Specific volume

This is the volume in cubic metres of one kilogram of dry air mixed with g kilograms of water vapour. In the mixture each constituent occupies the same volume and is at the same temperature, but each exerts its own partial pressure. By Dalton’s law the sum of these partial pressures is the total (barometric) pressure of the mixture. See Figure 2.4.

The general gas law, in the form of equation (2.8), may be transposed to express the specific volume:

V=mRT (2.17)

P

This equation could be used to refer to the dry air, or to the water vapour, independently if Dalton’s law is accepted. In doing so, the appropriate values for the mass, particular gas constant and partial pressure of the constituent considered must be used.

EXAMPLE 2.8

Calculate the specific volume of air at a dry-bulb temperature of 20°C and a moisture content of 0.007 34 kg per kg dry air at a barometric pressure of 95 kPa.

Answer

Considering the water vapour alone and adopting the subscript ‘s’ to indicate this we have

Y _ ms *, Ts 8 " Ps

From the answer to Example 2.6 we know that the partial pressure of the water vapour in the mixture is 1.1080 kPa and from section 2.6 we know that Rs is 461 J kg-1K-1, hence

0.00734 x 461 x (273 + 20) s 1108.0 u. ayom

Alternatively the dry air could be considered and, adopting the subscript ‘a’ to denote this, we have:

Y __ ^a ^a

Pa

We know from section 2.6 that the particular gas constant for dry air is 287 J kg_1K_1 and hence

= 1 X 287 x (273 + 20) 3

A (95 000- 1108.0) uoyom

A more exact agreement would be obtained if the influences of intermolecular forces were taken into account. This is discussed in section 2.19.

The thermodynamic properties of dry air and steam are well established. Hence the general principle followed by ASHRAE (1997), Goff (1949) and CIBSE (1986) in the expression of the volume of moist air is to add to the volume of 1 kg of dry air (va), a proportion of the difference between the volume of saturated air (vs), and that of dry air (va). This gives rise to the following equation:

V = va + n(vs — va)/100 (2.18)

Where (a, is the percentage saturation.

EXAMPLE 2.9

Calculate the specific volume of moist air at a dry-bulb temperature of 20°C, 50 per cent saturation and a barometric pressure of 101.325 kPa.

Answer

From CIBSE psychrometric tables (or, less accurately, from a chart) the specific volume of dry air, va, is 0.8301 m3 per kg dry air and the specific volume of saturated air, vs, is

0. 8497 m3 per kg dry air. Then, by equation (2.18), the specific volume of moist air at 50 per cent saturation is

V = 0.8301 + 50(0.8497 — 0.8301)/100 = 0.8399 m3 per kg dry air.

CIBSE psychrometric tables quote the same value.

Posted in Engineering Fifth Edition