Energy Efficient Steam Systems
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
watersteam 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
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 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:
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
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.orgCut-away view of water-tube
boilerSource: 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.
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 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:
Inverted bucket.Float + thermostaticThermostaticThermodynamic
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.
Well-insulated steam pipesUninsulated 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 Ta4) (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 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 Ta4) (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 walls 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 Ta4) = 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.
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:
Calculations of current heat loss and thermal resistance of the
pipe, Rp, are:
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 Ta4) =
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:
Thus, the heat loss, Qsav, and fuel, Qf,sav, savings from adding
insulation would be:
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:
Calculations of current heat loss and thermal resistance of the
wall, Rw, are:
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 Ta4) =
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:
Thus, the heat loss, Qsav, and fuel savings, Qf,sav, from adding
insulation would be:
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 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.
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 -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:
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 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, Qin, 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)
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.
Thus, mass flow rate of combustion air is 94 lb/hr and the mass
flow rate of the combustion gasses is 98 lb/hr.
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.
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:
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:
Qf,2 = Qp / Eff2
The fuel use savings, Qf,sav, would be:
Qf,sav = Qf,1 - Qf,2
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.
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:
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:
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
after the efficiency is improved. The new fuel use, Qf2, with the
higher efficiency, Eff2, is:
Qf,2 = Qp / Eff2
The fuel use savings, Qf,sav, would be:
Qf,sav = Qf,1 - Qf,2
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
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:
The new combustion efficiencies, Eff2, if the excess air was
held to 10% across the firing range would be:
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
after the efficiency is improved. The new fuel use, Qf2, with the
higher efficiency, Eff2, is:
Qf,2 = Qp / Eff2
The fuel use savings, Qf,sav, would be:
Qf,sav = Qf,1 - Qf,2
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:
The new efficiency after descaling 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:
Qf,2 = Qp / Eff2
The fuel use savings, Qf,sav, would be:
Qf,sav = Qf,1 - Qf,2
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.
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, 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.
Next, calculate the heat transfer from the hot exhaust gasses,
h, to the cold feedwater, c, using the heat exchanger effectiveness
method.
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:
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 $10
/mmBtu. If the boiler were operated in modulation mode, calculate
the fuel savings from eliminating purge losses (mmBtu/yr), the fuel
savings from improving combustion efficiency (mmBtu/yr), and the
overall cost savings ($/yr)
The mass flow rate of exhaust gasses at full fire is:
The combustion efficiency at full fire is:
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:
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:
The combustion efficiency at less than full fire is:
The savings would be:
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.
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 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
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:
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:
Qf,2 = Qp / Eff2
The fuel use savings, Qf,sav, would be:
Qf,sav = Qf,1 - Qf,2
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:
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) 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
1Makeup water
Th1
Th2
Tc1
Tc2
Qact