Minimum work associated with separating nitrogen from air: An exergy analysis

Introduction

Nitrogen is essential for a variety of industries, including heat treatment, laser cutting, fire protection, and food packaging. The nitrogen requirements for a process are usually specified by purity (expressed as the percentage form of the mole fraction) and flow rate at a certain pressure. Most processes require nitrogen purities in the range of 95–99.999%. The three main processes for obtaining nitrogen from ambient air are cryogenic distillation, membrane separation, and pressure-swing adsorption (PSA). The selection of the process depends on the product flow rate, purity, and state (liquid or gas). Cryogenic distillation was first introduced by Carl von Linde in 1895 and patented in 1903.¹ PSA and membrane separation technologies became more popular in the 1980s, and these technologies can provide sufficient nitrogen purity and flow rates for a wide range of applications at a much lower cost than cryogenics.²

Cryogenic distillation is the most energy intensive of the three main air separation processes; cryogenic systems typically have an energy consumption of 2.56 kWh/kg of liquid nitrogen, whereas PSA systems typically require 0.31–0.63 kWh/kg of nitrogen gas.³ This usually makes cryogenic distillation economically viable for large-scale production. However, cryogenic distillation can provide greater nitrogen purity (>99.999%⁴) than membrane separation and PSA. In addition, some industries require nitrogen in the liquid form, which cannot be obtained via membrane separation or PSA. Membrane systems operate via selective permeation, where differences in gas molecule diffusion rates drive separation through a packed container of hollow fibers. Membrane systems typically provide nitrogen with a purity range of 95–99.5%.⁴ PSA systems typically utilize two sieve beds filled with carbon molecular sieves (CMS), a material that selectively adsorbs gas molecules at pores on its surface. PSA systems can provide nitrogen purities in the range of 95–99.999%.⁵

Many examples of exergy analyses performed for gas separation via cryogenic distillation,^6–9 PSA,^10–12 and membrane systems^13–15 exist in the literature, most of which focus on specific realistic processes. Entropy generation (directly proportional to exergy destruction) has also been used to describe the performance of gas separation systems.¹⁶ Weber et al.¹⁷ utilized the concept of a “physical optimum” to compare the energy efficiencies of cryogenic and PSA processes for oxygen production. This “physical optimum” identifies avoidable and unavoidable losses as modifications of exergy analysis.

Regardless of the approach, a minimum theoretical amount of work input is required to obtain gas at different pressures and concentrations than under ambient conditions. This article presents a theoretical exergy analysis of air separation, indicating the minimum theoretical work input as a function of the product pressure, purity, and process recovery rate. This analysis considered an air separation system as a black box, with the input, output, and exhaust assumed to be ideal gas mixtures of nitrogen and oxygen at 15°C.

Methods

Mathematical modeling

Assumptions

The following assumptions will be made for the analysis:

• • Air is composed of 79% nitrogen and 21% oxygen (all other constituents were neglected).

• • Gas mixtures will be treated as ideal gases.

• • The feed (initial state) air was standard atmospheric air at zero gauge pressure (15°C and 101.325 kPa),¹⁸ representing the dead state.

• • The exhaust and product mixtures were equal to that of the dead state temperature.

• • The exhaust gas pressure will be equal to that of the dead state.

• • Kinetic and potential energy changes are negligible.

• • Only the initial and final states of gas mixtures are required.

Performance of air separation systems

For this analysis, the minimum theoretical work required to obtain nitrogen at the specified purities and pressures was determined by calculating the useful work potential (exergy) of the product and exhaust gas mixtures. Exergy describes how much work a reversible system could produce as the contents of the system proceed from a specified state to the dead state, a condition in which the contents are in thermodynamic equilibrium with the environment. Under reversible conditions, the work required to separate a gas mixture is equal to that required for the mixing of its individual components.

The efficiency of a work-consuming device (such as an air separation system) in terms of the second law of thermodynamics can be described by:

(1)

$${\mathrm{\eta }}_{\mathrm{II}}=\frac{{\mathrm{W}}_{\mathrm{rev}}}{{\mathrm{W}}_{\mathrm{a}}}$$

where ${\mathrm{\eta }}_{\mathrm{II}}$ is the second law efficiency, W_rev is the theoretical reversible work required for a process, and W_a is the actual work input to the process.

The purity and recovery of nitrogen in the product affect the amount of work required to perform the separation process. Purity (θ) is equivalent to the mole fraction of nitrogen in the product gas (Equation 2). The recovery (r) of nitrogen from the input air is given by Equation 3.

(2)

$$\mathrm{\theta }={\mathrm{y}}_{\mathrm{N}2,\mathrm{P}}=1-{\mathrm{y}}_{\mathrm{O}2,\mathrm{P}}$$

(3)

$$\mathrm{r}=\frac{{\mathrm{N}}_{\mathrm{N}2,\mathrm{P}}}{{\mathrm{N}}_{\mathrm{N}2,\mathrm{F}}}=\frac{{\mathrm{N}}_{\mathrm{N}2,\mathrm{P}}}{{\mathrm{N}}_{\mathrm{N}2,\mathrm{P}}+{\mathrm{N}}_{\mathrm{N}2,\mathrm{E}}}$$

N is the number of moles, and the subscripts F, P, and E represent the feed (input), product, and exhaust, respectively.¹¹ Equations 4 and 5 represent the component balances for the nitrogen and oxygen involved in the separation process, respectively.

(4)

$${\mathrm{N}}_{\mathrm{N}2,\mathrm{F}}={\mathrm{N}}_{\mathrm{N}2,\mathrm{P}}+{\mathrm{N}}_{\mathrm{N}2,\mathrm{E}}$$

(5)

$${\mathrm{N}}_{\mathrm{O}2,\mathrm{F}}={\mathrm{N}}_{\mathrm{O}2,\mathrm{P}}+{\mathrm{N}}_{\mathrm{O}2,\mathrm{E}}$$

Derivation of the exergy equation

Pal¹⁹ described the exergy of a gas mixture $\left(\mathrm{\phi }\right)$ as being composed of two parts, namely the thermo-mechanical exergy $\left({\mathrm{\phi }}_{\mathrm{TM}}\right)$ and chemical exergy $\left({\mathrm{\phi }}_{\mathrm{C}}\right)$:

(6)

$$\mathrm{\phi }={\mathrm{\phi }}_{\mathrm{TM}}+{\mathrm{\phi }}_{\mathrm{C}}$$

On a unit mass basis, the thermo-mechanical exergy of a closed system is expressed as²⁰:

(7)

$${\mathrm{\phi }}_{\mathrm{TM}}=\left(\mathrm{u}-{\mathrm{u}}_0\right)+{\mathrm{P}}_0\left(\mathrm{v}-{\mathrm{v}}_0\right)-{\mathrm{T}}_0\left(\mathrm{s}-{\mathrm{s}}_0\right)+\mathrm{ke}+\mathrm{pe}$$

where u is the specific internal energy (kJ/kg), P is the pressure (kPa), v is the specific volume (m³/kg), T is the temperature (K), s is the specific entropy (kJ/kg · K), ke is the specific kinetic energy, and pe is the potential energy. The subscript 0 refers to the dead state conditions. The kinetic and potential energy changes are assumed to be negligible:

(8)

$${\mathrm{\phi }}_{\mathrm{TM}}=\left(\mathrm{u}-{\mathrm{u}}_0\right)+{\mathrm{P}}_0\left(\mathrm{v}-{\mathrm{v}}_0\right)-{\mathrm{T}}_0\left(\mathrm{s}-{\mathrm{s}}_0\right)$$

The portion of Equation 8 concerning the specific volume can be written as (assuming an ideal gas):

(9)

$${\mathrm{P}}_0\left(\mathrm{v}-{\mathrm{v}}_0\right)={\mathrm{P}}_0\left(\frac{{\mathrm{R}}_{\mathrm{m}}\mathrm{T}}{\mathrm{P}}-\frac{{\mathrm{R}}_{\mathrm{m}}{\mathrm{T}}_0}{{\mathrm{P}}_0}\right)={\mathrm{R}}_{\mathrm{m}}{\mathrm{P}}_0\left(\frac{\mathrm{T}}{\mathrm{P}}-\frac{{\mathrm{T}}_0}{{\mathrm{P}}_0}\right)$$

where R_m (kJ/kg · K) is the gas constant of the mixture. The entropy change of an ideal gas with a constant specific heat approximation from a specified state to a dead state is:

(10)

$$\mathrm{s}-{\mathrm{s}}_0={\mathrm{c}}_{\mathrm{p},\mathrm{avg}}ln\left(\frac{\mathrm{T}}{{\mathrm{T}}_0}\right)-{\mathrm{R}}_{\mathrm{m}}ln\left(\frac{\mathrm{P}}{{\mathrm{P}}_0}\right)$$

Substituting Equation 9 and Equation 10 into Equation 8:

(11)

$${\mathrm{\phi }}_{\mathrm{TM}}=\left(\mathrm{u}-{\mathrm{u}}_0\right)+{\mathrm{R}}_{\mathrm{m}}{\mathrm{P}}_0\left(\frac{\mathrm{T}}{\mathrm{P}}-\frac{{\mathrm{T}}_0}{{\mathrm{P}}_0}\right)-{\mathrm{T}}_0\left[{\mathrm{c}}_{\mathrm{p},\mathrm{avg}}ln\left(\frac{\mathrm{T}}{{\mathrm{T}}_0}\right)-{\mathrm{R}}_{\mathrm{m}}ln\left(\frac{\mathrm{P}}{{\mathrm{P}}_0}\right)\right]$$

Following the assumption that the temperature of all states is equal to that of the dead state, the specific internal energy terms in Equation 11 are eliminated because the specific internal energy of an ideal gas is only a function of temperature. Simplifying Equation 11 with the constant temperature assumption:

(12)

$${\mathrm{\phi }}_{\mathrm{TM}}={\mathrm{R}}_{\mathrm{m}}{\mathrm{T}}_0\left(\frac{{\mathrm{P}}_0}{\mathrm{P}}-1\right)+{\mathrm{R}}_{\mathrm{m}}{\mathrm{T}}_{0}ln\left(\frac{\mathrm{P}}{{\mathrm{P}}_0}\right)={\mathrm{R}}_{\mathrm{m}}{\mathrm{T}}_0\left\{\frac{{\mathrm{P}}_0}{\mathrm{P}}+ln\left(\frac{\mathrm{P}}{{\mathrm{P}}_0}\right)-1\right\}$$

Pal¹⁹ expressed the molar chemical exergy of an ideal gas mixture as follows:

(13)

$${\overline{\mathrm{\phi }}}_{\mathrm{C}}=\overline{\mathrm{R}}{\mathrm{T}}_0\sum {\mathrm{y}}_{\mathrm{i}}ln\left({\mathrm{y}}_{\mathrm{i}}/{\mathrm{y}}_{\mathrm{i},0}\right)$$

where $\overline{\mathrm{R}}$ is the universal gas constant, y_i is the molar concentration of gas species in the mixture, and y_i,0 is the molar concentration of gas species in the reference environment. Assuming that nitrogen and oxygen were the only gases in the system:

(14)

$${\overline{\mathrm{\phi }}}_{\mathrm{C}}=\overline{\mathrm{R}}{\mathrm{T}}_0\sum {\mathrm{y}}_{\mathrm{i}}ln\left({\mathrm{y}}_{\mathrm{i}}/{\mathrm{y}}_{\mathrm{i},0}\right)=\overline{\mathrm{R}}{\mathrm{T}}_0\left\{{\mathrm{y}}_{\mathrm{N}2}ln\left(\frac{{\mathrm{y}}_{\mathrm{N}2}}{{\mathrm{y}}_{\mathrm{N}2,0}}\right)+{\mathrm{y}}_{\mathrm{O}2}ln\left(\frac{{\mathrm{y}}_{\mathrm{O}2}}{{\mathrm{y}}_{\mathrm{O}2,0}}\right)\right\}$$

Considering the process shown in Figure 1, one can observe the minimum work required to obtain the separated product and exhaust gas mixtures is the sum of their exergy values. Note that the exhaust gas mixture has no thermomechanical exergy, as it is at the dead state temperature and pressure. Therefore, in this case, Equation 6 becomes:

(15)

$$\mathrm{\phi }={\left[{\mathrm{\phi }}_{\mathrm{TM}}+{\mathrm{\phi }}_{\mathrm{C}}\right]}_{\mathrm{P}}+{\left[{\mathrm{\phi }}_{\mathrm{C}}\right]}_{\mathrm{E}}$$

Figure 1. Reversible separation and mixing processes.

Equation 15 is expanded to determine the total exergy (Φ) of the product and exhaust mixtures as follows:

(16)

$$\mathrm{\Phi }={\left[{\mathrm{m}}_{\mathrm{m}}{\mathrm{\phi }}_{\mathrm{TM}}+{\mathrm{N}}_{\mathrm{m}}{\overline{\mathrm{\phi }}}_{\mathrm{C}}\right]}_{\mathrm{P}}+{\left[{\mathrm{N}}_{\mathrm{m}}{\overline{\mathrm{\phi }}}_{\mathrm{C}}\right]}_{\mathrm{E}}$$

Substituting Equation 12 and Equation 14 into Equation 16:

(17)

$$\mathrm{\Phi }={\left[{\mathrm{m}}_{\mathrm{m}}{\mathrm{R}}_{\mathrm{m}}{\mathrm{T}}_0\left\{\frac{{\mathrm{P}}_0}{\mathrm{P}}+ln\left(\frac{\mathrm{P}}{{\mathrm{P}}_0}\right)-1\right\}+{\mathrm{N}}_{\mathrm{m}}\overline{\mathrm{R}}{\mathrm{T}}_0\left\{{\mathrm{y}}_{\mathrm{N}2}ln\left(\frac{{\mathrm{y}}_{\mathrm{N}2}}{{\mathrm{y}}_{\mathrm{N}2,0}}\right)+{\mathrm{y}}_{\mathrm{O}2}ln\left(\frac{{\mathrm{y}}_{\mathrm{O}2}}{{\mathrm{y}}_{\mathrm{O}2,0}}\right)\right\}\right]}_{\mathrm{P}}+{\left[{\mathrm{N}}_{\mathrm{m}}\overline{\mathrm{R}}{\mathrm{T}}_0\left\{{\mathrm{y}}_{\mathrm{N}2}ln\left(\frac{{\mathrm{y}}_{\mathrm{N}2}}{{\mathrm{y}}_{\mathrm{N}2,0}}\right)+{\mathrm{y}}_{\mathrm{O}2}ln\left(\frac{{\mathrm{y}}_{\mathrm{O}2}}{{\mathrm{y}}_{\mathrm{O}2,0}}\right)\right\}\right]}_{\mathrm{E}}$$

The following expression can be substituted into Equation 17 to obtain terms with similar units:

(18)

$${\mathrm{m}}_{\mathrm{m}}={\mathrm{N}}_{\mathrm{m}}\frac{\overline{\mathrm{R}}}{{\mathrm{R}}_{\mathrm{m}}}$$

Substituting:

(19)

$$\mathrm{\Phi }={\mathrm{N}}_{\mathrm{P}}\overline{\mathrm{R}}{\mathrm{T}}_0\left[\left\{\frac{{\mathrm{P}}_0}{\mathrm{P}}+ln\left(\frac{\mathrm{P}}{{\mathrm{P}}_0}\right)-1\right\}+{\mathrm{y}}_{\mathrm{N}2,\mathrm{P}}ln\left(\frac{{\mathrm{y}}_{\mathrm{N}2,\mathrm{P}}}{{\mathrm{y}}_{\mathrm{N}2,0}}\right)+{\mathrm{y}}_{\mathrm{O}2,\mathrm{P}}ln\left(\frac{{\mathrm{y}}_{\mathrm{O}2,\mathrm{P}}}{{\mathrm{y}}_{\mathrm{O}2,0}}\right)\right]+{\mathrm{N}}_{\mathrm{E}}\overline{\mathrm{R}}{\mathrm{T}}_0\left[{\mathrm{y}}_{\mathrm{N}2,\mathrm{E}}ln\left(\frac{{\mathrm{y}}_{\mathrm{N}2,\mathrm{E}}}{{\mathrm{y}}_{\mathrm{N}2,0}}\right)+{\mathrm{y}}_{\mathrm{O}2,\mathrm{E}}ln\left(\frac{{\mathrm{y}}_{\mathrm{O}2,\mathrm{E}}}{{\mathrm{y}}_{\mathrm{O}2,0}}\right)\right]$$

Equations 20 through 24 can be utilized to express Equation 19 in terms of product pressure, total volume (V), purity, and recovery:

(20)

$${\mathrm{N}}_{\mathrm{P}}=\frac{\mathrm{PV}}{\overline{\mathrm{R}}{\mathrm{T}}_0}$$

(21)

$${\mathrm{N}}_{\mathrm{E}}={\mathrm{N}}_{\mathrm{P}}\left(\frac{\mathrm{\theta }}{\mathrm{r}{\mathrm{y}}_{\mathrm{N}2,0}}-1\right)$$

(22)

$${\mathrm{N}}_{\mathrm{N}2,\mathrm{E}}=\mathrm{\theta }{\mathrm{N}}_{\mathrm{P}}\left(\frac{1}{\mathrm{r}}-1\right)$$

(23)

$${\mathrm{y}}_{\mathrm{O}2,\mathrm{E}}=1-{\mathrm{y}}_{\mathrm{N}2,\mathrm{E}}$$

(24)

$${\mathrm{y}}_{\mathrm{N}2,\mathrm{E}}=\frac{\mathrm{\theta }{\mathrm{y}}_{\mathrm{N}2,0}\left(1-\mathrm{r}\right)}{\mathrm{\theta }-\mathrm{r}{\mathrm{y}}_{\mathrm{N}2,0}}$$

Substituting into Equation 19 and simplifying:

(25)

$$\mathrm{\Phi }=\mathrm{PV}\left[\left\{\frac{{\mathrm{P}}_0}{\mathrm{P}}+ln\left(\frac{\mathrm{P}}{{\mathrm{P}}_0}\right)-1\right\}+\mathrm{\theta ln}\left(\frac{\mathrm{\theta }}{{\mathrm{y}}_{\mathrm{N}2,0}}\right)+\left(1-\mathrm{\theta }\right)ln\left(\frac{1-\mathrm{\theta }}{{\mathrm{y}}_{\mathrm{O}2,0}}\right)+\frac{\mathrm{\theta }}{{\mathrm{y}}_{\mathrm{N}2,0}}\left\{\left(1-\mathrm{r}\right)ln\left(\frac{\mathrm{\theta }\left(1-\mathrm{r}\right)}{\mathrm{\theta }-\mathrm{r}{\mathrm{y}}_{\mathrm{N}2,0}}\right)+\left(\frac{1}{{\mathrm{y}}_{\mathrm{N}2,0}}-\frac{\mathrm{r}}{\mathrm{\theta }}+\mathrm{r}-1\right)ln\left(\frac{1}{{\mathrm{y}}_{\mathrm{O}2,0}}-\frac{\mathrm{\theta }{\mathrm{y}}_{\mathrm{N}2,0}\left(1-\mathrm{r}\right)}{{\mathrm{y}}_{\mathrm{O}2,0}\left(\mathrm{\theta }-\mathrm{r}{\mathrm{y}}_{\mathrm{N}2,0}\right)}\right)\right\}\right]$$

The second-law efficiency of an actual air separation system can be determined via Equation 1 by utilizing the calculated values from Equation 25 as W_rev, provided the final state of the product from the actual process as a gas at the dead state temperature. Expressing Equation 25 in kJ/m³:

(26)

$$\mathrm{\Phi }\left(\mathrm{kJ}/{\mathrm{m}}^3\right)=\mathrm{P}\left[\left\{\frac{{\mathrm{P}}_0}{\mathrm{P}}+ln\left(\frac{\mathrm{P}}{{\mathrm{P}}_0}\right)-1\right\}+\mathrm{\theta ln}\left(\frac{\mathrm{\theta }}{{\mathrm{y}}_{\mathrm{N}2,0}}\right)+\left(1-\mathrm{\theta }\right)ln\left(\frac{1-\mathrm{\theta }}{{\mathrm{y}}_{\mathrm{O}2,0}}\right)+\frac{\mathrm{\theta }}{{\mathrm{y}}_{\mathrm{N}2,0}}\left\{\left(1-\mathrm{r}\right)ln\left(\frac{\mathrm{\theta }\left(1-\mathrm{r}\right)}{\mathrm{\theta }-\mathrm{r}{\mathrm{y}}_{\mathrm{N}2,0}}\right)+\left(\frac{1}{{\mathrm{y}}_{\mathrm{N}2,0}}-\frac{\mathrm{r}}{\mathrm{\theta }}+\mathrm{r}-1\right)ln\left(\frac{1}{{\mathrm{y}}_{\mathrm{O}2,0}}-\frac{\mathrm{\theta }{\mathrm{y}}_{\mathrm{N}2,0}\left(1-\mathrm{r}\right)}{{\mathrm{y}}_{\mathrm{O}2,0}\left(\mathrm{\theta }-\mathrm{r}{\mathrm{y}}_{\mathrm{N}2,0}\right)}\right)\right\}\right]$$

Equations 25 and 26 are valid within the following ranges for the purity and recovery:

$$\left\{\mathrm{\theta } | 0.79\le \mathrm{\theta }<1\right\}$$

$$\left\{\mathrm{r} | 0<\mathrm{r}<1\right\}$$

Results

A source code for plotting results for Equation 26 has been made available on Zenodo.²¹ Figure 2 and 3 show graphical depictions of Equation 26 for the product nitrogen gas at atmospheric pressure and 800 kPa, respectively. Figure 2 shows the condition in which the product gas has no thermomechanical exergy. It can be observed that no work input is required at 79% product purity, regardless of the recovery rate, because it was assumed that the ambient air is 79% nitrogen; therefore, the product gas has neither thermomechanical exergy nor chemical exergy. As shown in Figure 3, the work input is still required for a product purity of 79%, which essentially represents the reversible compression of ambient air. This work input value is constant at a product purity of 79% for any value of recovery and is determined using Equation 12.

Figure 2. Reversible work input to obtain nitrogen gas at 101.325 kPa.

Figure 3. Reversible work input to obtain nitrogen gas at 800 kPa.

Discussion

The work input for actual air separation systems will always be greater than the value determined by utilizing Equation 26, owing to irreversibility. Cryogenic distillation, membrane separation, and PSA all experience heat loss, friction, and unrestrained gas expansion, among other irreversibilities. Friction exists at all locations where there is relative motion between a fluid and a solid, and all gases exhausted to the atmosphere represent an unrestrained expansion of gas. One of the most notable sources of irreversibility is during compression, as heat rejection from a compressor is usually within 60–90% of the power input.²⁰

To compare realistic air separation processes to an ideal process, an average PSA system was assumed to provide 99.9% pure nitrogen at 800 kPa within a recovery range of 15–50%. As shown in Figure 4, the minimum required work input for this process was between 1,151–1,198 kJ/m³ (an average of 1,174.5 kJ/m³). It is also assumed that the typical PSA input power range of 0.31–0.63 kWh/kg provided by Syakdani et al.³ applies to these conditions. At 800 kPa, this input power range was equivalent to 10,446–21,228 kJ/m³. Utilizing Equation 1, this represents a second law efficiency of only 5.5–11.2%.

Figure 4. Reversible work input to obtain 99.9% nitrogen gas at 800 kPa.

Conclusions

A theoretical exergy analysis of air separation is presented in this article, indicating the minimum theoretical work input as a function of the product pressure, purity, and process recovery rate. The air separation system was considered a black box, with the input, output, and exhaust gases assumed to be binary mixtures of nitrogen and oxygen at 15°C. Equations 25 and 26 derived from this analysis are only valid for cases where the feed, product, and exhaust mixtures are in the gas phase and at a dead state temperature of 15°C. These equations are still valid for cryogenic distillation, even though the process involves phase changes, assuming that the feed to the system is ambient air and the product and exhaust mixtures are brought back to the gas phase. The following conclusions were drawn from this analysis: