Добавил:
Опубликованный материал нарушает ваши авторские права? Сообщите нам.
Вуз: Предмет: Файл:

01 POWER ISLAND / 01 CCPP / V. Ganapathy-Industrial Boilers and HRSG-Design (2003)

.pdf
Скачиваний:
43
Добавлен:
23.06.2022
Размер:
3.75 Mб
Скачать

Figure 8.6 Solid and serrated fins.

20 Btu=ft2 h F. A large fin density or a large ratio of external to internal surface area is justified in this case. As the ratio between the outside and inside coefficients decreases, the effectiveness of using a large ratio of external to internal surface areas decreases. For example, in superheaters or high pressure air heaters, where the tube-side coefficient could be in the range of 30– 300 Btu=ft2 h F, it does not pay to use a large fin surface; in fact, it is counterproductive, as will be shown later. A moderate fin density such as two or three fins per inch would be adequate, whereas for economizers or evaporators, five or even six fins per inch may be justified if cleanliness permits.

The other important fact to be kept in mind is that more surface area does not necessarily mean more energy transfer. It is possible, through poor choice of fin configuration, to have more surface area and yet transfer less energy. One has to look at the product of surface area and overall heat transfer coefficient and not at surface area alone. The overall heat transfer coefficient is significantly reduced as we increase the fin surface or use more fins per inch.

Finned tubes offer several advantages over bare tubes such as a compact design that occupies less space, lower gas pressure drop, lower tube-side pressure drop due to the fewer rows of tubes, and smaller overall weight and cost.

Solid fins offer slightly lower gas pressure drop than serrated fins, which have a higher heat transfer coefficient for the same fin density and configuration. Particulates, if present, are likely to accumulate on serrated finned tubes, which may be difficult to clean.

Copyright © 2003 Marcel Dekker, Inc.

8.19a

Q:

How are the heat transfer and pressure drop over finned tubes and tube and fin wall temperatures evaluated?

A:

The widely used ESCOA correlations developed by ESCOA Corporation [9] will be used to evaluate the heat transfer and pressure drop over solid and serrated finned tubes in in-line and staggered arrangements. The basic equation for heat transfer coefficient with finned tubes is given by Eq. (3).

The calculation for tube-side coefficient hi was discussed earlier. ho consists of two parts, a nonluminous coefficient hN ; which is computed as discussed in Q8.07, and hc, the convective heat transfer coefficient. Computation of hc involves an elaborate procedure and the solving of several equations, as detailed below.

Determination of hc

[9]

 

 

 

 

 

 

 

 

d

2 h

 

0:5

tg þ 460

0:25

hc ¼ C3C1C5

 

 

þ

 

 

 

 

 

 

 

d

 

ta þ 460

 

 

GCp mCp!

0:67

 

 

 

 

 

 

k

 

 

 

 

 

 

 

 

 

 

 

 

G ¼

 

 

 

Wg

 

 

 

 

 

 

 

½ðST =12Þ Ao&NwL

 

 

d

nbh

 

 

Ao ¼

 

 

þ

 

 

 

 

 

 

 

12

6

 

 

 

 

 

C1; C2; and C3 are obtained from Table 8.10.

Gd

Re ¼ 12 m

s ¼ 1n b

Fin Efficiency and Effectiveness

For both solid and serrated fins, effectiveness Z is

Z ¼ 1 ð1 EÞ Af

At

ð38Þ

ð39Þ

ð40Þ

ð41Þ

ð42Þ

ð43Þ

Copyright © 2003 Marcel Dekker, Inc.

TABLE 8.10a Factors C1–C6 for Solid and Serrated Fins in In-Line and Staggered Arrangements—old ESCOA Correlations.

Solid fins

C2 ¼ 0:07 þ 8 Re 0:45

 

C1 ¼ 0:25 Re 0:35

 

In-line

 

C4 ¼ 0:08ð0:15ST =dÞ 1:1ðh=sÞ0:15

 

C3 ¼ 0:2 þ 0:65e 0:25h=s

 

C5 ¼ 1:1 ð0:75 1:5e 0:7Nd Þe 2:0SL=ST

 

2

 

C6 ¼ 1:6 ð0:75 1:5e 0:7Nd Þe 0:2ðSL=ST Þ

 

 

Staggered

 

 

2 C4 ¼ 0:11ð0:15ST =dÞ 0:7ðh=sÞ0:20

 

C3 ¼ 0:35 þ 0:65e 0:25h=s

 

 

C5 ¼ 0:7 þ ð0:7 0:8e 0:15Nd

2 Þe 1:0SL=ST

 

2

Þe 0:6ðSL=ST Þ

C6 ¼ 1:1 þ ð1:8 2:1e 0:15Nd

Þe 2:0ðSL=ST Þ

ð0:7 0:8e 0:15Nd

Serrated fins

C2 ¼ 0:07 þ 8:0 Re 0:45

 

C1 ¼ 0:25 Re 0:35

 

In-line

 

C4 ¼ 0:08ð0:15ST =dÞ 1:1ðh=sÞ0:2

 

C3 ¼ 0:35 þ 0:5e 0:35h=s

 

C5 ¼ 1:1 ð0:75 1:5e 0:7Nd Þe 2:0SL=ST

 

2

 

C6 ¼ 1:6 ð0:75 1:5e 0:7Nd Þe 0:2ðSL=ST Þ

 

 

Staggered

 

 

2 C4 ¼ 0:11ð0:05ST =dÞ 0:7ðh=sÞ0:23

 

C3 ¼ 0:55 þ 0:45e 0:35h=s

 

 

C5 ¼ 0:7 þ ð0:7 0:8e 0:15Nd

2 Þe 1:0SL=ST

 

2

Þe 0:6ðSL=ST Þ

C6 ¼ 1:1 þ ð1:8 2:1e 0:15Nd

Þe 2:0ðSL=ST Þ

ð0:7 0:8e 0:15Nd

Source: Fintube Technologies, Tulsa, OK.

For solid fins,

Af ¼ pn 4dh þ 4h2 þ 2bd þ 4bh

24

At ¼ Af þ p dð1 nbÞ

12

E ¼ 1=f1 þ 0:002292 m2h2 ½ðd þ 2hÞ=d&0:5g

where

m ¼ ð24ho=KbÞ0:5

For serrated fins,

Af ¼ pdn 2hðws þ bÞ þ bws

12ws

At ¼ Af þ pd ð1 nbÞ

12

E ¼ tanh ðmhÞ mh

ð44Þ

ð45Þ

ð46Þ

ð47Þ

ð48Þ

ð49Þ

ð50Þ

Copyright © 2003 Marcel Dekker, Inc.

TABLE 8.10b Factors C1 C6 for Solid and Serrated Fins in In-line and Staggered Arrangements—Revised Correlations

Solid fins

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In-line

 

 

 

 

 

 

 

 

 

 

2:3 Re 0:21

 

 

 

 

 

 

 

 

 

 

1:4 Re 0:4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C1

¼

0:053 1:45

 

2:9SL=d

 

C2

¼

0:11

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ð

 

 

 

 

 

 

 

Þ

 

 

 

 

 

 

 

 

 

 

1:1þ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C3 ¼ 0:20 þ 0:65e 0:25h=s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

h=s 0:15

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C4 ¼ 0:08ð0:15ST =dÞ ð Þ

 

 

 

1:5e 0:7Nd

 

e 0:2ðSL=ST Þ

2

 

 

 

 

 

 

 

 

 

 

C5

¼

1:1

ð

0:75

 

1:5e 0:7Nd

e 2:0SL=ST

 

C6

¼

1:6

ð

0:75

 

Þ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0:5

 

 

Þ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¼

 

 

 

½ð þ

 

Þ

& ½ð

 

þ

 

Þ ð

 

þ

 

 

 

0:5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

J

C1C3C5

 

 

460

ta

 

0:Þ&

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d

 

 

2h =d

 

 

 

 

tg

 

 

=

 

460

25

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f ¼ C2C4C6½ðd þ 2hÞ=d&½ðtg þ 460Þ=ðta þ

460Þ&

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Staggered

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C1 ¼ 0:091 Re 0:25

 

C2 ¼ 0:075 þ 1:85 Re 0:3

 

0:7 h=s 0:20

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C3 ¼ 0:35 þ 0:65e 0:25h=s

 

 

 

2 C4 ¼ 0:11ð0:05ST =dÞ ð Þ

 

 

 

2:1e 0:15Nd

2

e 2:0ðSL=ST Þ

 

 

 

 

 

 

2

 

e 0:6ðSL=ST Þ

C

 

¼

0:7

þ ð

0:7

 

0:8e 0:15Nd

 

 

e 1:0SL=ST

&

 

C6

 

1:1

1:8

 

 

 

0:7

 

0:8e 0:15Nd

&

 

5

 

 

 

 

0:5

 

 

 

Þ½

 

 

 

 

 

 

0:5¼

 

 

 

þ ð

 

 

 

 

 

Þ

 

 

½

 

 

 

 

J ¼ C1C3C5½ðd þ 2h=d&

0:5½ðtg þ 460Þ=ðta þ 460Þ&

0:25

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f ¼ C2C4C6½ðd þ 2hÞ=d&

 

½ðtg þ 460Þ=ðta þ 460Þ&

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Serrated fins

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In-line

 

 

 

 

 

 

 

 

 

 

2:3 Re 0:21

 

 

 

 

 

 

 

 

 

 

1:4 Re 0:4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C1

¼

0:053 1:45

 

2:9SL=d

 

C2

¼

0:11

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ð

 

 

 

 

 

 

 

Þ

 

 

 

 

 

 

 

 

 

 

1:1 hþ

 

0:15

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C3 ¼ 0:25 þ 0:6e 0:26h=s

 

 

 

C4 ¼ 0:08ð0:15ST =dÞ ð

 

=s

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Þ

 

 

 

 

1:5e 0:7Nd

 

e 0:2ðSL=ST Þ

2

 

 

 

 

 

 

 

 

 

 

C5

¼

1:1

ð

0:75

 

1:5e 0:7Nd

e 2:0SL=ST

 

C6

¼

1:6

ð

0:75

 

Þ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

0:5

 

 

Þ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¼

 

 

 

½ð þ

 

Þ

& ½ð

 

þ

 

Þ ð

 

þ

 

 

 

0:5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

J

C1C3C5

 

 

460

ta

 

0:Þ&

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

d

 

 

2h =d

 

 

 

 

tg

 

 

=

 

460

25

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f ¼ C2C4C6½ðd þ 2hÞ=d&½ðtg þ 460Þ=ðta þ

460Þ&

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Staggered

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C1 ¼ 0:091 Re 0:25

 

C2 ¼ 0:075 þ 1:85 Re 0:3

 

0:7 h=s 0:2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

C3 ¼ 0:35 þ 0:65e 0:17h=s

 

 

 

2 C4 ¼ 0:11ð0:05ST =dÞ ð Þ

 

 

2:1e 0:15Nd

2

 

 

e 2:0ðSL=ST

Þ

 

 

 

 

 

0:8e 0:15Nd

2

e 0:6ðSL=ST Þ

C5

¼

0:7

þ ð

0:7

 

0:8e 0:15Nd

Þ

e 1:0SL=ST

 

C6

¼

1:1

 

þ ð

1:8

 

 

Þ

ð

0:7

 

 

 

 

 

 

 

 

 

0:5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Þ

 

J ¼ C1C3C5½ðd þ 2hÞ=d&0:5

 

½ðtg þ 460Þ=ðta þ

 

 

 

0:25

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

460Þ& 0:25

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f ¼ C2C4C6½ðd þ 2hÞ=d&

 

½ðtg þ 460Þ=ðta þ 460Þ&

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Source: Fintube Technologies, Tulsa, OK.

Copyright © 2003 Marcel Dekker, Inc.

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m

 

 

24 hoðb þ wsÞ

 

0:5

 

 

 

51

 

 

 

 

 

 

 

 

 

 

¼

 

 

 

 

 

 

 

 

 

 

 

 

ð

Þ

 

 

 

 

Kbws

 

 

 

 

 

 

 

 

Gas pressure drop DPg

is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

G2N

 

 

 

 

 

DPg ¼ ð f þ aÞ

 

 

 

 

 

 

 

 

d

 

 

ð52Þ

rg 1:083 109

 

where

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

f

 

C

C C

 

 

 

 

d þ 2h

 

0:5

for staggered arrangement

 

53

 

¼

 

 

 

 

 

ð

Þ

 

2

4 6

 

 

d

 

 

 

 

 

 

¼

C

C

C

6

d þ 2h

 

for in-line arrangement

ð

54

Þ

 

2

4

 

 

 

 

d

 

 

 

 

 

 

 

 

a

¼

1 þ B2

 

tg2 tg1

 

 

 

 

 

 

55

 

 

 

460 þ tg

 

 

 

 

 

ð

Þ

 

4Nd

 

 

 

 

 

 

 

 

 

 

free gas area

2

 

 

 

 

 

 

 

 

 

B ¼

 

 

 

 

 

 

 

 

 

 

 

ð56Þ

 

total area

 

 

 

 

 

 

 

 

C2; C4; C6 are given in Table 8.10 for solid and serrated fins.

Tube Wall and Fin Tip Temperatures

For solid fins the relationship between tube wall and fin tip temperatures is given by

 

tg tf

 

K1ðmreÞ I0ðmreÞ þ I1ðmreÞ K0ðmreÞ

ð

57

Þ

tg tb

¼ K1ðmreÞ I0ðmr0Þ þ K0ðmr0Þ I1ðmreÞ

 

The various Bessel functional data are shown in Table 8.11 for serrated fins,

treated as longitudinal fins:

 

 

 

 

tg tf

 

1

 

ð58Þ

tg tb

¼

coshðmbÞ

 

A good estimate of tf can also be obtained for either type of fin as follows:

 

 

tf ¼ tb þ ðtg tbÞ ð1:42 1:4 EÞ

ð59Þ

tb, the fin base temperature, is estimated as follows:

 

 

 

tb ¼ ti þ q ðR3 þ R4 þ R5Þ

ð60Þ

where R3; R4; and R5 are resistances to heat transfer of the inside film, fouling layer, and tube wall, respectively, and heat flux qo is given by

qo ¼ Uoðtg tiÞ

ð61Þ

The following example illustrates the use of the equations.

Copyright © 2003 Marcel Dekker, Inc.

TABLE 8.11 I0; I1; K0; and K1 Values for Various Arguments

x

I0ðxÞ

I1ðxÞ

K0ðxÞ

K1ðxÞ

0

1.0

0

8

8

0.1

1.002

0.05

2.427

9.854

0.2

1.010

0.10

1.753

4.776

0.3

1.023

0.152

1.372

3.056

0.4

1.040

0.204

1.114

2.184

0.5

1.063

0.258

0.924

1.656

0.6

1.092

0.314

0.778

1.303

0.7

1.126

0.372

0.66

1.05

0.8

1.166

0.433

0.565

0.862

0.9

1.213

0.497

0.487

0.716

1.0

1.266

0.565

0.421

0.602

1.2

1.394

0.715

0.318

0.434

1.4

1.553

0.886

0.244

0.321

1.6

1.75

1.085

0.188

0.241

1.8

1.99

1.317

0.146

0.183

2.0

2.28

1.591

0.114

0.140

2.2

2.629

1.914

0.0893

0.108

2.4

3.049

2.298

0.0702

0.0837

2.6

3.553

2.755

0.554

0.0653

2.8

4.157

3.301

0.0438

0.0511

3.0

4.881

3.953

0.0347

0.0402

3.2

5.747

4.734

0.0276

0.0316

3.4

6.785

5.670

0.0220

0.0250

3.6

8.028

6.793

0.0175

0.0198

3.8

9.517

8.140

0.0140

0.0157

4.0

11.30

9.759

0.0112

0.0125

4.2

13.44

11.70

0.0089

0.0099

4.4

16.01

14.04

0.0071

0.0079

4.6

19.09

16.86

0.0057

0.0063

4.8

22.79

20.25

0.0046

0.0050

5.0

27.24

24.34

0.0037

0.0040

 

 

 

 

 

Example

A steam superheater is designed for the following conditions.

Gas flow ¼ 225,000 pph

Gas inlet temperature ¼ 1050 F

Gas exit temperature ¼ 904 F

Gas analysis (vol%): CO2 ¼ 3, H2O ¼ 7, N2 ¼ 75, O2 ¼ 15

Copyright © 2003 Marcel Dekker, Inc.

Steam flow ¼ 50,000 pph

Steam temperature in ¼ 501 F (sat)

Steam exit temperature ¼ 758 F

Steam pressure (exit) ¼ 650 psig

Tubes used: 2 0.120 low alloy steel tubes; 18 tubes=row, 6 deep, 10 ft long, in-line arrangement with 4 in. square pitch and nine streams. Tube inner diameter ¼ 1.738 in.; outer diameter ¼ 2 in.

Fins used: solid stainless steel, 2 fins=in., 0.5 in. high and 0.075 in. thick. Fin thermal conductivity K ¼ 15 Btu=ft h F.

Determine the heat transfer coefficient and pressure drop.

Solution.

A

¼

 

2

þ

2 0:5 0:075

 

¼

0:17917 ft2

=ft

 

12

6

o

 

 

 

 

 

G ¼

 

 

 

225;000

 

 

 

¼ 8127 lb=ft2 h

 

 

 

18 10 ½ð4=12Þ 0:17917Þ&

The gas properties at the average gas temperature (from the Appendix) are

Cp ¼ 0:276; m ¼ 0:086; k ¼ 0:03172

 

Re

 

 

8127 2

15;750

 

 

 

 

 

¼

12

 

0:086 ¼

 

 

 

 

 

 

 

 

 

 

 

 

C1 ¼ 0:25 ð15;750Þ 0:35 ¼ 0:0085

 

s ¼ 1=2 0:075 ¼ 0:425

 

 

 

 

C3 ¼ 0:2 þ 0:65 e 0:25 0:5=0:425 ¼ 0:6843

C5 ¼ 1:1 ð0:75 1:5 e 0:7 6Þ ðe 2 4=4Þ ¼ 1:0015

Assume that the average fin temperature is 750 F. The average gas temperature ¼ 977 F, and steam temperature ¼ 630 F. The fin thermal conductivity K is assumed to be 15 Btu=ft h F. Then,

3

 

0:5

hc ¼ 0:0085 0:6843 1:0015

 

 

 

2

 

977 þ 460 0:25 8127 0:276 750 þ 460

 

0:276 0:086

0:67

¼ 20:29

 

0:03172

 

 

Copyright © 2003 Marcel Dekker, Inc.

Using methods discussed in Q8.07, we find hN ¼ 1.0. The beam length for finned tubes is computed as 3.4 volume=surface area. Hence

ho ¼ 20:29 þ 1:0 ¼ 21:29

 

 

 

 

 

 

 

 

 

 

 

 

 

 

m

 

 

24 21:29

 

 

0:5

 

 

21:31

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¼

15 0:075

 

 

 

¼

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E

 

¼

0:758

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

¼

1=ð1 þ 0:002292

21:31 21:31 0:5 0:5 p1:5Þ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Af ¼ 3:14 2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

4 2 0:5 þ 4 0:5 0:5 þ 2 0:075 2 þ 4 0:075 5

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

24

 

 

 

 

 

 

 

 

 

 

 

¼ 1:426

 

 

 

 

 

 

 

 

 

 

 

1 2 0:075

 

 

 

 

 

 

 

 

 

A

t ¼

1:426

þ

3:14

 

2

 

 

¼

1:871

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

12

 

 

 

 

 

 

 

 

 

 

Hence

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Z ¼ 1 ð1 0:758Þ

 

1:426

¼ 0:8156

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1:871

 

 

 

 

 

 

 

 

 

 

Let us compute hi

for steam. w ¼ 50,000=9 ¼ 5555 lb=h per tube. From

Table 8.2, factor C ¼ 0.34.

 

 

 

 

0:8

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

hi

¼ 2:44 0:34

 

ð5555Þ

 

¼ 303 Btu=ft2 h F

 

 

 

1:8

 

 

 

1

 

 

 

 

 

1

 

 

 

 

ð1:738Þ

 

 

 

 

 

1:871

 

 

 

 

 

 

 

¼

 

 

 

 

 

 

 

 

þ 12

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

U

21:29

 

0:816

303

 

3:14

 

1:738

 

 

 

 

 

 

 

 

 

þ

 

 

þ

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1:738

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3:14 1:738 þ

 

 

 

 

 

 

 

 

 

0:001

 

0:001

 

 

 

1:871

 

 

12

 

 

 

24 ln

 

2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1:871

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

24 20 3:14 1:738

¼0:0576 þ 0:01358 þ 0:001 þ 0:0041 þ 0:0032

¼0:0795 or U ¼ 12:58 Btu=ft2hF

Calculation of Tube Wall and Fin Tip Temperature

Heat flux q ¼ 12:58 ð977 630Þ ¼ 4365 Btu=ft2 h

tb ¼ 630 þ 4365 ð0:0032 þ 0:0041 þ 0:01358Þ ¼ 722 F

Copyright © 2003 Marcel Dekker, Inc.

Using the elaborate Bessel functions, from Table 8.11,

 

 

 

 

 

1:5

 

¼ 2:661 ft;

 

 

 

 

 

mre ¼ 21:29

 

 

mro ¼ 1:7742 ft

12

K0 ð2:661Þ ¼ 0:0517

K1 ð2:661Þ ¼ 0:061

 

 

I0ð2:661Þ ¼ 3:737;

I1ð2:661Þ ¼ 2:921

 

 

 

K0ð1:7742Þ ¼ 0:1515;

I0ð1:7742Þ ¼ 1:959

 

 

Hence,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

977 tf

 

0:061 3:737 þ 2:921 0:0517

¼

0:6743

 

 

 

977

 

722

¼ 0:061

 

1:959

þ

0:1515

 

2:921

 

 

 

 

 

 

tf ¼ 805 F

Using the approximation

tf ¼ tb þ ð1:42 1:4 0:758Þ ð977 722Þ ¼ 813 F

Note that this is only an average base and fin tip temperature. For material selection purposes one should look at the maximum heat flux, which occurs, for instance, at the gas inlet in a counterflow arrangement, and also consider the nonuniformity or maldistribution in gas and steam flow. A computer program can be developed to compute the tube wall and fin tip temperatures at various points along the tube length and the results used to select appropriate materials.

It can be noted from the above that there are a few ways to reduce the fin tip temperature:

1.Increase fin thickness. This reduces the factor m and hence tf .

2.Increase the thermal conductivity of the fin material. This may be difficult, because the thermal conductivity of carbon steels is higher

than that of alloy steels, and carbon steels can withstand temperatures only up to 850 F, whereas alloy steels can withstand up to 1300 F depending on the alloy composition.

3.Reduce ho or the gas-side coefficient by using a lower gas mass velocity.

4.Reduce fin height or density.

5.In designs where the gas inlet temperature is very high, use a combination of bare and finned rows. The first few rows could be bare, followed by tubes with a low fin density or height or increased thickness and then followed by tubes with higher fin density or height or smaller thickness to obtain the desired boiler performance. A row- by-row analysis of the finned bundle is necessary, which requires the use of a computer program.

Copyright © 2003 Marcel Dekker, Inc.

Computation of Gas Pressure Drop

C2 ¼ 0:07 þ 8 ð15;750Þ 0:45 ¼ 0:1734

C4 ¼ 0:08 ð0:15 2Þ 1:11ð0:5=0:425Þ0:15 ¼ 0:3107

C6 ¼ 1

f ¼ 0:1734 0:3107 1 32 ¼ 0:0808

 

 

 

 

 

 

0:17917

 

2

 

 

B2

¼

0:333

 

 

¼ 0:2134

0:333

 

a

¼

904 1050

1 þ 0:2134

¼

0:005

 

460

þ

977

24

 

 

 

 

6

DPg ¼ ð0:0808 0:0051Þ 8120 8120 0:0271 1:083 109 ¼ 1:02 in: WC

ðGas density ¼ 0:0271:Þ

Computer solution of the above system of equations saves a lot of time. However, I have developed a chart (Fig. 8.7) that can be used to obtain hc (or hg) and Z values for serrated fins and an in-line arrangement for various fin configurations and gas mass velocities for gas turbine exhaust gases at an average gas temperature of 700 F. Although a computer program is the best tool, the chart can be used to show trends and the effect of fin configuration on the performance of finned surfaces. The use of the chart is explained later with an example. The following points should be noted.

1.From Fig. 8.7, it can be seen that for a given mass velocity, the higher the fin density or height, the lower the gas-side coefficient or effec-

tiveness, which results in lower Uo. The amount of energy transferred in heat transfer equipment depends on the product of the overall heat transfer coefficient and surface area and not on the surface area alone. We will see later that one can have more surface area and yet transfer less duty due to poor choice of fin configuration.

2.Higher fin density or height results in higher DPg. Even after adjusting for the increased surface area per row, it can be shown that the higher the fin density or the greater the height, the higher the gas pressure drop will be for a given mass velocity.

Copyright © 2003 Marcel Dekker, Inc.