Mixtures

Figure 3.2 shows what happens when two airstreams meet and mix adiabatically. Moist air at state 1 mixes with moist air at state 2, forming a mixture at state 3. The principle of the conservation of mass allows two mass balance equations to be written:

Mai + mzi = md3 f°r the dry air and 8imai + 8ima2 = 8ima3 f°r the associated water vapour

Hence

(gi — 8i)ma = (83 — 82)ma2

Fig. 3.2 The adiabatic mixing of two airstreams.

Therefore

Gi -83 = 0ha_

G3 ~ gi

Similarly, making use of the principle of the conservation of energy,

Hi — _ ^a2_

H — h2 wal

From this it follows that the three state points must lie on a straight line in a mass — energy co-ordinate system. The psychrometric chart published by the CIBSE is such a system—with oblique co-ordinates. For this chart then, a principle can be stated for the expression of mixture states. When two airstreams mix adiabatically, the mixture state lies on the straight line which joins the state points of the constituents, and the position of the mixture state point is such that the line is divided inversely as the ratio of the masses of dry air in the constituent airstreams.

EXAMPLE 3.1

Moist air at a state of 60°C dry-bulb, 32.1°C wet-bulb (sling) and 101.325 kPa barometric pressure mixes adiabatically with moist air at 5°C dry-bulb, 0.5°C wet-bulb (sling) and

101.325 kPa barometric pressure. If the masses of dry air are 3 kg and 2 kg, respectively, calculate the moisture content, enthalpy and dry-bulb temperature of the mixture.

Answer

From CIBSE tables of psychrometric data, gj = 18.400 g per kg dry air

G2 = 2.061 g per kg dry air hi = 108.40 kJ per kg dry air h2 = 10.20 kJ per kg dry air

The principle of the conservation of mass demands that

Gl^al + 82^2 = Ј3™a3

=£з(»»аі + ma2)

Hence

_ gl^al + g2^a2

83 mal + ma2

_ 18.4 x 3 + 2.061 x 2 ~ 3 + 2

_ 59.322 5

= 11.864 g per kg dry air Similarly, by the principle of the conservation of energy,

A, mal + h2ma2

N3 — ———— ;

J mal + ma2

_ 108.40 x 3 + 10.20 x 2 ~ 3 + 2

= 69.12 kJ per kg dry air To determine the dry-bulb temperature, the following practical equation must be used h = (1.007* — 0.026) + g(2501 + 1.84*) (2.24)

Substituting the values calculated for moisture content and enthalpy, this equation can be solved for temperature:

H = 69.12 = (1.007*- 0.026) + 0.011 86(2501 + 1.84*)

T _ 39.48 _ io л°С’

1 ~ L029 " 38-4 C

On the other hand, if the temperature were calculated by proportion, according to the masses of the dry air in the two mixing airstreams, a slightly different answer results:

3×60 + 2×5 5

= 38°C

This is clearly the wrong answer, both numerically and by the method of its calculation. However, the error is small, considering that the values chosen for the two mixing states

Spanned almost the full range of the psychrometric chart. The error is even less for states

Likely to be encountered in everyday practice and is well within the acceptable limits of accuracy. To illustrate this, consider an example involving the mixture of two airstreams at states more representative of common practice.

EXAMPLE 3.2

A stream of moist air at a state of 21°C dry-bulb and 14.5°C wet-bulb (sling) mixes with another stream of moist air at a state of 28°C dry-bulb and 20.2°C wet-bulb (sling), the respective masses of the associated dry air being 3 kg and 1 kg. With the aid of CIBSE tables of psychrometric data calculate the dry-bulb temperature of the mixture (a) using the principles of conservation of energy and of mass and, (b), using a direct proportionality between temperature and mass.

Answer

The reader is left to go through the steps in the arithmetic, as an exercise, using the procedure adopted in example 3.1.

The answers are (a) 22.76°C, dry-bulb, (b) 22.75°C dry-bulb.

The conclusion to be drawn is that the method used to obtain answer (b) is accurate enough for most practical purposes.

Posted in Engineering Fifth Edition