What is 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. Thermal Engineering

Enthalpy in Intensive Units – Specific Enthalpy

Extensive vs. intensive thermodynamic properties — Extensive and intensive properties of medium in the pressurizer.

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 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.

Energy Balance in a Steam Generator

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.

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.

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:

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:

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

References:

Reactor Physics and Thermal Hydraulics:

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

What is Specific Enthalpy – Definition

Enthalpy in Intensive Units – Specific Enthalpy

Specific Enthalpy of Wet Steam

See also:

Enthalpy in Intensive Units – Specific Enthalpy

Specific Enthalpy of Wet Steam

See also:

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