calculating boiler and process heater thermal efficiency

8/18/2019 Calculating Boiler and Process Heater Thermal Efficiency

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Note: The source of the technical material in this volume is the ProfessionalEngineering Development Program (PEDP) of Engineering Services.

Warning: The material contained in this document was developed for Saudi Aramco and is intended for the exclusive use of Saudi Aramco’s employees. Any material contained in this document which is not already in the publicdomain may not be copied, reproduced, sold, given, or disclosed to thirdparties, or otherwise used in whole, or in part, without the written permissionof the Vice President, Engineering Services, Saudi Aramco.

Chapter : Mechanical For additional information on this subject, contactFile Reference: MEX-104.06 PEDD Coordinator on 874-6556

Engineering Encyclopedia

Saudi Aramco DeskTop Standards

CALCULATING BOILER AND PROCESSHEATER THERMAL EFFICIENCY


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Engineering Encyclopedia Introduction to Boilers

Calculating Boiler and ProcessHeater Thermal Efficiency

Saudi Aramco DeskTop Standards i

Section Page

INFORMATION............................................................................................................... 3

CALCULATING THERMAL EFFICIENCY USING THE INPUT/OUTPUT ORDIRECT METHOD................................................................................. 3

Thermal Efficiency ................................................................................................ 3

Example Problem 1.................................................................................... 5

Input/Output or Direct Method.................................................................... 7

Example Problem 2.................................................................................... 7

CALCULATING THERMAL EFFICIENCY USING THE HEAT LOSS METHOD............. 9

Excess Air............................................................................................................. 9

Example Problem 3 Calculation Of Excess Oxygen ................................ 10

Stack (Flue Gas) Temperature ................................................................ 12

Heater Efficiency Calculation ................................................................... 14

Combustion Efficiency Charts............................................................................. 14

Example Problem 4.................................................................................. 15

Simplified Equation............................................................................................. 16

Thermal Efficiency Improvement ............................................................. 16

Example Problem 5 ............................................................................................ 17

Reduce Excess Air ............................................................................................. 21

Reduce Stack Temperature................................................................................ 23

Reduce Other Losses......................................................................................... 24

EFFECTS OF FIRING RATE ON THERMAL EFFICIENCY.......................................... 25

WORK AIDS.................................................................................................................. 26

WORK AID 1: PROCEDURE FOR CALCULATING THERMAL EFFICIENCYUSING INPUT/OUTPUT METHOD....................................................... 26

WORK AID 2: PROCEDURE FOR CALCULATING THERMAL EFFICIENCYUSING HEAT LOSS METHOD ............................................................. 27

Work Aid 2A: Excess Air and Thermal Efficiency Using Short Cut Equations.... 27

Work Aid 2B: Procedures for Calculating Furnace Efficiency by Heat Loss

Method ............................................................................................................... 28

WORK AID 3: FLUE GAS OXYGEN (DRY BASIS) VS. EXCESS AIR......................... 30

WORK AID 4: HEAT ABSORBED CHARTS ................................................................ 31

GLOSSARY .................................................................................................................. 36

ADDENDUM ................................................................................................................. 37

API - RP - 532 PROCEDURE ....................................................................................... 38


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Saudi Aramco DeskTop Standards ii

REFERENCES.............................................................................................................. 56

List of Figures

Figure 1. Steam Boiler System....................................................................................... 5

Figure 2. Excess Oxygen ............................................................................................. 10

Figure 3. Flue Gas Oxygen Versus Excess Air ............................................................ 11

Figure 4. Typical Aspirating (High Velocity) Thermocouple.......................................... 13

Figure 5. Combustion Heat Available to Process ......................................................... 14

Figure 6. Steam Boiler System..................................................................................... 17

Figure 7. Furnace Air Leaks......................................................................................... 22

Figure 8. Flue Gas Oxygen Versus Excess Air ............................................................ 30

Figure 9. Heat Available from the Combustion of 1000 Btu/ft3 Refinery Gas............... 31

Figure 10. Heat Available from the Combustion of 1600 Btu/ft3 Refinery Gas............. 32

Figure 11. Heat Available from the Combustion of 5º API Fuel Oil............................... 33

Figure 12. Heat Available from the Combustion of 10º API Fuel Oil............................. 34

Figure 13. Heat Available from the Combustion of 15º API Fuel Oil............................. 35

Figure 1A. Typical Heater Arrangement....................................................................... 38

Figure 2A. Vapor Pressure of Water ............................................................................ 41

Figure 3A. Enthalpy of Flue Gas Components ............................................................. 42

Figure 4A. Enthalpy of Flue Gas Components.............................................................. 43

Figure 5A. Combustion Work Sheet............................................................................. 46

Figure 5A. Combustion Work Sheet, (cont’d) ............................................................... 52

List of Tables

Table 1. Furnace Fuel Savings .................................................................................... 19


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Saudi Aramco DeskTop Standards 3

INFORMATION

CALCULATING THERMAL EFFICIENCY USING THE INPUT/OUTPUT

OR DIRECT METHOD

Thermal Efficiency

Thermal efficiency is defined as the percentage of the absorbedenergy to the total energy input. Calculation of thermal efficiencyis based on an energy balance around the boiler or processheater.

In a boiler, although only the energy in the steam is usable, theheat absorbed in a boiler is the sum of the energy in the steamand the energy in the blowdown above that of the boiler feed

water. The energy in the stack gas above that of ambient air is aloss. The energy transferred from the boiler through theinsulation and refractory to the atmosphere is also a loss. In aprocess heater, heat losses are the same and include losses tothe stack gases and losses to the atmosphere through therefractory and insulation.

Factors that increase the losses will decrease the thermalefficiency. For example, operating with too much excess airreduces the thermal efficiency by increasing the stack heat lossbecause the excess air is heated from ambient to stack gastemperature.

The thermal efficiency for which a boiler or a process heater isdesigned is an economic evaluation involving the cost of fueland the cost of equipment to reduce the losses. Examples ofeconomic analyses include the amount of insulation orrefractory used to reduce heat losses to the atmosphere, theamount of heat transfer surface provided in the radiant andconvection sections to reduce the stack temperature, use of apreheater to reduce the stack gas temperature, types of burnersused (determines minimum excess air requirement) and the useof chemicals to reduce the blowdown requirement.


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The thermal efficiency can be calculated using either the higherheating value (HHV) or the lower heating value (LHV). The LHVis a better measure of achievable thermal efficiency since thelatent heat of vaporization of the water in the flue gas cannot be

recovered. The HHV efficiency is several percentage pointslower than the LHV efficiency. It is common practice in thefurnace industry to use the LHV in calculations while the boilerindustry uses the HHV efficiency. All calculations will be done ona LHV basis including boilers.

The Example Problem 1 shows the calculation of the thermalefficiency and the magnitude of the heat losses.


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Heat Balance: Water side/Process

Heat In: Feedwater = 910,000 lb/hr x 158 = 143.78 MBtu/hr (M = million)

Heat Absorbed = Q A Total = 143.78 + Q A

Heat Out: Steam = 805,000 x 1,237.6 = 996.27Blowdown = 105,000 x 343.5 = 36.07Total = 1,032.34 MBtu/hr

Heat In = Heat Out143.78 + Q A = 1,032.34

Heat absorbed = Q A = 1,032.34 - 143.78 = 888.56 M Btu/hr

Fuel Heat InputQP = 55,000 x 19,400 = 1067 million Btu/HR.

Pump Energy

( )( )( )

( )( )( )

Hp10790.701715

10812000

0.651715

∆PgpmHPp ===

HPp = 1079 x 2544 = 2.7 million Btu/hr

Fan Energy

Btu/hr million0.5=2544x200HP

Hp200HP

F

F

=

=

Total Energy Input

Qin. = 1067 + 2.7 + 0.5 = 1070.2

Total Energy Efficiency

( ) ( )83.0%

1070.2

100888.6

Q

100QEff.

M

A ===


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To simplify the thermal efficiency calculation the energy input from pump and fan areignored because these are relatively small and fairly constant. If this is done then:

( )

83.2%1067

100888.6

Eff.LHV==

Blowdown (Unusable Energy)

QBD = 105,000(343.5 – 158) = 19.5 million Btu/hr

( ) 1.8%=1001067

19.5=Loss

Heat Losses

To atmosphere = 2% given.To blowdown = 1.8%.

Flue gas loss = 100 - 83.2 - 2 - 1.8 = 13%

Input/Output or Direct

Method

The input/output or direct method is used whenever the heat absorbed by the boiler orprocess heater can be measured. This is the usual method for boilers and is used forprocess heaters only when there is a known amount of vaporization of the process fluid.

The energy balance on a boiler requires knowing all the rates on the boiler. Often theblowdown (BD) rate is not measured. Sometimes the boiler feed water (BFW) rate is notmeasured. The steam rate is always measured. Knowing the concentrations of oneimpurity in both the BFW and the BD allows the calculation of the material and energybalances. Example Problem 2 illustrates this calculation.

Example Problem 2

Calculate thermal efficiency of a boiler given the following data:

Steam production = 500,000 lb./hr at 434ºF and 140 psig

BFW chloride = 0.2 wppmBD chloride = 10 wppmFuel fired = 673,576 ft3/hr with 1005 Btu/ ft3


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Enthalpies

Enthalpy, Btu/lb

Stream Temp. ºF Psia HL H

V

Blowdown 370 174.7 343.5 1196.4

Steam 434 154.7 -- 1237.6

Feed water 190 -- 158.0 --

Solution:

Material Balance

lb/hr 10,204F0.02F

lb/hr 510,204F

500,000F0.98

F0.02500,000F

F0.02F10

0.2F

C

CF

F+CFC

F+FF

BFWBD

BFW

BFW

BFWBFW

BFWBFWBFW

BD

BFWBD

BDBDBFWBFW

BDsBFW

==

=

=

+=

=

=

=

=

=

Heat Absorbed, Q A

Heat In 106 Btu/hr Heat Out 106 Btu/hr

BFW = 510.204 (158) = 80.6 Steam = 500,000 x 123.7.6 = 618.8

Heat Absorbed Q A BD = 10,204 x 343.5 3.5

Total 80.6 + Q A Total 622.3

Heat In = Heat out

80.6 + Q A = 622.3

Q A = 622.3 – 80.6 = 541.7 million Btu/hr

Heat Fired

QF = 673,756 x 1005 = 677.1 million Btu/hr

Thermal Efficiency

( ) ( )80.0%

677.1

100541.7

Q

100QEff LHV

F

A ===


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CALCULATING THERMAL EFFICIENCY USING THE HEAT LOSS

METHOD

The heat loss method calculation is used when the heatabsorbed cannot be readily calculated such as most processheaters. The heat absorbed can be calculated by subtractingthe heat losses from the heat fired. In a boiler or process heaterthe primary heat loss is that lost to the stack gas. The heat lossin the stack is a function of the stack temperature, the amount ofexcess air and the carbon and hydrogen ratio in the fuel. Amaterial and energy balance can be calculated knowing theabove parameters.

Excess Air

The amount of excess air is defined as a percentage of the air inthe flue gas to the air that is required for complete combustion.Excess air and excess oxygen are numerically equivalentbecause the numerator and denominator are both multiplied bythe same constant to convert from one to the other.

Analysis from the lab will always be on a dry basis. Stack gasanalyzers that sample the stack gas will dry the stack gasbefore analysis. Stack gas analyzers that are in the stackmeasure on a wet basis but may be calibrated to report on a drybasis.

The calculation based on a dry flue gas analysis is outlined inFigure 2 and detailed in Example Problem 3.


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Normally there is no correction for incomplete combustionshown in step 3 of Figure 2 because the carbon monoxide (CO)concentration is usually negligible (10-50 ppm).

1. Obtain flue gas analyses CO2, CO, O2, N2.

2. From the percent N2, calculate the total O2 into the furnace.

3. Reduce the free O2 by the amount required to burn the CO to CO2.

The remaining free O2 is excess. (CO is usually negligible)

4. O2 required = (total in) less (excess)

5. ( )( )

( )x100

excess-total

excess=x100

Orequired

Oexcess=OexcessPercent

2

22

Figure 2. Excess Oxygen

Example Problem 3

Calculation Of Excess Oxygen

Lab Flue gas analysis: CO2 9.5

CO 1.8

O2 2.0

N2 86.7

100.0

Air composition: 21% O2, 79% N2

O2 into furnace = 86.7 x0.79

0.21 = 23.0 moles/100 moles flue gas

1.8 CO + 0.9O2 —→1.8 CO2

(Note: Usually CO is in parts per million and this correction can be ignored)

Net O2 = 2.0 - 0.9 = 1.1 moles/100 moles flue gas

Percent excess O2 =

( )1.123

1.1

−

x 100 = 5.02%

If there were no CO in the stack gas, the above analysis would have 11.3% CO2 and

the percent excess O2 would have been:

Percent excess O2 =( )

( )2.0231002.0

− x 100 = 5.02%


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Figure 3 (Work Aid 3) can also be used to calculate excess air (oxygen) once theoxygen has been adjusted for complete combustion. For 1.1% O2 Figure 3 gives an

excess air of 5%. For 2.0% O2 Figure 3 gives an excess air of 9%. This checks our

previous calculations.

Figure 3. Flue Gas Oxygen Versus Excess Air

Excess air and excess oxygen are numerically equal, because both numerator anddenominator are multiplied by the same constant to convert between the two. % O2 in

flue gas is not % excess O2. Considering these equal is a common error.

The following shortcut equations can also be used to estimate percent excess air.These equations assume complete combustion and a nominal carbon to hydrogen ratio.

When the flue gas analysis is on a wet basis:

%O-20.95

%Ox111.4 Air Excess

2

2=

where: %O2 = Percent oxygen in the flue gas.

For 2% O2 in the stack gas.

11.8%=18.95

222.8=

2-20.95

2x111.4 Air Excess =


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When the flue gas analysis is on a dry basis:

%O-20.95

%Ox91.2 Air Excess

2

2=

For 2% O2 in the stack gas.

9.6%= 18.95

182.4=

2-20.95

2x91.2 Air Excess =

Lab analysis is always on a dry basis because the water drops out as the gas samplecools. When the oxygen analyzer is located in the stack, the oxygen is measured on thewet basis but the analyzer may be calibrated using lab results so that it reports on a drybasis. When the flue gas is extracted from the stack and is transported to an analyzerthat is located some distance away, the analysis is on the dry basis.

The precise relationship between oxygen content and excess air is a function of thehydrogen-to-carbon ratio of the fuel. However, there is very little change in thisrelationship over a wide range of fuels at low excess air rates as shown in Figure 3(Work Aid 3).

Stack (Flue Gas)

Temperature

Another potential source of error in all efficiency calculations is an error in stacktemperature measurements. Ordinary stack temperature thermocouples can read lowby as much as 100ºF, depending upon their location and the flue gas temperature beingmeasured. If the thermocouple can "see" cold surroundings, such as the top of theconvection section or the sky, the indicator will likely read low. The higher the actualstack temperature, the higher the radiation losses and thus, the higher the error. Theaspirating thermocouples shown in Figure 4 minimizes any error due to radiation.


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6

6

5

43

6

6

2 2

A–A

2

2

A

7A

122

8

1

1. = Thermocouple junction.

2. = Thermocouple wires to temperature-indicating instrument.

3. = Outer thin-wall 310 stainless steel tube.

4. = Middle thin-wall 310 stainless steel tube.

5. = Center thin-wall 310 stainless steel tube.

6. = Centering tripods.

7. = Air or steam at 10 lb/sq in. gage or more in increments of 10 lb/sq in.until stable.

8. = Hot gas eductor.

From Furnace Operations, Third Edition by Robert Reed. Copyright © 1981 by Gulf Publishing Company,Houston, Texas. Used with permission. All rights reserved.

Figure 4. Typical Aspirating (High Velocity) Thermocouple


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Heater Efficiency

Calculation

API RP 532 specifies a detailed procedure for calculating the

thermal efficiency. This procedure is long and requires ananalysis of the fuel composition. This procedure is included inthe Addendum with an example problem and a blank calculationsheets.

The API RP 532 procedure is a detailed heat balance on thecombustion side of the furnace to determine the amount of heatlost up the stack.

Combustion Efficiency Charts

The API material and heat balance has been solved for anumber of cases and these cases plotted as heat availablecharts to simplify the calculations. These charts are attached asWork Aid 4. Work Aid 4 has charts for 1000 Btu/ft3 gas, 1600Btu/ft3 gas, 5º API fuel oil, 10º API fuel oil, and 15º API fuel oil.

All charts have the general relationship shown in Figure 5.Figure 5 shows that the heat available to the process is reducedas excess air is increased and a stack gas temperature isincreased.

Flue Gas Temperature

H e a t A v a i l a b l e f r o m F

l u e G a s

A b o v e 6 0 º F ,

B t u / l b F u e l

0% Excess Air

1020

Useful in furnace design.Useful in calculating furnace efficiency.

Figure 5. Combustion Heat Available to Process

Example Problem 4 illustrates the use of these charts incalculating thermal efficiency.


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Example Problem 4

Process heat absorbed = Q A = 353 MBtu/hr

Stack temperature = 600ºF (from stack TI)Percent excess air = 5%

Fuel = 1000 Btu/ft3 fuel gas LHV (Fuel rate not measured)19,700 Btu/lb LHV (from refinery utilities coordinator)

From Heat Available Curve: Work Aid 4 for 1000 Btu/ft3 gas.

H A = 17,100 Btu/lb fuel at 600ºF and 5% excess air

lb/hr 20,643Btu/lb17,100

Btu/hr 10x353

H

QF=FuelNet

6

A

AN ===

Assume furnace box losses QL are 2%. (Usually 2 - 3%)

Gross fuel = FG = 1.02 x 20,643 = 21,056 lb/hr

Heat fired = QF = 21,056 x 19,700 Btu/lb = 414.8 x 106 Btu/hr

( ) ( ) ( ) 85.1%100x414.8x10

353x10100

Q

Q100

firedheat

absorbedheat=efficiencyLHV

6

6

F

A ===

Given the heat absorbed, the heat loss method will calculate the fuel consumed. If a fuelmeter is available the calculated fuel rate should be rationalized with the fuel meterreadings.

If only the thermal efficiency is desired the calculation simplifies to the following:

From above we have:

H A = 17,100 Btu/lb. fuel at 600ºF and 5% excess air (heat absorbed from chart)

HF = 19,700 Btu/lb. fuel (heating value of fuel for chart used)

QL = 2% (Percent heat release/lost to atmosphere)

HL = Heat loss, decimal fraction

( )( )

( )( )( )

85.1%20,094

10017,100

0.02+119,700

10017,100

H1H

100H=efficiencyLHV

0.02100

2

100

QH

LF

A

LL

==

=

+

===


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Simplified Equation

A simplified (shortcut) equation can also be used to estimate LHV thermal efficiency.The simplified equation assumes a nominal heating value of the fuel (carbon to

hydrogen ratio).

( )( )( )( )[ ]

+−+−

L

ASTQ100

100TTEA0.0001890.0237100=efficiencyPercent

where: EA = Percent excess air.

TST = Stack temperature, ºF.

T A = Ambient air temperature, ºF.

QL = Casing heat loss, %.

For Example Problem 4 conditions and assuming the ambient temperature is 80ºF, thefurnace efficiency calculated by the shortcut formula is as follows:

( )( )( )( )[ ]

( )( )[ ]( ) 85.50.98045200.0246-100=efficiencyPercent2100

1008060050.000189+0.0237-100=efficiencyPercent

=

+−

This is a close check to the 85.1% calculated in Example Problem 4.

Thermal Efficiency

Improvement

Example Problem 5 calculates the thermal efficiency for a forced circulation boiler andthe changes in thermal efficiency that would result from reductions in stack temperature,blowdown rate, and excess air.


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Example Problem 5

Introduction:

In this example we will perform an energy balance around a boiler system and calculatethe fuel it requires. We will also examine methods of efficiency improvement.

Directions:

Calculate the fuel and boiler feedwater required for the boiler system shown in Figure 6.How can the furnace efficiency be improved?

• Use 2% for heat losses.

• Use 10% blowdown (BFW basis).

• For convenience, the required enthalpy data are given below:

Stream Temp. ºF psia HI HV

Feedwater 180 - - 148.0 --

Steam 700 600 -- 1351.8

Blowdown 492 633 478.5 1203.1

Figure 6. Steam Boiler System


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Answer:

Material Balance:

Feedrate = FBlowdown = 0.1 F

Steam product = 250,000

Material balance, F = 250,000 + 0.1 F

Feedrate F = lb/hr 277,7780.9

250,000=

Blowdown 0.1 F = 27,778 lb/hr

Heat in: Feed water = 277,778 x 148 = 41.11 MBtu/hr Absorbed Heat = Q A

Total = 41.11 + Q A

Heat out: Blowdown = 27,778 x 478.5 = 13.29 MBtu/hrSteam = 250,000 x 1351.8 = 337.95Total = 351.24

Heat in = Heat out 41.11 + Q A = 351.24

Heat absorbed Q A = 351.24 - 41.11 = 310.13 MBtu/hr

Heat loss QL = 2%

Fuel LHV LHV = 19,400 Btu/lb

Heat available H A = 16,725 Btu/lb at 600°F Stack and 20% excess

air from Work Aid 4 using 1000 Btu/ft3 gaschart.

Net fuel FN =16,725

10x310.13 6 = 18,543 lb/hr

Gross fuel FG = 1.02 x 18,543 = 18,914 lb/hr

Heat fired QF = 18,914 x 19,400 = 366.93 MBtu/hr

( ) 84.52%x10010z366.93

10x310.13=100

Q

Q= efficiencyLHV

6

6

F

A =


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Shortcut efficiency assuming the atmospheric air temperature is 100°F:

Efficiency = [(100 -(0.0237 + (0.000189)(EA)))(TST - T A )][100/(100 + QL)]

= [(100 -(0.0237 + (0.000189)(20)))(600 - 100)][100/(100 + 2)]

= [(100 - 13.74)][0.9804] = 84.5%

To Increase Efficiency:

• Lower stack temperature.

− Add more surface to convection section and increase boiler feedwaterpreheat.

− Add more surface to convection section and preheat another process stream. A 50ºF reduction in stack temperature would increase efficiency from 84.5%

to 85.9%.

• Reduce blowdown rate.

− If boiler feedwater quality allows, the blowdown rate can be reduced.

− Reduction of blowdown from 10% to 2% would not increase the efficiency, butwould directly reduce fuel use by decreasing the process heat absorbed.

• Reduce percent excess air.

− A reduction of excess air from 20% to 10% increases efficiency from 84.5% to85.4%. This might be accomplished by changing burners and closer control ofexcess air.

As shown by the table below in Table 1, the improvements are all of the same order ofmagnitude. Which one (or all) is used depends on economics of the specific boilerunder consideration.

Case BaseLowerStackTemp.

ReduceBlowdown

ReduceExcess Air

Percent blowdown 10 10 2 10

Heat absorbed, MBtu/hr 310.13 310.13 302.63 310.13

Stack temperature, ºF 600 550 600 600

Excess air, percent 20 20 20 10

Furnace efficiency, percent 84.52 85.91 84.52 85.40

Fuel savings, percent Base 1.62 2.42 1.04

Table 1. Furnace Fuel Savings


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Calculation for efficiency Improvement:

Case 1 2 3 4

Base

LowerStack Temp. Reduce Blowdown

Reduce %Excess Air

Heat in

277,778 x 148 = 41.11 41.11 255,103 x 148 = 37.76 41.11

Heat out

27,778 x 478.5 = 13.29 5,102 x 478.5 = 2.44

250,000 x 1,351.8 = 337.95 250,000 x 1,351.8 = 337.95

351.24 351.24 340.39 351.24

Heat absorbed 310.13 310.13 302.63 310.13

Stack 600 550 600 600

Percent excess air 20 20 20 20

Heat loss 2% 2% 2% 2%

Fuel LHV 19,400 19,400 19,400 19,400

Heat Avail.* 16,725 17,000 16,725 16,900

Net fuel 18,543 18,243 18,095 18,351

Gross fuel 18,914 18,608 18,457 18,718

Heat fired 366.93 360.99 358.06 363.13

LHV, percent eff. 84.52 85.91 84.52 85.40

Fuel savings Base 1.62% 2.41% 1.04%

*Maxwell p. 185


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Reduce Excess Air

All the air that enters a boiler or furnace is ultimately dischargedto the atmosphere at the stack temperature, and the energy it

contains is lost. The primary objective of efficient boiler andfurnace operations is to minimize airflow beyond that requiredfor good combustion. The air required for combustion shouldenter only through the burners. The following steps can betaken to reduce excess air:

1. Seal air leaks. This is particularly important in furnaces,which operate with a draft (negative pressure) throughoutthe furnace. These furnaces are more susceptible to airinfiltration. Figure 7 shows typical sources of air leaks into afurnace.

Since most boilers operate with a positive pressure throughmuch of the boiler, air leakage into boilers is much less aproblem.

2. Fire all burners at the same rate (close off idle burners).

3. Control furnace draft.

4. Determine excess air targets for each furnace through aseries of plant tests. These targets are the minimum excessair rates that are necessary for good combustion. Since notwo furnaces or boilers are exactly the same, there can be

different targets for each boiler and furnace in the plant.


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Inlet

Outlet

Leaky Coverson Observation

Doors

Clearance Around Tube

Penetration

CasingCorrosion

ConstructionJoint

Poor Seal on Access Door

IdleBurner

Figure 7. Furnace Air Leaks


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• Add improved combustion control systems.

- Automatic draft control on boilers and process heaters.

- Use closed loop oxygen and/or CO trim control.• Replace oversized burners. It is difficult to operate burners

efficiently at high turndown rates.

• Use high-capacity, high-intensity, or axial flow forced-draftburners for improved, low excess air combustion.

• Use low NOX burners for reduced emissions and low excess

air.

Reduce Stack Temperature

Fouling of the convection section tubes is the primary cause ofstack temperatures exceeding design. The extent of fouling canbe determined by visual inspection of the tubes or by observingan increase in stack temperature over time. A 40ºF increase instack temperature typically represents a loss of 1% in thermalefficiency.

Fouling can be reduced by operating sootblowers in boilers andfurnaces. Sootblowers should be provided for all boilers andfurnaces where heavy liquid fuels are fired. Units withoutsootblowers should be periodically cleaned during turnarounds.Fuel oil additives can be used to reduce deposits.

Reducing the stack temperature of a furnace or boiler that isoperating satisfactorily usually requires the addition of heattransfer surface. The following are means of reducing stacktemperature:

• Add heat transfer surface in convection section of processheaters.

• Add economizers on boilers to preheat the boiler feedwaterbefore entering the steam drum.

• Add combustion air preheaters. Air preheaters can transferheat from the flue gas leaving the stack, to the air used for

combustion. Depending upon the flue gas temperature, theincoming air can be heated several hundred ºF. The flue gastemperature should be kept above about 300ºF to preventcorrosion of the heat exchanger due to sulfuric acid in theflue gas.


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Reduce Other Losses

Although less important several other parameters listed belowcan improve boiler and process heater efficiency:

• Boiler blowdown should be controlled to the rate needed tomaintain boiler drum water impurities at the specifiedconcentration. Excess blowdown wastes heat and water.Heat can be recovered from the blowdown stream.

• Insulation should be maintained or improved to reduce heatlosses.

• Steam leaks should be repaired.


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EFFECTS OF FIRING RATE ON THERMAL EFFICIENCY

As the firing rate is increased the loss to the stack increasesprimarily because the heat transfer area is fixed. The increase in

heat loss is not necessarily proportional to the increase in firingrate. Increased loss will reduce thermal efficiency. Similarly adecrease in firing will slightly improve thermal efficiency. At verylow firing rates the heat losses to the atmosphere becomesignificant and the thermal efficiency may decrease.

Over firing a boiler or a process heater will reduce thermalefficiency.


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WORK AIDS

WORK AID 1: PROCEDURE FOR CALCULATING THERMAL

EFFICIENCY USING INPUT/OUTPUT METHODThis Work Aid is to assist in Exercise 1

Step 1. Calculate heat absorbed (Q A) by a heat balance.

Step 2. Calculate heat released from fuel combustion (QF) by using the fuel rate and

the heat of combustion.

Step 3. Calculate thermal efficiency

( )100Q

QEff.

F

A=


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Work Aid 2B: Procedures for Calculating Furnace Efficiency by

Heat Loss Method

This Work Aid will assist the Participant in Exercise 2B: Calculate Furnace Efficiencyusing Heat Loss Method.

To determine a furnace thermal efficiency, follow the steps listed below:

Step 1: Calculate oxygen to furnace, using the formula:

79

(21)gasfluemoles100

Nmoles

=

air of moles100

Nmoles

air of moles100

Omoles

gasfluemoles100

Nmoles

=gasfluemoles0furnace/10toO

2

2

22

2

Step 2: Calculate percent excess oxygen (air), using the formula:

( )

−

=

gasfluemoles100

furacefromOmoles

gasfluemoles100

furnacetoOmoles

100gasfluemoles100

furacefromOmoles

OexcessPercent22

2

2

Percent excess O2 = percent excess air.

Step 3: Determine heat available (H A) per lb of fuel from Work Aid 4.

Step 4: Calculate net fuel fired, FN (If fuel consumption desired):

A

AN

H

Q=F

Step 5: Calculate gross fuel fired, FG (If fuel consumption desired):

( )( )LNG

LL

L

H1FF

lossheat%Qwhere100

QH

+=

==


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Step 6: Calculate heat fired, QF, Btu/hr (If fuel consumption desired):

QF = (FG) (LHV fuel)

Step 7: Calculate furnace efficiency:

( )( ) ( )

chart.efficiencycombustionfromfuelLHV

H+1fuelLHV

H

Q

100Q=efficiency%

L

A

F

A =


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WORK AID 3: FLUE GAS OXYGEN (DRY BASIS) VS. EXCESS AIR

Figure 8. Flue Gas Oxygen Versus Excess Air


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WORK AID 4: HEAT ABSORBED CHARTS

Source: Maxwell, Data Book on Hydrocarbon, page 184.

Figure 9. Heat Available from the Combustion of 1000 Btu/ft3 Refinery Gas


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Figure 10. Heat Available from the Combustion of 1600 Btu/ft3 Refinery Gas


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Figure 11. Heat Available from the Combustion of 5º API Fuel Oil


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GLOSSARY

blowdown Water removed from the boiler to control the level of

dissolved impurities in the boiler water.

economizer A device for transferring heat from the flue gas to the boilerfeedwater (BFW) before the BFW enters the boiler drum.

excess air The percentage of air in excess of the stoichiometric amountrequired for combustion.

flue gas Gaseous products from the combustion of fuel.

higher heating value (HHV) The amount of heat released during complete combustion offuel when the water formed is considered as a liquid (credit

is taken for its heat of condensation.) Also called grossheating value.

lower heating value (LHV) The amount of heat released during complete combustion offuel when no credit is taken for heat of condensation of waterin the flue gas. Also called net heating value.

radiation heat loss A defined percentage of the net heat of combustion of thefuel to account for heat losses through the boiler or furnacewalls to the atmosphere.

stack heat loss The total sensible heat of the flue gas components, at the

temperature of flue gas, when it leaves the last heatexchange surface.

stack temperature The temperature of the flue gas when it leaves the last heatexchange surface

thermal efficiency The total heat absorbed divided by the total heat input.Usually expressed in percent.

total heat absorbed The total heat input minus the total heat losses.

total heat losses The sum of the radiation heat loss and the stack heat loss.


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ADDENDUM

API - RP - 532 PROCEDURE...................ERROR! BOOKMARK NOT DEFINED.


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API - RP - 532 PROCEDURE

The API RP 532 procedure is a detailed version of the stack loss method. In addition tothe data required by the Simple Efficiency Equation, an analysis of the fuel compositionis required.

All sources of heat inputs and losses need to be included to make a precise efficiencycalculation. These sources are illustrated in Figure 1A. This calculation requires thefollowing additional data.

• Relative humidity of the air.

• Temperature and specific heat of the fuel.

• Temperature and rate of atomizing steam when liquid fuel is fired.

If not known, it is usually satisfactory to estimate these data, based on typical local

conditions.

QS

TST

Q r

Ha a t T t = TaHV + H f + HmAmbient

Air Fuel

Source: API Recommended Practice 532, Measurement of the Thermal Efficiency of Fired Process Heaters, 1stEdition, August 1982. Reprinted courtesy of the American Petroleum Institute.

Figure 1A. Typical Heater Arrangement


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The work sheets required for the RP 532 procedure are attached. An example of how itis used to calculate the efficiency of a gas-fired furnace is attached.

This procedure consists of the following steps:

1. Using the Lower Heating Value Work Sheet, determine the lower heating value ofliquid fuel (if required). If the fuel is gas, or if typical liquid fuel properties areknown, it is not necessary to complete this work sheet.

2. Using the Combustion Work Sheet, determine flue gas properties for stoichiometriccombustion conditions.

3. Using the Excess Air and Relative Humidity Work Sheet, determine the amount ofwater vapor in the flue gas. The vapor pressure of water at the ambienttemperature can be determined from steam tables on Figure 2A.

4. Using the Stack Loss Work Sheet, determine the stack heat losses. The enthalpyof the flue gas components can be determined from Figures 3A and 4A.

5. The thermal efficiency can then be determined by the following equation:

( )( )4Eqn.

HHHLHV

QQ100-100=e

mf a

r s

+++

where: Cp = Specific heat, Btu/lb-˚F.

e = Net thermal efficiency, % (LHV).

Ha = Air sensible heat correction, Btu/lb of fuel.

= Cp(air)(Ta - Td)(pounds of air per pound of fuel).

LHV = Lower heating value of the fuel, Btu/lb of fuel.

Hf = Fuel sensible heat correction, Btu/lb of fuel.

= Cp(fuel)(Tf - Td).

hs = Enthalpy of atomizing steam, Btu/lb.

Hm = Atomizing medium (usually steam) sensible heat correction,

Btu/lb of fuel.

= Cp(medium)(Tm - Td)(pounds of medium per pound of fuel).

If steam, Hm = (Enthalpy difference)(lb of steam/lb of fuel).

= (hs - 1087.7)(lb of steam/lb of fuel).

Qr = Radiation heat losses, Btu/lb of fuel.


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Qs = Calculated stack heat losses (from Stack Loss Work Sheet),

Btu/lb of fuel.

Ta = Ambient air temperature, ˚F.

Td = Reference (or datum) temperature, ºF.

= 60ºF (usually).

Tf = Temperature of fuel, ºF.

Tm = Temperature of atomizing medium, ºF.

6. The gross thermal efficiency can be determined by the following equation:

egross =( )

mf a

s

HHH+HHV

heatlatentQ100100

++

+−

where: egross = Gross thermal efficiency, % (HHV).

Latent heat = (H2O formed by combustion of fuel) x1059.7.

7. The firing rate can be calculated, based on the heat absorbed in the boiler orfurnace, as follows:

( )6Eqn. e/100

QQ af =

where: Qf = Heat fired, MBtu/hr (LHV).

Qa = Heat absorbed, MBtu/hr.

e = Net thermal efficiency, %.

This procedure calculates the efficiency of boilers by both the Input/Output and StackLoss methods. It uses the HHV of the fuel and can be used for coal-fired boilers, as wellas gas- and oil-fired units. The forms for this procedure are attached. Line items onthese forms that do not apply to Saudi Aramco boilers have been crossed out.

Sample Calculation - RP 532 Procedure

The following sample calculation illustrates the use of the RP 532 calculation procedureto determine thermal efficiency. (Based on Par. 3.2.2 of RP 532.)


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Temperature, ºF

V a p o r P r e s s u r e o f W a t e r , p s i a

20 30 40 50 60 70 80 90 100 110 120 130

2.4

2.2

2.0

1.8

1.6

1.4

1.2

1.0

.8

.6

.4

.2

0

Source: Data taken from Steam Tables

Figure 2A. Vapor Pressure of Water


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Figure 3A. Enthalpy of Flue Gas Components


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Figure 4A. Enthalpy of Flue Gas Components


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Sample Problem:

Given:

Stack temperature TST = 300ºF Air temperature Ta = 28ºF

Specific heat of air Cp(air) = 0.24 Btu/lb- ºF

Relative humidity = 50 %

Oxygen content of flue gas = 3.5 % (wet basis)

Radiation losses Qr = 2.5 % of lower heating value of fuel

Fuel temperature Tf = 100ºF

Fuel specific heat Cp(fuel) = 0.525 Btu/lb- ºF

Fuel composition:

Methane = 75.41 vol. %

Ethane = 2.33

Ethylene = 5.08

Propane = 1.54

Propylene = 1.86

Nitrogen = 9.96

Hydrogen = 3.82

Solution:

1. Complete the following work sheets attached (completed copies attached).

Combustion Work Sheet.

Excess Air and Relative Humidity Work Sheet.

Stack Loss Work Sheet.

2. Determine Net Thermal Efficiency, as follows:

From Combustion Work Sheet, LHV = 18,120 Btu/lb

Radiation Loss Qr = 18,120 x 0.025

= 453.0 Btu/lb of fuel

From Stack Loss Work Sheet, Qs = 1162.1 Btu/lb of fuel

Data extracted from API Recommended Practice 532, Measurement of the Thermal Efficiency of Fired ProcessHeaters, 1st Edition, August 1982. Reprinted courtesy of the American Petroleum Institute.


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Figure 5A. Combustion Work Sheet


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Source: API Recommended Practice 532, Measurement of the Thermal Efficiency of Fired ProcessHeaters, 1st Edition, August 1982. Reprinted courtesy of the American Petroleum Institute.


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Calculate the thermal efficiency of a boiler or furnace, using the Stack Loss Method. Attached are calculation sheets you may require.

1. Determine Net Thermal Efficiency.

LHV = ________________Btu/lb

Radiation Loss Qr = LHV x %Qr /100

= (_______ )(_______) = __________ Btu/lb of fuel

Qs = _________ Btu/lb of fuel

Air required = ____________ (lb of air/lb of fuel)

Excess air = ____________ (lb of air/lb of fuel)

Total air rate = ____________ (lb of air/lb of fuel)

Sensible heat corrections:

Air: Ha = Cp(air) (Ta - Td)(total lb of air/lb of fuel)

= ___________(________ - 60)(___________)

= ___________ Btu/lb of fuel

Fuel: Hf = Cp(fuel) (Tf - Td)

= ___________(___________ - 60)

= ___________ Btu/lb of fuel

Atomizing medium Hm = Cp(medium) (Tm - Td)(lb of medium/lb of fuel)

If steam is used: Hm = (Enthalpy difference)(lb of steam/lb of fuel)

= (hs - 1087.7)(lb of steam/lb of fuel)

Atomizing steam temperature = ___________ºF

Steam enthalpy hs = ___________ Btu/lb

Hm = (__________ - 1087.7)(___________)

= ___________Btu/lb of fuel


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Thermal efficiency

( )

( )( )

( )LHV%

e

+ + +

+ 100100

HHHLHV

QQ100100 e

mf a

r s

=

−=

+++

+−=

2. H2O formed = ___________lb/lb of fuel

Latent heat = H2O formed x 1059.7

= (_________) x 1059.7

= __________Btu/lb of fuel

HHV = LHV + latent heat

= (_________) + (_________ ) = ____________Btu/lb

egross mf a

r s

HHH+HHV

heatlatentQQ100100

++

+−=

= 100 - 100 ( + + )

(_______ + ______ + ______ + ______ )

egross = ___________% (HHV)


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Figure 5A. Combustion Work Sheet, (cont’d)


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REFERENCES

Saudi Aramco Standards

SAES-F-001 Process Fired Heaters

API Standards

API-RP-532 Measurement of the Thermal Efficiency of Fired Processheaters (RP = Recommended Practice)

calculating boiler and process heater thermal efficiency

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