# Moisture content and humidity ratio

Moisture content is defined as the mass of water vapour in kilograms which is associated with one kilogram of dry air in an air-water vapour mixture. It is sometimes called specific humidity or humidity ratio.

Starting with the definition, we can write—

Moisture content = mass of water vapour per unit mass of dry air = ms/ma

By using Dalton’s law we can now apply the general gas law to each of the two constituents of moist air, just as though the other did not exist:

PV = mRT in general

Hence

PsVs = msRsTs for the water vapour

And

PaYsi = wa/?a7’a for the dry air

The general gas law may be rearranged so that mass is expressed in terms of the other variables:

_ Pv m RT

By transposition in the equations referring to water vapour and dry air, we can obtain an expression for moisture content based on its definition:

. psVsRaTa

Moisture content =

RsTspaVa

= R*Ps RsPz

Since the water vapour and the dry air have the same temperature and volume.

The ratio of Ra to Rs is termed the relative density of water vapour with respect to dry air and, as already seen, it depends on the ratio of the molecular mass of water vapour to that of dry air:

K__MJL_ 18.02 _ n.79 Rs Ma 28.97

Hence we may write moisture content as 0.622 ps/pa and hence

G = 0-622 P* (2.12)

(P at P s)

Since the vapour pressure of superheated steam mixed with dry air is proportional to the mass of the steam present, the above equation can be re-arranged to express such a vapour pressure:

= (aSETi) (2-13)

EXAMPLE 2.4

Calculate the moisture content of 1 kg of dry air at 20°C mixed with saturated steam for barometric pressures of (a) 101.325 kPa and (b) 95 kPa.

(a) By equation (2.10), or from steam tables, or from CIBSE psychrometric tables, the saturation vapour pressure, pss, is 2.337 kPa. Hence, using equation (2.12)

8 = 0.622 (iop3253?2337) = °’°14 68 kg per kg dry air

CIBSE psychrometric tables give a value of 0.014 75 kg per kg dry air. The discrepancy is because the tables use a slightly different, more correct, equation that takes account of intermolecular forces (see section 2.19).

(b) Similarly, because the saturation vapour pressure is independent of barometric pressure pss is still 2.337 kPa, hence

G = 0.622 (95^2337) = 001569 k8 Per kS dry ^

It is often convenient to express moisture content as g/kg dry air. The two above answers would then be 14.68 g/kg dry air and 15.69 g/kg dry air.

 Dry-bulb temperature Fig. 2.6 The effect of barometric pressure on percentage saturation. Note that although the saturation vapour pressure is independent of barometric pressure the moisture content of air at 100 per cent saturation is not. Figure 2.6 shows that the curve For air at 100 per cent saturation corresponds to higher moisture contents as the barometric pressure decreases.

Posted in Engineering Fifth Edition