Heat transfer to ducts

Heat transfer to ductsA 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

~U ~ 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 nT1 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 = (t — 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

Heat transfer to ducts

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 Air Conditioning Engineering


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