# The importance to know the real natural gas properties

Fuel consumption and fuel heat input belong to the most sensitive parameters in today's gas turbine operation. For large consumers and power plants 0.1% difference in heat rate can have hundreds of thousands of dollars impact on fuel costs within just one year. Because of this penalties to manufacturers are extraordinary high when not meeting the heat rate guarantees.

Gas turbine operators are well advised to have their own calculation tool, being able to crosscheck thermal performance, but also monthly bills coming from gas suppliers. Of course there are enough other reasons for engineers to stay up to date with natural gas properties.

Let's discuss the effects calculating real while using as an example gas# 4 from ISO 20765-1-2005 (BS ISO 20765-1-2009) with 100 bar and 76.85°C at the »limit of supply«. To be consistent with reference conditions across chemical energies and enthalpies for air and flue gas flows let's take also for the gas the reference conditions of 0°C and 1.01325 bar.

## Gas example

Gas 4 ISO 20765-1 2005(E) p = 10.00 MPa absolute gas pressure T = 330.00 K gas temperature H2 = 9.50 Mol% hydrogen He = 0.02 Mol% helium H2O = 0.01 Mol% water vapor CO = 1.00 Mol% carbon monoxide N2 = 10.00 Mol% nitrogen O2 = 0.01 Mol% oxygen H2S = 0.01 Mol% hydrogen sulfide Ar = 0.01 Mol% argon CO2 = 1.60 Mol% carbon dioxide CH4 = 73.50 Mol% ethane C2H6 = 3.30 Mol% ethane C3H8 = 0.74 Mol% propane i-C4H10 = 0.08 Mol% iso-butane n-C4H10 = 0.08 Mol% n-butane neo-C5H12 = 0.00 Mol% neo-pentane i-C5H12 = 0.04 Mol% iso-pentane n-C5H12 = 0.04 Mol% n-pentane n-C6H14 = 0.02 Mol% n-hexane n-C7H16 = 0.01 Mol% n-heptane n-C8H18 = 0.01 Mol% n-octane n-C9H20 = 0.01 Mol% n-nonane n-C10H22 = 0.01 Mol% n-decane

Results

Molar Mass = 17.3170 kg/kmol Gas Constant = 480.135 J/(kg*K) Low Heat Value = 39.729 MJ/kg = 17081. BTU/lb High Heat Value = 44.185 MJ/kg = 18996. BTU/lb Pressure 1.013bar 1.013bar 1.013bar 14.73psi Actual Temperature 0. degC 15. degC 25. degC 60. degF Actual ------------------------------------+---------+---------+---------+---------+ Wobbe Lower Index MJ/m3 39.747 37.668 Wobbe Upper Index MJ/m3 44.204 41.892 Low Heat Value MJ/m3 30.755 29.144 28.161 29.155 2480.543 Low Heat Value BTU/ft3 825.4 782.2 755.8 782.5 66575.3 High Heat Value MJ/m3 34.204 32.412 31.319 32.425 2758.736 High Heat Value BTU/ft3 918.0 869.9 840.6 870.2 74041.7 Density kg/m3 0.7741 0.7336 0.7088 0.7338 62.4360 Density lb/ft3 0.0483 0.0458 0.0443 0.0458 3.8979 Relative Density [-] 0.5987 0.5986 Compression Factor [-] 0.99803 0.99839 0.99859 0.99840 0.95309 Internal Energy kJ/kg -182.385 -159.342 -143.749 -158.483 -106.063 Enthalpy kJ/kg -51.494 -21.214 -0.799 -20.087 54.101 Entropy kJ/(kg*K) 0.2860 0.3939 0.4636 0.3967 -1.5311 Heat Capacity cp kJ/(kg*K) 2.0058 2.0321 2.0512 2.0331 2.5087 Heat Capacity cv kJ/(kg*K) 1.5206 1.5475 1.5671 1.5486 1.7374 Isentropic Exponent [-] 1.3165 1.3110 1.3071 1.3108 1.4123 Joule Thomson Coeff. K/bar 0.4627 0.4137 0.3845 0.4120 0.1996 Speed of Sound m/s 415.105 425.537 432.263 425.915 475.610 Dynamic Viscosity Pa*s*E-6 10.954 11.477 11.821 11.496 15.329 HeatCircuitCapacityW/(m*K) 0.0286 0.0306 0.0320 0.0307 0.0462 ISO 6976-1995 for Molar Mass and Calorific Values at 0degC ISO 20765-1 2005 for Thermodynamic Properties

## Real gas density is higher than ideal gas density

The effect of compression factor tells that the real gas density is higher than the ideal gas density, typically by 1...5% at operating conditions. Ideal gas density neglects the compression factor. The compression factor for the example is z = 0.95309, therefore the real natural gas density in this example is 4.69% higher.

The real gas density is

ρ = MR · pz · T = pR_{s} · z · T (1)

## Real gas enthalpy is lower than ideal gas enthalpy

Next to the chemical energy, which is expressed by the heat value, the real gas enthalpy is part of the overall heat input. The real natural gas enthalpy is smaller than the ideal gas enthalpy, and the overall heat input is

HI = m · (LHV + Δh) (2)

HI = m · [39729 + 54.101 - (-51.494)] kJ/kg = m · 39835 kJ/kg (3)

The ideal gas enthalpy would be calculated with just the temperature difference and a typical heat capacity of 2.3 kJ/(kg·K).

HI = m · (LHV + c_{p}ΔT) (4)

HI = m · [39729 + 2.3 · (76.85 - 0.0)] kJ/kg = m · 39906 kJ/kg (5)

The real heat input in this example is 0.18% lower.