Building Services Engineering 5th Edition Handbook

Heating 107
and,
( )
t ai − 5 82 − 1.3 tai
23 − (−2) = 85 − 23
where t ai is the expected internal air temperature on the test day. Then:
t ai − 5
25
( ) 82 − 1.3 tai
=
62
Assume a range of values for the internal air temperature of say 30 ◦ C, 25 ◦ C and 20 ◦ C.
Evaluate each side of the equation (Table 4.5) and plot against t ai (Fig. 4.25). The graphical solution
reveals that an internal air temperature of 26.6 ◦ C satisfies both equations. This can be verified
by substituting 26.6 ◦ C for t ai in the heat balance equation. Thus the measured temperature on
the test day is sufficiently close to the theoretically expected figure to say that the heating system
meets its design specification.
Table 4.5 Radiator heat output test in Example 4.3.
t ai ◦ C Heat loss Radiator heat output
(t ai − 5)/25 (82 − t ai )/62 [(82 − t ai )/62] 1.3
30.0 1.0 0.839 0.80
25.0 0.8 0.919 0.90
20.0 0.6 1.000 1.000
1.0
0.9
Radiator heat output
Heat flow
0.8
0.7
Room heat loss
0.6
20 25
26.6
Room air temperature t ai (°C)
30
4.25 Variation of room heat loss and radiator heat output with room air temperature in Example 4.3.