Cooling and dehumidification with reheat
As was seen in section 3.4, when a cooler coil is used for dehumidification, the temperature of the moist air is reduced, but it is quite likely that under these circumstances this reduced temperature is too low. Although (as will be seen later in chapter 6), we usually arrange that under conditions of maximum loads, both latent and sensible, the state of the air leaving the cooler coil is satisfactory. This is not so for partial load operation. The reason for this is that latent and sensible loads are usually independent of each other. Consequently, it is sometimes necessary to arrange for the air that has been dehumidified and cooled by the cooler coil to be reheated to a temperature consistent with the sensible cooling load; the smaller the sensible cooling load, the higher the temperature to which the air must be reheated.
Figure 3.10(a) shows, in diagrammatic form, the sort of plant required. Moist air at a state A passes over the finned tubes of a cooler coil through which chilled water is flowing. The amount of dehumidification carried out is controlled by a dew point thermostat, Cl, positioned after the coil. This thermostat regulates the amount of chilled water flowing through the coil by means of the threeway mixing valve R1. Air leaves the coil at state B, with a moisture content suitable for the proper removal of the latent heat gains occurring in the room being conditioned. The moisture content has been reduced from ga to gb and the cooler coil has a mean surface temperature of tc, Figure 3.10(A) illustrating the psychrometric processes involved.
If the sensible gains then require a temperature of rd, greater than tb, the air is passed over the tubes of a heater battery, through which some heating medium such as low temperature hot water may be flowing. The flow rate of this water is regulated by means of a twoport modulating valve, R2, controlled from a thermostat C2 positioned in the room actually being airconditioned. The air is delivered to this room at a state D. with the correct temperature and moisture content.
The cooling load is proportional to the difference of enthalpy between ha and hb, and the load on the heater battery is proportional to hd minus hb. It follows from this that part of the cooling load is being wasted by the reheat. This is unavoidable in the simple system illustrated, and is a consequence of the need to dehumidify first, and heat afterwards. It should be observed, however, that in general it is undesirable for such a situation to exist during maximum load conditions. Reheat is usually only permitted to waste cooling capacity under partial load conditions, that is, the design should be such that state В can adequately deal with both maximum sensible and maximum latent loads. These points are illustrated in the following example.
EXAMPLE 3.10
Moist air at 28°C drybulb, 20.6°C wetbulb (sling) and 101.325 kPa barometric pressure flows over a cooler coil and leaves it at a state of 10°C drybulb and 7.046 g per kg of dry air.
(a) If the air is required to offset a sensible heat gain of 2.35 kW and a latent heat gain of 0.31 kW in a space to be airconditioned, calculate the mass of dry air which must be supplied to the room in order to maintain a drybulb temperature of 21°C therein.


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Airflow





(a) 
9a 
9ь — 9d 
Fig. 3.10 (a) Plant arrangement for cooling and dehumidification with reheat, (b) Psychrometry for Cooling and dehumidification with reheat. 
(b) What will be the relative humidity in the room?
(c) If the sensible heat gain diminishes by 1.175 kW but the latent heat gain remains unchanged, at what temperature and moisture content must the air be supplied to the room?
Answer
(a) If ma kg per s of dry air are supplied at a temperature of 10°C, they must have a sensible cooling capacity equal to the sensible heat gain, if 21°C is to be maintained in the room. Thus assuming the specific heat capacity of air is 1.012 kJ kg1 K“1
Ma x ca x (21° — 10°) = 2.35 kW
Then,
Ma_ 1.012 x511° a211kgsl
(b) 0.211 kg s“1 of dry air with an associated moisture content of 7.046 g per kg of dry air must take up the moisture evaporated by the liberation of 0.31 kW of latent heat. Assuming a latent heat of evaporation of, say, 2454 kJ per kg of water (at about 21°C), then the latent heat gain corresponds to the evaporation of
0 31 kT s_I
—:———— = 0.000 126 3 kg of water per second
2454 kJ kg"1
The moisture associated with the delivery to the room of 0.211 kg s_1 of dry air will increase by this amount. The moisture picked up by each kg of dry air supplied to the room will be
= 0.599 g kg1
Thus, the moisture content in the room will be equal to 7.046 + 0.599, that is, 7.645 g kg1. The relative humidity at this moisture content and 21°C drybulb is found from tables of psychrometric data to be 49.3 per cent. If the relevant data are taken from a psychrometric chart, instead of from psychrometric tables, then similar but slightly different results are obtained, of adequate practical accuracy in most cases. The percentage saturation at 21°C drybulb is then just under 49 per cent.
(c) If 0.211 kg s“1 of dry air is required to absorb only 1.175 kW of sensible heat then, if 21°C is still to be maintained, the air must be supplied at a higher temperature:
1 175
Temperature rise = ^ 012’x 0 211 = ^’5 K
Thus, the temperature of the supply air is 15.5°C.
Since the latent heat gains are unrelated, the air supplied to the room must have the same ability to offset these gains as before; that is, the moisture content of the air supplied must still be 7.046 g per kg of dry air.
EXAMPLE3.il
(a) For sensible and latent heat gains of 2.35 and 0.31 kW, respectively, calculate the load on the cooler coil in example 3.10.
(b) Calculate the cooling load and the reheater load for the case of 1.175 kW sensible heat gains and unchanged latent gains.
Answer
(a) The load on the cooler coil equals the product of the mass flow of moist air over the coil and the enthalpy drop suffered by the air. Thus, as an equation:
Cooling load = 7«a x (h, d — hb)
The notation adopted is the same as that used in Figure 3.10. In terms of units, the equation can be written as
Kw= kg of dry air q
S kg of dry air
Using enthalpy values obtained from tables for the states A and B, the equation becomes
Cooling load = 0.211 x (59.06 — 27.81)
= 0.211 x 31.25 = 6.59 kW
(b) Since it is stipulated that the latent heat gains are unchanged, air must be supplied to the conditioned room at the same state of moisture content as before. That is, the cooler coil must exercise its full dehumidifying function, the state of the air leaving the coil being the same as in example 3.11(a), above. Hence the cooling load is still 6.59 kW. To deal with the diminished sensible heat gains it is necessary to supply the air to the room at a temperature of 15.5°C, as was seen in example 3.10(c). The air must, therefore, have its temperature raised by a reheater battery from 10°C to 15.5°C. Since the moisture content is 7.046 g kg1 at both these temperatures, we can find the corresponding enthalpies at states B and D directly from a psychrometric chart or by interpolation (in the case of state D) from psychrometric tables.
Hence, we can form an equation for the load on the reheater battery:
Reheater load = 0.211 x (33.41 — 27.81) = 1.18 kW
If the relative humidity in the room is unimportant there is no need to supply air to it at a controlled moisture content. This fact can be taken advantage of, and the running costs of the refrigeration plant which supplies chilled water to the cooler coil minimised. The method is to arrange for the room thermostat (Cl in Figure 3.11(a)) to exercise its control in sequence over the cooler valve and the reheater valve (Rla and Rib). As the sensible heat gain in the conditioned space reduces to zero, the threeway mixing valve, Rla, gradually opens its bypass port, reducing the flow of chilled water through the cooler coil and, hence, reducing its cooling capacity. If a sensible heat loss follows on the heels of the sensible heat gain, the need arises for the air to be supplied to the room at a temperature exceeding that maintained there, so that the loss may be offset. The plant shown in Figure 3.11(a) deals with this by starting to open the throttling valve, Rib, on the heater battery.
The consequences of this are shown in Figure 3.11 (ft), as far as a partial cooling load is concerned. Because its sensible cooling capacity has been diminished by reducing the flow of chilled water through its finned tubes, the cooler coil is able to produce an ‘off’ state B at the higher temperature required by the lowered sensible heat gain, without having to waste any of its capacity in reheat. There is a penalty to pay for this economy of control: because of the reduction in the flow rate of chilled water, the characteristics of the coil performance change somewhat and the mean coil surface temperature is higher (tc> is greater than tc), there is less dehumidification as well as less sensible cooling, and, as a result, the moisture content of the air supplied to the room is greater. In consequence, for the same latent heat gain in the room a higher relative humidity will be maintained, the room state becoming R’ instead of R.
EXAMPLE 3.12
If the plant in example 3.10 is arranged for sequence control over the cooler coil and heater battery, and if under the design conditions mentioned in that example the plant is able to maintain 21°C drybulb and 49.3 per cent relative humidity in the conditioned space, calculate the relative humidity that will be maintained there, under sequence control, if the sensible heat gains diminish to 1.175 kW, the latent gains remaining unaltered.
It is given that the mean coil surface temperature under the condition of partial load is 14°C. (See section 10.7 on the partial load performance of cooler coils.)



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R1b 
R1a9. 
□ C1
Conditioned
Room
(a)
O° <§> 
(0
*D
12.1 T O)
10.83
10.01 ® c Ф
7.645 о 7.046 cd
<0
28° 
14° 15.5° 21°
Drybulb temperature °С
(b)
Fig. 3.11 (a) Plant arrangement for heating and cooling in sequence, (b) Psychrometry for heating and cooling in sequence, for Example 3.12.
Answer
As was mentioned in section 3.4, the mean coil surface temperature (identified by the state point C), lies on the straight line joining the cooler coil ‘on’ and ‘off’ states (points A and B) where it cuts the saturation curve. Figure 3.11(b) illustrates this, and also shows that the
Moisture content of state B’ is greater than that of B. It is possible to calculate the moisture content of the supply air by assuming that the drybulb scale on the psychrometric chart is linear. By proportion, we can then assess a reasonably accurate value for the moisture content of the air leaving the cooler coil under the partial load condition stipulated in this example:
8a ~ 8b’ _ tq ~ tb’
8a ~ 8c’ tq ~ tc’
From tables, or from a psychrometric chart, ga = 12.10 g kg 1 and gc’ = 10.01 g kg1. Also, it was calculated in example 3.10(b) that the supply temperature must be 15.5°C dry — bulb for the partial sensible load condition.
Hence we can write
_ / w, (*a h’)
8bf 8a 8a 8c’ ) ^ ^ )
= 12.10 — (12.10 — 10.01) x = 10.23 g kg1
Since the moisture pickup has been previously calculated as 0.599 g kg1, the moisture content in the conditioned room will be 10.83 g kg1; at 21°C drybulb the relative humidity will be about 69 per cent.
Sometimes, to deal with the possibility of high relative humidities at partial load conditions, a high limit humidistat (not shown in Figure 3.11(a)) is located in the treated room, with a set point of, say, 60% ± 5%. Upon rise in humidity to above the upper limit, the humidistat breaks into the temperature control sequence and regulates the cooler coil capacity through Rla (Figure 3.11(a)) in order to dehumidify the air supplied to the room, while allowing temperature control to be retained by Cl through Rib. Thus, for a short time, until the room humidity falls to the lower limit of the humidistat setting, the system operates with conventional reheat. After the humidity has been reduced, the system reverts to the control of the cooler coil and heater battery in sequence.
Posted in Engineering Fifth Edition