Saturation vapour pressure

There are two requirements for the evaporation of liquid water to occur.

(i) Thermal energy must be supplied to the water.

(ii) The vapour pressure of the liquid must be greater than that of the steam in the


These statements need some explanation.

Molecules in the liquid state are comparatively close to one another. They are nearer to one another than are the molecules in a gas and are less strongly bound together than those in a solid. The three states of matter are further distinguished by the extent to which an individual molecule may move. At a given temperature, a gas consists of molecules which have high individual velocities and which are arranged in a random fashion. A liquid at the same temperature is composed of molecules, the freedom of movement of which is much less, owing to the restraining effect which neighbouring molecules have on one another, by virtue of their comparative proximity. An individual molecule, therefore, has less kinetic energy if it is in the liquid state than it does if in the gaseous state. Modern thought is that the arrangement of molecules in a liquid is not entirely random as in a gas, but that it is not as regular as it is in most, if not all, solids. However, this is by the way.

It is evident that if the individual molecular kinetic energies are greater in the gaseous state, then energy must be given to a liquid if it is to change to the gaseous phase. This explains the first stated requirement for evaporation.

As regards the second requirement, the situation is clarified if one considers the boundary between a vapour and its liquid. Only at this boundary can a transfer of molecules between the liquid and the gas occur. Molecules at the surface have a kinetic energy which has a value related to the temperature of the liquid. Molecules within the body of the gas also have a kinetic energy which is a function of the temperature of the gas. Those gaseous molecules near the surface of the liquid will, from time to time, tend to hit the surface of the liquid, some of them staying there. Molecules within the liquid and near to its surface will, from time to time, also tend to leave the liquid and enter the gas, some of them staying there.

It is found experimentally that, in due course, an equilibrium condition arises for which the gas and the parent liquid both have the same temperature and pressure. These are termed the saturation temperature and the saturation pressure. For this state of equilibrium the number of molecules leaving the liquid is the same as the number of molecules entering it from the gas, on average.

Such a state of thermal equilibrium is exemplified by a closed insulated container which has within it a sample of liquid water. After a sufficient period of time, the space above the liquid surface, initially a vacuum, contains steam at the same temperature as the remaining sample of liquid water. The steam under these conditions is said to be saturated.

Before this state of equilibrium was reached the liquid must have been losing molecules more quickly than it was receiving them. Another way of saying this is to state that the vapour pressure of the liquid exceeded that of the steam above it.

One point emerges from this example: since the loss of molecules from the liquid represents a loss of kinetic energy, and since the kinetic energy of the molecules in the liquid is an indication of the temperature of the liquid, then the temperature of the liquid must fall during the period preceding its attainment of thermal equilibrium.

It has been found that water in an ambient gas which is not pure steam but a mixture of dry air and steam, behaves in a similar fashion, and that for most practical purposes the relationship between saturation temperature and saturation pressure is the same for liquid water in contact only with steam. One concludes from this a very important fact: saturation vapour pressure depends solely upon temperature.

If we take the results of experiment and plot saturation vapour pressure against saturation temperature, we obtain a curve which has the appearance of the line on the psychrometric chart for 100 per cent saturation. The data on which this particular line is based can be found in tables of psychrometric information. Referring, for instance, to those tables published by the Chartered Institution of Building Services Engineers, we can read the saturation vapour pressure at, say, 20°C by looking at the value of the vapour pressure at 100 per cent saturation and a dry-bulb temperature of 20°C. It is important to note that the term ‘dry — bulb’ has a meaning only when we are speaking of a mixture of a condensable vapour and a gas. In this particular context the mixture is of steam and dry air, but we could have a mixture of, say, alcohol and dry air, which would have its own set of properties of dry — and wet-bulb temperatures.

According to the National Engineering Laboratory Steam Tables (1964), the following equation may be used for the vapour pressure of steam over water up to 100°C:

Log p = 30.590 51-8.2 log(f + 273.16) + 0.002 480 4(t + 273.16) —


Where t is temperature in °C and p is pressure in kPa. Note that in equation (2.10) the saturation vapour pressure is expressed in terms of the triple point of water, t + 273.16, not the absolute temperature, t + 273.15. However, this is not the case in equation (2.11).

ASHRAE (1997) uses a very similar equation for the expression of saturation vapour pressure, based on equations calculated by Hyland and Wexler (1983). The results are in close agreement with the answers obtained by equation (2.10), as Jones (1994) shows.

Over ice, the equation to be used, from the National Bureau of Standards (1955), is:

Log p = 12.538 099 7 — 266 391/(273.15 + t) (2.11)

Where p is pressure in Pa.

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