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

K ¼ attenuation factor, which depends on fuel type and presence of ash and its concentration. For a nonluminous flame it is

For a semiluminous flame, the ash particle size and concentration enter into the calculation:

where

dm ¼ the mean effective diameter of ash particles, in mm dm ¼ 13 for coals ground in ball mills

¼16 for coals ground in mediumand high-speed mills

¼20 for combustion of coals milled in hammer mills m ¼ ash concentration in g=N m3

Te ¼ furnace exit temperature, K

For a luminous oil or gas flame,

pw and pc are partial pressures of water vapor and carbon dioxide in the flue gas. The above equations give only a trend. A wide variation could exist due to the basic combustion phenomenon itself. Again, the flame does not fill the

furnace fully. Unfilled portions are subjected to gas radiation only, the emissivity of which (0.15–0.30) is far below that of the flame. Hence, ef decreases. Godridge reports that in a pulverized coal-fired boiler, emissivity varied as follows with respect to location [3]:

Also, furnace tubes coated with ferric oxide have emissivities, ew, of the order of 0.8, depending on whether a slag layer covers them. Soot blowing changes ew considerably. Thus, only an estimate of ef and ew can be obtained, which varies with type of unit, fuel, and operation regimes.

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To illustrate these concepts, a few examples are worked out. The purpose is only to show the effect of variables like excess air and heat release rates on furnace absorption and furnace exit gas temperature.

Example 1

Determine the approximate furnace exit gas temperature of a boiler when net heat input is about 2000 106 Btu=h, of which 1750 106 Btu=h is due to fuel and the rest is due to air. HHV and LHV of coals fired are 10,000 and 9000 Btu=lb, respectively, and a furnace heat release rate of 80,000 Btu=ft2 h (projected area basis) has been used. The values ew and ef may be taken as 0.6 and 0.5, respectively; 25% is the excess air used. Water-wall outer temperature is 600 F. Ash content in coal is 10%.

From combustion calculation methods discussed in Chapter 5, using 1 MM Btu fired basis, we have the following ratio of flue gas to fuel:

Ap=Wf ¼ LHV=80;000 ¼ 0:1125

Assume te ¼ 1900 F. Then Cpm ¼ 0:3 Btu=lb F

tg ¼ 1900 þ 300 ¼ 2200 F ¼ 2660 R

Let us see if the assumed te is correct. Substituting for Ap=Wf ; ew; ef ; s; Tg; Te in the above equation, we have (LHS ¼ left-hand side; RHS ¼ right-hand side)

LHS ¼ 0:1125 0:6 0:5 0:173

ð26:64 10:64Þ ¼ 2850

RHS ¼ ð9000 10:4 1900 0:3Þ ¼ 3072

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These do not tally, so we try te ¼ 1920 F. Neglect the effect of variation in Cpm:

LHS ¼ 0:1125 0:6 0:5 ð26:84 10:64Þ

0:173 ¼ 2938

RHS ¼ 9000 1920 0:3 10:4 ¼ 3009

These agree closely, so furnace exit gas temperature is around 1920 F. Note that the effect of external radiation to superheaters has been neglected in the energy balance. This may give rise to an error of 1.5–2.5% in te, but its omission greatly simplifies the calculation procedure. Also, losses occurring in the furnace were omitted to simplify the procedure. The error introduced is quite low.

Example 2

It is desired to use a heat loading of 100,000 Btu=ft2 h in the furnace in Example 1. Other factors such as excess air and emissivities remain unaltered. Estimate the furnace exit gas temperature.

ð27:64 10:64Þ ¼ 2664

RHS ¼ ð9000 10:4 2000 0:3Þ ¼ 2760

From this it is seen that te will be higher than assumed. Let

LHS ¼ 0:09 0:6 0:5 0:173

½ð27:9Þ4 ð10:6Þ4& ¼ 2771 RHS ¼ 9000 10:4 2030 0:3 ¼ 2667

Hence, te will lie between 2000 and 2030 F, perhaps 2015 F.

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The exercise shows that the exit gas temperature in any steam generator will increase as more heat input is given to it; that is, the higher the load of the boiler, the higher the exit gas temperature. Example 3 shows the effect of excess air on te.

Example 3

What will be the furnace exit gas temperature when 40% excess air is used instead of 25%, heat loading remaining at about 100,000 Btu=ft2 h in the furnace mentioned in earlier examples?

te ¼ 1950 F; Cpm ¼ 0:3 Btu=lb F

Tg ¼ 1950 þ 300 þ 460 ¼ 2710 R

LHS ¼ 0:09 0:6 0:5 0:173

½ð27:1Þ4 ð10:6Þ4& ¼ 2460

RHS ¼ 9000 ð11:54 1950 0:3Þ ¼ 2249

These nearly tally; hence, te is about 1950 F, compared to about 2030 F in Example 2. The effect of the higher excess air has been to lower te.

Example 4

If ew ef ¼ 0:5 instead of 0.3, what will be the effect on te when heat loading is 100,000 Btu=ft2 h and excess air is 40%?

Solution. Let

te ¼ 1800 F; Tg ¼ 1800 þ 300 þ 460 ¼ 2560 R LHS ¼ 0:09 0:5 0:173 ½ð25:6Þ4 ð10:6Þ4&

¼ 3245

RHS ¼ 9000 ð11:54 1800 0:3Þ ¼ 2768

Try

te ¼ 1700 F; Tg ¼ 2460 R

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Then

LHS ¼ 0:09 0:5 0:173 ½ð24:6Þ4 ð10:6Þ4&

¼ 2752

RHS ¼ 9000 ð11:54 1700 0:3Þ ¼ 3115

Hence, te will be around 1760 F. This example shows that when surfaces are cleaner and capable of absorbing more radiation, te decreases.

In practice, furnace heat transfer is not evaluated as simply as shown above because of the inadequacy of accurate data on soot emissivity, particle size, distribution, flame size, excess air, presence and effect of ash particles, etc. Hence, designers develop data based on field tests. Estimating te is the starting point for the design of superheaters, reheaters, and economizers.

Some boiler furnaces are equipped with tilting tangential burners, whereas some furnaces have only front or rear nontiltable wall burners. The location of the burners affects te significantly. Hence, in these situations, correlations with practical site data would help in establishing furnace absorption and temperature profiles. (See also p. 112, Chapter 3.)

A promising technique for predicting furnace heat transfer performance is the zone method of analysis. It is assumed that the pattern of fluid flow, chemical heat release, and radiating gas concentration are known, and equations describing conservation of energy within the furnace are developed. The furnace is divided into many zones, and radiation exchange calculations are carried out.

8.08b

Q:

How is heat transfer evaluated in unfired furnaces?

A:

Radiant sections using partially or fully water-cooled membrane wall designs are used to cool gas streams at high gas temperatures (Fig. 8.2). They generate saturated steam and may operate in parallel with convective evaporators if any. The design procedure is simple and may involve an iteration or two. The higher the partial pressures of triatomic gases, the higher will be the nonluminous radiation and hence the duty.

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Figure 8.2 Radiant furnace in a water tube boiler.

If a burner is used as in the radiant section of a furnace-fired HRSG, the emissivity of the flame must also be considered. As explained elsewhere [8], radiant sections are necessary to cool the gases to below the softening points of any eutectics present so as to avoid bridging or slagging at the convection section. They are also required to cool gases to a reasonable temperature at the superheater if it is used.

Example

200,000 lb=h of flue gases at 1800 F has to be cooled to 1600 F in a radiant section of a waste heat boiler of cross section 9 ft 11 ft. Saturated steam at 200 psig is generated. Determine the furnace length required. Flue gas analysis is (vol%) CO2 ¼ 8, H2O ¼ 18, N2 ¼ 72, O2 ¼ 2. Assume a length of 25 ft and that the furnace is completely water-cooled.

Surface area for cooling ¼ ð11 þ 9Þ 2 25 ¼ 1000 ft2

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Gas emissivity eg ¼ 0:9 ð1 e 0:2053 2:16Þ ¼ 0:3223

Let the average surface temperature of the furnace be 420 F (saturation temperature plus a margin). Then the energy transferred is

Qr ¼ 0:173 0:9 0:3223 ð21:64 8:84Þ 1000 ¼ 10:63 MM Btu=h Required duty ¼ 200;000 0:99 0:32 200 ¼ 12:67 MM btu=h

where 0.32 is the gas specific heat. Hence the furnace should be longer. The beam length and hence the gas emissivity will not change much with change in furnace length; therefore one may assume that the furnace length required ¼ (12.67=10.63) 25 ¼ 29.8 or 30 ft.

If the performances at other gas conditions are required, a trial-and-error procedure is warranted. First the exit gas temperature is assumed; then the energy transferred is computed as shown above and compared with the assumed duty.

8.09a

Q:

How is the distribution of external radiation to tube bundles evaluated? Discuss the effect of tube spacing.

A:

Tube banks are exposed to direct or external radiation from flames, cavities, etc., in boilers. Depending on the tube pitch, the energy absorbed by each row of tubes varies, with the first row facing the radiation zone receiving the maximum energy. It is necessary to compute the energy absorbed by each row, particularly in superheaters, because the contribution of the radiation can result in high tube wall temperatures.

The following formula predicts the radiation to the tubes [8].

where a is the fraction of energy absorbed by the first row. The second row would then absorb ð1 aÞa; the third row, f1 ½a þ ð1 aÞa&ga; and so on.

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Example

1 MM Btu=h of energy from a cavity is radiated to a superheater tube bank that has 2 in. OD tubes at a pitch of 8 in. If there are six rows, estimate the distribution of energy to each row.

It can be seen that the first row receives the maximum energy and the amount lessens as the number of rows increases. For a tube pitch S of 4 in., a ¼ 0.6575. The first row receives 0.6575 MM Btu=h; the second, 0.225 M Btu=h; and the third, 0.077 MM Btu=h. Hence if the tube pitch is small, a large amount of energy is absorbed within the first two to three rows, resulting in high heat flux in those tubes and consequently high tube wall temperatures. Hence it is better to use a wide pitch when the external radiation is large so that the radiation is spread over more tubes and the intensity is not concentrated within two or three tubes. Screen tubes in boilers and fired heaters perform this function.

8.09b

Q:

A soot blower lance is inserted in a boiler convection section where hot flue gases at 2000 F are flowing around the tubes. If the water wall enclosure is at 400 F, what will be the lance temperature? Assume that the heat transfer coefficient between the flue gas and the lance is 15 Btu=ft2 h F and the emissivity of the lance and the water wall tubes is 0.9.

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A:

The energy transferred between the flue gases and lance and from the lance to the water wall enclosure in Btu=ft2 h is given by

Q¼ hcð2000 TÞ

¼ 0:173 0:9 0:9 ½ðT þ 460Þ4 ð400 þ 460Þ4& 10 8

remaining the same, what will be the lance temperature? It can be shown to be about 970 F and Q ¼ 5150 btu=ft2h.

Hence just as a thermocouple reads a lower temperature due to the radiation to the enclosure, the lance also will not reach the gas temperature. Its temperature will be lower than that of the gas.

8.10

Q:

Determine the size of a fire tube waste heat boiler required to cool 100,000 lb=h of flue gases from 1500 F to 500 F. Gas analysis is (vol%) CO2 ¼ 12, H2O ¼ 12, N2 ¼ 70, and O2 ¼ 6; gas pressure is 5 in.WC. Steam pressure is 150 psig, and

side boiling heat transfer coefficient ¼ 2000 Btu=ft F. Assume that heat losses and margin ¼ 2% and blowdown ¼ 5%.

A:

Use Eq. (4) to compute the overall heat transfer coefficient, and then arrive at the size from Eq. (1).

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hi, the tube-side coefficient, is actually the sum of a convective portion hc plus a nonluminous coefficient hn: hc is obtained from Q8.04:

hc ¼ 2:44 w0:8 C:

di1 8

At the average gas temperature of 1000 F, the gas properties can be shown to be Cp ¼ 0.287 Btu=lb F, m ¼ 0.084 lb=ft h, and k ¼ 0.0322 Btu=ft h F. Hence,

C ¼ 0:287 0:4 ð0:0322Þ0:6 ¼ 0:208 0:084

Boiler duty Q ¼ 100;000 0:98 0:287 ð1500 500Þ ¼ 28:13 106 Btu=h

Enthalpies of saturated steam, saturated water, and feedwater from steam tables are 1195.5, 338, and 188 Btu=lb, respectively. The enthalpy absorbed by steam is then (1195.5 7 188) þ 0.05 (338 7 188) ¼ 1015 Btu=lb, where 0.05 is the blowdown factor corresponding to 5% blowdown.

Hence,

Steam generation ¼ 28:13 106 ¼ 27;710 lb=h 1015

In order to compute hi, the flow per tube w is required. Typically w ranges from 100 to 200 lb=h for a 2 in. tube. Let us start with 600 tubes; hence w ¼ 100,000=600 ¼ 167 lb=h.

The nonluminous coefficient is usually small in fire tube boilers because the beam length corresponds to the tube inner diameter. However, the procedure used in Q8.07 can also be used here. Let us assume that it is 0.45 Btu=ft2 h F. Then

hi ¼ 10:90 þ 0:45 ¼ 11:35 Btu=ft2 h F

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