Natural gas, assumed here to be pure methane, is liquefied in a Claude process. Compression is to 60 bar and precooling is to 300 K. The expander and throttle exhaust to a pressure of 1 bar. Recycle methane at this pressure leaves the exchanger system (point 15, Fig. 9.7) at 295 K. Assume no heat

Question

Natural gas, assumed here to be pure methane, is liquefied in a Claude process. Compression is to 60 bar and precooling is to 300 K. The expander and throttle exhaust to a pressure of 1 bar. Recycle methane at this pressure leaves the exchanger system (point 15, Fig. 9.7) at 295 K. Assume no heat leaks into the system from the surroundings, an expander efficiency of 75%, and an expander exhaust of saturated vapor. For a draw-off to the expander of 25% of the methane entering the exchanger system (x = 0.25), what fraction z of the methane is liquefied, and what is the temperature of the high-pressure stream entering the throttle valve?

Accepted Answer

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

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

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

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

[latex] T^{sat} = 111.5 K[/latex]

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

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

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

The enthalpy at the draw-off point between exchangers I and II, [latex] H_5 [/latex] , is required for solution of Eq. (9.7). The expander efficiency η is known, as is [latex] H_{12} [/latex], the enthalpy of the expander exhaust. The calculation of [latex] H_5 (=H_{11}) [/latex] , 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:

[latex] z=\frac{x\left(H_{12}-H_5 ight)+H_4-H_{15}}{H_9-H_{15}} [/latex] (9.7)

[latex] \Delta H=H_{12}-H_5=\eta(\Delta H)_S=\eta\left(H_{12}^{\prime}-H_5 ight) [/latex]

Solution for [latex] H_{12 }[/latex] yields:

[latex] H_{12}=H_5+\eta\left(H_{12}^{\prime}-H_5 ight) [/latex] (A)

where [latex] H_{12}^{\prime} [/latex] 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 [latex] T_5[/latex] leads to values for [latex]H_5 and S_5[/latex], from which [latex] H_{12}^{\prime} [/latex] can be found. All quantities in Eq. (A) are then known. If the equation is not satisfied, then a new value is chosen for [latex]T_5[/latex], and the process continues until Eq. (A) is satisfied. For example, at 60 bar and 260 K, the enthalpy and entropy are [latex]745.27 kJ⋅kg^{−1} and 4.033 kJ⋅kg^{−1} .K^{−1}[/latex] , respectively. The saturated liquid and vapor at 1 bar have [latex]S^ l = −0.005 and S^v = 4.579[/latex], 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:

[latex] H_{12}^{\prime}=H_9+0.8808\left(H_{12}-H_9 ight)=449.6 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} [/latex]

Using this value in Eq. (A) yields [latex]H_{12} = 508.8 kJ⋅kg^{−1}[/latex], which is below the known value of [latex]H_{12} = 510.6 kJ⋅kg^{−1}[/latex]. Thus, [latex]T_5[/latex] 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 [latex]T_5[/latex] shows that Eq. (A) is satisfied for:

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

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

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

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,

[latex] \dot{m}_4\left(H_5-H_4 ight)+\dot{m}_{15}\left(H_{15}-H_{14} ight)=0 [/latex]

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

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

Then,

[latex] \left.H_{14}=782.2 \mathrm{~kJ} \cdot \mathrm{kg}^{-1} \quad T_{14}=239.4 \mathrm{~K} ext { (at } 1 \mathrm{bar} ight) [/latex]

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

For exchanger II,

[latex] \dot{m}_7\left(H_7-H_5 ight)+\dot{m}_{14}\left(H_{14}-H_{12} ight)=0 [/latex]

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

[latex] H_7=H_5-\frac{1-z}{1-x}\left(H_{14}-H_{12} ight)=748.8-\frac{1-0.1187}{1-0.25}(782.2-510.6) [/latex]

Then

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

As the value of x increases, [latex]T_7[/latex] 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),

[latex] z=\frac{H_4-H_{15}}{H_9-H_{15}} [/latex] (9.8)

[latex] z=\frac{855.3-903.0}{-0.6-903.0}=0.0528 [/latex]

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:

[latex] H_7=H_4-(1-z)\left(H_{15}-H_{10} ight) [/latex]

Substitution of known values yields:

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

The corresponding temperature of the methane entering the throttle valve is [latex]T_7 = 202.1 K.[/latex]

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

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

Related Answered Questions

Verified Answer:

A house has a winter heating requirement of 30 kJ.s^−1 and a summer cooling requirement of 60 kJ.s^−1. Consider a heat-pump installation to maintain the house temperature at 20°C in winter and 25°C in summer. This requires circulation of the refrigerant through interior exchanger coils at ...

Verified Answer: