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

A:

Each pound of sulfur in fuel converts to 2 lb of SO2. Using natural gas at 20% excess air, S lb=MM Btu of SO2 is equivalent to

¼598S ppmvd

0.1lb=MM Btu of SOx is equivalent to 60 ppmv. [We are simply using Eq. (21) and substituting for MW and Y .]

Similarly, for No. 2 oil at 20% excess air;

6.26e

Q:

A gas turbine HRSG has the following data: Exhaust gas flow ¼ 500,000 lb=h at 900 F

Gas analysis vol%; CO2 ¼ 3; H2O ¼ 7; N2 ¼ 75; O2 ¼ 15. The exhaust gas has 9 lb=h of NOx and CO. The HRSG is fired to 1500 F using natural gas consisting of vol% methane ¼ 97, ethane ¼ 2, propane ¼ 1. Fuel input ¼ 90 MM LHV. HHV of fuel ¼ 23,790 Btu=lb, and LHV ¼ 21,439 Btu=lb. The burner contributes 0.05 lb=MM Btu of NOx and CO. Also see what happens when the burner contributes 0.1 lb=MM Btu of these pollutants. Flue gas analysis after combustion vol% CO2 ¼ 4:42, H2O ¼ 9:78, N2 ¼ 73:91, O2 ¼ 11:86, and flue gas flow ¼ 504,198 lb=h. Compute the NOx and CO levels in ppmvd corrected to 15% oxygen before and after the burner.

A:

We have to convert the mass flow of NOx and CO to volumetric units and correct for 15% oxygen dry basis.

In this example, the molecular weights of NOx ¼ 46, flue gas ¼ 28.38. The mass of CO remains the same, so ppmvd CO ¼ (46=28) 14.7 ¼ 24.2.

At the burner exit; the mass of NOx in the exhaust gases after combustion is

9 þ 90 23,79021,439 0:05 ¼ 14 lb=h

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Because the burner heat input is on LHV basis and emissions are on HHV basis, we correct the values using the above expression.

With 0.1 lb=MM Btu emissions from the burner, NOx ppmvd ¼ 19.5 and CO ppmvd ¼ 32.1 Thus both the burner contribution and the initial pollutant levels in the turbine exhaust gases affect the ppmv values after combustion. ppmvd values after the burner can be lower or higher than the inlet ppmvd values, though in terms of mass flow they will always be higher.

6.26f

Q:

Steam generator emissions are usually referred to 3% oxygen dry basis, and gas turbine or HRSG emissions are referred to 15% oxygen dry basis. However, in operation, different excess air rates are used that generate flue gases with different oxygen levels. What is the procedure for converting from actual to 3% oxygen basis?

A:

21 3 ppm (@ 3% dry) ¼ ppm (actual) 21 O2 ðactualÞ

If dry oxygen in flue gases is 1.7% and 12 ppm of a pollutant is measured, then at 3% oxygen,

Emission ¼ 12 21 3 ¼ 11:2 ppm 21 1:7

6.27a

Q:

In gas turbine cogeneration and combined cycle projects, the heat recovery steam generator may be fired with auxiliary fuel in order to generate additional steam. One of the frequently asked questions concerns the consumption of oxygen in the exhaust gas versus fuel quantity fired. Would there be sufficient oxygen in the exhaust to raise the exhaust gas to the desired temperature?

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A:

Gas turbine exhaust gases typically contain 14–16% oxygen by volume compared to 21% in air. Hence generally there is no need for additional oxygen to fire auxiliary fuel such as gas or oil or even coal while raising its temperature. (If the gas turbine is injected with large amounts of steam, the oxygen content will be lower, and we should refer the analysis to a burner supplier.) Also, if the amount of fuel fired is very large, then we can run out of oxygen in the gas stream. Supplementary firing or auxiliary firing can double or even quadruple the steam generation in the boiler compared to its unfired mode of operation [1]. The energy Q in Btu=h required to raise Wg lb=h of exhaust gases from a temperature of t1 to t2 is given by

Q ¼ Wg ðh2 h1Þ

where

h1; h2 ¼ enthalpy of the gas at t1 and t2, respectively

The fuel quantity in lb=h is Wf in Q=LHV, where LHV is the lower heating value of the fuel in Btu=lb.

If 0% volume of oxygen is available in the exhaust gases, the equivalent amount of air Wa in the exhaust is [9]

In this equation we are merely converting the moles of oxygen from volume to weight basis. A molecular weight of 29.5 is used for the exhaust gases, and 32 for oxygen. The factor 100=23 converts the oxygen to air.

Now let us relate the air required for combustion with fuel fired. From Q5.03– Q.5.05 we know that each MM Btu of fuel fired on HHV basis requires a constant amount A of air. A is 745 for oil and 730 for natural gas; thus, 106=HHV lb of fuel requires A lb of air. Hence Q=LHV lb of fuel requires

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Now for natural gas and fuel oils, it can be shown that LHV=ðA HHVÞ ¼ 0:00124. Substituting into Eq. (24), we get

This is a very important equation, because it relates the energy input by the fuel (on LHV basis) with oxygen consumed.

Example

It is desired to raise the temperature of 150,000 lb=h of turbine exhaust gases from 950 F to 1575 F in order to double the output of the waste heat boiler. If the exhaust gases contain 15 vol% of oxygen, and the fuel input is 29 MM Btu=h (LHV basis), determine the oxygen consumed.

Solution. From Eq. (24),

29 106

O ¼ 150,000 58:4 ¼ 3:32%

Hence if the incoming gases had 15 vol% of oxygen, even after the firing of 29 MM Btu=h we would have 15 7 3.32 ¼ 11.68% oxygen in the exhaust gases.

A more accurate method would be to use a computer program [9], but the above equation clearly tells us if there is likely to be a shortage of oxygen.

6.27b

Q:

150,000 lb=h of turbine exhaust gases at 900 F having a gas analysis (vol%) of CO2 ¼ 3; H2O ¼ 7; N2 ¼ 75 and O2 ¼ 15 enters a duct burner, and 35 MM Btu=h (LHV) of natural gas is fired. Determine the exhaust gas analysis after the burner. Use 100% methane as fuel gas analysis for illustrative purposes.

A:

From Table 6.3, the LHV ¼ 21,520 Btu=lb. Hence fuel fired ¼ 35 106= 21,520 ¼ 1626 lb=h.

From combustion basics,

CH4 þ 2O2 ! CO2 þ 2H2O

So 16 lb of methane requires 64 lb of oxygen and yields 44 lb of CO2 and 36 lb of water vapor, using molecular weights of 16 for methane, 32 for oxygen, 44 for carbon dioxide, and 18 for water vapor. Hence 1626 lb=h of methane will consume

1626 ð64=16Þ ¼ 6504 lb=h of oxygen

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Also, it will increase CO2 by 1626 ð44=16Þ ¼ 4471 lb=h

H2O will increase by

1626 ð36=16Þ ¼ 3659 lb=h

Convert the volume percent in incoming exhaust gases to weight percent basis as follows. The molecular weight of incoming gases is 0:03 44 þ 0:07 18 þ 0:75 28 þ 0:15 32 ¼ 28:38

Fraction by weight of CO2 ¼ 0:03 44=28:38 ¼ 0:0465

H2O ¼ 0:07 18=28:38 ¼ 0:0444

N2 ¼ 75 28=28:38 ¼ 0:74

O2 ¼ 0:15 32=28:38 ¼ 0:1691

The amounts of these gases in incoming exhaust gas in lb=h:

CO2 ¼ 150,000 0:0465 ¼ 6975 lb=h

H2O ¼ 150,000 0:0444 ¼ 6660 lb=h

N2 ¼ 150,000 0:74 ¼ 111,000 lb=h

O2 ¼ 150,000 0:1691 ¼ 25,365 lb=h

The final products of combustion will have

CO2 ¼ 6975 þ 4471 ¼ 11,446 lb=h

H2O ¼ 6660 þ 3659 ¼ 10,319 lb=h

N2 ¼ 111,000

O2 ¼ 25,365 6504 ¼ 18,861 lb=h

Total exhaust gas flow ¼ 11,446 þ 10,319 þ 111,000 þ 18,861

¼ 151,626 lb=h

which matches the sum of exhaust gas flow and the fuel gas fired.

To convert the final exhaust gas to vol% analysis, we have to obtain the number of moles of each constituent.

Moles of CO2 ¼ 11,446=44 ¼ 260:1

H2O ¼ 10,319=18 ¼ 573:2

N2 ¼ 111,000=28 ¼ 3964:3

O2 ¼ 18,861=32 ¼ 589:4

Total moles ¼ 5387

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Hence

CO2 ¼ 260:1=5387 ¼ 0:0483, or 4:83% by volume

Similarly,

H2O ¼ 573:2=5387 ¼ 0:1064, or 10:64 vol%

N2 ¼ 3964:2=5387 ¼ 0:7359; or 73:59 vol%

O2 ¼ 589:4=5387 ¼ 0:1094, or 10:94 vol%

agrees with the detailed calculations.

When possible, detailed combustion calculations should be done because they also reveal the volume percent of water vapor, which has increased from 7% to 10.64%. This would naturally increase the gas specific heat or its enthalpy and affect the heat transfer calculations.

Table 6.11 shows the exhaust gas analysis at various firing temperatures.

6.27c

Q:

Determine the final exhaust gas temperature after combustion in the example in Q6.27b.

A:

To arrive at the final gas temperature, the enthalpy of the exhaust gases must be obtained. A simplistic specific heat assumption can also give an idea of the temperature but will not be accurate.

TABLE 6.11 Effect of Firing Temperature on Exhaust Gas Analysis

150,000 lb=h of exhaust gases at 900 F. Exhaust gas analysis (vol%): CO2 ¼ 3, H2O ¼ 7, N2 ¼ 75, O2 ¼ 15. Natural gas: C1 ¼ 97 vol%, C2 ¼ 3 vol%.

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Using, say, 0.3 Btu=lb F for the average gas specific heat for the temperature range in consideration, the increase in gas temperature is

35 106=ð150,000 0:3Þ ¼ 777 F

or

Final gas temperature ¼ 900 þ 777 ¼ 1677 F

However, let us use gas enthalpy calculations, which are more accurate. Figure 6.3 shows the gas enthalpy for the turbine exhaust gas at various temperatures. (A program was used to compute these values based on the enthalpy of individual constituents.) Enthalpy of exhaust gas at 900 F ¼ 220 Btu=lb.

From an energy balance across the burner;

150,000 220 þ 35 106 ¼ 151,626 hg

where hg ¼ enthalpy of final products of combustion. hg ¼ 448.5 Btu=lb. From the chart, the gas temperature ¼ 1660 F.

A computer program probably gives more accurate results, because it can compute the gas temperature and enthalpy for any gas analysis and iterate for the actual enthalpy, whereas a chart can be developed only for a given exhaust gas analysis and a maximum firing temperature.

FIGURE 6.3 Enthalpy of turbine exhaust gas as a function of temperature.

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6.28

Q:

How can the fuel consumption for power plant equipment such as gas turbines and diesel engines be determined if the heat rates are known?

A:

The heat rate (HR) of gas turbines or engines in Btu=kWh refers indirectly to the efficiency.

Efficiency ¼ 3413HR

where 3413 is the conversion factor from Btu=h to kW. One has to be careful about the basis for the heat rate, whether it is on HHV or LHV basis. The efficiency will be on the same basis.

Example

If the heat rate for a gas turbine is 9000 Btu=kWh on LHV basis and the higher and lower heating values of the fuel are 20,000 and 22,000 Btu=lb, respectively, then

3413

Efficiency on LHV basis ¼ 9000 ¼ 0:379; or 37:9%

To convert this efficiency to HHV basis, simply multiply it by the ratio of the heating values:

Efficiency on HHV basis ¼ 37:9 20,00022,000 ¼ 34:45%

NOMENCLATURE

ATheoretical amount of air for combustion per MM Btu fired, lb

eEmission rate of sulfur dioxide, lb=MM Btu

EExcess air, %

KConstant used in Eq. (7)

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qHeat loss, Btu=ft2 h

QEnergy, Btu=h or kW

sSpecific gravity

SSulfur in fuel

wWeight of air, lb=lb fuel; subscript da stands for dry air; wa, wet air; wg, wet gas; dg, dry gas

WMoisture, lb=h

ZEfficiency; subscripts HHV and LHV denote the basis

rDensity, lb=cu ft; subscript g stands for gas, f for fuel

REFERENCES

1.V Ganapathy. Applied Heat Transfer. Tulsa, OK: PennWell Books, 1982, pp 14–24.

2.North American Combustion Handbook. 2nd ed. Cleveland, OH: North American Mfg. Co., 1978, pp 9–40.

3.Babcock and Wilcox. Steam: Its Generation and Use. 38th ed. New York, 1978, p 6–2.

4.V Ganapathy, Use chart to estimate furnace parameters. Hydrocarbon Processing, Feb 1982, p 106.

5.V Ganapathy. Figure particulate emission rate quickly. Chemical Engineering, July 26, 1982, p 82.

6.V Ganapathy. Nomogram estimates melting point of ash. Power Engineering, March 1978, p 61.

7.ASME. Power Test Code. Performance test code for steam generating units, PTC 4.1. New York: ASME, 1974.

8.V Ganapathy. Estimate combustion gas dewpoint. Oil and Gas Journal, April 1978, p 105.

9.V Ganapathy. Waste Heat Boiler Deskbook. Atlanta, GA: Fairmont Press, 1991.

Copyright © 2003 Marcel Dekker, Inc.

10.V Ganapathy. Converting ppm to lb=MM Btu; an easy method. Power Engineering, April 1992, p 32.

11.KY Hsiung. Predicting dew points of acid gases. Chemical Engineering, Feb 9, 1981, p 127.

12.C Baukal Jr. The John Zink Combustion Handbook. Boca Raton, FL: CRC Press, 2001.

13.Babcock and Wilcox, Steam, its generation and use, 40th ed. The Babcock and Wilcox Company, Barberton, Ohio, 1992.

14.AG Okkes. Get acid dew point of flue gas. Hydrocarbon Processing, July 1987.

Copyright © 2003 Marcel Dekker, Inc.