## 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 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 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 for 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 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: 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 the case of liquid to gas phase change, this amount of energy is 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 phase change. This energy breaks down the intermolecular attractive forces and 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), equal to **h _{lg} = h_{g} − h_{l}**.

The heat 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 the 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:**

A high-pressure steam turbine stage operates at a steady state with inlet conditions of 6 MPa, t = 275.6°C, x = 1 (point C). Steam leaves this turbine stage 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**