Book - School of Science and Technology

If the windows were weather stripped, the air-change rate might be reduced to 1 with aventilation allowance of 0:33 W/m 3 K. Thus,697 0:33 22 ˆ 5060 Wand, as before,Temperature difference 51(697 0:33)/419 ˆ 0:55 W/m 2 KIt is of interest to compare the magnitude of these two figures with the total conductionlosses for the uninsulated and insulated versions of this building, as calculated previously,i.e. 16 320 and 5280 W respectively. If it be assumed that the windows without weatherstripping are associated with the uninsulated building and vice versa, an overall saving of57% is revealed. Furthermore, the infiltration losses are shown to be 32 and 49% of thetwo respective totals, not insignificant proportions!As an example of application of the use of Table 2.11, take a typical private office,5m 4m 2:5 m high on the third floor of a ten-storey city building (say 25 m high). Thesingle outer wall has glazing on a 1 m module over the full width with a height of 1.5 m.The crack length of 14 m relates to two window modules, not weather stripped. The airto-airtemperature difference is 22 K.From the table:Basic heat requirement ˆ 1:4 W/m KAllowance for stack effect ˆ add 4%Thus, 14 1:4 1:04 22 ˆ 448 WThis requirement may be compared with that which would have resulted from use of anair-change basis:Room volume of 5 4 2:5 ˆ 50 m 3from Table 2.9, take 1 air change or a ventilation allowance of 0:33 W/m K and add 4%for stack effect. Thus,50 0:33 1:04 22 ˆ 378 WIt will be appreciated that a comparison such as this has no particular significance sinceaddition or reduction in the number of openable windows would not alter the roomvolume, nor would an increase or decrease in room depth affect the window crack length.In instances where some doubt may exist as to which method of assessment is the morevalid, both should be evaluated and the higher result used.Temperature differenceAs will have been noted in the various examples so far provided, the total heat required tomaintain a space at the chosen condition is calculated by multiplication of the conductionand air infiltration losses, both in W/K, by a `temperature' difference between inside andoutside. In each of these introductory examples, it was emphasised that air temperaturehad been used as a simplification.