Load diagrams

Diagrams can be drawn to give a useful picture of seasonal load variations for multi-

Load diagrams

(a)

Load diagrams

Tc tr

= tf

(b)

Fig. 8.10 Plant layout and psychrometry for a conventional double duct system.

Roomed buildings with several orientations. The sensible heat gain to a room comprises: transmission (7) by air-to-air temperature difference, heat gain from lights (L), heat gain from people (P), heat gain from business machines (A/) and solar heat gain through glass (S). For purposes of this simple analysis solar heat gains through walls are ignored. The construction of such load diagrams is conveniently illustrated by the two following examples.

EXAMPLE 8.5

Construct a load diagram for the west-facing module forming the subject of example 7.18, assuming that the natural infiltration rate stays constant at 0.5 air changes per hour throughout the year and that the design outside air temperature in January is -2°C. Take the room temperature as 22°C in July and 20°C in January. Make use of Tables 7.9 and 7.10, instead of CIBSE data, for solar gain through glass, Assume metal-framed single glazing, protected by internal Venetian blinds, and a building surface density of 150 kg m-2. For simplicity, and without significant error, ignore the effects of solar gain through the wall and hence use equation (7.17) for both the glass and the wall.

Answer

Table 5.2 shows that the maximum temperature is 26.9°C in July at 15.00 h, sun-time, with a diurnal range of: 21.8 — 13.5 = 8.3°. Allow 1.1° for the heat island effect in the middle of London (see section 5.11). Hence take the temperature as 28°C at 15.00 h sun-time in July. On the other hand, Table 7.9 shows that the maximum solar gain through glass is at

17.0 h sun-time in July. This indicates that the maximum sensible gain will almost certainly be at 17.00 h sum-time in July. (This could be verified by separate calculation.) On this basis it is necessary to determine the outside air temperature at 17.00 h, using equation (5.4):

T„ = 28 — (8.3/2)[l — sin{(1 7tt — 9rc)/12}]

= 27.4°C

Hence the sensible heat gains are calculated at 17.00 h in July:

Watts

Glass (equation (7.17)): 2.184 x 3.0 x (27.4 — 22)

= 35

Wall (equation (7.16)): (3.3 x 2.4 — 2.184) x 0.45 x (27.4 — 22)

= 14

Infiltration (equation (7.41)): 0.33 x 0.5 x (2.6 x 2.4 x 6.0) x (27.4

-22) = 33

T= 82

People: 2 x 90

P= 180

Lights: 17 x 2.4 x 6.0

L= 245

Machines: 20 x 2.4 x 6.0

M= 288

Solar (equation (7.41)): 2.184 x 1.0 x 268

S= 585

T + P + L + M + S = 1380

This allows the points 1, 2, 3, 4 and 5 to be plotted on Figure 8.11.

From T we can determine that ‘L(ALF) = 82/5.4 = 15.2 W K-1. Hence the heat loss in January is 15.2 x (20 + 2) = 334 W. This establishes the point 6 on the figure. (Note that the heat loss is less than was calculated in the answer to example 8.3 because that was based on an infiltration rate of one air change per hour in winter. As a simplification, the infiltration is kept constant at half an air change per hour in this example.) A line joining the points 1 and 6 is drawn and this represents the transmission load, T, against outside air temperature. The contributions of P, L and M are constants and load lines representing T+ P, T + P + L and T + P + L + M are drawn parallel to the line for T.

The contribution of the solar gain to the diagram needs some thought. The diagram is intended to show seasonal rather than diurnal variations of maximum load and hence the peak solar gains for each month of the year are relevant. The mean monthly maximum temperatures are not directly related to solar gains and they are not indicative of winter conditions, but they do offer a way of associating solar gains with outside air temperature and so drawing a line on the figure for the solar contribution. Reference to Tables 5.2 (for temperatures plus 1.1° for the heat island effect) and 7.9 leads to the following tabulation, using equation (5.4) to determine txl.

Month

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

?s

104

174

228

259

268

268

268

259

228

174

104

79

S

227

380

498

566

585

585

585

566

498

380

227

173

Anax

12.8

13.2

16.6

19.8

24.4

27.0

28.0

27.3

24.5

19.8

15.5

13.3

D

4.1

4.7

6.8

7.8

8.5

8.7

8.3

8.2

7.2

6.3

4.8

3.8

Hi

12.5

13.0

16.1

19.3

23.8

26.4

27.4

26.8

24.0

19.4

15.2

13.0

Where qs — specific solar gain (Table 7.9), W m-2

S = solar gain at 17.00 h sun-time for west-facing glass, W? max — mean monthly maximum temperatrue (Table 5.4), plus an allowance of 1.1° for the heat island effect, °C D = diurnal range (Table 5.2), K

R17 = outside air temperature at 17.00 h (equation (5.4)), °C

Plotting these values of the solar component at 17.00 h for a west-facing window gives the points through which a curved broken line runs in Figure 8.11. This is unsatisfactory. To deal with the difficulty it is suggested that the solar gain for January should be associated with the outside design temperature for January, namely, -2°C. Referring as necessary to the answer of example 8.5, the following establishes the point 7 in Figure 8.11.

P + L + M = 180 + 245 + 288 = 713

T (Jan) = 15.2 x (-2 — 20) = -334

5 (Jan) = 227

T+P + L + M + S (Jan) = 606 W net heat gain.

The points 5 and 7 are joined to give the load line for T + P + L + M + S.

Although clear skies and sunshine are associated with cold, dry weather in winter, the altitude of the sun is low and cloud cover prevails for much of the time, making an accurate forecast of the solar gain impossible. For this reason, the simple straight lines in Figure

8.11 Are proposed as a reasonable load diagram.

It is to be noted that there are three reasons why there is a significantly high sensible heat gain in January:

(i) The (7-values for the walls and glass are low.

(ii) The infiltration rate is only 0.5 air changes per hour (as might be expected with a tight building).

(iii) Business machines are liberating 20 W m’2. (This could easily be greater.) EXAMPLE 8.6

Construct a load diagram for the module in example 8.5, assuming a southerly orientation.

Load diagrams

Answer

Reference to Table 7.9 shows that the peak solar gain for a southern aspect is 284 W itT2, occurring at 12.00 h sun-time in October/February. In fact, for a lightweight building, the maximum solar gain nearly always occurs at mid-day, for a south-facing window. This complicates the construction of a load diagram and an investigation must be made of the heat gains by transmission (7) and solar gain (S). Since the room temperature is allowed to fall from 22°C in July to 20°C in January, account must be taken of this and it is proposed that a simple proportional change is used. Knowing that ‘L(AU) for the module is 15.2 W nT2 and adding 1.1° to the values of mean monthly maximum temperature taken from Table 5.2, the following tabulation is drawn up, using equation (5.4) as necessary to determine temperatures at mid-day.

Month

T,

°C

Mean monthly max. temp at 15.00 h °C + 1.1°

To

At 12.00 h

°C

T

W

9S

At 12.00 h W m-2

S

W

T+P+L+M+S

W

Dec

20.3

13.3

12.7

-116

240

524

1121

Jan

20

12.8

12.2

-119

262

572

1166

Feb

20.3

13.2

12.5

-119

284

620

1214

Mar

20.7

16.6

15.6

-78

268

585

1220

Apr

21

19.8

18.7

-35

234

511

1189

May

21.3

24.4

23.2

+29

180

393

1135

Jun

21.7

27.0

25.7

+61

158

345

1119

Jul

22

28.0

26.8

+73

180

393

1179

The points 1, 2, 3 and 4 are plotted in Figure 8.12, in the same way as before. The maximum value for T + P + L + M + S is 1179 W in July and this identifies the point 5, against an outside air temperature of 26.8°C at 12.00 h. If the contribution by P + L + M is smaller, the solar gain is more dominant and the maximum value for T + P + L + M + S is in February or March, when the solar gain, S, is a larger proportion of the total. This is seen in Figure 8.12 by the plot of solar gains alone, emphasised by the curved broken line.

Load diagrams

The lowest winter solar gain is in December. If it is assumed that, on a cold clear day in this month, the outside air temperature is -2°C, then the December solar gain can be added to the load line for T + P + L + M at —2°C, to give the point 7 on the diagram.

The peak value for the load diagram will occur when the maximum value of the solar gain is added to T + P + L + M. This is done in Figure 8.12 by adding the value of S, in February, to the value of T + P + L + M against an outside air temperature of 12.5°C (for February). In Figure 8.12 this identifies the point 8 (with a cooling load of 1214 W) and the load line for T + P + L + M + S is established by joining the points 5, 8 and 7.

Exercises

1. Air enters a cooler coil at a state of 32°C dry-bulb, 20.7°C wet-bulb (sling) and

Leaves at 11°C dry-bulb, 10.3°C wet-bulb (sling). The airflow rate at entry to the coil is 4.72 m3 s“1. Determine: (a) the cooling load in kW, (b) the rate of moisture condensation

In litres per minute, (c) the airflow rate leaving the coil in m3 s“1.

Answers

(a) 155.2 kW, (b) 0.942 litres per minute, (c) 4.37 m3 s“1.

2. An air conditioning plant supplies air to a room at a rate of 1.26 kg s“1, the supply air

Being made up of equal parts by weight of outdoor and recirculated air. The mixed air passes through a cooler coil fitted with a by-pass duct, face-and-by-pass dampers being used for humidity control. The air then passes through an after-heater and supply fan. In a test on the plant, the following conditions were measured:

Dry-bulb Moisture content Enthalpy

°C g kg“1 kJ kg“1

TOC o "1-5" h z Outdoor air 27 12.30 58.54

Room air 21 7.86 41.08

Supply air 16 7.07 33.99

Air leaving cooler coil 8 6.15 23.51

Make neat sketches showing the plant arrangement and the psychrometric cycle and identify corresponding points on both diagrams. Show on the plant arrangement diagram the controls required to maintain the desired room condition.

Ignoring any temperature rise through the supply fan calculate the mass flowrate of air through the cooler coil, and the cooler coil and heater battery loads, in kW.

Answers

0.963 kg s“1, 25.3 kW, 5.43 kW.

3. The air in a room is to be determined at 22°C dry-bulb and 50 per cent saturation by air supplied at a temperature of 12°C. The design conditions are as follows:

Sensible heat gain 6 kW Latent heat gain 1.2 kW

Outside condition 32° dry-bulb, 24° wet-bulb (sling)

The ratio of recirculated air to fresh air is fixed at 3:1 by weight. The plant consists of a direct expansion cooler coil, a reheater and a constant-speed fan. Allowing 1°C rise for fan power etc., calculate:

(a) The supply air quantity in m3 s“1 and its moisture content in g kg“1.

(.b) The load on the refrigeration plant in kW of refrigeration.

(c) The cooler coil contact factor.

Answers

(i) 0.478 m3 s“1, 7.53 g per kg, (ii) 11.96 kW, (iii) 0.86.

4. A room is air conditioned by a system which maintains 22°C dry-bulb with 50 per cent saturation inside when the outside state is 28°C dry-bulb with 19.5°C wet-bulb (sling), in

The presence of sensible and latent heat gains of 44 kW and 2.9 kW respectively. Fresh air flows over a cooler coil and is reduced in state to 10°C dry-bulb and 7.046 g per kg. It is then mixed with recirculated air, the mixture being handled by a fan, passed over another cooler coil and sensibly cooled to 13°C dry-bulb. The air is then delivered to the conditioned room.

If the fresh air is to be used for dealing with the whole of the latent gain and if the effects of fan power and duct heat gain are ignored, determine the following.

(a) The amount of fresh air handled in m3 s“1 at the outside state and in kg s’1.

(.b) The amount of air supplied to the conditioned space in m3 s”1 at the supply state and in kg s’1

(c) The dry-bulb temperature, enthalpy and moisture content of the air handled by the fan.

(id) The load, in kW, involved in cooling and dehumidifying the outside air.

(.e) The load in kW of refrigeration on the sensible cooler coil.

Answers

(a) 0.7725 m3 s’1, 0.891 kg s’1, (b) 3.91 m3 s’1, 4.76 kg s’1, (c) 19.8°C, 40.46 kJ kg’1, 8.118 g per kg, (d) 24.5 kW, (e) 34.1 kW.

Notation

Symbol

Description

Unit

A

Area

M2

D

Diurnal range

K

Fp

Factor for pump power and chilled water pipe heat gain

8b

Moisture content at state B

G kg’1

8c

Moisture content at state C

Gkg 1

8m

Moisture content at state M

G kg 1

Go

Moisture content at state 0

Gkg 1

8r

Moisture content at state R

G kg 1

8t’

Moisture content at state R’

Gkg“1

8s

Moisture content at state S

Gkg 1

8w

Moisture content at state W

Gkg 1

<?w’

Moisture content at state W’

Gkg 1

Hh

Enthalpy at state B

KJ kg“1

Hc

Enthalpy at state C

KJkg“1

K

Enthalpy at state M

KJ kg 1

K

Enthalpy at state S

KJ kg’1

L

Sensible heat gain from lighting

W

M

Sensible heat gain from business machines

W

P

Sensible heat gain from people

W

Qfa

Fresh air load

W

Q

Latent heat gain

KW

Qr

Design refrigeration load

KW

Qra

Return air load

KW

Qih

Reheat load

KW

Qs

Sensible heat gain

KW

Supply duct heat gain kW

Qsd

Qsf

9s

5

T

*c

*max

H

*r’

*s

H’

Hm

*sm’

Tw

Fw’

U

Vt

P

подпись: qsd
qsf
9s
5
t
*c
*max
h
*r'
*s
h'
hm
*sm'
tw
fw'
u
vt
p
Supply fan power kW

Specific solar heat gain W irf2

TOC o "1-5" h z sensible heat gain from solar radiation through glass W

Transmission heat gain through the building envelope, including infiltration, by virtue of air-to-air temperature difference W

Temperature at state C °C

Temperature at state M °C

Mean monthly maximum temperature plus 1.1 K for the heat island effect in a built-up area °C

Outside air temperature at state O °C

Room temperature at state R °C

Room temperature at state R’ °C

Supply air temperature at state S °C

Supply air temperature at state S’ °C

Mean coil surface temperature for state A °C

Mean coil surface temperature for state A’ °C

Off-coil temperature for state W °C

Off-coil temperature for state W °C

Overall thermal transmittance coefficient W irT2

Volumetric airflow rate at a temperature t m3 s-1 or

Litres s’1

Contact factor of a cooler coil —

Posted in Air Conditioning Engineering


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