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Use of enthalpies to calculate energy needed and / or recovered


Change of Water Properties with Temperature


As water heats up, the molecules gain thermal energy, and overcome the hydrogen-bonded network present in normal liquid water.  This results in some properties of water changing more than would usually be expected with temperature.  As the critical point is approached, some properties undergo very rapid change.  The following figure shows how several properties change along the saturation line, i.e. at sufficient pressure that liquid water is in equilibrium with the vapour (1 atm at 100°C). (Data for figures and following discussion taken from NBS/NRC Steam Tables, Lester Haar and George S. Kell, Taylor Francis 1984. ISBN 0-89116-354-9).



Enthalpy of Water

The energy needed to heat water is often calculated as:

Energy = (mass) x (temperature difference) x (specific heat capacity).

This would be correct if the heat capacity was constant with temperature, which is not actually the case.  Over fairly narrow temperature ranges under 100°C the variation is quite small, and errors correspondingly small.  For example, at atmospheric pressure, the specific heat capacity at constant pressure changes from 4.183 kJ / (kg K) at 20°C to 4.194 kJ / (kg K) at 80°C, a change of only 0.3%.  However, when superheated water is considered, the variation of heat capacity with temperature (and to a lesser extent pressure) should be accounted for.  At 350°C (200 bar) it is 8.138 kJ / (kg K), nearly twice that at 20°C and the same pressure (4.125 kJ / (kg K)).  The heat capacity increases with temperature partly because the extensive hydrogen bonding interactions break down and the extra energy needs to be put in to break the bonds.  A graph of heat capacity at constant pressure (250 bar) against temperature is shown below.  As the critical point is approached, the specific heat capacity at constant pressure rises towards infinity.  At constant pressure, the fluid expands when the temperature is raised, and work must be done against the intermolecular forces and the atmosphere.  In the critical region, a very large volume change is required to maintain the same pressure, and hence the energy goes towards expansion rather than raising the temperature.




Use of enthalpies to calculate energy requirements for superheated water plant

The following discussion will use a hypothetical plant using 1 tonne (1000 kg) per hour of water.  

Pressurising and heating

We wish to pressurise this water to 50 bar, and heat it to 250°C.  What are the energy requirements?

            The enthalpy, or amount of “heat energy” contained in water is shown in steam tables under different conditions.  The enthalpy of water changes very little by increasing the pressure.

1 bar, 25°C, enthalpy = 105 kJ / kg,

50 bar, 25°C, enthalpy = 109 kJ / kg. 

Therefore, at 50 bar, each kg of water only “stores” 4 kJ of energy, or it only requires 4 kJ to pressurise water to 50 bar.  For our hypothetical plant, this means a pump of 4000 kJ / hr, or 1.1kW.  The temperature of the water will be almost unchanged during the compression, as the pumping is isentropic, and entropy of liquid water changes very little with pressure.

            For the heating, the enthalpy at 250°C is 1085 kJ / kg, so the heat required is simply 1085 – 105 = 980 kJ / kg.  For our 1 tonne per hour plant, this means 272 kW of heating.  This is quite a large amount of heat.  However, this assumes the heat energy is not recovered, so in practice, as we will see later, this heat requirement can be significantly reduced. 



Now, if we wished to dry the product by evaporating this water, we could simply depressurise and allow the water to “flash” off couldn’t we?  Actually, in practice this is not the case due to the large latent heat of evaporation of water.  Using the steam tables again, we can find the energy released by depressurising the water and allowing it to use the heat energy to boil off the excess water.  At atmospheric pressure, the enthalpy of liquid water at 99°C is approximately 410 kJ / kg, and the enthalpy of steam at 100°C is 2676 kJ / kg.  This gives us the latent heat of vaporisation of 2266 kJ / kg.  The energy we release from cooling water from 250°C at 50 bar to 99°C at 1 bar is (1085 – 410) = 675 kJ / kg.  This is enough to vaporise only 300g / kg steam leaving 700 g / kg of water at 99°C.   Therefore “flashing” off the water is not a very effective method of drying.


Heat Recovery

However, all is not lost, as it is quite easy to recover most of the energy from the hot water by pre-heating the incoming water in a heat exchanger, thereby cooling the process water.   It is quite reasonable to recover 75% of the energy, leaving only 25% to be added by the heater (see the example here).  Therefore, once the plant is warmed up to operating temperature, the heating requirements drop from 272 kW to 68 kW.  Ideally the product will separate from the water as either a liquid or solid phase.  Most of the water can then be separated, and only a small amount of residual water need be evaporated to dry the product.


Calculation of Enthalpy

Methods for calculating the properties of water are continually reviewed by the International Association for the Properties of Water and Steam, who provide the equations for calculating steam tables (  For approximate calculations over the range 10 to 300°C the calculation below can be used for the enthalpy of water.  This is enough to provide an indication of the heat requirements; individual enthalpy values are within 1.5% of the steam table values. Enter the start temperature, the final temperature (Temperature 2) and the calculator will display the enthalpies at the two temperatures and the difference betweeen them.

Temperature  °C
Temperature 2  °C
Enthalpy 2
Enthalpy Difference


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A.A Clifford 04/01/2008