Cooling load offset by reheat

With a constant volume system, the room temperature will fall as the sensible heat gains diminish, unless the cooling capacity of the air supplied is also reduced: the sensible cooling capacity of the supply air must equal the sensible heat gain in the conditioned room. Similarly, with industrial applications, the moisture content of the supply air must be changed to achieve a match between the latent cooling capacity of the supply air and the latent gain in the room.

One simple, but inherently wasteful, method of achieving a match between sensible cooling capacity and sensible heat gain is to reheat the air leaving the cooler coil to a higher temperature, before it is supplied to the room.

EXAMPLE 8.2

If the sensible heat gains used in example 8.1 reduce to 500 W, other data remaining unchanged, calculate the necessary supply air temperature and reheat duty to maintain the room temperature at its design value.

Answer

The supply fan continues to handle 117.7 litres s_1 at a temperature of 13.5°C hence, by equation (6.6):

,- = 22-i#7 x(2733583’5) = I8-6°C The reheat load should be calculated by using the conventional method (refer to Figure 8.2):

Reheat duty = x (38.95 — 33.74) = 0.746 kW

Reheater

подпись: reheaterCooler

Coll

©

M

Cooling load offset by reheat

©

 

Fan

 

To room

 

10.65 Cooling load offset by reheatG kg“1

8.832 g kg"1

8.366 g kg"1

7.985 g kg-1

O o o o o o. .

Tf in in CD Odin CO 00

O i — CO

подпись: o i- coCMCNJ Tf CM oJ CO CM CM

Fig. 8.2 Psychrometry for example 8.2.

Alternatively, equation (6.6) could be used:

I, u ^ * 0.177 x (18.6 — 13.5) x 358

Reheat duty =———— (273TT33)————— = 0J50 tW

(Note that it is really unnecessary to carry out a calculation in this case since, with a reheat system, the reduction in sensible gain is made good by reheat, hence the reheat duty is 1250 W — 500 W = 750 W.)

The process is illustrated in Figure 8.2. The cooling load is unchanged and the surplus sensible cooling capacity is cancelled by wasteful reheating. It is seen that the room ratio line has a steeper slope for the partial load condition because the sensible heat gains have fallen while the latent heat gains are unchanged.

A single air handling plant can deliver air through multiple reheaters to several rooms. The common cooler coil in the air handling unit cools and dehumidifies all the air to a nominally constant value and temperature control is achieved independently in each room by thermostatic control over the individual reheaters. If the latent heat gains change in the rooms then there will be some variation in the humidity because the mosture content delivered to each room will be at a nominally constant value. For comfort conditioning this will not matter since the dry-bulb temperature is under control and humidity is not very important for human comfort.

For industrial conditioning, on the other hand, control over humidity as well as temperature may be important. In this case it becomes necessary to provide independent dry steam humidifiers for the air supply to each treated room, as well as reheaters. The cooler coil in the central air handling plant cools and dehumidifies all the air down to the lowest necessary moisture content (for supply to the room with the largest latent heat gain). Dry steam humidification is then provided for rooms where this supply air moisture content is too low to deal with the latent heat gains. Meanwhile, the reheaters control room temperatures, independently. The air conditioning process is wasteful in two respects, humidity and temperature, but this does not matter because the industrial process is the prime concern.

Where dry steam humidification is provided for individual rooms the steam should be delivered into the air stream in a place where the air is not near to saturation. This is usually as close as possible to the treated room itself, after any reheater.

There are many ways in which the required supply air state can be achieved for a partial load condition and some of these are considered in the following sections. It might be thought that the cooler coil capacity could itself be reduced so that the air leaving the coil had the correct temperature. This is possible and can be done, but simple conclusions cannot be drawn because the behaviour of cooler coils is complicated. This is dealt with in chapter 10. Figure 8.3 illustrates one important aspect of cooler coil performance (also discussed in section 3.4). The process of cooling and dehumidification from state M to state W’ in Figure 8.3(a) is not possible. For such a process to be possible the line MW’, if extended, must cut the saturation curve. That is to say, there must be an apparatus dew point, A. Figure 8.3(b) shows a process that is possible.

Reheat and humidification duties must be worked out for the most exacting condition. If the reheater is the only source of heating for the room then it must be able to cope with the heat loss for the design winter state. The humidifier must be able to make good the latent loss that can arise from infiltration in winter, no benefit being allowed for the latent heat gain from people.

EXAMPLE 8.3

Determine the design reheat and dry steam humidification loads for the module forming the subject of examples 7.18 and 8.1. Assume that motorised, automatically controlled dampers are used to provide a mixed air temperature of 11.5°C in winter design conditions, the cooler coil not then being in operation. The winter inside design condition is 20°C dry — bulb, 50 per cent saturation, 7.376 g kg“1, 38.84 kJ kg-1 and the outside design condition is -2°C saturated, 3.205 kJ kg-1, 5.992 kJ kg-1. A natural infiltration rate of one air change per hour is assumed.

Answer

See Figure 8.4. Using equations (7.17) and (7.41) the heat loss from the module is calculated as follows.

Cooling load offset by reheat

(a)

Re-heater

Battery

подпись: re-heater
battery
Cooling load offset by reheatCooler

Coil

M

подпись: mM

V

(b)

Fig. 8.3 Impossible and possible psychrometry.

= 144 W = 57 W = 272 W

подпись: = 144 w = 57 w = 272 wGlass: 2.184 x 3.0 x (20 + 2)

Wall: (3.3 x 2.4 — 2.184) x 0.45 x (20 + 2)

Infiltration: 0.33 x 1.0 x (2.4 x 2.6 x 6.0) x (20 + 2)

Total:

подпись: total:473 W

The supply air temperature needed to offset this is determined from equation (6.6):

Cooling load offset by reheat

CM

I

In 10 T™ CO

O LO CVJ CVJ © CVJ

Fig. 8.4 Psychrometry for example 8.3. The enthalpy at R is almost the same as at C by coincidence.

‘•=20° + mi X = 23.2.c

The motorised mixing dampers in the air handling plant adjust the proportions of fresh and recirculated air to give a mixed temperature of 11.5°C. The fan power then causes this to rise to 13.5°C. This value is used for the temperature in the bracket in the above calculation because the reheater is after the fan. Hence the reheater must warm the air from 13.5°C to 23.2°C.

Referring to Figure 8.4, it is necessary to establish the enthalpies of the states B and C, and the specific volume at state B (for the airflow handled by the supply fan).

— 3 205 I (7-376 — 3.205) X (11.5 + 2)

8c gb 8m J-AO + (2Q ^ + ^

= 5.736 g kg“’

Since the temperature of B is 13.5°C its enthalpy can be determined from tables, or from a chart, or by equation (2.24):

Hb = (1.007 x 13.5 — 0.026) + 0.005 736(2501 + 1.84 X 13.5)

= 28.06 kJ kg“1

The temperature at state C is 23.2° and the enthalpy is established in a similar way:

Hc = (1.007 x 24.1 — 0.026) + 0.005 736(2501 + 1.84 x 23.2)

= 38.84 kJ kg“1

By coincidence, this happens to be almost the same as the enthalpy of the air at the room state and we shall regard it as such, as a simplification.

From psychrometric tables or a chart the specific volume at В is 0.8191 m3 kg“1. This is for the volumetric airflow rate of 117.7 litres s_1, handled by the fan.

The reheater duty is then

X (38.84 — 28.06) = 1.549 kW

At an outside winter design condition of -2°C, saturated, the moisture content is 3.205 g kg-1. One air change per hour of infiltrating air at this moisture content will represent a latent heat loss, it being necessary to assume the absence of people under the most exacting conditions. At the winter design state, the moisture content in the rooms is 7.376 g kg-1.

Hence, the duty of the dry steam humidifier is to raise the moisture content of the supply air to a value higher than 7.376 g kg“1, in order to counter the latent loss, which must be calculated.

By equation (7.42)

Latent loss = 0.8 x 1 x (2.4 x 2.6 x 6.0) x (7.376 — 3.205)

= 125 W

Then, by equation (6.8), the necessary supply air moisture content is

125 (273 + 13.5) _i

Gs = 7.376 + уYPj x——————————————— 856 = 7/731 g k§

The moisture content of the mixed air, at 11.5°C in the air handling plant, has been determined as 5.736 g kg“1. Hence the duty of the dry steam humidifier is to raise the moisture content of air supplied to the room from 5.736 g kg-1 to 7.731 g kg-1.

Assuming that dry steam humidification occurs up a dry-bulb line (see section 3.7), the state of the supply air, S, is 23.2°C dry-bulb and 7.731 g kg-1. Its enthalpy can be found from psychrometric tables, or from the psychrometric chart, or by equation (2.24):

A, = (1-007 x 23.2 — 0.026) + 0.007 731(2501 + 1.84 x 23.2)

= 43.00 kJ kg“1

The dry steam humidification duty is

0.1177

0.8191

подпись: 0.8191(43.00 — 38.84) = 0.598 kW

Posted in Air Conditioning Engineering


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