Steam System Efficiency. Bill Lumsden Leidos Engineering

Steam System Efficiency Bill Lumsden Leidos Engineering

Steam System Efficiency

Steam System Efficiency Key Take-aways: Review of the properties of ice, water, and steam Learn the basics of steam trap application, types, modes and cost of failure Learn how to calculate the value of recovering flash steam Learn how to calculate the value of condensate Learn two approaches to recovering heat from the boiler stack Learn how to recover heat from the boiler blowdown water Other common steam system best practice Insulation Burner tune-up and O 2 trim High turndown burners

Physical Properties and Terms Used When Talking About Water in Its Various States

Terms Used In the Subject of Heat Transfer Specific Heat, C p : The amount of heat per unit mass required to raise the mass temperature by 1 degree water C p = 1 Btu/lb ᵒF ice C p = 0.50 Btu/lb ᵒF steam C p = 0.50 Btu/lb ᵒF Sensible Heat: Heat absorbed or removed during a change in temperature without a change in state Latent Heat: Heat absorbed or removed during a change of state at a constant temperature (examples: ice to water or water to steam) Latent heat of ice at 32 degree, h = 144 Btu/lb Latent heat of water at 212 degrees, h = 970 Btu/lb

Property Terms Used In Heat Transfer Enthalpy (h): Heat content of a substance (Btu/lb) In the Steam Tables, water at 32 F by definition has no heat content, h f = 0 Btu/lb Water at the boiling point of 212 F has heat content, h f = 1 lb x 1 Btu/lb ᵒF x (212 32)ᵒF = 180 Btu/lb 212 F Steam at atmospheric pressure has total heat content, h g = 1,150 Btu/lb (180 Btu/lb sensible heat + 970 Btu/lb latent heat)

Sensible, Latent and Change of State Liquid to Steam Change of State 212 ºF Latent Heat = Liquid to Gas 32 to 212 ºF Sensible Heat = Liquid 32 ºF Latent Heat = Solid to Liquid Solid to Liquid Change of State 20 to 32 ºF Sensible Heat = Solid Block of Ice 20ºF Heat Source

Super Heating Steam (vapor) Vapor @ 212 ºF Vapor & Water @ 212 ºF Water @ 212 ºF Water @ 62 ºF Superheated Steam 212 ºF Latent Heat 62 to 212 ºF Sensible Heat Heat Source

100 lbs 20ᵒF Ice HEAT 100 lbs Ice to Water HEAT 100 lbs Water HEAT 100 lbs Water to Steam HEAT Sensible Heat Temperature Change Add Heat: 20 ºF to 32 ºF Solid Ice = Sensible q = (100 lb) x (0.50 Btu/lb o F) x (32 o F-20 o F) = 600 Btu Latent Heat No Temperature Change Add Heat: 32 ºF ice to 32 ºF Liquid Water = Latent Heat q = (100 lb) x (144 Btu/lb) = 14,400 Btu Sensible Heat Temperature Change Add Heat: 32 ºF to 212 ºF Liquid Water = Sensible q = (100 lb) x (1 Btu/lb o F) x (212 o F-32 o F) = 18,000 Btu Latent Heat No Temperature or Pressure Change Add Heat: 212 ºF Water to 212 ºF Steam = Latent q = ( 100 lb) x (970 Btu/lb) = 97,000 Btu

Steam Trap Basics

Steam Trap Functions 1. Hold Steam in Correct Place Until Latent Heat is Absorbed by the Heat Exchanger 2. Drain Away Condensate and Discharge Air and CO 2 (while preventing the escape of steam)

Things That Cause Steam Traps to Fail or Loose Performance 1. Normal wear occurs on all mechanical devices over time 2. Dirt can plug orifice or impede linkages 3. Pressure surges can damage the trap 4. Water hammer caused by a slug of water being pushed rapidly down stream ahead of steam causing damage, especially on bellows and float & thermostatic traps 5. Back pressure will make any steam trap loose capacity and can cause high cycling on disk traps 6. Freezing sometimes occurs and can damage a steam trap. Bellows and floats can be crushed 7. Excessive cycling for whatever the reason accelerates wear

Steam Traps in a Steam System pocket & cleanout Take-offs from top of header air vent treated M.U. water FW tank boiler steam trap steam trap pressure reduction C t LPS process equip. steam trap air vent HPS process equip. air vent steam trap flash vent steam trap flash vent condensate tank with pump

Steam Trap Installation on a Steam Header sediment clean out valve steam trap check valve

Steam Trap Types Mechanical - Inverted Bucket - Float with Thermostatic Air Vent Thermostatic - Expansion - Bimetallic Thermodynamic - Disc

Mechanical Trap, Float with Thermostatic Air Vent Steam Trap (A Modulating Steam Trap) Steam, condensate and air in Bellows Thermostatic Valve Sealed Float Condensate and air out Float Valve Air Line Derived from: Spirax Sarco www.wermac.org

Thermostatic Trap (Bellows or Bimetallic) Open Position Hot or sub-cooled liquid condensate (and some air) Bellows filled with alcohol/water mixture is collapsed, opening the valve Condensate and some air in Liquid condensate & air out Valve Seat Source: Review of Orifice Plate Steam Traps By C.B.Oland, Oak Ridge National Lab

Thermostatic Steam Trap (Bimetallic or Bellows) Closed Position Steam Bellows filled with alcohol/water mixture is expanded, closing the valve Steam in Valve closed Source: Review of Orifice Plate Steam Traps By C.B.Oland, Oak Ridge National Lab

Thermodynamic Disk Trap, Fully Open Stage Thermodynamic Steam Trap, Disk Type P1 > P2, disk pushed up by condensate Condensate in Liquid condensate out Copyright 2015, Armstrong International, Inc., Reproduced with permission

Thermodynamic Disk Trap, Fully Closed Stage Thermodynamic Steam Trap, Disk Type Steam on top of the disk holds it closed until steam cools at which time pressure on top drops No flow until the disk lifts from the seat again No flow until the disk lifts from the seat again Copyright 2015, Armstrong International, Inc., Reproduced with permission

Steam Trap Failure Two Typical Failure Modes Are: 1. Leaking or Blowing a) Energy and water treatment chemicals can be wasted. b) Reduce the function of the affected heat exchanger. c) Cause high pressure in the condensate line. d) Cause water hammer resulting in system damage 2. Plugged a) Back up condensate into the equipment causing the heat exchanger to quite working. b) Back up condensate to the steam line header, causing steam cooling and also risk of water hammer

Calculating Losses from Blowing Steam Trap It is difficult to know the steam loss rate from a blowing or leaking steam trap because there are a number of unknowns: a) difficult to know for sure whether the orifice is fully or partially open (that is, open or leaky) b) has no way to know the percentage of orifice blockage caused by the water that goes through the orifice with the steam c) may not know the magnitude of backpressure on the low pressure side

Napier s Formula for Steam Flow Through an Orifice The following is a conservative method for estimating steam loss through a blowing steam trap (guards from overstating losses) using Napier s Formula and adjusting for unknowns: Napier s Formula: Steam Flow (lb/hr) = 51.43 x Orifice Area (in 2 ) x (Gauge Pressure +14.7) x COD x 50% Where: - COD is the orifice coef. of discharge (0.72 is a reasonable estimate) - fifty percent (50%) takes into account unknowns stated in the previous slide

Calculating Energy Loss & Energy Value of Lost Steam Energy Loss: Energy Loss (MMBtu/yr) = Steam Flow x (h steam h M.U. water ) x Hrs 1,000,000 Where: - h steam is the total energy of the steam on the high pressure side - Hrs are the annual hours the steam trap operates - 1,000,000 is the conversion factor: 1,000,000 Btu per MMBtu Energy Value: Energy Value ($/yr) = Energy Loss Boiler Efficiency x Average Cost of Fuel

Example Steam Trap Repair Problem Example Of 30 identical 1/16 orifice steam traps surveyed, 10 were found failed open and blowing steam to the condensate return pipe system which drains to a condensate tank vented to the outdoors. The steam pressure is 100 psig, annual operating hours are 8,400, gas cost is $6.00 / MMBTU and boiler efficiency is 80%. Boiler make-up water is supplied at 55 ᵒF. What is the energy value of the lost steam? Solve for steam flow: Steam Flow (lb/hr) = 51.43 x Orifice Area (in 2 ) x (Gauge Pressure +14.7) x COD x 50% = 51.43 x.00306 x (100 +14.7) x.72 x 50% x 10 traps = 65 lb/hr Solve for energy value: From steam tables: h steam is 1,190 Btu/lb and h M.U. water is 23 Btu/lb Energy Value = Stm. Flow x (h steam h M.U. water ) x Hrs 1,000,000 eff. x $/MMBtu = 65 x (1,190 23) x 8,400 80% 1,000,000 x $6.00 = $4,790 per year

Flash Steam What is Flash Steam? Flash steam is the name given to saturated steam formed from saturated condensate when the pressure is reduced

Flash Steam What happens to the condensate when it passes through the stream trap orifice?

Flash Steam Forms Flash steam forms when condensate passes to the low pressure side of the steam trap s orifice condensate or steam flash steam condensate high pressure side of steam trap orifice low pressure side of steam trap orifice

Flash Steam What happens to the flash steam down stream of the steam trap?

Condensate Collection Tank Flash Steam is Commonly Vented to the Outdoors Flash Steam Vented To Outdoors Condensate return line Condensate Pumped to Boiler Room

How Much Flash Steam is There? % Flash Steam = [(h f )@P 1 - (h f )@P 2 ] / [ (h fg )@P 2 ] x 100% Where: - (h f )@ P 1 is the enthalpy of saturated water at the high pressure - (h f )@P 2 is the enthalpy of saturated water at the low pressure - (h fg )@P 2 is the latent heat of steam at the low pressure

Example Flash Steam Value Problem Example: Steam from a 100 psi header is delivered to a process heat exchanger through a control valve where pressure on average is dropped to 50 psi. Condensate from the steam trap serving the process drains to a condensate collection system with the condensate tank vented to the outdoors. The total steam consumed by the process averages 1,500 pounds per hour and operates for 8,400 hours per year. Gas cost is $6.00 / MMBtu and boiler efficiency is 80%. What is the energy value of the flash steam? Solve: From steam tables, (h f )@ 50 psi = 267 Btu/lb, (h f )@ 0 psi = 180 Btu/lb, (h fg )@ 0 psi = 970 Btu/lb, (h M.U.water )@ 0 psi = 23 Btu/lb, (h g )@ 0 psi = 1,150 Btu/lb % Flash Steam = [(h f )@P 1 - (h f )@P 2 ] [ (h fg )@P 2 ] x 100% = (267 180) 970 x 100% = 9.0% Total Steam Flow (lb/hr) = 1,500 pph x 8,400 hours/year = 12,600,000 pounds per year Energy Value ($/yr) = % Flash x Ann. Stm. Flow x (h g@0psi h M.U. water ) 1,000,000 eff. x $/MMBtu = 9% x (12,600,000) x (1,150 23) 1,000,000 80% x $6.00 = $9,585 per year Note: This example assumes all flash steam makes it back to the recovery tank which in application may not be the case.

What can you do to recover the flash steam?

Vent Condenser Installed on Condensate Tank Trace of Flash Steam to Outdoors Cold Fluid Flash Steam Condensed Flash Steam Heated Fluid Condensate return line Condensate Pumped to Boiler Room

Vent Condenser Installation pressure relief valve vapor outlet vapor inlet condensate outlet

Vent Condenser Installation cold fluid in and out

Condensate Receiver Vent (BEFORE & AFTER)

Recover Condensate

Energy Value of Condensate Sometimes condensate is not recovered from some areas of the plant for reasons such as first cost concerns associated with installing a condensate recovery system. Energy Value of Condensate ($/yr) = (Annual Steam Flow - Flash Steam) x % Condensate Not Recovered x (h f )@ 0 psig eff. 1,000,000 Btu/MMBtu x $/MMBtu Note: This formula assumes that the increased condensate recovered is returned to a condensate return tank that is vented to the atmosphere and the tank does not use a vent condenser.

Energy Cost for Making Up Un-recovered Condensate Example: The total steam consumed by the process averages 15,000 pph, steam pressure is 100 psig, annual operating hours are 8,400, gas cost is $6.00 / MMBtu and boiler efficiency is 80%. 75% of condensate is currently recovered. What is the cost of energy to make-up the 25% that is not recovered? Solve: From steam tables, (h f )@ 100 psi = 309 Btu/lb, (h f )@ 0 psi = 180 Btu/lb, (h fg )@ 0 psi = 970 Btu/lb, (h M.U.water )@ 0 psi = 23 Btu/lb % Flash Steam = [(h f )@P 1 - (h f )@P 2 ] [ (h fg )@P 2 ] x 100% = (309 180) 970 x 100% = 13.3% Energy Value of Condensate ($/yr) = (Annual Steam Flow - Flash Steam) x %Condensate not Recovered x (h f )@ 0 psi 1,000,000 eff. x $/MMBtu Energy Cost to Make-up Condensate ($/yr) = 25% x 15,000 x 8400 x (1-.133) x (180-23) 1,000,000 80% x $6.00 = $ 32,158 per year

Steam Trap Incentives Steam Trap Survey Incentives: HVAC Steam Traps 15 psig = $30/trap Process Steam Traps > 15 psig = $50/trap Repair or Replacement Incentives: HVAC Steam Traps (< 15 psig & dry cleaning apps) = $100/trap Process Steam Traps (< 15 psig) = $100/trap (15-29 psig) = $100/trap (30-74 psig) = $150/trap (75-124 psig) = $200/trap (125-174 psig) = $300/trap (175-249 psig) = $400/trap ( 250 psig) = $500/trap Note: Any steam trap project with incentive over $10,000 requires pre-approval before a purchase order for the survey or steam trap repair/replacement is made.

Recover Heat from Boiler Stack

Heat Recovery From Natural Gas-fired Boiler Stack Natural Gas -fired Source: DOE Improving Steam Performance A Source Book for Industry, 2 nd Edition

Conventional Stack Gas Heat Recovery Recovered heat often used for feedwater heating or combustion air pre-heat Typical increase in boiler efficiency 2% to 3% Due to corrosion damage conventional recovery limited by o The minimum allowable flue gas temperature, typically 250 F and o The minimum allowable inlet feedwater temperature, typically 210 F

Stack Gas Condensing Heat Recovery (CHR) CHR units recover 10% to 15% of the 18% or so that is normally lost out the boiler stack on natural gas-fired boilers They do this because they cool the flue gas below the water dew point temperature of 135 F If flue gas temperature is 300 F or greater, then this boiler is a candidate for either conventional or CHR. Best candidates for CHR are those where the cold-side water is cold (50 to 60 F), where there is a continuous demand for heated water (20 gpm or more), where there is substantial boiler capacity (total all boilers on-line equals 350 boiler HP or more), and there are sufficient annual production hours (5000 hrs/yr or more) Typical deliverable water temperature up to about 140 F for direct contact units and about 185 F for indirect contact units

Recover Heat from Blowdown Water

Blowdown Heat Recovery Flash Steam Boiler Blowdown Flash Tank Deaerator Feedwater to boiler Blowdown enters the flash tank. Flash from the tank used in the deaerator. Blowdown water heats the make-up water. Heat Exchanger Drain Make-up Water

Insulate Pipes and Fittings

Insulate Steam Pipes and Fittings In most cases, all steam pipes should be insulated with simple returns of less than two years common Generally, condensate return pipes should be insulated if they are 2 or larger and the condensate temperature is 180ᵒ F or higher Removable insulation blankets can be used for valves and fittings Example: Insulation At a gas cost of $0.65 per therm, a 10-foot length of un-insulated 6-inch 125 psi (353 F) steam pipe will increase gas cost by about $1,000 per year compared to the same size insulted pipe. An un-insulated 6-inch steam valve will cost $300 per year. NAIMA s 3EPlus software is a great tool for insulation calculations. Get a free download at: (http://www.pipeinsulation.org/pages_v4/download.html)

Regular Boiler Tune-ups and O 2 Trim Control

Reduce Excess Air by Tuning Burner & by Adding O 2 Trim Tune Boilers - Good excess air percentages are 9.5% to 15% (2% to 3% O2) O 2 trim can be used to automatically adjust burner s fuel-to-air mixture for best O 2 over the boiler firing ranges and during seasonal ambient changes. When considering O 2 trim, the first step is to optimize the boiler with the normal tune-up, then utilize O 2 trim to gain additional efficiency (typically ranging from 1/2 to 2 percent boiler efficiency gain)

High Turndown Burner

High Turndown Burners A boiler that cycles on and off wastes energy due mainly to pre- and postfire purging losses. In addition, natural convection stack losses are greater for boilers that cycle. A rule-of-thumb: -2 cycles/hour = 2% of energy input is lost -5 cycles/hour = 8% of energy input is lost -10 cycles/hour = 30% of energy input is lost Basis: Equal time between on and off, purge time is 1 minute, stack temperature is 400 F and air flow through the boiler with fan off is 10% of the fan forced air flow. Example: With a 4:1 burner, cycling will occur below 25% firing rate. With a 10:1 burner, cycling will occur below 10% firing rate.

Custom Incentives

Custom Incentives For Natural Gas Projects that Are Not Covered Under a Standard Category, Custom Incentives Are Available: Notes: $0.90 Per Therm N.G. Saved Over One Year 1) Pre-approval is required before the Custom project is initiated 2) Payback must be between 1 and 10 years to qualify for the custom incentive

Steam Trap Basics Bill Lumsden Sr. Technical Support Engineer Leidos Engineering William.J.Lumsden@Leidos.com