The Ellingham Diagram: How to Use It in Heat-Treat-Process Atmosphere Troubleshooting

FEATURE | Industrial Gases/Combustion

The Ellingham Diagram: How to Use it in Heat-Treat-Process Atmosphere Troubleshooting Michael J. Stempo – Air Liquide International U.S. LP; Houston, Texas The Ellingham diagram is not always thought of as a heat-treater’s tool. This article demonstrates its usefulness.

H

a arold Johann Thomas Ellingham (1897-1975) was a Brith ish physical chemist and is best b known for the diagrams named after him that plot the change in standard free energy with respect to temperature for reactions like the formation of oxides, sulfides and chlorine of various elements, such as: 2x

⁄y M + O2 ‹ 2 ⁄y MxOy

The oxide plot tends to be the most common and will be highlighted here as its applicability to heat-treating processes is the most relevant. This phenomenon was known before Ellingham’s time, but Ellingham demonstrated it more clearly and made it more accessible to industry as a tool. His diagram and its variants help to select the best reducing agent for various ores in the extractive-metal process. Ellingham discovered that by normalizing the thermodynamic functions to a given reaction with one mole of oxygen he was able to compare the temperature stability of many different oxides on the same diagram. In particular, and this reaction is critical to metal reduction systems that use carbon dioxide, he could show graphically that carbon becomes a stronger reducing agent as the temperature increases. The reduction of metal oxides with carbon (or carbon monoxide) to form the free, reduced metals is of immense industrial importance (blast furnace reduction of iron ores), and Ellingham diagrams show the lowest temperature at which the reaction will occur for each metal.

Let’s restate and go stepwise through the Ellingham diagram to set up and make clearer how metal heat treaters can use it. We can adapt the diagram from its original use as a higher-temperature extractivemetallurgy tool to one where we can predict the effects of protective atmospheres and common atmosphere impurities and their impact on heat-treated product. See Fig. 1 for the classic Ellingham diagram that would typically be used for heat-treating atmosphere processes. Background An Ellingham diagram is a plot of 6G (change in Gibbs free energy) versus temperature, which, for our purposes, would be a temperature in a continuous furnace by zone or in a batch furnace by time in the cycle. The Ellingham diagram shown is for metals reacting to form oxides. Since any explanation of how to calculate and plot an Ellingham diagram is more about the mechanics of the derivation process and less about how to use the tool, which is the intent of this article, the concept is covered very briefly here. Enthalpy (6H) and entropy (6S) are essentially constant with temperature. Unless a phase change occurs, the free energy (6G) versus temperature plot can be drawn as a series of straight lines, where 6S is the slope and 6H is the y-intercept. 6G = 6H – T 6S The slopes of those plotted lines change when any of the materials involved melt (M) or vaporize/boil (B). In a heat-treat

system this is (for the most part) irrelevant, except in the brazing and sintering processes where those phase changes (melting) can indeed occur. The free energy of formation is negative for most metal oxides, which means the reaction can proceed without further influence. Therefore, the diagram is drawn with 6G=0 at the top of the diagram, and the values of 6G shown are all negative numbers. Temperatures where either the metal or oxide melt (M) or vaporize (B) are marked on the diagram. Note that the majority of the plots for metals slope upward because both the metal and the oxide exist as condensed solid or liquid phases. The oxygen partial pressure is taken as 1 atmosphere, and all of the reactions are normalized then plotted to represent consumption of one mole of O2. Interestingly, there are two plots that do not look like all of the others. C + O2 => CO2 Carbon, a solid, reacts with one mole of oxygen and produces one mole of carbon dioxide (CO2), which results in little change of entropy – an almost horizontal plot. The other has a distinct negative slope: 2C + O2 => 2CO In this reaction, a solid once more reacts with one mole of gas but produces two moles of gas – carbon monoxide (CO). This causes a substantial increase in entropy, and the plot has a distinct IndustrialHeating.com - April 2011 55

FEATURE | Industrial Gases/Combustion

Ellingham Diagram for Metals Processing When plotting whether a given metal system will oxidize, reduce or remain as an oxide or pure metal, the Ellingham diagram does not indicate the quantitative rate of the reaction, only the probability of it occurring based on a given set of conditions. One can make the assumption that the reaction will qualitatively occur more rapidly as temperature increases or as the conditions for reducing or oxidation deviate farther from neutral-atmosphere conditions. We can use the diagram to determine the relative ease by which a metal can be oxidized or an oxide can be reduced. Metals plotted high up on the diagram are easier to reduce (noble metals) than those plotted lower on the diagram, which naturally tend to exist in very stable oxide forms. For example, Ag, or silver, is very hard to oxidize, while Ca, or calcium, does 56 April 2011 - IndustrialHeating.com

not naturally exist in its elemental form, indicating a very stable oxide form. The metal/metal oxide plots also interrelate to each other. A metal plotted below another metal can reduce the oxide of the one plotted higher up the diagram. Hence titanium, not the oxide, can reduce the oxide of chromium, which is plotted higher on the diagram. We can also use the diagram to determine the following at a given temperature: • The ratio of hydrogen to water/dewpoint (PH2/H2O) that can reduce a metal oxide to metal or prevent a metal from oxidizing • The ratio of CO to CO2 (PCO/CO2) that can reduce a metal oxide to metal or prevent a metal from oxidizing • The partial pressure of oxygen that will be in equilibrium with a metal oxide

Determining the Equilibrium Partial Pressure of Oxygen For a PO2 higher than the equilibrium value at a given temperature, the metal will be oxidized. Conversely, for an oxygen partial pressure that is lower than the equilibrium value at a given temperature, that metal will be reduced. Use the scale or nomograph to determine the equilibrium PO2 by the following method. Using a straightedge and knowing the temperature or series of temperatures you wish to use (and the metal/ metal-oxide system), put the one side of the straightedge on the upper left-hand corner of the diagram that is labeled “O” (near the Ag to silver-oxide plot). Next, position the straightedge from that anchor point to the temperature point where the metal in question intersects that temperature value. Now, continue across

PH 2 /PH2O O

o

10-8 10-7

PCO/PCO2

10-14 10-12 10-1010-910-8 10-7 10-6

-100 -200 -300

10-6

Cu 2O O =2 4Cu + 2 2C + O2 = 2CO

-400

10-5

10-5 10-4

-500

C -600

r + O2 4/3C

-700

-900

-1100

nO = 2M

M 10-14 M

M

l + O2 4/3A

Al 2O 3 = 2/3

gO = 2M + O 2 2CaO g M 2 = + O2 2Ca 0 200 PO2,(atm)

400 10-100

B

B

M

800

10

10-1 1 10

102 103

102

1000 10-60

1200 10-50

104

105

10-20

107

106

10-22

108

107

10-30 1400

109 1010

PCO/PCO2

1011

10-34

1010 1011

1013 1300 ˚C

109

1013

1014

PH2/PH2O

108

1012

1600 10-38

104

106

10-28

10-42

103

10-18

10-24

M M Melting point of metal B Boiling Point of metal M Melting point of oxide

10-80

1

105

M

600

10-2

10-1

10-16

10-26

Temp. ˚C

-1200

O Kelvin

M

SiO 2 O = Si + 2 TiO 2 = O Ti + 2

-800

-1000

+ O2 2Mn

Cr 2O 3 = 2/3

10-3 10-2

10-12

3/2Fe + O2 = 1/3Fe3O4

H

10-4

10-3 1 10-1 10-2 M 10-3 O4 -4 2Fe 3 10 M = M O2 + e iO oO 6F 10-5 M = 2N = 2C + O2 2H 2O + O2 = 2Ni o O O C + 2 2 2 2H 2 = 2C M + O2 10-8 2Co M C + O2 = CO2 10-10 M

4Ag + O2 = 2Ag2O

6G˚ = RTInpO 2,k j

negative slope. The severe negative slope of this carbon reaction results in an increasingly more powerful reducing agent as temperature increases. For example, at 1500˚C (2732˚F), the carbon reaction crosses under that for silicon dioxide (SiO2), resulting in the potential ability to reduce the highly stable compound of SiO2 to silicon under oxygen-starved conditions. Note on the diagram there is a scale on the right and along the bottom of the graph illustrating PO2 (atm). PO2 is plotted showing partial-pressure values from 1 to 10-100. It will be shown how to use this scale in the next section, along with the scales immediately to the right of this one showing the relationship of PH2/H2O and PCO/CO2 to the various metals and their oxides as a function of temperature. These partial-pressure values and their associated ratios can be plotted on the Ellingham diagram and can be obtained by sampling the process atmosphere at various temperatures, or at specific times (temperatures) in a cycle, using familiar and readily available atmosphere analytical tools (oxygen, hydrogen, dewpoint/moisture, CO and CO2).

1012

the straightedge plot to the value on the PO2 scale. This is the equilibrium partial pressure of oxygen. For this diagram, it is expressed in atmospheres. Once again, any oxygen partial pressure that is lower than the one derived will cause metallic reduction. Any partial pressure above will cause oxidation. For a real-life example, find the equilibrium partial pressure of oxygen for chromium at a temperature of 1300˚C (2372˚F). Put the straightedge on the “O” dot in the upper left hand corner of the diagram. Find the chromium plot and where it intersects the temperature of 1300˚C. While anchoring the one end of the straightedge on the “O” dot, move the other end of the straightedge so that it runs through that intersection point and continue down the straightedge to the PO2 scale. Read off the value, which should be 10-16 expressed in atmospheres. This is an extremely small amount of oxygen than can be present in

this high-temperature system before the deleterious effects of oxygen will adversely oxidize the metallic chromium. The effect of chromium oxidation by even the smallest amounts of oxygen can be countered by using a strong atmosphere reducing agent such as hydrogen. Let’s go next to determining the equilibrium ratio (PH2/H2O) or the hydrogen-to-water (dewpoint) ratio. Determining the Equilibrium PH2/H2O Ratio For a PH2/H2O, or hydrogen-to-water ratio, that is higher than the equilibrium value – relatively more hydrogen to water – at a given temperature, the metal will be reduced. For a PH2/H2O that is lower than the equilibrium value (relatively less hydrogen to water) at a given temperature, the metal will be oxidized. Using essentially the same straightedge method that we used above to plot the

equilibrium partial pressure of oxygen, we now plot from the “H” dot on the left side of the diagram through our desired metal and desired temperature and now read from the PH2/H2O plot on the right side of the diagram. A real-life example would be to find the equilibrium hydrogen-to-water ratio for chromium at a temperature of 1300˚C. Locate the straightedge left side on the “H” dot, pass through the chromium equation where it intersects at 1300˚C and read off the equilibrium hydrogen-to-water ratio value on the PH2/H2O scale. The value should be between 102 and 103. This represents a value of between 100 and 1,000 to 1 ratio of hydrogen needed to water or dewpoint level. For this exercise, we can estimate the hydrogen-to-water ratio to be 400 to 1. This is a high level of hydrogen that is necessary to counter the oxidizing effects of water or dewpoint on chromium,

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FEATURE | Industrial Gases/Combustion

hence the reason why most stainless steel (chromium present as an alloy) hightemperature processing is usually done in very high levels of hydrogen – often 100%. At the processing temperature of 1300˚F, in the presence of hydrogen, oxygen species will convert to water and become a part of whatever background moisture or dewpoint that might be in the process-atmosphere system. Enough hydrogen must be present in the atmosphere system to counter oxidation of chromium. With too low of a hydrogen ratio to a given moisture level, chromium oxidation will occur. With no hydrogen in such an atmosphere system (vacuumfurnace processes excluded), it is literally impossible to exclude enough oxygen from the system to prevent chromium oxidation. Determining the Equilibrium PCO/CO2 Ratio The PCO/CO2, or CO/CO2 ratio, is used with more frequency in extractive-metallurgy systems for all metals to determine whether this reaction will reduce or oxidize a given metal system at a given temperature. In heat-treated metals, the ratio is usually used for carbon-bearing iron alloys to determine whether the metal will decarburize or carburize. It is the backbone of carburizing and hardening and all of its derivative processes. The topic of carburization and its derivative processes is reserved for articles far beyond the space allotted and the scope of this article on the Ellingham diagram. There is far more to consider than merely plotting the equilibrium value for the CO/CO2 ratio. In these systems, we are often interested in depth of the carbon in the iron, activities, diffusion rates, time at temperature, degree of surface oxidation and its removal, carbon concentration profiles, avoiding intergranular oxidation (IGO), etc. to name but a few variables. However, we can get a rough idea of what the equilibrium ratio should be so that conditions can be set to promote deliberate decarburization in the case of electrical steels. In the case of 58 April 2011 - IndustrialHeating.com

carburization of steel, we can at least see what equilibrium ratio we must roughly exceed to cause the carburization process to occur. Summary Harold Ellingham’s work made clearer and more accessible, in an easy-tounderstand graphical format, the wellknown concepts of standard free energy with respect to temperature for reactions like the formation of oxides, sulfides and chlorides. In the case of his metal and metal-oxide diagram, by normalizing the thermodynamic functions to reaction with one mole of oxygen, Ellingham was able to compare the temperature stability of many different oxides on the same diagram. The result was his relatively easy-to-use Ellingham Diagram to determine reduction and oxidation conditions without resorting to painstaking efforts to interrelate all metal systems’ free-energy-based calculations from scratch. The addition of the original nomograph tool that plots, in a continuous scale, the values of PO2 for ease of calculation by the user is the work of F. D. Richardson and Jeffes. Since CO/CO2 and H2/H2O ratios are often used in conjunction with determining equilibrium partial pressure of oxygen, L.S. Darken and R.W. Gurry added their calculations as nomograph scales for these additional ratios. These diagrams have subsequently been referred to as Ellingham, Ellingham-Richardson, Darken and Gurry, or modified Ellingham diagrams. IH For more information: Contact Michael J. Stempo, P.E., business development/applications engineering, Air Liquid Industrial U.S. LP, 2700 Post Oak Blvd., Suite 1800, Houston, TX 77056; tel: 610-997-0936; fax: 713-803-7344; e-mail: [email protected]; web: www.us.airliquide.com

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