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

To estimate the gas density, its molecular weight must be obtained (see Q5.05).

MW ¼ PðMWi yiÞ

¼ 28 71:3 þ 18 17:7 þ 32 2:5 þ 44 8:5 100

¼ 27:7

Hence, from Eq. (6),

14:7

rg ¼ 27:7 492 359 760 14:7 ¼ 0:05 lb=cu ft

The gas pressure was assumed to be 14.7 psia. In the absence of flue gas analysis, we can obtain the density as discussed in Q5.03.

40

rg ¼ 760 ¼ 0:052 lb=cu ft

6.10a

Q:

Discuss the basis for the million Btu method of combustion calculations.

A:

Each fuel such as natural gas, coal, or oil requires a certain amount of stoichiometric air per MM Btu fired (on HHV basis). This quantity does not vary much with the fuel analysis and has therefore become a valuable method of evaluating combustion air and flue gas quantities produced when fuel gas analysis is not available.

For solid fuels such as coal and oil, the dry stoichiometric air wda in lb=lb fuel can be obtained from

where C; H2; O2, and S are carbon, hydrogen, oxygen, and sulfur in the fuel in fraction by weight.

For gaseous fuels, wda is given by

wda ¼ 2:47 CO þ 34:34 H2 þ 17:27 CH4

þ13:3 C2H2 þ 14:81 C2H4

þ16:12 C2H6 4:32 O2

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Solution. From (4a),

HHV ¼ 17,887 þ 57:5 28 102:2 0

¼ 19,497 Btu=lb

The amount of air in lb=lb fuel from the above equation is

wda ¼ 11:53 0:875 þ 34:34 0:125 ¼ 14:38 lb=lb fuel

1 MM Btu of fuel fired requires (1 106)=19,497 ¼ 51.28 lb of fuel. Hence, from the above, 51.28 lb of fuel requires

51:28 14:38 ¼ 737 lb of dry air

Table 6.4 shows a range of 735–750. To this must be added excess air; the effect of moisture in the air should also be considered.

Example 2

Let us take the case of natural gas with the following analysis: methane ¼ 83.4%, ethane ¼ 15.8%, and nitrogen ¼ 0.8%.

Solution. Converting this to percent weight basis, we have

Let us compute the air required in lb=lb fuel.

From Table 6.3,

Air required ¼ 17:265 0:7289 þ 16:119 0:2589

¼ 16:75 lb=lb fuel

HHV of fuel ¼ 0:7289 23,876 þ 0:2589 22,320

¼ 23,181 Btu=lb

where 23,876 and 22,320 are HHV of methane and ethane from Table 6.3.

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fired would need 722 lb of dry air; this is close to the value indicated in Table 6.4. Let us take the case of 100% methane and see how much air it needs for combustion. From Table 6.3, air required per pound of methane is 17.265 lb, and its heating value is 23,879 Btu=lb. In this case 1 MM Btu is equivalent to (1 106)=23,879 ¼ 41.88 lb of fuel, which requires 41.88 17.265 ¼ 723 lb of

dry air.

Taking the case of propane, 1 lb requires 15.703 lb of air.

1 MM Btu ¼ 1 106 ¼ 46:17 lb fuel 21,661

This would require 46.17 15.703 ¼ 725 lb of air.

Thus for all fossil fuels we can come up with a good estimate of theoretical dry air per MM Btu fired on HHV basis, and gas analysis does not affect this value significantly. The amount of air per MM Btu is termed A and is shown in Table 6.4 for various fuels.

6.10b

Q:

A fired heater is firing natural gas at an input of 75 MM Btu=h on HHV basis. Determine the dry combustion air required at 10% excess air and the amount of flue gas produced if the HHV of fuel is 20,000 Btu=lb.

A:

From Table 6.4, A is 730 lb=MM Btu. Hence the total air required is

Wa ¼ 75 1:1 730 ¼ 60,200 lb=h

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The flue gas produced is

106

Wg ¼ Wa þ Wf ¼ 60,200 þ 20,000 ¼ 60,250 lb=h

These values can be converted to volume rates at any temperature using the procedure described in Chapter 5.

The MM Btu method is quite accurate for engineering purposes such as fan selection and sizing of ducts and air and gas systems. Its advantage is that fuel analysis need not be known, which is generally the case in power and process plants. The efficiency of heaters and boilers can also be estimated using the MM Btu method of combustion calculations.

6.10c

Q:

A coal-fired boiler is firing coal of HHV ¼ 9500 Btu=lb at 25% excess air. If ambient conditions are 80 F, relative humidity 80%, and flue gas temperature 300 F, estimate the combustion air in lb=lb fuel, the volume of combustion air in cu ft=lb fuel, the flue gas produced in lb=lb fuel, and the flue gas volume in cu ft=lb fuel.

A:

Because the fuel analysis is not known, let us use the MM Btu method. From Table 6.4, A ¼ 760 for coal. 1 MM Btu requires 760 1.25 ¼ 950 lb of dry air. At 80% humidity and 80 F, air contains 0.018 lb of moisture per pound of air

¼ 0:0736 lb=cu ft ðsee Chap: 5; Q5:03Þ:

Hence

Volume of air ¼ 9:21 ¼ 125 cu ft=lb fuel 0:0736

rg ¼ density of flue gas ¼ 76040 ¼ 0:0526 lb=cu ft

wdg ¼ dry flue gas in lb=lb fuel ¼ 950 þ 105 ¼ 10:05 105

Volume of flue gas, cu ft=lb fuel ¼ 10:05 ¼ 191 0:0526

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6.11

Q:

Is there a way to figure the excess air from flue gas CO2 readings?

A:

Yes. A good estimate of excess air E in percent can be obtained from the equation

%CO2 is the percent of carbon dioxide in dry flue gas by volume, and K1 is a constant depending on the type of fuel, as seen in Table 6.5. For example, if %CO2 ¼ 15 in flue gas in a coal-fired boiler, then for bituminous coal (K1 ¼ 18.6),

E ¼ 100 1815:6 1 ¼ 24%

6.12

Q:

Discuss the significance of %CO2 and %O2 in flue gases.

A:

Excess air levels in flue gas can be estimated if the %CO2 and %O2 in dry flue gas by volume are known. The higher the excess air, the greater the flue gas quantity and the greater the losses. Plant engineers should control excess air levels to help control plant operating costs. The cost of operation with high excess

Source: Ref. 1.

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where O2; CO, and N2 are the oxygen, carbon monoxide, and nitrogen in dry flue gas, vol%, and E is the excess air, %.

where K2 is a constant that depends on the type of fuel (see Table 6.6).

6.13

Q:

In a natural gas boiler of capacity 50 MM Btu=h (HHV basis), the oxygen level in the flue gas is reduced from 3.0% to 2.0%. What is the annual savings in operating costs if fuel costs $4=MM Btu? The HHV of the fuel is 19,000 Btu=lb. The exit gas temperature is 500 F, and the ambient temperature is 80 F.

A:

The original excess air is 90 3=ð21 3Þ ¼ 15% (see Q6.12). The excess air is now

Source: Ref. 1.

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With 9.47% excess air,

Air required ¼ 50 746 1:0947 ¼ 40,832 lb=h

¼ 43,463 lb=h

Reduction in heat loss ¼ ð45,526 43,463Þ 0:25 ð500 80Þ

¼ 0:22 MM Btu=h

This is equivalent to an annual savings of 0.22 4 300 24 ¼ $6336. (We assumed 300 days of operation a year.) This could be a significant savings considering the life of the plant. Hence plant engineers should operate the plant realizing the implications of high excess air and high exit gas temperature. Oxygen levels can be continuously monitored and recorded and hooked up to combustion air systems in order to operate the plant more efficiently. (It may be noted that exit gas temperature will also be reduced if excess air is reduced. The calculation above indicates the minimum savings that can be realized.)

6.14

Q:

Fuels are often interchanged in boiler plants because of relative availability and economics. It is desirable, then, to analyze the effect on the performance of the system. Discuss the implications of burning coal of 9800 Btu=lb in a boiler originally intended for 11,400 Btu=lb coal.

A:

Let us assume that the duty does not change and that the efficiency of the unit is not altered. However, the fuel quantity will change. Combustion air required, being a function of MM Btu fired, will not change, but the flue gas produced will increase. Let us prepare a table.

We can use the same fans, because the variation in flue gas produced is not significant enough to warrant higher gas pressure drops. We must look into other

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aspects, such as the necessity of higher combustion air temperature (due to higher moisture in the fuel), ash concentration, and fouling characteristics of the new fuel. If a different type of fuel is going to be used, say oil, this will be a major change, and the fuel-handling system’s burners and furnace design will have to be reviewed. The gas temperature profiles will change owing to radiation characteristics, and absorption of surfaces such as superheaters and economizers will be affected. A discussion with the boiler design engineers will help.

6.15

Q:

What is meant by combustion temperature of fuels? How is it estimated?

A:

The adiabatic combustion temperature is the maximum temperature that can be attained by the products of combustion of fuel and air. However, because of dissociation and radiation losses, this maximum is never attained. Estimation of temperature after dissociation requires solving several equations. For purposes of estimation, we may decrease the adiabatic combustion temperature by 3–5% to obtain the actual combustion temperature.

From an energy balance it can be shown that

where

LHV; HHV ¼ lower and higher calorific value of fuel, Btu=lb A ¼ theoretical air required per million Btu fired, lb a ¼ excess air factor ¼ 1 þ E=100

ta; tc ¼ temperature of air and combustion temperature, F

Cpa; Cpg ¼ specific heats of air and products of combustion, Btu=lb F

Cpa and Cpg were taken as 0.25 and 0.32, respectively.

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6.16a

Q:

How is the ash concentration in flue gases estimated?

A:

Particulate emission data are needed to size dust collectors for coal-fired boilers. In coal-fired boilers, about 75% of the ash is carried away by the flue gases and 25% drops into the ash pit. The following expression may be derived using the MM Btu method of combustion calculation [5]:

where

Ca ¼ ash concentration, grains=cu ft

E ¼ excess air, %

T ¼ gas temperature, R

HHV ¼ higher heating value, Btu=lb

Example

If coals of HHV ¼ 11,000 Btu=lb having 11% ash are fired in a boiler with 25% excess air and the flue gas temperature is 850 R, determine the ash concentration.

Solution. Substituting into Eq. (12a), we have

240,000 0:11

Ca ¼ 850 ð7:6 10 6 11,000 125 þ 1 0:11Þ

¼ 2:75 grains=cu ft

6.16b

Q:

How do you convert the ash concentration in the flue gas in wt% to grains=acf or grains=scf?

A:

Flue gases from incineration plants or solid fuel boilers contain dust or ash, and often these components are expressed in mass units such as lb=h or wt%, whereas engineers involved in selection of pollution control equipment prefer to work in terms of grains=acf or grains=scf (actual and standard cubic feet). The relationship is

Ca ¼ 0:01 A 7000 r ¼ 70A ð12bÞ

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where

r ¼ gas density, lb=cu ft ¼ 39.5=(460 þ t) t ¼ gas temperature, F

Ca ¼ ash content, grains=acf or grains=scf depending on whether density is computed at actual temperature or at 60 F

A ¼ ash content, wt%

The expression for density is based on atmospheric flue gases having a molecular weight of 28.8 (see Q5.03).

Flue gases contain 1.5 wt% ash. The concentration in grains=acf at 400 F is

39:5

Ca ¼ 70 1:5 860 ¼ 4:8 grains=acf

and at 60 F,

Ca ¼ 70 1:5 39520:5 ¼ 7:98 grains=scf

6.17

Q:

Discuss the importance of the melting point of ash in coal-fired boilers. How is it estimated?

A:

In the design of steam generators and ash removal systems, the ash fusion temperature is considered an important variable. Low ash fusion temperature may cause slagging and result in deposition of molten ash on surfaces such as superheaters and furnaces. The furnace will then absorb less energy, leading to higher furnace exit gas temperatures and overheating of superheaters.

A quick estimate of ash melting temperature in C can be made using the expression [6]

tm ¼ 19 Al2O3 þ 15 ðSiO2 þ TiO2Þ

þ10 ðCaO þ MgOÞ

þ6 ðFe2O3 þ Na2O þ K2OÞ

where tm is the fusion temperature in C, and the rest of the terms are percent ash content of oxides of aluminum, silicon, titanium, calcium, magnesium, iron, sodium, and potassium.

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