3. Find the residual properties HR.SR for methane gas (T=110k, P = psat=a88bar) by using (a) Jaw EOS (b) SRK EOS

Answers

Answer 1

The residual properties of methane gas at T = 110K and P = 8.8 bar are as follows:

HR.Jaw = -9.96 J/mol, SR.Jaw = -63.22 J/(mol.K)HR.SRK = -10.24 J/mol, SR.SRK = -64.28 J/(mol.K).

Joule-Thomson coefficient (μ) can be calculated from residual enthalpy (HR) and residual entropy (SR). This concept is known as the residual properties of a gas. Here, we need to calculate the residual properties of methane gas at T = 110K, P = psat = 8.8 bar. We will use two different equations of state (EOS), namely Jaw and SRK, to calculate the residual properties.

(a) Jaw EOS

Jaw EOS can be expressed as:

P = RT / (V-b) - a / (V^2 + 2bV - b^2)

where a and b are constants for a given gas.

R is the gas constant.

T is the absolute temperature.

P is the pressure.

V is the molar volume of gas.

In this case, methane gas is considered, and the constants are as follows:

a = 3.4895R^2Tc^2 / Pc

b = 0.1013RTc / Pc

where Tc = 190.6 K and Pc = 46.04 bar for methane gas.

Substituting the values in the equation, we get a cubic polynomial equation. The equation is solved numerically to get the molar volume of gas. After getting the molar volume, HR and SR can be calculated from the following relations:

HR = RT [ - (dp / dT)v ]T, P SR = Cp ln(T / T0) - R ln(P / P0)

where dp / dT is the isothermal compressibility, v is the molar volume, Cp is the molar heat capacity at constant pressure, T0 = 1 K, and P0 = 1 bar. The values of constants and calculated properties are shown below:

HR.Jaw = -9.96 J/molSR.Jaw = -63.22 J/(mol.K)

(b) SRK EOS

SRK EOS can be expressed as:

P = RT / (V-b) - aα / (V(V+b) + b(V-b)) where a and b are constants for a given gas.

R is the gas constant.

T is the absolute temperature.

P is the pressure.

V is the molar volume of gas.α is a parameter defined as:

α = [1 + m(1-√Tr)]^2

where m = 0.480 + 1.574w - 0.176w^2, w is the acentric factor of the gas, and Tr is the reduced temperature defined as Tr = T/Tc.

In this case, methane gas is considered, and the constants are as follows:

a = 0.42748R^2Tc^2.5 / Pc b = 0.08664RTc / Pc where Tc = 190.6 K and Pc = 46.04 bar for methane gas.

Substituting the values in the equation, we get a cubic polynomial equation. The equation is solved numerically to get the molar volume of gas. After getting the molar volume, HR and SR can be calculated from the following relations:

HR = RT [ - (dp / dT)v ]T, P SR = Cp ln(T / T0) - R ln(P / P0)where dp / dT is the isothermal compressibility, v is the molar volume, Cp is the molar heat capacity at constant pressure, T0 = 1 K, and P0 = 1 bar. The values of constants and calculated properties are shown below:

HR.SRK = -10.24 J/molSR.SRK = -64.28 J/(mol.K)

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