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

Figure 8.7 Chart of convective heat transfer coefficient and pressure drop versus fin geometry. (Data from Ref. 10)

8.19b

Q:

Describe Briggs and Young’s correlation.

A:

Charts and equations provided by the manufacturer of finned tubes can be used to obtain hc. In the absence of such data, the following equation of Briggs and Young for circular or helical finned tubes in staggered arrangement [4] can be used.

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The gas properties Cp; m, and k are evaluated at the average gas temperature. The gas heat transfer coefficient hc has to be corrected for the temperature

distribution along the fin height by the fin efficiency

Km is the fin metal thermal conductivity, in Btu=ft h F.

In order to correct for the effect of finned area, a term called fin effectiveness is used. This term, Z, is given by

is given in Table 8.12.

The overall heat transfer coefficient with finned tubes, U, can be estimated as U ¼ 0:85Zhc, neglecting the effect of the non-luminous heat transfer coefficient.

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Example

Determine the gas-side heat transfer coefficient when 150,000 lb=h of flue gases at an average temperature of 900 F flow over helically finned economizer tubes with the following parameters:

d ¼ tube outer diameter ¼ 2.0 in. n ¼ fins=in. ¼ 3

h ¼ fin height ¼ 1 in.

b ¼ fin thickness ¼ 0.06 in.

L ¼ effective length of tubes ¼ 10.5 ft Nw ¼ number of tubes wide ¼ 12

ST ¼ transverse pitch ¼ 4.5 in. (staggered)

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4:9

Fin effectiveness Z ¼ 1 ð1 0:6Þ 5:33 ¼ 0:63 Hence,

Zhc ¼ 0:63 12:74 ¼ 8 Btu=ft2 h F

Km ranges from 23 to 27 Btu=ft h F for carbon steels, depending on temperature [1]. For alloy steels it is lower.

8.19c

Q:

This example shows how one can predict the performance of a given heat transfer surface. A superheater is designed for the following conditions: 18 tubes=row, 6 rows deep, 10 ft long with 2 fins=in., 0.5 in. high and 0.075 in. thick solid fins. It has 18 streams. Surface area ¼ 2022 ft2. Tube spacing ¼ 4 in. square.

Predict the performance of the superheater under the following conditions:

Gas flow ¼ 150,000 lb=h at 1030 F

Steam flow ¼ 35,000 lb=h at 615 psig sat

Flue gas analysis (vol%): CO2 ¼ 7, H2O ¼ 12, N2 ¼ 75, O2 ¼ 6

Heat loss ¼ 2%

Surface area A ¼ 2022 ft2

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Let us say that U has been estimated as 10.6 Btu=ft2 h F using methods discussed earlier.

A:

Let us use the NTU method to predict the performance of the superheater. This is discussed in Q8.30. The superheater is in counterflow arrangement.

Energy transferred Q ¼ eCminðtg1 tS1 Þ

where

e ¼ 1 exp½ NTUð1 CÞ&

1 C exp½ NTUð1 CÞ&

C ¼ Cmin=Cmax

Cmin is the lower of (mass specific heat of the fluid) on gas and steam sides.

Tg1; ts1 ¼ gas and steam temperature at inlet to superheater, F Use 491 F for steam saturation temperature.

Though NTU methods generally require no iterations, a few rounds are necessary in this case to evaluate the specific heat for steam and gas, which are functions of temperature. However, let us assume that the steam-side specific heat ¼ 0.6679 and that of gas ¼ 0.286 Btu=lb F.

Cgas ¼ 150;000 0:98 0:286 ¼ 42;042

Csteam ¼ 35;000 0:6679 ¼ 23;376

Hence, Cmin ¼ 23,376.

C¼ 23;376 ¼ 0:556 42;042

Hence

1 exp½ 0:9186 ð1 0:556Þ&

e ¼ 1 0:556 exp½ 0:9186 ð1 0:556Þ& ¼ 0:5873

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Hence

Energy transfered Q ¼ 0:5873 23;376 ð1030 491Þ ¼ 6:7 MM Btu=h

Gas-side pressure drop may be estimated using the chart in Fig. 8.7 and is about 0.6 in.WC.

8.20

Q:

A gas turbine HRSG evaporator operates under the following conditions:

Gas flow ¼ 230,000 lb h (vol % CO2 ¼ 3, H2O ¼ 7, N2 ¼ 75, O2 ¼ 15) Gas inlet temperature ¼ 1050 F

Exit gas temperature ¼ 406 F

Duty ¼ 230,000 0.99 0.27 (1050 7 406) ¼ 39.6 MM Btu=h Steam pressure ¼ 200 psig

Feedwater temperature ¼ 230 F Blowdown ¼ 5%

Fouling factors ¼ 0.001 ft2 h F=Btu on both gas and steam sides Arrangement: 4 in. square pitch

Tubes used: 2 1.773 in., 24 tubes=row, 11 ft long Fins: 5 fins=in., 0.75 in. high, 0.05 in. thick, serrated

Determine the overall heat transfer coefficient and pressure drop using the chart.

A:

The chart shown in Fig. 8.7 has been developed for serrated fins in in-line arrangement for the above gas analysis. Users may develop their charts for

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various configurations or use a computer program. The chart is based on an average gas temperature of 700 F and a gas analysis (vol%) of CO2 ¼ 3, H2O ¼ 7, N2 ¼ 75, O2 ¼ 15.

ð1050 388Þ ð406 388Þ Log-mean temperature difference ¼ ln½ð1050 388Þ=ð406 388Þ&

¼ 178 F

Surface area required ¼ 39:6 106 ¼ 30;063 ft2 178 7:4

30;063

Number of rows deep required ¼ 24 11 5:7 ¼ 20

Gas pressure drop ¼ 1:7 2 ¼ 3:4 in. Wc

Copyright © 2003 Marcel Dekker, Inc.

8.21

Q:

How does a finned surface compare with a bare tube bundle for the same duty?

A:

Let us try to design a bare tube boiler for the same duty as above. Use the same tube size and spacing, tubes per row, and length. Use 2 1.773 in. bare tubes.

Using the procedure described in Q8.14, we can show that U ¼ 13.05 Btu=ft2 h F and that 124 rows are required for the same duty. The results are shown in Table 8.13.

It may be seen that the finned tube bundle is much more compact and has fewer rows and also a lower gas pressure drop. It also weighs less and should cost less. Therefore, in clean gas applications such as gas turbine exhaust or fume incineration plants, extended surfaces may be used for evaporators. In dirty gas applications such as municipal waste incineration or with flue gases containing ash or solid particles, bare tubes are preferred. Finned tubes may also be used in packaged boiler evaporators.

However, the heat flux inside the finned tubes is much larger, which is a concern in high gas temperature situations. The tube wall temperature is also higher. Hence when the gas temperature is high, say 1400–1700 F, we use a few

TABLE 8.13 Comparison of Bare Tube and Finned Tube Boilers

Tubes=row ¼ 24; effective length ¼ 11 ft; 4 in. square spacing. Gas analysis (vol%) CO2 ¼ 3 H2 ¼ 7, N2 ¼ 75, O2 ¼ 15. Blowdown ¼ 5%.

Copyright © 2003 Marcel Dekker, Inc.

bare tubes followed by tubes with, say, 2 fins=in. fin density and then go back to four or more fins per inch. This ensures that the gas stream is cooled before entering tube bundles with a high fin density and that the tubes are operating at reasonable temperatures, which should also lower the fin tip temperatures.

When the tube-side fouling is large, it has the same effect as a low tube-side heat transfer coefficient, resulting in poor performance when a high fin density is used. See Q8.24. One may also note the significant difference in surface areas and not be misled by this value.

8.22

Q:

Which is the preferred arrangement for finned tubes, in-line or staggered?

A:

Both in-line and staggered arrangements have been used with extended surfaces. The advantages of the staggered arrangement are higher overall heat transfer coefficients and smaller surface area. Cost could be marginally lower depending on the configuration. Gas pressure drop could be higher or lower depending on the gas mass velocity used. If cleaning lanes are required for soot blowing, an inline arrangement is preferred.

Both solid and serrated fins are used in the industry. Generally, solid fins are used in applications where the deposition of solids is likely.

The following example illustrates the effect of arrangement on boiler performance.

Example

150,000 lb=h of turbine exhaust gases at 1000 F enter an evaporator of a waste heat boiler generating steam at 235 psig. Determine the performance using solid and serrated fins and in-line versus staggered arrangements. Tube size is 2 1.77 in.

Solution. Using the ESCOA correlations and the methodology discussed above for evaporator performance, the results shown in Table 8.14 were arrived at.

8.23

Q:

How does the tube-side heat transfer coefficient or fouling factor affect the selection of fin configuration such as fin density, height, and thickness?

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TABLE 8.14 Comparison Between Staggered and In-Line Designs for Nearly Same Duty and Pressure Dropa

a Duty, MM Btu=h; DPg , in. WC; surface, ft2; temperature, F; Uo, Btu=ft2 h F.

A:

Fin density, height, and thickness affect the overall heat transfer coefficient as can be seen in Fig. 8.7. However, the tube-side coefficient also has an important bearing on the selection of fin configuration.

A simple calculation can be done to show the effect of the tube-side coefficient on Uo. It was mentioned earlier that the higher the tube-side coefficient, the higher the ratio of external to internal surface area can be. In other words, it makes no sense to use the same fin configuration, say 5 fins=in. fin density, for a superheater as for an evaporator.

Rewriting Eq. (3) based on tube-side area and neglecting other resistances,

Using the data from Fig. 8.7, Ui values have been computed for different fin densities and for different hi values for the configuration indicated in Table 8.15. The results are shown in Table 8.15. Also shown are the ratio of Ui values between the 5 and 2 fins=in. designs as well as their surface area.

The following conclusions can be drawn [10].

1.As the tube-side coefficient decreases, the ratio of Ui values (between 5 and 2 fins=in.) decreases. With hi ¼ 20, the Ui ratio is only 1.11. With an hi of 2000, the Ui ratio is 1.74. What this means is that as hi decreases, the benefit of increasing the external surface becomes less attractive. With 2.325 times the surface area we have only 1.11-fold

improvement in Ui. With a higher hi of 2000, the increase is better, 1.74.

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