Steam injection

As in water injection, steam injection may be dealt with by a consideration of a mass and energy balance. If ms kg of dry saturated steam are injected into a moving airsteam of mass flow 1 kg of dry air per second, then we may write

Gb = ga + ms

And

Hb = ha + hs

If the initial state of the moist airstream and the condition of the steam is known, then the

Final state of the air may be determined, provided none of the steam is condensed. The change of state takes place almost along a line of constant dry-bulb temperature between limits defined by the smallest and largest enthalpies of the injected steam, provided the steam is in a dry, saturated condition. If the steam is superheated then, of course, the dry — bulb temperature of the airstream may increase by any amount, depending on the degree of superheat. The two limiting cases for dry saturated steam are easily considered. The lowest possible enthalpy is for dry saturated steam at 100°C. It is not possible to use steam at a lower temperature than this since the steam must be at a higher pressure than atmospheric if it is to issue from the nozzles. (Note, however, that steam could be generated at a lower temperature from a bath of warm water, which was not actually boiling.)

The other extreme is provided by the steam which has maximum enthalpy; the value of this is 2803 kJ kg-1 of steam and it exists at a pressure of about 30 bar and a temperature of about 234°C.

What angular displacement occurs between these two limits, and how are they related to a line of constant dry-bulb temperature? These questions are answered by means of two numerical examples.

EXAMPLE 3.8

Dry saturated steam at 100°C is injected at a rate of 0.01 kg s“1 into a moist airstream moving at a rate of 1 kg of dry air per second and initially at a state of 28°C dry-bulb,

11. 9°C wet-bulb (sling) and 101.325 kPa barometric pressure. Calculate the leaving state of the moist airstream.

Answer

From psychrometric tables, ha = 33.11 kJ per kg dry air,

Ga = 0.001 937 kg per kg

From NEL steam tables, hs = 2675.8 kJ per kg steam.

Gb = 0.001 937 + 0.01

= 0.011 937 kg per kg dry air, by the mass balance hb = 33.11 + 0.01 x 2675.8 = 59.87 kJ per kg dry air

Hence,

59.87 = (1.007f„ — 0.026) + 0.011937(2501 + 1.84fb) tb = 29.2°C

EXAMPLE 3.9

Dry saturated steam with maximum enthalpy is injected at a rate of 0.01 kg s“1 into a moist airstream moving at a rate of 1 kg of dry air per second and initially at a state of 28°C dry — bulb, 11.9°C wet-bulb (sling) and 101.325 kPa barometric pressure. Calculate the leaving state of the moist airstream.

Answer

The psychrometric properties of state A are as for the last example. The moisture content at state B is also as in example 3.8.

From NEL steam tables, hs = 2803 kJ per kg of steam, at 30 bar and 234°C saturated.

Consequently,

Hb = 33.11 +0.01 x2803 = 61.14 kJ kg“1 gb = 0.001 937 + 0.01

= 0.011 937 kg kg-1 dry air Hence, as before:

61.14 = (1.007?b — 0.026) + 0.011 937(2501 + 1.84fb) tb = 30.4°C

Figure 3.9 illustrates what occurs. It can be seen that for the range of states considered, the change in dry-bulb value is not very great. In fact, the angular displacement between the two condition lines for the last two examples of steam injection is only about 3 or 4 degrees. We can conclude that, although there is an increase in temperature, it is within the accuracy usually required in practical air conditioning to assume that the change of state following steam injection is up a line of constant dry-bulb temperature.

In the injection of superheated steam, every case should be treated on its merits, as the equation for the dry-bulb temperature resulting from a steam injection process shows:

(3.9)

Hb + 0.026 — 2501 g h ~ 1.007 + 1.84g