What is Enthalpy

thermodynamic potentials - enthalpy

Four common thermodynamic potentials. H = U+ pV

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 = 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

Extensive vs. intensive thermodynamic properties

Extensive and intensive properties of medium in the pressurizer.

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 temperature and pressure:

dH = CpdT + V(1-αT)dp

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

dH = CpdT

The first law of thermodynamics in terms of enthalpy show us, why engineers use the enthalpy in thermodynamic cycles (e.g. Brayton cycle or Rankine cycle).

The classical form of the law 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 piston-cylinder 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 = H2 – H1

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 = H2 – H1

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.

Let assume the ideal Brayton cycle that describes the workings of a constant pressure heat engine. Modern gas turbine engines and airbreathing jet engines also follow the Brayton cycle. This cycle consist of four thermodynamic processes:

  1. first law - example - brayton cycle

    Ideal Brayton cycle consist of four thermodynamic processes. Two isentropic processes and two isobaric processes.

    isentropic compression – ambient air is drawn into the compressor, where it is pressurized (1 → 2). The work required for the compressor is given by WC = H2 – H1.

  2. 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 constant-pressure process, since the chamber is open to flow in and out. The net heat added is given by Qadd = H3 - H2
  3. isentropic expansion – the heated, pressurized air then expands on turbine, gives up its energy. The work done by turbine is given by WT = H4 – H3
  4. isobaric heat rejection – the residual heat must be rejected in order to close the cycle. The net heat rejected is given by Qre = H4 - H1

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

Enthalpy - Example - A frictionless piston

Calculate the final temperature, if 3000 kJ of heat is added.

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 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 the 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 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 hL = 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.

Steam Generator - vertical

Steam Generator – vertical

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.

hI, inlet = 1516 kJ/kg

=> ΔhI = -206 kJ/kg

hI, 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.

hII, inlet = 991 kJ/kg

=> ΔhII = 1789 kJ/kg

hII, 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

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.

Enthalpy of Vaporization

Latent heat of vaporization - water at 0.1 MPa, 3 MPa, 16 MPa

The heat of vaporization diminishes with increasing pressure, while the boiling point increases. It vanishes completely at a certain point called the critical point.

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 ∆Hvap; 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)

hlg = 2257 kJ/kg

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

hlg = 1795 kJ/kg

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

hlg = 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 hlg = hg − hl.

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 hsl = hl − hs, where hs is the enthalpy of saturated solid and hl is the enthalpy of saturated liquid.

Phase changes - enthalpy of vaporization

Latent heat of vaporization – water at 0.1 MPa. Dominant part of heat absorbed.

Specific Enthalpy of Wet Steam

wet-steam-vapor-liquid-mixture-minThe 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:

hwet = hs x + (1 – x ) hl              

where

hwet = enthalpy of wet steam (J/kg)

hs = enthalpy of “dry” steam (J/kg)

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

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

Example:

engineering thermodynamics

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:

h1, wet = 2785 kJ/kg

h2, wet = h2,s x + (1 – x ) h2,l  = 2782 . 0.87 + (1 – 0.87) . 790 = 2420 + 103 = 2523 kJ/kg

Δh = 262 kJ/kg

Reactor Physics and Thermal Hydraulics:

  1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).
  2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.
  3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.
  4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317
  5. Todreas Neil E., Kazimi Mujid S. Nuclear Systems Volume I: Thermal Hydraulic Fundamentals, Second Edition. CRC Press; 2 edition, 2012, ISBN: 978-0415802871
  6. Zohuri B., McDaniel P. Thermodynamics in Nuclear Power Plant Systems. Springer; 2015, ISBN: 978-3-319-13419-2
  7. Moran Michal J., Shapiro Howard N. Fundamentals of Engineering Thermodynamics, Fifth Edition, John Wiley & Sons, 2006, ISBN: 978-0-470-03037-0
  8. Kleinstreuer C. Modern Fluid Dynamics. Springer, 2010, ISBN 978-1-4020-8670-0.
  9. U.S. Department of Energy, THERMODYNAMICS, HEAT TRANSFER, AND FLUID FLOW. DOE Fundamentals Handbook, Volume 1, 2 and 3. June 1992.

See above: