Chemical Reactions That Produce Heat and Light Easy to Make Chemical Reactions That Produce a Gas

Thermochemistry

In Practical Chemical Thermodynamics for Geoscientists, 2013

B Exothermic and endothermic reactions

Exothermic reactions are chemical reactions that produce heat. In Section II-B of Chapter 3, we defined the heat flow q as negative when heat flows from the system to the surroundings. Thus, exothermic reactions have a negative ΔH of reaction. The word exothermic is derived from exo, the Greek word for outside, and therme, the Greek word for heat. Table 5-1 gives examples of several exothermic reactions. Endothermic reactions are chemical reactions that take heat from the surroundings. In Chapter 3, we defined the heat flow q as positive when a system gains heat from the surroundings. Thus, endothermic reactions have a positive ΔH of reaction. The word endothermic is derived from endo, the Greek word for inside, and therme. Table 5-2 gives examples of several endothermic reactions. The description of chemical reactions as exothermic or endothermic dates to 1869, when these terms were introduced by the French thermochemist Marcellin Berthelot (see sidebar).

Table 5-1. Examples of Exothermic Reactions

Reaction	Δr H o (kJ mol−1)
½ H2 (g)   +   ½ Cl2 (g)   =   HCl (g)	−92.31
formation of hydrogen chloride gas from the elements
C (graphite)   +   2 H2 (g)   =   CH4 (g)	−74.81
formation of methane from the elements
NH3 (g)   +   H2S (g)   =   NH4SH (s)	−90.32
reaction of NH3 and H2S to form solid ammonium hydrosulfide
KOH (c)   +   ∞ H2O (l)   =   K+ (aq)   +   OH− (aq)   +   ∞ H2O (l)	−57.46
potassium hydroxide dissolving in water to make an infinitely dilute solution
CaO (lime)   +   H2O (l)   =   Ca(OH)2 (portlandite)	−65.40
hydration of lime
H2O2 (aq)   =   H2O (l)   +   ½ O2 (g)	−94.66
decomposition of aqueous hydrogen peroxide to water and O2 (g)
2 NO2 (g)   =   N2O4 (g)	−57.36
formation of nitrogen tetroxide from nitrogen dioxide

Enthalpy values at 298.15   K from Appendix 1.

Table 5-2. Examples of Endothermic Reactions

Reaction	Δ r H o (kJ mol−1)
½ H2 (g)   +   ½ I2 (c)   =   HI (g) formation of hydrogen iodide gas from the elements	26.50
½ N2 (g)   +   ½ O2 (g)   =   NO (g) formation of nitric oxide from the elements	91.27
NH4Cl (c)   +   ∞ H2O (l)   =   NH4 + (aq)   +   Cl− (aq)   +   ∞ H2O (l) NH4Cl dissolving in water to make an infinitely dilute solution	14.06
NaCl (c)   +   ∞ H2O (l)   =   Na+ (aq)   +   Cl− (aq)   +   ∞ H2O (l) table salt dissolving in water to make an infinitely dilute solution	3.88
C (graphite)   =   C (diamond) transformation of graphite into diamond	1.85
S (rhombic)   =   S (monoclinic) transformation of rhombic to monoclinic sulfur	0.36
H2O (ice)   =   H2O (water) ice melting to water (273.15 K)	6.0095
Ga (metal)   =   Ga (l) gallium metal melting in your hand (302.15 K)	5.585
Hg (l)   =   Hg (g) vaporization of liquid mercury to Hg (g)	61.38
H2O (l)   =   H2O (g) water vaporizing to steam (373.15 K)	40.656
C3H6O (l)   =   C3H6O (g) acetone vaporization at its boiling point (329.3 K)	29.09
CO (g)   =   C (g)   +   O (g) dissociation of carbon monoxide to the atoms	1076.4
CaCO3 (limestone)   =   CaO (lime)   +   CO2 (g) calcination of limestone during cement manufacture	179.0
I2 (c)   =   I2 (g) sublimation of iodine	62.42

Enthalpy values at 298.15   K unless noted otherwise. Data are from Appendix 1.

PIERRE MARCELLIN BERTHELOT (1827–1907)

Berthelot was a French chemist whose work confirmed several important ideas. He used bomb calorimetry to measure the enthalpies of hundreds of reactions. He thereby verified Hess' law, although he was not the first to propose it. He coined the terms endothermic and exothermic. He thought that the heat evolved by reactions drove them; however, we now know that some endothermic reactions proceed spontaneously.

His work in organic chemistry led to some very important conclusions. Berthelot synthesized many organic compounds, thus showing that biological processes are not required to form them. He believed that chemical processes were nothing mystical, but that they were subject to the same physical laws that apply to everything in the universe. While experimenting with the synthesis of organic compounds, Berthelot noticed that his reactions never went to completion. Instead, they stopped at some definite midway point. The products were present in some proportion to the unused reactants which was the same regardless of how much material he started with. This was the beginning of the study of chemical equilibria.

Berthelot had interests beyond his organic chemistry experiments. Active in the government, he was elected as a senator for the first time in 1871, despite the fact that he had not run for the office. He was chosen by voters because of his work as the president of the Comité Scientifique pour la Défense de Paris during the Franco-Prussian war. He also studied the history of chemistry, which is commonly thought to have originated in Egypt. He felt that alchemy arose out of a misinterpretation of the practical knowledge of Egyptian metal smiths. By analyzing metal artifacts from Egypt and Mesopotamia, he created the new field of chemical archaeology. Berthelot intertwined his interests in science and philosophy. He felt there was no limit to possible scientific attainments and that a Utopia could be achieved through scientific principles by the year 2000.

Happily married for 45 years, Berthelot died within an hour after his wife. They had six children.

Home heating with natural gas, which contains >80% methane in most U.S. cities, is an example of an exothermic reaction. When natural gas is burned in air, for example, in a furnace that is heating a house or apartment building, CH₄ combustion occurs by the reaction

(5-11) $2{\text{CH}}_4+4{\text{O}}_2=2{\text{CO}}_2+4{\text{H}}_2\text{O}\text{(1)}\mathrm{\Delta }{H}_{298}=-1780.72\text{kJ}$

where ΔH ₂₉₈ stands for the ΔH value at 298.15   K. Of course, the gas is being burned at a much higher temperature than this. Later in this chapter we will describe how ΔH values change with temperature and how to calculate flame temperatures. For now we will discuss and compare ΔH values at 298   K. The combustion of ethane (C₂H₆) or propane (C₃H₈), the next most abundant hydrocarbons in natural gas, is also exothermic, for example,

(5-12) $2{\text{C}}_2{\text{H}}_6+7{\text{O}}_2=4{\text{CO}}_2+6{\text{H}}_2\text{O}\text{(1)}\Delta H_{298}=-3119.6\text{kJ}$

(5-13) ${\text{C}}_3{\text{H}}_8+5{\text{O}}_2=3{\text{CO}}_2+4{\text{H}}_2\text{O}\text{(1)}\mathrm{\Delta }{H}_{298}=-2220.0\text{kJ}$

The ΔH value for combustion of one mole of a hydrocarbon or any other combustible material in air or any other gas is the molar enthalpy of combustion (Δ _cH_m ) or molar heat of combustion for the material and is written with the name of the combusted material, its phase, and the temperature following in parenthesis. Thus, Δ _cH_m (CH₄, g, 298)   =   −890.36   kJ mol⁻¹ is the molar enthalpy of combustion for methane gas at 298.15   K, and Δ _cH_m (C₃H₈, g, 298)   =   −2220.0   kJ mol⁻¹ is the molar enthalpy of combustion for propane gas at 298.15   K. Liquid water is a product in reactions (5-11) to (5-13) because enthalpy of combustion values are experimentally measured at 100% relative humidity to ensure its formation. The enthalpy of combustion is important industrially for evaluating the caloric value and the cost of fuels such as coal, gas, and oil. The enthalpy of combustion is also important scientifically for determining enthalpy of formation values (see the following discussion), which cannot be directly measured because of experimental difficulties.

Several examples of endothermic reactions are given in Table 5-2. Two well-known endothermic reactions are the formation of hydrogen iodide gas (HI) and the formation of nitric oxide gas (NO) from the constituent elements:

(5-14) $\frac{1}{2}{\text{H}}_2(\text{g})+\frac{1}{2}{\text{I}}_2(\text{c})=\text{HI}\text{(g)}\mathrm{\Delta }{H}_{298}=+26.50\text{kJ}{\text{mol}}^{-1}$

(5-15) $\frac{1}{2}{\text{N}}_2(\text{g})+\frac{1}{2}{\text{O}}_2(\text{g})=\text{NO}\text{(g)}\mathrm{\Delta }{H}_{298}=+91.27\text{kJ}{\text{mol}}^{-1}$

The ΔH value in each case is the molar enthalpy of formation (Δ _fH_m ) because the HI or NO are formed from their constituent elements. Likewise, the ΔH values for hydrogen chloride (HCl) gas and CH₄ in Table 5-1 are the molar enthalpies of formation of these two compounds.

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Material Performance and Corrosion/Waste Materials

J. Fachinger , in Comprehensive Nuclear Materials, 2012

5.21.6.3 The Russian 'Self-Propagating High-Temperature Synthesis SHS'

Graphite is homogenously mixed aluminum and titanium dioxide. The amounts are related to the following reaction:

$3\text{C}+4\text{Al}+3{\text{TiO}}_2\to 2{\text{Al}}_2{\text{O}}_3+3\text{TiC}$

The exothermic reaction is self-propagating and only an initial start is required. The formed stable titanium carbide contains ¹⁴C and the other radionuclides incorporated into the corundum and titanium carbide lattice. Additional confining additives can be added to the reaction mixture, for example, zirconium, which build even more stable crystalline phases with selected radionuclides such as uranium and plutonium. Furthermore, additives are used to improve the final product quality and to minimize the volatilization of radionuclides.

Therefore, this process is also suitable for graphite contaminated with actinides from the Russian production reactors. The process requires a carefully controlled regime to minimize the radionuclide release.

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Process and Operations

Mukesh Doble , Anil Kumar Kruthiventi , in Green Chemistry and Engineering, 2007

Annular Tubular Reactor

Endothermic and exothermic reactions are combined in an annular tubular reactor to achieve tremendous energy savings (see Fig. 6.39). Dehydrogenation of ethyl benzene to form styrene is an endothermic reaction (ΔH = 188 kJ/m). Side products such as CH⁴, ethylene, benzene, and toluene are also formed during the process. The catalyst for dehydrogenation is Fe oxide promoted by K²O and CrO. This reaction is carried out in a radial flow reactor where the core of the packed bed has an oxidative catalyst. Steam and oxygen are introduced at the center and will react with H² to generate heat that flows out of the core to the annulus. The shell side has the dehydrogenation catalyst, where the conversion of ethyl benzene to styrene takes place, capturing the heat flowing out.

FIGURE 6.39.. Annular tubular reactor (radial heat flow reactor).

Conducting endothermic and exothermic reactions in parallel in the same reactor can be extended to the manufacture of synthesis gas from natural gas. The reaction CH⁴ + H²O → CO + 3H² is endothermic (ΔH = 206 kJ/mol), while CH⁴ + 0.5 O² ⇌ CO + 2H² is exothermic (ΔH = –38 kJ/mol). When these two reactions are coupled in a radial flow, considerable savings in energy can be achieved.

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Reactors in Process Engineering

Gary L. Foutch , Arland H. Johannes , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

IV.K.1 Description

For rapid exothermic reactions that require continuous stirred-tank operating conditions for good reaction control, a jet tube reactor can be used. This reactor gives thorough mixing despite the extremely short residence times involved in these reactions. One reactant is injected into the other through a jet, orifice, or venturi. This gives high turbulence to insure a well-mixed condition. Large-scale testing is needed to select the reactor conditions accurately, since minor errors in kinetic constants are magnified due to the high activation energies of the reactions. Jets can handle both gas and liquid feed and are usually built in multiple jet configurations.

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Coal Conversion

Michael A. Nowak , ... Adrian Radziwon , in Encyclopedia of Energy, 2004

2 Pyrolysis and Coking

Although combustion is an exothermic reaction involving stoichiometric amounts of oxygen and achieving flame temperatures as high as 1650°C, coal pyrolysis is an endothermic process. When coal is heated, it undergoes thermal decomposition that results in the evolution of gases, liquids, and tars, leaving behind a residue known as char. Coal pyrolysis is an important process for making metallurgical coke. Coal pyrolysis is a very old technique based on relatively simple technology that dates back to the 18th century. Most pyrolysis systems used in the late 1800s to the early 1900s were located in Europe, where the objective was the production of a smokeless fuel (char) for domestic heating and cooking purposes. It was soon realized in the manufacture of char for fuel that the coal tar fraction contained valuable chemical products that could be separated through refining. However, as inexpensive petroleum appeared in the mid-1900s, interest in coal pyrolysis and its by-products faded.

Coal pyrolysis processes are generally classified as low temperature (<700°C), medium temperature (700–900°C), or high temperature (>900°C). Coal undergoes many physical and chemical changes when heated gradually from ambient temperature to approximately 1000°C. Low-temperature changes include loss of physically sorbed water, loss of oxygen, and carbon–carbon bond scission. At higher temperatures (375–700°C), thermal destruction of the coal structure occurs, as reflected by the formation of a variety of hydrocarbons, including methane, other alkanes, polycyclic aromatics, phenols, and nitrogen-containing compounds. In this temperature range, bituminous coals often soften and become plastic (thermoplastic) to varying degrees. At temperatures between 600 and 800°C, the plastic mass undergoes repolymerization and forms semicoke (solid coke containing significant volatile matter). At temperatures exceeding 600°C, semicoke hardens to form coke with the evolution of methane, hydrogen, and traces of carbon oxides. Pyrolysis of coal is essentially complete at approximately 1000°C.

During pyrolysis, the yield of gaseous and liquid products can vary from 25 to 70% by weight, depending on a number of factors, including coal type, pyrolysis atmosphere, heating rate, final pyrolysis temperature, and pressure. Although certain operating conditions may increase product yield, achieving these conditions may result in increased costs. Coal rank is the predominant factor in determining pyrolysis behavior. Higher rank coal, particularly high volatile A bituminous coals, produce the highest yield of tar. However, higher rank coals give products that tend to be more aromatic (i.e., they have a lower hydrogen:carbon ratio).

Other factors that can improve pyrolysis yields are lower pyrolysis temperatures, utilization of smaller particle size coal, and reduced pressure or employing a reducing (i.e., hydrogen) atmosphere. A reducing atmosphere also improves the yield of liquid and lighter products. Heating rate also affects yield distribution, with rapid heating providing a higher liquid:gas ratio. Pyrolysis in the presence of other gases, particularly steam, has been investigated and is reported to improve liquid and gas yields, but little information is available.

Process conditions also affect the product char. At temperatures higher than 1300°C, the inorganic components of coal can be separated from the products as slag. Rapid heating, conducted in a reactive atmosphere, produces a char with higher porosity and reactivity. The use of finer coal particles, lower temperatures resulting in longer process times, and the complexities of using vacuum, pressure, or hydrogen all add to the cost of coal pyrolysis.

The uses of coal pyrolysis liquids can be divided into two broad categories: (i) direct combustion, requiring little or no upgrading, and (ii) transportation fuels and chemical, requiring extensive upgrading. Much attention with regard to low-temperature tar processing has been devoted to hydroprocessing techniques, such as hydrotreating and hydrocracking, with the primary objectives of reducing viscosity, reducing polynuclear aromatics, and removing heteroatoms (sulfur, nitrogen, and oxygen) to produce usable fuels and chemicals. The cost of hydrogen is the primary impediment to tar upgrading. The tar fraction can be used as a source for chemicals, such as phenolics, road tars, preservatives, and carbon binders, but these uses do not constitute a large enough market to support a major coal pyrolysis industry.

Probably the most common application of coal pyrolysis is in the production of coke. The production of metals frequently requires the reduction of oxide-containing ores, the most important being production of iron from various iron oxide ores. Carbon in the form of coke is often used as the reducing agent in a blast furnace in the manufacture of pig iron and steel. The blast furnace is basically a vertical tubular vessel in which alternate layers of iron ore, coke, and limestone are fed from the top. Coal cannot be fed directly at the top of a blast furnace because it does not have the structural strength to support the column of iron ore and limestone in the furnace while maintaining sufficient porosity for the hot air blast to pass upward through the furnace.

Not all coals can produce coke that is suitable for use in a blast furnace. The property that distinguishes coking coals is their caking ability. Various tests are performed to identify suitable coal for conversion to coke through pyrolysis. Frequently, several coals are blended together to achieve the necessary coal properties to produce a suitable coke. Commercial coke-making processes can be divided into two categories: nonrecovery coke making and by-product coke making.

In nonrecovery coke plants, the volatile components released during coke making are not recovered but, rather, are burned to produce heat for the coke oven and for auxiliary power production. One of the earliest nonrecovery units was the beehive oven, which for many years produced most of the coke used by the iron and steel industry. With these ovens, none of the by-products produced during coking were recovered. Because of their low efficiency and pollution problems, beehive ovens are no longer in use in the United States.

Nonrecovery coking takes place in large, rectangular chambers that are heated from the top by radiant heat transfer and from the bottom by conduction through the floor. Primary air for the combustion of evolved volatiles is controlled and introduced through several ports located above the charge level. Combustion gases exit the chamber through down comers in the oven walls and enter the floor flue, thereby heating the floor of the oven. Combustion gases from all the chambers collect in a common tunnel and exit via a stack that creates a natural draft for the oven. To improve efficiency, a waste heat boiler is often added before the stack to recover waste heat and generate steam for power production.

At the completion of the coking process, the doors of the chamber are opened, and a ram pushes the hot coke (approximately 2000°F) into a quench car, where it is typically cooled by spraying it with water. The coke is then screened and transported to the blast furnace.

The majority of coke produced in the United States comes from by-product coke oven batteries. By-product coke making consists of the following operations: (i) Selected coals are blended, pulverized, and oiled for bulk density control; (ii) the blended coal is charged to a number of slot type ovens, each oven sharing a common heating flue with the adjacent oven; (iii) the coal is carbonized in a reducing atmosphere while evolved gases are collected and sent to the by-product plant for by-product recovery; and (iv) the hot coke is discharged, quenched, and shipped to the blast furnace.

After the coke oven is charged with coal, heat is transferred from the heated brick walls to the coal charge. In the temperature range of 375–475°C the coal decomposes to form a plastic layer near the walls. From 475 to 600°C, there is marked evolution of aromatic hydrocarbons and tar, followed by resolidification into semicoke. At 600–1100°C, coke stabilization occurs, characterized by contraction of the coke mass, structural development of coke, and final hydrogen evolution. As time progresses, the plastic phase moves from the walls to the center of the oven. Some gas is trapped in the plastic mass, giving the coke its porous character. When coking is complete, the incandescent coke mass is pushed from the oven and wet or dry quenched prior to being sent to the blast furnace. Modern coke ovens trap the emissions released during coke pushing and quenching so that air pollution is at a minimum.

Gases evolving during coking are commonly termed coke oven gas. In addition to hydrogen and methane, which constitute approximately 80 volume percent of typical coke oven gas, raw coke oven gas also contains nitrogen and oxides of carbon and various contaminants, such as tar vapors, light oil vapors (mainly benzene, toluene, and xylene), naphthalene, ammonia, hydrogen sulfide, and hydrogen cyanide. The by-product plant removes these contaminants so that the gas can be used as fuel. The volatiles emitted during the coking process are recovered as four major by-products: clean coke oven gas, coal tar, ammonium sulfate, and light oil. Several processes are available to clean coke oven gas and separate its constituents.

In the past, many products valuable to industry and agriculture were produced as by-products of coke production, but today most of these materials can be made more economically by other techniques. Therefore, the main emphasis of modern coke by-product plants is to treat the coke oven gas sufficiently so that it can be used as a clean, environmentally friendly fuel. Coke oven gas is generally used in the coke plant or a nearby steel plant as a fuel.

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Construction of experimental liquid-metal facilities

J. Pacio , ... T. Wetzel , in Thermal Hydraulics Aspects of Liquid Metal Cooled Nuclear Reactors, 2019

3.3.1.2.2 Chemical interactions

Alkali metals such as sodium present exothermic reactions with many common substances, such as water, air (oxygen and humidity), CO ₂, and some organic liquids as alcohol (Addison, 1984). The interaction with water involves also the production of hydrogen, resulting in potentially large energy releases. Thus, sodium facilities must be constructed in a way that these reactions do not occur. Practical guidelines for sodium handling have been developed; see, for example, Kottowski (1981). In particular, the construction of a liquid-metal facility must consider the following issues:

•

A sump tank with capacity for the complete liquid inventory acts as a final vessel for emergency draining of the facility, for example, in case of a leakage in order to reduce its consequences. Accordingly, filling and draining procedures must be defined.

•

Avoiding direct contact with air during operation, a cover gas system (e.g., argon) is used.

•

Although not critical for safety, dissolved impurities such as oxygen can play a key role for corrosion; see next section. Depending on the application, it might be necessary to remove or control the amount of dissolved impurities during operation. Taking advantage of the temperature-dependent solubility, eventually, a cold trap can be used.

Although LBE does not present safety concerns related to violent exothermic reactions, the toxicity of lead must be taken into account, and the oxygen content plays a major role. Thus, the above-listed considerations usually apply also to heavy liquid-metal facilities.

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Transition Metal Groups 9–12

L.J. Yellowlees , K.G. Macnamara , in Comprehensive Coordination Chemistry II, 2003

6.2.7.1.3 Alkyls and aryls

Benzene addition to Ir(PⁱPr₃)₂ Cl is an exothermic reaction (22  kcal mol⁻¹), while addition to Ir(PⁱPr₃)₂(CO)Cl is endothermic (−5   kcal mol⁻¹). ⁵⁰¹ The reaction enthalpies of substitution reactions to complexes containing the Ir(PⁱPr₃)₂Cl fragment are supplied. Reaction of Ir(PⁱPr₃)₂Cl with 2-pyridyl esters gives a μ² (C,O)-bound ketene, ( 307 ), where R²  =   R¹  =   aryl or R²  =   aryl, R′   =   alkyl. ⁵⁰² ( 307 ) (R¹  =   R²  =   Ph) reacts with alkynes to form a ketene–alkyne complex, ( 308 ), which ultimately yields the five-coordinate Ir^III species ( 309 ). ⁵⁰³

Reaction of hexafluorobenzene with MeIr(PEt₃)₃ results in CF and PC bond cleavage and PF bond formation, according to reaction Scheme 28. ⁵⁰⁴ An X-ray crystallographic study of (310) reveals a distorted square-planar arrangement of Ir.

Scheme 28.

A theoretical evaluation of CX (X   =   F, Cl, Br, I) bond activation in CX₄ by an Ir^I complex is detailed by Su and Chu. ⁵⁰⁵ Reaction of C₂(CO₂Me)₂ with trans-MeIr(CO)(PPh₃)₂ affords a kinetic isomer ( 311 ), identified by NMR and IR spectroscopy, which converts to the thermodynamic isomer ( 312 ), characterized by ³¹P and ¹H NMR, IR spectroscopy, and X-ray crystallography. ⁵⁰⁶

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STEADY MICROSCOPIC BALANCES WITH GENERATION

Ismail Tosun , in Modeling in Transport Phenomena (Second Edition), 2007

9.2.1 Conduction in Rectangular Coordinates

Consider one-dimensional transfer of energy in the z-direction through a plane wall of thickness L and surface area A as shown in Figure 9.5. Let ℜ be the position-dependent rate of energy generation per unit volume within the wall.

Figure 9.5. Conduction through a plane wall with generation.

Since T = T (z), Table C.4 in Appendix C indicates that the only nonzero energy flux component is e _z , and it is given by

(9.2-2) $e_z=q_z=-k\frac{dT}{dz}$

For a rectangular volume element of thickness Δz as shown in Figure 9.5, Eq. (9.2-1) is expressed as

(9.2-3) $q_z{|}_{z}A-q_z{|}_{z+\mathrm{\Delta }z}A+\Re A\mathrm{\Delta }z=0$

Dividing each term by A Δz and taking the limit as Δz → 0 give

(9.2-4) $\underset{\mathrm{\Delta }z\to 0}{\text{lim}}\frac{q_z{|}_z-q_z{|}_{z+\mathrm{\Delta }z}}{\mathrm{\Delta }z}+\Re =0$

or,

(9.2-5) $\frac{d{q}_z}{dz}=\Re$

Substitution of Eq. (9.2-2) into Eq. (9.2-5) gives the governing equation for temperature as

(9.2-6) $-\frac{d}{dz}\left(k\frac{dT}{dz}\right)=\Re$

Integration of Eq. (9.2-6) gives

(9.2-7) $k\frac{dT}{dz}=-{\displaystyle {\int }_0^z\Re \left(u\right)du+C_1}$

where u is a dummy variable of integration and C ₁ is an integration constant. Integration of Eq. (9.2-7) once more leads to

(9.2-8) ${\displaystyle {\int }_0^{T}k\left(T\right)dT=-{\displaystyle {\int }_0^z\left[{\displaystyle {\int }_0^z\Re \left(u\right)du}\right]dz+C_{1}z+C_2}}$

Evaluation of the constants C ₁ and C ₂ requires the boundary conditions to be specified. The solution of Eq. (9.2-8) will be presented for two types of boundary conditions, namely, Type I and Type II. In the case of the Type I boundary condition, the temperatures at both surfaces are specified. On the other hand, the Type II boundary condition implies that while the temperature is specified at one of the surfaces the other surface is subjected to a constant wall heat flux.

Type I boundary condition

The solution of Eq. (9.2-8) subject to the boundary conditions

(9.2-9a) $\begin{array}{ccc}\text{at}& z=0& T=T_o\end{array}$

(9.2-9b) $\begin{array}{ccc}\text{at}& z=L& T=T_L\end{array}$

is given by

(9.2-10) ${\displaystyle {\int }_{T_o}^{T}k\left(T\right)dT=-{\displaystyle {\int }_0^z\left[{\displaystyle {\int }_0^{z}R\left(u\right)du}\right]dz+\left\{{\displaystyle {\int }_{T_o}^{T_L}k\left(T\right)dT+{\displaystyle {\int }_0^L\left[{\displaystyle {\int }_0^{z}R\left(u\right)du}\right]dz}}\right\}\frac{z}{L}}}$

Note that, when ℜ = 0, Eq. (9.2-10) reduces to Eq. (G) in Table 8.1. Equation (9.2-10) may be further simplified depending on whether the thermal conductivity and/or energy generation per unit volume are constant.

■ Case (i) k = constant

In this case, Eq. (9.2-10) reduces to

(9.2-11) $k\left(T-T_o\right)=-{\displaystyle {\int }_0^z\left[{\displaystyle {\int }_0^z\Re \left(u\right)du}\right]dz+\left\{k\left(T_L-T_o\right)+{\displaystyle {\int }_0^L\left[{\displaystyle {\int }_0^z\Re \left(u\right)du}\right]dz}\right\}\frac{z}{L}}$

When ℜ = 0, Eq. (9.2-11) reduces to Eq. (H) in Table 8.1.

■ Case (ii) k = constant; ℜ = constant

In this case, Eq. (9.2-10) simplifies to

(9.2-12) $T=T_o+\frac{\Re L^2}{2k}\left[\frac{z}{L}-{\left(\frac{z}{L}\right)}^2\right]-\left(T_o-T_L\right)\frac{z}{L}$

The location of the maximum temperature can be obtained from dT/dz = 0 as

(9.2-13) ${\left(\frac{z}{L}\right)}_{T=T_{\text{max}}}=\frac{1}{2}-\frac{k}{\Re L^2}\left(T_o-T_L\right)$

Substitution of Eq. (9.2-13) into Eq. (9.2-12) gives the value of the maximum temperature as

(9.2-14) $T_{\text{max}}=\frac{T_o+T_L}{2}+\frac{\Re L^2}{8k}+\frac{k{\left(T_o-T_L\right)}^2}{2\Re L^2}$

The representative temperature profiles depending on the values of T _o and T _L are shown in Figure 9.6.

Figure 9.6. Representative temperature distributions in a rectangular wall with constant generation.

Type II boundary condition

The solution of Eq. (9.2-8) subject to the boundary conditions

(9.2-15a) $\begin{array}{ccc}\text{at}& z=0& -k\frac{dT}{dz}=q_o\end{array}$

(9.2-15b) $\begin{array}{ccc}\text{at}& z=L& T=T_L\end{array}$

is given by

(9.2-16) ${\displaystyle {\int }_{T_L}^{T}k\left(T\right)dT={\displaystyle {\int }_z^L\left[{\displaystyle {\int }_0^z\Re \left(u\right)}du\right]}}dz+q_{o}L\left(1-\frac{z}{L}\right)$

When ℜ = 0, Eq. (9.2-16) reduces to Eq. (G) in Table 8.2. Further simplifications of Eq. (9.2-16) depending on whether k and/or ℜ are constant are given below.

■ Case (i) k = constant

In this case, Eq. (9.2-16) reduces to

(9.2-17) $k\left(T-T_L\right)={\displaystyle {\int }_z^L\left[{\displaystyle {\int }_0^z\Re \left(u\right)}du\right]dz+q_{o}L\left(1-\frac{z}{L}\right)}$

When ℜ = 0, Eq. (9.2-17) reduces to Eq. (H) in Table 8.2.

■ Case (ii) k = constant; ℜ = constant

In this case, Eq. (9.2-16) reduces to

(9.2-18) $T=T_L+\frac{\Re L^2}{2k}\left[1-{\left(\frac{z}{L}\right)}^2\right]+\frac{q_{o}L}{k}\left(1-\frac{z}{L}\right)$

9.2.1.1 Macroscopic equation

The integration of the governing equation, Eq. (9.2-6), over the volume of the system gives

(9.2-19) $-{\displaystyle {\int }_0^L{\displaystyle {\int }_0^{\mathcal{W}}{\displaystyle {\int }_0^H\frac{d}{dz}\left(k\frac{dT}{dz}\right)dxdydz=}}}{\displaystyle {\int }_0^L{\displaystyle {\int }_0^{\mathcal{W}}{\displaystyle {\int }_0^H\Re dxdydz}}}$

Integration of Eq. (9.2-19) yields

(9.2-20) $\underset{\text{Net\hspace{0.17em}rate\hspace{0.17em}of\hspace{0.17em}energy\hspace{0.17em}out}}{\underbrace{\mathcal{W}H\left[{\left(-k\frac{dT}{dz}\right)}_{z=L}+{\left(k\frac{dT}{dz}\right)}_{z=0}\right]}}=\underset{\begin{array}{l}\text{Rate\hspace{0.17em}of\hspace{0.17em}energy}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}generation}\end{array}}{\underbrace{\mathcal{W}H{\displaystyle {\int }_0^L\Re }dz}}$

which is simply the macroscopic energy balance under steady conditions by considering the plane wall as a system. Note that energy must leave the system from at least one of the surfaces to maintain steady conditions. The "net rate of energy out" in Eq. (9.2-20) implies that the rate of energy leaving the system is in excess of the rate of energy entering it.

It is also possible to make use of Newton's law of cooling to express the rate of heat loss from the system. If heat is lost from both surfaces to the surroundings, Eq. (9.2-20) can be written as

(9.2-21) $〈h_A〉\left(T_o-T_A\right)+〈h_B〉\left(T_L-T_B\right)={\displaystyle {\int }_0^L\Re dz}$

where T _o and T _L are the surface temperatures at z = 0 and z = L, respectively.

Example 9.1

Energy generation rate as a result of an exothermic reaction is 1 × 10 ⁴ W/m³ in a 50 cm thick wall of thermal conductivity 20 W/ $\text{m}\cdot \text{K}$ . The left face of the wall is insulated while the right side is held at 45 °C by a coolant. Calculate the maximum temperature in the wall under steady conditions.

Solution

Let z be the distance measured from the left face. The use of Eq. (9.2-18) with q _o = 0 gives the temperature distribution as

(1) $T=T_L+\frac{\Re L^2}{2k}\left[1-{\left(\frac{z}{L}\right)}^2\right]=45+\frac{\left(1\times {10}^4\right){\left(0.5\right)}^2}{2\left(20\right)}\left[1-{\left(\frac{z}{0.5}\right)}^2\right]$

Simplification of Eq. (1) leads to

(2) $T=107.5-250z^2$

Since dT/dz = 0 at z = 0, the maximum temperature occurs at the insulated surface and its value is 107.5 °C.

Example 9.2

Consider a composite solid of materials A and B, shown in the figure below. An electrical resistance heater embedded in solid B generates heat at a constant volumetric rate of ℜ (W/m³). The composite solid is cooled from both sides to avoid excessive heating.

a)

Obtain expressions for the steady temperature distributions in solids A and B.

b)

Calculate the rate of heat loss from the surfaces located at z = −L _A and z = L _B .

c)

For the following numerical values

$\begin{array}{cccc}{T}_1=-5^{\circ }\text{C}& T_2={25}^{\circ }\text{C}& 〈h_1〉=500\text{W}/{\text{m}}^2\cdot \text{K}& 〈h_2〉=10\text{W}/{\text{m}}^2\cdot \text{K}\end{array}$

$\begin{array}{cccc}{k}_A=180\text{W}/\text{m}\cdot \text{K}& k_B=1.2\text{W}/\text{m}\cdot \text{K}& L_A=36\text{cm}& L_B=3\text{cm}\end{array}$

calculate the value of ℜ to keep the surface temperature of the wall at z = −L _A constant at 15 °C.

d)

Obtain the temperature distribution in solid A when the thickness of solid B is very small, and draw the electrical analog. A practical application of this case is the use of a surface heater, i.e., a very thin plastic film containing electrical resistance, to clear condensation and ice from the rear window of your car or condensation from the mirror in your bathroom.

Solution

a)

Since area is constant, the governing equation for temperature in solid A can be easily obtained from Eq. (8.2-5) as

(1) $\begin{array}{ccc}\frac{d{q}_z^A}{dz}=0& \Rightarrow & \frac{d^2{T}_A}{d{z}^2}\end{array}=0$

The solution of Eq. (1) gives

(2) $T_A=C_{1}z+C_2$

The governing equation for temperature in solid B is obtained from Eqs. (9.2-5) and (9.2-6) as

(3) $\begin{array}{ccc}-\frac{d{q}_z^B}{dz}+\Re =0& \Rightarrow & \frac{d^2{T}_B}{d{z}^2}\end{array}=-\frac{\Re }{k_B}$

The solution of Eq. (3) yields

(4) $T_B=-\frac{\Re }{2k_B}{z}^2+C_{3}z+C_4$

Evaluation of the constants C ₁, C ₂, C ₃, and C ₄ requires four boundary conditions. They are expressed as

(5) $\begin{array}{ccc}\text{at}& z=-L_A& k_A\frac{d{T}_A}{dz}\end{array}=〈h_1〉\left(T_A-T_1\right)$

(6) $\begin{array}{ccc}\text{at}& z=L_B& -k_B\frac{d{T}_B}{dz}\end{array}=〈h_2〉\left(T_B-T_2\right)$

(7) $\begin{array}{ccc}\text{at}& z=0& T_A\end{array}=T_B$

(8) $\begin{array}{ccc}\text{at}& z=0& k_A\frac{d{T}_A}{dz}\end{array}=k_B\frac{d{T}_B}{dz}$

Application of the boundary conditions leads to the following temperature distributions within solids A and B

(9) $T_A=T_1+\left[\frac{\Re L_B\left(\frac{1}{〈h_2〉}+\frac{L_B}{2k_B}\right)+T_2-T_1}{k_A\left(\frac{1}{〈h_1〉}+\frac{L_A}{k_A}+\frac{L_B}{k_B}+\frac{1}{〈h_2〉}\right)}\right]\left(z+\frac{k_A}{〈h_1〉}+L_A\right)$

(10) $T_B=T_1-\frac{\Re }{2k_B}{z}^2+\left[\frac{\Re L_B\left(\frac{1}{〈h_2〉}+\frac{L_B}{2k_B}\right)+T_2-T_1}{k_A\left(\frac{1}{〈h_1〉}+\frac{L_A}{k_A}+\frac{L_B}{k_B}+\frac{1}{〈h_2〉}\right)}\right]\left(\frac{k_A}{k_B}z+\frac{k_A}{〈h_1〉}+L_A\right)$

b)

The rate of heat transfer per unit area through the surface at z = −L _A is given by

(11) $\frac{{\dot{Q}|}_{z=-L_A}}{A}=k_A{\frac{d{T}_A}{dz}|}_{z=-L_A}=\frac{\Re L_B\left(\frac{1}{〈h_2〉}+\frac{L_B}{2k_B}\right)+T_2-T_1}{\frac{1}{〈h_1〉}+\frac{L_A}{k_A}+\frac{L_B}{k_B}+\frac{1}{〈h_2〉}}$

On the other hand, the rate of heat transfer per unit area through the surface at z = L _B is given by

(12) $\frac{{\dot{Q}|}_{z=L_B}}{A}=-k_B{\frac{d{T}_B}{dz}|}_{z=L_B}=\Re L_B-\frac{\Re L_B\left(\frac{1}{〈h_2〉}+\frac{L_B}{2k_B}\right)+T_2-T_1}{\frac{1}{〈h_1〉}+\frac{L_A}{k_A}+\frac{L_B}{k_B}+\frac{1}{〈h_2〉}}$

Note that the addition of Eqs. (11) and (12) results in

(13) $\underset{\text{Rate\hspace{0.17em}of\hspace{0.17em}energy\hspace{0.17em}out}}{\underbrace{{\dot{Q}|}_{z=-L_A}+{\dot{Q}|}_{z=L_B}}}=\underset{\begin{array}{l}\text{Rate\hspace{0.17em}of\hspace{0.17em}energy}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}generation}\end{array}}{\underbrace{\Re A{L}_B}}$

which is nothing more than the steady-state macroscopic energy balance by considering a composite solid as a system.

c)

Evaluation of Eq. (9) at z = −L _A leads to

(14) ${T_A|}_{z=-L_A}=T_1+\left[\frac{\Re L_B\left(\frac{1}{〈h_2〉}+\frac{L_B}{2k_B}\right)+T_2-T_1}{〈h_1〉\left(\frac{1}{〈h_1〉}+\frac{L_A}{k_A}+\frac{L_B}{k_B}+\frac{1}{〈h_2〉}\right)}\right]$

Solving Eq. (14) for ℜ leads to

$\begin{array}{lll}\Re \hfill & =\hfill & \frac{\left({T_A|}_{z=-L_A}-T_1\right)〈h_1〉\left(\frac{1}{〈h_1〉}+\frac{L_A}{k_A}+\frac{L_B}{k_B}+\frac{1}{〈h_2〉}\right)+T_1-T_2}{L_B\left(\frac{1}{〈h_2〉}+\frac{L_B}{2k_B}\right)}\hfill \\ \hfill & =\hfill & \frac{\left(15+5\right)\left(500\right)\left(\frac{1}{500}+\frac{0.36}{180}+\frac{0.03}{1.2}+\frac{1}{10}\right)-5-25}{0.03\left[\frac{1}{10}+\frac{0.03}{2\left(1.2\right)}\right]}=3.73\times {10}^5\text{W}/{\text{m}}^3\hfill \end{array}$

d)

When the thickness of solid B is very small, then it is possible to assume that the temperature in solid B is constant and equal to the temperature in solid A at z = 0. Moreover, the heat generation is expressed in terms of the heat generation rate per unit area, i.e., $\overline{\Re }$ = ℜL _B . Thus, Eq. (9) becomes

(15) $T_A=T_1+\left[\frac{\overline{\Re }+〈h_2〉\left(T_2-T_1\right)}{k_A\frac{〈h_2〉}{〈h_1〉}+L_A〈h_2〉+k_A}\right]\left(z+\frac{k_A}{〈h_1〉}+L_A\right)$

The electrical circuit analog of this case is shown in the figure below:

Comment: When Eq. (3) is integrated in the z-direction, the result is

(16) ${\displaystyle {\int }_0^{L_B}{k}_B\frac{d^2{T}_B}{d{z}^2}}dz+{\displaystyle {\int }_0^{L_B}\Re dz=0}$

or,

(17) $\begin{array}{cc}\underset{-〈h_2〉\left({T_A|}_{z=0}-T_2\right)}{\underbrace{k_B{\frac{d{T}_B}{dz}|}_{z=L_B}}}& \underset{k_A{\frac{d{T}_A}{dz}|}_{z=0}}{\underbrace{-k_B{\frac{d{T}_B}{dz}|}_{z=0}}}\end{array}+\underset{\overline{\Re }}{\underbrace{\Re L_B=0}}$

Thus, the solution of Eq. (1) with the following boundary conditions

(18) $\begin{array}{ccc}\text{at}& z=-L_A& k_A\frac{d{T}_A}{dz}=\end{array}〈h_1〉\left(T_A-T_1\right)$

(19) $\begin{array}{ccc}\text{at}& z=0& -k_A\frac{d{T}_A}{dz}+\overline{\Re }=\end{array}〈h_2〉\left(T_A-T_2\right)$

also results in Eq. (15).

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Hydrothermal Activity☆

R.P. Lowell , ... P.A. Rona , in Reference Module in Earth Systems and Environmental Sciences, 2014

Chemical heat-serpentinization of peridotite

It has long been recognized that hydration of peridotite is an exothermic reaction that produces heat ( Fyfe and Lonsdale, 1981; Macdonald and Fyfe, 1985), that alters the chemistry of the rocks as well as the resulting hydrothermal fluid (Janecky and Seyfried, 1986; Wetzel and Shock, 2000). Serpentinization also leads to an increase in rock volume by up to ~   40% (Coleman, 1971; O'Hanley, 1992). Serpentinization is commonly observed in ultramafic rocks recovered from the seafloor and exposed in slices of ancient oceanic mantle exposed on land as ophiolites. This reaction yields distinctive chemical solutions characterized by high alkalinity and abiogenic generation of methane (CH₄) and hydrogen (H₂) gas (e.g., Früh-Green et al., 2004; Seyfried et al., 2004; Kelley et al., 2005; Charlou et al., 2010; Cannat et al., 2010). The heat released, depending on the volume and rate of serpentinization, may be sufficient to drive hydrothermal circulation over a range of fluid temperatures, typically low to intermediate (degrees to tens of degrees Celsius), and possibly up to several hundred degrees Celsius (Lowell and Rona, 2002).

Serpentinization is favored by conditions that facilitate access of water to large volumes of peridotite rocks. In ocean basins the conditions include a low magma budget, which produces thin ocean crust, and tectonic extension and volume expansion that creates permeability through fractures and faults and that exposes rocks of the upper mantle on the seafloor. Such conditions generally occur at sections of slow-spreading ocean ridges in the Atlantic, Indian and Arctic oceans. For example, fluids with the chemical signatures of serpentinization reactions are common along the Mid-Atlantic Ridge where several high-temperature (to 360   °C) seafloor hydrothermal fields (Logatchev at 14°45'N, 44°58'W and Rainbow, 36°14'N, 33°54'W) at least partially situated in serpentinized ultramafic rocks of the upper mantle have been found. The Lost City field, near 30°N, 42°W is located about 15   km west of the eastern intersection of the rift valley with the Atlantis Fracture Zone, where the crust is apparently isolated from magmatic heat sources (Kelley et al., 2001). There serpentinzation derived fluids are discharging at temperatures up to 75   °C and precipitating calcium carbonate and magnesium hydroxide chimneys, which have grown up to 60   m high. Thermal and chemical fluxes from such seafloor hydrothermal systems have yet to be determined, but are expected to be a significant fraction of global hydrothermal mass and heat budgets.

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TECTONICS | Hydrothermal Activity

R.P. Lowell , P.A. Rona , in Encyclopedia of Geology, 2005

Chemical heat

It has long been recognized that hydration of peridotite is an exothermic reaction that produces heat, that alters the chemistry of the rocks and hydrating solutions involved, and that expands the volume of the rocks (∼40%). It is only now emerging how widespread this process called the "serpentinization reaction" may be beneath ocean basins and possibly continents. The reaction involves peridotite, the characteristic ultramafic rock type of the Earth's upper mantle, and either seawater or meteoric water. Serpentinization is commonly observed in ultramafic rocks recovered from the seafloor and in slices of ancient oceanic mantle exposed on land as ophiolites. This reaction yields distinctive chemical solutions characterized by high alkalinity, high ratios of Ca to Mn and other metals, and abiogenic generation of methane (CH ₄) and hydrogen (H₂) gas. The heat released, depending on the volume and rate of serpentinization, may be sufficient to drive hydrothermal circulation over a range of fluid temperatures, typically low to intermediate (degrees to tens of degrees Celsius), and possibly up to several hundred degrees Celsius.

Serpentinization is favoured by conditions that facilitate access of water to large volumes of the upper mantle. In ocean basins the conditions include a low magma budget, which produces thin ocean crust, and tectonic extension and volume expansion that creates permeability through fractures and faults and that exposes rocks of the upper mantle on the seafloor. Such conditions generally occur at sections of slow-spreading ocean ridges in the Atlantic, Indian, and Arctic oceans. For example, fluids with the chemical signatures of serpentinization reactions are common along the mid-Atlantic ridge where several high-temperature (to 360°C) seafloor hydrothermal fields (Logatchev at 14°45′   N, 44°58′   W and Rainbow, 36°14′   N, 33°54′   W) at least partially situated in serpentinized ultramafic rocks of the upper mantle have been found. Only one of these fields appears to be an end member of a hydrothermal system entirely driven by serpentinization reactions (Lost City field, near 30°   N, 42°   W) located about 15   km west of the eastern intersection of the rift valley with the Atlantis Fracture Zone, where the field is apparently isolated from magmatic heat sources. There serpentinization-derived fluids are discharging at temperatures up to 75°C and precipitating calcium carbonate and magnesium hydroxide chimneys, which have grown up to 60   m high. Thermal and chemical fluxes from such serpentinization-driven seafloor hydrothermal systems have yet to be determined, but may be a significant fraction of global hydrothermal mass and heat budgets. Seawater and upper mantle rocks are ubiquitous in ocean basins, although the sites of serpentinization may be localized.

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