Cooling Load
Examples 6.11 and 6.12 showed how a cooling load is calculated for a simple system and how the load can be broken down into its component parts and the answer checked. In carrying out the check it is important to do so in a way that is as different as possible from the method used to determine the cooling load in the first place. This is straightforward for the sensible and latent heat gains because they were calculated in a different way prior to determining the cooling load. Equation (6.6) can be used to evaluate the contributions by the supply and extract fans, the supply duct gain, and any heat gain to the extracted air as it flows through ventilated luminaires. However it is not possible to assess the fresh air load except by using the enthalpy difference between the ouside air and the room air, applied to the fresh air flowrate. A less satisfactory way, because it is directly duplicating the method used to determine the original cooling load, is to use the difference in enthalpy between the mixed air and the room air, applied to the supply air flowrate. To appreciate the relative proportions of the various components in the cooling load, in a practical case, the design data and results in example 7.18 can be used. This example referred to the west — facing module on an intermediate floor of an office block under typical summer design conditions in London.
EXAMPLE 8.1
Using the assumptions and results of example 7.18(a) calculate the design cooling load. Check the answer and establish the relative percentages of the elements comprising the load. Assume a temperature rise of 2° for the supply fan power and duct gain, and 0.25° for the extract fan power.
Answer
Identify on a psychrometric chart the following state points:
O (outside air): 28°C drybulb, 19.5°C wetbulb (sling)
R (room air): 22°C drybulb, 50 per cent saturation, 8.366 g kg1 R’ (recirculated air): 22.25°C drybulb, 8.366 g kg1
The states O and R’ on the chart are joined by a straight line.
From example 7.18(a):
10.65 g kg
8.832 G kg“1
8.366 g kg“1
7.985 g kg1
J_l____ I__________ I__ 11________ L

Fig. 8.1 Psychrometry for example 8.1. 
Sensible heat gain: 1250 W Latent heat gain: 134 W
Sensible/total heat gain ratio: 1250/(1250 + 134) = 0.90
Mark the value of the sensible/total ratio on the chart protractor and draw the room ratio line parallel to this through the room state, R.
Bearing in mind the methods outlined in chapter 6, an inspection of the psychrometric chart suggests that a supply air temperature of 13.5°C drybulb could be a satisfactory choice. On this basis the following calculations are carried out.
By equation (6.6):
• _ 1250 (273 + 13.5) t
Vl35 _ (22 — 13.5) X 358 — 117.7 litres s
The supply air moisture content is found from equation (6.8):
134 w (273 + 13.5) nnoc, gs = 8.366 — yyyy x ^ = 7.985 g kg 1
Identify the supply air state S, at 13.5°C drybulb and 7.985 g kg1, on the room ratio line.
Identify the offcoil state W, at 11.5°C drybulb and 7.985 g kg1, on the psychrometric chart.
Allow 12 litres s"1 of fresh air for each of the two persons assumed to be in a module. This represents a total of 24 litres s_1 of fresh air. It is convenient to assume that this is measured at the supply air state in order to simplify the calculation of the proportions of fresh and recirculated air handled. Hence:
Fresh air fraction = ^77 = 0204
Recirculated air fraction = 0.796
The mixture state, M, which is also the oncoil state, must now be determined:
Tm = 0.796 x 22.25 + 0.204 x 28 = 23.42°C gm = 0.796 x 8.366 + 0.204 x 10.65 = 8.832 g kg"1
From the chart or, more accurately, by equation (2.24):
Hm = [(1.007 x 23.42 — 0.026) + 0.008 832(2501 + 1.84 x 23.42)]
= 46.03 kJ kg"1
The state M is identified on the chart and M is joined to W by a straight line, which is extended to cut the saturation curve at the apparatus dew point, A. The temperature of A is found to be 10.4°C. The practicality of the proposed choice of the supply state S is now established by determining the contact factor of the cooler coil, using equation (3.3):
(23.42 — 11.5)
P (23.42 — 10.4)
This implies a sixrow coil, with 315 fins per metre (8 per inch) and a face velocity of about 2.5 m s1 (see section 10.4). This is quite practical and so the choice of 13.5°C drybulb for the supply air temperature is accepted.
The specific volume at the supply state, S, is found from the chart or, more accurately, from psychrometric tables, to be 0.8220 m3 kg1. Similarly, the enthalpy at the offcoil state, W, is determined as 31.71 kJ kg1. The design cooling load can now be calculated.
Design cooling load = 0^? (46.03 — 31.71) = 2.050 kW

The cooling load is analysed into its component parts and, by evaluating the contribution of each, a check on the answer is obtained:
Sensible heat gain:
Latent heat gain:
Supply fan power and duct gain (equation (6.6))
= (2 x 117.7)(358)/(273 + 13.5) =
Recirculation fan power (equation (6.6))
= (0.25 x 117.7 x 0.796) x 358/(273 + 13.5) =
Fresh air load = (0.024/0.8220)(55.36 — 43.39) x 1000 =
Total
Although the sensible heat gain dominates the load a sizeable part is due to the fresh air component. This provides a good indication of the time of the day and month of the year for which the cooling load will be a maximum. The fresh air load is greatest when the outside air enthalpy is at its highest value and, in most countries in the northern hemisphere, this is at about 15.00 h, suntime, in July. There are occasional exceptions but it is rare for
The maximum load to be earlier than 13.00 h, or later than 17.00 h. Sometimes the maximum is in August.
The end of example 7.18 compared the solar gain through glass from four different sources and the above example used data from Tables 7.9 and 7.10. If, instead, solar gain data in Tables 7.13 and 7.14 from the CIBSE Guide A2 (1999) is used, the sensible heat gain increases by about 8 per cent from 1250 W to 1352 W, requiring a similarly larger supply airflow rate of 127.3 litres s~*. This has no effect on the fresh air load and the latent heat gain. Consequently, the cooling load does not increase as much, rising only by about 6 per cent in the simple case considered, because of the extra fan power needed for the increased airflow rate. Such an increase is not insignificant because it affects the selection of the refrigeration plant and its ancillaries, the size of air handling plants, duct systems and the amount of building space used. Capital and running costs will increase.
Posted in Air Conditioning Engineering
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