Calculation of heat gain through a wall or roof

The use of analytical methods to determine the heat transfer rate through a wall into a room, in the presence of unsteady heat flow into the outer wall surface, presents considerable difficulties. No exact solution is possible but the CIBSE Guide A5 (1999) offers a method

Table 7.8 Sol-air temperatures at Bracknell (latitude 51°33’N) for the month of July. D and L signify dark and light coloured surfaces, respectively

Sun­

Air

Sol-air

Temperatures, °C

Time

Temperature

Horizontal

North

East

South

West

D

L

D

L

D

L

D

L

D

L

01

14.8

11.5

11.5

12.3

12.3

12.3

12.3

12.3

12.3

12.3

12.3

02

14.2

10.7

10.7

11.5

11.5

11.5

11.5

11.5

11.5

11.5

11.5

03

13.6

10.1

10.1

10.9

10.9

10.9

10.9

10.9

10.9

10.9

10.9

04

13.2

9.6

9.6

10.4

10.4

10.4

10.4

10.4

10.4

10.4

10.4

05

13.3

11.1

10.6

11.8

11.3

11.9

11.3

11.5

11.1

11.5

11.1

06

14.3

18.1

15.0

24.5

19.2

39.9

28.0

14.8

13.8

14.8

13.8

07

15.9

25.3

19.7

23.9

19.9

49.6

34.6

18.4

16.8

18.4

16.8

08

17.7

32.5

24.8

22.1

20.0

53.9

38.2

23.5

20.7

22.1

20.0

09

19.5

38.5

28.8

25.2

22.8

53.6

39.0

33.0

27.2

25.2

22.8

10

21.0

42.7

31.9

27.6

25.0

49.2

37.3

40.4

32.2

27.6

25.0

11

22.5

45.5

34.2

29.6

27.0

42.8

34.4

45.4

35.9

29.6

27.0

12

23.4

46.7

35.3

30.8

28.0

35.0

30.4

47.7

37.6

30.8

28.0

13

24.2

48.0

36.4

31.7

29.0

31.7

29.0

49.0

38.8

36.1

31.4

14

24.8

46.8

35.9

31.7

29.1

31.7

29.1

46.8

37.7

44.3

36.2

15

25.3

44.4

34.8

31.3

29.0

31.3

29.0

42.7

35.3

50.5

39.8

16

25.4

40.5

32.7

30.1

28.1

30.1

28.1

36.5

31.7

53.4

41.3

17

25.0

35.8

29.9

28.2

26.6

28.2

26.6

29.2

27.2

53.1

40.7

18

24.4

30.6

26.7

30.0

27.1

26.1

24.9

26.1

24.9

48.5

37.6

19

23.3

25.2

23.3

29.5

26.2

23.3

22.7

23.3

22.7

39.6

31.9

20

21.7

20.1

19.8

20.6

20.2

20.4

20.1

20.4

20.1

20.6

20.3

21

20.1

17.8

17.6

18.1

18.1

18.1

18.1

18.1

18.1

18.1

18.1

22

18.9

18.1

16.1

16.7

16.7

16.7

16.7

16.7

16.7

16.7

16.7

23

17.5

14.4

14.4

15.2

15.2

15.2

15.2

15.2

15.2

15.2

15.2

24

16.7

13.4

13.4

14.2

14.2

14.2

14.2

14.2

14.2

14.2

14.2

24 hr

Mean

19.6

27.4

22.6

22.4

20.7

27.8

23.8

25.8

22.6

26.5

23.0

Reproduced by kind permission of the CIBSE from their Guide A5 (1999) Thermal response and plant sizing.

Involving the use of sol-air temperatures and related to the concepts of environmental temperature, as well as dry resultant and air temperatures. On an average basis, the mean flow of heat, Qm, through a wall into a room conditioned at a constant temperature, tT, is given by

Qm = AU(tem-tT) (7.22)

Wherein fem is the mean sol-air temperature over 24 hours. If thermal capacity effects were ignored the instantaneous heat gain to the room at time 0 would be given by

Qe = AU(teo — fr) (7.23)

If thermal capacity is considered then a simplified picture of the heat flow into the room at some later time, 0 + (|), where (]) is the time lag, is shown by

(7.24)

подпись: (7.24)2e+* — ?r) + AU(tc o tem)f
where teo is the sol-air temperature at time 0 and in which / is a decrement factor (see Figure 7.16(b)).

Up to 1200 kg/m3 1200 — 1800 kg/m3 1800 — 2400 kg/m3

подпись: up to 1200 kg/m3 1200 - 1800 kg/m3 1800 - 2400 kg/m3

Wall thickness (mm)

Wall thickness (mm)

(b)

(Reproduced by kind permission from the CIBSE Guide)

Fig. 7.16 Approximate time and decrement factors for use with sol-air temperatures.

подпись: wall thickness (mm)
 
wall thickness (mm)
(b)
(reproduced by kind permission from the cibse guide)
fig. 7.16 approximate time and decrement factors for use with sol-air temperatures.
Since most building materials have specific heats of about 0.84 kJ kg-1 K-1 their thermal capacities depend largely on their density and thickness. The CIBSE Guide A5 (1999) derives equations for the calculation of decrement factor, time lag and admittance and tabulates values for a range of common building materials. Figures 7.16(a) and (b) show simple curves allowing an approximate determination of time lags and decrement factors

Which may be used in equation (7.24) to yield answers of practical value.

The CIBSE Guide pursues a more complicated approach because a distinction is made between systems that are controlled to keep dry resultant and air temperatures constant. For all practical purposes of control dry resultant temperature can be ignored, although the CIBSE Guide does not do so. If, as is invariably the case for real-life air conditioning systems, the air temperature is held constant, then the CIBSE Guide introduces two dimensionless factors, Fau and Fay, defined by

(7.25)

(7.26)

подпись: (7.25)
(7.26)
Fm = 4.5X(A)/[4.5 X(A) + I(At/)] Fay = 4.5l(A)/[4.5 1(A) + S(AF)]

In which 4.5 is a heat transfer coefficient and Y is the admittance of a surface.

When the air temperature is held constant in an air conditioned room the equation for the heat gain through a wall or a roof, exposed to the outside, becomes:

(7.27)

подпись: (7.27)Qiho 1-MiA^em 0 F&yAU(te0 tcu]) f

In words, this could be stated as the mean heat transfer over 24 hours, plus the variation about the mean.

For the case of heat transfer through a glass window, which has no thermal inertia and where the heat gains by solar radiation are calculated separately (see section 7.20), equation

(7.17) applies. Similarly, for the case of heat flow through a floor overhanging an open space, such as a car park, the same simple equation applies.

In equation (7.27) Qq+<^ is the heat gain into the room at the time 9 + <j), tem is the 24-hour mean sol-air temperature, tr is the room temperature (dry-bulb) and is assumed to be held at a constant value, feo is the sol-air temperature at a time 0, when the heat entered the outside surface of the wall, (|> is the time lag of the wall and / is its decrement factor.

Note: When using equations (7.25) and (7.26), 1A is the sum of all the internal surface areas of the room. YAU refers to the surface under consideration through which heat is flowing, but Y.(AY) refers to all the internal surfaces of the room.

Equations (7.24) and (7.27) show that Qq+§ is the sum of the mean heat flow over 24 hours and the cyclic input from the inner surface of the wall to the room at the time 0 + <)>. Since this cyclic input depends on the external conditions at the earlier time, 0, its sum with the mean heat flow may be positive or negative, depending on the size and sign of the earlier heat flow into the outer surface. If the wall is very thick, say over 600 mm, with a density of more than about 1000 kg m-3, the decrement factor will be very small and the influence of the second term in equations (7.24) and (7.27) slight. Equation (7.22) then gives a good approximation of the answer. If the wall is very thin and has negligible thermal capacity the heat gain will vary considerably over 24 hours because <|) is small and /is large. For such a case (as with the roof of a factory of lightweight construction) (j> is best taken as zero and f as unity.

EXAMPLE 7.13

Calculate the heat gain at 15.00 h sun-time in July through a light-coloured east-facing wall which is 2.4 m wide and 2.6 m high (floor-to-ceiling). The room is 6.0 m deep and is held at a constant temperature of 22°C dry-bulb. Make use of the following information:

Window: area 2.184 m2, {/-value 5.6 W m 2 K ‘, F-value (admittance) 5.6 W nT2 K_1

Wall: thickness 150 mm, density 1200 kg m 3, {7-value 0.45 W m 2K ’, F-value 3.7 W nT2 K-1 Ceiling: K-value 2.5 W m-2 Kr1 Floor: F-value 2.5 W m“2 K_1

Answer

From Figures 7.16(a) and (b), the time lag is 5 hours and the decrement factor is 0.65. From Table 7.8 the sol-air temperature is 37.3°C at 10.00 h (five hours earlier than the time at which the heat gain is to be calculated) and the 24-hour mean sol-air temperature is 23.8°C. The area of the wall, through which heat is flowing, is 2.4 x 2.6 — 2.184 = 4.056 m2.

1A = (2.4 + 6.0 + 2.4 + 6.0)2.6 + (2.4 x 6.0)2 = 72.48 m2 1(AU) = 4.056 x 0.45 + 2.184 x 5.6 = 14.06 W m“2 K“1

I (AY) = 4.056 x 3.7 + 2.184 x 5.6 + (2.4 x 6.0) x 2 x 2.5

+ [(6.0 + 2.4 + 6.0) x 2.6] x 3.7 = 237.77 W m“2 K"1

By equation (7.25)

Fau = 4.5 x 72.48/(4.5 x 72.48 + 14.06) = 0.959

By equation (7.26)

Fay = 4.5 x 72.48/(4.5 x 72.48 + 237.77) = 0.578

By equation (7.27)

Ј210+5 = 4.056 x 0.45[0.959(23.8 — 22) + 0.578(37.3 — 23.8)0.65] = 12.4 W

Relative to the treated floor area of 14.4 m2 this is 0.9 W irf2. Typically, the maximum sum of all sensible heat gains is about 80 or 90 W nT2 for the sort of office module considered in example 7.13.

Using the simplified version, namely equation (7.24) yields an answer of 19.3 W. Although this is 56 per cent greater than the answer given by equation (7.27) the value of the apparently greater accuracy is questionable. There is probably some uncertainty in the U — values, decrement factor and time lag. There is much additional calculation involved and, furthermore, the heat gain through the wall is usually only about one or two per cent of the total sensible gain. The conclusion could be that the refinement introduced by the inclusion of the factors Fau and Fay is scarcely worth while and that the answers obtained by equation

(7.24) are good enough.

For a roof of lightweight construction and of large plan area, as, for example, that of a hypermarket, the heat gain can be a larger proportion of the total sensible heat gain and hence of more significance. For such a case, a safer answer would be to assume a time lag of zero and a decrement factor of unity. The heat gain is then instantaneous and equation

(7.24) simplifies to (7.23), which could be used.

Note that if the design maximum air temperature is higher than the value tabulated with the sol-air temperatures, the difference should be added to the tabulated sol-air temperatures. The definition of sol-air temperature in equations (7.20) and (7.21) shows the dependence on outside air temperature.

EXAMPLE 7.14

Calculate the heat gain through a light-coloured flat roof of negligible mass at 1300 h sun­time in July, given that the design outside air temperature is 28°C at 1600 h sun-time, the room temperature is 22°C and the [/-value of the roof is 0.45 W m-2 K-1.

Answer

Assume a decrement factor of 1.0 and a time lag of 0 h, because the roof has negligible mass. Add 2.6° to all sol-air temperatures taken from Table 7.8 because the design temperature of 28°C is 2.6° higher than the tabulated value of 25.4°C at 16.00 h. Hence teo at 13.00 h is 36.4° + 2.6° = 39.0°C and tem is 22.6° + 2.6° = 25.2°C.

By equation (7.23)

<2i3 = 1 x 0.45(39.0 — 22) = 7.65 W nT2

It is to be noted that the values of/and <|> represented by the curves in Figure 7.16 are based on work done by Danter (1960) that take account of the characteristics of different wall and roof structures by adopting load curves of different shapes, as indicated by Figure 7.17.

The early analytical work of Mackey and Wright (1944), modified and developed by Stewart (1948), is the basis of the current methods adopted by ASHRAE, involving the use of transfer functions (Mitelas and Stephenson (1967), Stephenson and Mitelas (1967) and Mitelas (1972)). A transfer function is a set of coefficients relating heat transfer into the outer surface of a wall or roof with that entering the room at a later time. ASHRAE propose several methods for the calculation of heat gain through a wall or roof, most being impractical without the use of a computer. One method in the ASHRAE Handbook (1993b), however,

Calculation of heat gain through a wall or roof

Time

Fig. 7.17 Cooling loads for walls of different masses.

Is amenable to manual calculation and expresses heat gains in terms of cooling load temperature differences. Such equivalent temperature differences are tabulated for various typical structures (according to American building practice) and would be applied, as factors, to the product of area and (/-value to yield the heat flow into the room at a specified time.

Posted in Air Conditioning Engineering


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