Heat transfer to ducts
A heat balance equation establishes the change of temperature suffered by a ducted airstream under the influence of a heat gain or loss:
(7.29)
7.21 Heat transfer to ducts 191 Table 7.12 Storage load factors, solar heat gain through glass. 12 hour operation, constant space temperature
Mass per Sun time
Unit area —————
Exposure of floor
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From Handbook of Air Conditioning System Design, by Carrier Air Conditioning Co., copyright 1965 by McGraw-Hill Inc. Used with permission of McGraw-Hill Book Company.
Where Q = heat transfer through the duct wall in W,
P = external duct perimeter in m,
L = duct length in m,
U = overall thermal transmittance in W m-2 K _1,
Table 7.13 Solar cooling loads for SE England (Bracknell, latitude 51°33’N) for fast-responding buildings with the reference glazing type (single clear glass, internal blinds, used intermittently) Month and Solar cooling load at stated sun-time
|
( Contd) |
Month and facade |
Solar cooling load at stated sun-time |
||||||||||
07.30 |
08.30 |
09.30 |
10.30 |
11.30 |
12.30 |
13.30 |
14.30 |
15.30 |
16.30 |
17.30 |
|
SE |
252 |
317 |
341 |
326 |
285 |
224 |
88 |
140 |
101 |
83 |
71 |
S |
46 |
203 |
213 |
263 |
291 |
303 |
288 |
244 |
179 |
65 |
99 |
SW |
51 |
63 |
75 |
96 |
250 |
248 |
304 |
328 |
315 |
357 |
180 |
W |
52 |
64 |
75 |
87 |
95 |
103 |
226 |
240 |
284 |
272 |
220 |
NW |
45 |
57 |
68 |
80 |
88 |
90 |
90 |
89 |
77 |
236 |
148 |
Horiz |
105 |
305 |
244 |
278 |
296 |
305 |
295 |
264 |
217 |
87 |
171 |
Reproduced by kind permission of the CIBSE from their Guide A5 (1999) Thermal response and plant sizing. |
T = initial air temperature in the duct in °C, t2 = final air temperature in the duct in °C, tr = ambient air temperature in °C.
Also, by equation (6.6):
Q = flow rate (m3 s“1) x (+^ ± t2) x
The factor in equation (7.29), which is of considerable importance, is the f/-value. This is defined in the usual way by
Tf ~ rsi + X + rso (7.30)
In this equation the thermal resistance of the metal is ignored and the symbols have the following meanings:
Rsi = thermal resistance of the air film inside the duct, in m2 K W-1 rso = thermal resistance of the air film outside the duct, in m2 K W-1
I = thickness of the insulation on the duct, in metres A = thermal conductivity of the insulation in W m"1 K_1
The value of X is usually easily determined but it is customary to take a value between 0.03 and 0.07. We have here selected a value 0.045 as typical. Small alterations in the value of X are not significant within the range mentioned, but changing the thickness is, of course, very influential in altering the heat gain. Values of rso are difficult to establish with any certainty; the proximity of the duct to a ceiling or wall has an inhibiting effect on heat transfer and tends to increase the value of rso. A value of 0.1 is suggested.
The internal surface resistance, on the other hand, may be calculated with moderate accuracy if the mean velocity of airflow in the duct is known. Theoretical considerations suggest that the value of rsi is a function of the Reynolds number, and experimental evidence suggests that
/)0.25
Rsi = 0.286—jjg-, for circular ducts (7.31)
N ^0J2AB/(A + B)]025 „ rsi = 0.286 Qg, for rectangular ducts (7.32)
Table 7.14 Correction factors for Table 7.13
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Reproduced by kind permission of the CIBSE from their Guide A5 (1999) Thermal response and plant sizing.
Where D = internal duct diameter in m,
V = mean air velocity in a duct in m s-1,
A, B = internal duct dimensions in m.
The change of temperature suffered by the air as it flows through the duct is of prime importance and ASHRAE (1993d) gives an expression for this which yields a positive answer for the case of duct heat loss but a negative answer for duct heat gain:
F _ h (y ~ 1) +
H — (y+ 1} (7-33)
Where
Y = 503pDv/UL for circular ducts (7.34)
Y = 2010pAv/UPL for rectangular ducts (7.35)
In which p is the density of air in kg m-3 and A is the internal cross-sectional area of the duct in m2.
The value of rsi is not sensitive to changes of air velocity for the range of duct sizes and velocities in common use; for example, rsi is 0.0284 m2 K W-1 for 8 m s_1 in a 75 mm diameter duct and 0.0243 m2 K W-1 for 20 m s~’ and 750 mm. Hence the [/-value of a lagged duct is almost independent of the air velocity. From equation (7.31) the U values are 1.47,0.81 and 0.56 W m-2 KT1 for lagging with thicknesses of 25,50 and 75 mm, respectively, assuming rsi is 0.026 m2 K W-1 and X is 0.045 W m-1 KT1. Then by equation (7.34) y has values of 400 DV, 726 DV and 1046 DV, respectively, for the three thicknesses mentioned, if we take p = 1.165 kg m-3 (as typical for air at 30°C) and L = 1 metre.
Equation (7.33) can be rewritten to give the temperature drop or rise per metre of duct length: At
A‘ = »-‘’ = 17TTT <7’36)
Then, for most practical purposes:
At = (?| — tr)/200 Dv for 25 mm lagging (7.37)
At = (t- fr)/363 Dv for 50 mm lagging (7.38)
At = (?] — tr)/523 Dv for 75 mm lagging (7.39)
These equations show that it is important to assess properly the ambient temperature, tr,
And that the temperature drop along the duct is inversely proportional to Dv, being independent of the method of duct sizing adopted.
The equations are easy to use and they indicate that the rate of temperature drop is considerable once the value of Dv falls below about 1.5 m2 s-1 It is usually impractical to attempt to keep Dv above this value by sizing or by increasing the airflow; it follows that the last few lengths of ductwork will suffer a considerable temperature drop—particularly a difficulty with any system using warm air for heating. The only sure way of virtually stopping the drop is to use 75 mm of lagging. Figure 7.20 illustrates the relative merits of different thicknesses of insulation.
EXAMPLE 7.17
Calculate the air temperature rise in a 10 m length of 500 mm diameter duct, lagged with
Dv = duct dia. x mean air velocity Fig. 7.20 Duct heat gain: the effect of lagging in reducing temperature change. |
A thickness of 25 mm, when conveying air at (a) 10 m s-1 and (b) 5 m s’1, given that the air within the duct is initially at 12°C and the ambient air is at 22°C.
Answer
Using equation (7.37) with temperatures reversed because it is a heat gain:
(a) Dv is 0.5 x 10 = 5.0 and
At = 10 x (22 — 12)/200 x 0.5 x 10 = 0.1°
(b) Dv is 0.5 x 5 = 2.5 and
At = 10 x (22 — 12)/200 x 0.5 x 5 = 0.2°
These are not insignificant temperature rises over 10 m of duct. For a long length of duct,
The calculations should be done in short increments of, say 10 m, because the temperature difference, inside to outside the duct, changes with respect to the duct length.
Posted in Engineering Fifth Edition