Electrical Systems
And
Electric Energy
Management
Main Topics Discussed
• Electric Rates
• Electrical system utilization
• Power quality
• Harmonics
• Power factor improvement
Session 14.2
2
Review of Electric Cost Components
Energy cost - e.g. $0.05/kWh
Demand cost - e. g. $6.50/kW/mo
Fuel adjustment - e.g. $0.005/kWh
Power factor penalty - e.g. $6.50/kVA/mo
or
kW billed = kW u (0.85/PF)
Ratchet clause - e.g. Maximum of (kW this
month, or 70% of maximum kW in last 11
months)
Session 14.3
3
Power Computation Formulas
Single-phase system
P = —1 u V u I u PF
Where PF = power factor
Three-phase system
P = —3 u V u I u PF
—3 = 1.732
Session 14.4
4
Examples
a) For a 10 ampere, 120 volt, electric space
heater
P = —1 x 120 u 10 u 1.0 = 1200 watts
b) For a three phase 460 volt, 20 ampere motor
with power factor of 90% at full load
P = —3 u 0.460 u 20 u 0.9 = 14.34 kW
5
Session 14.5
Electric Motor Equations
Pr [kW]=
—3 u kV u I u PF
Ps[kVA]=
—3 u kV u I
PF
kW/kVA =
=
Pr [kW]=
cos)
HP u 0.746 kW u Load Factor
Efficiency
Session 14.6
6
Power Quality
• Power Quality is related to how well a
bus voltage—usually our facility load
bus voltage—maintains a pure sinusoidal waveform at
rated voltage and frequency.
• PQ issues involve all momentary phenomena including
spikes, notches and outages; as well as harmonics and
power factor.
• Modern electronic equipment both causes and is
affected by the problem.
• Power Quality is becoming one of the most important
issues in energy management today.
Session 14.7
7
Harmonics
• Harmonics are a multiple of the fundamental frequency.
If the fundamental frequency is 60 hertz, the 2nd harmonic
is 120 Hz, the 3rd is 180 Hz, the 4th is 240 Hz, etc.
• Harmonics are usually generated by solid-state-based
equipment such as switching power supplies in PCs, DC
drives, variable frequency drives (VFDs), electronic
ballasts, arc welders and ovens.
Session 14.8
8
Importance of Grounding
• Up to 80 percent of PQ problems in facilities
today may be caused by wiring and grounding
systems that met the NEC at the time, but do
not meet the needs of today's sensitive
electronic equipment.
• The first step taken to deal with PQ problems
should be to inspect the wiring and grounding,
and clean and tighten all connections. Loose
connections come from vibration, oxidation,
corrosion, and age.
Session 14.9
9
What Problems Occur Because of
Harmonics?
• Circuit breakers tripping
• Neutrals overheating (smoke, fire)
• Panel or transformer overheating
• RFI – Radio Frequency Interference
• Errors/damage in Electronic Equipment
• Digital clocks running fast
• Failures in power factor correction capacitors
Session 14.10
10
Mitigation of Harmonic Problems
• Derate equipment (symptom treatment)
50% Transformers
70% Load centers
Circuit breakers
Neutrals
• Install preventive equipment
Inductors
Harmonic filters
Isolation transformers
Locate near drive if possible
Connect back to "strongest" point of
power system – the load center
11
Session 14.11
Power Triangle
kW
)
kVAR
kVA
Session 14.12
12
Schematic arrangement showing how capacitors reduce total
kVa by supplying magnetizing requirements locally.
13
Session 14.13
Sample Power Factor Example
A facility is operating with a demand of 2000
kW. The 2500 kVa transformer is fully loaded.
How many kVARS are required to bring the
power factor back to unity?
kW2 + kVAR2=kVa2
kVAR2= kVa2-kW2
kVAR
2500 2 2000 2
Session 14.14
1500
14
Sample Power Factor Problem
During my last energy audit I saw a 100 HP
electric motor that had the following information
on the nameplate: 460 volts; 114 amps; three
phase; 95% efficient – all at full load . What is
the power factor of this motor?
[(100 hp)*(0.746 kW/HP)/0.95] =[(1.73)*(0.46 kV)*(114 amp)*(PF)]
78.52 kW = (90.72 kVa)*(PF)
PF = (78.52 kW / 90.72 kVa) = 0.866
Session 14.15
Transformer
M
15
Where to Put Power Factor
Correction Capacitors
Session 14.16
16
Sample CEM Test Question
• A facility is operating at a power factor of 70%
with a real power load of 2000 kW. How much
corrective capacitance in kVAR is needed to
improve the facility power factor to 90%?
• kVAR = Table Factor x Real power load in kW
Session 14.17
17
Short Power Factor Table
Session 14.18
18
CEM Review Questions
1. If power factor correction capacitors are
located at the utility meter, but on the
customer’s side of the meter, the power factor
in the customer’s facility will not be improved.
A. True
B. False
2. A facility has a 100 kW electric resistance oven
for drying parts. What is the power factor of
the oven?
A. 0 %
B. 50%
C. 90% D. 100%
Session 5.1.19
19
3. A facility has a motor that draws 200 kVA and
has a power factor of 70.7%. How many kW and
how many kVAR does it draw?
4. A facility has a motor that draws 200 kVA and
has a power factor of 80%. How many kW and
how many kVAR does it draw?
Session 5.1.20
20
Electric Motors
and Motor Management
Session 15.1
Electric Motor Management
Why Bother?
• Electric motors use over ½ all U.S. electricity
• Motor driven systems use over 70% electric
energy for many plants
• Motor driven systems cost about $90 billion to
operate per year
• A heavily used motor can cost 10 times its first
cost to run one year
Session 15.2
Electric Motor Management
Why So Difficult?
• Load on most driven systems is unknown at
least on retrofits
• Very difficult to determine load accurately
through measurements
• Electric motor management is FULL of
surprises
• Yet, savings can be large (small percentage
of a big number is a big number)
• Important note: Often oversized wiring
(above code) is cost effective in heavily used
systems as it reduces I2R losses. (CDA and
Southwire Corp.) Session 15.3
Electric Motor Management
Types of Motors
• AC Synchronous motors
– One to two percent or so (but larger)
– Large HP and slow speed applications typical
– Similar in construction to induction motors,
but more expensive
– More efficient, can be run at leading PF
– Can generate or absorb reactive power
Session 15.4
Electric Motor Management
Types of Motors
• DC Motors
– Good for precise speed control and strong torque
properties
– Not efficient, (historically) high maintenance, and
higher down time (commutator and brushes need
inspection and maintenance)
– Newer brushless motors much better
– Less than 5% of the motors today are DC
– Replacing with a VFD driven AC motor may be cost
effective especially if down time is reduced
Session 15.5
Electric Motor Management
Motor Types Cont.
• AC Induction motors
– AC induction motors work on electromagnetic
field principles
– Lagging power factor
– Many different types (ODP, TEFC, etc.)
– Approximately 95% of motors today are
induction
– Concentration for this discussion
– Run on single- or three- phase power (most)
Session 15.6
Electric Motor Basics
Name Plates
•
•
•
•
•
•
•
•
•
HP ____ (shaft power
design- output)
NLRPM (synchronous
speed)
FLRPM (running RPM at
design load)
LRA (starting amps – 1 sec?)
FLA (amps at design load
and voltage)
Volts (design voltage)
Max. capacitor
Efficiency (test vs.
guaranteed)
Service factor
Session 15.7
Name Plate Example One
Session 15.8
Session 15.9
Terminology (NEMA)
• NEMA (National Electrical Manufacturers
Association)
• Standard Efficiency Motors: Pre Epact 92 motor
• Energy Efficient Motors: Motors meeting
EPACT requirements
• Premium Efficiency Motors: Motors exceeding
EPACT requirements (extra efficient, ultra high
efficient, but these terms are not used by NEMA)
Session 15.10
Electric Motor Basics
Motor Speeds
• Alternating current, thus speed will vary with pole pairs
(inside motor, pole pairs between stator and rotor)
• One pole pair (2 poles) - one RPM per cycle (60 cps or
Hertz); two pole pairs – ½ rpm per cycle, etc.
• Thus
cycles
sec
u 60
sec
min
number of pole pairs
60
SPEED
• SPEED = 3600, 1800, 1200, 900, 720, etc.
(no other choices) for 60 Hz power
Session 15.11
Electric Motor Basics
Slip
•
•
•
•
% Load = (True Slip)/(Design Slip)
Design Slip = (NLRPM – FLRPM)
True Slip = (NLRPM – RPM measured)
Perfect indicator but very difficult to
measure accurately (+- 1% typical)
• Many don’t use this - why?
– Difficult to measure accurately
– Large motors are more efficient than small
motors (more later)
Session 15.12
Electric Motor Basics
Slip Example
•
•
•
•
•
•
•
•
•
•
FLRPM = 1760 (off name plate)
Design HP = 50 (off name plate)
Measured RPM = 1776
NLRPM = ? (obviously 1800)
Design slip = 1800 – 1760 = 40
True slip = 1800 – 1776 = 24
% load = 24/40 = 0.6 or 60%
True load = 50HP(0.6) = 30 HP
See plot next slide, motors run very well at 60% load
This motor will run very cool, is not causing a problem,
why bother!!
Session 15.13
Efficiency and PF vs. Load:
%
% Rated Load
Session 15.14
Session 5.2.15
Session 15.16
Efficiency and PF (Cos phi) vs. Size
%
Rating
Source: Electrical and Energy Management, IEES, Ga,Tech., Atlanta, GA.
Session 15.17
Session 5.2.18
Electric Motor Management—Losses in Motors
Note: This graph explains the efficiency drop off with load reduction Source:
“Electrical Motors and Energy Conservation”, Ronald Cota, Specifying Engineer, July,
1978.
Session 15.19
Motor Performance as
Supply Voltage Varies
Session 5.2.20
Voltage Imbalance
• Problems can occur because of voltage
imbalance between the three phases. This
can be a serious problem in motors.
• Percent voltage imbalance is found as the
ratio of the largest phase voltage
difference from average, divided by the
average voltage.
• For example, if we have 220, 215 and 210
volts, the voltage imbalance is 5/215 =
.023, or 2.3 percent.
Session 15.21
Voltage Imbalance Impact
Source: Electrical and Energy Management, IEES, Ga,Tech., Atlanta, GA.
Session 15.22
Approach to Motors
• Leave existing motors alone until they fail except:
– Exceptionally oversized motors (25% loading or so)
– Sizes that are needed elsewhere (requires inventory)
• When they fail, maybe buy new energy efficient
motors (EPACT or Premium) instead of paying for
rewind (much more on this later)
• If financial incentives are available, much more may be
done
• Premium efficient motors need economic help in much of
the country (PUC, Utility, motor mfgs.)
Session 15.23
Motor Basics-Motor Rewinds
• Most rewind motors over ___ HP
• Typical rewinds cost 60+% of a new motor
• New motor could be an energy efficient
motor
• Motor efficiency often suffers during
rewind. Average drop about 1% according
to one study and sometimes significantly
more.
• If efficiency drops, losses increases, motor
runs hotter and won’t last as long
Session 15.24
Electric Motor Management
Energy Efficient Motors
• Energy efficient motor characteristics
– More efficient, and often higher power factor
– Save energy and reduce demand
– Reduce load on cables, transformers, etc.
(note double whammy with higher efficiency
and higher PF)
– Speed is slightly higher (can be critical)
– Significantly larger inrush (LRA)
Session 15.25
Energy Efficient Motors
Calculating Savings
• Power and energy savings depends of
efficiency of standard vs. energy efficient
motor
Power savings kWe
§ HP u 0.746 u LF ·
§ HP u 0.746 u LF ·
¸
¸ ¨
¨
EFF
EFF
¹ EE
¹Stan ©
©
• Energy savings = Power savings x Time
= kWe X Operating hours
Session 15.26
Electric Motor Principles Review
ELECTRICAL MOTOR PRINCIPLES
A three phase 50-hp motor with a load factor of 0.8 has an efficiency
of 90%, what is the kW electrical power input?
kW
u 0.80
Hp
0.90
50 Hp u 0.746
1. kW
33.15 kW
33.15 kW
Session 15.27
Electric Motor Principles Review
2. For the motor in 1. If the PF = cos ș = 0.7 and voltage is 480 V,
what is the kVA and what is the amp draw?
PF = 0.7 = kW/kVA
0.7 = 33.15/kVA or
kVA = 33.15/0.7 = 47.36 kVA
Also kVA = ¥3(kV)I = (¥3)(0.480) I = 47.36 kVA
33.15 kW
I = 47.36 / (¥3 x 0.480) = 56.96 amps
Session 15.28
47.36
Electric Motor Principles Review
3. Next, we want to correct the PF to 0.90. What size capacitor is needed
and what is the impact on the amperage?
¨kVAR = 33.15 (tan cos-1 0.7 – tan cos-1 0.9) = 17.77 kVAR
You will find the quantity in ( ) above in PF table (see Appendix)
New kVA = 33.15/0.9 = 36.83
kVA = 36.83 = (kV)I¥3
I = 36.83/(0.480¥3) = 44.30 amps
Thus, PF correction dropped amperage
(upstream of the capacitor) from 56.96
to 44.30 amps or 22%
33.15 kW
36.83
47.36
¨ kVAR=17.77
Also, new kVA = ¥3 kV I = 36.83 kVA
Session 15.29
Tools to Help
•
The following software packages are available free from OIT of DoE.
Contact OIT Clearinghouse 800-862-2086 or clearinghouse@ee.doe.gov
They are also downloadable from the DoE web site.
– MotorMaster: An energy-efficient motor selection and management
tool. Motor inventory management, maintenance log tracking, efficiency
analysis, savings evaluation, energy accounting, and environmental
reporting
– Pump System Assessment Tool (PSAT): Efficiency of pumping system
operations. Pump performance and potential energy and other cost
savings
– ASD Master: Adjustable speed drive evaluation methodology and
application software. Available from EPRI also.
– Steam Sourcebook: Guide to improved steam system performance.
Session 15.30
Motor Sample Problem
• A recent advertisement said a premium
efficiency 50 hp motor is available at 94.5%. It
would replace a motor that presently runs at
90.7%. Given the parameters below, calculate
the cost of operating both motors and the
savings for conversion:
–
–
–
–
Motor runs 8760 hours/year
Demand cost is $10 per kW month
Energy cost is $0.06/kWh
Motor runs at 80% load all the time
Session 15.31
Motor Sample Problem
• Cost to operate existing motor
– Demand
– Energy
– Total
Session 15.32
Motor Sample Problem
• Cost to operate premium efficiency motor
– Demand
– Energy
– Total
• Savings
Session 15.33
Session 5.2.34
Motors, Drives
and Air Compressors
Session 16.1
Electric Motor Management
Drives
• Motors are fixed speed devices likely running
between NLRPM and FLRPM
• Other speeds on the driven end have to be
engineered (which will affect the load on the
motor)
• Because of the “fan” laws (pumping or blowing)
centrifugal devices are desired applications for
varying CFM or GPM
Session 16.2
Electric Motor Management
Fan Laws
(Centrifugal Devices ONLY)
• CFM2 = CFM1(RPM2/RPM1)
1st law
• SP2 = SP1(RPM2 /RPM1)2
2nd law
• HP2 = HP1 (RPM2/RPM1)3
3rd law
Session 16.3
Electric Motor Management
Fan Laws Example
• A 40 HP centrifugal blower is on a forced draft
cooling tower. It is basin temperature controlled
but conversion to a variable speed drive is being
considered. When the blower is running at ½
speed, what is the impact on the CFM and what
is the HP requirement?
Session 16.4
Electric Motor Management
Fan Laws Example
• New CFM is __________old CFM
• New HP requirement is:
• These type savings are why variable speed
drives are so popular today
Session 16.5
Electric Motors
Variable Volume Options
• Outlet damper control (see sketch, location 1)
• Inlet vane control (see sketch, location 2)
• Magnetic clutching (see sketch, location 3)
– Eddy current clutch
– Permanent magnetic clutch
• Variable Frequency Drives (see sketch,
location 4)
• Hydraulic drives, variable sheaves, etc.
Session 16.6
Electric Motors
Variable Volume Options Sketch
4
Session 16.7
Variable Speed Drive
Alternatives Performance
• The next page shows performance expectations
from an older EPRI report
• The page after that shows performance from a
more recent PNL test
• The third page shows an “average” VAV loading
profile. It can be used as a default loading if
better figures are not available. Quick Fan from
DoE presents another default possibility.
Session 16.8
Typical Power Consumption of Various Control Systems
Session 16.9
Session 16.10
VFD: Default Loading Profile
Session 16.11
Variable Frequency Drive
Example
• A large (50 HP) blower with inlet vane control
drives a VAV system operating 6500 hours per
year. Energy costs $0.04/kWh. What is the total
savings per year for removing the inlet vane
control and replacing it with a VFD?
– Assume the performance figures in slide 9 apply
– Assume the loading figures in slide 11 apply.
– Construct an Excel spread sheet to do the
calculations (will be done for you on next page)
Session 16.12
Variable Frequency Drive Example
Profit Improvement With Variable Frequency Drives
Annual Savings for a Large Air Handler
Session 16.13
Variable Frequency Drive
Example
• Calculation for 50% load row in Spread
Sheet:
(50HP)(0.746kW/HP)(0.72-0.20)(0.23)(6500hr/yr)($0.04/kWH) = $1159
Spread Sheet repeats this for all rows
Session 16.14
Electric Motor Management
Selection of Best Option
• Magnetic clutches (permanent magnet or eddy
current)
– Bulky and heavy on motor shaft
– No harmonics
– Close to same savings as VFDs, but less
Session 16.15
Electric Motor Management
Other Drives
• Variable sheaves
– Very closely approach fan laws
– Repeatability and maintenance often a
problem
– Many lock blades and install VFDs
• Hydraulics
– Effective
– Expensive
– Not often used as discussed here
Session 16.16
Electric Motor Management
Axial and Reciprocating
• Centrifugal laws do not apply
• More difficult to predict savings
• If linear, no “real energy savings” over present
on/off operation (certainly improved soft start
operations and perhaps control)
• Obviously, savings if converting from constant
volume to variable volume
Session 16.17
$$$$
Variable Speed Drive Applications
• Any large centrifugal blower or pump that runs a
lot!
– Constant volume? Convert to var. volume
– Variable volume with inlet or outlet control
• Chilled water pumps, large campus
• Cooling water pumps
• VAVs using inlet vane
• Forced draft (blower) cooling towers
Session 16.18
Industrial Systems
Compressed Air Management
• Most expensive utility for many companies
• Large cost reduction potential (20 to 30%
common)
• Management
– Demand side
– Supply side
Session 17.2
Compressed Air Management
• Each 100 hp compressor costs approximately
$25,000 per year to operate
• Try it: 8000 hours/year, $80/kW-yr, $0.05/kWh,
running at 70% load.
– Cost:
• Typical first cost is $30,000 to $50,000
Session 17.3
Session 17.4
Air Systems Components Mgmt.
• Demand side: should be managed first
– Air leaks: large cost (see diagram), often 20%
or more of capacity goes to leaks (artificial
demand). Air leak detection and repair is
extremely cost effective.
– Air motors: great torque properties, small
handprint, will not spark, but VERY ENERGY
INEFFICIENT (7 hp air vs. 1 hp electric
common)
Session 17.5
Air Systems Components Mgmt.
• Pressure supplied: proper level critical
– Approximately 1% energy savings for each 2
psig drop (increase) in air pressure
– Productivity cost can greatly exceed energy
savings so be careful
– Distribution drop should not be more than
about 10% (90 psig at tank to supply 80 psig
to tool) so receivers, better looping, etc. may
enable pressure to be dropped without
affecting tool
Session 17.6
System Components
• Supply Side
– Air Compressor
– Aftercooler
– Receiver tank (storage)
– Dryer
– Distribution lines
– Users (demand side)
Session 17.7
Air System Components Mgmt.
• Air Compressors
– Screw: convenient, fairly efficient, skid
mounted, most popular (100 HP or so)
– Reciprocating: usually large, expensive, very
efficient, part load well, often two stage
– Centrifugal: usually quite large, modulates
efficiently down to a point (surge), often VFD
– Others: other types exist but these cover the
vast majority
Session 17.8
Air Systems Components Mgmt.
• Intake air should be as cool and dry as possible
• Intake air filter should be large with low
pressure drop (2 psig), changed frequently
• Multi stage with interstage cooling helps
• Drivers can be electric motor (most), gas engine
(load management or hybrid), or steam
• Significant waste heat available (250,000
Btu/100 HP.) Easy to recover on packaged
screw air cooled machines.
Session 17.9
Air Systems Components Mgmt.
• Sequencers: For multiple compressor
dispatching
– Use a remote (not in compressors) PID
control loop to dispatch multiple compressors
– Dramatically reduces control differential and
smoothes operation
– Can be extremely cost effective
Session 17.10
Air Systems Components Mgmt.
• After coolers: Air leaves a compressor at
approximately 100% RH.
– Cooling removes significant moisture; must be
trapped and drained (exhausted)
– Air cooled or water cooled and they are quite
cost effective; save energy in dryer and
provide dryer air.
– Water cooled more effective but water should
be used elsewhere or reused through small
cooling tower
Session 17.11
Air Systems Components Mgmt.
• Receivers: storage devices to smooth demand
on compressor
– Further cooling removes more water (traps)
– Can significantly reduce demand fluctuation
on compressor controls
– 2 to 4 gallons per CFM (but this varies a great
deal)
– Located before dryer (wet), after dryer (dry),
throughout plant, and at large loads
Session 17.12
Air Systems Components Mgmt.
• Dryers: Further reduce dew point to avoid
moisture in plant equipment
– Refrigerated: cools air to about 40 F and reheats by
precooling incoming air. Thus, dew point around 40
F. Not dry enough for many plants.
– Desiccant: chemically removes moisture saturating
media which must be dried before reusing. Usually
two stacks. Expensive but dew point around -10 F.
– Deliquescent: similar to desiccant but wet material is
removed and replaced.
Session 17.13
Air Systems Components Mgmt.
• Distribution lines: get air where needed and
provide further storage
– Large lines are good for systems operating
full time (often 4 inch)
– All lines should be “looped” so air comes at
any need from two directions
• Traps and drains: traps and drains located
throughout system to remove moisture. Can be
a large source of air leaks.
Session 17.14
Waste Heat Recovery
• Successful waste heat recovery requires
– Sufficient quantity
– Sufficient quality (sufficiently high
temperature)
– Appropriate use
– Timely production (or storage)
– Economic use
Session 17.15
Waste Heat Recovery
• How much waste heat is available?
• Conduct a waste heat survey
– Temperature of waste stream
– Quantity of waste stream
– Heat available is
q MC P (Tinitial Tfinal )
Session 17.16
Waste Heat Recovery
• Other concerns
– How close is location where heat is needed?
– Is the waste heat available when needed?
– Is the waste heat compatible with a suitable
heat exchanger?
• Fouling
• Plugging
• Corrosion
Session 17.17
Waste Heat Recovery
• Heat exchanger effectiveness, ɽ
İ
actual heat transfer to cold fluid
maximum possible heat transfer
– Effectiveness generally increases with heat
exchanger area but not linearly
– At some point a large area increase produces only a
small effectiveness increase
– 50% < ɽ < 90%
Session 17.18
Waste Heat Recovery—Equipment Types
• Shell and tube—for liquids and gases at all
temperatures
*B.L. Capehart, et. al., Guide to Energy Management, 5th ed., Fairmont Press, 2003, p. 320.
Session 17.19
Waste Heat Recovery—Equipment Types
• Radiation
recuperator—for
gases from
medium to high
temperatures
*. *B.L. Capehart, et. al., Guide to Energy Management, 5th ed., Fairmont Press, 2003, p. 319.
Session 17.20
Waste Heat Recovery—Equipment Types
• Rotary heat wheel—for gases at all
temperatures
*B.L. Capehart, et. al., Guide to Energy Management, 5th ed., Fairmont Press, 2003, p. 321.
Session 17.21
Waste Heat Recovery—Equipment Types
• Heat pipes (capillary tubes)—for liquids and
gases at all temperatures
*B.L. Capehart, et. al., Guide to Energy Management, 5th ed., Fairmont Press, 2003, p. 253.
Session 17.22
Industrial Insulation
• Industrial insulation problems
– Walls of vats, tanks, pipes or machines that
were never insulated.
– Damaged insulation.
– Insulation removed for a repair and never
replaced.
Session 17.23
Industrial Insulation
• Uninsulated walls result in the following
temperature profile:
Hot source
at Th
wall
Twall, unins
Tair
inside
surface
outside
surface
of wall
Session 17.24
Industrial Insulation
• With insulation the profile changes:
Hot source
at Th
wall
insulation
Inside
surface
outside
surface of
insulation
Twall, ins=Twall, unins
Touter ins
Tair
outside
surface
of wall
Session 17.25
Industrial Insulation
• From the Building Envelope Section recall• At steady-state, the heat loss in Btu/ft2·°F·hr is
q
A u 'T
ȈR
ª Btu º
«¬ h »¼
• Where q is heat loss rate and ȈR is the sum of the
resistances to the heat flow
Session 17.26
Air Film Resistances (Industrial)
*B.L. Capehart, et. al., Guide to Energy Management, 5th ed., Fairmont Press, 2003, p. 392.
Session 17.27
Industrial Insulation
• Thus for the uninsulated case with metal walls,
q unins
A(Th Tair )
R surface
And both Th and Tair are easily measured.
Session 17.28
Industrial Insulation
• For the insulated metal wall case,
¦R
R wall R insulation R surface
• And again, the resistance of the wall can be neglected so
that
¦R
R insulation R surface
Session 17.29
Industrial Insulation
• Rinsulation is given by the general formula for R.
R insulation
t
k
• In general, R is proportional to the thickness of the
substance resisting the heat flow.
Session 17.30
k* for common materials
Tm is mean temperature of Thot & Tsurf. P/C is pipe covering.
*B.L. Capehart, et. al., Guide to Energy Management, 5th ed., Fairmont Press, 2003, p. 401.
Session 17.31
Industrial Insulation
• Personnel safety is often a concern.
• Determine insulation thickness to bring the Touter ins
to the “safe touch value”
• 130 ” Tsafe touch, ºF ” 150*
• For a specified surface temperature, the thickness
of insulation needed is
t insulation
ª Th Touter insulation º
k insulation R surface «
»
T
T
«¬ outer insulation
air »
¼
*W. C. Turner, Energy Management Handbook, 5th Edition, Fairmont Press, 2005, p. 457.
Session 17.32
Industrial Insulation
• The North American Insulation Manufacturers
Association (NAIMA) has a useful insulation program, 3E
Plus
• 3E Plus calculates
– Insulation thicknesses
– Energy loss and gain
– Economic thicknesses
– CO2, NOx, and carbon equivalent (CE) reductions
• Useful software can be downloaded from
http://www.pipeinsulation.org
Session 17.33
Typical Pump System*
*Drawing Figure 1 from Energy Reduction in Pumps
and Pumping Systems, Hydraulic Institute. Photo
courtesy Oklahoma State University Industrial
Assessment Center.
Session 17.34
Pump and System Curves
Pump Curve (larger impeller)
Head, H
Pump Curve (smaller impeller)
Velocity and Friction Head
Elevation Head
Pressure Head
Volume Flow Rate, Q
Session 17.35
Pump Summary
1.
2.
20% Savings on Normal Systems
Keep Head to Minimum
a. minimize capacity
b. reduce process pressures
c. lower outlet tanks
d. use siphons
e. reduce nozzle velocities
f. use larger pipes
g. use lower loss fittings
h. eliminate throttle valves
Session 17.36
System Curve
Conclusions (cont.)
3.
Avoid excessive safety margins
4.
Select proper pump
5.
Utilize VSDs
6.
Utilize multiple pumps in parallel
7.
Use pumps as HPRT
Session 17.37
Conclusions (cont.)
8. Maintain, Maintain, Maintain!
Session 17.38
Bibliography
1. IMPROVING PUMPING SYSTEM PERFORMANCE, A
SOURCEBOOK FOR INDUSTRY, The Hydraulic Institute,
(DOE Motor Challenge Program), Parsippany, NJ, 1997
2. Chelette, Garry, “A Pump Primer”, ENGINEERED SYSTEMS,
August 2001, page 72
3. Rishel, James B., “Wire to Water Efficiency of Pumping
Systems” ASHRAE JOURNAL, page 40, April 2001
Session 17.39
Appendix
Session 17.40
Compressed Gas Systems
• Quantifying air leaks
– Compute the volume of all the receivers and
major air headers.
– Small air lines (under about 1 inch diameter)
can be neglected.
– Leaving some volume out makes the
calculation conservative.
Session 17.41
Compressed Gas Leaks
• Quantifying air leaks
– Apply the following formula to find standard
cubic feet per minute (SCFM) lost
R
V (P1 - P2 )
('T) 14.7
R
average leakage rate (Scfm)
V
P1
system volume (ft 3 )
initial pressure (psig)
P2
final pressure (psig)
'T
Session 17.42
time interval over which leaks are measured (minutes)
Compressed Gas Systems
• Quantifying air leaks
– Find the compressor specific efficiency in
BHP/SCFM from the curve on the next page.
– Then multiply
[R x specific efficiency x 0.746 kW/Hp x operating
hours x cost of energy (in $/kWh)]/Ș
to arrive at annual cost of leaks.
Session 17.43
Compressed Gas Systems
*Compressed Air Systems, Varigas Research, Inc., Timinium, Maryland,1984 DOEICSI 40520
– T2, page 61.
Session 17.44
Compressed Gas Systems
• Quantifying air leaks
– The figure is for 100 psig. Other pressures
require a correction factor.*
Pressure,
psig
Correction
factor
130
1.15
120
1.11
110
1.05
100
1.00
90
0.942
80
0.880
*Nutter, DW, Britton AJ, and Heffington WM, “Five Common Energy Conservation Projects in Small and Medium-Sized
Industrial Plants”, Fifteenth National Industrial Energy Technology Conference, Houston, Texas, Mar 24-25, 1993. pp 112-120.
Session 17.45
Compressed Gas Systems
• Quantifying air leaks
– Other methods include measuring
compressor run times and amp draws.
– Drawbacks to all these methods include lack
of location of leaks.
– One method includes identifying the locations
and trying to assign the size of an equivalent
round hole to the leaks.
– Then a chart such as the one on the next
page is applied.
Session 17.46
Compressed Gas Systems
• Quantifying air leaks
– Leak rate in SCFM through equivalent area
round holes*
*M.D. Oviatt and R.K. Miller, Industrial Pneumatic Systems, Fairmont Press, 1981, p. 64.
Session 17.47
Compressed Gas Systems
• Other energy conservation techniques (cont):
Relocate air intakes to cooler positions*
Air Intake Temp, °F
Power Savings**, %
30
7.5
50
3.8
70
0
90
(3.8)
110
(7.6)
*MD Oviatt and RK Miller, Industrial Pneumatic Systems, Fairmont Press, 1981, p. 49.
** Relative to 70ºF
Session 17.48
Compressed Gas Systems
• Other energy conservation techniques (cont):
– Recover heat for personnel comfort from air compressor
Heat recovered from air compressors*
Compressor Size, Hp
Heat Available, Btu/min
40
1870
100
4660
125
5830
150
6990
200
300
9330
14,000
400
18,700
*From Sullair screw compressor data
Session 17.49
Compressed Gas
• AIRMaster+
– Software tool sponsored by US Department of Energy
– Estimates potential savings from selected energy
efficiency measures and calculates paybacks
•
•
•
•
•
•
•
Reduce air leaks
Improve use efficiency
Reduce air pressure
Use unloading controls
Adjust set points
Reduce operating time
Add receiver volume
Session 5.3.50
Compressed Gas
• AIRMaster+ available for download from
http://www1.eere.energy.gov/industry/bestpractices/software.html
Session 5.3.51
Industrial Insulation Example
• Consider the wall of a mild steel tank with these
conditions based on measurements or
observations at a plant
– thickness = 0.5 inch
– finish = dull metal
– temperature of contents Th = 180 ºF
– outside air temperature Tair = 80 ºF
– air velocity = 0 (still air)
Session 17.52
Industrial Insulation Example
• Find these properties in tables
• Rsurface = 0.46 h·ft2·°F/Btu (for dull metal)
• Basic equation yields
Th Tair
ȈR
q
A
(180 80) o F
hr ˜ ft 2 ˜o F
0.46
Btu
220
Btu
hr ˜ ft 2
Session 17.53
Industrial Insulation Example
• Add 1 inch of fiberglass insulation with a 3/32 inch shiny
aluminum protective cover.
• The resistance of the aluminum will be negligible, just
as the resistance of the steel wall is negligible.
However, the shiny surface affects the surface film
coefficient
• R for the fiberglass (at the correct mean temp):
R insulation
t
k
1.0 inch
Btu ˜ inch
0.33
hr ˜ ft 2 ˜o F
Session 17.54
hr ˜ ft 2 ˜o C
3
Btu
Industrial Insulation Example
• Now Rinsulation= 3 hr·ft2·ºF/Btu
Rsurface = 0.9 hr·ft2·ºF/Btu (for shiny
aluminum where the surface temperature
will now be near that of the surrounding air)
• The basic equation yields
q
Th Tair
ȈR
(180 80) o F
hr ˜ ft 2 ˜o F
(3.0 0.9)
Btu
26
Btu
hr ˜ ft 2
Session 17.55
Industrial Insulation Example
• Savings = quninsulated-qinsulated = 194 Btu/hr·ft2
• If energy costs $8/million Btu, and the heat is
supplied with an efficiency of 80%, what are the
savings? The savings are $16.99/yr·ft2.
Session 17.56
Boilers and Steam System
Session 18.1
Temperature vs. Enthalpy
Properties of Steam
(@ 14.7 psia = 0.0 psig)
Hf
Latent Heat of
Vaporization (Work) Hfg
350
Temperature (F)
300
1192.6
180
250
1150.9
200
150
Hg
100
100
50
50
0
0
0
200
400
600
800
Enthalpy (h)
Session 18.2
1000
1200
1400
Boiler System
Session 18.3
Firetube Boilers
Session 18.4
Watertube Boilers
Session 18.5
Boiler System
Session 18.6
Balance Diagram
Session 18.7
Boiler and Fired System Survey
• Demand Analysis
• Combustion Optimization
• Waste Heat Recovery
–
–
–
–
Condensate Return
Blowdown
Recouperator
Economizer
• Steam Traps
Session 18.8
Combustion Analysis
Session 18.9
Session 6.2.10
Boiler Efficiency Calculation
• A check on the combustion system indicates a
stack temperature rise of 700°F and an
excess Oxygen level of 5.5%. What is the
impact of trimming the excess Oxygen to 3%?
% Savings
NewEfficiency OldEfficiency
NewEfficiency
Session 18.11
Session 6.2.12
Boiler Efficiency Calculation
• Result
% Savings
NewEfficiency OldEfficiency
NewEfficiency
% Savings
77 75
77
2.6%
That means the fuel consumption will be reduced by 2.6%
Now, If we add an Economizer to drop the stack
temperature rise to 500°F, how much more can we reduce
the fuel consumption?
Session 18.13
Session 6.2.14
Boiler Efficiency Calculation
• Result
% Savings
NewEfficiency OldEfficiency
NewEfficiency
% Savings
81 77
81
4.9%
The fuel consumption will be further reduced by 4.9%
Session 18.15
Waste Heat Recovery
1. Return more Condensate
2. Economizer preheats Make-up/Feed Water
3. Shell & Tube HX for Skimmer Blowdown
4. Recouperator preheats Combustion Air
5. Pressurize Condensate Return Tank
2
4
5
CR Tank
1
Feedwater
Tank
3
Blowdown
Session 18.16
Steam Traps – Inverted Bucket
Session 18.17
Steam Traps – Float & Thermostatic
Session 18.18
Steam Traps –Thermostatic
Condensate
Session 18.19
Steam Traps – Disc
Session 18.20
Steam
Steam Trap Maintenance
• Purposes:
– Reject Condensate
– Reject Non-condensables (Air)
– Hold Back Steam
• A poorly maintained system has 20 – 30% of
the traps failed open.
• Maintenance Methods:
–
–
–
–
Sight (watch the discharge)
Sound (listen to the operation)
Temperature (delta T across trap)
Conductivity
Session 18.21
Flash Steam Calculation
• Condensate at steam pressure has too much
heat to exist as saturated liquid at lower
pressures, causing a portion of the liquid to
“Flash” back to steam.
% Flash
H f steam Pr essure H f condensate Pr essure
H fg condensate Pr essure
Session 18.22
Flash Steam Calculation
• 1000 lb/hr of blowdown at 60 psia is sent to
an unpressurized tank. How much water is
lost?
% Flash
% Flash
H f 60 psia H f 14.7 psia
H fg 14.7 psia
262.2btu / lb 180.2btu / lb
970.3btu / lb
LostMass
0.0841000lb / hr 8.4%
84lb / hr
Session 18.23
Flash Steam Calculation
• If the tank is pressurized to 20 psia and the
steam is recovered, how much can be
produced?
% Flash
% Flash
H f 60 psia H f 20 psia
H fg 20 psia
262.2btu / lb 196.3btu / lb
960.1btu / lb
LostMass
0.0691000lb / hr Session 18.24
6.9%
69lb / hr
Steam Tables
Session 18.25
Session 6.2.26
Session 6.2.27
Combined Heat and Power,
Distributed Generation, and
Renewable Energy
Session 19.1
Distributed
Generation
Renewable
Energy
Combined
Heat and Power
Session 19.2
Combined Heat and Power
• Also known as:
– CHP
– Cogeneration
– Combined cooling, heating and power (CCHP)
– Building cooling, heating and power (BCHP)
• Definition:
– Simultaneous production and use of useful mechanical and
useful thermal energy
– Mechanical energy is frequently used to turn a generator
producing electrical energy
– Thermal energy can be used to generate cooling (i.e., absorption
chiller)
Session 19.3
Why Cogeneration?
• CHP has the opportunity to:
– Improve system efficiency (as compared to typical
power generation without useful heat recovery)
– Reduce total operating costs (compared to
purchasing or generating electricity and heat energy
in separate systems)
– Improve system reliability and availability (when CHP
is used a primary and the utility systems are used as
a back-up source)
Session 19.4
CHP Energy Balance
Source: EPA Combined Heat and Power Partnership (www.epa.gov/chp)
Session 19.5
Types of Cycles
• Three primary types of cycles:
– Topping cycle
– Bottoming cycle
– Combined cycle (which is usually a dual topping
cycle)
• Why is the type of cycle important?
– Regulations apply differently based on type of cycle
Session 19.6
Topping Cycle
• Primary energy first produces mechanical
energy and residual thermal energy is recovered
and used
– Example 1: High-pressure boiler steam is used to
power a turbine. The resulting shaft power turns a
motor or generator. In addition, steam out of the
turbine provides useful heat energy to a process.
– Example 2: Diesel engine turns a generator producing
electric power. Waste heat recovery is applied to the
exhaust gas and engine coolant producing useful hot
water.
Session 19.7
Topping Cycle Example 1
Intermediate-Pressure
Steam
High-Pressure
Steam
Steam
Turbine
Generator
Process/
Heating
Very Low-Pressure
Steam
Steam
Generator
Condenser
Low-Pressure
Steam
DA
Feedwater Tank
Condensate
Return
Make-Up Water
Session 19.8
Topping Cycle Example 2
Return
IC Engine
Useful Heat Supply
Radiator
Engine
Coolant
Heat
Recovery
Generator
Return
Exhaust Gas
Heat
Recovery
Exhaust Gas
Useful Heat Supply
Session 19.9
Bottoming Cycle
• Primary energy first satisfies a thermal demand,
such as a furnace, and residual thermal energy
is recovered and used to produce useful
mechanical or electrical power.
– Example: A large combustion process, such as a heat
treating furnace, where the exhaust is used in a waste
heat boiler to develop steam that is used to power a
turbine. The resulting shaft power turns a motor or
generator.
Session 19.10
Bottoming Cycle Example
High-Pressure
Steam
Steam
Turbine
Process/
Heating
Generator
Very LowPressure Steam
Steam
Generator
or
Furnace
Low-Pressure
Steam
Condenser
Condensate
Return
DA
Feedwater Tank
Make-up Water
Session 19.11
Combined Cycle
• Cycle produces useful mechanical energy at two
different stages within the process. Residual thermal
energy is utilized at least once in the process.
– Example: A combustion (gas) turbine creates shaft power, which
powers a generator (a topping cycle). The exhaust gas (perhaps
with supplemental firing) is used in a waste heat recovery boiler
to develop steam, which is used to power a steam turbine. The
shaft power from the turbine is used to power a motor or
generator (normally a topping cycle but used here as a
bottoming cycle). In addition, steam out of the turbine, either as
extraction or back-pressure steam, provides useful heat energy
to a process. (the useful thermal energy makes this CHP).
Session 19.12
Combined Cycle Example
Steam
Turbine
High-Pressure
Steam
Fuel
Compressed
Air
Burner
Exhaust
Gases
Generator
Intermediate-Pressure
Steam
Waste heat
Boiler
Process/
Heating
Generator
Air
Low Pressure
Steam
Gas Turbine
Condensate
Return
DA
Feedwater Tank
Make-up Water
Session 19.13
PURPA
• Public Utility Regulatory Policy Act (PURPA)
gave qualifying facilities (QF), for both CHP and
small power producers (SPP), certain regulatory
advantages if they met certain efficiency and
ownership requirements.
• Regulatory advantages granted
– QFs and SPPs not regulated as a utility
– Utility must buy excess electricity (at the utility’s
avoided cost)—repealed with EPACT 2005 if access
is available to competitive wholesale market.
– Utility must provide back-up power (at a nondiscriminatory rate) )—repealed with EPACT 2005 if
access if appropriate market access is available.
Session 19.14
Other EPACT 2005 PURPA Changes
• FERC is to establish new standards that
prevent QFs from being built and operated
as “PURPA machines.”
• These new standards will require that the
thermal and electric output be used
fundamentally for industrial, commercial or
institutional purposes; and not primarily for
sale to electric utilities.
Session 19.15
Intent of PURPA
• Promote and support highly efficient
systems (CHP)
• Promote and support renewable energy
systems (SPP)
• People still argue advantages and
limitations of PURPA
Session 19.16
Sample Problem
• A CHP facility consumes 81,900 Mcf/yr natural gas
[Assume gas HHV is 1,050 Btu/ft3 and the cost is
$10/Mcf]
• The CHP generates 6,000,000 kWh/yr plus 300,000
therms/yr of useful heat energy.
• Electricity generated offsets electricity that would be
purchased at $0.08/kWh
• Useful thermal energy recovered offsets a gas-fired
boiler. Assume gas cost is the same as above and the
net boiler efficiency is 80%.
• Estimate the net reduction in annual energy cost
delivered by the CHP.
Solution to Sample Problem
• CHP Gas Consumed
= - (81,900 Mcf/yr)*($10/Mcf)
= - $819,000/yr
• Electricity Cost Offset
=(+6,000,000 kWh/yr)*($0.08/kWh)
=$480,000/yr
• Boiler Gas Cost Offset
=(+300,000 therm/yr)*(100,000 Btu/therm)
*(1 ft3/1050 Btu)*(1 Mcf/1000 ft3)*($10/Mcf)
/(.80)
=$357,143/yr
• Net Reduction in Energy Cost
=(-$819,000/yr)+($480,000/yr)+($357,143/yr)
=$18,143/yr
Other Items
• CHP is (typically) a type of DG
• CHP can work well with
– District heating systems
– Thermal energy storage systems
– Gas cooling systems
• Energy security and surety issues are
giving CHP and DG more justification
Session 19.19
Distributed Generation
• Also known as:
– Distributed Generation (DG)
– Distributed Energy (DE)
– Distributed Energy Resources (DER), although DER can include
more that DG (flywheels, batteries, etc.)
– Self generation
• Definition: (and there are several)
– Any small-scale power generation that provides electric power at
a site closer to the end user than central generation, and is
usually interconnected to the distribution system or directly to the
end user’s facility [Reference 2]
– Any method of producing power that will be used on or near the
site at which it is generated [Reference 5]
Session 19.20
DG Technologies
• Internal combustion engines
– Fuel can include natural gas, diesel, biogas, gasoline, propane,
and more
– Available in sizes typically from 30 kW to 3,000 kW. Some
systems are available as low as 1 kW for home energy systems,
including CHP.
– Efficiencies up to 37% electric, over 80% when heat recovery
added
– Basic equipment costs around $300 to $600/kW, without CHP
• Combustion turbines
– Fuels are typically gas (natural gas, biogas, etc.) but liquid
systems are available
– Most efficient systems are greater than 40 MW but systems as
low as 500 kW are available
Session 19.21
DG Technologies
• Wind-Powered Generators
– Large systems (600 kW and above) are becoming cost effective
in select locations
– Smaller systems (2 kW to 500 kW) are commercially available
but more expensive
– Wind is an intermittent source, so another power source is
frequently required for a stable supply
• Photovoltaics (PV)
– Still very expensive but the cost continues to come down.
Equipment around $10,000/kW today
– PV is an intermediate source, so another power source (i.e.,
battery) is frequently required for a stable supply
– Can be cost effective in remote locations
Session 19.22
DG Technologies
• Microturbines
– Fuels are typically gas but liquid systems are being developed
– Size range is limited because technology is still being developed. 30
kW, 60 kW, 70 kW and 250 kW systems are available
– Electrical efficiency is low but emission levels are attractive
• Fuel Cells
– Fuel cells use a chemical reaction rather than a combustion process.
They require hydrogen as a fuel source.
– Fuel processors extract hydrogen from other fuels
– Emission levels are excellent because of non combustion reaction
– Technology is still developmental and very (very) expensive
– Fuel processor, maintenance costs, and fuel cell stack life are current
concerns
Session 19.23
Renewable Energy
• Definition: Energy that comes from a
renewable source
• What is a renewable energy source?
• Renewable energy is energy from natural
resources, which are naturally replenished
in the short term, typically within a year or
so.
Session 19.24
Renewable Energy
• The definition gets political
– High-head hydro, which can disrupt stream flows and
fish habitat, is frequently excluded from the definition.
– Ground-source heat pumps, which consume
conventional electric energy but can be more efficient
because of heat sink/source temperatures, are
frequently included.
– Biomass, or the burning of agricultural products,
increases local emissions, but is included because we
assume the emissions (CO2 and mineral ash) support
the growth of new agricultural products.
Session 19.25
Renewable Energy-Electric
• Photovoltaic (fixed or tracking)
• Wind-power generators
– Horizontal axis
– Vertical axis
• Hydropower
– High-head
– Low-head and “kinetic” hydropower
• Ocean Energy
– Surface wave or wave column
– Tidal and current power
Session 19.26
Renewable EnergyElectric or Thermal
• Concentrating solar thermal
– Tower or dish, usually tracking
• Dish Stirling
• Geothermal (usable heat from below ground)
• Biomass and bagasse
• Waste-to-energy
• Landfill gas
Session 19.27
Renewable-Thermal
• Solar thermal panels
• Concentrating solar thermal
• Transpired solar collectors (solar air preheaters)
• Thermal mass systems (Trombe wall)
• Ocean energy
– Thermal gradient
– Ocean thermal energy conversion (OTEC) uses the
temperature difference that exists between deep and
shallow waters to run a heat engine
Session 19.28
Net Metering
• In general, the utility bills you for the “net” energy
consumed.
• This means that excess electricity generated is valued at
the retail price, provided you are a net consumer of
electricity (not a net generator).
• Any excess energy you generate goes into the electric
grid and creates a “credit” for future energy consumed.
• “Net” may be defined as a billing period (monthly) or
annually, depending on the utility.
Session 19.29
Net Zero Energy
• Several organizations have the goal of developing net
zero energy buildings
• Net zero energy buildings are highly efficient but still
consume energy
• Energy needs are met through self generation and
interconnection to the utility grid and utilize net metering
• “Net” zero is typically defined on an annual basis
Session 19.30
Net Zero Energy
• Efficiency is still “job one”
• Reducing energy requirements through energy efficiency
is generally less expensive than renewable energy
• Make the building as efficient as possible until renewable
energy resources become cost effective
• General rule of thumb: 75% EE & 25% RE
Session 19.31
Power-Purchase Agreements
Popular for Renewable Energy
• 3rd Party finances project installation
• 3rd Party sells you the solar energy produced on your
site (at a known price) for 15-25 years.
– They like it because it will likely payback for them in
10 years or less.
• You get “green” power and a known future energy cost
(lower risk)
Session 19.32
References
1.
Kowlanowski, Bernard F. Small Scale Cogeneration Handbook,
The Fairmont Press, Inc. Atlanta, GA, 2000.
2.
McKinley, Sarah, “Untapped,” Energy Decisions, JanuaryFebruary 2000, pages 34-38.
3.
Parks, William, et al. Reliable and Economic Natural Gas
Distributed Generation Technologies, US Department of Energy,
Washington DC.
4.
Petursson, Gestur, Reducing Operating Costs Through On-Site
Generation of Electricity, Working Paper, Oklahoma Industrial
Assessment Center, Oklahoma State University, Stillwater, OK.
5.
Sturdevant, Nicole, “Getting On Track with On-Site Power,”
Building Operating Management, July 2000, pages 79-88.
6.
Wong, Jorge B. and Kovacik, John M., “Cogeneration,” Energy
Management Handbook (Chapter 12), 5th edition, The Fairmont
Press, Inc., Atlanta, GA.
References
7.
Landreth, Michael, “On-Site Power Generation: Items to
Consider,” Proceedings Strategic Energy Forum, May 18, 2000.
8.
Blazewicz, Stan and Walker, Stow, “Distributed Generation: What
Will it Take to Deliver Grid Reliability?” Power Value, July-August
2000, page 12.
9.
Gas Research Institute, “Natural Gas-Fueled Reciprocating
Engines; Fastest-growing Prime Movers for Distributed
Generation,” Natural Gas Application in Industry, GRI.
10.
US Department of Energy-Federal Energy Management Program,
“Using Distributed Energy Resources—A How-to Guide for
Federal Energy Managers,” Cogeneration and Competitive Power
Journal, Vol. 17, No. 4, The Fairmont Press, Atlanta, GA, Fall
2002, pages 37-68.
Summary Table of Typical Cost and Performance Characteristics by CHP Technology1
Steam Turbine2
Recip. Engine
Gas Turbine
Microturbine
Fuel Cell
Power efficiency (HHV)
15-38%
22-40%
22-36%
18-27%
30-63%
Overall efficiency (HHV)
80%
70-80%
70-75%
65-75%
55-80%
Effective electrical efficiency
75%
70-80%
50-70%
50-70%
55-80%
Typical capacity (MWe)
0.5-250
0..01-5
0.5-250
0.03-0.25
0.005-2
Typical power to heat ratio
0.1-0.3
0.5-1
0.5-2
0.4-0.7
1-2
ok
ok
poor
ok
good
430-1,100
1,100-2,200
970-1,300
(5-40 MW)
2,400-3,000
5,000-6,500
0.032-0.038
Technology
Part-load
CHP Installed costs ($/kWe)
O&M costs ($/kWhe)
Availability
Hours to overhauls
Start-up time
<0.005
0.009-0.022
0.004-0.011
0.012-0.025
near 100%
92-97%
90-98%
90-98%
>95%
>50,000
25,000-50,000
25,000-50,000
20,000-40,000
32,000-64,000
1 hr - 1 day
10 sec
10 min - 1 hr
60 sec
3 hrs - 2 days
Fuel pressure (psig)
n/a
1-45
100-500
(compressor)
50-80
(compressor)
0.5-45
Fuels
all
natural gas, biogas,
propane, landfill gas
natural gas, biogas,
propane, oil
natural gas, biogas,
propane, oil
hydrogen, natural gas,
propane, methanol
Noise
Uses for thermal output
Power Density (kW/m2)
high
High
Moderate
moderate
low
LP-HP steam
hot water, LP steam
heat, hot water, LPHP steam
heat, hot water, LP
steam
hot water, LP-HP steam
>100
35-50
20-500
5-70
5-20
NOx ( lb/MMBtu)
(not including SCR)
Gas 0.1-.2
Wood 0.2-.5
Coal 0.3-1.2
0.013 rich burn 3-way cat.
0.17 lean burn
0.036-0.05
0.015-0.036
0.0025-.0040
lb/MWhTotalOutput
(not including SCR)
Gas 0.4-0.8
Wood 0.9-1.4
Coal 1.2-5.0.
0.06 rich burn 3-way cat.
0.8 lean burn
0.17-0.25
0.08-0.20
0.011-0.016
1. Data are illustrative values for typically available systems. All costs are in 2007 dollars.
2. For steam turbine, not the entire boiler package.
Source: Catalog of CHP Technologies, EPA, 2008.
35
Performance Comparisons
Technology
Size Range
(kW)
Installed Cost
($/kW)
(2)
Heat Rate
(Btu/kWhe)
Approx.
Efficiency
(%)
Variable
O&M
($/kWh)
Diesel Engine
1-10,000
350-800
7,800
45
0.025
0.017
1.7
Natural Gas Engine
1-5,000
450-1,100
9,700
35
0.025
0.0059
0.97
Natural Gas Engine w/CHP (3)
1-5,000
575-1,225
9,700
35
0.027
0.0059
0.97
Dual-Fuel Engine
1-10,000
625-1,000
9,200
37
0.023
0.01
1.2
15-60
950-1,700
12,200
28
0.014
0.00049
1.19
Microturbine
Microturbine w/CHP (3)
Emissions (1)
(lb/kWh)
NOx
CO2
15-60
1,100-1,850
12,200
28
0.014
0.00049
1.19
Combustion Turbine
300-10,000
550-1,700
11,000
31
0.024
0.0012
1.15
Combustion Turbine w/CHP (3)
300-10,000
700-2,100
11,000
31
0.024
0.0012
1.15
100-250
5,500++
6,850
50
0.01-0.05
0.000015
0.85
Photovoltaic
0.01-8
8,000-13,000
--
N/A
0.002
0.0
0.0
Wind Turbine
0.2-5,000
1,000-3,000
--
N/A
0.010
0.0
0.0
Battery
1-1,000
1,100-1,300
--
70
0.010
(4)
(4)
Flywheel
2-1,600
400
--
70
0.004
(4)
(4)
SMES
750-5,000
600
--
70
0.02
(4)
(4)
Hybrid System
1-10,000
(6)
(5)
(5)
(5)
(5)
(5)
Fuel Cell
(1) Nationwide utility averages for emissions from generating plants are 0.0035 lb/kWh of NOx and 1.32 lb/kWh of CO2.
(2) The high end of the range indicates costs with NOx controls for the most severe emissions limits (internal combustion technologies only).
(3) Although the electric conversion efficiency of the prime mover does not change much, CHP significantly improves the fuel utilization efficiency of a DER
system.
(4) Storage devices have virtually no emissions at the point of use. However, the emissions associated with the production of the stored energy will be those
from the generation source.
(5) Same as generation technology selected.
(6) Add cost of component technologies.
SOURCE; DOE-FEMP. Reference 10
Comparison of DG Technologies
Comparison
Factor
Diesel
Engine
Gas Engine
Simple Cycle
Gas Turbine
Microturbine
Fuel Cell
Photovoltaic
Product
Availability
Commercial
Commercial
Commercial
1999-2005
1996-2010
Commercial
Size Range
(kW/unit)
20 to
10,000+
50 to 5,000+
1,000 to
30,000
20 to 200
50 to 1000+
1+
200 to 2,000
300 to 3,000
1,000 to
10,000
20 to 100
50 to 200
1 to 5
Efficiency
(HHV)
36 to 43%
28 to 42%
21 to 40%
25 to 30%
35 to 54%
n.a.
Genset
Package Cost
($/kW)
125 to 300
250 to 600
300 to 600
300 to 600
1,500 to 3,000
n.a.
Turnkey CostWith no heat
recovery ($/kW)
350 to 500
600 to 1,000
650 to 900
650 to 900
1,900 to 3,500
5,000 to
10,000
Heat Recovery
Added Cost
($/kW)
100 to 200
75 to 150
100 to 200
75 to 350
Included
n.a.
0015 to
0.010
0.007 to 0.015
0.003 to 0.008
0.005 to 0.010
0.005 to 0.010
0.001 to 0.004
Typical DG
Range (kW/unit)
O&M Cost
($/kWh)
Source: Gas Research Institute (2000)
Comparison of CHP Technologies