic feet) of gas at 1000 psi and 104 F for a gas of the following composition assuming : a) Real gas behavior b) Ideal gas behavior 6

7 Component CH C 2 H C 3 H Solution: Component CH C 2 H C 3 H At standard conditions: at reservoir conditions: n at standard conditions = n at reservoir conditions. 7

8 a) For real gas behavior: b) For ideal gas behavior ( ): 8

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