What is Enthalpy
In thermodynamics, the enthalpy is a measurement 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 steam tables along with specific volume and 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 = H_{2} – H_{1}) or work other than expansion work. For a variablepressure 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 H_{2} – H_{1}.
There are expressions in terms of more familiar variables such as temperature and pressure:
dH = C_{p}dT + V(1αT)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_{p}dT
The classical form of the is the following equation:
dU = dQ – dW
In this equation dW is equal to dW = pdV and is known as the boundary work.
Boundary work occurs because the mass of the substance contained within the system boundary causes a force, the pressure times the surface area, to act on the boundary surface and make it move. Boundary work (or pΔV Work) occurs when the volume V of a system changes. It is used for calculating piston displacement work in a closed system. This is what happens when steam, or gas contained in a pistoncylinder device expands against the piston and forces the piston to move.
Since H = U + pV, therefore dH = dU + pdV + Vdp and we substitute dU = dH – pdV – Vdp into the classical form of the law:
dH – pdV – Vdp = dQ – pdV
We obtain the law in terms of enthalpy:
dH = dQ + Vdp
or
dH = TdS + Vdp
In this equation the term Vdp is a flow process work. This work, Vdp, is used for open flow systems like a turbine or a pump in which there is a “dp”, i.e. change in pressure. There are no changes in control volume. As can be seen, this form of the law simplifies the description of energy transfer. At constant pressure, the enthalpy change equals the energy transferred from the environment through heating:
Isobaric process (Vdp = 0):
dH = dQ → Q = H_{2} – H_{1}
At constant entropy, i.e. in isentropic process, the enthalpy change equals the flow process work done on or by the system:
Isentropic process (dQ = 0):
dH = Vdp → W = H_{2} – H_{1}
It is obvious, it will be very useful in analysis of both thermodynamic cycles used in power engineering, i.e. in Brayton cycle and Rankine cycle.

isentropic compression – ambient air is drawn into the compressor, where it is pressurized (1 → 2). The work required for the compressor is given by W_{C} = H_{2} – H_{1}.
 isobaric heat addition – the compressed air then runs through a combustion chamber, where fuel is burned and air or another medium is heated (2 → 3). It is a constantpressure process, since the chamber is open to flow in and out. The net heat added is given by Q_{add} = H_{3 } H_{2}
 isentropic expansion – the heated, pressurized air then expands on turbine, gives up its energy. The work done by turbine is given by W_{T} = H_{4} – H_{3}
 isobaric heat rejection – the residual heat must be rejected in order to close the cycle. The net heat rejected is given by Q_{re} = H_{4 } H_{1}
As can be seen, we can describe and calculate (e.g. thermodynamic efficiency) such cycles (similarly for Rankine cycle) using enthalpies.
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 m^{3 } at 500 K. Calculate the final temperature, if 3000 kJ of heat is added.
Solution:
Using 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/m^{3}, 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 = H_{2} − H_{1}, then the resulting enthalpy of steam will be:
H_{2} = H_{1} + 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/m^{3}, it is obvious that it has expanded by about 2.2/1.31 = 1.67 (+67%). Therefore the resulting volume is 2 m^{3} x 1.67 = 3.34 m^{3} and ∆V = 3.34 m^{3} – 2 m^{3} = 1.34 m^{3}.
The p∆V part of enthalpy, i.e. the work done is:
W = p∆V = 500 000 Pa x 1.34 m^{3} = 670 kJ
Enthalpy in Intensive Units – Specific Enthalpy
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 specific internal energy of the system plus the product of pressure and 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 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.
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.
Enthalpy in Chemical Reactions
Since most of the chemical reactions in laboratory are constantpressure 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.
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
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:
A highpressure 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
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