Energy-Efficient Steam Systems
Introduction
Air, 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 increase 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 Systems
Energy Balance Approach
The 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 Systems
These 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 Components
Boilers and Steam Generators
Steam 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.
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.
Deaerator Tanks
Makeup 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 Valves
Boilers 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 Systems
Steam 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 Traps
As 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 +
thermostatic
• Thermostatic
• Thermodynamic
Condensate Return Tanks
Condensate 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 Metering
Steam 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 Surfaces
Uninsulated 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 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) + s A e (Ts4 – Ta4)
(2)
where h is the convection coefficient, s is the Stefan-Boltzman constant (0.1714 ·
10-8 Btu/ft2-hr-R4, or 5.67 · 10-8
W/m2-K4), e 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 DT is temperature difference
between the surface, Ts, and the surrounding air, Ta (F).
Laminar: L3
DT < 63 Turbulent: L3 DT
> 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 (DT
/ L) 0.25; htur
= 0.22 (DT) 0.33
Tilted/Vertical
Surfaces:
Horizontal Pipes and Cylinders:
hlam = 0.27 (DT
/ D) 0.25; htur =
0.18 (DT) 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 A
(Ti – Ta) + s
A e (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 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) - s A e
(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)
- s A e
(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%.
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)
- s A e
(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 Tanks
In 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.
Fix Steam Traps
Steam 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 Traps
The 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/yr
110 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:
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 Blow down Controls
Blow down 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 blow down rates
range from 4% to 8% of boiler feed-water. Blowdown may be manual or
automatic. Manual blow down relies on
intuition or periodic testing to determine when the concentration of
contaminants is high enough to warrant blow down. Manual blow down virtually always results in
either excess blow down that wastes energy or insufficient blow down that
creates excess scale on heat transfer surfaces and reduces boiler
efficiency. Automatic blow down controls
monitor the conductivity of the water in the boiler and open the blow down
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 Efficiency
Boilers
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) Ã
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, h, 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, h
is:
h = [{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
h = [{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 Controls
Most 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.
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 Efficiency
Scale 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 Economizer
An 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 Method
Energy 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) / mcpc
Th2 = 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 Cycling
Most 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)
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)
Blowdown Heat Recovery
Blowdown 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.
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.
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 Boiler
Steam
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.
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 Applications
Some 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 Models
Fuel energy savings, DFuel,
can be estimated by calculating the reduction in energy loss through a given
pathway, DEnergy, by the overall
efficiency of the steam system, Eff,sys.
DFuel = DEnergy / 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
If the hole is big enough, the pressure in that cell of the heat exchanger can become positive, and push the flames out of the heat exchanger into the area where normally there is only secondary combustion air.
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