Calculating Boiler and Process Heater Thermal Efficiency

Engineering Encyclopedia Saudi Aramco DeskTop Standards

CALCULATING BOILER AND PROCESS HEATER THERMAL EFFICIENCY

Note: The source of the technical material in this volume is the Professional Engineering 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 public domain may not be copied, reproduced, sold, given, or disclosed to third parties, or otherwise used in whole, or in part, without the written permission of the Vice President, Engineering Services, Saudi Aramco.

Chapter : Mechanical File Reference: MEX-104.06

For additional information on this subject, contact PEDD Coordinator on 874-6556

Engineering Encyclopedia

Introduction to Boilers Calculating Boiler and Process Heater Thermal Efficiency

Section

Page

INFORMATION ............................................................................................................... 3 CALCULATING THERMAL EFFICIENCY USING THE INPUT/OUTPUT OR DIRECT 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 EFFICIENCY USING INPUT/OUTPUT METHOD....................................................... 26 WORK AID 2: PROCEDURE FOR CALCULATING THERMAL EFFICIENCY USING 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

Saudi Aramco DeskTop Standards

i



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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INFORMATION CALCULATING THERMAL EFFICIENCY USING THE INPUT/OUTPUT OR DIRECT METHOD Thermal Efficiency Thermal efficiency is defined as the percentage of the absorbed energy to the total energy input. Calculation of thermal efficiency is based on an energy balance around the boiler or process heater. In a boiler, although only the energy in the steam is usable, the heat absorbed in a boiler is the sum of the energy in the steam and the energy in the blowdown above that of the boiler feed water. The energy in the stack gas above that of ambient air is a loss. The energy transferred from the boiler through the insulation and refractory to the atmosphere is also a loss. In a process heater, heat losses are the same and include losses to the stack gases and losses to the atmosphere through the refractory and insulation. Factors that increase the losses will decrease the thermal efficiency. For example, operating with too much excess air reduces the thermal efficiency by increasing the stack heat loss because the excess air is heated from ambient to stack gas temperature. The thermal efficiency for which a boiler or a process heater is designed is an economic evaluation involving the cost of fuel and the cost of equipment to reduce the losses. Examples of economic analyses include the amount of insulation or refractory used to reduce heat losses to the atmosphere, the amount of heat transfer surface provided in the radiant and convection sections to reduce the stack temperature, use of a preheater to reduce the stack gas temperature, types of burners used (determines minimum excess air requirement) and the use of chemicals to reduce the blowdown requirement.


3



The thermal efficiency can be calculated using either the higher heating value (HHV) or the lower heating value (LHV). The LHV is a better measure of achievable thermal efficiency since the latent heat of vaporization of the water in the flue gas cannot be recovered. The HHV efficiency is several percentage points lower than the LHV efficiency. It is common practice in the furnace industry to use the LHV in calculations while the boiler industry uses the HHV efficiency. All calculations will be done on a LHV basis including boilers. The Example Problem 1 shows the calculation of the thermal efficiency and the magnitude of the heat losses.


4



Example Problem 1 Figure 1 is a schematic of a forced recirculation steam boiler system.

Figure 1. Steam Boiler System

Steam Boiler System: Calculate the thermal efficiencies for the boiler in Figure 1 if firing 55,000 lb/hr of fuel gas. The enthalpy data from a steam table that is needed for entering and exiting streams is shown below: Enthalpy, Btu/lb Steam

Temp. ºF

psia

HT

HV

Blowdown

370

174.7

343.5

1196.4

Steam

434

154.7

--

1237.6

Feed water

190

--

158.0

--

Fan-200 HP


5



Heat Balance: Water side/Process Heat In:

Feedwater = 910,000 lb/hr x 158 = 143.78 MBtu/hr (M = million) Heat Absorbed = QA Total = 143.78 + QA

Heat Out: Steam Blowdown Total Heat In 143.78 + QA Heat absorbed

= 805,000 x 1,237.6 = 105,000 x 343.5

= 996.27 = 36.07 = 1,032.34 MBtu/hr

= Heat Out = 1,032.34 = QA = 1,032.34 - 143.78

= 888.56 M Btu/hr

Fuel Heat Input QP = 55,000 x 19,400 = 1067 million Btu/HR. Pump Energy

(gpm )(∆P ) = (12000 )(108 ) = 1079 Hp 1715 (0.65 ) 1715 (0.70 )

HPp =

HPp = 1079 x 2544 = 2.7 million Btu/hr Fan Energy HPF = 200 Hp HPF = 200 x 2544 = 0.5 million Btu/hr Total Energy Input Qin. = 1067 + 2.7 + 0.5 = 1070.2 Total Energy Efficiency

Eff. =

Q A (100 ) 888.6(100 ) = = 83.0% QM 1070.2


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

LHV Eff. =

888.6(100 ) = 83.2% 1067

Blowdown (Unusable Energy) QBD = 105,000(343.5 – 158) = 19.5 million Btu/hr Loss =

19.5 (100) = 1.8% 1067

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 or process heater can be measured. This is the usual method for boilers and is used for process 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 the blowdown (BD) rate is not measured. Sometimes the boiler feed water (BFW) rate is not measured. The steam rate is always measured. Knowing the concentrations of one impurity in both the BFW and the BD allows the calculation of the material and energy balances. Example Problem 2 illustrates this calculation. Example Problem 2

Calculate thermal efficiency of a boiler given the following data: Steam production BFW chloride BD chloride Fuel fired

= = = =


500,000 lb./hr at 434ºF and 140 psig 0.2 wppm 10 wppm 673,576 ft3/hr with 1005 Btu/ ft3

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Enthalpies Enthalpy, Btu/lb Temp. ºF

Psia

HL

HV

Blowdown

370

174.7

343.5

1196.4

Steam

434

154.7

--

1237.6

Feed water

190

--

158.0

--

Stream

Solution: Material Balance FBFW = Fs + FBD CBFW FBFW = CBD + FBD C   0.2  FBD =  BFW  FBFW =  FBFW = 0.02 FBFW  10   CBD  FBFW = 500,000 + 0.02 FBFW 0.98 FBFW = 500,000 FBFW = 510,204 lb/hr FBD = 0.02 FBFW = 10,204 lb/hr Heat Absorbed, QA 6

Heat In BFW = 510.204 (158) Heat Absorbed Total

=

10 Btu/hr 80.6 QA 80.6 + QA

6

Heat Out Steam = 500,000 x 123.7.6 BD

= 10,204 x 343.5 Total

Heat In 80.6 + QA

= Heat out = 622.3

QA

= 622.3 – 80.6 = 541.7 million Btu/hr

=

10 Btu/hr 618.8 3.5 622.3

Heat Fired QF = 673,756 x 1005 = 677.1 million Btu/hr Thermal Efficiency LHV Eff =

Q A (100 ) 541.7 (100 ) = = 80.0% QF 677.1


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CALCULATING THERMAL EFFICIENCY USING THE HEAT LOSS METHOD The heat loss method calculation is used when the heat absorbed cannot be readily calculated such as most process heaters. The heat absorbed can be calculated by subtracting the heat losses from the heat fired. In a boiler or process heater the primary heat loss is that lost to the stack gas. The heat loss in the stack is a function of the stack temperature, the amount of excess air and the carbon and hydrogen ratio in the fuel. A material and energy balance can be calculated knowing the above parameters.

Excess Air The amount of excess air is defined as a percentage of the air in the flue gas to the air that is required for complete combustion. Excess air and excess oxygen are numerically equivalent because the numerator and denominator are both multiplied by the same constant to convert from one to the other. Analysis from the lab will always be on a dry basis. Stack gas analyzers that sample the stack gas will dry the stack gas before analysis. Stack gas analyzers that are in the stack measure on a wet basis but may be calibrated to report on a dry basis. The calculation based on a dry flue gas analysis is outlined in Figure 2 and detailed in Example Problem 3.


9



Normally there is no correction for incomplete combustion shown 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. 5.

O2 required = (total in) less (excess) Percent excess O2 =

(excess O2 ) x100 = (excess ) x100 (required O2 ) total - excess

Figure 2. Excess Oxygen Example Problem 3 Calculation Of Excess Oxygen

Lab Flue gas analysis: CO2

9.5

CO O2

1.8 2.0

N2

86.7 100.0

Air composition: 21% O2, 79% N2 O2 into furnace = 86.7 x

0.21 = 23.0 moles/100 moles flue gas 0.79

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 =

1.1 x 100 = 5.02% (23 − 1.1)

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 =

2.0(100 ) x 100 = 5.02% (23 − 2.0 )


10



Figure 3 (Work Aid 3) can also be used to calculate excess air (oxygen) once the oxygen 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 and denominator 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: Excess Air = where:

111.4 x %O2 20.95 - %O2

%O2 = Percent oxygen in the flue gas.

For 2% O2 in the stack gas. Excess Air =

111.4 x 2 222.8 = = 11.8% 20.95 - 2 18.95


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When the flue gas analysis is on a dry basis: Excess Air =

91.2 x %O2 20.95 - %O2

For 2% O2 in the stack gas. Excess Air =

91.2 x 2 182.4 = = 9.6% 20.95 - 2 18.95

Lab analysis is always on a dry basis because the water drops out as the gas sample cools. When the oxygen analyzer is located in the stack, the oxygen is measured on the wet basis but the analyzer may be calibrated using lab results so that it reports on a dry basis. When the flue gas is extracted from the stack and is transported to an analyzer that 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 the hydrogen-to-carbon ratio of the fuel. However, there is very little change in this relationship 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 stack temperature measurements. Ordinary stack temperature thermocouples can read low by as much as 100ºF, depending upon their location and the flue gas temperature being measured. If the thermocouple can "see" cold surroundings, such as the top of the convection section or the sky, the indicator will likely read low. The higher the actual stack temperature, the higher the radiation losses and thus, the higher the error. The aspirating thermocouples shown in Figure 4 minimizes any error due to radiation.


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6 3

6

4 5

2

2 1

6 6

A

1

A–A 2

2 2

2

A

7

8

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 an analysis of the fuel composition. This procedure is included in the Addendum with an example problem and a blank calculation sheets. The API RP 532 procedure is a detailed heat balance on the combustion side of the furnace to determine the amount of heat lost up the stack.

Combustion Efficiency Charts

Heat Available from Flue Gas Above 60ºF, Btu/lb Fuel

The API material and heat balance has been solved for a number of cases and these cases plotted as heat available charts to simplify the calculations. These charts are attached as Work Aid 4. Work Aid 4 has charts for 1000 Btu/ft3 gas, 1600 Btu/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 reduced as excess air is increased and a stack gas temperature is increased.

0% Excess Air 20

10

Flue Gas Temperature

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 in calculating thermal efficiency.


14



Example Problem 4

Process heat absorbed

= QA = 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.

HA = 17,100 Btu/lb fuel at 600ºF and 5% excess air Net Fuel = FN =

Q A 353 x 10 6 Btu/hr = = 20,643 lb/hr HA 17,100 Btu/lb

Assume furnace box losses QL are 2%. (Usually 2 - 3%) Gross fuel Heat fired

= FG = 1.02 x 20,643 = 21,056 lb/hr = QF = 21,056 x 19,700 Btu/lb = 414.8 x 106 Btu/hr

LHV efficiency =

6 heat absorbed (100 ) = Q A (100 ) = 353x10 6 x (100 ) = 85.1% heat fired QF 414.8x10

Given the heat absorbed, the heat loss method will calculate the fuel consumed. If a fuel meter is available the calculated fuel rate should be rationalized with the fuel meter readings. If only the thermal efficiency is desired the calculation simplifies to the following: From above we have: HA HF QL HL

= = = =

17,100 Btu/lb. fuel at 600ºF and 5% excess air (heat absorbed from chart) 19,700 Btu/lb. fuel (heating value of fuel for chart used) 2% (Percent heat release/lost to atmosphere) Heat loss, decimal fraction

QL 2 = = 0.02 100 100 H (100 ) LHV efficiency = A HF (1 + HL ) HL =

17,100 (100 ) 19,700 (1+ 0.02 ) 17,100 (100 ) = = 85.1% 20,094 =


15



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).  100   Percent efficiency = [100 − (0.0237 + (0.000189 )(EA )(TST − TA ))] 100 Q + L  

where: EA = Percent excess air. TST = Stack temperature, ºF. TA

= Ambient air temperature, ºF.

QL

= Casing heat loss, %.

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

 100  Percent efficiency = [100 - (0.0237 + (0.000189 )(5 )(600 − 80 ))]   100 + 2  Percent efficiency = [100 - (0.0246 )(520 )](0.9804 ) = 85.5 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 and the changes in thermal efficiency that would result from reductions in stack temperature, blowdown rate, and excess air.


16



Example Problem 5 Introduction:

In this example we will perform an energy balance around a boiler system and calculate the 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


17



Answer:

Material Balance: Feedrate

= F

Blowdown

= 0.1 F

Steam product

= 250,000

Material balance, F = 250,000 + 0.1 F Feedrate

F =

Blowdown Heat in:

250,000 = 277,778 lb/hr 0.9

0.1 F = 27,778 lb/hr

Feed water = 277,778 x 148 Absorbed Heat Total

Heat out: Blowdown Steam Total Heat in = Heat out

= 41.11 MBtu/hr = QA = 41.11 + QA

= 27,778 x 478.5 = 250,000 x 1351.8

41.11 + QA

= 13.29 MBtu/hr = 337.95 = 351.24

= 351.24

Heat absorbed

QA

= 351.24 - 41.11

Heat loss

QL

= 2%

Fuel LHV Heat available

LHV = 19,400 Btu/lb HA = 16,725 Btu/lb at 600°F Stack and 20% excess air from Work Aid 4 using 1000 Btu/ft3 gas chart.

Net fuel Gross fuel Heat fired LHV efficiency

310.13 x 10 6 FN = 16,725 FG = 1.02 x 18,543 QF = 18,914 x 19,400

=

= 310.13 MBtu/hr

= 18,543 lb/hr = 18,914 lb/hr = 366.93 MBtu/hr

6 QA (100 ) = 310.13 x 106 x100 = 84.52% QF 366.93 z 10


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Shortcut efficiency assuming the atmospheric air temperature is 100°F: Efficiency = [(100 -(0.0237 + (0.000189)(EA)))(TST - TA )][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 feedwater preheat. − 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, but would 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% to 85.4%. This might be accomplished by changing burners and closer control of excess air.

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

Lower Stack Temp.

Reduce Blowdown

Reduce Excess Air

10

10

2

10

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

Case Percent blowdown Heat absorbed, MBtu/hr

Table 1. Furnace Fuel Savings


19



Calculation for efficiency Improvement: Case

1 Base

Heat in 277,778 x 148

=

41.11

Heat out 27,778 x 478.5 250,000 x 1,351.8

= =

13.29 337.95 351.24

Heat absorbed Stack Percent excess air Heat loss Fuel LHV Heat Avail.* Net fuel Gross fuel Heat fired

2 Lower Stack Temp.

41.11

3

4 Reduce % Excess Air

Reduce Blowdown

255,103 x 148

=

37.76

41.11

5,102 x 478.5 250,000 x 1,351.8

= =

351.24

2.44 337.95 340.39

351.24

310.13 600 20 2% 19,400 16,725

310.13 550 20 2% 19,400 17,000

302.63 600 20 2% 19,400 16,725

310.13 600 20 2% 19,400 16,900

18,543 18,914 366.93

18,243 18,608 360.99

18,095 18,457 358.06

18,351 18,718 363.13

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 discharged to the atmosphere at the stack temperature, and the energy it contains is lost. The primary objective of efficient boiler and furnace operations is to minimize airflow beyond that required for good combustion. The air required for combustion should enter only through the burners. The following steps can be taken to reduce excess air: 1. Seal air leaks. This is particularly important in furnaces, which operate with a draft (negative pressure) throughout the furnace. These furnaces are more susceptible to air infiltration. Figure 7 shows typical sources of air leaks into a furnace. Since most boilers operate with a positive pressure through much of the boiler, air leakage into boilers is much less a problem. 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 a series of plant tests. These targets are the minimum excess air rates that are necessary for good combustion. Since no two furnaces or boilers are exactly the same, there can be different targets for each boiler and furnace in the plant.


21



Inlet

Construction Joint

Clearance Around Tube Penetration

Poor Seal on Access Door

Casing Corrosion

Leaky Covers on Observation Doors Idle Burner

Outlet

Figure 7. Furnace Air Leaks


22



•

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-draft burners 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 of stack temperatures exceeding design. The extent of fouling can be determined by visual inspection of the tubes or by observing an increase in stack temperature over time. A 40ºF increase in stack temperature typically represents a loss of 1% in thermal efficiency. Fouling can be reduced by operating sootblowers in boilers and furnaces. Sootblowers should be provided for all boilers and furnaces where heavy liquid fuels are fired. Units without sootblowers 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 is operating satisfactorily usually requires the addition of heat transfer surface. The following are means of reducing stack temperature: •

Add heat transfer surface in convection section of process heaters.

•

Add economizers on boilers to preheat the boiler feedwater before entering the steam drum.

•

Add combustion air preheaters. Air preheaters can transfer heat from the flue gas leaving the stack, to the air used for combustion. Depending upon the flue gas temperature, the incoming air can be heated several hundred ºF. The flue gas temperature should be kept above about 300ºF to prevent corrosion of the heat exchanger due to sulfuric acid in the flue gas.


23



Reduce Other Losses Although less important several other parameters listed below can improve boiler and process heater efficiency: •

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

•

Insulation should be maintained or improved to reduce heat losses.

•

Steam leaks should be repaired.


24



EFFECTS OF FIRING RATE ON THERMAL EFFICIENCY As the firing rate is increased the loss to the stack increases primarily because the heat transfer area is fixed. The increase in heat loss is not necessarily proportional to the increase in firing rate. Increased loss will reduce thermal efficiency. Similarly a decrease in firing will slightly improve thermal efficiency. At very low firing rates the heat losses to the atmosphere become significant and the thermal efficiency may decrease. Over firing a boiler or a process heater will reduce thermal efficiency.


25



WORK AIDS WORK AID 1:

PROCEDURE FOR CALCULATING THERMAL EFFICIENCY USING INPUT/OUTPUT METHOD

This Work Aid is to assist in Exercise 1

Step 1.

Calculate heat absorbed (QA) 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

Eff. =

QA (100 ) QF


26



WORK AID 2:

PROCEDURE FOR CALCULATING THERMAL EFFICIENCY USING HEAT LOSS METHOD

Work Aid 2A:

Excess Air and Thermal Efficiency Using Short Cut Equations

Excess Air, EA Dry Basis, O2 in stack gas Excess Air =

91.2 x % O2 20.95 - % O2

Wet Basis, O2 in stack gas Excess Air =

111.4 x % O2 20.95 - % O2

Thermal Efficiency  100   LHV efficiency = [100 − (0.0237 + (0.000189 (EA )(TST − TA )))]  100 + QL 


27



Work Aid 2B:

Procedures for Calculating Furnace Efficiency by Heat Loss Method

This Work Aid will assist the Participant in Exercise 2B: Calculate Furnace Efficiency using Heat Loss Method. To determine a furnace thermal efficiency, follow the steps listed below: Step 1: Calculate oxygen to furnace, using the formula:   moles N2 moles O2     100 moles flue gas  100 moles of air   O2 to furnace/10 0 moles flue gas = moles N2      100 moles of air    moles N2  (21) 100 moles flue gas   = 79

Step 2: Calculate percent excess oxygen (air), using the formula:  moles O2 from furace   (100 ) 100 moles flue gas   Percent excess O2 =  moles O2 to furnace   moles O2 from furace    −    100 moles flue gas   100 moles flue gas  Percent excess O2 = percent excess air. Step 3: Determine heat available (HA) per lb of fuel from Work Aid 4. Step 4: Calculate net fuel fired, FN (If fuel consumption desired): FN =

QA HA

Step 5: Calculate gross fuel fired, FG (If fuel consumption desired): QL where QL = % heat loss 100 FG = (FN )(1 + HL ) HL =


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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: % efficiency =

Q A (100 ) HA = (LHV fuel) (1+ HL ) QF

LHV fuel from combustion efficiency chart.


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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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Source: Maxwell, Data Book on Hydrocarbon, page 185.

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 boiler feedwater (BFW) before the BFW enters the boiler drum.

excess air

The percentage of air in excess of the stoichiometric amount required for combustion.

flue gas

Gaseous products from the combustion of fuel.

higher heating value (HHV) The amount of heat released during complete combustion of fuel when the water formed is considered as a liquid (credit is taken for its heat of condensation.) Also called gross heating value. lower heating value (LHV)

The amount of heat released during complete combustion of fuel when no credit is taken for heat of condensation of water in the flue gas. Also called net heating value.

radiation heat loss

A defined percentage of the net heat of combustion of the fuel to account for heat losses through the boiler or furnace walls 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 heat exchange surface.

stack temperature

The temperature of the flue gas when it leaves the last heat exchange 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 to the data required by the Simple Efficiency Equation, an analysis of the fuel composition is required. All sources of heat inputs and losses need to be included to make a precise efficiency calculation. These sources are illustrated in Figure 1A. This calculation requires the following 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 T ST

Qr

LHV + H f + H m

Source:

H a at T t = T a Am bient Fuel Air

API Recommended Practice 532, Measurement of the Thermal Efficiency of Fired Process Heaters, 1st Edition, 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 it is 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 of liquid fuel (if required). If the fuel is gas, or if typical liquid fuel properties are known, it is not necessary to complete this work sheet.

2.

Using the Combustion Work Sheet, determine flue gas properties for stoichiometric combustion conditions.

3.

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

4.

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

5.

The thermal efficiency can then be determined by the following equation: e = 100 -

where:

100(QsQr ) LHV + Ha + Hf + Hm

(Eqn. 4)

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).

6.

Tf

= Temperature of fuel, ºF.

Tm

= Temperature of atomizing medium, ºF.

The gross thermal efficiency can be determined by the following equation: egross where: egross

= 100 −

100(Qs + latent heat ) HHV + Ha + Hf + Hm

= 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 or furnace, as follows: Qf = where: Qf

Qa e/100

(Eqn. 6) = 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 Stack Loss methods. It uses the HHV of the fuel and can be used for coal-fired boilers, as well as gas- and oil-fired units. The forms for this procedure are attached. Line items on these 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 procedure to determine thermal efficiency. (Based on Par. 3.2.2 of RP 532.)


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2.4 2.2

Vapor Pressure of Water, psia

2.0 1.8 1.6 1.4 1.2 1.0 .8 .6 .4 .2 0 20

30

40

50

60

70

80

90

100

110

120

130

Temperature, ºF 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) Relative humidity Oxygen content of flue gas Radiation losses Qr Fuel temperature Tf

= = = =

Fuel specific heat

= 0.525 Btu/lb- ºF

Cp(fuel)

0.24 Btu/lb- ºF 50 % 3.5 % (wet basis) 2.5 % of lower heating value of fuel

= 100ºF

Fuel composition: Methane Ethane Ethylene Propane Propylene Nitrogen Hydrogen

= = = = = = =

75.41 vol. % 2.33 5.08 1.54 1.86 9.96 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 Radiation Loss Qr

= 18,120 Btu/lb = 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 Process Heaters, 1st Edition, August 1982. Reprinted courtesy of the American Petroleum Institute.


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Sensible heat corrections: Pounds of air/pound of fuel is obtained by adding the total from column 7 of the Combustion Work Sheet with the pounds of dry excess air per pound of fuel from the Excess Air and Relative Humidity Work Sheet. Air:

Ha

Fuel:

Hf

Atomizing medium Hm

= p(air) (Ta - Td)(pounds of air/pound of fuel) = 0.24 (28 - 60)(14.322 + 3.191) = -134.5 Btu/lb of fuel = p(fuel) (Tf - Td) = 0.525 (100 - 60) = 21.0 Btu/lb of fuel = 0 (no atomizing steam required)

Using Eqn. 4: e = 100 − e = 100 −

3.

100(Qs + Qr ) LHV + Ha + Hf + Hm

100(1162.1 + 453.0 ) = 91.03 % (LHV ) (18120 − 134.5 + 210 )

Determine Gross thermal efficiency, as follows: From Combustion Work Sheet, H2O formed = 1.784 lb/lb of fuel. Latent heat = H2O formed x 1059.7

HHV

= = = =

1.784 x 1059.7 1890.5 Btu/lb of fuel LHV + latent heat 18120 + 1890.5 = 20010 Btu/lb.

Using Eqn. 5:

egross = 100 − egross = 100 −

100(Qs + Qr + latent heat ) HHV + Ha + Hf + Hm

100(1062.1 + 453.0 + 1890.5 ) = 82.83% (HHV ) 20010 - 134.5 + 21.0

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


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Source:


Figure 5A. Combustion Work Sheet


46



Source:



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Source:



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Source:



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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.

Radiation Loss

LHV

= ________________Btu/lb

Qr

= LHV x %Qr/100 = (_______)(_______) = __________ Btu/lb of fuel

Qs

= _________ Btu/lb of fuel Air required

= ____________ (lb of air/lb of fuel)

Excess air


Total air rate


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 = 100 −

e

= 100 −

100(Qs + Qr ) LHV + Ha + Hf + Hm

(

100(

+

+

+

)

% (LHV )

e= 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

= 100 −

100 Qs + Qr latent heat HHV + Ha + Hf + Hm

= 100 - 100 (

+

+

)

(_______ + ______ + ______ + ______) egross

= ___________% (HHV)


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Source:


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


52



Source:



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Source:



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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 Process heaters (RP = Recommended Practice)


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Calculating Boiler and Process Heater Thermal Efficiency

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