Kellogg process (medium pressure)

Fig. 5.25 shows a flowsheet for the Kellogg process. In this figure the different sections can easily be recognized. Natural gas is the typical feedstock. After sulfur removal, we have primary and secondary reformers followed by high-temperature and low-temperature sweet shift. Once the syngas is obtained, CO₂ is removed. Selexol or MDEA processes are among the selections for the Kellogg design. The traces of CO and CO₂ are removed in a methanator. The synthesis occurs in their proprietary converter designs using ruthenium catalyst (Kaap), which has recently substituted the Fe-based ones (after 90 years). The reaction operates at 300°C and 100–130 atm. Next, ammonia is separated. There are some rules that determine the type of separation, condensation, or absorption, depending on the operating conditions. Condensation is recommended above 100 atm in spite of the high fix costs. Absorption typically follows Raoults’ Law (see Chapter 2: Chemical processes).

Haber–Bosch process (medium pressure)

This is one of the most extended processes for the production of ammonia. There are versions that use natural gas, but coal was commonly the raw material. Fig. 5.26 shows the scheme of the process. The syngas is produced using steam, air, and coke. There are a few preparation stages including the scrubber, WGSR, and CO₂ capture. The converter operates at 200–350 atm and 550°C. The ammonia concentration in the outlet stream from the converter is from 13% to 15%, and the ratio between the recycle and the feed is 6–7 to 1. The converter is 12 m tall, with an inner diameter of 0.8 m. The wall is 0.16 m thick and uses 6 t of double-promoted iron catalyst (see Fig. 5.27). The purge is around 5%. The ammonia is absorbed in water that is distilled to recover the ammonia (Lloyd, 2011).

Claude’s process (high pressure)

George Claude’s process is characterized by a high operating pressure at the reactor (900–1000 atm), and temperatures in the range of 500–650°C using a promoted iron catalyst. No recirculation is present due to the high conversion achieved—higher than 40%. Since there is no recycle and no purge, the unreacted hydrogen is lost. The catalyst is iron oxidized in magnesia containing 5–10% calcium oxide. Magnesium oxide also dissolves in the mixture. The reactor includes a guard bed to absorb poisons and convert residual CO into methanol. As a result of the pressure, water can be used to condense the produced ammonia out of the product stream. The drawbacks of the process are the compression costs and shorter converter life. Thirty percent of US plants used the Claude process. The reactor is 5 m tall, with an inner diameter of 0.5 m and a wall thickness of 0.3 m. Fig. 5.28 shows a scheme of the original reactor (Lloyd, 2011).

Uhde’s process

The first stages of this process are common to others (see Fig. 5.29). The raw material is typically natural gas that is processed to remove the sulfur compounds. Next, it is mixed with steam to produce the syngas over nickel catalysts. The first reformer operates at 40 atm and 800–850°C to save compression energy later on. This reformer is top-fired, and its tubes are made of a stainless steel alloy for better reliability. In the second reformer, air is added. After gas cooling, the CO is transformed into CO₂ using the sweet shift combining HTS and LTS. The process uses ethanolamines for the removal of the CO₂, while the traces of CO and CO₂ are eliminated via methanation. This process is characterized by the use of two converters with three catalytic beds operating under radial flow conditions; see Fig. 5.30 for a scheme of the reactor. The first reactor consists of an intercooled three bed reactor operating at 110 atm. The second one, operating at 210 atm, is responsible for two thirds of the total ammonia production. The use of this system allows a limited pressure drop. The energy generated is used to produce high-pressure steam. The ammonia is recovered by condensation. The purge is processed to recover the hydrogen, and the rest of the gas is used as flue gas.

Fauser process (old process)

In this process the hydrogen is produced via electrolysis while the nitrogen comes from air fractionation Ernest et al. (1925). Both streams are mixed and compressed up to 200–300 atm after processing to remove oil and oxygen in a deoxo-type reactor.

${\mathrm{H}}_2+{\mathrm{O}}_2\to {\mathrm{H}}_2\mathrm{O}$

Water is condensed and removed before feeding the stream to the reactor. The reactor uses iron-promoted catalysts and operates at 500°C, achieving a conversion of 20%. It is 14 m tall with an interdiameter of 0.85 m and a wall thickness of 0.16 m. The ratio between the recycle and the feed is 4.5–5 to 1, and a part of the recycle is purged to avoid build-up. Fig. 5.31 shows a scheme of the reactor, and Fig. 5.32 the process.

Texaco process

The Texaco process is characterized by the technology used in the production of the syngas, the partial oxidation of hydrocarbons. It is called the Texaco syngas generation process (TSGP). Next, WGSR is carried out. Subsequently, the CO₂ is captured using the Rectisol process. Fig. 5.33 shows the stages for syngas production in the Texaco process.

Gasification

It is also possible to gasify coal or other carbonous mass for syngas production and later use it to produce ammonia. Gasification was responsible for 90% of the share of ammonia production by the 1940s.

Casale process

The main characteristic of this process, developed by Luigi Casale, is the reactor; a scheme can be seen in Fig. 5.22. It operates at 600 atm and 500°C, and the control of temperature in the reaction is achieved by recycling the gas so that 2–3% ammonia in the feed of the converter is allowed. As a result, the rate of ammonia production is reduced, and with it, the energy generation. The high operating pressures allow the use of cooling water to condense and separate the ammonia. The reactor is typically 10 m tall, with an inner diameter of 0.6 m and a wall thickness of 0.25 m. The ammonia concentration at the reactor exit is around 20% and the ratio between the recycle and the feed is 4–5 to 1 (Vancini, 1961, Lloyd, 2011).

Gas generator/water gas process

This process is characterized by the production of the syngas using the combination of the generator gas and water gas cycles that were presented in the beginning of the chapter. After the gas is generated, sulfur is removed using hydrodesulfurization and later H₂S removal stages. Next, WGS is carried out to remove the CO and increase the production of hydrogen. Subsequently, CO₂ is captured and the traces of CO are removed using copper. Finally, ammonia is synthesized. Fig. 5.34 shows the flowsheet.

Haldor–Topsoe process

The main feature of this process is that the raw material can be either natural gas or liquid hydrocarbons. After hydrodesulfurization and H₂S removal, primary and secondary reforming are performed. Air is injected in the second reformer. Next, HTS and LTS are carried out, and after CO₂ removal using Selexol or MDEA, methanation removes the traces of CO and CO₂. The main feature of the process is also the use of proprietary radial flow reactors operating at 140 atm. Ammonia can be condensed using water, and 20% of the unconverted gas is lost. Fig. 5.35 shows the flowsheet.

Example 5.7

In an ammonia production facility, the converter is fed with a stoichiometric mixture of syngas N₂+3H₂. The conversion at the reactor is 25% to ammonia. Assume that all the ammonia is separated in a condensation stage. The unconverted gases are recycled to the reactor. The initial feedstock contains 0.2 moles of Ar per 100 mol of mixture. At the entrance of the reaction, a maximum of 5 moles of ammonia per 100 mol of syngas is allowed. Compute the purge.

Solution

The synthesis loop is shown in the figure below. We formulate the mass balance from point 1 onwards, assuming that R is the syngas recycle (Fig. 5E7.1).

For the balance to the converter, the reaction taking place is as follows:

${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$

Note that we can consider the syngas a mixture, and that per 4 moles of reactants, 2 moles of ammonia are produced. Using this consideration, only three species are evaluated. Let “Recy” be the recycled syngas (Tables 5E7.1 and 5E7.2):

Table 5E7.1

Balance to Converter

	In	Reacts	Out
N2+3H2	100+Recy	−(100+Recy)·0.25	(100+Recy)·0.75
NH3		(2/4) (100+ Recy)·0.25	(100+Recy)·0.125
Ar	(100+Recy)·0.05		(100+Recy)·0.05

Table 5E7.2

Balance to Separator

	In	Ammonia Removal	Out
N2+3H2	(100+ Recy)·0.75		(100+Recy)·0.75
NH3	(100+ Recy)·0.125	(100+ Recy)·0.125
Ar	(100+ Recy)·0.05		(100+Recy)·0.05

The balance to the gas–liquid separator is covered in Table 5E7.2.

For the balance to the divider (the purge) we assume that a fraction of the recycle, α, is purged (Table 5E7.3).

Table 5E7.3

Balance to Divider

	In	Purge	Recycle
N2+3H2	(100+Recy)·0.75	(100+Recy)·0.75·α	(100+Recy)·0.75 (1−α)
NH3
Ar	(100+Recy)·0.05	(100+Recy)·0.05·α	(100+Recy)·0.05·(1−α)

Under steady-state conditions, the Ar fed to the system must be eliminated through the purge. Furthermore, the recycle, assumed to be R, is also a function of the purge. Thus, we have a system of two equations and two variables:

$\begin{array}{l}0.2=(100+\mathrm{Recy})·0.05\alpha \hfill \\ \mathrm{Recy}=(1-\alpha )·(100+\mathrm{Recy})·0.75\hfill \end{array}$

Solving the system we have R=288 and α=0.0103. The compositions of the streams at the four positions along the flowsheet are given in Table 5E7.4.

Table 5E7.4

Results of the Mass Balance

	1	2	3	4
N2+3H2	388	291	291	3
NH3	0	48.5	0	0
Ar	19.4	19.4	19.4	0.2

Example 5.8

The gas exiting the reactor at 400°C and 300 atm contains ammonia in a concentration that corresponds to 30% of the equilibrium value under those conditions. Assume that the pressure does not change across the facility. Assuming humid air calculations with ammonia as the moisture:

a. Compute the composition of the gases at the reactor exit.

b. Determine the gas composition after a cooling stage at 5°C.

c. Compute the gas composition after a second cooling at −25°C

Hint: The equilibrium constant at 400°C and 300 atm is 0.0153.

Solution

As presented in the section describing the equilibrium, the constant is as follows:

$\begin{array}{l}{K}_{\mathrm{p}}=\frac{x}{P_{\mathrm{T}}{\left({\displaystyle \frac{1-x}{4}}\right)}^{0.5}{\left({\displaystyle \frac{3}{4}}(1-x)\right)}^{1.5}}=\frac{4^2}{3^{1.5}}\frac{x}{P_{\mathrm{T}}{(1-x)}^2}=3.0792\frac{x}{P_{\mathrm{T}}{(1-x)}^2}=0.0153\hfill \\ x=0.45\hfill \end{array}$

Since only 30% of the conversion at equilibrium is achieved,

$\begin{array}{l}{x}_{\mathrm{real}}=0.135\hfill \\ BC=100\mathrm{kmol}\hfill \\ \underset{75-3y}{3{\mathrm{H}}_2}+\underset{25-y}{{\mathrm{N}}_2}\leftrightarrow \underset{2y}{2{\mathrm{NH}}_3}\hfill \\ \frac{2y}{100-2y}=0.135\hfill \end{array}$

$\begin{array}{l}y=5.947\hfill \\ {\mathrm{H}}_2=57.159\hfill \\ {\mathrm{N}}_2=19.053\hfill \\ {\mathrm{NH}}_3=11.895\hfill \end{array}$

the molar gas composition is as follows:

$\begin{array}{l}{\mathrm{H}}_2=64.87\%\hfill \\ {\mathrm{N}}_2=21.63\%\hfill \\ {\mathrm{NH}}_3=13.50\%\hfill \end{array}$

Ammonia vapor pressure at the two temperatures, from Antoine correlation,

$P_{\mathrm{v}}=\exp \left(16.9471-\frac{2132.5}{T(\mathrm{K})-32.98}\right)$

is given in Table 5E8.1:

Table 5E8.1

Vapor Pressures of Ammonia at the Example Pressures

P (atm)	1.5	5
T (°C)	−25	5

After the condensation, the gas is saturated, and thus:

$\begin{array}{l}{P}_{{\mathrm{NH}}_3}=P_{\mathrm{T}}·y_{{\mathrm{NH}}_3}\hfill \\ 5=300·y_{{\mathrm{NH}}_3}\hfill \\ y_{{\mathrm{NH}}_3}=0.0167\hfill \end{array}$

Thus, the uncondensed moles are computed as follows:

$\begin{array}{l}0.0167=\frac{n_{{\mathrm{NH}}_3}}{19.053+57.159+n_{{\mathrm{NH}}_3}}\hfill \\ n_{{\mathrm{NH}}_3}=1.2944\hfill \end{array}$

Therefore, 10.6 moles have condensed. After the second condensation, we proceed similarly to compute the ammonia in the gas:

$\begin{array}{l}{P}_{{\mathrm{NH}}_3}=P_{\mathrm{T}}·y_{{\mathrm{NH}}_3}\hfill \\ 1.5=300·y_{{\mathrm{NH}}_3}\hfill \\ y_{{\mathrm{NH}}_3}=0.005\hfill \\ 0.005=\frac{n_{{\mathrm{NH}}_3}}{19.053+57.159+n_{{\mathrm{NH}}_3}}\hfill \\ n_{{\mathrm{NH}}_3}=0.383\hfill \end{array}$

Therefore, 0.91 kmol have been condensed. In total, 11.51 kmol have been recovered out of the 11.8945, around 97% recovery.

Example 5.9

The stream from the reactor consists of 281 kmol of syngas, 9.12 kmol of Ar, and 62 kmol of ammonia. Ammonia is cooled and condenses. Assuming a condensation pressure of 50 atm, compute the temperature of the cooling for recovering 75% of the ammonia produced. Use flash calculations to compute the separation.

Solution

The problem is formulated as a gas–liquid equilibrium using the following nomenclature:

L: Liquid exit (kmol)

V: Vapor exit (kmol)

K_i: Equilibrium constant for noncondensables

K_j: Equilibrium constant for condensables

x_i: Liquid molar fraction

y_i: Vapor molar fraction

A global balance and a component balance are used to compute the molar fraction of the gas and liquid phases:

$\begin{array}{l}F=V+L\\ F{z}_i=V{y}_i+L{x}_i\\ y_i=K_i\cdot x_i\end{array}$

Combining the equations:

$\begin{array}{l}{y}_i=\frac{z_i}{\frac{V}{F}+\left(1-\frac{V}{F}\right)·\left(\frac{1}{K_i}\right)}\hfill \\ x_i=\frac{z_j}{(K_j-1)·\left(\frac{V}{F}\right)+1}\hfill \end{array}$

$\begin{array}{ll}{K}_i>>1==>V·y_i\approx F\cdot z_i\hfill & \hfill \\ \hfill & ==\to \hfill \\ K_j<<1==>L\cdot x_i\approx F\cdot z_j\hfill & \hfill \\ V\approx \sum f_i;L\approx \sum f_j\hfill & \hfill \end{array}$

$y_i=\frac{f_i}{V}=\frac{f_i}{\sum f_i}\mathrm{and}{x}_i=\frac{y_i}{K_i}=\frac{f_i}{K_i·\sum f_i}$

The vapor compositions are given by the following equations:

$l_i=L·x_i=\frac{f_i·\sum f_j}{K_i·\sum f_i}$

$v_i=f_i-l_i=f_i\left(1-\frac{\sum f_j}{K_i\sum f_i}\right)$

The liquid compositions are computed as follows:

$v_j=V\cdot y_j=V\cdot K_j\cdot x_j=\frac{K_j\cdot f_j\cdot \sum f_i}{\sum f_j}$

$l_j=f_j-v_j=f_j\cdot \left(1-\frac{K_j\cdot \sum f_i}{\sum f_j}\right)$

where K_i or K_j is equal to P_v/P_total. Thus, we fix l_j to be 75% of f_j for ammonia and solve the system of equations. Alternatively, we can just solve v_j=0.25 f_j to obtain K_j, and from that the temperature. Table 5E9.1 presents the results. We see that the Ks are within the error of the approximation.

Table 5E9.1

Results of the Gas–Liquid Separation

Example 5.10

Compute the liquid flow required to recover 98% of the ammonia content in the product gases of the converters using an absorption column. The total gas flow is 10 kmol/s with 20% ammonia by volume. The stream is at 10 atm and 25°C. The gas flow is 50% of the minimum required. Calculate the number of stages.

Solution

Assume that Raoult’s law holds. Therefore:

$x{P}_{\mathrm{v}}=y{P}_{\mathrm{T}}\leftarrow \to y=Hx$

The vapor pressure of ammonia is computed using the Antoine correlation as follows:

$P_{\mathrm{v}}=\exp \left(16.9471-\frac{2132.5}{T(\mathrm{K})-32.98}\right)$

We formulate a mass balance to the ammonia using as control volume that of the scrubber:

$\begin{array}{l}{G}_{\mathrm{out}}{y}_{\mathrm{out}}+L_{\mathrm{out}}·x_{\mathrm{out}}=L_{\mathrm{in}}·x_{\mathrm{in}}+G_{\mathrm{in}}·y_{\mathrm{in}}\hfill \\ 10·0.2=2·0.02+L_{\mathrm{out}}·x_{\mathrm{out}}\hfill \end{array}$

The concentration of ammonia in the liquid is in equilibrium with that of the gas in the inlet, y=0.2.

$\begin{array}{l}x{P}_{\mathrm{v}}=0.2\cdot 10\cdot 760\hfill \\ x=\frac{0.2\cdot 7600}{\exp \left(16.9471-\frac{2132.5}{298-32.98}\right)}=0.207\hfill \end{array}$

L_in=7.72 kmol/s. Thus, the flow used is 11.58 kmol/s. y_i is the composition of the exiting gas consisting of 8 kmol/s of air and 0.04 kmol of ammonia, y_i=0.005.

$\begin{array}{ll}{N}_{\min }\hfill & =\frac{\log \left[\left(\frac{y_{\mathrm{n}+1}-H{x}_{\mathrm{o}}}{y_1-H{x}_{\mathrm{o}}}\right)\left(\frac{A-1}{A}\right)+\frac{1}{A}\right]}{\log A}=\frac{\log \left[\frac{0.2}{0.005}\left(\frac{A-1}{A}\right)+\frac{1}{A}\right]}{\log (1.20)}\hfill \\ & =11.1=12(\mathrm{say})\hfill \\ A\hfill & =\frac{L}{V\cdot H}=\frac{L}{V·\frac{P_{\mathrm{v}}(T)}{P_{\mathrm{T}}}}=\frac{11.58}{10\cdot \frac{7336}{7600}}=1.20\hfill \end{array}$