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

Example

Analysis of a given ash indicates the following composition:

Al2O3 ¼ 20%; SiO2 þ TiO2 ¼ 30%

Fe2O3 þ Na2O þ K2O ¼ 20%; CaO þ MgO ¼ 15%

Find the fusion temperature.

Solution. Substituting into Eq. (13), we find that tm ¼ 1100 C.

6.18a

Q:

What is the emission of SO2 in lb=MM Btu if coals of HHV ¼ 11,000 Btu=lb and having 1.5% sulfur are fired in a boiler?

A:

The following expression gives e, the emission of SO2 in lb=MM Btu:

where S is the percent sulfur in the fuel.

If an SO2 scrubbing system of 75% efficiency is installed, the exiting SO2 concentration will be 0.25 2.73 ¼ 0.68 lb=MM Btu.

6.18b

Q:

What is the SO2 level in ppm (parts per million) by volume if the coals in Q6.18a are fired with 25% excess air?

A:

We have to estimate the flue gas produced. Using the MM Btu method,

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Let the molecular weight be 30, which is a good estimate in the absence of flue gas analysis. Then,

Moles of flue gas ¼ 104130 ¼ 34:7 per MM Btu fired

Moles of SO2 ¼ 264:73

¼ 0:042 (from Q6.18a and Table 5.1)

(64 is the molecular weight of SO2. Dividing weight by molecular weight gives the moles.)

Hence ppm of SO2 in flue gas will be 0.042 106=34.7 ¼ 1230 ppm.

6.18c

Q:

If 5% of the SO2 gets converted to SO3, estimate the ppm of SO3 in the flue gas.

A:

Moles of SO3 ¼ 0:05 280:73 ¼ 0:0017 per MM Btu

Hence

ppm by volume of SO3 ¼ 0:0017 106 ¼ 49 ppm 34:7

(80 is the molecular weight of SO3.)

6.19a

Q:

How is the efficiency of a boiler or a fired heater determined?

A:

The estimation of the efficiency of a boiler or heater involves computation of several losses such as those due to flue gases leaving the unit, unburned fuel, radiation losses, heat loss due to molten ash, and so on. Readers may refer to the ASME Power Test Code [7] for details. Two methods are widely used, one based on the measurement of input and output and the other based on heat losses. The latter is preferred, because it is easy to use.

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There are two ways of stating the efficiency, one based on HHV and the other on LHV. As discussed in Q6.01,

ZHHV HHV ¼ ZLHV LHV

4.Radiation loss, L4. The American Boiler Manufacturers Association (ABMA) chart [7] may be referred to to obtain this value. A quick estimate of L4 is

5.To losses L1–L4 must be added a margin or unaccounted loss, L5. Hence efficiency becomes

Note that combustion calculations are a prerequisite to efficiency determination. If the fuel analysis is not available, plant engineers can use the MM Btu method to estimate wdg rather easily and then estimate the efficiency (see Q6.20).

The efficiency can also be estimated on LHV basis. The various losses considered are the following.

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(Cp, gas specific heat, will be in the range of 0.26–0.27 for wet flue gases.)

2.Radiation loss (see Q6.23)

3.Unaccounted loss, margin

Then

ZLHV ¼ 100 ðsum of the above three lossesÞ

One can also convert ZHHV to ZLHV using Eq. (3b) (see Q6.01).

6.19b

Q:

Coals of HHV ¼ 13,500 Btu=lb and LHV ¼ 12,600 Btu=lb are fired in a boiler with 25% excess air. If the exit gas temperature is 300 F and ambient temperature is 80 F, determine the efficiency on HHV basis and on LHV basis.

A:

From the MM Btu method of combustion calculations, assuming that moisture in air is 0.013 lb=lb dry air,

(760 is the constant obtained from Table 6.4.) Hence wet flue gas loss ¼ 100 14:0 0:26

300 80

12,600

¼6:35%

Let radiation and unaccounted losses be 1.3%. Then

ZLHV ¼ 100 ð6:35 þ 1:3Þ ¼ 92:34%

12,600

ZHHV ¼ 92:34 13,500 ¼ 86:18%

(Radiation losses vary from 0.5% to 1.0% in large boilers and may go up to 2.0% in smaller units. The major loss is the flue gas loss.)

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6.19c

Q:

Determine the efficiency of a boiler firing the fuel given in Q6.09a at 15% excess air. Assume radiation loss ¼ 1%, exit gas temperature ¼ 400 F, and ambient temperature ¼ 70 F. Excess air and relative humidity are the same as in Q6.09a (15% and 80%).

A:

Results of combustion calculations are already available.

Dry flue gas ¼ 18 lb=lb fuel

Moisture in air ¼ 19:52 19:29 ¼ 0:23 lb=lb fuel Water vapor formed due to combustion of fuel ¼ 20:4 18 0:23 ¼ 2:17 lb=lb fuel

HHV ¼ 83:4 1013:2 þ 15:8 1792 ¼ 1128 Btu=cu ft 100

Fuel density at 60 F ¼ 18.3=379 ¼ 0.483 lb=cu ft, so HHV ¼ 01128:0483 ¼ 23,364 Btu=lb

The losses are

1.Dry gas loss,

2.Loss due to combustion of hydrogen and moisture in fuel,

L2 ¼ 100 2:17 1080 þ 0:46 400 70 23,364

¼11:1%

3.Loss due to moisture in air,

4.Radiation loss ¼ 1.0%

5.Unaccounted losses and margin ¼ 0%

Total losses ¼ 6:1 þ 11:1 þ 0:15 þ 1:0 ¼ 18:35%

Hence

Efficiency on HHV basis ¼ 100 18:35 ¼ 81:65%

One can convert this to LHV basis after computing the LHV.

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6.19d

Q:

How do excess air and boiler exit gas temperature affect the various losses and boiler efficiency?

A:

Table 6.7 shows the results of combustion calculations for various fuels at different excess air levels and boiler exit gas temperatures. It also shows the amount of CO2 generated per MM Btu fired.

It can be seen that natural gas generates the lowest amount of CO2.

Oil (wt%): C ¼ 87.5, H2 ¼ 12.5, API ¼ 32; HHV ¼ 19,727 Btu=lb; LHV ¼ 18,512 Btu=lb. Gas (vol%): CH4 ¼ 97; C2H6 ¼ 2, C3H8 ¼ 1; HHV ¼ 23,789 Btu=lb; LHV ¼ 21,462 Btu=lb.

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(The above is obtained by converting the volumetric analysis to weight basis using the molecular weights of CO2 and the flue gas.) For oil, CO2 generated ¼ 162.4 lb, and for coal, 202.9 lb.

6.20

Q:

A fired heater of duty 100 MM Btu=h (HHV basis) firing No. 6 oil shows the following dry flue gas analysis:

CO2 ¼ 13:5%; O2 ¼ 2:5%; N2 ¼ 84%

The exit gas temperature and ambient temperature are 300 F and 80 F, respectively. If moisture in air is 0.013 lb=lb dry air, estimate the efficiency of the unit on LHV and HHV basis. LHV ¼ 18,400 Btu=lb and HHV ¼ 19,500 Btu=lb.

A:

Because the fuel analysis is not known, let us estimate the flue gas produced by the MM Btu method. First, compute the excess air, which is

2:5

E ¼ 94:5 21 2:5 ¼ 12:8%

The factor 94.5 is from Table 6.6 (see Q6.12). The wet flue gas produced is

Hence

Wet gas loss ¼ 100 17:6 0:26 300 80 ¼ 5:47% 18,400

The radiation loss on HHV basis can be approximated by Eq. (15d):

Radiation loss ¼ 100:62 0:42 log Q ¼ 0:60%

Q ¼ 100 MM Btu=h

Let us use 1.0% on LHV basis, although this may be a bit high. Hence the efficiency on LHV basis is 100 7 6.47 ¼ 93.53%. The efficiency on HHV basis would be [Eq. (3b)]

ZHHV HHV ¼ ZLHV LHV

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or

18,400

ZHHV ¼ 95:53 19,500 ¼ 88:25

Thus, even in the absence of fuel ultimate analysis, the plant personnel can check the efficiency of boilers and heaters based on operating data.

6.21

Q:

How is the loss due to incomplete combustion such as the formation of CO determined?

A:

Efforts must be made by the boiler and burner designers to ensure that complete combustion takes place in the furnace. However, because of various factors such as size of fuel particles, turbulence, and availability of air to fuel and the mixing process, some carbon monoxide will be formed, which means losses. If CO is formed from carbon instead of CO2, 10,600 Btu=lb is lost. This is the difference between the heat of reaction of the two processes

C þ O2 ! CO2 and C þ O2 ! CO

The loss in Btu=lb is given by [1]

CO

L ¼ CO þ CO2 10,160 C

where C is the carbon in the fuel, fraction by weight, and CO and CO2 are vol% of the gases.

Example

Determine the losses due to formation of CO if coal with HHV of 12,000 Btu=lb is fired in a boiler, given that CO and CO2 in the flue gas are 1.5% and 17% and the fuel has a carbon content of 56%.

Solution. Substituting into the equation given above,

or L ¼ 3.8% on HHV basis (dividing loss in Btu=lb by HHV).

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6.22

Q:

Is there a simple formula to estimate the efficiency of boilers and heaters if the excess air and exit gas temperature are known and the fuel analysis is not available?

A:

Boiler efficiency depends mainly on excess air and the difference between the flue gas exit temperature and the ambient temperature. The following expressions have been derived from combustion calculations for typical natural gas and oil fuels. These may be used for quick estimations.

For natural gas:

For fuel oils:

ZHHV, % ¼ 92:9 ð0:001298 þ 0:01905 EAÞ DT

ZLHV, % ¼ 99:0 ð0:001383 þ 0:0203 EAÞ DT

where

EA ¼ excess air factor (EA ¼ 1.15 means 15% excess air)

DT ¼ difference between exit gas and ambient temperatures

Example

Natural gas at 15% excess air is fired in a boiler, with exit gas temperature 280 F and ambient temperature 80 F. Determine the boiler efficiency. EA ¼ 1:15 and DT ¼ 280 80 ¼ 200 F.

Solution.

ZHHV ¼ 89:4 ð0:001123 þ 0:0195 1:15Þð280 80Þ ¼ 84:64%

ZLHV ¼ 99:0 ð0:001244 þ 0:0216 1:15Þð280 80Þ ¼ 93:78%

The above equations are based on 1% radiation plus unaccounted losses.

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6.23

Q:

The average surface temperature of the aluminum casing of a gas-fired boiler was measured to be 180 F when the ambient temperature was 85 F and the wind

casing ¼ 0.1.

A:

This example shows how radiation loss can be obtained from the measurement of casing temperatures. The wind velocity is 5 mph ¼ 440 fpm. From Q8.51 we see that the heat loss q in Btu=ft2 h will be

¼ 252 Btu=ft2 h

The total heat loss will be 2500 252 ¼ 0.63 106 Btu=h. The radiation loss on LHV basis will be 0.63 106 100=(50,000 1075) ¼ 1.17%. If the HHV of the fuel were 1182 Btu=scf, the radiation loss on HHV basis would be 0.63 1182=1075 ¼ 1.06%.

6.24

Q:

How does the radiation loss vary with boiler duty or load? How does this affect the boiler efficiency?

A:

The heat losses from the surface of a boiler will be nearly the same at all loads if the ambient temperature and wind velocity are the same. Variations in heat losses can occur owing to differences in the gas temperature profile in the boiler, which varies with load. However, for practical purposes this variation can be considered minor. Hence the heat loss as a percent will increase as the boiler duty decreases.

The boiler exit gas temperature decreases with a decrease in load or duty and contributes to some improvement in efficiency, which is offset by the increase in radiation losses. Hence there will be a slight increase in efficiency as the load increases, and after a certain load, efficiency decreases.

The above discussion pertains to fired water tube or fire tube boilers and not waste heat boilers, which have to be analyzed for each load because the gas flow

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