Boiler Efficiency Workshop

Boiler Efficiency Workshop May 30, 2013 Presented by: Aqeel Zaidi, P.Eng., CEM, CMVP Energy Solutions Manager Questions and follow-up should be directed to: Aqeel.Zaidi@enbridge.com 416.495.6531

Safety Moment: Heat Stress On hot days it is possible for anyone to be susceptible to heat exhaustion or in severe cases heat stroke The ambient conditions in boiler rooms are generally hot and the heat gets worse in the summer. The ambient temperature could easily reach over 100 F Avoid heat stress by: Drinking plenty of fluids Increasing the frequency and duration of rest breaks Watching out for each other 2

Safety Moment: Heat Stress During hot days it is possible for anyone to be susceptible to heat exhaustion or in severe cases heat stroke The ambient conditions in boiler rooms are generally hot. It gets worse in the Summer. The ambient temperature can easily reach over 100 F Drink plenty of fluids Increase frequency & duration of rest breaks Watch out for each other 3

Today s Workshop Will Cover: Steam Basics Combustion basics Definition of boiler efficiency How to identify and quantify combustion improvement opportunities How to identify and quantify heat recovery opportunities o Blowdown heat recovery o Combustion air preheating o Feedwater economizers o Condensing economizers Practical exercises 4

Building our Workshop Steam Plant Boilers: 3 X 500 BHP Plant Operating Hours: 8,000 hrs. Natural Gas Cost: $1.1 million/yr. Each Boiler Runs For 5,000 hrs. Stack Flue Gas Sat. Steam @ 100 psig Total Steam 15,000 pph Process Steam DA Steam 10% Natural Gas 4.4 million m 3 /yr. Boiler 1 Boiler 2 Boiler 3 Blowdown 4% Boiler Feedwater 8 psig DEAERATOR Cond. Tank Condensate Return 50% 180 F Make-Up Water 50% 5

Today s Objective Improve boiler and steam plant efficiency enough to reduce natural gas (NG) costs by 20% Boiler Plant Specifications: Number of Boilers 3 Boiler Rating 500 HP Plant Hours 8,000 hr/yr Boiler Hours 5,000 hr/yr Natural Gas Cost (0.25 $/m 3 ) $1.1 million/yr Rated Input per Boiler 20.9 MMBtu/hr Avg. Firing Rate 50% Hourly Gas Consumption per Boiler 10.5 MMBtu/hr 6

To achieve our 20% savings objective, we must: 1. Identify savings opportunities 0% Saved 2. Quantify the opportunities This will require a review of some basic principles of steam and combustion.

Basic Concepts of Steam

Steam Basics Generated by adding heat energy to water to bring it to its boiling temperature Adding more energy transforms water from liquid to vapor (steam) Steam is used to carry heat energy from one location to another Heat energy is expressed in British Thermal Units (Btu) I Btu = amount of heat required to raise the temperature of 1 pound of water by 1 F Heat Energy is also called Enthalpy (h): Energy due to both Temp. and Pressure 9

Steam Basics Increasing Pressure increases boiling point (saturated water temp.) 450 Saturation Temperature vs. Pressure 0 psig 212 F 400 Saturation Temp ( F) 350 300 250 200 150 100 psig 338 F 100 0 50 100 150 200 Pressure (psig) 10

Steam Basics Sensible vs. Latent Heat Heat required to make steam has two components: o Sensible heat (Btu/lb) o Latent heat (Btu/lb) Sensible Heat o Amount of heat required to raise the temperature of water from 32 F to its boiling point (saturated water) o Adding sensible heat raises the temperature o It can be detected with a thermometer o Sensible heat of water at 32 F is taken as zero 11

Steam Basics Sensible Heat More Sensible Heat is required to boil water at a higher pressure o Increasing the temperature of water going into the boiler reduces the amount of Sensible Heat required to boil the water 400 350 Sensible Heat vs. Pressure 0 psig 212 F Sensible Heat (Btu/lb) 300 250 200 150 100 1 lb of saturated water contains 180 Btu 100 psig 50 0 0 50 100 150 200 Pressure (Psig) 338 F 1 lb saturated water contains 309 Btu 12

Steam Basics Latent Heat Amount of heat required to change saturated water to steam Adding Latent Heat does not raise the temperature: the boiling water and steam has the same temperature for a given pressure Latent heat added to the boiler is what is transferred to the load Removing latent heat at the load creates condensate Returning maximum amount of condensate reduces heat energy required by the boiler 13

Steam Basics Latent Heat vs. Pressure 1000 Latent Heat vs. Pressure 950 900 Latent Heat (Btu/lb) 850 800 750 0 50 100 150 200 Pressure (psig) 14

Steam Basics Total Heat of Steam Total Heat of Steam = Sensible Heat + Latent Heat Total Heat of Steam = hg Sensible Heat = hf Latent Heat = hfg For 100 psig steam: hf (saturated liq. Enthalpy) = 309 Btu/lb hfg (latent heat) = 880 Btu/lb hg (saturated vapour enthalpy)= 1189 Btu/lb Heat (Btu/hr) Total heat of Steam at 100 psig 1400 1200 1000 800 600 400 200 0 338 F 338 F 32 F 880 309 100 psig Pressure (psig) Latent Heat Sensible Heat 15

Steam Basics Enthalpy Explained Courtesy of: Spirax Sarco 16

Basic Concepts of Combustion

Perfect Combustion: Ideal Air:Fuel Ratio CO 2 + 2H 2 O + 7.52 N 2 Heat Released =1012 Btu/ft 3 = 23,000 Btu/lb Natural Gas (CH 4 ) 1 ft 3 16 lbs. Air( 2O 2 + 7.52 N 2 ) 9.52 ft 3 275 lbs. O 2 contributes to combustion, while N 2 absorbs heat. Air : Fuel Ratio (by vol.) = 9.5 : 1 or 10:1 (approx.) Air : Fuel Ratio (by wt.) = 17 : 1 (approx.) 18

Practical Air:Fuel Ratio for Complete Combustion All boiler burners use excess air to avoid risk of CO and un-burned CxHy Excess air varies with firing rate CO 2 + 2H 2 O + x 2O 2 + (1 + x) 7.52 N 2 Excess O 2 Low Fire: 8-12 % High Fire: 3-5% Natural Gas (CH 4 ) 1 ft3 16 lbs. Air(1+x) (2O 2 + 7.52 N 2 ). Excess Air e.g. x= 0.2 for 20% A:F by Volume 18:1 @ 10% O 2 12:1 @ 3% O 2 19

Excess Air and Excess O 2 Simple Correlation For 100 CH 4 Excess Air (for dry O 2 ) = 8.52 x %O 2dry /(2 9.52 x %O 2dry ) Excess Air (for wet O 2 ) = 10.52 x %O 2wet /(2 9.52 x %O 2wet ) Example: Flue Gas Analyzer reads 5% excess O 2 (dry) Excess Air = 8.52 x 0.05 / (2 9.52 x 0.05) = 0.426 / 1.524 = 28% 20

Heating Values of Natural Gas: HHV, LHV Higher Heating Value (HHV) CO 2 + 2H 2 O + 2O 2 + 7.52 N 2 o Takes into account the latent heat of vaporization o Assumes all the heat in product of combustion can be put to use o HHV of N.Gas = 1012 Btu/ft 3 = 37.7 MJ/m 3 NG Lower Heating Value (LHV) Air o Determined by subtracting the heat of vaporization from the HHV o Assumes latent heat of vaporization of water is not put to use o LHV of N. Gas = 912 Btu/ft 3 = 33.9 MJ/m 3 21

Thermal Efficiency Seasonal Efficiency Boiler Operation Efficiency Combustion Efficiency What is Boiler Efficiency? Boiler Energy Efficiency Net Plant Efficiency Net Efficiency Fuel-to-steam Efficiency Fuel Efficiency

Steam Boiler Efficiency To identify energy saving opportunities, we need to properly understand what Boiler Efficiency means: We ll focus on four commonly used efficiencies for Steam Boilers: 1. Combustion efficiency 2. Fuel efficiency, ASME PTC 4-2008 3. Fuel-to-steam efficiency 4. Net plant efficiency 23

Combustion Efficiency Combustion Efficiency = Input Energy Stack Losses Input Energy Combustion Efficiency =(100 21)/100 = 79% Stack Losses = 21.0 Q ng = 100 Natural Gas

Combustion Efficiency 79% of Fuel Energy is Delivered to Boiler Stack Losses = 21.0 Q ng = 100 Natural Gas 79.0 Available

Combustion Efficiency Combustion Efficiency = Input Energy Stack Losses Input Energy There is no need to know gas consumption or steam production to calculate combustion efficiency. Stack Losses = 21.0 But how do we measure stack losses? Q ng = 100 Natural Gas 79.0 Available

Flue Gas Analyzers Are Used To Measure Stack Loss Enbridge Can Help! 27

What do Flue Gas (FG) Analyzers Measure? Excess O 2 CO FG Temp. Comb. Air Temp. NO x CO 2 + 2H 2 O + X 2O 2 + (1 + x) 7.52 N 2 H 2 O is removed in portable analyzers O 2 is recorded on dry basis Natural Gas (CH 4 ) Air(1+X) (2O 2 + 7.52 N 2 ). Excess Air e.g. X= 0.2 for 20% 28

Types of Stack Losses Dry flue gas loss o Heat lost in the Dry products of combustion (CO 2, O 2, N 2 ) which carry only sensible heat Loss in water from burning Hydrogen o Hydrogen component of fuel exits in the form of water vapour o Most of its enthalpy is in the form of heat of vaporization o Approximately 10% of natural gas energy is lost CO loss o CO is a fuel - any CO in the flue gas represents a loss Unburned Hydrocarbons (C x H y ) loss o Unburned combustibles (C x H y ) have the same Higher Heating Value (HHV) as natural gas - any C x H y in the flue gas represents fuel loss 29

Combustion Efficiency Stack Losses (Calculated as a % of fuel) Dry flue gas (CO 2, O 2, N 2 ) Water from burning H 2 CO Unburned Hydrocarbons, C x H y Natural Gas Combustion Efficiency (%) = 100 Σ losses

Exercise # 1 Use the Combustion Efficiency Chart to determine the combustion efficiency based on the following parameters: Parameters Excess O 2 = 8% Flue Gas Temp. T fg = 460 F Combustion Air Temp. T air = 80 F Calculate Delta T = (460 80) F = 380 F Excess, % FG Temperature - Combustion Air O2 340 360 380 400 420 7.00 80.6 80.1 79.5 79.0 78.4 7.50 80.4 79.8 79.2 78.6 78.0 8.00 80.0 79.4 78.9 78.3 77.7 8.50 79.7 79.1 78.5 77.9 77.2

Fuel Efficiency ASME PTC 4-2008 (% of fuel energy transferred to boiler feedwater) There are two methods of calculating Fuel Efficiency: 1. Input-Output (Direct) Method: 2. Energy Balance (Indirect) Method: 32

Fuel Efficiency Boiler Heat Balance Q sl Stack loss Q ra Radiation Loss Q ng Natural Gas Q st Steam Q fw Feedwater Q bd Blowdown Q ng + Q fw = Q sl + Q ra + Q st + Q bd Q ng - Q sl - Q ra = Q st + Q bd -Q fw Output for Indirect Method Output for Direct Method 33

Fuel Efficiency ASME PTC 4-2008 1. Input-Output (Direct) Method: Fuel Efficiency (Q st + Q bd Q fw) Q ng x100 Q ng Natural Gas Q sl Stack loss Q ra Radiation Loss Q st Steam 2. Energy Balance (Indirect) Method: Fuel Efficiency Q ng (Q sl + Q ra ) Q ng x100 Q fw Feedwater Q bd Blowdown 34

ASME PTC Input Output (Direct) Method Fuel Efficiency (Q st + Q bd Q fw) Q ng x 100 Measurements Required: 1. Fuel flow rate 2. Steam flow rate 3. Feedwater flow rate 4. Steam pressure and temp. 5. Blowdown flow rate Considerations Q ng Natural Gas Q fw Feedwater Q st Steam Q bd Blowdown Most small to medium steam plants do not have flow meters, making it impossible to calculate efficiency using this method If steam flow meters are not maintained, their readings will not be accurate Often results in unrealistic efficiency values 35

ASME PTC Energy Balance (Indirect) Method Fuel Efficiency Q ng (Q sl + Q ra ) Q ng x100 Fuel Efficiency (%) = 100 Σ losses as a % of Q ng (Stack, Radiation, Unaccounted for) Measurements Required: 1. Measure stack losses (%) with a flue gas analyzer 2. Calculate radiation losses (%) Typically 0.5 to 1% of boiler rating 3. In some cases, include unaccounted for losses (%) 0.1% of input 4. Subtract from 100% Considerations Needs accurate measurements of losses Does not require gas and steam flow measurements Generally more accurate than Input-Output Method Preferred method by ASME PTC 4-2008 36

Fuel-to-Steam Efficiency Practical efficiency Reflects the amount of fuel energy converted to steam No standard definition like Fuel Efficiency in ASME PTC But what is the "? Steam, yes, but what is the energy absorbed by steam? 37

Fuel-to-Steam Efficiency Input-Output Method = (Q st Q fw ) Fuel-to-Steam Efficiency (Q st Q fw ) Q ng Q ng Natural Gas Steam Q st Considerations Feedwater Need to measure: o Steam flow rate o Feedwater flow rate o Gas consumption Can we use the Energy Balance Method instead? Yes. 38

Fuel-to-Steam Efficiency Energy Balance Method Losses Q ng Q sl Stack Q ra Radiation F-t-S Q ng Q sl Q ra Q bd ) Q ng Natural Gas Steam = 100 Σ losses (% of Q ng ) Q Feedwater bd Considerations Blowdown Measure losses with a flue gas analyzer Notice: BD is considered a loss, ASME PTC considers it an output Very good tool to calculate steam flow rate Will do an exercise to calculate steam flow when Fuel-to-Steam efficiency is known 39

Fuel-to-Steam Efficiency Radiation loss (%of fuel input) is constant at all firing rates o A higher percentage of fuel input is lost at low firing rates Efficiency is reduced at low firing rates due to high excess O 2 and high radiation loss Optimum efficiency to operate boiler at is generally higher than 50% firing rate Efficiency (%) Combustion Effiency Fuel-to-Steam Efficiency Fuel Input (MBTU/HR) 40

Exercise # 2 Fuel-to-Steam Efficiency Calculate the fuel-to-steam efficiency using the Energy Balance Method. Fuel-to-Steam Efficiency = (Q ng Q sl Q ra - Q bd ) / Q ng Stack Loss (Q sl ) = (1- Ƞ combustion ) x Q ng Radiation Loss (Q ra ) = 1% of Boiler Rated Input Q sl = (1 0.789) x 10.5 = 2.22 MMBtu/hr Q rad = 0.01 x 21 = 0.21 MMBtu/hr Parameters: Natural Gas (Q ng ) = 10.5 MMBtu/hr Blowdown (Q bd ) = 0.1 MMBtu/hr Boiler Input Rating = 21 MMBtu/hr Ƞ combustion = 78.9% Fuel-to-Steam Efficiency = (10.5 2.22-0.21-0.1) / 10.5 = 76 %

Exercise # 3 Steam Flow Calculate the steam flow using the following parameters: Steam flow = Ƞ fuel-to-steam * Q in * 1,000,000 / ((hg-hfw) (%BD x hfw)) Q in Natural Gas Stack Feedwater Steam flow Blowdown (BD) (4%) Parameters: Indirect Ƞ fuel-to-steam = 76 % Natural Gas Input (Q in ) = 10.5 MMBtu/hr Blowdown (BD) % = 4% (0.04) Steam Enthalpy Gas (hg) = 1,189 Btu/lb FW enthalpy (hfw) = 198 Btu/lb Ƞ = Efficiency Steam Flow =( 0.76 x 10.5 x1,000,000) / ((1,189-198) - (4% x 198) = 8,118 pph

Exercise # 3 Steam Flow Most common mistake made in calculating steam flow is the use of Combustion Efficiency instead of Fuel-to-Steam Efficiency Ƞ combustion = 78.9 % Steam flow = Ƞ combustion * Q in * 1,000,000 / ((hg-hfw) (%BD x hfw)) = 0.789 * 10.5 * 1,000,000 / {(1,189-198) (0.04 x 198)} = 8,428 pph (as compared to 8,118)

Net Plant Efficiency Takes into Account Steam to DA Net Plant Efficiency = 68.3% Stack Flue Gas Sat. Steam @ 100 psig Total Steam 15,139 pph Process Steam 13,617 pph DA Steam 10% Natural Gas 4.4 million m 3 /yr. Boiler 1 Boiler 2 Boiler 3 Blowdown 4% Boiler Feedwater 8 psig DEAERATOR Cond. Tank Condensate Return 50% 180 F F Make Up Water 50% 44

Summary of Efficiency Definitions Combustion Efficiency: % of fuel energy delivered to boiler Fuel Efficiency: % of fuel energy picked up by the boiler feedwater Fuel-to-Steam Efficiency: % of fuel energy used to produce steam Net plant Efficiency: % of fuel energy used to deliver steam out of steam plant 45

Cost of Steam When determining the cost of steam, three important variables must be considered: 1. At what point will you be evaluating the cost? Generation (Point A) Out of steam plant - line steam (Point B) Point of use (Point C) Natural Gas Steam A Steam to load B 2. What is to be included in the total cost? Varies (in-house use, sold to 3 rd party, etc.) DA Tank 8 psig 3. The total operating cost of generation: Fuel cost Water treatment costs Fan and pump electricity costs Water and sewage costs Maintenance and labour costs C 46

Exercise # 4 Fuel Cost of Steam ($/1,000 lb) Calculate the cost of steam generated at Point A using the following parameters: Fuel Steam Cost($/1,000 lb) x Ƞ fuel to steam Parameters: Steam Enthalpy(hg) = 1,189 Btu/lb Feedwater Enthalpy (hfw) = 198 Btu/lb Ƞ fuel-to-steam = 76% Fuel Cost = $7.0MMBtu Fuel Steam Cost = $7.0 x (1,189 198)Btu/lb/ (1,000 x 76%) = $9.1 to produce 1,000 lb of steam

Exercise # 4 Fuel Cost of Steam ($/1000 lb) Cost Components 1 Fuel Cost of Steam Generation (FC) 9.10 Other Cost Factors * 2 Electricity consumption 0.325 3 Water 0.128 4 Water treatment 0.11 5 Labour 1.027 6 Maintenance 0.474 Total of Other Costs (2 6) 2.064 Total Cost of Generation (CG) Cost ( $/1,000 lb) $11.16/1,000 lb Based on an Enbridge study of 25 manned water tube boiler plants by: Bob Griffin Approximate Cost of Generation CG = FC * ( 1 + 0.30) Source: US DOE Steam Technical Brief How to Calculate True Cost of Steam by Kumana & Associate and Steam Technical Subcommittee

Natural Gas Savings Opportunities

Gas Savings Opportunity Stack Loss Stack loss is a major source of heat loss Reducing stack loss will help us reduce our gas consumption and achieve our savings target Q ng = 100 Q sl = 21.0 Stack Q ra = 2.5 (1% of boiler rating) Q st Steam flow Natural Gas Q fw Feedwater Q bd = 0.7 (4% of FW) Blowdown 50

Gas Savings Opportunity Stack Loss Stack loss can be reduced by: CO 2 + 2H 2 O + X 2O 2 + (1 + x) 7.52 N 2 1. Improving combustion Reduce excess air (O 2 ) Reduce/eliminate CO and C x H y NG 2. Reducing flue gas temperature Recover heat from hot flue gases (40 F drop = 1% efficiency improvement) Air 3. Increasing combustion air temperature Consider air pre-heater Draw air from a high point in the boiler room 51

Reducing Excess O 2 Excess air is required to achieve proper combustion Excess air wastes heat as air enters at ambient temp. and leaves at stack temp. o 79% of air goes for a free ride How much excess air is required? Depends on burner design, boiler configuration, air/fuel control, etc.: o Older coil-tube boilers without Linkageless Controls (LLC) have high excess O 2 : (5% 12%) o Fire-tube boilers generally have lower excess O 2 : (4% 9%) o Large water-tube boilers can achieve lower excess O 2 : (2% - 6%)

How to Find the Optimum Level of O 2 1. Set up a flue gas analyzer 2. Bring boiler to the minimum firing or high firing rate 3. Reduce O 2 until CO starts to appear 4. Increase air slightly to give a safety margin 5. Lock this pin-position for this air:fuel ratio 6. Increase/decrease firing rate 7. Repeat process for each pin position to develop an O 2 characterization curve

Typical Excess O 2 Characterization Curves 54

Reducing Excess O 2 Increases Efficiency 5% O 2 = 77.8% efficiency 12% O 2 = 70% efficiency 55

Reducing FG Temperature Increases Efficiency 6% O 2 at 400 F = 79.5% efficiency 6% O 2 at 600 F = 74% efficiency 56

Exercise # 5a : Excess O 2 Cost Savings Use the Combustion Efficiency Chart to calculate the cost savings associated with reducing excess O 2 levels from 8% to 5%. Ƞ old = 78.9 % Calculate new ƞ Ƞ new = 80.6 % Excess, %FG Temperature - Combustion Air O2 340 360 380 400 420 4.00 82.0 81.5 81.1 80.6 80.1 4.50 81.8 81.3 80.8 80.4 79.9 5.00 81.6 81.1 80.6 80.1 79.6 5.50 81.4 80.9 80.4 79.9 79.3 6.00 81.2 80.6 80.1 79.6 79.0 6.50 80.9 80.4 79.8 79.3 78.7 7.00 80.6 80.1 79.5 79.0 78.4 7.50 80.4 79.8 79.2 78.6 78.0 8.00 80.0 79.4 78.9 78.3 77.7 8.50 79.7 79.1 78.5 77.9 77.2 57

Exercise # 5b : Excess O 2 Cost Savings Calculate the cost savings associated with reducing excess O 2 levels from 8% to 5%. Energy Saving = Q in x {1 - (Ƞ old /Ƞ new )} Energy Saving = 10.5 x (1-78.9/80.6) = 0.221 MMBtu/hr Annual Saving = 0.221 x 5,000 = 1,105 MMBtu Conversion Factor = 1 MMBtu x (1,000,000Btu/1MMBtu) x (1ft 3 /1,012 Btu) x (1m 3 /35.314 ft 3 ) = 27.982m 3 Parameters: FG temp. Comb Air = 380 F Old excess O 2 = 8% New excess O 2 = 5% Ƞ old = 78.9% Ƞ new = 80.6% Q in = 10.5 MMBtu/hr Hours = 5,000 hrs Cost of Gas = $0.25/m 3 1 MMBtu = 1,000,000 Btu 1ft 3 Natural Gas = 1,012 Btu 1m 3 = 35.314 ft 3 Annual Savings m 3 = 1,105 x 27.982 = 30,920 m 3 Gas Cost Savings = $7,730 /yr per boiler 58

Linkageless Controls 59

Linkageless (LLC) Controls To properly assess the value of linkageless controls, we must first understand: 1. The purpose of combustion controls 2. The types of boiler combustion controls today we will focus on: Linkages Linkageless Controls 3. How LLCs improve combustion efficiency 60

Purpose of Combustion Controls: Maintain optimum air:fuel ratio at all firing rates to run boilers safely at optimum combustion efficiency Monitor process boiler temperature and pressure and quickly respond to changes in load 61

Linkage Combustion Controls Mechanical system using cams, linkages and jackshafts to characterize the air:fuel ratio A single actuator motor adjusts its jackshaft arm according to master load (demand) signal As the actuator motor moves the jackshaft, the arms connected to the fuel valve and air fan damper move with it Air:fuel ratio is set by adjusting the cam Primary Fuel Valve Second Fuel Valve Calibrating involves combustion tests in which actuator is positioned to various settings, usually at least 10, and at each setting setscrews are adjusted to achieve the desired O 2 level in flue gas Firing Rate Actuator Prime Motor Fan 62

Issues with Linkage Controls: Hysteresis or drift caused by wear, tear and slop in linkages Control devices do not return to the same position during boiler ramp-up or turn-down 63

Issues with Linkage Controls: Air : fuel ratio is typically set high to compensate for hysteresis Air : fuel ratio generally drifts after tune-up These issues result in: o Reduced efficiency o Safety concerns (High CO and combustibles) Most small boilers do not have an in-situ flue gas analyzer, and as a result, high CO can not be detected until tune-up. 64

Linkageless Combustion Controls Obviously no linkages Linkageless Combustion Control System Individual servomotors attached to gas valve and air damper. Position of each motor is programmed independently P Steam Pressure A programmable control unit provides precise air:fuel ratio over entire range No hysteresis for a properly tuned and maintained LLC system Microprocessor Controller Natural Gas Combustion Air Damper with Servo Motor Drive Gas Control Valve with Servo Motor Drive Additional controller can be added to provide O 2 trim Combustion Air Blower Burner 65

Linkagelss Controls Excess O2 10.00 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 Boiler 3 Oxygen Curve with LLC Ramping Up Turning Down 0% 20% 40% 60% 80% 100% 120% Firing Rate 66

Honeywell and Siemens LLC Systems Honeywell Siemens 67

Power Flame Boiler Burner Linkages Linkageless with Honeywell LLC 68

Gas Savings Opportunity: Linkageless Controls 1. No hysteresis o Function of Base Case burner age 2. Improved combustion efficiency due to: o Reduced excess air o Accurate characterization of air:fuel ratio o Accurate control of firing rate o Depends on how much excess O 2 can be reduced 3. Reduced cycling due to improved turn-down o Savings are a function of Base Case burner turndown 4. Additional savings due to O 2 trim 69

We re here to help! What your ESC uses to calculate and quantify LLC Savings Opportunities 70

Gas Savings Opportunity: Install LLC Reduce O 2 Maintain optimum air/fuel all the time Savings Initiative Savings Removal of hysteresis 0.50 % Improved combustion 2.19 % Increased turndown 0.13 % Total 2.82 % Annual gas savings /boiler 41,427 m 3 /yr Gas cost saving / boiler $10,357 Gas savings for 3 boilers 124,281 m 3 Gas cost savings for 3 boilers $31,070 2.8% Saved 71

Personal experiences with LLC Some vendors claim 10-15% savings, based on unrealistic assumptions Savings possible if burner is in very bad condition with very high O 2, CO, Combustibles, Cycling etc. Need to establish base case performance Your ESC will help you establish a base line to calculate realistic savings Important to estimate savings based on a real base case LLC installation for a less than 100 HP (4.2 MMBtu/hr input) boiler would typically not be cost effective 72

Blowdown Loss 73

Gas Savings Opportunity: Blowdown (BD) Loss Blowdown loss can be reduce by: 1. Reducing amount of BD 2. Recovering BD heat Flash tank Heat exchanger 74

Blowdown Basics When water is boiled, steam is generated Solids are left behind: o o Suspended solids form sludge which degrades heat transfer Dissolved solids promote foaming and water carryover Water is discharged to keep solids within acceptable limits o o Bottom blowdown from mud drum removes suspended solids (once/twice a day) Surface blowdown removes dissolved solids, concentrated near liquid surface (continuous) Insufficient BD leads to carryover and deposits Excessive BD leads to wasted energy, water and chemicals Blowdown water temp. is same as steam Typical range is 3% 6% of feedwater 75

Blowdown Heat Recovery 1. Flash Tank 2. Heat Exchanger Stack Flue Gas Sat. Steam @ 100 psig Total Steam 15,139 pph Process Steam 13,617 pph DA Steam 10% Natural Gas 4.4 million m 3 /yr. Boiler 1 Boiler 2 Boiler 3 Blowdown 4% Flash Tank 10 psig Flash Stream Heat Exchanger Boiler Feedwater 8 psig DEAERATOR Cond. Tank To Drain Make-Up Water 50% Condensate Return 50% 180 F 230 F 76

Exercise # 6a Flash Tank Savings Use the Flash Steam Table to find the percentage of flash steam and then calculate the blowdown heat recovery savings associated with installing a flash tank. Parameters: Blowdown Pressure = 100 psig Flash Tank Pressure = 10 psig Steam Pressure Percent of Blowdown / Condensate Flashed Flash Tank Pressure / Low Steam Pressure psig 0 2 3 5 10 15 90 12.6 11.9 11.6 11.1 10.0 8.8 100 13.3 12.6 12.3 11.9 10.7 9.5 110 14.0 13.4 13.1 12.6 11.4 10.3 BD Pressure 100 psig 10 psig Flash Tank Flash Steam Heat Exchanger Make-Up Water 50% % Flash Steam = 10.7 % To Drain 77

Exercise # 6a Flash Tank Savings Flash Steam Flow = %Flash x BD Flow Rate Flash Steam Savings = Flash Steam Flow Rate x Enthalpy Annual Flash Steam Savings = (Flash Steam Savings x Hours)/(1,000,000 x Ƞ fuel-to-steam ) Flash Steam Flow = 10.7 % x 630 lb/hr = 67.41 lb/hr Flash Steam Savings = 67.41 lb/hr x 962 Btu/lb = 64,848.42 Btu/hr Annual Flash Steam Savings = (64,848.42 x 8,000)/(1,000,000 x 0.76) = 682.61 MMBtu/yr Gas Savings = 682.61 x 27.982 = 19,101 m 3 /yr Parameters: % Flash Steam = 10.7% Ƞ fuel-to-steam = 76.8% Boiler Blowdown = 4% (0.04) Enthalpy = 962 Btu/lb FW Flow Rate = 15,760 lbs/hr BD Flow Rate = 630 pph BD Pressure = 100 psig Flash Tank Pressure = 10 psig Plant Hours = 8,000 hrs 1 MMBtu = 27.982m 3 78

Exercise # 6b Heat Exchanger Savings Calculate the savings associated with installing a heat exchanger. Blowdown Liquid Flow = Blowdown Flow Rate Flash Steam Flow Rate Energy Savings = Blowdown Liquid Flow x Heat Capacity x (T inlet T outlet ) Annual Energy Savings = (Energy Savings x Hours)/(1,000,000 x Ƞ fuel-to-steam ) Blowdown Liquid Flow = 630 67.41 = 563 lb/hr Energy Savings = 563 lb/hr x 1 Btu/lb F x (239 65) F = 97,962 Btu/hr Annual Energy Savings = (97,962 Btu/hr x 8,000)/(1,000,000 x 0.76) = 783,696,000/760,000 = 1,031.18 MMBtu x 27.982m 3 = 28,854 m 3 Parameters: Ƞ fuel-to-steam = 76% Blowdown Flow Rate = 630 pph Heat Capacity = 1 Btu/hr F Inlet Temp. (T inlet ) = 239 F Outlet Temp. (T out ) = 65 F Plant hours = 8,000 hours 1 MMBtu = 27.982m 3 79

Exercise # 6c Total Blowdown Gas Savings Total BD Gas Savings = Flash Steam Savings + Heat Exchanger Savings Total Gas Savings = 19,101 + 28,854 = 47,955 m 3 /yr = 1.1% Gas Cost Savings = 47,955 m3 x $0.25/m3 = $11,989 Parameters: Cost of Gas = $0.25/m 3 3.9% Saved 80

Insulation of Pipes and Valves 81

Gas Savings Opportunity: Insulation of Pipes and Valves Reasons for installing and monitoring insulation on steam valves and pipes: Bare pipes and valves lose a significant amount of heat Insulation is required to prevent heat loss through valves and pipes Lack of proper insulation poses a safety hazard to plant workers Insulation is often removed or damaged during maintenance without being replaced Any surface over 120 F should be insulated 82

Gas Savings Opportunity: Insulation of Pipes and Valves Insulation increases the amount of steam energy available for end uses Insulation can reduce heat loss by 90% Valves have a large surface area: e.g. a 6 gate valve may have over 6 square feet of surface area Removable and reusable insulation covers that are easy to remove and replace are available Your ESC can provide savings estimates 3EPLUS software is available 83

Gas Savings Opportunity: Insulation of Pipes and Valves Source: US DOE Steam tip Sheet #17 84

Gas Savings Opportunity: Insulation of Pipes and Valves Prepared by Enbridge, based on data from US DOE Steam tip Sheet #17 85

Steam Plant: Un-Insulated Pipe and Valves Sat. Steam @ 100 psig Process Steam Boiler 1 Boiler 2 Boiler 3 86

Sample Project: Installing Insulation on Pipes Calculate the savings associated with insulating a 150 ft long X 4 dia pipe Heat Loss (no insulation)= 1,072 MMBtu/yr Heat Loss (1 insulation)= 120 MMBtu/yr Heat Saved = 952 MMBtu/yr = (89%) Gas Savings = 35,319m 3 /yr Parameters: Steam Pressure = 100 psig Operating Temp. = 338 F Hours = 8,000 Hrs Ƞ fuel-to-steam = 76% Cost of Gas = $0.25/m 3 1 MMBtu = 27.982m 3 1 MMBtu = 1 Mil. Btu Gas Cost Savings = $8,830 per yr 87

Exercise # 7 : Installing Insulation on Valves Calculate the savings associated with insulating three 4 valves and three 6 valves. Net Savings = Σ Energy Savings from All Valves Gas Savings = Net Savings/ Ƞ fuel-to-steam Net energy savings = (2,718 x 3) + (4,250 x 3) = 20,904 Btu/hr x 8,000 = 167,232,000 Btu/hr/1,000,000 = 167.2 MMBtu/yr Gas savings (valves) = 167.2 MMBtu/hr / 0.76 = 220 MMBtu/yr x 27.982 = 6,156 m 3 Gas savings (150 pipe) = 35,319m 3 Parameters: 4.8% Saved 4 Savings = 2,718 Btu/hr 6 Savings = 4,250 Btu/hr Steam Pressure = 100 psig Operating Temp. = 338 F Hours = 8,000 Hrs Ƞ fuel-to-steam = 76% Cost of Gas = $0.25/m 3 1 MMBtu = 27.982m 3 1 MMBtu = 1 Mil. Btu Total gas savings = 41,475m 3 Gas cost savings = $10,369 88

2013 Insulation Survey Campaign Enbridge will fund 100% of the cost of an Insulation Survey up to a maximum of $2,500.00 per customer!! Special promotion beginning right here, right now Limited space available - first come, first serve basis Scope of Work must be approved by an ESC Application forms due by October 1, 2013 Flyer in package has additional information Speak with your ESC to find out if this campaign is right for you!

Flue Gas Heat Recovery 90

Gas Savings Opportunity: Flue Gas Heat Recovery Flue gas heat recovery can be achieved through: Air Preheaters Feedwater Economizers Condensing Economizers 91

Air Preheaters Flue Gas Inlet Preheat combustion air Generally used on large water tube boilers, not common on small boilers Two Common Types: 1. Tubular 2. Heat wheel (Ljungstom) Air Outlet Baffle Tubes Baffle Air Inlet Flue Gas Outlet Combustion Air (to boiler) Flue Gas (from boiler) Inlet Air To stack 92

Gas Savings Opportunity: Feedwater Economizers Purpose: Preheat boiler feedwater Feedwater Stack 300 F 230 F 288 F FW Economizer 460 F Steam flow Natural Gas Blowdown DEAERATOR 8 psig 93

Types of Feedwater Economizers 1.In-line Cylindrical o Fire tube and coil tube boilers Rectangular o Large water tube boilers 2.Integral Built-in as part of boiler. e.g. Miura, Clayton 94

Feed Water Economizer Considerations T fg out of boiler should be 100 150 F greater than T sat. Steam o e.g. T fg > 440 F (338 + 100) for 100 psig steam A lower temperature may indicate integral economizer T stack > 250 F to avoid condensation in the economizer (unless made out of stainless steel) and stack For existing economizers confirm if they are working or by-passed. o Take T fg at inlet and outlet 95

Major Suppliers of Feedwater Economizers Thermogenics Cain Kentube E-Tech Heatspong Cleaver Brooks Combustion and Energy Canon 96

Exercise # 8 : Installing a Feedwater Economizer Use your Combustion Efficiency Chart to identify the new efficiency level after installing a feedwater economizer and then calculate the associated savings. Parameters Ƞ w/o econo. = 80.6 % O 2 = 5% T out = 300 F T air = 80 F Delta T = 220 F Ƞ new = 84.6 % 230 F 288 F 300 F Excess, FG Temperature - Combustion Air Temp,Degree F O2 200 220 240 260 280 300 320 340 360 380 4.50 85.3 84.8 84.3 83.8 83.3 82.8 82.3 81.8 81.3 80.8 5.00 85.1 84.6 84.1 83.6 83.1 82.6 82.1 81.6 81.1 80.6 5.50 85.0 84.5 84.0 83.5 82.9 82.4 81.9 81.4 80.9 80.4 460 F 97

Exercise # 8 : Installing a Feedwater Economizer Energy Savings = Q in x {1 - (Ƞ old /Ƞ new )} 9.6% Saved Energy Savings = 10.5 x (1-80.6/84.6) = 0.5 MMBtu/hr Annual Savings = 0.5 MMBtu/hr x 5,000 Hrs = 2,500 MMBtu/yr x 27.982m 3 = 69,955 m 3 Gas Cost Savings = 69,955 m 3 x 0.25m 3 = $17,489/yr per boiler Gas Cost Savings for 3 Boilers = $17,489 x 3 boilers =$52,467/yr Parameters Ƞ old = 80.6% Ƞ new = 84.6% O 2 = 5% T out = 300 F T air = 80 F Delta T = 220 F Q in = 10.5 MMBtu/hr Hours = 5,000 Hrs Cost of Gas = $0.25/m 3 1 MMBtu = 27.982m 3 98

Gas Savings Opportunities: Condensing Economizers Condensing economizers help maximize boiler heat recovery Hot gas 300 F+

Condensing Economizer Overview 1. Review basic concepts of condensing heat recovery 2. Types of condensing economizers 3. Energy savings potential 4. Case studies 5. Potential applications 6. Key considerations 7. Exercises

Basic Concept of Condensing Heat Recovery When one molecule of CH 4 is burned, it produces two molecules of H 2 O CH 4 + 2O 2 + 7.52 N 2 CO 2 + 2H 2 O + 7.52 N 2 16 lb 1 lb 36 lb 2.25 lb One lb of CH 4 produces 2.25 lb of water One lb of Natural Gas produces 2.14 lb of water

Basic Concepts of Condensing Heat Recovery Water in products of combustion is vaporized due to heat of combustion Water vapours absorb about 10% of fuel input (Hydrogen Loss in Combustion Efficiency) Energy is lost to atmosphere with exhaust gases through stack Heat of vaporization can be recovered if flue gases are cooled below water dew point When water vapour condenses, it releases heat of vaporization Condensing economizer recovers both: 1. Heat of condensation (latent heat) 2. Sensible heat

Types of Condensing Economizers Indirect Contact Direct Contact 200 F 135 F Source: DOE Condensing Economizers Tip Sheets

Sensible and latent heat As flue gas temperature decreases, efficiency increases Condensation starts below dew point at about 137 F Sensible heat only Excess O2 = 5% 104

Energy Savings with Condensing Economizers Savings come from steam reduction in DA tank to heat make-up water Main Stack 300 F FW Econo. Makeup Water 60 F 200 F Condensing Economizer New Stack Before Natural Gas Steam Steam to load Cond. Tank Cond Return. DA Tank 7 psig Cond. Tank Cond. return Make-up Water 105

Requirements of Flue Gas and Condensate Passageways Make-Up Water Flue gas Condensate Make-Up Water Condensate Flue gas and condensate must flow in the same direction (parallel flow) Flue gas 106

Available Heat Varies with FG Temp. leaving Economizer 1.6 Heat available from one boiler Tfg-in = 300 F Heat Available (MMBTU/hr) 1.4 1.2 1 0.8 0.6 0.4 0.2 0 1.39 0.82 0.56 1.11 0.61 0.50 0.61 0.17 0.44 0.38 0.38 Latent Heat Sensible Heat 75 F 100 F 125 F 150 F Flue Gas Temp. Leaving Condensing Economizer ( F)

Recovered Heat Depends on Heat Sink Size 1.6 1.4 1.39 1.39 1.39 1.39 Heat (MMBtu/hr) 1.2 1 0.8 0.6 0.4 0.28 0.57 0.85 1.13 Heat Available @ 75 F Exit Temp., 62 F MUW Temp. Heat Recovered at various MUW Rates 0.2 0 25% 50% 75% 100% Make-up Water Rate %

Case Study: Indirect Condensing Economizer Two 350 HP coil tube boilers, 105 psig saturated steam 100% boiler make-up water 5 day x 24 hr operation ConDex condensing economizer system installed on roof Pull exhaust from two stacks into one common duct feeding Condex

Case Study: Indirect Condensing Economizer System heats boiler make-up water from 65 F up to 165 F At maximum load the system recovers 2,541,000 Btu/hr Projected Gas savings: $190,000 Payback: 1.4 yrs.

Typical Applications Industries with steam boilers, requiring a large amount of hot water such as make-up, washing, process, DHW Best Candidates: o Textile, commercial laundries o Food and beverage o Breweries o Non-integrated paper mills o Chemicals o District heating o Large hospitals o Greenhouses

Key Considerations Establish how much heat is available o Existing FW economizer, Flue gas temp., excess O 2, steam production, gas consumption, hours of operation, etc. Is there sufficient heat sink available? o Boiler make-up water o Domestic hot water o Process water Entering temperature of heat sink must be below dew point to cause condensation Evaluate impact on existing system including blowdown, flash steam, DA, water treatment etc.

Key Considerations Space for installation, stacks, icing due to plume impingement, indoor/outdoor installation, etc. Cost savings, installation costs, payback Direct versus indirect? o Site specific o Customer preference o Cost o Temperature requirement o Application o Heat sink, etc.

Key Considerations Not an off-the-shelf technology for many applications. Often an engineered solution is required. Requires a good understanding of the technology and its application. Needs a suitable heat sink. Small amounts of make-up water (25%) capture only a small portion of available heat. Need to include condensing economizer as part of a standard steam plant assessment. Condensing heat recovery is a proven, commercially available cost-effective technology

Case Study: In-line Condensing Economizer For individual boilers (100 500 BHP), a new in-line condensing stack economizer is available: 115

Major Manufacturers Indirect Contact Economizers Combustion and Energy System, Toronto, Canada E-Tech, Tulsa, Oklahoma Benz Air, Las Vegas, NV Direct Contact Economizers Sofame, Montreal, Canada Thermal Energy System, Ottawa, Canada Direct Contact Inc. Renton, WA Kemco System Sidel System USA, California CHX Corporation, Clifton Park, NY

Exercise # 9 : Indirect Contact Condensing Economizer 300 F 300 F 62 F 200 F 117

Exercise # 9 : Indirect Contact Condensing Economizer Calculate the savings associated with installing a condensing economizer to heat make-up water. Heat Available in Flue Gases = Q in x (1- Ƞ new combustion ) Heat Recovered = Make-Up Water Flow x Heat Capacity x (Water T out Water T in ) Gas Savings = Heat Recovered/ Ƞ new fuel-to-steam 62 F 200 F 126 F 300 F Parameters Q in = 9.8 MMBtu/hr Ƞ new combustion = 84.6 % Ƞ new fuel-to-steam = 81.5% Water T in = 62 F Water T out = 200 F Make-up Water Flow = 4,041 lb/hr Heat Capacity = 1 Btu/hr F Hours = 5,000 Hrs Cost of Gas = $0.25/m 3 1 MMBtu = 27.982m 3 1 MMBtu = 1 mil. Btu # of Boilers = 3 118

Exercise # 9 : Indirect Contact Condensing Economizer Step 1: Calculate heat available in flue gases Heat available in stack = 9.8 x (1-0.846) = 1.5 MMBtu/hr Step 2: Calculate heat recovered by 50% make-up water Heat recovered = 4,041 x 1 x (200 62) = 557,658 Btu/hr/1,000,000 = 0.558 MMBtu/hr Step 3: Check if Heat Recovered is less than Heat Available Parameters Q in = 9.8 MMBtu/hr Ƞ new combustion = 84.6 % Ƞ new fuel-to-steam = 81.5% Water T in = 62 F Water T out = 200 F Make-up Water Flow = 4,041 lb/hr Heat Capacity = 1 Btu/hr F Hours = 5,000 Hrs Cost of Gas = $0.25/m 3 1 MMBtu = 27.982m 3 1 MMBtu = 1 mil. Btu # of Boilers = 3 119

Exercise # 9 : Indirect Contact Condensing Economizer Step 4 : Calculate gas savings Gas Savings = 0.558 / 81.5% = 0.6847 MMBtu/hr (6.7% of energy in) 16.1% Saved Annual Saving = 0.6847 x 5,000 = 3,423 MMBtu x 27.982m 3 = 95,782m 3 Gas cost Savings = $0.25/m 3 x 95,782m 3 = $23,946/yr/boiler Gas Cost Savings for 3 Boilers = $71,838/yr 120

Condensing Economizers to Heat Process Water Stack Flue Gas Sat. Steam @ 100 psig Total Steam 15,139 pph Process Steam 13,617 pph Boiler 1 Boiler 2 Boiler 3 Steam 1705 pph 140 F Currently process water is heated with steam Can we heat process water with flue gases? Condensate 65 F Process Water 40 gpm 20,000 pph 121

Add a Second Coil to Heat Process Water Flue Gases 300 F 300 F MUW 200 F 62 F 120 F 126 F From boiler Heat water to 120 F with FG, then top it up to 140 F with steam Steam 455 pph 140 F Process Water 40 gpm 20,000 pph 65 F Process Water Flue Gases 120 F 122

Exercise # 10 : Heating Process Water Calculate the savings associated with adding a second coil to the condensing economizer to heat process water. Heat Recovered = Process Water Flow x Heat Capacity x (T out T in ) Gas Savings = Heat Recovered/ Ƞ new fuel-to-steam Heat Recovered = 20,000 lb/hr x 1Btu/hr F x (120-65) F = 1,100,000 Btu/hr/1,000,000 = 1.1 MMBtu/hr Gas Savings = 1.1 MMBtu/hr/81.5% = 1.35 MMBtu/hr Annual Gas Savings = 1.35 MMBtu/hr x 5,000 = 6,750 MMBtu/yr x 27.982m 3 = 188,879 m 3 /yr Cost Savings = 188,879m 3 /yr x $0.25/m 3 = $47,220/yr Parameters Q ng x 3 boilers = 29.4 MMBtu/hr Ƞ new fuel-to-steam = 81.5 % Water T in = 65 F Water T out = 120 F Process Water Flow = 20,000 lb/hr Heat Capacity = 1 Btu/hr F Hours = 5,000 Hrs Cost of Gas = $0.25/m 3 1 MMBtu = 27.982m 3 1 MMBtu = 1 mil. Btu 123

How does our plant look with all the improvements? Did we achieve our goal of 20% savings?

Energy Efficient Steam Plant Stack Flue Gas FW Economizer Condensing Economizer Process Water Boiler 1 BD Boiler 2 Boiler 3 Flue Gases Natural Gas Boiler Feedwater 8 psig DEAERATOR Cond. Tank Flash Tank Heat Exchanger Condensate Return Make-Up Water 125 To Drain

Exercise # 11 : Summary of Savings Project Summary Improve Combustion with Linkageless Controls Net Gas Savings (m 3 ) Net Costs Savings ($) 124,281 m 3 $31,070 Recover Heat from Boiler Blowdown 47,955 m 3 $11,989 Insulate Valves 41,475 m 3 $10,369 Heat Feedwater with a Feedwater Economizer Heat Make-Up Water with a Condensing Economizer Heat Process Water with a Condensing Economizer 209,865 m 3 $52,467 287,346 m 3 $71,383 188,879 m 3 $47,220 Total Savings 899,801 m 3 $224,498 Total Gas Costs = $1.1 million/yr Total Cost Savings = $224,498/yr Percentage Savings = 20% 126

20% Saved 127

At Enbridge, We re here to help: Arm our customers with information. Action and Implementation Analyzing data and Monetizing Savings. Engineering Analysis Knowledge Development Contact your ESC to learn more about our free services and financial incentives! Through testing and energy use analysis. Opportunity Identification Quantify key energy inputs Measurement

At Enbridge, We re here to help: There are many things you can do to ensure your boiler plant is as efficient as possible. Each boiler plant is different and opportunities should be assessed on a case by case basis. Your Enbridge Energy Solutions Consultant is available to help you: Evaluate the efficiency of your boiler plant Quantify key energy inputs and outputs Identify energy savings opportunities Monetize energy savings opportunities Connect you with third party vendors Identify any Enbridge incentives you may qualify for Develop an implementation plan 129

Knowledge Development Action and Implementation THANK YOU! QUESTIONS? Opportunity Identification Engineering Analysis Measurement