Supersaturation

When hot, humid air passes over a cooler coil with a sufficiently low mean coil surface temperature it is possible for the state of the air leaving the coil to be momentarily supersaturated, as point B in Figure 3.15 shows. Such a state is unstable and the presence of condensation nuclei causes an immediate reversion to a stable, saturated state (point E in Figure 3.15). No heat is supplied to or rejected from the system and no work is done. Hence the enthalpy at a notional state D equals that at B. However, moisture is condensed out of the airstream and lost, representing a liquid enthalpy drop, above a datum of 0°C. Hence the enthalpy of the saturated air at state E must be less than that at D, by this amount. Thus we have

— 21.47 g kg“1

— 9b_9d ; 9e

— 5.422 g kg’1

5° fe’d 34°

Fig. 3.15 Supersaturation.

Hb = hd = he + (gb — ge) x с x (te — 0) whence

He = hb — (gb — ge)cte (3.10)

Where с is the specific heat capacity of liquid water.

The reduction in moisture content from gb to ge appears as a fog of liquid droplets, conveyed by the airstream leaving the cooler coil. Operational conditions for this exist in suitably warm and humid climates, as in the Gulf or on the West African coast, where room air conditioning units may sometimes be seen emitting plumes of mist.

EXAMPLE 3.16

Air at 34°C dry-bulb, 28°C wet-bulb (sling) enters a cooler coil having a mean coil surface temperature of 5°C and a contact factor of 0.75. Determine the unstable, supersaturated state, the moisture condensed as a visible mist and the stable, saturated state of the air leaving the coil.

Answer

Plotting states A and С on a psychrometric chart shows that, with a contact factor of 0.75, state В is supersaturated. Hence we may calculate:

Hb = 0.25ha + 0.75hc = 0.25 x 89.24 + 0.75 x 18.64 = 36.29 kJ kg’1 gb = 0.25 x 21.47 + 0.75 x 5.422 = 9.434 g kg’1 tb = 0.25 x 34 + 0.75 x 5 = 12.25°C

At the notional state B, hd = hb = 36.29 kJ kg-1. Interpolating in psychrometric tables at 100 per cent saturation yields td = 12.830°C and gd = 9.262 g kg-1. As a first approximation we may assume gc = gd and te = td. Then using equation (3.10) and taking the specific heat capacity of water to be 4.168 kJ kg-1 K_1 we have

He = 36.29 — (0.009 434 — 0.009 262) x 4.168 x (12.83 — 0)

= 36.28 kJ kg"1

Interpolating again at saturation in psychrometric tables yields te = 12.826°C and ge = 9.260 g kg-1. Hence, the moisture condensed as a mist is

Gb — ge = 9.434 — 9.260 = 0.174 g kg“1

As a reasonable approximation.

Posted in Engineering Fifth Edition