Question 9.3: Natural gas, assumed here to be pure methane, is liquefied i...

Data for methane are available in the NIST WebBook  ,^7   from which the following values were obtained:

H_4=855.3 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}            (at 300 K and 60 bar)

H_{15}=903.0 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}           (at 295 K and 1 bar)

For saturated liquid and vapor, at a pressure of 1 bar:

H_9=-0.6 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}           ( saturated liquid )

H_{12}=510.6 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}              ( saturated vapor)

S_{12}=4.579 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \cdot \mathrm{K}^{-1}       ( saturated vapor )

The enthalpy at the draw-off point between exchangers I and II, H_5 , is required for solution of Eq. (9.7). The expander efficiency η is known, as is H_{12} , the enthalpy of the expander exhaust. The calculation of H_5  (=H_{11}) , the expander inlet enthalpy, is less straightforward than the usual calculation of the exhaust enthalpy from the entrance enthalpy. The equation defining expander efficiency can be written:

z=\frac{x\left(H_{12}-H_5\right)+H_4-H_{15}}{H_9-H_{15}}          (9.7)

\Delta H=H_{12}-H_5=\eta(\Delta H)_S=\eta\left(H_{12}^{\prime}-H_5\right)

Solution for H_{12 } yields:

H_{12}=H_5+\eta\left(H_{12}^{\prime}-H_5\right)          (A)

where H_{12}^{\prime} is the enthalpy at 1 bar as the result of isentropic expansion from point 5. This enthalpy is readily found once the conditions at point 5 are known. Thus a trial calculation or iterative solution is required. A trial value of temperature T_5 leads to values for H_5  and  S_5, from which H_{12}^{\prime} can be found. All quantities in Eq. (A) are then known. If the equation is not satisfied, then a new value is chosen for T_5, and the process continues until Eq. (A) is satisfied. For example, at 60 bar and 260 K, the enthalpy and entropy are 745.27  kJ⋅kg^{−1}  and  4.033  kJ⋅kg^{−1} .K^{−1} , respectively. The saturated liquid and vapor at 1 bar have S^ l = −0.005  and  S^v = 4.579, respectively. Using these values, isentropic expansion from 260 K and 60 bar to 1 bar would give a vapor fraction of 0.8808. This would give:

H_{12}^{\prime}=H_9+0.8808\left(H_{12}-H_9\right)=449.6 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

Using this value in Eq. (A) yields H_{12} = 508.8  kJ⋅kg^{−1}, which is below the known value of H_{12} = 510.6  kJ⋅kg^{−1}. Thus, T_5  must be higher than the assumed value of 260 K. Repeating this process (in an automated fashion using a spreadsheet or computer program) for other values of T_5 shows that Eq. (A) is satisfied for:

T_5=261.2 \mathrm{~K} \quad H_5=748.8 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} ( at 60 bar )

Substitution of values into Eq. (9.7) now yields:

z=\frac{0.25(510.6-748.8)+855.3-903.0}{-0.6-903.0}=0.1187

Thus 11.9% of the methane entering the exchanger system is liquefied.

The temperature at point 7 depends on its enthalpy, which is found from energy
balances on the exchanger system. Thus, for exchanger I,

\dot{m}_4\left(H_5-H_4\right)+\dot{m}_{15}\left(H_{15}-H_{14}\right)=0

With \dot{m}_{15}=\dot{m}_4-\dot{m}_9 \text { and } \dot{m}_9 / \dot{m}_4=z , this equation may be rearranged to give:

H_{14}=\frac{H_5-H_4}{1-z}+H_{15}=\frac{748.8-855.3}{1-0.1187}+903.0

Then,

\left.H_{14}=782.2 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \quad T_{14}=239.4 \mathrm{~K} \text { (at } 1 \mathrm{bar}\right)

where T_{14} is found by evaluating H for methane at 1 bar and varying the temperature to match the known T_{14} .

For exchanger II,

\dot{m}_7\left(H_7-H_5\right)+\dot{m}_{14}\left(H_{14}-H_{12}\right)=0

With \dot{m}_7=\dot{m}_4-\dot{m}_{12} \text { and } \dot{m}_{14}=\dot{m}_4-\dot{m}_9 and with the definitions of z and x, this equation upon rearrangement becomes:

Then

H_7=429.7 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \quad T_7=199.1 \mathrm{~K} ( at 60 bar )

As the value of x increases, T_7 decreases, eventually approaching the saturation temperature in the separator, and requiring an exchanger II of infinite area. Thus x is limited on the high side by the cost of the exchanger system.

The other limit is for x = 0, the Linde system, for which by Eq. (9.8),

z=\frac{H_4-H_{15}}{H_9-H_{15}}       (9.8)

z=\frac{855.3-903.0}{-0.6-903.0}=0.0528

In this case only 5.3% of the gas entering the throttle valve emerges as liquid. The temperature of the gas at point 7 is again found from its enthalpy, calculated by the energy balance:

H_7=H_4-(1-z)\left(H_{15}-H_{10}\right)

Substitution of known values yields:

H_7=855.3-(1-0.0528)(903.0-510.6)=483.6 \mathrm{~kJ} \cdot \mathrm{kg}^{-1}

The corresponding temperature of the methane entering the throttle valve is T_7 = 202.1  K.

^7E. W. Lemmon, M. O. McLinden, and D. G. Friend, op. cit., http://webbook.nist.gov.