The choice of a suitable supply state

The choice of the supply air state is constrained by four practical issues:

(i) The desire to minimise the air quantity handled

The supply air temperature should be as low as possible, consistent with providing draught- free air distribution in the occupied space and the required conditions of temperature and humidity. In practice this means that the supply air temperature is usually about 8° to 110 less than the room temperature, although larger differences are sometimes used. (With special attention to air distribution in the conditioned room, supply air temperatures as much as 15° less than the room air temperature have been successfully used in order to minimise the airflow rate.)

Equation (6.6) shows that, to deal with a given sensible heat gain, the supply airflow rate is inversely related to the difference between the room and supply air temperatures. Choosing a lower supply air temperature gives a smaller airflow rate, smaller air distribution system and smaller air handling plant. Fan power will be less with lower running costs, less space will be occupied by ducts and the system will be easier to install.

(ii) Cooler coil selection

The effectiveness of a cooler coil is expressed in terms of its contact factor (see sections

3.4 and 10.4) and this depends on the number of rows of tubes and other factors. An effective cooler coil must therefore be chosen that achieves the design cooling and dehumidification duty (see Figure 6.6) in an economical and practical manner. Air handling plants in the UK commonly use four or six rows for this purpose but, in a hotter and more humid climate, six or eight rows are generally needed. More than eight rows is unnecessary for comfort conditioning because the cooling duty of additional downstream rows progressively decreases, cooler coils being piped for counterflow heat exchange.

The choice of a suitable supply state

<0

Fig. 6.6 The choice of supply air state (see also Example 6.12).

A practical contact factor is between 0.8 and about 0.95, depending principally on the number of rows, although the fin spacing and the face velocity of airflow also play a part (see section 10.4). Reference to Figure 6.6 shows that the contact factor is the ratio of the distance M-W to the distance M-A. W will lie closer to the saturation curve if the contact factor is larger.

(iii)Fan power and duct gain

If the supply fan is located on the downstream side of the cooler coil, as in a so-called draw-through plant, there is a temperature rise after the cooler coil because of fan power. The air will also rise in temperature, before it reaches the conditioned room, because of heat gain to the supply air ducting. Referring to Figure 6.6 it is seen that the supply air temperature, ts, is greater than the temperature leaving the cooler coil, tw, for these two reasons. Similar considerations apply when the fan blows air through a cooler coil, except that fan power then warms the air before it enters the coil and duct gains warm it after leaving the coil.

The supply fan total pressure depends on the velocity of air distribution in the ducts. Conventional, low velocity systems are likely to have fan total pressures of about 600 Pa to 750 Pa, medium velocity about 1000 Pa and high velocity about 2000 Pa, with corresponding temperature rises, as given by equations (6.11) and (6.12).

Duct heat gains often receive inadequate attention and can be surprisingly large (see section 7.21). It is difficult to quote typical values for temperature rise, except to say several degrees are possible.

The air extracted from the conditioned room can also rise in air temperature (tr to tT> in Figure 6.6) and this will affect the cooler coil performance if, as is usually the case, some of the extracted air is recirculated. The temperature rise can be caused by air exhausted from the room through extract-ventilated luminaires (see section 7.23) and also by the extract fan power. A typical temperature rise through extract-ventilated light fittings is about 1° but such fittings are not always desirable with some types of fluorescent tube. Extract duct systems are invariably simpler than supply systems and airflow is usually at low velocity. There are generally no plant items for air treatment. Extract fan total pressure is therefore likely to be about 200 to 300 Pa. Hence a typical temperature rise in the recirculated air is about 0.2° to 1.3°K.

Heat gains to extract ducts can usually be ignored.

(iv) The freezing temperature of water occurring at 0°C

When chilled water is used as the cooling medium the behaviour of the coil depends on the flow temperature of the water and the temperature rise as it flows through the tubes. The mean coil surface temperature of the coil, rsm, is the same as the temperature of the apparatus dew point, ta (see sections 3.4 and 10.3). It follows that the chilled water flow temperature, /chw, will be less than the mean coil surface temperature, as Figure 6.6 shows. A refrigeration plant must therefore be selected that can provide the chilled water flow temperature required, in a stable and safe manner, under all conditions of load. If reciprocating plant, controlled by cylinder unloading from return chilled water temperature, is used (see section 12.10), then a suitable chilled water flow temperature is 6.5°C. If a centrifugal, screw, or other machine is used, controlled from its flow temperature under proportional, plus integral, plus derivative control (see section 13.11), then a suitable flow temperature is 5°C. Manufacturers’ advice should be sought. The lowest, practical dry-bulb temperature leaving a cooler coil is about 10°C to 11 °C, depending on the chilled water flow temperature available. If chilled brine is used, lower temperatures are possible but the use of brine or glycol introduces significant complications, with a possible reduction in heat transfer coefficient.

Figure 6.6 shows the performance of a cooler coil. The ratio of the sensible to total heat gains in the conditioned room is known, so the slope of the design room ratio line can be drawn on the protractor and then transferred to run through the room state, R. If a supply state, S, is to be chosen in conformance with the four principles outlined above, the following procedure is suggested. It is usually carried out for the chosen summer design conditions, at 15.00 h sun time in July.

(1) Knowing the occupancy of the conditioned room, calculate the design minimum supply rate of outside air.

(2) Identify summer design state points O and R on the psychrometric chart.

(3) Knowing the sensible and latent heat gains for the conditioned room, calculate the ratio of the sensible to total heat gain. This is the slope of the summer design room ratio line.

(4) Mark the value of the slope on the psychrometric chart protractor and, using a parallel rule or a pair of set squares, draw the design room ratio line to run through the room state, R. Referring to Figure 6.6, the line 1-2 on the protractor is parallel to the room ratio line through R.

(5) Make a reasonable estimate of the temperature rise due to extract fan power and the use of any extract-ventilated luminaires. Identify the state point R’ on the chart, with Si = 8r’- J°in the points R’ and O by a straight line.

(6) Make a reasonable estimate of the temperature rise due to supply fan power and supply duct heat gain.

(7) Make a first, arbitrary choice of state S and identify this on the design room ratio line. Clearly, S must be to the right of the saturation curve and should be a reasonable choice, bearing in mind the four considerations mentioned earlier. The degree of arbitrariness will depend on experience, but the temperature of S is likely to be 8° or 9° less than the room dry-bulb temperature.

(8) Using equation (6.6) calculate the supply airflow rate.

(9) Knowing the minimum outside airflow rate and the supply airflow rate, determine the proportions of fresh and recirculated air and hence identify the mixture state, M, on the straight line O-R’.

(10) Knowing the temperature rise due to supply fan power and duct heat gain identify the point W on the chart, with gw = Јs-

(11) Join the points M and W by a straight line and extend this to cut the saturation curve at A.

(12) Calculate the contact factor by means of equation (3.3):

O ____ ‘m ‘w

T — t ’

1 m tsm

(13) If the contact factor is reasonable (in the vicinity of 0.85 if a four-row coil is to be used or 0.93 if a six-row coil is likely) accept the choice of S made in (7), above. If the contact factor is not reasonable, or if other possibilities are to be explored, go back to (7) and make another choice of S, using a dry-bulb temperature half a degree warmer or cooler, to suit the circumstances.

(14) When a satisfactory choice of S has been made, calculate the supply air moisture content, gs, by means of equation (6.8). This establishes the moisture content of the air leaving the cooler coil, equal to the supply air moisture content.

EXAMPLE 6.12

The sensible and latent heat gains to a room are 10 kW and 1 kW, respectively. The occupancy of the room is 12 people and the minimum fresh air allowance is 12 litres s-1 for each person. The outside summer design state is: 28°C dry-bulb, 19.5°C wet-bulb (sling), 10.65 g kg-1 dry air, 55.36 kJ kg-1 dry air. The room state is: 22°C dry-bulb, 50 per cent saturation, 8.366 g kg-1 dry air, 43.39 kJ kg-1 dry air. Select a suitable supply air state, specify the states on and off the cooler coil and calculate the design cooling load. Analyse and check the cooling load. The temperature rise through the extract system is estimated as 0.5°C. A low velocity supply system is to be adopted and the temperature rise for supply fan power and duct heat gain is estimated at 2°.

Answer

Following the steps suggested above:

(1) Minimum fresh air = 12 persons x 12 litres s’1 each

= 144 litres s“1

For convenience later (in step 9) it is assumed that this is expressed at the supply air state.

(2) Identify O and R on the psychrometric chart (Figure 6.6).

(3) Sensible/total heat ratio = 10/(10 + 1) = 0.91.

(4) The room ratio line is located on the chart with this slope on the protractor and transferred to run parallel to this through state R.

(5) Extract system temperature rise = 0.5°C. The point R’ is located on the chart with a temperature of 22.5°C and a moisture content of 8.366 g kg-1. O is joined to R’ on the chart.

(6) Supply system temperature rise = 2.0°C.

(7) Make a first arbitrary choice of 14°C dry-bulb for state S.

(8) By equation (6.6)

. 10 000 (273 + 14)

Vl4 = (22 — 14) X———— 358—- = S at

Fresh airflow rate _ 144 _ni/M K } Supply airflow rate “ 1002 “

On-coil state = mixture state M:

Tm = 0.144 x 28 + 0.856 x 22.5 = 23.3°C

Gm = 0.144 x 10.65 + 0.856 x 8.366 = 8.695 g kg"1

By equation (2.24)

Hm = (1.007 x 23.3 — 0.026) + 0.008 695 [(2501 + 1.84 x 23.3)]

= 45.56 kJ kg"1

(10) Off-coil dry-bulb temperature:

Fw = 14° — supply system temperature rise = 12°C

(11) The process line M-W, when projected, cuts the saturation curve on the chart at the apparatus dew point, A, for which the mean coil surface temperature is read from the chart as 10.7°C.

(12) The contact factor = 2^2-1 =

(13) This is a practical contact factor, needing four or six rows of tubes, depending on the fin spacing and the face velocity of airflow. Hence accept the choice of state S, made in (7), above.

„ 1000 v (273 + 14) _ ,

G„ — 8.366 1002 x g56 8.031 g kg

Cooling Load Calculation:

On coil state (M): 23.3°C dry-bulb, 16.32°C wet-bulb (sling), 8.695 g kg-1,

45.56 kJ kg"1

Off coil state (W): 12°C dry-bulb, 11.26°C wet-bulb (sling), 8.031 g kg-1, 32.32 kJ kg-1 (determined from equation (2.24))

From psychrometric tables or, less accurately, from a chart, the specific volume at state S is 0.8235 m3 kg“1.

Cooling load = x [45.56 — 32.32] = 16.11 kW

U. oiJj

10.00

KW

1.00

KW

2.50

KW

0.53

KW

2.09

KW

16.12

KW

подпись: 10.00 kw
1.00 kw
2.50 kw
0.53 kw
2.09 kw
16.12 kw
Analysis and check

Sensible heat gain:

Latent heat gain:

1.2

Supply system gain: Extract system gain:

подпись: supply system gain: extract system gain:X 2 x 358

(273 + 14)

1.2 X 0.856 x 0.5 x 358 (273 + 14)

Fresh air load: n°Q*4i x [55.36 — 43.39] U. o-ZJj

Posted in Air Conditioning Engineering


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