## 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:**

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**

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