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 (north lat.) Kg m 2 6 7 8 9 10 11 N 1 2 3 4 5 North and 500 0.98 0.98 0.98 0.98 0.98 0.98 0.98 0.98 0.98 0.98 0.98 0.98 Shade 150 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 NE 500 0.59 0.68 0.64 0.52 0.35 0.29 0.24 0.23 0.20 0.19 0.17 0.15 150 0.62 0.80 0.75 0.60 0.37 0.25 0.19 0.17 0.15 0.13 0.12 0.11 E 500 0.52 0.67 0.73 0.70 0.58 0.40 0.29 0.26 0.24 0.21 0.19 0.16 150 0.53 0.74 0.82 0.81 0.65 0.43 0.25 0.19 0.16 0.14 0.11 0 09 _ SE 500 0.18 0.40 0.57 0.70 0.75 0.72 0.63 0.49 0.34 0.28 0.25 0.21 1 150 0.09 0.35 0.61 0.78 0.86 0.82 0.69 0.50 0.30 0.20 0.17 0.13 | S 500 0.26 0.22 0.38 0.51 0.64 0.73 0.79 0.79 0.77 0.65 0.51 0.31 S- 150 0.21 0.29 0.48 0.67 0.79 0.88 0.89 0.83 0.56 0.50 0.24 0.16 §- SW 500 0.33 0.28 0.25 0.23 0.23 0.35 0.50 0.64 0.74 0.77 0.70 0.55 150 0.29 0.21 0.18 0.15 0.14 0.27 0.50 0.69 0.82 0.87 0.79 0.60 W 500 0.67 0.33 0.28 0.26 0.24 0.22 0.20 0.28 0.44 0.61 0.72 0.73 150 0.77 0.34 0.25 0.20 0.17 0.14 0.13 0.22 0.44 0.67 0.82 0.85 NW 500 0.71 0.31 0.27 0.24 0.22 0.21 0.19 0.18 0.23 0.40 0.58 0.70 150 0.82 0.35 0.25 0.20 0.18 0.15 0.14 0.13 0.19 0.41 0.64 0.80 North and 500 0.81 0.84 0.86 0.89 0.91 0.93 0.93 0.94 0.94 0.95 0.95 0.95 Shade 150 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 NE 500 0.35 0.45 0.50 0.49 0.45 0.42 0.34 0.30 0.27 0.26 0.23 0.20» 150 0.40 0.62 0.69 0.64 0.48 0.34 0.27 0.22 0.18 0.16 0.14 0.12 3 E 500 0.34 0.44 0.54 0.58 0.57 0.51 0.44 0.39 0.34 0.31 0.28 CfQ 0.24 p 150 0.36 0.56 0.71 0.76 0.70 0.54 0.39 0.28 0.23 0.18 0.15 0.12 1 SE 500 0.29 0.33 0.41 0.51 0.58 0.61 0.61 0.56 0.49 0.44 0.37 0.33 g 150 0.14 0.27 0.47 0.64 0.75 0.79 0.73 0.61 0.45 0.32 0.23 0.18 g S 500 0.44 0.37 0.39 0.43 0.50 0.57 0.64 0.68 0.70 0.68 0.63 0.53 — 150 0.28 0.19 0.25 0.38 0.54 0.68 0.78 0.84 0.82 0.76 0.61 0.42 $ SW 500 0.53 0.44 0.37 0.35 0.31 0.33 0.39 0.46 0.55 0.62 0.64 0.60 " 150 0.48 0.32 0.25 0.20 0.17 0.19 0.39 0.56 0.70 0.80 0.79 0.69 W 500 0.60 0.52 0.44 0.39 0.34 0.31 0.29 0.28 0.33 0.43 0.51 0.57 150 0.77 0.56 0.38 0.28 0.22 0.18 0.16 0.19 0.33 0.52 0.69 0.77 NW 500 0.54 0.49 0.41 0.35 0.31 0.28 0.25 0.23 0.24 0.30 0.39 0.48 150 0.75 0.53 0.36 0.28 0.24 0.19 0.17 0.15 0.17 0.30 0.50 0.66

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 Faзade 07.30 08.30 09.30 10.30 11.30 12.30 13.30 14.30 15.30 16.30 17.3( May N 109 111 116 121 126 129 129 125 120 114 109 NE 263 227 107 163 148 140 139 135 130 122 111 E 344 355 325 255 116 163 150 140 136 127 116 SE 251 301 323 310 265 197 88 122 122 109 98 S 70 80 253 217 248 254 236 195 74 143 79 SW 80 93 104 117 205 222 280 312 310 280 223 W 96 109 120 128 133 141 269 280 334 354 330 NW 92 105 116 123 129 132 132 139 260 243 226 Horiz 349 293 347 382 396 398 390 368 324 269 124 June N 134 128 131 135 140 142 143 140 136 133 129 NE 303 255 124 182 161 152 152 150 146 138 129 E 388 385 343 268 128 173 160 153 149 142 132 SE 272 314 328 311 264 195 90 123 127 116 107 S 74 77 239 209 240 248 229 187 73 137 79 SW 85 97 106 117 198 220 277 309 309 283 229 W 107 119 128 136 140 148 276 289 345 369 350 NW 106 117 127 134 139 141 142 152 284 264 291 Horiz 251 311 360 391 404 406 397 375 334 280 218 July N 117 116 120 125 131 135 135 131 126 121 115 NE 272 233 113 169 153 144 145 140 136 128 116 E 350 354 319 251 119 165 153 144 139 131 120 SE 250 294 310 296 253 107 204 122 123 111 100 S 66 74 231 201 230 238 220 181 69 132 73 SW 80 92 102 115 114 328 267 296 297 271 219 W 96 109 119 127 133 142 265 271 325 346 327 NW 94 107 117 125 131 135 135 142 265 245 269 Horiz 355 294 342 373 385 389 382 360 321 269 126 August N 74 85 91 99 104 108 108 106 99 91 84 NE 225 193 88 130 126 121 122 120 112 104 91 E 314 332 303 238 104 148 135 127 120 111 99 SE 244 298 320 313 270 203 86 127 116 101 88 S 64 84 281 233 262 267 249 210 79 160 83 SW 67 81 92 108 212 224 278 309 299 253 187 W 75 88 99 108 113 122 248 257 300 298 254 NW 75 88 100 108 114 117 118 125 216 197 201 Horiz 278 250 301 340 356 357 346 323 276 214 96 September N 39 51 60 71 79 81 81 76 69 60 51 NE 180 175 158 87 100 97 97 93 86 77 66 E 294 318 288 219 88 127 113 102 95 86 75

( 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 Glazing/blind arrangement (inside-to-outside) Correction factor for stated response Fast Slow Absorbing/blind 0.52 0.55 Blind/clear 1.00 1.03 Blind/reflecting 0.69 0.71 Blind/absorbing 0.75 0.76 Blind/clear/clear 0.95 0.94 Blind/clear/reflecting 0.62 0.62 Blind/clear/absorbing 0.66 0.66 Blind/clear/clear/clear 0.86 0.86 Blind/clear/clear/reflecting 0.55 0.55 Blind/clear/clear/absorbing 0.57 0.56 Blind/low E/clear 0.92 0.92 Blind/low E/reflecting 0.59 0.60 Blind/low E/absorbing 0.63 0.62 Blind/low E/clear/clear 0.84 0.83 Blind/low E/clear/reflecting 0.53 0.53 Blind/low E/clear/absorbing 0.55 0.55 Clear/blind 0.73 0.81 Clear/blind/clear 0.69 0.72 Clear/clear/blind 0.57 0.61 Clear/blind/reflecting 0.47 0.48 Clear/reflecting/blind 0.37 0.38 Clear/blind/absorbing 0.50 0.51 Clear/absorbing/blind 0.38 0.40 Clear/clear/blind/clear 0.56 0.58 Clear/clear/clear/blind 0.47 0.51 Clear/clear/blind/reflecting 0.37 0.38 Clear/clear/reflecting/blind 0.30 0.33 Clear/clear/blind/absorbing 0.39 0.39 Clear/clear/absorbing/blind 0.32 0.34 Low E/clear/blind 0.56 0.58 Low E/reflecting/blind 0.36 0.39 Low E/absorbing/blind 0.39 0.41 Low E/clear/blind/clear 0.55 0.57 Low E/clear/clear/blind 0.48 0.55 Low E/clear/blind/reflecting 0.37 0.38 Low E/clear/reflecting/blind 0.32 0.35 Low E/clear/blind/absorbing 0.39 0.37 Reflecting/blind 0.50 0.53 Air point correction: Internal blind 0.91 0.89 Mid pane blind 0.87 0.83 External blind 0.88 0.85

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