Energy Efficient Steam Systemsacademic.udayton.edu/.../EnergyEfficientSteamSystems.docx · Web...

Energy-Efficient Steam Systems

IntroductionAir, water and steam are three media commonly used to distribute heat to process loads. However, steam has several advantages compared to hot air and hot water. These advantages include.

• the heat carrying capacity of steam is much greater than air or water• steam provides its own locomotive force.• steam provides heat at a constant temperature

To illustrate these advantages, consider the quantities of air, hot water and steam required to transfer 1,000,000 Btu/hr of heat to a process. If 100 psig steam were condensed in a heat exchanger, the mass flow rate of steam required to transfer 1,000,000Btu/hr of heat would be about:

Msteam = Q / hfg = 1,000,000 Btu/hr / 881 Btu/lb = 1,135 lb/hr

If the temperature of hot water dropped by 100 F as it passed through a heat exchanger, the mass flow rate of water to transfer the same amount of heat would be about nine times as much as steam:

Mwater = Q / (cp x dt) = 1,000,000 Btu/hr / (1 Btu/lb-F x 100 F) = 10,000 lb/hr

If the temperature of hot air dropped by 100 F as it passed through a heat exchanger, the mass flow rate of air to transfer the same amount of heat with the same temperature difference would be about 34 times as much as steam:

Mair = Q / (cp x dt) = 1,000,000 Btu/hr / (0.26 Btu/lb-F x 100 F) = 38,500 lb/hr

The higher flow rates required by water and air require pipes and ducts with larger diameters than steam pipes, which increases first cost and heat loss. In addition, air and water do not propel themselves. Thus, hot air and water distribution systems require

1

fans or pumps, whereas a steam distribution system does not require any additional propulsion for outgoing steam and a very small pumping system for returning the condensate to the boiler. Finally, because steam condenses at a constant temperature, 100-psig steam could heat a process stream to a maximum temperature of 338 F which is the temperature of the steam. On the other hand, the temperature of water and air decrease as heat is transferred; thus, if the heat in these examples was delivered by a cross-flow heat exchanger, the maximum temperature of the process stream would be 100 F less than the incoming temperature of the air or water. Because of these advantages, steam is the most widely used heat-carrying medium in the world.

Principles of Energy-Efficient Steam SystemsEnergy Balance ApproachThe figure below shows the primary energy flows into and out of a steam system.

Figure 1. Basic steam system.

Heat from combustion of fuel, Qfuel, is added to the boiler, which generates saturated steam at a discharge pressure P2. Some steam is discharged from boiler as blowdown to reduce the concentration of minerals in the steam. The boiler losses some heat, Qb, through the shell. The steam may pass through an adiabatic throttling valve to reduce the pressure of steam to P3. Some heat is lost heat from the steam pipes, Qsp. As the steam delivers heat to one or more processes, Qprocess, the steam vapor condenses. The steam trap discharges condensate, 5, into the condensate return line. Some steam may bypass the process and flow directly into the condensate return line, 4L, if steam trap(s) fail open. Some heat is lost from the condensate pipes, Qcp. The deareator tank pressure, Pda, is generally maintained slightly above ambient pressure. As the pressure of condensate is reduced to the pressure of the deareator tank, some condensate vaporizes and is lost as flash vapor, av. The remaining liquid condensate is mixed with makeup water, 0, in the deaerator tank. Some heat is lost from the deaerator tank, Qda. The pressure of the feed water from the dearator tank, al, is raised to the pressure

2

of the boiler by the feed water pump. The feedwater may be preheated by an economizer, which reclaims heat from exhaust gasses, before entering the boiler, 1e.

Thermodynamic state points of the steam in this system are shown below on a temperature versus entropy diagram. In the diagram below feed water pump work and heat loss from steam pipes, condensate pipes, deaerator tank and boiler are assumed to be negligible. The steam leaves the boiler as 200 psia saturated vapor at 2. The pressure is reduced to 100 psia at constant enthalpy at 3 by the throttling valve. Steam condenses at constant pressure in the process heat exchanger and leaves the steam trap at 5 as a saturated liquid. The dearator operates at 20 psia. The condensate at 7 losses pressure in a constant enthalpy process to become some combination, Xc, of liquid and vapor. Flash vapor leaves the deaerator tank at av. Makeup water enters the deaerator at 0, mixes with the liquid condensate and leaves leaves the deaerator tank at al. Pre-heated feed water leaves the economizer at 1e, before entering the boiler.

Thus, energy enters a steam system as:

Fuel and combustion air Makeup water Pump work

Energy leaves a steam system as:

3

Useful heat to the process Exhaust air Blowdown Condensate loss Flash vapor loss Heat loss from the boiler, steam pipes, condensate pipes and deaerator tank.

Fuel use is reduced by reducing these losses.

Opportunities for Improving the Energy-Efficiency of Steam SystemsThese principles can be organized using the inside-out approach, which sequentially reduces end-use energy, distribution energy, and primary conversion energy. Combining the energy balance and inside-out approach, common opportunities to improve the energy efficiency of steam systems are: End Use

Improve process control to reduce steam demand Insulate hot surfaces Insulate open tanks Employ counter-flow rinsing Employ counter-flow heat exchange

Distribution Insulate steam pipes and condensate pipes Throttle steam to minimum pressure required by each end-use to reduce flash

loss and conductive heat loss Fix steam traps Close condensate return to reduce flash loss

Conversion Turn off and valve off boiler(s) when not in use Insulate deaerator tank and boiler to reduce heat loss Reduce steam pressure to increase efficiency and reduce heat loss Improve water treatment to reduce scaling and improve efficiency Descale boiler to improve efficiency Reduce excess air across firing range Control combustion air based on oxygen in exhaust Operate multiple boilers at even firing rates Avoid on/off firing Eliminate stack effect loss by installing stack damper on atmospheric boilers Employ automatic blowdown control Reclaim blowdown flash Preheat feedwater with economizer

4

The remainder of the chapter describes steam system components, discusses fundamentals needed to quantify these savings opportunities, describes individual savings opportunities, and introduces an integrated steam system model to capture synergistic effects.

Steam System ComponentsBoilers and Steam GeneratorsSteam boilers are broadly classified as “fire-tube” or “water-tube” boilers. In fire tube boilers, the boiler shell contains the water/steam and hot combustion gasses pass through the tubes to heat the water/steam. In water-tube boilers, the water/steam passes through tubes and the hot combustion gasses pass through shell of the boiler.

Schematic of fire-tube boiler. Source : http://www.energysolutionscenter.org Cut-away view of water-tube boiler

Source : http://www.energysolutionscenter.org

Fire-tube boilers are popular for smaller applications requiring saturated steam at less than 150 psig because of their low first cost and durability. The large volume of water/steam serves as thermal mass which enhances steady operation. However, because the steam is generated on the shell side, the shell itself is a pressure vessel, making it difficult to generate steam at high pressures. In addition, the large surface area causes relatively large heat loss, which varies from about 0.5% of input energy at full-fire to a much higher fraction at low loads.

For high-pressure applications, it is easier to construct small diameter tubes to handle the high pressures of the steam than an entire boiler shell. In addition, the tubes can be configured to pass through high-temperature combustion gasses before exiting the boiler to create superheated steam. Thus, most high-pressure applications like power generation, which benefits from dry, high-temperature, super-heated steam at pressures up to 3,000 psig, use water-tube boilers.

5

http://www.energysolutionscenter.org/

http://www.energysolutionscenter.org/

Steam generators are like water-tube boilers, except that they are made from light- weight materials. In many jurisdictions, the lack of a dedicated pressure vessel enables steam generators to be used without a boiler operator. The light weight materials and absence of a large holding tank allow steam generators to come up to pressure quickly in a manner of minutes. This enables steam-generators to be turned on and off as needed, reducing standby losses. Installing the water-tubes in a counter-flow configuration to the path of the combustion gasses increases thermal efficiency.

Source: http://www.claytonindustries.com

Deaerator TanksMakeup water and condensate contain dissolved oxygen, carbon dioxide and ammonia. These dissolved gasses reduce the conductivity of the steam and hence its ability to transfer heat. More importantly, oxygen is highly corrosive and leads to pitting and possible system failure. Economizers are particularly susceptible to oxygen pitting. For these reasons, oxygen is typically removed from steam systems by a deaerator.

A deaerator works by spraying makeup water into a steam environment and heating the makeup water to within about 5 F of saturation temperature. At this temperature, the solubility of oxygen is low and the makeup water contains very little oxygen. Oxygen and flash vapor are vented to atmosphere. To function effectively, the pressure of the dearator can only be a few psi above ambient pressure, or else the oxygen will be forced back into the water.

Thottling ValvesBoilers are generally designed to operate at a specific pressure. For safety reasons, boilers should never be operated above the rated pressure. If the pressure of steam needed at the application is less than the rated pressure of the boiler, the boiler can be operated at less than the design pressure or the boiler can be operated at the design pressure and the pressure of steam reduced through a valve located between the boiler

6

and the application. Operating at a lower pressure will slightly increase the efficiency of the boiler because of the decreased steam temperature and subsequent boiler skin losses. However, it may also cause problems such as raising the level of water in the boiler and reduced boiler heating capacity. A primary advantage for operating the boiler at the design pressure and then reducing the pressure through a valve is that the steam exiting the valve will be slightly super heated, resulting in less water in the steam lines and heat exchangers. Because of this, some consultants recommend that steam boilers be operated at their design pressure, even if the steam is to be used at lower pressures in the plant.

Steam Piping SystemsSteam is generally distributed to the plant through one or more large steam “mains” which connect to smaller branch pipes. Condensate is produced and carried along with the steam as steam condenses on the inside surface of the pipes. Excess condensate can block steam flow and cause serious pipe erosion. Thus, “drip stations” need to be installed at all low points and ends of all “mains” at intervals of about 100 feet along the main. A drip station consists of a vertical section of pipe at least 18 inches long installed on the underside of the main and connected to a steam trap. Strainers should also be installed along the piping system to filter out scale and solid contaminants.

The velocity of steam out of the boiler is determined by the outlet nozzle. It is common practice to design piping systems for space-heating applications for a velocity of about 6,000 ft/min and piping systems for process-heating applications for a velocity of about 10,000 ft/min. Lower velocities reduce pressure loss, pipe erosion, water hammer and noise as well as providing more efficient condensate drainage.

As steam condenses on a cold surface a thin film of condensate is produced and any air entrained with the steam is released. Air in a steam system steam causes two major problems. First, even a thin layer of air on a heat transfer surface, dramatically reduces the heat transfer across the surface (See figure below). For example a layer of air 0.04 inches thick adds the same thermal resistance as a layer of water 1 inch thick or a layer of iron 4.3 feet thick. Second, when air is absorbed into condensate carbolic acid is produced. This acid can attack piping and heat exchange surfaces. To reduce air in the piping system, thermostatic air vents should be installed at high points, the end of steam mains and on all heat exchange equipment.

Steam TrapsAs steam delivers heat through a heat exchanger, the steam vapor condenses to a liquid. Steam traps are located downstream of heat exchangers and discharge the condensate into the condensate return line while preventing steam vapor from passing through.

The four most common types of steam traps are:

7

• Inverted bucket.• Float + thermostatic• Thermostatic• Thermodynamic

Source: Grainger Catalog, 2000-2001

Condensate Return TanksCondensate return tanks collect condensate discharged from steam traps. In open condensate return systems, the condensate return tank is open to the environment and condensate is pumped back to the boiler. The enthalpy of condensate at atmospheric pressure is substantially less than the enthalpy of condensate at the operating pressure of a steam system. Thus, the energy released as the pressure of condensate falls to atmospheric pressure, vaporizes some of the condensate into “flash” steam, which is discharged to atmosphere.

In closed condensate return systems, steam pressure forces the condensate all the way back to the deaerator tank. Thus, in closed systems, flash steam is created as the pressure of condensate falls to the pressure of the deaerator tank, and is discharged to atmosphere from the deaerator rather than the condensate return tank.

Steam MeteringSteam metering is expensive, but gives valuable information for managing a steam system. Most steam meters work by measuring the pressure difference across a pressure reduction valve and comparing the output to calibrated values. High-quality steam metering devices for a 4-inch steam pipe cost about $4,000.

Insulate Pipes, Tanks and Hot SurfacesUninsulated steam pipes, condensate return pipes, condensate return tanks, deaerator tanks and process tanks lose heat to the surrounding by convection and radiation. Insulating these surfaces reduces steam use and the risk of burns.

8

Well-insulated steam pipes Uninsulated condensate return tank

The method that follows describes how to calculate energy savings from insulating hot surfaces, while explicitly taking radiation and the dependence of the convection coefficient on surface temperature into account. The required input variables for this procedure are easily measured in the field.

To calculate heat loss savings, the heat loss from both the uninsulated surface and from the surface with the proposed insulation must be calculated. The savings from adding insulation are difference between the uninsulated and insulated heat loss.

Qsavings = Quninsulated – Qinsulated (1)

Hot surfaces lose heat to the surroundings via convection and radiation. The equation for heat loss, Q, to the surroundings at Ta, from a hot surface at Ts, with area A is:

Quninsulated = h A (Ts – Ta) + A (Ts4 – Ta

4) (2)

where h is the convection coefficient, is the Stefan-Boltzman constant (0.1714 · 10-8 Btu/ft2-hr-R4, or 5.67 · 10-8 W/m2-K4), is the emissivity of the surface. Very shiny surfaces have emissivities of about 0.1; dark or rough surfaces have emissivities of about 0.9. The flow of air over warm surfaces is due to the buoyancy of warm air next to the surface compared to the cooler surrounding air. For surfaces a few degrees warmer than the surrounding air, the natural convection coefficient is about 1.5 Btu/ft2-hr-F (8.5 W/m2-K) (Mitchell, 1983 [9]). However, as the surface temperature increases, so does the buoyancy effect and convection coefficient. To account for this effect, the value of the convection coefficient can be approximated as a function of the orientation and vertical dimension of the surface, and the temperature difference between the surface and the surrounding air (ASHRAE Fundamentals, 1989). The appropriate relation depends on whether the air flow is laminar or turbulent. Dimensional approximations for determining whether the flow is laminar or turbulent are shown in Equation 3. In

9

these relations, L is the characteristic length (ft) in the vertical direction and T is temperature difference between the surface, Ts, and the surrounding air, Ta (F).

Laminar: L3 T < 63 Turbulent: L3 T > 63 (3)

After the nature of the flow is determined, the convection coefficient can be estimated using the relations in Equation 4 (ASHRAE Fundamentals, 1989 [1]). In these relations, L is the length (ft) in the vertical direction, D is the diameter (ft), B is tilt angle of the surface from horizontal, and h is convection coefficient (Btu/hr-ft2-F). For use with SI units, the proper conversion would need to be made (1 Btu/hr-ft2-F = 5.678 W/m2-K).

Horizontal Surfaces Losing Heat Upwards:hlam = 0.27 (T / L) 0.25; htur = 0.22 (T) 0.33

Tilted/Vertical Surfaces:hlam =0.29 [T (sin B) / L] 0.25; htur =0.19 [T (sin B)]0.33 Horizontal Pipes and Cylinders:hlam = 0.27 (T / D) 0.25; htur = 0.18 (T) 0.33 (4)

Using these relations, Equation 2 can be solved for Quninsulated to calculate the current heat loss. Similarly, heat loss from the insulated surface can be calculated from:

Qinsulated = h ATi – Ta A (Ti4 – Ta

4) (5)

where Ti is the temperature of the outside surface of the insulation. Unfortunately, in Equation 5, the values of Ti and h are not known. To determine Ti and h, the first step is to determine the thermal resistance of the current wall, Rc, based on the temperature of the fluid inside the heating system, Tf, and the current surface temperature Ts. Thermal resistance of the current wall includes both the conduction thermal resistance through the wall and the convection thermal resistance at the wall’s inner surface.

Quninsulated = A (Tf – Ts) / Rc (6)

Next, an equation can be written from a steady-state energy balance on the surface of the insulation:

Qcnd,in – Qcnv,out – Qrad,out = A (Tf – Ti) / (Rc +Ri) - h A (Ti – Ta) - A (Ti4 – Ta

4) = 0 (7)

where Ri is the thermal resistance of the insulation. The relations for convection coefficient as a function of the temperature difference between Ti and Ta form a second equation. Thus, this system has two equations (Equation 4 and Equation 7) and two unknowns and can be solved to determine Ti and h.

10

An easy way to solve this system of equations is to guess a value for Ti, calculate the convection coefficient h using Equation 4, then substitute Ti and h into Equation 7. The left side of Equation 7 will evaluate to 0 when Ti is correct. Hence, the system of equations can be solved by repeating this process with guesses for Ti until Equation 7 converges to close to 0. The final values of Ti and h can then be substituted into Equation 5 to find Qinsulated. The heat loss savings, Qsav is the difference between Quninsulated and Qinsulated.

Example

The surface temperature of 100 ft of 0.5 ft diameter un-insulated pipe carrying condensate at 200 F is 180 F. The pipe is located in a room with air and surroundings at 70 F. The surface emissivity of the pipe is 0.70. Calculate convection, radiation and total heat loss from the pipe (Btu/hr). The pipe is insulated with 2 inches on insulation with thermal resistance R = 2 hr-ft2-F/Btu per inch. The surface emissivity of the insulation is 0.70. Calculate convection, radiation and total heat loss from the insulated pipe (Btu/hr). Calculate the heat loss and fuel savings from insulating the pipe (Btu/hr) if the efficiency of the steam system is 70%.

Input data are:

INPUTSTa (F) = ambient temperature 70Tf (F) = temperature fluid inside pipe 200Tp (F) = temperature pipe 180Dp(ft) = diameter pipe 0.5Lp (ft) = length pipe 100ep = emissivity pipe 0.7Proposed InsulationRi,per inch (hr-ft2-F/Btu-in) = R value insulation per inch 2dxi (in) = thickness insulation 2ei = emissivity insulation 0.7Eff,sys 0.7

CONSTANTSsigma (Btu/hr-ft2-R4) 1.714E-09

Calculations of current heat loss and thermal resistance of the pipe, Rp, are:

11

Ap (ft2) = pi Dp Lp 157 L^3 dT = Dp^3 (Tp-Ta) 14 hp (Btu/hr-ft2-F) = .27 ((Tp-Ta)/Dp) .̂25 or .18 ((Tp-Ta))^.33 1.040Qcnv,p (Btu/hr) = hp Ap (Tp - Ta) 17,967 Qrad,p (Btu/hr) = ep sigma Ap ((Tp+460)^4 - (Ta+460)^4) 16,748 Qtot,p (Btu/hr) = Qcnv,p + Qrad,p 34,716 Qtot,p (kW) = Qtot,p / 3,412 10.81 Rp (hr-ft2-F/Btu-in) = Ap (Tf-Tp)/Qtot,p 0.09

Note that radiation loss is approximately equal to convection heat loss; thus, neglecting radiation loss significantly underestimates total heat loss.

To calculate the heat loss with insulation, an iterative method is used in which the surface temperature of the insulation, Ti, is guessed until the energy balance Equation 7 is satisfied. Equation 7 is satisfied when:

EB(Ti) = A (Tf – Ti) / (Rp +Ri) - h A (Ti – Ta) - A (Ti4 – Ta

4) = 0

In the calculations below, Ti = 89.9 F gave EB(Ti) = 0.22, which is close to zero. After Ti is known, the heat loss can be calculated as:

GUESS Ti to make EB(Ti) = 0 89.9

EB(Ti) = Qcnd,i - Qcnv,i - Qrad,i should be zero 0.22 Ai (ft2) = pi (Dp + 2 Ti) Lp 262 Ri (hr-ft2-F/Btu) = Ri,per inch x dxi 4L^3 dT = (Dp+ 2Ti)^3 (Ti-Ta) 12 hi (Btu/hr-ft2-F) = .27 ((Ti-Ta)/Dp) .̂25 or .18 ((Ti-Ta))^.33 0.597Qcnd,i (Btu/hr) = Ai (Tf-Ti)/(Rp+Ri) 7,047 Qcnv,i (Btu/hr) = hi Ai (Ti - Ta) 3,110 Qrad,i (Btu/hr) = ei sigma Ai ((Ti+460)^4 - (Ta+460)^4) 3,937 Qtot,i (Btu/hr) = Qcnv,i + Qrad,i 7,047 Qtot,i (kW) = Qtot,i / 3,412 2.19

Thus, the heat loss, Qsav, and fuel, Qf,sav, savings from adding insulation would be:

Qsav (Btu/hr) = Qtot,p - Qtot,i 27,669 Qsav (kW) = Qsav /3,412 8.11 Qf,sav (Btu/hr) = Qsav / Eff,sys 39,527 Qf,sav (kW) = Qf,sav /3,412 11.58

12

The same method can be used to calculate heat loss, and the savings from insulating, walls of steam-heated tanks. The only modifications required are when calculating the convection coefficient. When determining whether the flow of air is laminar of turbulent, the effective length is the wall height instead of pipe diameter, and the relation for convection coefficient is for vertical surfaces instead of pipes.

Example

The surface temperature of a steam-heated, un-insulated rectangular tank with four walls with height 4 ft and length 8 ft is 160 F. The temperature of fluid in the tank is 180 F, and the temperature of the air and surroundings is 70 F. The surface emissivity of the tank is 0.70. Calculate convection, radiation and total heat loss from the tank walls (Btu/hr). The tank walls are insulated with 1 inch on insulation with thermal resistance R = 2 hr-ft2-F/Btu per inch. The surface emissivity of the insulation is 0.70. Calculate convection, radiation and total heat loss from the insulated tank walls (Btu/hr). Calculate the heat loss and fuel savings from insulating the tank walls (Btu/hr) if the efficiency of the steam system is 75%.

Input data are:

INPUTSTa (F) = ambient temperature 70Tf (F) = temperature fluid inside wall 180Tw (F) = temperature wall 160Htw(ft) = height wall 4Lw (ft) = length wall 32ew = emissivity wall 0.70Proposed InsulationRi,per inch (hr-ft2-F/Btu-in) = R value insulation per inch 2dxi (in) = thickness insulation 1ei = emissivity insulation 0.70Eff,sys 0.75

CONSTANTSsigma (Btu/hr-ft2-R4) 1.714E-09

Calculations of current heat loss and thermal resistance of the wall, Rw, are:

13

CurrentAw (ft2) = Htw Lw 128 L^3 dT = Htw^3 (Tw-Ta) 5,760 hw (Btu/hr-ft2-F) = .29((Tw-Ta)/Htw)^.25 or .19 ((Tw-Ta))^.33 0.839Qcnv,w (Btu/hr) = hw Aw (Tw - Ta) 9,663 Qrad,w (Btu/hr) = ep sigma Aw ((Tw+460)^4 - (Ta+460)^4) 10,575 Qtot,w (Btu/hr) = Qcnv,w + Qrad,w 20,238 Qtot,w (kW) = Qtot,w / 3,412 5.93 Rw (hr-ft2-F/Btu-in) = Aw (Tf-Tw)/Qtot,w 0.13

Note that radiation loss is less than convection heat loss at these relatively low temperature differences between the surface and air. To calculate the heat loss with insulation, an iterative method is used in which the surface temperature of the insulation, Ti, is guessed until the energy balance Equation 7 is satisfied. Equation 7 is satisfied when:

EB(Ti) = A (Tf – Ti) / (Rw +Ri) - h A (Ti – Ta) - A (Ti4 – Ta

4) = 0

In the calculations below, Ti = 99.3 F gives EB(Ti) = 3.53, which is close to zero. After Ti is known, the heat loss can be calculated as:

GUESS Ti to make EB(Ti) = 0 98.45

EB(Ti) = Qcnd,i - Qcnv,i - Qrad,i should be zero 0.96 Aw (ft2) = Htw Lw 128 Ri (hr-ft2-F/Btu) = Ri,per inch x dxi 2L^3 dT = (Htw)^3 (Ti-Ta) 2,058 hi (Btu/hr-ft2-F) = .29((Ti-Ta)/Htw)^.25 or .19 ((Ti-Ta)) .̂33 0.574Qcnd,i (Btu/hr) = Ai (Tf-Ti)/(Rw+Ri) 4,909 Qcnv,i (Btu/hr) = hi Ai (Ti - Ta) 2,089 Qrad,i (Btu/hr) = ei sigma Ai ((Ti+460)^4 - (Ta+460)^4) 2,819 Qtot,i (Btu/hr) = Qcnv,i + Qrad,i 4,908 Qtot,i (kW) = Qtot,i / 3,412 1.53

Thus, the heat loss, Qsav, and fuel savings, Qf,sav, from adding insulation would be:

Qsav (Btu/hr) = Qtot,p - Qtot,i 15,330 Qsav (kW) = Qsav /3,412 4.49 Qf,sav (Btu/hr) = Qsav / Eff,sys 20,440 Qf,sav (kW) = Qf,sav /3,412 5.99

14

Cover Open TanksIn open tanks, the total heat loss is the sum of heat loss through convection, radiation and evaporation. These losses can be significantly reduced by adding a cover or floats to the tank.

Convection, radiation and evaporation heat loss is reduced by covering open tanks.

Fix Steam TrapsSteam traps are automatic valves that discharge condensate from a steam line without discharging steam. If the trap fails open, steam escapes into the condensate return pipe without being utilized in the process. If it fails closed, condensate fills the heat exchanger and chokes-off heat to process. Fixing failed steam traps is usually highly cost-effective.

Steam traps are designed to operate about 10 years, but can fail sooner due to contamination, improper application, and other reasons. Steam traps can fail “open” or “closed”. If a steam trap fails “open”, it allows steam to pass through the trap; hence the energy value of the steam is completely wasted. If a trap fails “closed”, condensate will back up into the piping (which reduces steam flow, inhibits valve function and causes pipe erosion) and/or flood the heat exchanger (which reduces or eliminates effective heat transfer). Because of these problems, it is recommended that all traps be tested at least once per year. The most common methods of identifying failed-open steam traps are:

Ultrasonic sensor Temperature sensor Excess flash

Ultrasonic Sensor: Ultrasonic sensors amplify high frequency noise from steam and condensate flow into the audible spectrum. Thus, an analyst can determine whether steam and condensate is being discharged through the trap by listening to the condensate side of a steam trap. If the discharge is continuous, it could indicate that

15

the trap has failed open. If no discharge can be sensed, it may indicate that the trap has failed closed.

Properly functioning inverted bucket, IB, and thermodynamic, TD, traps discharge condensate intermittently. Thus, a continuous discharge indicates that these types of traps have failed open. Properly functioning float and thermostatic, FT, and thermostatic, TS, traps discharge condensate continuously. Thus, the failure of these types of traps cannot be diagnosed by listening for continuous discharge. The four types of steam traps can be identified by their distinctive shapes and nameplates.

Temperature Sensor: Infrared temperature sensors can detect the temperature on the steam and condensate sides of steam traps. Properly functioning traps are generally warm on both sides, but hotter on the steam side than the condensate side. A trap that is equally hot on both sides may have failed open. A trap which is cold on both sides may have failed closed and be flooded with water.

Flash: The enthalpy of condensate at atmospheric pressure is substantially less than the enthalpy of condensate at the operating pressure of a steam system. Thus, the energy released as the pressure of condensate falls to atmospheric pressure, vaporizes some of the condensate into “flash” steam. The quantity of condensate “flashed” to vapor dramatically increases when live steam enters the condensate return system. Thus, increased flash from the condensate return or deaerator tank is an indicator of failed-open steam traps.

Estimating Savings from Repairing Steam TrapsThe rate of steam loss through a leaking trap depends on the size of the condensate orifice in the trap. Orifice size is a function of the size of the trap and the differential pressure between the steam and condenstate lines that the trap was designed for. Orifice sizes for Sprirax Sarco float+thermostatic and inverted-bucket traps are listed below. Orifice sizes for thermostatic and thermodynamic traps are generally not specified; however the effective orifice size is similar to the orifice size for inverted bucket and float+thermostatic traps.

Cast Iron Float and Thermostatic Steam Traps (FT, FTI and FTB) Spriax Sarco Product Manual, 2001, pg 386 dP(psi) / NPT (in) .5, .75, 1 1.25 1.5 2

15 0.2180 0.3120 0.5000 0.6250 30 0.2180 0.2280 0.3900 0.5000 75 0.1660 0.3120 0.3120 0.4210 125 0.1250 0.2460 0.2460 0.3220 200 0.1000

16

Cast Iron Inverted Bucket Steam Traps (B Series) Spriax Sarco Product Manual, 2001, pg 436 dP(psi) / NPT (in) .5, .75 0.75 1 1.25 2

15 0.2500 0.3750 0.5000 0.6250 1.0625 30 0.1875 0.3125 0.3750 0.5000 0.7500 75 0.1563 0.2500 0.2813 0.3750 0.5625

125 0.1250 0.2031 0.2500 0.3438 0.5000 180 0.0938 0.1563 0.2188 0.2813 0.4375 250 0.0700 0.1406 0.1875 0.2500 0.3750

The rate of steam loss through an orifice is given by:

Steam flow (lb/hr) = 24.24 lb/(hr-psia-in2) x P psia x [D inch]2 x C

where P is the pressure of the steam, D is the diameter of the orifice and C is the fraction of the orifice that is open (Design of Fluid Systems: Hook-ups, Spirax-Sarco, 2000, pg. 57).

In many cases, leaking steam traps are identified using an ultrasonic sensor and/or by measuring temperatures on both sides of the trap. Large leaks typically make more noise and create higher downstream temperatures than small leaks. Thus, experienced personnel often estimate the fraction of the orifice that is open using these indicators.

Example

Calculate savings from replacing a failed 0.5-inch inverted bucket trap rated at 180 psi if actual steam pressure is 120 psig. The orifice is estimated to be 50% open. The steam system operates 6,000 hours per year and the cost of fuel is $10 /mmBtu. 100% of the condensate is returned at 200 F. The overall efficiency of the boiler is 80%. From the table above, the orifice size for this trap is 1/32-inch. Assuming that the orifice is 50% open, the steam loss through the leaking trap is about:

24.24 lb/(hr-psia-in2) x )120 + 14.7) psia x [0.0938 inch]2 x 50% = 14.36 lb/hr

The latent heat of steam at 120 psig is about 872 Btu/lb and the saturation temperature is about 350 F. The natural gas savings from fixing the steam trap would be about:

14.36 lb/hr [872 Btu/lb + 1 Btu/lb-F (350 – 200) F] 6,000 hr/yr / 80% = 110 mmBtu/yr110 mmBtu/yr x $10 / mmBtu = $1,100 /yr

An inverted-bucket steam trap for ½-inch pipe connections with a maximum operating pressure of 125 psig costs about $100. If the labor cost of installing a new trap is $50, the simple payback would be about:

17

SP = $150 / $1,100 /yr x 12 months/yr = 1.63 months

Reduce Steam Pressure Generating steam at unnecessarily high pressures decreases boiler efficiency, increases heat loss from steam pipes and increases flash loss. Reducing boiler pressure to match the highest required process temperature decreases these losses. Moreover, reducing steam pressure to match the local required process temperature reduces flash loss. Thus, always produce and supply steam at the minimum pressure required to meet the process temperature requirement.

Install Automatic Blowdown ControlsBlowdown is the practice of expelling steam to reduce contaminant build ups. Blow down can occur from the surface and/or bottom of the boiler. Typical blowdown rates range from 4% to 8% of boiler feed-water. Blowdown may be manual or automatic. Manual blowdown relies on intuition or periodic testing to determine when the concentration of contaminants is high enough to warrant blowdown. Manual blowdown virtually always results in either excess blowdown that wastes energy or insufficient blow down that creates excess scale on heat transfer surfaces and reduces boiler efficiency. Automatic blowdown controls monitor the conductivity of the water in the boiler and open the blowdown valve as needed to maintain the conductivity within a specified range. Optimizing the quantity of blow down using automatic controls reduces energy, water and water treatment costs.

Combustion EfficiencyBoilers typically employ combustion to covert fuel energy into high temperature thermal energy. This section describes natural gas combustion and how to calculate combustion air flow, combustion temperature and the efficiency of the process. These results are used extensively throughout this chapter.

The minimum amount of air required for complete combustion is called the “stoichiometric” air. Air consists of about 1 mole of oxygen to 3.76 moles of nitrogen. Assuming that natural gas is made up of 100% methane, the equation for the stoichiometric combustion of natural gas with air is:

CH4 + 2 (O2 + 3.76 N2) CO2 + 2 H2O +7.52 N2 (17)

The ratio of the mass of air required to completely combust a given mass of fuel is called the stoichiometric air to fuel ratio, AFs. AFs can be calculated using the molecular

18

masses of the air and fuel at stoichiometric conditions. For combustion of natural gas in air, AFs is about:

AFs = Mair,s / Mng,s = 2[ (2 x 16) + (3.76 x 2 x 14)] / [12 + (4 x 1)] = 17.2

The quantity of air supplied in excess of stoichiometric air is called excess combustion air, ECA. Excess combustion air can be written in terms of the stoichiometric air to fuel ratio, AFs, the combustion air mass flow rate, mca, and natural gas mass flow rate, mng.

ECA = [(mca / mng) / AFs] – 1 (18)

Large quantities of excess air dilute combustion gasses and lower the temperature of the gasses, which results in decreased efficiency. The energy input, Q in, to a combustion chamber is the product of the natural gas mass flow rate, mng, and the higher heating value of natural gas, HHV, which is about 23,900 Btu/lbm.

Qin = mng HHV (19)

The mass flow rate of the combustion gasses, mg, is the sum of the natural gas mass flow rate, mng, and combustion air mass flow rate, mca.

mg = mng + mca (20)

The temperature of combustion, Tc, can be calculated from an energy balance on the combustion chamber, where the chemical energy released during combustion is converted into sensible energy gain of the gasses. The energy balance reduces to the terms of inlet combustion air temperature, Tca, lower heating value of natural gas (21,500 Btu/lbm), excess combustion air, ECA, stoichiometric air fuel ratio, AFs, and specific heat of combustion gasses, Cpg (0.30 Btu/lbm-F). Combustion temperature, Tc, is calculated in terms of these easily measured values as:

Tc = Tca + LHV / [{1 + (1 + ECA) AFs} Cpg] (21)

The combustion efficiency, is the ratio of energy delivered to the system to the total fuel energy supplied. The energy delivered to the system is the energy loss of combustion gasses. The energy loss of the combustion gasses can be expressed as the product of the mass flow rate, specific heat and temperature drop of the gasses. The total energy fuel energy supplied is the higher heating value of the fuel. Using this approach, the combustion efficiency, is:

= [{1 + (1 + ECA) AFs} Cpg (Tc – Tex)] / HHV (22)

19

The dew-point temperature of products of combustion is about 140 F. If the products of combustion leave the process at temperature of less than the dew-point temperature the water vapor will condense to a liquid and release energy. To include this effect, the efficiency equation can be written:

If Tex > 140 F then hfg = 0 Else hfg = HHV – LHV = [{1 + (1 + ECA) AFs} Cpg (Tc – Tex) + hfg] / HHV (22b)

The three required input values for computing combustion efficiency, entering combustion air temperature, Tca, exhaust gas temperature, Tex, and excess combustion air, ECA, can be measured using a combustion analyzer. The quantity of excess air in the combustion gasses is sometimes expressed as fraction oxygen. For methane (natural gas) the conversion between fraction oxygen, FO2, and excess combustion air, ECA, are:

FO2 = 2 ECA / (10.52 + 9.52 ECA) ECA = 10.52 FO2 / (2 – 9.52 FO2) (23)

Example

A boiler consumes 100,000 Btu/hr of natural gas. An analysis of the exhaust gasses finds that the fraction of excess air is 30% and the temperature of the exhaust gasses is 500 F. Calculate combustion air flow (lb/hr), exhaust gas flow (lb/hr), combustion temperature (F) and the combustion efficiency.

COMBUSTION AIR FLOW

INPUTSQfuel (Btu/hr) 100,000 EA = excess air (0=stoch, 0.1 = optimum) 0.30

CONSTANTS (FOR NATURAL GAS)HHVng (Btu/lb) 23,900AFs = air/fuel mass ratio at stochiometric conditions 17.2p (at 68 F) (lb/ft3) 0.075

CALCULATIONSMng = Qfuel / HHVng (lb/hr) 4 Ma = Mng AFs (1+EA) (lb/hr) 94 Mtot (lb/hr) 98 Vtot = Mtot / p (scfm) 21

Thus, mass flow rate of combustion air is 94 lb/hr and the mass flow rate of the combustion gasses is 98 lb/hr.

20

COMBUSTION EFFICIENCY

INPUTSEA = excess air (0=stoch, 0.1 = optimum) 0.30Tca = temperature combustion air before burner (F) 70Tex = temperature exhaust gasses (F) 500

CONSTANTS (FOR NATURAL GAS)LHV = lower heating value (Btu/lb) 21,500HHV = higher heating value (Btu/lb) 23,900cpp = specific heat of products of exhaust (Btu/lb-F) 0.300Tdpp = dew point temp of H20 in exhaust (F) 140AFs = air/fuel mass ratio at stochiometric conditions 17.20

CALCULATIONSTc = temp combustion (F) = Tca+LHV/[(1+(1+EA)(Afs))cpp] 3,138hfg = water vapor latent energy (Btu/lb) = (if Tex<140 then hfg=HHV-LHV else hfg = 0) 0Effi ciency = {hfg+[1 + (1+EA)(AFs)]*cpp*(Tc-Tex)}/HHV 77.3%

Thus, the combustion efficiency is 77.3%.

The relationship between excess air, exhaust temperature and combustion efficiency using this method is shown in the graph below. Efficiency decreases with increasing excess air and increasing exhaust air temperature.

70%

75%

80%

85%

90%

0 10 20 30 40 50 60 70 80 90 100

Excess Air (%)

Effic

ienc

y Ts=300F

Ts=400FTs=500F

21

Reduce Excess Air by Adjusting Combustion Air Linkages Most boilers use linkages that connect natural gas supply valves with combustion air inlet dampers. As the natural gas valve closes, the mechanical linkages close dampers on the combustion air supply to attempt to maintain a constant air/fuel ratio. If the exhaust gasses contain too much excess air, the linkages can be adjusted to tune the air/fuel ratio so that the exhaust gasses contain about 10% excess air.

Mechanical linkages vary the position of the inlet air damper with natural gas supply.

Example

A boiler burns 2,000 mmBtu of natural gas per year at a cost of $10 /mmBtu. The average temperature of the incoming combustion air is 70 F and the average temperature of the exhaust is 450 F. The fraction excess air in the exhaust is measured to be 50%, but is reduced to 10% by adjusting the inlet combustion air dampers. Calculate a) the projected annual cost savings ($/yr) and b) the projected savings as a percent of current annual natural gas use.

The initial efficiency is:

22




The new efficiency is:




The heating energy delivered to the process, Qp, is the product of the initial fuel use, Qf1, and the initial efficiency, Eff1.

Qp = Qf,1 x Eff1

The heating energy delivered to the process remains constant. The new fuel use, Qf2, with the higher efficiency, Eff2, is:

23

Qf,2 = Qp / Eff2

The fuel use savings, Qf,sav, would be:

Qf,sav = Qf,1 - Qf,2

INPUTQf,1 (mmBtu/yr) 2,000Cf ($/mmBtu) 10

CALCULATIONSQp (mmBtu/yr) = Qf,1 Eff1 1,543Qf,2 (mmBtu/yr) = Qp / Eff2 1,918Qf,sav (mmBtu/yr) = Qf1-Qf2 82Cf,sav ($/yr) = Qf,sav Cf 816FracRed = Qf,sav / Qf1 0.041

Unfortunately, the linkages between the fuel valve and combustion air dampers seldom function perfectly. Thus, the air/fuel ratio is seldom held constant over the firing range. For example, the figure below shows that excess air varies from 120% at low fire to 38% at mid file to 42% at high fire. This indicates that the linkages were incapable of sufficiently reducing combustion air to match fuel supply at low fire. The high level of excess air at low fire causes the efficiency of the boiler to drop, even though the lower exhaust temperature should drive the efficiency higher. In cases like this, it is often very difficult to adjust the linkages so that excess air is constant at 10% at all firing levels. However, it is usually possible to adjust the linkages so that the minimum level of excess air is about 10%, and the excess air at other firing rates drops by about the same percentage.

24

Example

A boiler operates 4,000 hours per year at low fire, 2,000 hours per year at mid fire, and 2,000 hours per year at high fire with excess air and exhaust temperatures shown in the figure above. Boiler fuel consumption is 4 mmBtu/hr at low fire, 12 mmBtu/hr at mid fire, and 20 mmBtu/hr at high fire. Ambient temperature is 70 F. Calculate annual fuel energy savings (mmBtu/year) from adjusting the linkages so the minimum excess air is 10%, and the excess air at other firing rates is decreased by the same percentage.

The initial combustion efficiencies, Eff1, are:

25

INPUTS Low Fire Mid Fire High FireEA = excess air (0=stoch, 0.1 = optimum) 1.12 0.38 0.42Tca = temperature combustion air before burner (F) 70 70 70Tex = temperature exhaust gasses (F) 240 300 310

CONSTANTS (FOR NATURAL GAS)LHV = lower heating value (Btu/lb) 21,500 21,500 21,500HHV = higher heating value (Btu/lb) 23,900 23,900 23,900cpp = specific heat of products of exhaust (Btu/lb-F) 0.300 0.300 0.300Tdpp = dew point temp of H20 in exhaust (F) 140 140 140AFs = air/fuel mass ratio at stochiometric conditions 17.20 17.20 17.20

CALCULATIONSTc = temp combustion (F) = Tca+LHV/[(1+(1+EA)(Afs))cpp] 1,983 2,967 2,889hfg = water vapor latent energy (Btu/lb) = (if Tex<140 then hfg=HHV-LHV else hfg = 0) 0 0 0Effi ciency = {hfg+[1 + (1+EA)(AFs)]*cpp*(Tc-Tex)}/HHV 82.0% 82.8% 82.3%

Minimum excess air is 38% at mid-fire. If the linkages were adjusted so the excess air was 10% at mid-fire, the reduction in excess air would be 28%. If the excess air at all firing rates was reduced by 28%, the new levels of excess air would be 84% at low fire, 10% at mid fire and 14% at high fire. The new combustion efficiencies, Eff2, at these firing rates and temperatures would be:





Qp = Qf,1 x Eff1

The heating energy delivered to the process remains constant after the efficiency is improved. The new fuel use, Qf2, with the higher efficiency, Eff2, is:

Qf,2 = Qp / Eff2

26



INPUT Low Fire Mid Fire High FireQf,1 (mmBtu/hr) 4 12 20HPY (hr/yr) 4000 2000 2000

CALCULATIONS Low Fire Mid Fire High Fire TotalQf,1 (mmBtu/yr) = Qf,1 HPY 16,000 24,000 40,000 80,000Qp (mmBtu/yr) = Qf,1 Eff1 13,114 19,876 32,920 65,910Qf,2 (mmBtu/yr) = Qp / Eff2 15,802 23,604 39,307 78,713Qf,sav (mmBtu/yr) = Qf1-Qf2 198 396 693 1,287FracRed = Qf,sav / Qf1 0.012 0.017 0.017 0.016

Thus, in this example, simply reducing excess air by adjusting the linkages reduced fuel use by 1.6%.

Install O2 Trim ControlsMost boilers use linkages that connect natural gas supply valves with combustion air inlet dampers. Unfortunately, the linkages do not function perfectly, and the air/fuel ratio is seldom held constant over the firing range. O2 trim combustion controls measure the oxygen in the exhaust gasses to regulate combustion intake air to maintain about 10% excess air across the entire firing range. O2 trim controls cost about $30,000 and require periodic calibration which costs about $2,000 per year. Thus, O2 trim combustion controls are most cost-effective for boilers that operate all year long.

O2 trim system to continually vary combustion air

27

Example

A boiler operates 4,000 hours per year at low fire, 2,000 hours per year at mid fire, and 2,000 hours per year at high fire with excess air and exhaust temperatures shown in the figure below. Boiler fuel consumption is 4 mmBtu/hr at low fire, 12 mmBtu/hr at mid fire, and 20 mmBtu/hr at high fire. Ambient temperature is 70 F. Calculate annual fuel energy savings (mmBtu/year) from installing an O2 trim system so the minimum excess air is 10% across the firing range.

The initial combustion efficiencies, Eff1, are:

28




The new combustion efficiencies, Eff2, if the excess air was held to 10% across the firing range would be:





Qp = Qf,1 x Eff1

The heating energy delivered to the process remains constant after the efficiency is improved. The new fuel use, Qf2, with the higher efficiency, Eff2, is:

Qf,2 = Qp / Eff2


29


INPUT Low Fire Mid Fire High FireQf,1 (mmBtu/hr) 4 12 20HPY (hr/yr) 4000 2000 2000

CALCULATIONS Low Fire Mid Fire High Fire TotalQf,1 (mmBtu/yr) = Qf,1 HPY 16,000 24,000 40,000 80,000Qp (mmBtu/yr) = Qf,1 Eff1 13,114 19,876 32,920 65,910Qf,2 (mmBtu/yr) = Qp / Eff2 15,301 23,604 39,210 78,115Qf,sav (mmBtu/yr) = Qf1-Qf2 699 396 790 1,885FracRed = Qf,sav / Qf1 0.044 0.017 0.020 0.024

In the previous example, adjusting the linkages reduced fuel use by 1.6%. In this example, installing an O2 trim system reduced fuel use by 2.4%.

Descale Boiler to Improve EfficiencyScale buildup from hard water increases the thermal resistance between the hot combustion gasses and the steam, which increases exhaust temperature and decreases boiler efficiency. Mechanical and/or chemical descaling can significantly reduce exhaust gas temperature and increase boiler efficiency.

Example

A boiler burns 3,000 mmBtu of natural gas per year at a cost of $10 /mmBtu. The average temperature of the incoming combustion air is 70 F. The fraction excess air in the exhaust is measured to be 20%. The exhaust temperature from the boiler increases from 380 F to 450 F over a 14 month period due to scale buildup. Calculate a) the projected annual cost savings ($/yr) and b) the projected savings as a percent of current annual natural gas use from descaling the boiler.

The initial efficiency before descaling is:



30

The new efficiency after descaling is:






Qp = Qf,1 x Eff1


Qf,2 = Qp / Eff2



31


CALCULATIONSQp (mmBtu/yr) = Qf,1 Eff1 2,389Qf,2 (mmBtu/yr) = Qp / Eff2 2,930Qf,sav (mmBtu/yr) = Qf1-Qf2 70Cf,sav ($/yr) = Qf,sav Cf 700FracRed = Qf,sav / Qf1 0.023

Preheat Boiler Feed-water with EconomizerAn economizer is a heat exchanger that preheats feed-water to the boiler using heat from the exhaust gasses. Economizers are most cost effective in process boilers that operate all year. The energy reclaimed by the economizer can be modeled as a function of the effectiveness of the economizer.

Economizer pre-heating boiler feed water.

Heat Exchanger Effectiveness MethodEnergy savings from reclaiming heat can be calculated using the heat exchanger effectiveness method. Heat exchangers transfer heat from a hot stream with entering and exiting temperatures of Th1 and Th2 to a cold stream with entering and exiting temperatures of Tc1 and Tc2. The product of the mass flow rate and specific heat of the hot and cold streams are called the mass capacitances, mcph and mcpc. A schematic of a counterflow heat exchanger with these temperatures is shown below.

32

Heat exchanger effectiveness, e, is the ratio of the actual heat transfer, Qact, to maximum heat transfer, Qmax.

e = Qact / Qmax

The actual heat transfer is the product of the mass capacitance and the temperature rise of either the hot or cold stream. The mass capacitance, mcp, is the product of the mass flow rate, m, and the specific heat, cp.

Qact = mcph (Th1 – Th2) = mcpc (Tc2 – Tc1)

In an infinitely long heat exchanger, the exit temperature of the hot stream would reach the entering temperature of the cold stream. Similarly, the exit temperature of the cold stream would reach the entering temperature of the hot stream. The maximum heat transfer would be limited only by the capacity of the each stream to absorb the heat. Thus, the maximum heat transfer would be:

Qmax = mcp,min (Th1 – Tc1)

Thus, the heat exchanger effectiveness is:

e = Qact / Qmax = Qact / mcp,min (Th1 – Tc1)

If the heat exchanger effectiveness, mass capacitances and entering temperatures are known, this equation can be solved to determine the actual heat transfer, Qact, and exit temperatures of each stream.

Qact = e mcp,min (Th1 – Tc1)

Tc2 = Tc1 + e mcp,min (Th1 – Tc1) / mcpcTh2 = Th1 - e mcp,min (Th1 – Tc1) / mcph

Heat exchangers are typically designed with sufficient heat transfer area such that the effectiveness of the heat exchanger is between about 0.6 and 0.9. At higher levels of heat exchanger effectiveness, the cost of the required surface area frequently outweighs the additional performance benefits. Heat exchanger designers must also ensure that the pressure drop on each side of the heat exchanger is acceptably small,

33

and that the materials can withstand the temperatures, fouling and corrosiveness of the fluids involved.

Example

Consider reclaiming heat from boiler exhaust at 400 F to preheat boiler feedwater at 80 F. The boiler consumes 1,000,000 Btu/hr of natural gas and produces 900 lb/hr of steam. The fraction excess air in the exhaust is measured to be 20%. Calculate the heat and fuel savings, and the temperatures of the two streams leaving the economizer, if the economizer is 50% effective and the overall efficiency of the steam system is 75%.

The first step is to calculate the flow rate of exhaust gasses using combustion relations.

COMBUSTION AIR FLOWINPUTSQng (Btu/hr) 1,000,000 HHVng (Btu/lb) 23,900 AFRstoch (lba/lbng) 17.2EA 0.20CALCULATIONSMng = Qng / HHVng (lb/hr) 42 Ma = Mng AFRs (1+EA) (lb/hr) 864 Mtot = (Mng + Ma) (lb/hr) 905

Next, calculate the heat transfer from the hot exhaust gasses, h, to the cold feedwater, c, using the heat exchanger effectiveness method.

ECONOMIZER HEAT EXCHANGERINPUTSTh1 (F) 400mh (lb/hr) 905cph (Btu/lb-F) 0.30Tc1 (F) 80mc (lb/hr) 900cpc (Btu/lb-F) 1.00e 0.500CALCULATIONSCh = mh * cph (Btu/hr-F) 272Cc = mc * cpc (Btu/hr-F) 900Cmin = min(Ch, Cc) (Btu/hr-F) 272Qsav = e*Cmin*(Th1-Tc1) (Btu/hr) 43,461 Th2 = Th1-Q/Ch (F) 240 Tc2 = Tc1 + Q/Cc (F) 128

34

Note that in this example the feedwater was pre-heated from 80 F to 130 F, and the exhaust gasses were cooled from 400 F to 240 F. The dewpoint temperature of water vapor in exhaust is about 140 F. So the 240 F exhaust gas temperature is still hot enough to prevent condensation in the exhaust pipe.

The fuel savings would be:

FUEL SAVINGSEff,sys 0.75 Qf,sav (Btu/hr) = Qsav / Eff,sys 57,948

Run Boiler in Modulation Mode to Avoid On/Off CyclingMost boilers are designed for peak load, but operate at less than full load most of the time. To meet part load conditions using “on/off control”, the burner intermittently fires at full-fire then turn offs. To meet part load conditions using “modulation control”, fuel and combustion air are to the burner are modulated down and the burner fires at less than full fire. Modulation control is more efficient that on/off control for two reasons. First, each time a boiler cycles on and off, it purges natural gas from inside the boiler by blowing the combustion air fan. These purge cycles remove heat from the steam and increase fuel use. In addition, boilers are more efficient at low or medium fire than at full fire because the combustion gasses have more time to transfer heat to the steam as they pass through the boiler. Thus, it is advantageous to control the boiler with modulation control and avoid cycling.

In most boilers with on/off control, it is possible to upgrade to modulation control. In addition, modulating burners typically have a minimum firing rate of 25% to 33% of maximum output. If steam demand is less than the minimum firing rate, the boiler cycles on and off. Installing a burner with a smaller minimum firing rate can eliminate the on/off cycling and reduce fuel use.

Example

A boiler operating with on/off control consumes 6,300,000 Btu/hr at full fire. At full fire, the temperature of the exhaust gasses are 450 F and the excess air in the exhaust gasses is 20%. The temperature of combustion air entering the boiler is 70 F. The boiler operates 8,400 hours per year and fires at full fire 70% of the time. The boiler cycles off two times per hour, and purges natural gas from inside the boiler for 1 minute after cycling off and for 1 minute before reigniting. The saturation temperature of steam in the boiler is 335 F. The cost of natural gas is $10 /mmBtu. If the boiler were operated in

35

modulation mode, calculate the fuel savings from eliminating purge losses (mmBtu/yr), the fuel savings from improving combustion efficiency (mmBtu/yr), and the overall cost savings ($/yr)

The mass flow rate of exhaust gasses at full fire is:

COMBUSTION AIR FLOW AT FULL FIREINPUTSQf,rated (Btu/hr) 6,300,000 EA 0.20

CONSTANTS (FOR NATURAL GAS)HHVng (Btu/lb) 23,900 AFRstoch (lba/lbng) 17.2

CALCULATIONSMng = Qf / HHVng (lb/hr) 264 Ma = Mng AFRs (1+EA) (lb/hr) 5,441 Mtot = (Mng + Ma) (lb/hr) 5,704 The combustion efficiency at full fire is:

COMBUSTION EFFICIENCY AT FULL FIREINPUTSEA = excess air (0=stoch, 0.1 = optimum) 0.20Tca = temperature combustion air before burner (F) 70Tex = temperature exhaust gasses (F) 450


CALCULATIONSTc = temp combustion (F) = Tca+LHV/[(1+(1+EA)(Afs))cpp] 3,382hfg = water vapor latent energy (Btu/lb) = (if Tex<140 then hfg=HHV-LHV else hfg = 0)0Eff = {hfg+[1 + (1+EA)(AFs)]*cpp*(Tc-Tex)}/HHV 79.6%

To calculate purge losses, first calculate the energy delivered to the steam, Qsteam, and heat exchanger effectiveness of the boiler, e1, at full fire. Using this effectiveness, the heat loss during the purge cycle can be calculated as:

36

PURGE LOSSINPUTSQf,rated (Btu/hr) 6,300,000 Eff 79.6%Ma (lb/hr) 5,441 Mtot (lb/hr) 5,704 cpp = specific heat of products of exhaust (Btu/lb-F) 0.30Tc = temperature combustion (F) 3,382Tca = temperature combustion air before burner (F) 70Tsteam (F) 335Num cycles per hour = N 2Post-fire purge time (min) = PstP 1Pre-fire purge time (min) = PreP 1

CALCULATIONSQsteam (Btu/hr) = Qf Eff 5,017,077 e1 = Qsteam/(mtot cp (Tc-Ts) 0.962FracPurge = N (PstP+PreP) / 60 0.067Qpurge (Btu/hr) = FracPurge e1 ma cpp (Tsteam - Tca) 27,747 Qf,sav,purge (Btu/hr) = Qpurge/Eff 34,842

When boilers operate at less than full-fire, the velocity of exhaust gasses travelling through the boiler decreases, resulting in greater heat transfer and lower exhaust temperature. To calculate the lower exhaust temperature, first solve the relation for heat exchanger effectiveness, e1, for condensing heat exchangers such as boilers

e1 = 1 – exp(-UA/Cmin)

for the UA of the boiler. Next, calculate the heat exchanger effectiveness at the lower flow rate, e2, by solving the equation for heat delivered to the steam.

Qsteam = e2 m,ex cp,ex (Tc – Tex) = m,ex, cp,ex (Tc - Tex)

Using the method, the reduced exhaust temperature at less than full fire is:

37

EXHAUST TEMPERATURE WHEN MODULATINGINPUTSFFF = FracFullFire 0.7Mtot (lb/hr) 5,704 cpp = specific heat of products of exhaust (Btu/lb-F) 0.30Tc = temp combustion (F) 3,382e1 = Qsteam/(mtot cp (Tc-Ts) 0.962Tsteam (F) 335

CALCULATIONSUA (Btu/hr-F) = -Mtot cpp ln(1-e1) 5,608 mcp2 (Btu/hr-F) = FFF Mtot cpp 1,198 e2 = 1-exp(-UA/mcp2) 0.991 Tex2 = Tc - e2(Tc-Tsteam) 363

The combustion efficiency at less than full fire is:

COMBUSTION EFFICIENCY WHEN MODULATINGINPUTSEA = excess air (0=stoch, 0.1 = optimum) 0.20Tca = temperature combustion air before burner (F) 70Tex = temperature exhaust gasses (F) 363


CALCULATIONSTc = temp combustion (F) = Tca+LHV/[(1+(1+EA)(Afs))cpp] 3,382hfg = water vapor latent energy (Btu/lb) = (if Tex<140 then hfg=HHV-LHV else hfg = 0)0Eff = {hfg+[1 + (1+EA)(AFs)]*cpp*(Tc-Tex)}/HHV 82.0%

The savings would be:

38

SAVINGSINPUTHPY (hr/yr) 8400Cf ($/mmBtu) 10

CALCULATIONSQf,1 (mmBtu/yr) = Qfr FFF HPY / 1,000,000 37,044Qp (mmBtu/yr) = Qf,1 Eff1 29,500Qf,2 (mmBtu/yr) = Qp / Eff2 35,979Qf,sav,eff (mmBtu/yr) = Qf1-Qf2 1,065 Qf,sav,purge (mmBtu/yr) = Qf,sav,purge HPY / 1,000,000 293 Qf,sav,tot (mmBtu/yr) = Qf,sav,eff + Qf,sav,purge 1,357 Cf,sav ($/yr) = Qf,sav Cf 13,575FracRed = Qf,sav / Qf1 0.029

Blowdown Heat RecoveryBlowdown removes impurities that inevitably accumulate because makeup water is never 100% pure and the steam leaving the boiler is a distilled vapor with no impurities. Most boilers employ two types of blowdown: surface and bottom. Surface blowdown remove dissolved solids which tend to accumulate near the top of the boiler where steam is formed. Bottom blowdown removes sludge that accumulates on the bottom of the boiler. Total blowdown rates vary with the quality and quantity of boiler makeup water; however total rate of blowdown is typically between 4% and 8% of the steam generation rate.

Up to 80% of the thermal energy in the blowdown can be recovered. A schematic of a flash + condensate blowdown recovery system is shown below. Blowdown flash vapor and condensate are separated in a flash tank. Blowdown flash vapor is piped into the deaerator. Blowdown condensate flows through a plate heat exchanger to warm make-up water.

39

Source: http://www.spiraxsarco.com/

Install Stack Damper on Atmospheric Boilers Schematics of typical atmospheric and forced air hot-water boilers are shown below. In both types of boiler, hot combustion gasses transfer heat to the water as they move upward then out the exhaust flue. In on/off burner control, the burner fires whenever the water temperature drops to the low-temperature set point and turns off when the water temperature rises to the high-temperature set point.

Source: 2008 ASHRAE Handbook HVAC Systems and Equipment.

When open atmospheric boilers are not firing, air is drawn upward through the interior of the boiler as it warms and becomes more buoyant. This air pulls heat out of the water and reduces the overall efficiency of the boiler. This ‘chimney’ effect is exaggerated when the outlet of the exhaust flue is higher than the inlet to the base of the boiler. To reduce this loss, the exhaust flue can be equipped with a stack damper that closes when the burners are not firing. To be completely effective, the stack damper must be located below the exhaust flue hood. Closed forced-draft boilers

40

minimize this effect by sealing the combustion area with a fan that stops inlet air flow when the burner is not firing. However, to the extent that these losses still occur, they reduce the overall or total efficiency of the boiler.

Replace Conventional Hot Water Boiler with Condensing BoilerSteam boilers generate steam at 212 F and higher as steam pressure increases. Hot water boilers generate hot water at lower temperatures, and hence have the potential of operating at higher efficiencies than steam boilers. In addition, because of the low operating pressure, hot water boilers do not require dedicated boiler operators.

In HVAC applications, high-temperature hot-water boiler systems typically operate at about 180 F. Low-temperature systems operate at about 120 F. Low-temperature systems are more fuel efficient because the temperature difference between the water and hot combustion gasses is greater, which results in greater heat transfer and lower exhaust gas temperature. Efficiency increases significantly when water vapor condenses out of the exhaust gasses. To condense water vapor, the temperature of water returning from the building and entering the boiler must be 125 F or less. The graph below shows, how combustion efficiency increases with decreasing inlet water temperature.

Source: 2008 ASHRAE Handbook HVAC Systems and Equipment.

Example

41

A traditional hot-water boiler burns 1,000 mmBtu of natural gas per year at a cost of $10 /mmBtu. The average temperature of the incoming combustion air is 70 F. The average temperature of the exhaust is 300 F. The fraction excess air in the exhaust is measured to be 10%. It is proposed to 1) install a larger process heat exchanger that reduces the temperature of the return water from 150 F to 110 F, and 2) install a new condensing boiler. The average temperature of the exhaust from the condensing boiler is 120 F. Calculate a) the projected annual cost savings ($/yr) and b) the projected savings as a percent of current annual natural gas use.

The initial efficiency, Eff1, is:




A efficiency of the condensing boiler, Eff2, would be:

42



CALCULATIONSTc = temp combustion (F) = Tca+LHV/[(1+(1+EA)(Afs))cpp] 3,668hfg = water vapor latent energy (Btu/lb) = (if Tex<140 then hfg=HHV-LHV else hfg = 0) 2,400Effi ciency = {hfg+[1 + (1+EA)(AFs)]*cpp*(Tc-Tex)}/HHV 98.7%


Qp = Qf,1 x Eff1


Qf,2 = Qp / Eff2




CALCULATIONSQp (mmBtu/yr) = Qf,1 Eff1 842Qf,2 (mmBtu/yr) = Qp / Eff2 853Qf,sav (mmBtu/yr) = Qf1-Qf2 147Cf,sav ($/yr) = Qf,sav Cf 1,473FracRed = Qf,sav / Qf1 0.147

43

Use Direct Contact Water Heater for Direct Inject and Hot Water ApplicationsSome industrial processes, such a food processing, require large volumes of hot water that cannot be returned to the system. In these cases, make-up water can enter the boiler at near ambient temperatures. Direct contact hot water heaters capitalize on low incoming water temperatures, counter flow design and large heat exchange areas between the combustion gasses and water droplets to generate efficiencies of up to 99%.

High-efficiency direct-contact water heater in the food processing industry.

Energy Savings and Steam System ModelsFuel energy savings, Fuel, can be estimated by calculating the reduction in energy loss through a given pathway, Energy , by the overall efficiency of the steam system, Eff,sys.

Fuel = Energy / Eff,sys

The energy efficiency of any system is the ratio of useful energy delivered to required energy input. For steam systems, the energy efficiency is the ratio of useful heat delivered to the process to the pump and fuel energy input.

Eff,sys = Qprocess / (Epump + Efuel)

However, because pump energy is quite small compared to fuel energy, pump energy is neglected and the efficiency is typically calculated as:

44

Eff,sys = Qprocess / Efuel

However, in an integrated system like a steam system, changes in one part of the system affect other parts of the system. The simplistic method of estimating savings shown above does not account for these synergistic effects between system components. A more accurate way to calculate expected fuel savings is to use an integrated model of the steam system, calibrate it to baseline fuel use, change parameters to model energy efficiency opportunities, and compare baseline versus energy-efficient fuel use.

One model of a steam system is called SteamSim. SteamSim is a thermodynamic model of the steam system shown below. The steam system is modeled from the following readily obtainable input data using known state points, energy balances, and mass balances and the methods described above.

SteamSim required input data are:

Qprocess : heat delivered to process (Btu/hr) P2 : steam pressure at exit to boiler (psia) P3 : steam pressure at exit from throttling valve (psia) Pda : steam pressure of deaerator tank (psia) T0 : temperature of makeup water (F)

45

Fbd : Fraction of input water discharged as blowdown Fcl : Fraction of condensate lost Eecon: effectiveness of economizer EA: excess air in combustion exhaust Tca: temperature of combustion air entering boiler (F) Tex : temperature of combustion exhaust from boiler before economizer (F) Mstl : Mass flow rate of steam leaking through steam traps (lb/hr) Dsp, Lsp, Rsp : diameter (ft), length (ft) and thermal resistance (hr-ft2-F/Btu) of

steam pipes Dcp, Lcp, Rcp : diameter (ft), length (ft) and thermal resistance (hr-ft2-F/Btu) of

condensate pipes Db, Lb, Rb : diameter (ft), length (ft) and thermal resistance (hr-ft2-F/Btu) of

boiler Dda, Lda, Rda : diameter (ft), length (ft) and thermal resistance (hr-ft2-F/Btu) of

dearator tank

SteamSim output are:

Qfuel : fuel energy input to Boiler (Btu/hr) Fqp : fraction of fuel energy delivered to process Qexhaust, Fqex : energy lost in combustion exhaust (Btu/hr) and fraction of fuel

energy lost in combustion exhaust Qbd, Fqbd : energy lost in blowdown (Btu/hr) and fraction of fuel energy lost in

blowdown Qflash, Fqflash : energy lost in flash steam(Btu/hr) and fraction of fuel energy

lost in flash steam Qecon, Fqecon : energy reclaimed by economizer (Btu/hr) and fraction of fuel

energy reclaimed by economizer Qcl, Fqcl : energy lost in condensate loss (Btu/hr) and fraction of fuel energy lost

in condensate loss Tsp, Qsp, Fqsp : temperature of steam pipe (F), heat loss from steam pipe

(Btu/hr), fraction of fuel input lost from steam pipe Tcp, Qcp, Fqcp : temperature of condensate pipe (F), heat loss from condensate

pipe (Btu/hr), fraction of fuel input lost from condensate pipe Tda, Qda, Fqda : temperature of dearator tank (F), heat loss from dearator tank

(Btu/hr), fraction of fuel input lost from dearator tank Tb, Qb, Fb : temperature of boiler (F), heat loss from boiler (Btu/hr), fraction of

fuel input lost from boiler

46

Date post:	08-Mar-2018
Category:	Documents
Upload:	dinhkien
View:	213 times
Download:	0 times

Energy Efficient Steam Systemsacademic.udayton.edu/.../EnergyEfficientSteamSystems.docx · Web...

Documents