Nick Connor

Units of Entropy

The SI unit for entropy is J/K. According to Clausius, the entropy was defined via the change in entropy S of a system. The change in entropy S, when an amount of heat Q is added to it by a reversible process at constant temperature, is given by:

Here Q is the energy transferred as heat to or from the system during the process, and T is the temperature of the system in kelvins during the process. If we assume a reversible isothermal process, the total entropy change is given by:

∆S = S₂ – S₁ = Q/T

In this equation the quotient Q/T is related to the increase in disorder. Higher temperature means greater randomness of motion. At lower temperatures adding heat Q causes a substantial fractional increase in molecular motion and randomness. On the other hand if the substance is already hot, the same quantity of heat Q adds relatively little to the greater molecular motion.

Example: Entropy change in melting ice

Calculate the change in entropy of 1 kg of ice at 0°C, when melted reversibly to water at 0°C.

Since it is an isothermal process, we can use:

∆S = S₂ – S₁ = Q/T

therefore  the entropy change will be:

∆S = 334 [kJ] / 273.15 [K] = 1.22 [kJ/K]

where 334  kilojoules of heat are required to melt 1 kg of ice (latent heat of fusion = 334 kJ/kg) and this heat is transferred to the system at 0°C (273.15 K).

Specific Entropy

The entropy can be made into an intensive, or specific, variable by dividing by the mass. Engineers use the specific entropy in thermodynamic analysis more than the entropy itself. The specific entropy (s) of a substance is its entropy per unit mass. It equals to the total entropy (S) divided by the total mass (m).

s = S/m

where:

s = specific entropy (J/kg)

S = entropy (J)

m = mass (kg)

Entropy quantifies the energy of a substance that is no longer available to perform useful work. Because entropy tells so much about the usefulness of an amount of heat transferred in performing work, the thermodynamics/steam-tables/”>steam tables include values of specific entropy (s = S/m) as part of the information tabulated.

In general, specific entropy is a property of a substance, like pressure, temperature, and volume, but it cannot be measured directly. Normally, the entropy of a substance is given with respect to some reference value. For example, the specific entropy of water or steam is given using the reference that the specific entropy of water is zero at 0.01°C and normal atmospheric pressure, where s = 0.00 kJ/kg. The fact that the absolute value of specific entropy is unknown is not a problem, however, because it is the change in specific entropy (∆s) and not the absolute value that is important in practical problems.

See also:

Entropy

Boltzmann’s entropy formula

In statistical physics, entropy is a measure of the disorder of a system. What disorder refers to is really the number of microscopic configurations, W, that a thermodynamic system can have when in a state as specified by certain macroscopic variables (thermodynamics/thermodynamic-properties/what-is-volume-physics/”>volume, thermodynamics/what-is-energy-physics/”>energy, thermodynamics/thermodynamic-properties/what-is-pressure-physics/”>pressure, and thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature). By “microscopic states”, we mean the exact states of all the molecules making up the system.

Mathematically, the exact definition is:

Entropy = (Boltzmann’s constant k) x logarithm of number of possible states

S = k_B logW

This equation, known as the Boltzmann’s entropy formula, relates the microscopic details, or microstates, of the system (via W) to its macroscopic state (via the entropy S). It is the key idea of statistical mechanics. In a closed system, entropy never decreases, so in the Universe entropy is irreversibly increasing. In an open system (for example, a growing tree), entropy can decrease and order can increase, but only at the expense of an increase in entropy somewhere else (e.g. in the Sun).

Order is decreasing.

Entropy is increasing.

Irreversibility of Natural Processes

According to the second law of thermodynamics:

The entropy of any isolated system never decreases. In a natural thermodynamic process, the sum of the entropies of the interacting thermodynamic systems increases.  

This law indicates the irreversibility of natural processes. Reversible processes are a useful and convenient theoretical fiction, but do not occur in nature. From this law follows that it is impossible to construct a device that operates on a cycle and whose sole effect is the transfer of heat from a cooler body to a hotter body. It follows, perpetual motion machines of the second kind are impossible.

Entropy at Absolute Zero

According to third law of thermodynamics:

The entropy of a system approaches a constant value as the temperature approaches absolute zero.

Based on empirical evidence, this law states that the entropy of a pure crystalline substance is zero at the absolute zero of temperature, 0 K and that it is impossible by means of any process, no matter how idealized, to reduce the temperature of a system to absolute zero in a finite number of steps. This allows us to define a zero point for the thermal energy of a body.

Absolute zero is the coldest theoretical temperature, at which the thermal motion of atoms and molecules reaches its minimum.  This is a state at which the enthalpy and entropy of a cooled ideal gas reaches its minimum value, taken as 0. Classically, this would be a state of motionlessness, but quantum uncertainty dictates that the particles still possess a finite zero-point energy. Absolute zero is denoted as 0 K on the Kelvin scale, −273.15 °C on the Celsius scale, and −459.67 °F on the Fahrenheit scale.

See also:

Entropy

What is Enthalpy

In thermodynamics, the enthalpy is the measure of energy in a thermodynamic system. It is the thermodynamic quantity equivalent to the total heat content of a system. The enthalpy is defined to be the sum of the internal energy E plus the product of the pressure p and volume V. In many thermodynamic analyses the sum of the internal energy U and the product of pressure p and volume V appears, therefore it is convenient to give the combination a name, enthalpy, and a distinct symbol, H.

The enthalpy is the preferred expression of system energy changes in many chemical, biological, and physical measurements at constant pressure. It is so useful that it is tabulated in the thermodynamics/steam-tables/”>steam tables along with thermodynamics/thermodynamic-properties/what-is-specific-volume/”>specific volume and thermodynamics/what-is-energy-physics/internal-energy-thermal-energy/specific-internal-energy/”>specific internal energy. It is due to the fact, it simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating (Q = H₂ – H₁) or work other than expansion work. For a variable-pressure process, the difference in enthalpy is not quite as obvious.

Enthalpy in Extensive Units

H = U + pV

Enthalpy is an extensive quantity, it depends on the size of the system, or on the amount of substance it contains. The SI unit of enthalpy is the joule (J). It is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields. It is the thermodynamic quantity equivalent to the total heat content of a system.

On the other hand, energy can be stored in the chemical bonds between the atoms that make up the molecules. This energy storage on the atomic level includes energy associated with electron orbital states, nuclear spin, and binding forces in the nucleus.

Enthalpy is represented by the symbol H, and the change in enthalpy in a process is H₂ – H₁.

There are expressions in terms of more familiar variables such as thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature andthermodynamics/thermodynamic-properties/what-is-pressure-physics/”> pressure:

dH = C_pdT + V(1-αT)dp

Where C_p is the heat capacity at constant pressure and α is the coefficient of (cubic) thermal expansion. For ideal gas αT = 1 and therefore:

dH = C_pdT

Example: Frictionless Piston – Heat – Enthalpy

A frictionless piston is used to provide a constant pressure of 500 kPa in a cylinder containing steam (superheated steam) of a volume of 2 m³  at 500 K. Calculate the final temperature, if 3000 kJ of heat is added.

Solution:

Using thermodynamics/steam-tables/”>steam tables we know, that the specific enthalpy of such steam (500 kPa; 500 K) is about 2912 kJ/kg. Since at this condition the steam has density of 2.2 kg/m³, then we know there is about 4.4 kg of steam in the piston at enthalpy of 2912 kJ/kg x 4.4 kg = 12812 kJ.

When we use simply Q = H₂ − H₁, then the resulting enthalpy of steam will be:

H₂ = H₁ + Q = 15812 kJ

From steam tables, such superheated steam (15812/4.4 = 3593 kJ/kg) will have a temperature of 828 K (555°C). Since at this enthalpy the steam have density of 1.31 kg/m³, it is obvious that it has expanded by about 2.2/1.31 = 1.67 (+67%). Therefore the resulting volume is 2 m³ x 1.67 = 3.34 m³ and ∆V = 3.34 m³ – 2 m³ = 1.34 m³.

The p∆V part of enthalpy, i.e. the work done is:

W = p∆V = 500 000 Pa x 1.34 m³ = 670 kJ

What is Entropy

In thermodynamics and statistical physics, entropy is a quantitative measure of disorder, or of the energy in a system to do work.

In statistical physics, entropy is a measure of the disorder of a system. What disorder refers to is really the number of microscopic configurations, W, that a thermodynamic system can have when in a state as specified by certain macroscopic variables (thermodynamics/thermodynamic-properties/what-is-volume-physics/”>volume, thermodynamics/what-is-energy-physics/”>energy, thermodynamics/thermodynamic-properties/what-is-pressure-physics/”>pressure, and thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature). By “microscopic states”, we mean the exact states of all the molecules making up the system.

Mathematically, the exact definition is:

Entropy = (Boltzmann’s constant k) x logarithm of number of possible states

S = k_B logW

This equation, which relates the microscopic details, or microstates, of the system (via W) to its macroscopic state (via the entropy S), is the key idea of statistical mechanics. In a closed system, entropy never decreases, so in the Universe entropy is irreversibly increasing. In an open system (for example, a growing tree), entropy can decrease and order can increase, but only at the expense of an increase in entropy somewhere else (e.g. in the Sun).

Units of Entropy

The SI unit for entropy is J/K. According to Clausius, the entropy was defined via the change in entropy S of a system. The change in entropy S, when an amount of heat Q is added to it by a reversible process at constant temperature, is given by:

Here Q is the energy transferred as heat to or from the system during the process, and T is the temperature of the system in kelvins during the process. If we assume a reversible isothermal process, the total entropy change is given by:

∆S = S₂ – S₁ = Q/T

In this equation the quotient Q/T is related to the increase in disorder. Higher temperature means greater randomness of motion. At lower temperatures adding heat Q causes a substantial fractional increase in molecular motion and randomness. On the other hand if the substance is already hot, the same quantity of heat Q adds relatively little to the greater molecular motion.

Example

Calculate the change in entropy of 1 kg of ice at 0°C, when melted reversibly to water at 0°C.

Since it is an isothermal process, we can use:

∆S = S₂ – S₁ = Q/T

therefore  the entropy change will be:

∆S = 334 [kJ] / 273.15 [K] = 1.22 [kJ/K]

where 334  kilojoules of heat are required to melt 1 kg of ice (latent heat of fusion = 334 kJ/kg) and this heat is transferred to the system at 0°C (273.15 K).

T-s Diagrams

In general, the phases of a substance and the relationships between its properties are most commonly shown on property diagrams. A large number of different properties have been defined, and there are some dependencies between properties.

A Temperature-entropy diagram (T-s diagram) is the type of diagram most frequently used to analyze energy transfer system cycles. It  is used in thermodynamics to visualize changes to temperature and specific entropy during a thermodynamic process or cycle.

This is because the work done by or on the system and the heat added to or removed from the system can be visualized on the T-s diagram. By the definition of entropy, the heat transferred to or from a system equals the area under the T-s curve of the process.

dQ = TdS

An isentropic process is depicted as a vertical line on a T-s diagram, whereas an isothermal process is a horizontal line. In an idealized state, compression is a pump, compression in a compressor and expansion in a turbine are isentropic processes. Therefore it is very useful in power engineering, because these devices are used in thermodynamic cycles of power plants.

Note that, the isentropic assumptions are only applicable with ideal cycles. Real thermodynamic cycles have inherent energy losses due to inefficiency of compressors and turbines.

See also:

Energy

What is Entropy

In statistical physics, entropy is a measure of the disorder of a system. What disorder refers to is really the number of microscopic configurations, W, that a thermodynamic system can have when in a state as specified by certain macroscopic variables (thermodynamics/thermodynamic-properties/what-is-volume-physics/”>volume, thermodynamics/what-is-energy-physics/”>energy, thermodynamics/thermodynamic-properties/what-is-pressure-physics/”>pressure, and thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature). By “microscopic states”, we mean the exact states of all the molecules making up the system.

Mathematically, the exact definition is:

Entropy = (Boltzmann’s constant k) x logarithm of number of possible states

S = k_B logW

This equation, which relates the microscopic details, or microstates, of the system (via W) to its macroscopic state (via the entropy S), is the key idea of statistical mechanics. In a closed system, entropy never decreases, so in the Universe entropy is irreversibly increasing. In an open system (for example, a growing tree), entropy can decrease and order can increase, but only at the expense of an increase in entropy somewhere else (e.g. in the Sun).

Order is decreasing.

Entropy is increasing.

Units of Entropy

The SI unit for entropy is J/K. According to Clausius, the entropy was defined via the change in entropy S of a system. The change in entropy S, when an amount of heat Q is added to it by a reversible process at constant temperature, is given by:

Here Q is the energy transferred as heat to or from the system during the process, and T is the temperature of the system in kelvins during the process. If we assume a reversible isothermal process, the total entropy change is given by:

∆S = S₂ – S₁ = Q/T

In this equation the quotient Q/T is related to the increase in disorder. Higher temperature means greater randomness of motion. At lower temperatures adding heat Q causes a substantial fractional increase in molecular motion and randomness. On the other hand if the substance is already hot, the same quantity of heat Q adds relatively little to the greater molecular motion.

Example: Entropy change in melting ice

Calculate the change in entropy of 1 kg of ice at 0°C, when melted reversibly to water at 0°C.

Since it is an isothermal process, we can use:

∆S = S₂ – S₁ = Q/T

therefore  the entropy change will be:

∆S = 334 [kJ] / 273.15 [K] = 1.22 [kJ/K]

where 334  kilojoules of heat are required to melt 1 kg of ice (latent heat of fusion = 334 kJ/kg) and this heat is transferred to the system at 0°C (273.15 K).

Specific Entropy

The entropy can be made into an intensive, or specific, variable by dividing by the mass. Engineers use the specific entropy in thermodynamic analysis more than the entropy itself. The specific entropy (s) of a substance is its entropy per unit mass. It equals to the total entropy (S) divided by the total mass (m).

s = S/m

where:

s = specific entropy (J/kg)

S = entropy (J)

m = mass (kg)

Entropy quantifies the energy of a substance that is no longer available to perform useful work. Because entropy tells so much about the usefulness of an amount of heat transferred in performing work, the thermodynamics/steam-tables/”>steam tables include values of specific entropy (s = S/m) as part of the information tabulated.

In general, specific entropy is a property of a substance, like pressure, temperature, and volume, but it cannot be measured directly. Normally, the entropy of a substance is given with respect to some reference value. For example, the specific entropy of water or steam is given using the reference that the specific entropy of water is zero at 0.01°C and normal atmospheric pressure, where s = 0.00 kJ/kg. The fact that the absolute value of specific entropy is unknown is not a problem, however, because it is the change in specific entropy (∆s) and not the absolute value that is important in practical problems.

Temperature-entropy Diagrams – T-s Diagrams

In general, the phases of a substance and the relationships between its properties are most commonly shown on property diagrams. A large number of different properties have been defined, and there are some dependencies between properties.

A Temperature-entropy diagram (T-s diagram) is the type of diagram most frequently used to analyze energy transfer system cycles. It  is used in thermodynamics to visualize changes to temperature and specific entropy during a thermodynamic process or cycle.

This is because the work done by or on the system and the heat added to or removed from the system can be visualized on the T-s diagram. By the definition of entropy, the heat transferred to or from a system equals the area under the T-s curve of the process.

dQ = TdS

An isentropic process is depicted as a vertical line on a T-s diagram, whereas an isothermal process is a horizontal line. In an idealized state, compression is a pump, compression in a compressor and expansion in a turbine are isentropic processes. Therefore it is very useful in power engineering, because these devices are used in thermodynamic cycles of power plants.

Note that, the isentropic assumptions are only applicable with ideal cycles. Real thermodynamic cycles have inherent energy losses due to inefficiency of compressors and turbines.

Irreversibility of Natural Processes

According to the second law of thermodynamics:

The entropy of any isolated system never decreases. In a natural thermodynamic process, the sum of the entropies of the interacting thermodynamic systems increases.  

This law indicates the irreversibility of natural processes. Reversible processes are a useful and convenient theoretical fiction, but do not occur in nature. From this law follows that it is impossible to construct a device that operates on a cycle and whose sole effect is the transfer of heat from a cooler body to a hotter body. It follows, perpetual motion machines of the second kind are impossible.

Entropy at Absolute Zero

According to third law of thermodynamics:

The entropy of a system approaches a constant value as the temperature approaches absolute zero.

Based on empirical evidence, this law states that the entropy of a pure crystalline substance is zero at the absolute zero of temperature, 0 K and that it is impossible by means of any process, no matter how idealized, to reduce the temperature of a system to absolute zero in a finite number of steps. This allows us to define a zero point for the thermal energy of a body.

Absolute zero is the coldest theoretical temperature, at which the thermal motion of atoms and molecules reaches its minimum.  This is a state at which the enthalpy and entropy of a cooled ideal gas reaches its minimum value, taken as 0. Classically, this would be a state of motionlessness, but quantum uncertainty dictates that the particles still possess a finite zero-point energy. Absolute zero is denoted as 0 K on the Kelvin scale, −273.15 °C on the Celsius scale, and −459.67 °F on the Fahrenheit scale.

See also:

Energy

Enthalpy in Extensive Units

H = U + pV

Enthalpy is an extensive quantity, it depends on the size of the system, or on the amount of substance it contains. The SI unit of enthalpy is the joule (J). It is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields. It is the thermodynamic quantity equivalent to the total heat content of a system.

On the other hand, energy can be stored in the chemical bonds between the atoms that make up the molecules. This energy storage on the atomic level includes energy associated with electron orbital states, nuclear spin, and binding forces in the nucleus.

Enthalpy is represented by the symbol H, and the change in enthalpy in a process is H₂ – H₁.

There are expressions in terms of more familiar variables such as thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature andthermodynamics/thermodynamic-properties/what-is-pressure-physics/”> pressure:

dH = C_pdT + V(1-αT)dp

Where C_p is the heat capacity at constant pressure and α is the coefficient of (cubic) thermal expansion. For ideal gas αT = 1 and therefore:

dH = C_pdT

Example: Frictionless Piston – Heat – Enthalpy

A frictionless piston is used to provide a constant pressure of 500 kPa in a cylinder containing steam (superheated steam) of a volume of 2 m³  at 500 K. Calculate the final temperature, if 3000 kJ of heat is added.

Solution:

Using thermodynamics/steam-tables/”>steam tables we know, that the specific enthalpy of such steam (500 kPa; 500 K) is about 2912 kJ/kg. Since at this condition the steam has density of 2.2 kg/m³, then we know there is about 4.4 kg of steam in the piston at enthalpy of 2912 kJ/kg x 4.4 kg = 12812 kJ.

When we use simply Q = H₂ − H₁, then the resulting enthalpy of steam will be:

H₂ = H₁ + Q = 15812 kJ

From steam tables, such superheated steam (15812/4.4 = 3593 kJ/kg) will have a temperature of 828 K (555°C). Since at this enthalpy the steam have density of 1.31 kg/m³, it is obvious that it has expanded by about 2.2/1.31 = 1.67 (+67%). Therefore the resulting volume is 2 m³ x 1.67 = 3.34 m³ and ∆V = 3.34 m³ – 2 m³ = 1.34 m³.

The p∆V part of enthalpy, i.e. the work done is:

W = p∆V = 500 000 Pa x 1.34 m³ = 670 kJ

Example: Energy Balance in a Steam Generator

Calculate the amount of primary coolant, which is required to evaporate 1 kg of feedwater in a typical steam generator. Assume that there are no energy losses, this is only idealized example.

Balance of the primary circuit

The hot primary coolant (water 330°C; 626°F; 16MPa) is pumped into the steam generator through primary inlet. The primary coolant leaves (water 295°C; 563°F; 16MPa) the steam generator through primary outlet.

h_{I, inlet} = 1516 kJ/kg

=> Δh_I = -206 kJ/kg

h_{I, outlet} = 1310 kJ/kg

Balance of the feedwater

The feedwater (water 230°C; 446°F; 6,5MPa) is pumped into the steam generator through the feedwater inlet. The feedwater (secondary circuit) is heated from ~230°C 446°F to the boiling point of that fluid (280°C; 536°F; 6,5MPa). Feedwater is then evaporated and the pressurized steam (saturated steam 280°C; 536°F; 6,5 MPa) leaves the steam generator through steam outlet and continues to the steam turbine.

h_{II, inlet} = 991 kJ/kg

=> Δh_II = 1789 kJ/kg

h_{II, outlet} = 2780 kJ/kg

Balance of the steam generator

Since the difference in specific enthalpies is less for primary coolant than for feedwater, it is obvious that the amount of primary coolant will be higher than 1kg. To produce of 1 kg of saturated steam from feedwater, about 1789/206 x 1 kg =  8.68 kg of primary coolant is required.

Example: Combustion of Hydrogen

Consider the combustion of hydrogen in air. In a flame of pure hydrogen gas, burning in air, the hydrogen (H₂) reacts with oxygen (O₂) to form water (H₂O) and releases energy.

Energetically, the process can be considered to require the energy to dissociate the H₂ and O₂, but then the bonding of the H₂O returns the system to a bound state with negative potential. It is actually more negative than the bound states of the reactants, and the formation of the two water molecules is therefore an exothermic reaction, which releases 5.7 eV of energy. In words of enthalpy, the enthalpy of combustion is −286 kJ/mol:

2H₂(g) + O₂(g) → 2H₂O(g)

In words of enthalpy, the enthalpy of combustion is −286 kJ/mol (energy per mol of molecular hydrogen):

2H₂(g) + O₂(g) → 2H₂O(l) +572 kJ

The balance of energy before and after the reaction can be illustrated schematically with the state in which all atoms are free taken as the reference for energy.

Example: Enthalpy of Wet Steam

” width=”300″ height=”277″ /> Rankine Cycle – thermodynamics as Energy Conversion Science

A high-pressure stage of steam turbine operates at steady state with inlet conditions of  6 MPa, t = 275.6°C, x = 1 (point C). Steam leaves this stage of turbine at a pressure of 1.15 MPa, 186°C and x = 0.87 (point D). Calculate the enthalpy difference between these two states.

The enthalpy for the state C can be picked directly from thermodynamics/steam-tables/”>steam tables, whereas the enthalpy for the state D must be calculated using vapor quality:

h_{1, wet} = 2785 kJ/kg

h_{2, wet} = h_2,s x + (1 – x ) h_2,l  = 2782 . 0.87 + (1 – 0.87) . 790 = 2420 + 103 = 2523 kJ/kg

Δh = 262 kJ/kg

See also:

Enthalpy

Enthalpy Formulas in Extensive Units

H = U + pV

thermodynamics/what-is-energy-physics/what-is-enthalpy/”>Enthalpy is an extensive quantity, it depends on the size of the system, or on the amount of substance it contains. The SI unit of enthalpy is the joule (J). It is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields. It is the thermodynamic quantity equivalent to the total heat content of a system.

On the other hand, energy can be stored in the chemical bonds between the atoms that make up the molecules. This energy storage on the atomic level includes energy associated with electron orbital states, nuclear spin, and binding forces in the nucleus.

Enthalpy is represented by the symbol H, and the change in enthalpy in a process is H₂ – H₁.

There are enthalpy formulas in terms of more familiar variables such as thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature andthermodynamics/thermodynamic-properties/what-is-pressure-physics/”> pressure:

dH = C_pdT + V(1-αT)dp

Where C_p is the heat capacity at constant pressure and α is the coefficient of (cubic) thermal expansion. For ideal gas αT = 1 and therefore:

dH = C_pdT

Enthalpy Formulas in Intensive Units – Specific Enthalpy

The enthalpy can be made into an intensive, or specific, variable by dividing by thethermodynamics/thermodynamic-properties/what-is-mass-and-weight/what-is-mass/”> mass. Engineers use the specific enthalpy in thermodynamic analysis more than the enthalpy itself. The specific enthalpy (h) of a substance is its enthalpy per unit mass. It equals to the total enthalpy (H) divided by the total mass (m).

h = H/m

where:

h = specific enthalpy (J/kg)

H = enthalpy (J)

m = mass (kg)

Note that the enthalpy is the thermodynamic quantity equivalent to the total heat content of a system. The specific enthalpy is equal to the thermodynamics/what-is-energy-physics/internal-energy-thermal-energy/specific-internal-energy/”>specific internal energy of the system plus the product of thermodynamics/thermodynamic-properties/what-is-pressure-physics/”>pressure and thermodynamics/thermodynamic-properties/what-is-specific-volume/”>specific volume.

h = u + pv

In general, enthalpy is a property of a substance, like pressure, temperature, and volume, but it cannot be measured directly. Normally, the enthalpy of a substance is given with respect to some reference value. For example, the specific enthalpy of water or steam is given using the reference that the specific enthalpy of water is zero at 0.01°C and thermodynamics/thermodynamic-properties/what-is-pressure-physics/atmospheric-pressure/”>normal atmospheric pressure, where h_L = 0.00 kJ/kg. The fact that the absolute value of specific enthalpy is unknown is not a problem, however, because it is the change in specific enthalpy (∆h) and not the absolute value that is important in practical problems.

Formulas for Specific Enthalpy of Wet Steam

The specific enthalpy of saturated liquid water (x=0) and dry steam (x=1) can be picked from steam tables. In case of wet steam, the actual enthalpy can be calculated with the vapor quality, x, and the specific enthalpies of saturated liquid water and dry steam:

h_wet = h_s x + (1 – x ) h_l              

where

h_wet = enthalpy of wet steam (J/kg)

h_s = enthalpy of “dry” steam (J/kg)

h_l = enthalpy of saturated liquid water (J/kg)

As can be seen, wet steam will always have lower enthalpy than dry steam.

Example:

” width=”300″ height=”277″ /> Rankine Cycle – thermodynamics as Energy Conversion Science

A high-pressure stage of steam turbine operates at steady state with inlet conditions of  6 MPa, t = 275.6°C, x = 1 (point C). Steam leaves this stage of turbine at a pressure of 1.15 MPa, 186°C and x = 0.87 (point D). Calculate the enthalpy difference between these two states.

The enthalpy for the state C can be picked directly from steam tables, whereas the enthalpy for the state D must be calculated using vapor quality:

h_{1, wet} = 2785 kJ/kg

h_{2, wet} = h_2,s x + (1 – x ) h_2,l  = 2782 . 0.87 + (1 – 0.87) . 790 = 2420 + 103 = 2523 kJ/kg

Δh = 262 kJ/kg

See also:

Enthalpy

Enthalpy in Intensive Units – Specific Enthalpy

The thermodynamics/what-is-energy-physics/what-is-enthalpy/”>enthalpy can be made into an intensive, or specific, variable by dividing by thethermodynamics/thermodynamic-properties/what-is-mass-and-weight/what-is-mass/”> mass. Engineers use the specific enthalpy in thermodynamic analysis more than the enthalpy itself. The specific enthalpy (h) of a substance is its enthalpy per unit mass. It equals to the total enthalpy (H) divided by the total mass (m).

h = H/m

where:

h = specific enthalpy (J/kg)

H = enthalpy (J)

m = mass (kg)

Note that the enthalpy is the thermodynamic quantity equivalent to the total heat content of a system. The specific enthalpy is equal to the thermodynamics/what-is-energy-physics/internal-energy-thermal-energy/specific-internal-energy/”>specific internal energy of the system plus the product of thermodynamics/thermodynamic-properties/what-is-pressure-physics/”>pressure and thermodynamics/thermodynamic-properties/what-is-specific-volume/”>specific volume.

h = u + pv

In general, enthalpy is a property of a substance, like pressure, temperature, and volume, but it cannot be measured directly. Normally, the enthalpy of a substance is given with respect to some reference value. For example, the specific enthalpy of water or steam is given using the reference that the specific enthalpy of water is zero at 0.01°C and thermodynamics/thermodynamic-properties/what-is-pressure-physics/atmospheric-pressure/”>normal atmospheric pressure, where h_L = 0.00 kJ/kg. The fact that the absolute value of specific enthalpy is unknown is not a problem, however, because it is the change in specific enthalpy (∆h) and not the absolute value that is important in practical problems.

See also: thermodynamics/steam-tables/”>Steam Tables

Enthalpy of Vaporization

In general, when a material changes phase from solid to liquid, or from liquid to gas a certain amount of energy is involved in this change of phase. In case of liquid to gas phase change, this amount of energy is known as the enthalpy of vaporization, (symbol ∆H_vap; unit: J) also known as the (latent) heat of vaporization or heat of evaporation. Latent heat is the amount of heat added to or removed from a substance to produce a change in phase. This energy breaks down the intermolecular attractive forces, and also must provide the energy necessary to expand the gas (the pΔV work). When latent heat is added, no temperature change occurs. The enthalpy of vaporization is a function of the pressure at which that transformation takes place.

Latent heat of vaporization – water at 0.1 MPa (atmospheric pressure)

h_lg = 2257 kJ/kg

Latent heat of vaporization – water at 3 MPa (pressure inside a steam generator)

h_lg = 1795 kJ/kg

Latent heat of vaporization – water at 16 MPa (pressure inside a pressurizer)

h_lg = 931 kJ/kg

The heat of vaporization diminishes with increasing pressure, while the boiling point increases. It vanishes completely at a certain point called the critical point. Above the critical point, the liquid and vapor phases are indistinguishable, and the substance is called a supercritical fluid.

The heat of vaporization is the heat required to completely vaporize a unit of saturated liquid (or condense a unit mass of saturated vapor) and it equal to h_lg = h_g − h_l.

The heat that is necessary to melt (or freeze) a unit mass at the substance at constant pressure is the heat of fusion and is equal to h_sl = h_l − h_s, where h_s is the enthalpy of saturated solid and h_l is the enthalpy of saturated liquid.

Specific Enthalpy of Wet Steam

The specific enthalpy of saturated liquid water (x=0) and dry steam (x=1) can be picked from steam tables. In case of wet steam, the actual enthalpy can be calculated with the vapor quality, x, and the specific enthalpies of saturated liquid water and dry steam:

h_wet = h_s x + (1 – x ) h_l              

where

h_wet = enthalpy of wet steam (J/kg)

h_s = enthalpy of “dry” steam (J/kg)

h_l = enthalpy of saturated liquid water (J/kg)

As can be seen, wet steam will always have lower enthalpy than dry steam.

Example:

” width=”300″ height=”277″ /> Rankine Cycle – thermodynamics as Energy Conversion Science

A high-pressure stage of steam turbine operates at steady state with inlet conditions of  6 MPa, t = 275.6°C, x = 1 (point C). Steam leaves this stage of turbine at a pressure of 1.15 MPa, 186°C and x = 0.87 (point D). Calculate the enthalpy difference between these two states.

The enthalpy for the state C can be picked directly from steam tables, whereas the enthalpy for the state D must be calculated using vapor quality:

h_{1, wet} = 2785 kJ/kg

h_{2, wet} = h_2,s x + (1 – x ) h_2,l  = 2782 . 0.87 + (1 – 0.87) . 790 = 2420 + 103 = 2523 kJ/kg

Δh = 262 kJ/kg

See also:

Enthalpy

Enthalpy in Chemical Reactions

The enthalpy is widely used also in chemistry. Chemical reactions are determined by the laws of thermodynamics. In thermodynamics, the internal energy of a system is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields. The enthalpy of a chemical reaction is defined as the enthalpy change observed in a constituent of a thermodynamic system when one mole of substance reacts completely.

Since most of the chemical reactions in laboratory are constant-pressure processes, we can write the change in enthalpy (also known as enthalpy of reaction) for a reaction. The enthalpy of reaction can be positive or negative or zero depending upon whether the heat is gained or lost or no heat is lost or gained. In an endothermic reaction, the products have more stored chemical energy than the reactants. In an exothermic reaction, the opposite is true. The products have less stored chemical energy than the reactants. The excess energy is generally released to the surroundings when the reaction occurs.

In chemical reactions, energy is stored in the chemical bonds between the atoms that make up the molecules. Energy storage on the atomic level includes energy associated with electron orbital states. Whether a chemical reaction absorbs or releases energy, there is no overall change in the amount of energy during the reaction. That’s because of the law of conservation of energy, which states that:

Energy cannot be created or destroyed. Energy may change form during a chemical reaction.

Combustion of Hydrogen

Consider the combustion of hydrogen in air. In a flame of pure hydrogen gas, burning in air, the hydrogen (H₂) reacts with oxygen (O₂) to form water (H₂O) and releases energy.

Energetically, the process can be considered to require the energy to dissociate the H₂ and O₂, but then the bonding of the H₂O returns the system to a bound state with negative potential. It is actually more negative than the bound states of the reactants, and the formation of the two water molecules is therefore an exothermic reaction, which releases 5.7 eV of energy. In words of enthalpy, the enthalpy of combustion is −286 kJ/mol:

2H₂(g) + O₂(g) → 2H₂O(g)

In words of enthalpy, the enthalpy of combustion is −286 kJ/mol (energy per mol of molecular hydrogen):

2H₂(g) + O₂(g) → 2H₂O(l) +572 kJ

The balance of energy before and after the reaction can be illustrated schematically with the state in which all atoms are free taken as the reference for energy.

See also:

Enthalpy

Enthalpy in Extensive Units

H = U + pV

thermodynamics/what-is-energy-physics/what-is-enthalpy/”>Enthalpy is an extensive quantity, it depends on the size of the system, or on the amount of substance it contains. The SI unit of enthalpy is the joule (J). It is the energy contained within the system, excluding the kinetic energy of motion of the system as a whole and the potential energy of the system as a whole due to external force fields. It is the thermodynamic quantity equivalent to the total heat content of a system.

On the other hand, energy can be stored in the chemical bonds between the atoms that make up the molecules. This energy storage on the atomic level includes energy associated with electron orbital states, nuclear spin, and binding forces in the nucleus.

Enthalpy is represented by the symbol H, and the change in enthalpy (delta H) in a process is H₂ – H₁.

There are expressions in terms of more familiar variables such as thermodynamics/thermodynamic-properties/what-is-temperature-physics/”>temperature andthermodynamics/thermodynamic-properties/what-is-pressure-physics/”> pressure:

dH = C_pdT + V(1-αT)dp

Where C_p is the heat capacity at constant pressure and α is the coefficient of (cubic) thermal expansion. For ideal gas αT = 1 and therefore:

dH = C_pdT

Example: Frictionless Piston – Heat – Enthalpy

A frictionless piston is used to provide a constant pressure of 500 kPa in a cylinder containing steam (superheated steam) of a volume of 2 m³  at 500 K. Calculate the final temperature, if 3000 kJ of heat is added.

Solution:

Using thermodynamics/steam-tables/”>steam tables we know, that the specific enthalpy of such steam (500 kPa; 500 K) is about 2912 kJ/kg. Since at this condition the steam has density of 2.2 kg/m³, then we know there is about 4.4 kg of steam in the piston at enthalpy of 2912 kJ/kg x 4.4 kg = 12812 kJ.

When we use simply Q = H₂ − H₁, then the resulting enthalpy of steam will be:

H₂ = H₁ + Q = 15812 kJ

From steam tables, such superheated steam (15812/4.4 = 3593 kJ/kg) will have a temperature of 828 K (555°C). Since at this enthalpy the steam have density of 1.31 kg/m³, it is obvious that it has expanded by about 2.2/1.31 = 1.67 (+67%). Therefore the resulting volume is 2 m³ x 1.67 = 3.34 m³ and ∆V = 3.34 m³ – 2 m³ = 1.34 m³.

The p∆V part of enthalpy, i.e. the work done is:

W = p∆V = 500 000 Pa x 1.34 m³ = 670 kJ

Enthalpy in Intensive Units – Specific Enthalpy

The enthalpy can be made into an intensive, or specific, variable by dividing by thethermodynamics/thermodynamic-properties/what-is-mass-and-weight/what-is-mass/”> mass. Engineers use the specific enthalpy in thermodynamic analysis more than the enthalpy itself. The specific enthalpy (h) of a substance is its enthalpy per unit mass. It equals to the total enthalpy (H) divided by the total mass (m).

h = H/m

where:

h = specific enthalpy (J/kg)

H = enthalpy (J)

m = mass (kg)

Note that the enthalpy is the thermodynamic quantity equivalent to the total heat content of a system. The specific enthalpy is equal to the thermodynamics/what-is-energy-physics/internal-energy-thermal-energy/specific-internal-energy/”>specific internal energy of the system plus the product of thermodynamics/thermodynamic-properties/what-is-pressure-physics/”>pressure and thermodynamics/thermodynamic-properties/what-is-specific-volume/”>specific volume.

h = u + pv

In general, enthalpy is a property of a substance, like pressure, temperature, and volume, but it cannot be measured directly. Normally, the enthalpy of a substance is given with respect to some reference value. For example, the specific enthalpy of water or steam is given using the reference that the specific enthalpy of water is zero at 0.01°C and thermodynamics/thermodynamic-properties/what-is-pressure-physics/atmospheric-pressure/”>normal atmospheric pressure, where h_L = 0.00 kJ/kg. The fact that the absolute value of specific enthalpy is unknown is not a problem, however, because it is the change in specific enthalpy (∆h) and not the absolute value that is important in practical problems.

Enthalpy Changes in Chemical Reactions

Consider the combustion of hydrogen in air. In a flame of pure hydrogen gas, burning in air, the hydrogen (H₂) reacts with oxygen (O₂) to form water (H₂O) and releases energy.

Energetically, the process can be considered to require the energy to dissociate the H₂ and O₂, but then the bonding of the H₂O returns the system to a bound state with negative potential. It is actually more negative than the bound states of the reactants, and the formation of the two water molecules is therefore an exothermic reaction, which releases 5.7 eV of energy. In words of enthalpy, the enthalpy of combustion is −286 kJ/mol:

2H₂(g) + O₂(g) → 2H₂O(g)

In words of enthalpy, the enthalpy of combustion is −286 kJ/mol (energy per mol of molecular hydrogen):

2H₂(g) + O₂(g) → 2H₂O(l) +572 kJ

The balance of energy before and after the reaction can be illustrated schematically with the state in which all atoms are free taken as the reference for energy.

Enthalpy of Vaporization

In general, when a material changes phase from solid to liquid, or from liquid to gas a certain amount of energy is involved in this change of phase. In case of liquid to gas phase change, this amount of energy is known as the enthalpy of vaporization, (symbol ∆H_vap; unit: J) also known as the (latent) heat of vaporization or heat of evaporation. Latent heat is the amount of heat added to or removed from a substance to produce a change in phase. This energy breaks down the intermolecular attractive forces, and also must provide the energy necessary to expand the gas (the pΔV work). When latent heat is added, no temperature change occurs. The enthalpy of vaporization is a function of the pressure at which that transformation takes place.

Latent heat of vaporization – water at 0.1 MPa (atmospheric pressure)

h_lg = 2257 kJ/kg

Latent heat of vaporization – water at 3 MPa (pressure inside a steam generator)

h_lg = 1795 kJ/kg

Latent heat of vaporization – water at 16 MPa (pressure inside a pressurizer)

h_lg = 931 kJ/kg

The heat of vaporization diminishes with increasing pressure, while the boiling point increases. It vanishes completely at a certain point called the critical point. Above the critical point, the liquid and vapor phases are indistinguishable, and the substance is called a supercritical fluid.

The heat of vaporization is the heat required to completely vaporize a unit of saturated liquid (or condense a unit mass of saturated vapor) and it equal to h_lg = h_g − h_l.

The heat that is necessary to melt (or freeze) a unit mass at the substance at constant pressure is the heat of fusion and is equal to h_sl = h_l − h_s, where h_s is the enthalpy of saturated solid and h_l is the enthalpy of saturated liquid.

See also:

Enthalpy

Enthalpy in Intensive Units – Specific Enthalpy

The thermodynamics/what-is-energy-physics/what-is-enthalpy/”>enthalpy can be made into an intensive, or specific, variable by dividing by thethermodynamics/thermodynamic-properties/what-is-mass-and-weight/what-is-mass/”> mass. Engineers use the specific enthalpy in thermodynamic analysis more than the enthalpy itself. The specific enthalpy (h) of a substance is its enthalpy per unit mass. It equals to the total enthalpy (H) divided by the total mass (m).

h = H/m

where:

h = specific enthalpy (J/kg)

H = enthalpy (J)

m = mass (kg)

Note that the enthalpy is the thermodynamic quantity equivalent to the total heat content of a system. The specific enthalpy is equal to the thermodynamics/what-is-energy-physics/internal-energy-thermal-energy/specific-internal-energy/”>specific internal energy of the system plus the product of thermodynamics/thermodynamic-properties/what-is-pressure-physics/”>pressure and thermodynamics/thermodynamic-properties/what-is-specific-volume/”>specific volume.

h = u + pv

In general, enthalpy is a property of a substance, like pressure, temperature, and volume, but it cannot be measured directly. Normally, the enthalpy of a substance is given with respect to some reference value. For example, the specific enthalpy of water or steam is given using the reference that the specific enthalpy of water is zero at 0.01°C and thermodynamics/thermodynamic-properties/what-is-pressure-physics/atmospheric-pressure/”>normal atmospheric pressure, where h_L = 0.00 kJ/kg. The fact that the absolute value of specific enthalpy is unknown is not a problem, however, because it is the change in specific enthalpy (∆h) and not the absolute value that is important in practical problems.

Specific Enthalpy of Wet Steam

The specific enthalpy of saturated liquid water (x=0) and dry steam (x=1) can be picked from steam tables. In case of wet steam, the actual enthalpy can be calculated with the vapor quality, x, and the specific enthalpies of saturated liquid water and dry steam:

h_wet = h_s x + (1 – x ) h_l              

where

h_wet = enthalpy of wet steam (J/kg)

h_s = enthalpy of “dry” steam (J/kg)

h_l = enthalpy of saturated liquid water (J/kg)

As can be seen, wet steam will always have lower enthalpy than dry steam.

Example:

” width=”300″ height=”277″ /> Rankine Cycle – thermodynamics as Energy Conversion Science

A high-pressure stage of steam turbine operates at steady state with inlet conditions of  6 MPa, t = 275.6°C, x = 1 (point C). Steam leaves this stage of turbine at a pressure of 1.15 MPa, 186°C and x = 0.87 (point D). Calculate the enthalpy difference between these two states.

The enthalpy for the state C can be picked directly from steam tables, whereas the enthalpy for the state D must be calculated using vapor quality:

h_{1, wet} = 2785 kJ/kg

h_{2, wet} = h_2,s x + (1 – x ) h_2,l  = 2782 . 0.87 + (1 – 0.87) . 790 = 2420 + 103 = 2523 kJ/kg

Δh = 262 kJ/kg

See also:

Enthalpy