Page 40 - Characterization and Properties of Petroleum Fractions - M.R. Riazi

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20 CHARACTERIZATION AND PROPERTIES OF PETROLEUM FRACTIONS
temperature units are related through the following relations:
this reason, to express amount of gases in terms of volume,
normally some SC are defined. The SC in the metric SI units
(1.9) T(K) = T( C) + 273.15 are 0 C and 1 atm and in the English system are 60 F and
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1 atm. Under these conditions molar volume of any gas is
(1.10) T( R) = T( F) + 459.67
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equivalent to 22.4 L/mol (in SI) and 379 scf/lbmol (in English
(1.11) T(K) = T( C) units). In reservoir engineering calculations and petroleum
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industry in general, the SC in the SI units are also set at 60 F
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(1.12) T( R) = T( F)
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(15.5 C or 289 K) and 1 atm. The choice of standard temper-
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(1.13) T( R) = 1.8T(K) ature and pressure (STP) varies from one source to another.
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In this book when the standard T and P are not specified the
(1.14) T( F) = 1.8T( C) + 32
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STP refers to 289 K and 1 atm, which is equivalent to the STP
in English unit system rather than SI system (273 K and 1
As an example, absolute temperature of 100 K is equivalent to
1.8 × 100 or 180 R. Therefore, the conversion factors between atm). However, for liquid systems the volume is less affected
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K and R are as follows: by pressure and for this reason specification of temperature
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alone is sufficient.
1K = 1.8 R (for absolute temperature T and the temperature --`,```,`,``````,`,````,```,,-`-`,,`,,`,`,,`---
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difference, T) 1.7.12 Units of Volumetric and Mass Flow Rates
1 C = 1.8 F (only for the temperature difference, T)
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Most processes in the petroleum industry are continuous and
usually the volume or mass quantities are expressed in the
1.7.11 Units of Volume, Specific Volume, form of rate defined as volume or mass per unit time. One
and Molar Volume—The Standard Conditions particular volumetric flow rate used for liquids in the English
system is gallon (U.S.) per minute and is known as GPM.
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Volume (V) has the dimension of cubic length (L ) and thus Some of the conversion factors for these quantities are
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in SI has the unit of m and in English its unit is cubic feet 3 3 4
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(cf or ft ). Some units particularly used for liquids in the SI 1m /s = 1 × 10 L/s = 1.5851 × 10 GPM 5 3
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system are liter (L), cm (cc), or milliliter (mL) and in English = 5.4345 × 10 bbl/d = 1.27133 × 10 ft /h
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units are gallon (in U.S. or Imperial) and barrel (bbl). Volume 1ft /h = 7.86558 × 10 −4 m /s = 0.12468 GPM
of one unit mass of a fluid is called specific volume and the = 4.27466 bbl/d
volume of 1 mol of a fluid is called molar volume. Some of 1 GPM = 2.228 × 10 −3 ft /s = 8.0205 m /h = 34.285 bbl/d
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the conversion factors are as follows.
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1 bbl/d = 2.9167 × 10 −2 GPM = 1.8401 × 10 −4 m /s
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1m = 10 cm = 1000 L = 35.315 ft = 264.18 gallon (U.S.) = 0.23394 ft /h
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= 35.316 ft = 6.29 bbl The conversion factors for the mass rates are as follows:
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1ft = 2.8316 × 10 −2 m = 28.316 L = 7.4805 gallon (U.S.) 1 kg/s = 7.93656 × 10 lb/h = 3.5136 × 10 ton/year
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1 bbl = 42 gallon(U.S.) = 158.98 L = 34.973 gallon (Imperial) 1 lb/s = 1.63295 × 10 kg/h = 39.1908 ton/d
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1 gallon (U.S.) = 0.8327 gallon (Imperial)
= 0.023809 bbl = 3.7853 L The same conversion factors apply to molar rates.
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1mL = 1cm = 10 −3 L = 10 −6 m = 0.061024 in. 3
1.7.13 Units of Density and Molar Density
For the molar volumes some of the conversion factors are Density shown by d or ρ is defined as mass per unit volume
given as follows: and it is reciprocal of specific volume. The conversion factors
can be obtained from reversing those of specific volume in
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1m /kmol = 1L/mol = 0.001 m /mol = 1000 cm /mol Section 1.7.11.
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= 16.019 ft /lbmol 1 kg/m = 6.24259 × 10 −2 lb/ft = 1 × 10 −3 g/cm 3
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1ft /lbmol = 6.24259 × 10 −2 m /kmol = 8.3455 × 10 −3 lb/gal
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= 6.24259 × 10 −5 m /mol = 62.4259 cm /mol 1 lb/ft = 16.019 kg/m = 1.6019 × 10 −2 g/cm 3
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1cm /mol = 1mL/mol = 1L/kmol = 0.001 m /kmol = 0.13368 lb/gal
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= 1.6019 × 10 −2 ft /lbmol 1 g/cm = 1 kg/L = 10 kg/m = 62.4259 lb/ft 3
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It should be noted that the same conversion factors apply = 8.3455 lb/gal
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to specific volumes. For example, 1 lb/gal = 1.19825 × 10 kg/m = 7.4803 lb/ft 3
= 0.119825 g/cm 3
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1ft /lb = 6.24259 × 10 −2 m /kg = 62.4259 cm /g
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Density may also be presented in terms of number of moles
Since volume and specific or molar volumes depend on tem- per unit volume, which is called molar density and is recipro-
perature and pressure of the system, values of volume in any cal of molar volume. It can be obtained by dividing absolute
unit system are meaningless if the conditions are not spec- density to molecular weight.The conversion factors for molar
ified. This is particularly important for gases in which both density are exactly the same as those for the absolute density
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temperature and pressure strongly influence the volume. For (i.e., 1 mol/cm = 62.4259 lbmol/ft ). In practical calculations
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