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# Haber process

The Haber process,[1] also called the Haber–Bosch process, is an artificial nitrogen fixation process and is the main industrial procedure for the production of ammonia today.[2][3] It is named after its inventors, the German chemists Fritz Haber and Carl Bosch, who developed it in the first decade of the 20th century. The process converts atmospheric nitrogen (N2) to ammonia (NH3) by a reaction with hydrogen (H2) using a metal catalyst under high temperatures and pressures:

${\displaystyle {\ce {N2 + 3 H2 -> 2 NH3}}\quad \Delta H^{\circ }=-91.8~{\text{kJ/mol}}}$

Before the development of the Haber process, ammonia had been difficult to produce on an industrial scale,[4][5][6] with early methods such as the Birkeland–Eyde process and Frank–Caro process all being highly inefficient.

Although the Haber process is mainly used to produce fertilizer today, during World War I it provided Germany with a source of ammonia for the production of explosives, compensating for the Allied Powers' trade blockade on Chilean saltpeter.

## History

Throughout the 19th century the demand for nitrates and ammonia for use as fertilizers and industrial feedstocks had been steadily increasing. The main source was mining niter deposits and guano from tropical islands.[7] At the beginning of the 20th century it was being predicted that these reserves could not satisfy future demands,[8] and research into new potential sources of ammonia became more important. Although atmospheric nitrogen (N2) is abundant, comprising nearly 80% of the air, it is exceptionally stable and does not readily react with other chemicals. Converting N2 into ammonia posed a challenge for chemists globally.

Haber, with his assistant Robert Le Rossignol, developed the high-pressure devices and catalysts needed to demonstrate the Haber process at laboratory scale.[9][10] They demonstrated their process in the summer of 1909 by producing ammonia from air, drop by drop, at the rate of about 125 mL (4 US fl oz) per hour. The process was purchased by the German chemical company BASF, which assigned Carl Bosch the task of scaling up Haber's tabletop machine to industrial-level production.[5][11] He succeeded in 1910. Haber and Bosch were later awarded Nobel prizes, in 1918 and 1931 respectively, for their work in overcoming the chemical and engineering problems of large-scale, continuous-flow, high-pressure technology.[5]

Ammonia was first manufactured using the Haber process on an industrial scale in 1913 in BASF's Oppau plant in Germany, reaching 20 tonnes per day the following year.[12] During World War I, the production of munitions required large amounts of nitrate. The Allies had access to large deposits of sodium nitrate in Chile (Chile saltpetre) controlled by British companies. Germany had no such resources, so the Haber process proved essential to the German war effort.[5][13] Synthetic ammonia from the Haber process was used for the production of nitric acid, a precursor to the nitrates used in explosives.

Today, the most popular catalysts are based on iron promoted with K2O, CaO, SiO2, and Al2O3. Earlier, molybdenum was also used as a promoter. The original Haber–Bosch reaction chambers used osmium as the catalyst, but it was available in extremely small quantities. Haber noted uranium was almost as effective and easier to obtain than osmium. Under Bosch's direction in 1909, the BASF researcher Alwin Mittasch discovered a much less expensive iron-based catalyst, which is still used today.

During the interwar years, alternative processes were developed, the most notably different being the Casale process and Claude process. Luigi Casale and Georges Claude proposed to increase the pressure of the synthesis loop to 80–100 MPa (800–1,000 bar; 12,000–15,000 psi), thereby increasing the single-pass ammonia conversion and making nearly complete liquefaction at ambient temperature feasible. Georges Claude even proposed to have three or four converters with liquefaction steps in series, thereby omitting the need for a recycle. Nowadays, most plants resemble the original Haber process (20 MPa (200 bar; 2,900 psi) and 500 °C (932 °F)), albeit with improved single-pass conversion and lower energy consumption due to process and catalyst optimization.

A major contributor to the elucidation of this mechanism was Gerhard Ertl.[14][15][16][17]

## Process

A historical (1921) high-pressure steel reactor for production of ammonia via the Haber process is displayed at the Karlsruhe Institute of Technology, Germany

This conversion is typically conducted at pressures above 10 MPa (100 bar; 1,450 psi) and between 400 and 500 °C (752 and 932 °F), as the gases (nitrogen and hydrogen) are passed over four beds of catalyst, with cooling between each pass for maintaining a reasonable equilibrium constant. On each pass only about 15% conversion occurs, but any unreacted gases are recycled, and eventually an overall conversion of 97% is achieved.[3]

The steam reforming, shift conversion, carbon dioxide removal, and methanation steps each operate at pressures of about 2.5–3.5 MPa (25–35 bar; 360–510 psi), and the ammonia synthesis loop operates at pressures ranging from 6 to 18 MPa (60 to 180 bar; 870 to 2,610 psi), depending upon which proprietary process is used.[3]

### Sources of hydrogen

The major source of hydrogen is methane from natural gas. The conversion, steam reforming, is conducted with steam in a high-temperature and -pressure tube inside a reformer with a nickel catalyst, separating the carbon and hydrogen atoms in the natural gas. Other fossil fuel sources include coal, heavy fuel oil and naphtha, while hydrogen is also produced from biomass and from electrolysis of water.

### Reaction rate and equilibrium

Nitrogen gas (N2) is very unreactive because the atoms are held together by strong triple bonds. The Haber process relies on catalysts that accelerate the scission of this triple bond.

Two opposing considerations are relevant to this synthesis: the position of the equilibrium and the rate of reaction. At room temperature, the equilibrium is strongly in favor of ammonia, but the reaction doesn't proceed at a detectable rate due to its high activation energy. Because the reaction is exothermic, the equilibrium constant becomes 1 around 150–200 °C (302–392 °F) (see Le Châtelier's principle).[3]

K(T) for N
2
+ 3 H
2
⇌ 2 NH
3
[18]
Temperature (°C) Kp
300 4.34 × 10−3
400 1.64 × 10−4
450 4.51 × 10−5
500 1.45 × 10−5
550 5.38 × 10−6
600 2.25 × 10−6

Above this temperature, the equilibrium quickly becomes quite unfavorable for the reaction product at atmospheric pressure, according to the van 't Hoff equation. Lowering the temperature is also unhelpful because the catalyst requires a temperature of at least 400 °C to be efficient.[3]

Increased pressure does favor the forward reaction because there are 4 moles of reactant for every 2 moles of product, and the pressure used (15–25 MPa (150–250 bar; 2,200–3,600 psi)) alters the equilibrium concentrations to give a substantial ammonia yield. The reason for this is evident in the equilibrium relationship, which is

${\displaystyle K={\frac {y_{{\ce {NH3}}}^{2}}{y_{{\ce {H2}}}^{3}y_{{\ce {N2}}}}}{\frac {{\hat {\phi }}_{{\ce {NH3}}}^{2}}{{\hat {\phi }}_{{\ce {H2}}}^{3}{\hat {\phi }}_{{\ce {N2}}}}}\left({\frac {P^{\circ }}{P}}\right)^{2},}$

where ${\displaystyle {\hat {\phi }}_{i}}$ is the fugacity coefficient of species ${\displaystyle i}$, ${\displaystyle y_{i}}$ is the mole fraction of the same species, ${\displaystyle P}$ is the pressure in the reactor, and ${\displaystyle P^{\circ }}$ is standard pressure, typically 1 bar (0.10 MPa).

Economically, pressurization of the reactor is expensive: pipes, valves, and reaction vessels need to be strengthened, and there are safety considerations when working at 20 MPa. In addition, running compressors takes considerable energy, as work must be done on the (very compressible) gas. Thus, the compromise used gives a single-pass yield of around 15%[3]

While removing the product (i.e., ammonia gas) from the system would increase the reaction yield, this step is not used in practice, since the temperature is too high; it is removed from the equilibrium mixture of gases leaving the reaction vessel. The hot gases are cooled enough, whilst maintaining a high pressure, for the ammonia to condense and be removed as liquid. Unreacted hydrogen and nitrogen gases are then returned to the reaction vessel to undergo further reaction.[3] While most ammonia is removed (typically down to 2–5 mol.%), some ammonia remains in the recycle stream to the converter. In academic literature, more complete separation of ammonia has been proposed by absorption in metal halides and by adsorption on zeolites. Such a process is called a absorbent-enhanced Haber process or adsorbent-enhanced Haber-Bosch process.

## Catalysts

First reactor at the Oppau plant in 1913
Profiles of the active components of heterogeneous catalysts; the top right figure shows the profile of a shell catalyst.

The Haber–Bosch process relies on catalysts to accelerate the hydrogenation of N2. The catalysts are "heterogeneous", meaning that they are solid that interact on gaseous reagents.[19]

The catalyst typically consists of finely divided iron bound to an iron oxide carrier containing promoters possibly including aluminium oxide, potassium oxide, calcium oxide, potassium hydroxide,[20] molybdenum,[21] and magnesium oxide.

### Production of iron-based catalysts

In industrial practice, the iron catalyst is obtained from finely ground iron powder, which is usually obtained by reduction of high-purity magnetite (Fe3O4). The pulverized iron metal is burnt (oxidized) to give magnetite or wüstite (FeO, ferrous oxide) of a defined particle size. The magnetite (or wüstite) particles are then partially reduced, removing some of the oxygen in the process. The resulting catalyst particles consist of a core of magnetite, encased in a shell of wüstite, which in turn is surrounded by an outer shell of iron metal. The catalyst maintains most of its bulk volume during the reduction, resulting in a highly porous high-surface-area material, which enhances its effectiveness as a catalyst. Other minor components of the catalyst include calcium and aluminium oxides, which support the iron catalyst and help it maintain its surface area. These oxides of Ca, Al, K, and Si are unreactive to reduction by the hydrogen.[3]

The production of the required magnetite catalyst requires a particular melting process in which the used raw materials must be free of catalyst poisons and the promoter aggregates must be evenly distributed in the magnetite melt. Rapid cooling of the magnetite melt, which has an initial temperature of about 3500 °C, produces the precursor desired highly active catalyst. Unfortunately, the rapid cooling ultimately forms a catalyst of reduced abrasion resistance. Despite this disadvantage, the method of rapid cooling is often preferred in practice.[3]

The reduction of the catalyst precursor magnetite to α-iron is carried out directly in the production plant with synthesis gas. The reduction of the magnetite proceeds via the formation of wüstite (FeO), so that particles with a core of magnetite surrounded by a shell of wüstite are formed. The further reduction of magnetite and wüstite leads to the formation of α-iron, which forms together with the promoters the outer shell.[22] The involved processes are complex and depend on the reduction temperature: At lower temperatures, wüstite disproportionates into an iron phase and a magnetite phase; at higher temperatures, the reduction of the wüstite and magnetite to iron dominates.[23]

The α-iron forms primary crystallites with a diameter of about 30 nanometers. These form crystallites a bimodal pore system with pore diameters of about 10 nanometers (produced by the reduction of the magnetite phase) and of 25 to 50 nanometers (produced by the reduction of the wüstite phase).[22] With the exception of cobalt oxide, the promoters are not reduced.

During the reduction of the iron oxide with synthesis gas, water vapour is formed. This water vapor must be considered for high catalyst quality as contact with the finely divided iron would lead to premature aging of the catalyst through recrystallization, especially in conjunction with high temperatures. The vapour pressure of the water in the gas mixture produced during catalyst formation is thus kept as low as possible, target values are below 3 gm−3. For this reason, the reduction is carried out at high gas exchange, low pressure and low temperatures. The exothermic nature of the ammonia formation ensures a gradual increase in temperature.[3]

The reduction of fresh, fully oxidized catalyst or precursor to full production capacity takes four to ten days.[3] The wüstite phase is reduced faster and at lower temperatures than the magnetite phase (Fe3O4). After detailed kinetic, microscopic and X-ray spectroscopic investigations it was shown that wüstite reacts first to metallic iron. This leads to a gradient of iron(II) ions, whereby these diffuse from the magnetite through the wüstite to the particle surface and precipitate there as iron nuclei.

In industrial practice, pre-reduced, stabilised catalysts have gained a significant market share. They are delivered showing the fully developed pore structure, but have been oxidized again on the surface after manufacture and are therefore no longer pyrophoric. The reactivation of such pre-reduced catalysts requires only 30 to 40 hours instead of the usual time periods of several days. In addition to the short start-up time, they also have other advantages such as higher water resistance and lower weight.[3]

Typical catalyst composition[24] % iron % potassium % aluminium % calcium % oxygen
volume composition 40.5 00.35 02.0 1.7 53.2
Surface composition before reduction 08.6 36.1 10.7 4.7 40.0
Surface composition after reduction 11.0 27.0 17.0 4.0 41.0

### Catalysts other than iron

Since the industrial launch of the Haber–Bosch process, many efforts have been made to improve it. Many metals were intensively tested in the search for suitable catalysts: The requirement for suitability is the dissociative adsorption of nitrogen (i. e. the nitrogen molecule must be split upon adsorption into nitrogen atoms). At the same time the binding of the nitrogen atoms must not be too strong, otherwise the catalyst would be blocked and the catalytic ability would be reduced (i. e. self-poisoning). The elements in the periodic table at the left of the iron group show such a strong bond to nitrogen. The formation of surface nitrides makes for example chromium catalysts ineffective. Metals to the right of the iron group, in contrast, adsorb nitrogen too weakly to be able to activate it sufficiently for ammonia synthesis. Haber himself initially used catalysts based on osmium and uranium. Uranium, however, reacts to its nitride during catalysis and osmium oxide is rare.[25]

Due to the comparatively low price, high availability, easy processing, lifespan and activity, iron was ultimately chosen as catalyst. The production of for example 1800 tons ammonia per day requires a gas pressure of at least 130 bar, temperatures of 400 to 500 °C and a reactor volume of at least 100 m³. According to theoretical and practical studies, further improvements of the pure iron catalyst are limited. It was only in 1984 that the activity of iron catalysts were increased noticeably by inclusion of cobalt.

### Second generation catalysts

Ruthenium forms highly active catalysts. Allowing milder operating pressures and temperatures, Ru-based materials are referred to as second-generation catalysts. Such catalysts are prepared by decomposition of triruthenium dodecacarbonyl on graphite.[3] A drawback of activated-carbon-supported ruthenium-based catalysts is the methanation of the support in the presence of hydrogen. Their activity is strongly dependent on the catalyst carrier and the promoters. A wide range of substances can be used as carriers, including carbon, magnesium oxide, aluminum oxide, zeolites, spinels, and boron nitride.[26]

Ruthenium-activated carbon-based catalysts have been used industrially in the KBR Advanced Ammonia Process (KAAP) since 1992.[27] The carbon carrier is partially degraded to methane; however, this can be mitigated by a special treatment of the carbon at 1500 °C, thus prolonging the catalysts lifetime. In addition, the finely dispersed carbon poses a risk of explosion. For these reasons and due to its low acidity, magnesium oxide has proven to be a good alternative. Carriers with acidic properties extract electrons from ruthenium, make it less reactive, and undesirably bind ammonia to the surface.[26]

### Catalyst poisons

Catalyst poisons lower the activity of the catalyst. They are usually impurities in the synthesis gas (a raw material). Sulfur compounds, phosphorus compounds, arsenic compounds, and chlorine compounds are permanent catalyst poisons. Water, carbon monoxide, carbon dioxide and oxygen are temporary catalyst poisons.[3]

Although chemically inert components of the synthesis gas mixture such as noble gases or methane are not catalyst poisons in the strict sense, they accumulate through the recycling of the process gases and thus lower the partial pressure of the reactants, which in turn has a negative effect on the conversion.[28]

## Industrial production

### Synthesis parameters

Change of the equilibrium constant Keq as a function of temperature[29]
temperature (°C) Keq
300 4.34 × 10−3
400 1.64 × 10−4
450 4.51 × 10−5
500 1.45 × 10−5
550 5.38 × 10−6
600 2.25 × 10−6

The formation of ammonia occurs from nitrogen and hydrogen according to the following equation:

${\displaystyle \ \mathrm {N_{2}} +3\ \mathrm {H_{2}} \quad \rightleftharpoons \quad \ 2\ \mathrm {NH_{3}} \quad \quad {\Delta H=-92.28\;\mathrm {kJ} }\ ({\Delta H_{298K}=-46.14\;\mathrm {kJ/mol} })}$[30]

The reaction is an exothermic equilibrium reaction in which the gas volume is reduced. The equilibrium constant Keq of the reaction (see table) is obtained from the following equation:

${\displaystyle K_{eq}={\frac {p^{2}\mathrm {(NH_{3})} }{p\mathrm {(N_{2})} \cdot p^{3}\mathrm {(H_{2})} }}}$

Since the reaction is exothermic, the equilibrium of the reaction shifts at lower temperatures to the side of the ammonia. Furthermore, four volumetric parts of the raw materials produce two volumetric parts of ammonia. According to Le Chatelier's principle, a high pressure therefore also favours the formation of ammonia. In addition, a high pressure is necessary to ensure sufficient surface coverage of the catalyst with nitrogen.[31] For this reason, a ratio of nitrogen to hydrogen of 1 to 3, a pressure of 250 to 350 bar, a temperature of 450 to 550 °C and α iron are used as catalysts.

The catalyst ferrite (α-Fe) is produced in the reactor by the reduction of magnetite with hydrogen. The catalyst has its highest efficiency at temperatures of about 400 to 500 °C. Even though the catalyst greatly lowers the activation energy for the cleavage of the triple bond of the nitrogen molecule, high temperatures are still required for an appropriate reaction rate. At the industrially utilized reaction temperature of 450 to 550 °C an optimum between the decomposition of ammonia into the starting materials and the effectiveness of the catalyst is achieved.[32] The formed ammonia is continuously removed from the system. The volume fraction of ammonia in the gas mixture is about 20%.

The inert components, especially the noble gases such as argon, should not exceed a certain content in order not to reduce the partial pressure of the reactants too much. To remove the inert gas components, part of the gas is removed and the argon is separated in a gas separation plant. The extraction of pure argon from the circulating gas is carried out using the Linde process.[33]

### Large-scale technical implementation

Modern ammonia plants produce more than 3000 tons per day in one production line. The following diagram shows the set-up of a Haber–Bosch plant:

 primary reformer  air feed  secondary reformer  CO conversion  washing tower  ammonia reactor  heat exchanger  ammonia condenser

Depending on its origin, the synthesis gas must first be freed from impurities such as hydrogen sulphide or organic sulphur compounds, which act as a catalyst poison. High concentrations of hydrogen sulphide, which occur in synthesis gas from carbonization coke, are removed in a wet cleaning stage such as the Sulfosolvan process, while low concentrations are removed by adsorption on activated carbon.[34] Organosulfur compounds are separated by pressure swing adsorption together with carbon dioxide after CO conversion.

To produce hydrogen by steam reforming, methane reacts with water vapor using a nickel oxide-alumina catalyst in the primary reformer to form carbon monoxide and hydrogen. The energy required for this, the enthalpy ΔH, is 206 kJ/mol.[35]

${\displaystyle \mathrm {CH_{4(g)}} +\mathrm {H_{2}O_{(g)}} \quad \rightarrow \quad \mathrm {CO_{(g)}} +\mathrm {3\ H_{2(g)}} \qquad {\Delta H=+206\ \mathrm {kJ/mol} }}$

The methane gas reacts in the primary reformer only partially. In order to increase the hydrogen yield and keep the content of inert components (i. e. methane) as low as possible, the remaining methane gas is converted in a second step with oxygen to hydrogen and carbon monoxide in the secondary reformer. The secondary reformer is supplied with air as oxygen source. Also the required nitrogen for the subsequent ammonia synthesis is added to the gas mixture.

${\displaystyle 2\ \mathrm {CH_{4(g)}} +\mathrm {O_{2(g)}} \quad \rightarrow \quad 2\ \mathrm {CO_{(g)}} +4\ \mathrm {H_{2(g)}} \qquad {\Delta H=-71\ \mathrm {kJ/mol} }}$

In a third step, the carbon monoxide is oxidized to carbon dioxide, which is called CO conversion or water-gas shift reaction.

${\displaystyle \mathrm {CO_{(g)}} +\mathrm {H_{2}O(g)} \quad \rightarrow \quad \mathrm {CO_{2(g)}} +\mathrm {H_{2(g)}} \qquad {\Delta H=-41\ \mathrm {kJ/mol} }}$

Carbon monoxide and carbon dioxide would form carbamates with ammonia, which would clog (as solids) pipelines and apparatus within a short time. In the following process step, the carbon dioxide must therefore be removed from the gas mixture. In contrast to carbon monoxide, carbon dioxide can easily be removed from the gas mixture by gas scrubbing with triethanolamine. The gas mixture then still contains methane and noble gases such as argon, which, however, behave inertly.[28]

The gas mixture is then compressed to operating pressure by turbo compressors. The resulting compression heat is dissipated by heat exchangers; it is used to preheat raw gases.

The actual production of ammonia takes place in the ammonia reactor. The first reactors were bursting under the high pressure because the atomic hydrogen in the carbonaceous steel partially recombined to methane and produced cracks in the steel. Bosch therefore developed tube reactors consisting of a pressure-bearing steel tube in which a low-carbon iron lining tube was inserted filled with the catalyst. Hydrogen that diffused through the inner steel pipe escaped to the outside via thin holes in the outer steel jacket, the so-called Bosch holes.[30] A disadvantage of the tubular reactors was the relatively high pressure loss, which had to be applied again by compression. The development of hydrogen-resistant chromium-molybdenum steels made it possible to construct single-walled pipes.[36]

Modern ammonia reactor with heat exchanger modules: The cold gas mixture is preheated to reaction temperature in heat exchangers by the reaction heat and cools in turn the produced ammonia.

Modern ammonia reactors are designed as multi-storey reactors with low pressure drop, in which the catalysts are distributed as fills over about ten storeys one above the other. The gas mixture flows through them one after the other from top to bottom. Cold gas is injected from the side for cooling. A disadvantage of this reactor type is the incomplete conversion of the cold gas mixture in the last catalyst bed.[36]

Alternatively, the reaction mixture between the catalyst layers is cooled using heat exchangers, whereby the hydrogen-nitrogen mixture is preheated to reaction temperature. Reactors of this type have three catalyst beds. In addition to good temperature control, this reactor type has the advantage of better conversion of the raw material gases compared to reactors with cold gas injection.

The reaction product is continuously removed for maximum yield. The gas mixture is cooled to 450 °C in a heat exchanger using water, freshly supplied gases and other process streams. The ammonia also condenses and is separated in a pressure separator. Unreacted nitrogen and hydrogen are than compressed back to the process by a circulating gas compressor, supplemented with fresh gas and fed to the reactor.[36] In a subsequent distillation, the product ammonia is purified.

## Mechanism

### Elementary steps

The mechanism of ammonia synthesis contains the following seven elementary steps:

1. transport of the reactants from the gas phase through the boundary layer to the surface of the catalyst.
2. pore diffusion to the reaction center
4. reaction
5. desorption of product
6. transport of the product through the pore system back to the surface
7. transport of the product into the gas phase

Transport and diffusion (the first and last two steps) are fast compared to adsorption, reaction and desorption because of the shell structure of the catalyst. It is known from various investigations that the rate-determining step of the ammonia synthesis is the dissociation of nitrogen.[3] In contrast, exchange reactions between hydrogen and deuterium on the Haber–Bosch catalysts still take place at temperatures of −196 °C (−320.8 °F) at a measurable rate; the exchange between deuterium and hydrogen on the ammonia molecule also takes place at room temperature. Since the adsorption of both molecules is rapid, it cannot determine the speed of ammonia synthesis.[37]

In addition to the reaction conditions, the adsorption of nitrogen on the catalyst surface depends on the microscopic structure of the catalyst surface. Iron has different crystal surfaces, whose reactivity is very different. The Fe(111) and Fe(211) surfaces have by far the highest activity. The explanation for this is that only these surfaces have so-called C7 sites - these are iron atoms with seven closest neighbours.[3]

The dissociative adsorption of nitrogen on the surface follows the following scheme, where S* symbolizes an iron atom on the surface of the catalyst:[22]

N2 → S*–N2 (γ-species) → S*–N2–S* (α-species) → 2 S*–N (β-species, surface nitride)

The adsorption of nitrogen is similar to the chemisorption of carbon monoxide. On a Fe(111) surface, the adsorption of nitrogen first leads to an adsorbed γ-species with an adsorption energy of 24 kJmol−1 and an N-N stretch vibration of 2100 cm−1. Since the nitrogen is isoelectronic to carbon monoxide, it adsorbs in an on-end configuration in which the molecule is bound perpendicular to the metal surface at one nitrogen atom.[16][38][3] This has been confirmed by photoelectron spectroscopy.[39]

Ab-initio-MO calculations have shown that, in addition to the σ binding of the free electron pair of nitrogen to the metal, there is a π binding from the d orbitals of the metal to the π* orbitals of nitrogen, which strengthens the iron-nitrogen bond. The nitrogen in the α state is more strongly bound with 31 kJmol−1. The resulting N-N bond weakening could be experimentally confirmed by a reduction of the wave numbers of the N-N stretching oscillation to 1490 cm−1.[38]

Further heating of the Fe(111) area covered by α-N2 leads to both desorption and emergence of a new band at 450 cm−1. This represents a metal-nitrogen oscillation, the β state. A comparison with vibration spectra of complex compounds allows the conclusion that the N2 molecule is bound "side-on", with an N atom in contact with a C7 site. This structure is called "surface nitride". The surface nitride is very strongly bound to the surface.[39] Hydrogen atoms (Hads), which are very mobile on the catalyst surface, quickly combine with it.

Infrared spectroscopically detected surface imides (NHad), surface amides (NH2,ad) and surface ammoniacates (NH3,ad) are formed, the latter decay under NH3 release (desorption).[30] The individual molecules were identified or assigned by X-ray photoelectron spectroscopy (XPS), high-resolution electron energy loss spectroscopy (HREELS) and IR spectroscopy.

Drawn reaction scheme

On the basis of these experimental findings, the reaction mechanism is believed to involve the following steps (see also figure):[40]

1. N2 (g) → N2 (adsorbed)
3. H2 (g) → H2 (adsorbed)
6. NH3 (adsorbed) → NH3 (g)

Reaction 5 occurs in three steps, forming NH, NH2, and then NH3. Experimental evidence points to reaction 2 as being the slow, rate-determining step. This is not unexpected, since the bond broken, the nitrogen triple bond, is the strongest of the bonds that must be broken.

As with all Haber–Bosch catalysts, nitrogen dissociation is the rate determining step for ruthenium activated carbon catalysts. The active center for ruthenium is a so-called B5 site, a 5-fold coordinated position on the Ru(0001) surface where two ruthenium atoms form a step edge with three ruthenium atoms on the Ru(0001) surface.[41] The number of B5 sites depends on the size and shape of the ruthenium particles, the ruthenium precursor and the amount of ruthenium used.[26] The reinforcing effect of the basic carrier used in the ruthenium catalyst is similar to the promoter effect of alkali metals used in the iron catalyst.[26]

### Energy diagram

Energy diagram

An energy diagram can be created based on the enthalpy of reaction of the individual steps. The energy diagram can be used to compare homogeneous and heterogeneous reactions: Due to the high activation energy of the dissociation of nitrogen, the homogeneous gas phase reaction is not realizable. The catalyst avoids this problem as the energy gain resulting from the binding of nitrogen atoms to the catalyst surface overcompensates for the necessary dissociation energy so that the reaction is finally exothermic. Nevertheless, the dissociative adsorption of nitrogen remains the rate determining step: not because of the activation energy, but mainly because of the unfavorable pre-exponential factor of the rate constant. Although hydrogenation is endothermic, this energy can easily be applied by the reaction temperature (about 700 K).[3]

## Economic and environmental aspects

Industrial fertilizer plant

When first invented, the Haber process competed against another industrial process, the cyanamide process. However, the cyanamide process consumed large amounts of electrical power and was more labor-intensive than the Haber process.[5]:137–143

As of 2018, the Haber process produces 230 million tonnes of anhydrous ammonia per year.[42] The ammonia is used mainly as a nitrogen fertilizer as ammonia itself, in the form of ammonium nitrate, and as urea. The Haber process consumes 3–5% of the world's natural-gas production (around 1–2% of the world's energy supply).[4][43][44][45] In combination with advances in breeding, herbicides and pesticides, these fertilizers have helped to increase the productivity of agricultural land:

With average crop yields remaining at the 1900 level[,] the crop harvest in the year 2000 would have required nearly four times more land[,] and the cultivated area would have claimed nearly half of all ice-free continents, rather than under 15% of the total land area that is required today.[46]

The energy-intensivity of the process contributes to climate change and other environmental problems:

leaching of nitrates into ground water, rivers, ponds and lakes; expanding dead zones in coastal ocean waters, resulting from recurrent eutrophication; atmospheric deposition of nitrates and ammonia affecting natural ecosystems; higher emissions of nitrous oxide (N2O), now the third most important greenhouse gas followed by CO2 and CH4.[46]

The Haber–Bosch process is one of the largest contributors to a buildup of reactive nitrogen in the biosphere, causing an anthropogenic disruption to the nitrogen cycle.[47]

Since nitrogen use efficiency is typically less than 50%,[48] farm runoff from heavy use of fixed industrial nitrogen disrupts biological habitats.[4][49]

Nearly 50% of the nitrogen found in human tissues originated from the Haber–Bosch process.[50] Thus, the Haber process serves as the "detonator of the population explosion", enabling the global population to increase from 1.6 billion in 1900 to 7.7 billion by November 2018.[51]

• Other nitrogen fixation processes
• Birkeland-Eyde process
• Cyanamide process
• Other contemporary nitrogen sources
• Hydrogen production
• Industrial gas

## References

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text:
# Diagonal intersection

Diagonal intersection is a term used in mathematics, especially in set theory.

If $\displaystyle\delta$ is an ordinal number and $\displaystyle\langle X_\alpha \mid \alpha<\delta\rangle$ is a sequence of subsets of $\displaystyle\delta$, then the diagonal intersection, denoted by

$\displaystyle\Delta_{\alpha<\delta} X_\alpha,$

is defined to be

$\displaystyle\{\beta<\delta\mid\beta\in \bigcap_{\alpha<\beta} X_\alpha\}.$

That is, an ordinal $\displaystyle\beta$ is in the diagonal intersection $\displaystyle\Delta_{\alpha<\delta} X_\alpha$ if and only if it is contained in the first $\displaystyle\beta$ members of the sequence. This is the same as

$\displaystyle\bigcap_{\alpha < \delta} ( [0, \alpha] \cup X_\alpha ),$

where the closed interval from 0 to $\displaystyle\alpha$ is used to avoid restricting the range of the intersection.

format: plain


text:
Janus kinase-2 (JAK2) supports breast cancer growth, and clinical trials testing JAK2 inhibitors are under way. In addition to the tumor epithelium, JAK2 is also expressed in other tissues including immune cells; whether the JAK2 mRNA levels in breast tumors correlate with outcomes has not been evaluated. Using a case–control design, JAK2 mRNA was measured in 223 archived breast tumors and associations with distant recurrence were evaluated by logistic regression. The frequency of correct pairwise comparisons of patient rankings based on JAK2 levels versus survival outcomes, the concordance index (CI), was evaluated using data from 2,460 patients in three cohorts. In the case–control study, increased JAK2 was associated with a decreasing risk of recurrence (multivariate P = 0.003, n = 223). Similarly, JAK2 was associated with a protective CI (<0.5) in the public cohorts: NETHERLANDS CI = 0.376, n = 295; METABRIC CI = 0.462, n = 1,981; OSLOVAL CI = 0.452, n = 184. Furthermore, JAK2 was strongly correlated with the favorable prognosis LYM metagene signature for infiltrating T cells (r = 0.5; P < 2 × 10−16; n = 1,981) and with severe lymphocyte infiltration (P = 0.00003, n = 156). Moreover, the JAK1/2 inhibitor ruxolitinib potently inhibited the anti-CD3–dependent production of IFN-γ, a marker of the differentiation of Th cells along the tumor-inhibitory Th1 pathway. The potential for JAK2 inhibitors to interfere with the antitumor capacities of T cells should be evaluated. Cancer Immunol Res; 2(4); 301–6. ©2014 AACR.

Janus kinase-2 (JAK2) is essential for the signaling of a variety of cytokine receptors, including receptors for erythropoietin during erythropoiesis and for prolactin during mammary differentiation (1, 2). JAK2 has emerged as an important target in myeloproliferative disorders, and increasingly, in solid tumors such as breast cancer. Recent studies have implicated JAK2 in interleukin (IL)-6–dependent breast cancer stem cell self-renewal (3), and in both IL-6- and IL-8–dependent growth of triple-negative breast cancers (4). Furthermore, JAK2 signaling has been implicated as a mechanism of escape from other targeted breast cancer therapies (5). Thus, JAK2 inhibitors are being evaluated in patients with breast cancer (6). JAK2 is also expressed in diverse cell types, including immune cells, and whether the overall JAK2 mRNA levels in breast tumors are associated with clinical outcomes has not been evaluated.

Studies of mRNA levels in primary breast tumors have been useful for classifying breast cancers into subtypes that correlate with prognosis and drug responsiveness (7–9), for predicting recurrence (10, 11), and for delineating gene expression signatures that correlate with prognosis (12–15). Here, we evaluated the association between tumor mRNA levels of JAK2 and clinical outcomes in a novel case–control study and in three public cohorts. Outcomes included distant metastatic recurrence in a matched case–control study (n = 223); recurrence-free survival in the Netherlands Cancer Institute cohort that was used to develop the MammaPrint recurrence risk test (n = 295; ref. 11); overall and disease-specific survival in METABRIC, currently the largest collection of gene expression and copy-number data linked to long-term breast cancer outcomes (n = 1,981; ref. 14); and overall survival in OSLOVAL, a recent cohort that, along with METABRIC, formed the basis of the Sage Bionetworks DREAM breast cancer prognosis challenge (n = 184; refs. 12, 15).

Selection of cases and controls

The protocol to use a breast cancer research database for case selection, to access institution-archived leftover tumor tissue, and to undertake molecular biology studies of the tissue was approved by the Institutional Review Boards of the Fred Hutchinson Cancer Research Center (File 6643) and the Swedish Medical Center (File 4924C-10). Patient consent was not required. The breast cancer research database at the Swedish Cancer Institute contains patient, tumor, treatment, and outcomes data collected since 1989 for more than 12,000 patients. The dataset was reduced to women followed for at least 2 years with invasive carcinoma with T1–3 primary tumors and treated by partial mastectomy plus breast irradiation or total mastectomy, sentinel node biopsy or axillary dissection, and adjuvant chemotherapy. Patients with multiple primaries, T4 primaries or distant metastases, and those receiving neoadjuvant chemotherapy, were excluded. Matching variables included extranodal extension of metastasis, lymphovascular invasion, estrogen receptor (ER)/progesterone receptor (PR)/human epidermal growth factor receptor-2 (HER2) status, T-stage, and N-stage. A T–N interaction term allowed for the fact that tumor size is more important for women without positive nodes than for women with positive nodes. Within each matched pair, diagnosis dates of the recurring and nonrecurring patients were no more than 2 years apart. Propensity scoring was used to match 112 cases of distant recurrence following surgery to 112 nonrecurring controls using the “Optmatch” package (16) and R (17).

Quantitative Reverse Transcriptase PCR

RNA was extracted from 4 × 10 μm sections using the Absolutely RNA FFPE System (Stratagene). The amount of tumor versus normal tissue in each section was greater than 50% for 84% of samples and greater than 90% for 47% of samples as determined by pathologists' inspection of hematoxylin and eosin–stained slides. cDNA was synthesized using random hexamers and SuperScript III (Invitrogen) and was preamplified for 14 cycles using the Taqman preamplification system (Applied Biosystems). All probes bound exon junctions to prevent genomic DNA amplification (Supplementary Table S1). Diluted cDNA was used to seed triplicate real-time PCR reactions for each Taqman assay using standard cycling conditions. Cycle threshold (Ct) values were determined using Sequence Detection Software (Applied Biosystems). Relative quantification was calculated as |$2^{-{\rm \Delta} C_{\rm t}}$|, where ΔCt values were calculated by subtracting the indicated control gene mean Ct value from the target gene mean Ct value.

JAK2 mRNA levels and distant recurrence in the case–control study

Associations between JAK2 mRNA and the likelihood of recurrence were evaluated by logistic regression. In the continuous model, coefficients were calculated as estimates of the change in the log of the odds that an individual experienced a recurrence for every 2-fold increase in JAK2 levels, in which a negative coefficient indicates that increasing JAK2 levels are associated with a decreasing likelihood of recurrence. Because the participants had one (n = 184), two (n = 26), or three tumor specimens (n = 14), generalized estimating equations were used to account for varying numbers of tissues per individual (“Geepack” package; ref. 18). Regressions were also performed using only primary tumors. For this, among the 26 individuals with both a primary and node specimen, only the primary tissue was included, and data from 31 individuals with only a node specimen were excluded. In the dichotomous model, coefficients were calculated with above-median versus below-median JAK2 levels as a predictor of recurrence, in which multiple specimens were averaged to one value per individual. Multivariate analysis adjusted for clinical factors with which JAK2 expression was significantly correlated.

JAK2 mRNA levels and survival outcomes in the public cohorts

The inclusion criteria, clinical characteristics, and follow-up of the NETHERLANDS, METABRIC, and OSLOVAL cohorts were described (11, 12, 14). Data are available via Sage Bionetworks (www.synapse.org) under the following identifiers: doi:10.7303/syn4517.1, doi:10.7303/syn1688369, and doi:10.7303/syn1688370. Concordance index (CI) values were determined as described (12, 19).

Case–control study of JAK2 mRNA levels and distant recurrence

We previously optimized methods for measuring JAK2 mRNA in formalin-fixed, paraffin-embedded tumors by quantitative reverse transcriptase PCR (qRT-PCR) despite the degradation that characterizes RNA extracted from these samples (20). We applied this approach to tumor specimens from 112 women who underwent surgery for breast cancer and subsequently experienced a distant metastatic recurrence and 112 women who did not. With the exception of a borderline significant increase in the number of ER tumors among recurrences, there were no significant differences in clinical characteristics between cases and controls (Supplementary Table S2). Of note, although ER was a matching variable, it was not the only variable, which accounts for the residual effect of this strong prognostic factor even after propensity score matching. Sufficient RNA was available in 223 tumor specimens for JAK2 mRNA determinations. The validity of our mRNA measurements was confirmed by (i) the strong correlation in values obtained using probes for both JAK2 exon8/9 and exon23/24; (ii) the reproducible mRNA levels across three separate tumor specimens for 14 tumors for which this comparison was possible; and (iii) the strong concordance in mRNA levels of ESR1, PGR, ERBB2, and the corresponding clinical immunohistochemistry results (Supplementary Fig. S1).

JAK2 mRNA levels were significantly higher in tumors from women who experienced no distant recurrence compared with those who experienced a distant recurrence (Fig. 1A). This association was significant for both JAK2 exon8/9 and exon23/24 probes in logistic regression when JAK2 mRNA was treated as a continuous or dichotomous variable (Supplementary Table S3). Furthermore, with the exception of the JAK2 exon8/9 probe in the dichotomous model, significance was maintained in multivariate analysis. The association between increasing JAK2 mRNA and decreasing distant recurrence was also significant when analysis was restricted to primary tumors. A receiver operator curve revealed that the association between higher JAK2 mRNA and reduced recurrence was maximal when tumor samples with the top 40% to 50% of JAK2 expression level were defined as high JAK2 (Fig. 1B).

Figure 1.

JAK2 mRNA is associated with reduced distant breast cancer recurrence. A, JAK2 mRNA was measured by qRT-PCR using RNAs extracted from 223 breast tumor samples. Values for the JAK2 exon23/24 junction probe are shown normalized to the endogenous control gene HMBS. Box plots depict the distribution of normalized JAK2 mRNA values in quartiles. Circle in the box, mean value; horizontal line, median value; triangles, outliers. The P value was calculated using the t test. RQ, relative quantification. B, JAK2 mRNA was modeled in a receiver operator curve as a predictor of reduced recurrence across different thresholds for defining high JAK2. The true-positive rate versus false-positive rate of distant recurrence for each model versus actual outcomes is plotted for each threshold of defining high JAK2. The percentage of samples that are defined as high JAK2 is shown for each model (top %). The P value for the area under the curve (AUC) was calculated using the Z test.

Figure 1.

JAK2 mRNA is associated with reduced distant breast cancer recurrence. A, JAK2 mRNA was measured by qRT-PCR using RNAs extracted from 223 breast tumor samples. Values for the JAK2 exon23/24 junction probe are shown normalized to the endogenous control gene HMBS. Box plots depict the distribution of normalized JAK2 mRNA values in quartiles. Circle in the box, mean value; horizontal line, median value; triangles, outliers. The P value was calculated using the t test. RQ, relative quantification. B, JAK2 mRNA was modeled in a receiver operator curve as a predictor of reduced recurrence across different thresholds for defining high JAK2. The true-positive rate versus false-positive rate of distant recurrence for each model versus actual outcomes is plotted for each threshold of defining high JAK2. The percentage of samples that are defined as high JAK2 is shown for each model (top %). The P value for the area under the curve (AUC) was calculated using the Z test.

Close modal

JAK2 mRNA levels and survival outcomes in the public cohorts

Next, we evaluated the association between JAK2 mRNA and outcomes in the NETHERLANDS, METABRIC, and OSLOVAL cohorts. We used the CI (19), which provides a convenient measure of the strength and direction of an association between a single gene and outcomes, to score submissions in the Sage Bionetworks DREAM breast cancer prognosis challenge (12). The CI is the relative frequency of correct pairwise comparisons of patient rankings based on gene expression levels versus survival outcomes. A CI > 0.5 indicates that higher expression is associated with shorter survival, whereas a value of <0.5 indicates that higher expression is associated with longer survival. For example, using disease-specific survival data in METABRIC, the single-gene mRNA with the poorest prognosis was previously found to be CDCA5 with a CI of 0.651, indicating that if 2 patients were randomly selected, the patient with the higher CDCA5 level will have shorter survival 65.1% of the time (12). Conversely, the single most protective gene was FGD3 with a CI of 0.352, indicating that if 2 patients were randomly selected, the patient with the higher FGD3 level will have the longer survival 64.8% (100%–35.2%) of the time.

JAK2 mRNA exhibited a protective CI in all three datasets (Fig. 2). The strongest effect was observed in the NETHERLANDS cohort, in which the CI of 0.376 indicates that if 2 patients were randomly selected, the patient with the higher tumor JAK2 mRNA level would have the longer recurrence-free survival 62.4% (100%–37.6%) of the time. Similarly, JAK2 mRNA was consistently protective, albeit to a lesser extent, in the METABRIC and OSLOVAL cohorts. Because METABRIC provided a sufficient sample size, we also evaluated the CI for JAK2 mRNA in ER and ER/PR/HER2 (triple-negative) subtypes; JAK2 mRNA was even more protective for both overall and disease-specific survival in these subtypes.

Figure 2.

JAK2 mRNA is associated with a protective CI in the NETHERLANDS, OSLOVAL, and METABRIC cohorts. The CI in patient rankings based on JAK2 mRNA versus survival outcomes is shown for each indicated cohort. For comparison, the CIs for the least and most protective single genes in METABRIC (CDCA5 and FDG3) are shown.

Figure 2.

JAK2 mRNA is associated with a protective CI in the NETHERLANDS, OSLOVAL, and METABRIC cohorts. The CI in patient rankings based on JAK2 mRNA versus survival outcomes is shown for each indicated cohort. For comparison, the CIs for the least and most protective single genes in METABRIC (CDCA5 and FDG3) are shown.

Close modal

JAK2 mRNA and protein levels

To explore the mechanism by which JAK2 mRNA is associated with favorable prognosis, we investigated the relationship between JAK2 mRNA and protein levels. The specificity of a total JAK2 antibody was validated by the strong correlation between JAK2 mRNA and protein levels (as measured by Western blotting) in a panel of 18 human breast cancer cell lines and by the ability to discriminate between JAK2-deficient γ2A cells (21) and JAK2-transfected γ2A cells in immunohistochemistry (Supplementary Figs. S2 and S3). We then measured 10 tumors from the case–control study randomly selected from the highest quartile of JAK2 mRNA expression and 10 tumors from the lowest quartile. Immunohistochemical staining ranged from absent (0) to robust (3+) and was prominent in the tumor epithelial cells. However, JAK2 antibody staining levels in the tumor epithelial cells were not correlated with overall tumor JAK2 mRNA levels (r = −0.17, n = 20; Supplementary Table S4). This finding suggests that the association between higher JAK2 mRNA levels and favorable outcomes in breast cancer may not be a result of JAK2 protein function in breast tumor epithelial cells, and that other cell types in the primary tumors that are associated with prognosis contribute to overall JAK2 mRNA levels.

JAK2 mRNA levels and tumor-infiltrating T cells

JAK2 is expressed in immune cells, and tumor-infiltrating lymphocytes, especially T cells, have been associated with favorable breast cancer prognosis (14, 22). We, therefore, tested whether breast tumor JAK2 mRNA levels correlate with the T-cell transcript-enriched LYM metagene signature. The LYM metagene is associated with favorable prognosis in breast cancer, in particular in ER breast cancer and even more so in the absence of multiple positive lymph nodes, and recently formed part of the winning prognostic model in the Sage Bionetworks DREAM breast cancer prognosis challenge (12, 15). The LYM metagene was recently defined with increased accuracy, following mining from data sets from multiple cancer types available from The Cancer Genome Atlas (23). Indeed, there was a highly significant correlation between JAK2 mRNA levels and the LYM metagene in tumor samples from METABRIC (Fig. 3A). In contrast, the LYM metagene had an inverse correlation with the breast epithelial-associated transcript ESR1 (Fig. 3B). Furthermore, JAK2 mRNA levels were correlated strongly with levels of infiltrating lymphocytes as determined by pathologic assessment in a subset of 156 tumors for which these data were available (Fig. 4). These tumor samples belonged to METABRIC integrative cluster 4, which was previously associated with favorable prognosis and a strong adaptive immune response signature (14). Finally, consistent with a functional role for JAK2 in supporting cytokine receptor signaling during T-cell activation, we found that the JAK1/2 inhibitor ruxolitinib markedly inhibited the anti-CD3–dependent production of IFN-γ, a marker of the differentiation of Th cells along the tumor-inhibitory Th1 pathway (Supplementary Fig. S4; 24). These results suggest that the consistently protective effect of JAK2 mRNA is related, at least in part, to the levels of infiltrating T cells.

Figure 3.

JAK2 mRNA correlates with the LYM metagene signature. A, the average expression of the top-ranked genes of the LYM metagene signature (SASH3, CD53, and NCKAP1L) in each tumor sample from METABRIC is shown relative to JAK2 mRNA. B, the scatter plot for ESR1, which is restricted to epithelial cells, is shown for comparison.

Figure 3.

JAK2 mRNA correlates with the LYM metagene signature. A, the average expression of the top-ranked genes of the LYM metagene signature (SASH3, CD53, and NCKAP1L) in each tumor sample from METABRIC is shown relative to JAK2 mRNA. B, the scatter plot for ESR1, which is restricted to epithelial cells, is shown for comparison.

Close modal
Figure 4.

JAK2 mRNA correlates with levels of tumor-infiltrating lymphocytes. JAK2 mRNA levels are shown relative to levels of tumor-infiltrating lymphocytes in tumor samples from the favorable prognosis METABRIC integrative cluster 4 that is enriched for an adaptive immune response signature. Box plots, distribution of JAK2 mRNA values in quartiles. Circle in the box, mean value; horizontal line, median value. The P value was calculated using ANOVA.

Figure 4.

JAK2 mRNA correlates with levels of tumor-infiltrating lymphocytes. JAK2 mRNA levels are shown relative to levels of tumor-infiltrating lymphocytes in tumor samples from the favorable prognosis METABRIC integrative cluster 4 that is enriched for an adaptive immune response signature. Box plots, distribution of JAK2 mRNA values in quartiles. Circle in the box, mean value; horizontal line, median value. The P value was calculated using ANOVA.

Close modal

To our knowledge, this is the first time that a consistent association between increasing JAK2 mRNA levels and improved breast cancer outcomes has been demonstrated. This association was strongest in the case–control study that matched for variables associated with recurrence. Although the influence of JAK2 mRNA on survival outcomes in the unmatched public cohorts was predictably not as strong, the remarkably consistent association between higher JAK2 mRNA and favorable survival is unexpected because JAK2 proteins collaborate with a variety of cytokine receptors that were shown to promote breast cancer growth (3, 4). The association of JAK2 mRNA with favorable prognosis may reflect a lack of concordance between the levels of JAK2 mRNA and total JAK2 protein and the active phospho-JAK2 in breast tumor epithelial cells. The association between JAK2 mRNA and favorable prognosis also likely reflects the presence of additional JAK2-expressing cell types in the tumor specimens. Consistent with this finding, our previous laser capture microdissection studies demonstrated that JAK2 mRNA was expressed in both breast tumor epithelial and nonepithelial fractions, but was 8.3 (1.5–44.6)-fold higher in nonepithelial fractions (25). Indeed, we observed a strong correlation between JAK2 mRNA and levels of tumor-infiltrating lymphocytes and the favorable prognosis LYM metagene signature. Our finding that a single gene correlates with a larger biomolecular metagene that is associated with prognosis is reminiscent of the frequent association between single genes and the PCNA and CIN metagene signatures for proliferation and chromosomal instability (12).

In addition to the present demonstration that the JAK1/2 inhibitor ruxolitinib inhibits CD3-dependent Th1 differentiation, support for a role of JAK2 in T cells is provided by studies demonstrating that JAK2 is involved in the signaling of IL-12 and IFN-γ, key regulators of the tumor-inhibitory Th1 response (24, 26, 27). Furthermore, JAK inhibitors have been shown to impair production of these Th1 cytokines (28) and to inhibit IFN-γ–dependent T-cell trafficking in murine preclinical studies (29). Determining how specific inhibition of the individual JAK family members influences the repertoire and antitumor activities of tumor-infiltrating T cells represents an important area for future investigation. Such studies will provide insights into whether the benefits of inhibiting JAK2 in breast tumor cells outweigh the potential risks of inhibiting tumor-infiltrating T cells.

No potential conflicts of interest were disclosed.

Conception and design: C.P. Miller, J.D. Thorpe, J.D. Beatty, N.D. Urban, C.A. Blau

Development of methodology: C.P. Miller, J.D. Beatty, N.D. Urban

Acquisition of data (provided animals, acquired and managed patients, provided facilities, etc.): C.P. Miller, A.N. Kortum, J.D. Beatty, C.A. Blau

Analysis and interpretation of data (e.g., statistical analysis, biostatistics, computational analysis): C.P. Miller, J.D. Thorpe, A.N. Kortum, C.M. Coy, W.-Y. Cheng, T.-H.O. Yang, D. Anastassiou, J.D. Beatty, N.D. Urban, C.A. Blau

Writing, review, and/or revision of the manuscript: C.P. Miller, J.D. Thorpe, A.N. Kortum, C.M. Coy, D. Anastassiou, J.D. Beatty, N.D. Urban, C.A. Blau

Administrative, technical, or material support (i.e., reporting or organizing data, constructing databases): J.D. Thorpe, C.M. Coy, J.D. Beatty

Study supervision: C.P. Miller, J.D. Beatty, N.D. Urban, C.A. Blau

The authors thank Hailing Lu, Brigham Mecham, and Michèl Schummer for insightful discussions, which improved the interpretation of the results; Mary Atwood for providing patient clinical characteristics; Desirée Iriarte and Kathy O'Briant for acquisition of pathology specimens; Ronald Tickman and Sean Thornton for tumor-block selection; Brian Johnson and Peggy Porter for immunohistochemistry and interpretation; Jeanne Lee for assistance in harvesting tumor tissue; Dan Herendeen, Ekram Gad, and Benjamin Curtis for providing murine splenocytes; Hsuan-Ni Lee, Nancy Zhang, and Galen Chen for assistance in analyzing data; and Erica Jonlin for assistance with human subjects.

This work was supported by grants from the National Cancer Institute (1R01CA135357 to C.A. Blau), Avon (P50CA083636-S2 to N.D. Urban), and the American Cancer Society (117682-MRSG-09-268-01-CCE to C.P. Miller).

The costs of publication of this article were defrayed in part by the payment of page charges. This article must therefore be hereby marked advertisement in accordance with 18 U.S.C. Section 1734 solely to indicate this fact.

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text:
# What is the reason for this apparent paradox caused by a massless spring with a locking mechanism?

If the spring locks when the ball is at rest in the lab frame, then by the arguments you give it follows that the spring must not be compressed at all. This is indeed the case.

As the ball slows down, the block begins to speed up. Eventually they are traveling at the same speed, at which point the spring has reached its maximum compression. As the spring begins to expand, the block's velocity becomes greater than that of the ball. When the spring attains its uncompressed length, the ball comes to rest and the block is traveling with speed $$v$$.

This can be shown directly. Let $$x(t)$$ be the position of the ball and $$y(t)$$ be the position of the block, and let us consider left to be the positive direction in accordance with your figure. At time $$t=0$$ the ball makes contact with the spring. Let the initial position and velocity of the ball be $$x(0)=0$$ and $$x'(0) = v$$, and the initial position and velocity of the block be $$y(0)=L$$ and $$y'(0) = 0$$ where $$L$$ is the (unimportant) unstretched length of the spring.

The dynamics of the system are governed by the equations

$$m x'' = k(y-x-L)$$ $$m y'' = -k(y-x-L)$$

We can define the auxiliary variables $$u = \frac{x+y}{2}$$ and $$w = \frac{x-y+L}{2}$$ to obtain $$u'' = 0 \implies u(t)= \frac{L+vt}{2}$$ $$w'' = -\frac{k}{m} w \implies w(t)= \frac{v}{2\omega}\sin(\omega t)$$ where $$\omega=\sqrt{k/m}$$ and I've applied the initial conditions stated above. We can invert these relations to find $$x$$ and $$y$$ to be $$x(t) = u+w-\frac{L}{2} = \frac{vt}{2}+\frac{v}{2\omega}\sin(\omega t)$$ $$y(t) = u-w+\frac{L}{2} = L + \frac{vt}{2} - \frac{v}{2\omega}\sin(\omega t)$$

The ball comes to rest when $$x'(t) = \frac{v}{2}(1+\cos(\omega t)) = 0 \implies \omega t = \pi$$. However, at this time we have that $$y-x = L$$ and $$y'= v$$.

An alternative is that the spring locks when the maximum compression is achieved, i.e. when $$y'=x'$$. This occurs when $$\cos(\omega t)=0 \implies \omega t = \pi/2$$. At this moment, the velocity of the ball and the block are both $$v/2$$, in accordance with the conservation of momentum in a completely inelastic collision.

If the spring locks when the deformation is maximum the collision is not ellastic. At that point the relative velocity of the two objects is zero so they continue to move with the same velocity. This is a model of non-elastic collision. The collisions are ellastic when the spring does not lock and restore its potential energy into kinetic. For actual balls the spring is played by the ellasticity of the material.

The problem is exactly how billiard balls work (in 1 dimension, it's a much easier game like that). The spring is replaced by the elasticity of the ball, which doesn't need to lock, because that's how billiards works: all the cue ball's E and p go to the target ball, leaving the cue ball at rest.


text:
# ______ increases the steady-state accuracy.

#### SOLUTION

Lag compensation increases the steady-state accuracy.

Detailed Explanation

Lead compensation and lag compensation are the most commonly used methods of compensation in the conventional frequency domain design of control systems.

### Lag compensator

The transfer function of a lag compensator is given by

$G\left( s \right) = \frac{{\tau s + 1}}{{\beta \tau s + 1}}$

Where

$\beta = \frac{{{Z_c}}}{{{P_c}}} > 1$

• Both pole and zero lie in LHS of the real plane but PC < ZC i.e. zero is farther away from the origin.
• Lag Compensation adds a pole at origin (or for low frequencies).
• It helps to reduce the steady-state error of the system.
• An unstable system is a system that has at least one pole at the right side of the s-plane.
• Even if we add a lag compensator to an unstable system, it will remain unstable.

### Effects of phase-lag compensator:

1. Since the lag compensator allows high gain at low frequencies, it is basically a low pass filter. Therefore, it improves the steady-state performance.
2. In lag compensation, the attenuation characteristics are used for the compensation, whereas the phase lag characteristics are of no use in compensation.
3. Since the attenuation due to a lag compensator shifts the gain crossover frequency to a lower frequency point, the bandwidth of the system is reduced. Hence a slower response occurs. Therefore, rise time and settling time are usually longer and hence the transient response lasts for a longer time.
4. The system is very sensitive to parameter variation.
5. Since a lag compensator approximately acts as a proportional plus integral controller, it thus tends to make a system less stable.

The integrator produces an output proportional to the time integral of error, any finite error will cause the integrator output to keep on increasing and forcing the error to zero. Once the error has reached zero, the integrator output stops increasing but holds the value necessary to remove the error. Phase lag compensation is an integrator, as it reduces the steady-state error and increases steady-state accuracy.

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text:
Sample Problem Solving In Math

This, of course, means looking for clue words as outlined in the previous section.Once you’ve identified your clue words, highlight or underline them.Problem 4A farming field can be ploughed by 6 tractors in 4 days.

This, of course, means looking for clue words as outlined in the previous section.Once you’ve identified your clue words, highlight or underline them.Problem 4A farming field can be ploughed by 6 tractors in 4 days.

Tags: Fraction Homework Year 3Kids EssayExample Of Business Plan ProposalEndangered Birds Of EssayOrganization Of A Research PaperWriting Names Of Books In Essays

If he sold 360 kilograms of pears that day, how many kilograms did he sell in the morning and how many in the afternoon? This must be equal to 360.$3x = 360$$x = \frac$$x = 120$ Therefore, the salesman sold 120 kg in the morning and $2\cdot 120 = 240$ kg in the afternoon. Together the three of them picked 26 kg of chestnuts. So $x 2x x 2=26$$4x=24$$x=6$ Therefore, Peter, Mary, and Lucy picked 6, 12, and 8 kg, respectively.

Solution: Let $x$ be the number of kilograms he sold in the morning. Problem 2 Mary, Peter, and Lucy were picking chestnuts. Solution: Let $x$ be the total number of pages in the book, then she finished $\frac\cdot x$ pages.

Then she has $x-\frac\cdot x=\frac\cdot x$ pages left.

$\frac\cdot x-\frac\cdot x=90$ $\frac\cdot x=90$ $x=270$So the book is 270 pages long.

Many problems are multistep and require some type of systematic approach.

There are a couple of things you need to do when solving problems.

This will let you know what kind of problem you’re dealing with.

Then do the following: Read the problem carefully, and decide on a method to solve the problem.

These techniques may involve reorganizing a problem into simpler terms, breaking a problem into steps, making a plan about how to proceed, determining the best way to solve a problem, pulling out key ideas, etc.

Being systematic in problem solving requires students to: Systematic problem solving often involves "step-wisdom," knowing that the best way to solve a particular problem may be to break it up into a series of logical steps, rather than to try to solve it all at once.

Comments Sample Problem Solving In Math

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• Step word problems practice Khan Academy

Solve two-step word problems with addition, subtraction, multiplication, and. CCSS Math 3. Practice Represent 2-step word problems with equations.…

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Jul 10, 2019. A multistep math problem-solving plan involves looking for clues, developing a game plan, solving the. Problem-solving requires practice.…

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Jun 2, 2019. We will articulate some useful problem solving strategies, but no such list will ever be. Click on this link to see an example of “Guess and Test”.…


text:
# Transformer power loss

annamal
Homework Statement:
A transformer on a utility pole steps the rms down from 12kV to 240V. If the input current to the transformer is 2 A, what would the power loss have been if there were no transformer
Relevant Equations:
##I_S = \frac{N_P}{N_S}I_P##
##P_{loss} = I_S^2R##
##\frac{N_S}{N_P} = \frac{240}{12\times 10^3} = \frac{1}{50}##
##I_S = \frac{N_P}{N_S}I_P = \frac{N_P}{N_S}I_P = 50\times 2 = 100## Amps
##P_{loss} = I_S^2R##, but I don't know how to find R

## Answers and Replies

Homework Helper
Gold Member
Sorry I don't completely understand the asked question. What does it mean by if there were no transformer because taking it literally, if there is no transformer, there is no problem at all lol.

alan123hk
Homework Helper
Gold Member
Btw, the equation you using to calculate ##I_S## it holds only in the case of an ideal transformer, so the power loss is zero in that case.

Gold Member
What? Are they asking about the power lost to the magnetizing inductance of the transformer, i.e. assuming there is no load.

I am getting really fed up with instructors that ask students to answer really poorly worded questions. Of course you had to ask for help, I would too.

In any case, did you really describe the original question exactly as asked?

Still, I'll say zero. Zero power loss in a component that doesn't exist. The transformer that isn't connected will have zero voltage, zero current, and zero power dissipation.

Or maybe they're asking what the power loss is when you put 12KV into a 240V distribution network. I don't know, but I'll guess that it doesn't last for very long.

vela, alan123hk and Delta2
annamal
If you think it is poorly worded, it is the last part (part d) of this problem

Gold Member
The phrase "if there were no transformer, the power would be sent at 240 V to work for these houses" was omitted from your original post. It was crucial to an understanding of the situation. The entire thread was derailed because you failed to post a complete statement of the problem you wanted solved.

Last edited:
nasu and DaveE
Gold Member
OK, that makes sense. I have no idea where R=6000Ω came from, maybe earlier in the text?

Gold Member
but I don't know how to find R
##R=\frac{V}{I}##

Gold Member
OK, that makes sense. I have no idea where R=6000Ω came from, maybe earlier in the text?
You have 12 000 volts across the primary coil, which carries a current of 2.0 amps.

Gold Member
You have 12 000 volts across the primary coil, which carries a current of 2.0 amps.
They are asking for the power loss in the transmission line. Not the power delivered to the load.

annamal
You have 12 000 volts across the primary coil, which carries a current of 2.0 amps.
The resistance should be 200 ohms though...

Homework Helper
Gold Member
@annamal, From now on, please post problem statements exactly as they are and not according to your own interpretation.

Because these two are not necessarily the same, for example here, the problem asks for the power loss in the transmission line while you asked for the power loss in the transformer, and the phrase "if there were no transformer" is accompanied by "and the power would be sent at 240V to homes" which is critical for our understanding.

nasu
Homework Helper
Gold Member
I guess the book implies here that when the power is sent directly @ 240V to house, then the power loss in the transmission line is still much bigger even if we shortened the length of the transmission line (hence its resistance drops from 6000 to 200 Ohm). Or I can't make any other usefull interpretation.

Homework Helper
Gold Member
2022 Award
The solution given to part c is clearly nonsense. There is not enough info to determine an answer. The calculated 24,000W is the power consumed by the transformer plus load, i.e. everything except the transmission line.
@annamal , please cite the source of the question.

Delta2
Homework Helper
Gold Member
No I don't find it complete nonsense, 2A is the current in the transmission line that feeds the primary. 6000ohm might be the total ohmic resistance of a typical power transmission line ...

Homework Helper
Gold Member
2022 Award
No I don't find it complete nonsense, 2A is the current in the transmission line that feeds the primary. 6000ohm might be the total ohmic resistance of a typical power transmission line ...
It could be, but we are not told it is. What we are told is that the transmission line feeds 2A into the transformer at 12kW, so it is supplying 24kW.

DaveE
Homework Helper
Gold Member
It could be, but we are not told it is. What we are told is that the transmission line feeds 2A into the transformer at 12kW, so it is supplying 24kW.
Ok I admit I am not aware of the full analysis of the power loss in a transmission line (which might be a bit complicated, since the current varies along the transmission line, there is the the skin effect which increases the ohmic resistance e.t.c) but I 've seen this simplification that ##P_{loss}=I^2R## in many books, where ##I## the current at the end of the transmission line and ##R## its total ohmic resistance.

Not sure but I think ##I^2R## is a good lower bound for the power loss in the transmission line.

Last edited:
annamal
@annamal, From now on, please post problem statements exactly as they are and not according to your own interpretation.

Because these two are not necessarily the same, for example here, the problem asks for the power loss in the transmission line while you asked for the power loss in the transformer, and the phrase "if there were no transformer" is accompanied by "and the power would be sent at 240V to homes" which is critical for our understanding.
I couldn't because if you notice in the word problem, there is no explicit question asked for it, but the question is answered

Delta2
Homework Helper
Gold Member
I couldn't because if you notice in the word problem, there is no explicit question asked for it, but the question is answered
Yes I understand that, but still you could have mention that it is the power loss in the transmission line, and not in the transformer.

annamal
Yes I understand that, but still you could have mention that it is the power loss in the transmission line, and not in the transformer.
In my original post, I don't mention if it is the power loss in the transmission line or transformer. I just say "what would the power loss have been if there were no transformer"

Homework Helper
Gold Member
In my original post, I don't mention if it is the power loss in the transmission line or transformer. I just say "what would the power loss have been if there were no transformer"
Ok but your solution attempt indicates that you are looking for the power loss in the secondary of the transformer.

alan123hk
Parts C and D are a bit hard to understand in my humble opinion.

In part C, 6000 ohms seems to be the ohmic resistance of the transmission line, but it is strange why this value is exactly equal to the input impedance ##~\frac {12kV}{2A} ~##of the transformer ?

In addition, part D does not explain how the value of 200 ohms is obtained. This is not the ohmic resistance of the transmission line in part C, nor the ohmic resistance inside the transformer. It is really difficult to guess what it is.

Delta2
Gold Member
Ok I admit I am not aware of the full analysis of the power loss in a transmission line (which might be a bit complicated, since the current varies along the transmission line, there is the the skin effect which increases the ohmic resistance e.t.c) but I 've seen this simplification that ##P_{loss}=I^2R## in many books, where ##I## the current at the end of the transmission line and ##R## its total ohmic resistance.

Not sure but I think ##I^2R## is a good lower bound for the power loss in the transmission line.
It's really not so much about your formulas R=V/I or P=I2R, we all know those. It's all about which voltage, what resistance and where did it come from.

Gold Member
The key phrase here is "power loss" There is no reasonable interpretation of this problem that would mix that up with the power sourced or delivered to the load. If you can't see that distinction, perhaps you should be asking questions instead of providing solutions. The options are simple in this case:

1) There isn't the necessary information given to answer this question. We can't know who is responsible, we just know that it isn't there.
2) The line resistance is given as 6000Ω, or 200Ω in a subtle manner with no explanation about why.

Homework Helper
Gold Member
2) The line resistance is given as 6000Ω, or 200Ω in a subtle manner with no explanation about why.
Yes I think that's exactly the problem here. I tried to find more info in the accompanying text that explains the physics of an Ideal(!?) transformer and says some things about power distribution network but couldn't find anything there either.

alan123hk
The line resistance is given as 6000Ω, or 200Ω in a subtle manner with no explanation about why.
This may be a subtle way, but it may not be suitable for the average beginner because they may not be able to tell if the question contains all the information needed to find the answer. They may think the values are calculated based on some given data, so they may be confused.

Delta2
Homework Helper
Gold Member
2022 Award
I 've seen this simplification that ##P_{loss}=I^2R## in many books, where ##I## the current at the end of the transmission line and ##R## its total ohmic resistance.
You are missing the point. We have no information regarding the ohmic resistance of the transmission line. The ##6000\Omega## calculated appears to be the effective resistance of the transformer + downstream load.

Homework Helper
Gold Member
You are missing the point. We have no information regarding the ohmic resistance of the transmission line. The ##6000\Omega## calculated appears to be the effective resistance of the transformer + downstream load.
ok, fine if that's your point I agree.

Homework Helper
Gold Member
2022 Award
part D does not explain how the value of 200 ohms is obtained
Perhaps we were meant to be given 200 Ohms as the transmission line resistance (still high, but at least feasible) and the part c solution should have used that value too.

I have contacted the site.

Delta2
Gold Member
OK, only slightly off-topic...

I am so G%$D%$@ tired of students being confused and asking questions that are solely the result of the apparent fact that their instructor is an idiot or too lazy to write comprehensible questions. This seems to be about 80% of the EE questions I see here. I think it's really unfair to confuse people that are new to a subject. They just don't have the background to question the quality of their course work; they think they don't understand, when, in fact, it's not understandable.

Sorry, I'm frustrated and needed to vent. Y'all can go back to this seemingly unending discussion about missing information.

Delta2
annamal
Ok but your solution attempt indicates that you are looking for the power loss in the secondary of the transformer.
That's what the answer key says the solution is, which I am trying to ask how they figured it out.

Homework Helper
Gold Member
2022 Award
That's what the answer key says the solution is, which I am trying to ask how they figured it out.
I think we have consensus that the given solution is nonsense.

My best guess is that the problem statement was supposed to tell you that the resistance of the transmission line is 200 Ohms (as assumed in the solution to part d), and that that value should have been used in part c, not 6000 Ohms. 200 still seems a bit high to me, but 6000 for a transmission line is ridiculous.

This is the message I sent to OpenStax:
Message: I wish to draw your attention to some misinformation on your website at https://openstax.org/books/university-physics-volume-2/pages/15-6-transformers. Part c of the example 15.6, Step-Down Transformer, is nonsense.
The question is unanswerable with the given data because there is no way to determine the resistance of the transmission line. The 6000Ohms used in the solution appears to be the combined resistance of the transformer and the downstream load. (I believe typical transmission lines have resistances of a few tens of Ohms at most.)

annamal
alan123hk
My best guess is that the problem statement was supposed to tell you that the resistance of the transmission line is 200 Ohms (as assumed in the solution to part d), and that that value should have been used in part c, not 6000 Ohms. 200 still seems a bit high to me, but 6000 for a transmission line is ridiculous.

The ohmic resistance of a transmission line is really hard to estimate and depends on many factors.
In this example, the input power of the transformer is ##12KV\cdot 2A=24KW ##. If the ohmic resistance of the transmission line is 6000 ohms, then its ##~2^2\cdot6000=24kW~##power loss is as high as the input power of the transformer. I don't think the power company will accept it.
Also, this question is really confusing. If you get a reply from OpenStax, please share the answer to this unsolved mystery.


text:
# Math Help - Quadratic equation

I don't know the steps, pls help me

Thanks.

2. Note that $\frac{\alpha}{2}$ is a root of $2x^2+x+1$. This means:
$2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+ 1=0$
$\frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0$
$\frac{\alpha^2}{2}=-\frac{\alpha}{2}-1$
$\alpha^2=-\alpha-2$

Use this, along with $\alpha^4=(\alpha^2)^2$, to get your result.

--Kevin C.

3. Originally Posted by TwistedOne151
Note that $\frac{\alpha}{2}$ is a root of $2x^2+x+1$. This means:
$2\left(\frac{\alpha}{2}\right)^2+\frac{\alpha}{2}+ 1=0$
$\frac{\alpha^2}{2}+\frac{\alpha}{2}+1=0$
$\frac{\alpha^2}{2}=-\frac{\alpha}{2}-1$
$\alpha^2=-\alpha-2$

Use this, along with $\alpha^4=(\alpha^2)^2$, to get your result.

--Kevin C.
I've tried that before,
$\begin{array}{l}
\alpha ^2 = - 2 - \alpha \\
\alpha ^4 = \left( { - 2 - \alpha } \right)^2 = \alpha ^2 + 4\alpha + 4 \\
\alpha ^2 + 4\alpha + 4 = p\alpha + q \\
????? \\
\end{array}$

4. Good job getting that far.
You still have an $\alpha^2$. Use again to get the answer

You still have an $\alpha^2$. Use again to get the answer
oh ya! stupid me, I didn't realize I can still sub $\alpha^2$ again!