# Latent heat removal

If the air in a room is not at saturation, then water vapour may be liberated in the room and cause the moisture content of the air in the room to rise. Such a liberation of steam is effected by any process of evaporation as, for example, the case of insensible perspiration and sweating on the part of the people present. Since it is necessary to provide heat to effect a process of evaporation, it is customary to speak of the addition of moisture to a room as kW of latent heat rather than as kg s_1 of water evaporated.

The heat gains occurring in a room can be considered in two parts: sensible gains and latent gains. The mixture of dry air and associated water vapour supplied to a room has therefore a dual role; it is cool enough initially to suffer a temperature rise up to the room dry-bulb temperature in offsetting the sensible gains, and its initial moisture content is low enough to permit a rise to the value of the room moisture content as latent heat gains are offset. Figure 6.1 illustrates this.

100% Offsetting sensible heat gains Fig. 6.1 Dealing with sensible and latent heat gains. |

If m kg s_1 of dry air with an associated moisture content of g kg per kg of dry air is supplied at state S to a conditioned space wherein there are only sensible heat gains, then its temperature will rise from 13°C to 22°C as it diffuses through the room, offsetting the sensible gains. The resultant room condition would be typified by state point A. If there were then a latent gain in the room caused by some evaporation taking place (say from people’s skin) without any further increase in sensible heat gain, then the state in the room would change from A to R, more or less up a line of constant dry-bulb temperature (see section 3.7). The exact amount that the moisture content of R exceeds ga (= gs) would depend on the latent heat gain. If the mass of dry air supplied and its associated moisture content is known, then it is possible to calculate the rise in room moisture content corresponding to given latent heat gains:

Latent heat gain = mass flow rate of supply air in kg dry air per second x moisture pick-up in kg moisture per kg dry air x latent heat of evaporation in kJ per kg moisture

= mx(gr — gs) x hfg

Where gT and gs are the moisture contents of the room and supply air, respectively, and hfg is the latent heat of evaporation. As with the derivation of equation (6.6)

M = v, x p0 x (273 + ?0)/(273 + t)

Therefore

TOC o "1-5" h z Latent heat gain = [ v, x p0 x (273 + f0)/(273 + 0] x h(g x (gr — gs)

If p0 = 1.191 kg m-3 at 20°C dry-bulb and 50 per cent saturation and if hfg = 2454 kJ kg-1

At 20°C then

, _ Latent heat gain w (273 + t) /r ox

where v, is in m3 s_1, the latent gain is in kW and (gr — gs) is in g kg-1 dry air.

It is emphasised that the air quantity calculated in m3 s-1 at temperature t is the same air quantity calculated by means of equation (6.6). We calculate the necessary air quantity by means either of equation (6.6), or equation (6.8), or by first principles. It is usual to use sensible gains to establish the required supply air quantity, by means of equation (6.6) first of all and then to establish by means of equation (6.8) the rise in moisture content which will result from supplying this air, with its associated moisture content (gs) to the room. Alternatively, we may use equation (6.8) to establish what the moisture content of the supply air must be in order to maintain a certain moisture content in the room in the presence of known latent heat gains. The temperature in the room would then be the secondary consideration.

EXAMPLE 6.6

A room measures 20 mx 10mx3m high and is to be maintained at a state of 20°C dry — bulb and 50 per cent saturation. The sensible and latent heat gains to the room are 7.3 kW and 1.4 kW, respectively.

*(a) *Calculate from first principles, the mass and volume of dry air that must be supplied at 16°C to the room each second. Also calculate its moisture content. Take the specific heats of dry air and superheated steam as 1.012 and 1.890 kJ kg-1 K_1, respectively, the density of air as 1.208 kg m~3 at 16°C and the latent heat of evaporation as 2454 kJ kg-1 of water.

(.b) Making use of equations (6.6) and (6.8), calculate the supply air quantity in m’3 s-1 and its moisture content in g kg-1.

Answer

From CIBSE tables, the moisture content in the room is found to be 7.376 g kg-1 of dry air at 20°C dry-bulb and 50 per cent saturation.

, v • __________________ 73_________________

*(a) *M (20 — 16) x (1 x 1.012 + 0.007 376 x 1.89)

= 1.779 kg dry air per second v16 = j^|= 1.473 m3 s-1 at 16°C

Latent gain = (1.779 kg dry air per s) x (moisture pick-up in kg per kg dry air) x (2454 kJ per kg moisture)

1.4 = 1.779 x (0.007 376 — gs) x 2454 gs — 0.007 055 kg per kg dry air

*(b) *By equation (6.6)

V _ __7.3__ „ __(273 + 16)__

16 " (20 — 16) 358

= 1.473 m3 s“1 at 16°C By equation (6.8)

„ — 7 !-4 w (273 + 16)

Gs — 7.376 L473 x 856

= 7.376-0.321 = 7.055 g per kg dry air

Posted in Engineering Fifth Edition