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Demand and Manufacturing
Process Improvement
DOE Industrial Assessment Center (IAC)
Guorong Chen
07/2012
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Agenda
1. Electricity Usage and Charge
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1.1 Demand Charge
1.2 Power factor
1.3 Load factor
1.4 Load Shifting
1.5 Heating/Cooling Degree Days
2. Manufacturing Process Improvement
– 2.1 Reduce Setup/Changeover Time
– 2.2 Waiting for More Unfinished Goods
– 2.3 Production System Improvement
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Total
Consumption
1. Electricity Usage and Charge (for
example)
1,521,686 kWh Customer Demand
5,475.80 kW
$1.007/kW
On-Peak
Consumption
595,915 kWh
Customer Demand
Rate
Off-Peak
Consumption
925771 kWh
On-Peak Demand
Rate
$10.882/kW
On-Peak Demand
On-Peak Energy
4,419,800 kW
Rate
$0.06985/kWh
Off-Peak Demand
3,767,000 kW
Off-Peak Energy
Rate
$0.04974/kWh
On Peak Period
10am - 10pm
Heating Degree
Days
1,142
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1.1 Demand Charge
• "Demand": total amount of electricity being used by a consumer at
any one time, measured in kilowatts (kW).
• If a time-of-use rate is applied, there will be two peak demands -- one
for on peak and one for off peak.
• The on-peak demand time, 10am – 10pm in the example, may vary
by season. Electricity used outside of these time periods, before
10am or after 10pm in this example, is considered off peak.
• Peak demand: the highest average kW measured in a 15-minute
interval during the period.
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1.1 Demand Charge (cont.)
How does the on-peak demand affect the customer’s electric bill?
1. A customer runs a 50 horsepower (hp) pump for only five hours
from 10am to 3pm during this month:
Demand Charge = 50 hp x .746 kW/hp x $10.882/kW = $405.90
Energy Charge = 50 hp x .746 kW/hp x 5 Hr x $0.06985/kWh =
$6.34
2. Same customer runs a 50 hp pump constantly through the entire
month:
Demand Charge = 50 hp x .746 kW/hp x $10.882/KW = $405.90
Energy Charge = 50 hp x .746 kW/hp x (372 Hr x
$0.06985/kWh+372 Hr x $0.04974/kWh) = $1,659.38
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1.1 Demand Charge (cont.)
Energy companies reserve the right on how to calculate the average demand.
Peak demand for this 45-minute period will be 1,600 kW, not 1,800 kW.
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1.1 Demand Charge (cont.)
• It is guarantee to reduce demand only when the electric devices run
continuously during on peak hours on every work day, since the
peak demand will occur at any time and any day.
• Electric devices, such as light bulbs, run 10 hours during on-peak
hours (there are 12 on-peak hours in our example)?
• How about if they run 24 hours a day but only 4 days a week
(assume the plant works 5 days a week)?
• Demand may be reduced, but not guaranteed.
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1.1 Demand Charge (cont.)
There are many ways to manage demand, ranging from manual controls
and time-clocks to sophisticated automatic units that program buildings
and processes.
• Motor is the correct size for your pump. An oversized motor could
increase demand and cost money. A newer, more efficient pump and
motor combination may be available that would save on demand and
energy.
• Pump and motor combination is the correct size for your system. If you
are using a 75 hp pump when a 50 hp pump would do the job, you are
wasting 25 hp or 18.7 kW of demand. (25 hp x .746 kW/hp = 18.7 kW)
• Worn pumps, motors, nozzles, and leaks are not increasing flow to the
point where demand has increased.
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•
1.1 Demand Charge (cont.)
Considerable savings can be realized by monitoring power use
and turning off or reducing non-essential loads during high power
use periods.
• Maximum Demand Controller is a device designed to meet the
need of industries conscious of the value of load management.
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1.1 Demand Charge (cont.)
Alarm is sounded when demand approaches a preset value.
If corrective action is not taken, the controller switches off non-essential loads in
a logical sequence.
This sequence is predetermined by the user and is programmed jointly by the
user and the supplier of the device.
The plant equipment selected for the load management are stopped and
restarted as per the desired load profile.
Demand control scheme is implemented by using suitable control contactors.
Audio and visual annunciations could also be used.
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1.1 Demand Charge (cont.)
How much will demand charge increase if the1,800 kW continues until
10:45am?
The peak demand will be 1,800 kW rather than 1,600 kW.
The peak demand increase will be 200 kW and monthly demand charge
will increase by 200×10.882=$2,176.40.
In another word, turning off non-essential machines for 7.5 minutes can
reduce demand charge by $2,176 in one month.
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1.1 Demand Charge (cont.)
• Be aware of when your meter is read each month. If you only run your
pump for a day or two during a billing period (for example, at the
beginning or end of the year) your demand cost could far exceed your
energy cost. If possible, run equipment that is operated infrequently
during a billing month when you’ve already incurred a demand charge
for that month.
For example, you want to test your irrigation system in the spring,
instead of waiting until you need it on a hot day in July, only to discover
that it's not running properly. You know that your meter is read on the
25th of each month. If you haven't used the system during the first part
of the month, you may want to wait until after the meter is read for the
month to test the system. While you’re testing your system, and running
your pump motor, you will be billed for the energy used by the irrigation
system, plus a demand charge for the entire month. By waiting a few
days, you could move that demand charge into a month when you'll be
using the irrigation system anyway, saving an extra month's demand
charges.
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1.1 Demand Charge (cont.)
• For the average commercial building, the best control over electrical
demand may not be in the electric system, but in the building itself.
• Good thermal design-tight construction, good window design, and
appropriately sized ventilation systems- is the key to limiting demand
and avoiding demand charges.
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1.2 Power Factor
Power factor: a unitless quantity, the ratio of the real power flowing to
the load to the apparent power in the circuit.
To understand power factor, visualize a horse pulling a railroad car
down a railroad track.
Because of the angle of the horse’s pull, not all of the horse’s effort is
used to move the car down the track. The sideways pull of the horse is
wasted effort or nonworking (reactive) power.
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1.2 Power Factor (cont.)
Because of the costs of larger equipment and wasted energy,
electrical utilities will usually charge a higher cost to industrial or
commercial customers if power factor is less than a specific level,
such as 0.95.
Uncorrected power factor will cause power losses in distribution
system, which may result in voltage drop. Excessive voltage drops
can cause overheating and premature failure of motors and other
inductive equipment.
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1.2 Power Factor (cont.)
• Causes that may be attributed for low power factor:
1. Inductive loads. Especially lightly loaded induction motors, and
transformers.
2. Induction Furnaces
3. Arc Lamps and arc furnaces with reactors.
4. Fault limiting reactors
5. High Voltage
• Opportunities to improve power factor:
1. Add consumers of reactive power in the system like Capacitors or
Synchronous Motors.
2. Using fully loading induction motors and transformers, and higher rpm
machines.
3. Using automatic tap changing system in transformers.
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1.2 Power Factor (cont.)
Various types of automatic power factor controls are available with
relay/microprocessor logic.
Two of the most common controls are: Voltage Control and kVAr Control.
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1.3 Load Factor
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1.4 Load Shifting
Load shifting involves moving load from on-peak to off-peak periods.
Popular applications include: use of storage water heaters, storage
space heaters, cool storage and customer load shifts to take
advantage of time-of-use or other special rates.
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1.4 Load Shifting
For example, one of the
frequently considered
examples in IAC assessments
is to charge forklift batteries
during off-peak hours.
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1.5 Heating/Cooling Degree Days
Degree days are essentially a simplified representation of outside airtemperature data.
"Heating degree days" are a measure of how much (in degrees), and for
how long (in days), outside air temperature was lower than a specific
"base temperature" (or "balance point"). They are used for calculations
relating to the energy consumption required to heat buildings.
"Cooling degree days“ are a measure of how much (in degrees), and for
how long (in days), outside air temperature was higher than a specific
base temperature. They are used for calculations relating to the energy
consumption required to cool buildings.
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2. Manufacturing Process Improvement
IAC assessments could be more effective if additional efforts were put
into direct productivity issues. Implementation of most ARs is expected
to be immediate.
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2.1 Reduce Setup/Changeover Time
Example:
After installing a rotating nozzle carousel, a
company is expected to save $60,750 per
year while the implementation cost is merely
$4,200 and payback period is one month.
The company makes metering pumps about
15 pumps an hour with sales $85,000,000.
The pump models differ by size and the
nozzle has to be changed and repositioned
every time there is a change of a pump size,
which takes an operator about 5 minutes. The
fix of the problem consisted of rotary mounting
hardware making it adaptable to different
types of pump cylinders.
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2.1 Reduce Setup/Changeover Time (cont.)
ACS = Annual Cost Savings = Net Sales Increase
ACS = (# of pieces/hr) x (value per piece - material cost per piece) x
(time saved) x (# of changes/yr)
ACS = (15 pieces/hr) x ($950/piece - $500/piece) x (4.5 minutes) x
(1/60 hr/min) x 10 x 12
ACS = $60,750/year
Implementation requires the manufacture and installation of the nozzle
carousel ($1,200). The cost of material was estimated at $800 and
labor at $400. It was recommended to buy an extra set of nozzles as a
backup in case of failure of existing ones, which are estimated to be
$1,800. Total implementation cost will be $4,200.
There are other ways to reduce setup/changeover time, such as add
more operators to set up machines.
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2.2 Waiting for More Unfinished Goods
If a high power consumption process works faster than the upstream
process, it may benefit to wait for more unfinished goods from
upstream and turn it off or to a low power consumption state.
Example:
B (20 parts/hr)
Total:
150
parts
30
parts
A (25 parts/hr)
0
1.5
7.5
time
(hr)
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2.2 Waiting for More Unfinished Goods
(cont.)
Then 1.5 hours of A’s running time will be saved.
AES = 200×1.5×5×50 = 75,000 kWh
ACS = 75,000×0.05 = $3,750 (assume average electricity rate is
$0.05/kWh).
While machines can be idle, workers must not be because the cost of
manpower is generally far higher than the cost of amortizing machines.
Therefore the cost savings will be much higher if manpower saving is
also considered.
However, pay attention to the warm-up/set-up time, space to store
upstream parts, and feasibility of waiting in practice.
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2.3 Production System Improvement
In this production system, behavior
is affected by RANDOM events:
machine breakdowns.
The following is one basic type of
production system, where the
circles represent the producing
units and rectangles are the
material handling devices:
The producing units may be individual machines or operators.
The material handing devices may be boxes, or conveyors, or
automated guided vehicles, or even empty space.
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2.3.1 Analysis and Improvement
Using the mathematical model of a production system, the following
problems can be solved:
(1) Evaluation of production rate, throughput, work-in-process,
blockage, starvation, etc.
(2) Investigation of “what if” scenarios: how will the system
performance change if some parameters are changed?
(3) Direction for improvement: provide improvement recommendation
based on the consideration of more effectiveness, low
implementation cost, and feasibility.
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2.3.2 Steps of Modeling
(1) Layout investigation
(2) Structural modeling
(3) Machine parameter identification
(4) Buffer parameter identification
(5) Model validation
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2.3.3 Data to Collect
1. For all the machines, the duration of each occurrence of up time
and down time must be recorded during a sufficiently long period
of time.
2.
The number of parts produced during this period and the cycle
time of the machines.
3. The buffer capacity should also be estimated based, for instance,
on carrier size, the number of carriers, available space, conveyor
speed, etc.
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2.3.4 Model Validation
To validate the model, we can compare the production rate calculated
using the model with that measured on the factory floor.
If the error is within 5-6%, the model is considered validated.
Otherwise, the above process should be repeated anew.
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2.3.5 Case Study
(1) Automotive ignition coil processing system
(2) Model Validation
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2.3.5 Case Study (cont.)
(3) “What-IF” Scenario Analysis
(a) Eliminating starvations by pallets (sufficient pallets will be
provided to operation 1):
Conclusion: Since the improvement is just about 1%, the system
modification by adjusting the number of pallets is not an effective way
for continuous improvement.
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2.3.5 Case Study (cont.)
(b) Increasing buffer capacities
Conclusion: Increasing the capacity by 50% leads to about 6% - 7%
of throughput improvement, while further increases have practically no
effect. Thus, increasing (perhaps, only some of the) buffer capacities
may be an effective way for system improvement.
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2.3.5 Case Study (cont.)
(c) Increasing efficiency of m9-10
Conclusion: increasing the efficiency of machine m9-10 leads to a
reasonable improvement in system productivity.
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2.3.5 Case Study (cont.)
(d) Increase machine uptime or downtime
Performance with increased
machine uptime
Performance with decreased
machine downtime
Conclusion: decreasing Tdown is more effective than increasing Tup.
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2.3.6 PSE Toolbox
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Thank you!
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Reference
1. http://www.mge.com/about/electric/glossary.htm#l
2. http://www.wrcc.dri.edu/cgi-bin/cliGCStH.pl?casjos
3. Productivity Manual.
4. J. Li and S. M. Meerkov, Production Systems Engineering.
Springer, 2009.
5. en.wikipedia.org
6. http://www.advancedenergy.org/progressenergy/usage_
vs_demand.html
7. http://www.em-ea.org/Guide%20Books/book-3/Chapter%203.10ENERGY%20EFFICIENT%20TECHNOLOGIES.pdf
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Supplement
1. What is reactive demand?
For example, “this provision requires that a customer be billed for the
highest 15-minute integrated kVA of lagging reactive demand
established during the month less 1/3 of the highest 15-minute kW
demand during that month. In simpler terms, a customer is billed for
having a power factor that is less than 95% at the rate of $0.83/RkVA.”
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2. Power Factor Measurement
Power factor in a single-phase circuit (or balanced three-phase circuit)
can be measured with the wattmeter-ammeter-voltmeter method,
where the power in watts is divided by the product of measured
voltage and current. The power factor of a balanced polyphase circuit
is the same as that of any phase. The power factor of an unbalanced
polyphase circuit is not uniquely defined.
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3. What are inductive loads and capacitive loads?
Inductive loads such as transformers and motors (any type of wound
coil) consume reactive power with current waveform lagging the voltage.
Capacitive loads such as capacitor banks or buried cable generate
reactive power with current phase leading the voltage. Both types of
loads will absorb energy during part of the AC cycle, which is stored in
the device's magnetic or electric field, only to return this energy back to
the source during the rest of the cycle.
For example, to get 1 kW of real power, if the power factor is unity, 1
kVA of apparent power needs to be transferred (1 kW ÷ 1 = 1 kVA). At
low values of power factor, more apparent power needs to be
transferred to get the same real power. To get 1 kW of real power at 0.2
power factor, 5 kVA of apparent power needs to be transferred (1 kW ÷
0.2 = 5 kVA). This apparent power must be produced and transmitted to
the load in the conventional fashion, and is subject to the usual
distributed losses in the production and transmission processes.
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4. What is time-of-use pricing?
Time-of-use pricing (TOU pricing): electricity prices are set for a
specific time period on an advance or forward basis, typically not
changing more often than twice a year. Prices paid for energy
consumed during these periods are pre-established and known to
consumers in advance, allowing them to vary their usage in response
to such prices and manage their energy costs by shifting usage to a
lower cost period or reducing their consumption overall.
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5. How to calculate heating degree days?
Heating Degree Day units are computed as the difference between the
base temperature and the daily average temperature. (Base Temp. Daily Ave. Temp.)
Base temperature is the outside temperature above which a building
needs no heating. One unit is accumulated for each degree Fahrenheit
the average temperature is below the base temperature. Negative
numbers are discarded.
Example: If the day’s high temperature was 65 and the low
temperature was 31, the base 50 heating degree day units is 50 - ((65
+ 31) / 2) = 2. This is done for each day of the month and summed.
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6. What is Synchronous Motor?
A synchronous electric motor is an AC motor in which the rotation rate
of the shaft is synchronized with the frequency of the AC supply
current; the rotation period is exactly equal to an integral number of AC
cycles. Synchronous motors contain electromagnets on the stator of
the motor that create a magnetic field which rotates in time with the
oscillations of the line current. The rotor turns in step with this field, at
the same rate.
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Another way of saying this is that the motor does not rely on "slip"
under usual operating conditions, and as a result produces torque at
synchronous speed. Synchronous motors can be contrasted with
induction motors, which must slip in order to produce torque. The
speed of the synchronous motor is determined by the number of
magnetic poles and the line frequency.
Synchronous motors are available in sub-fractional self-excited sizes
to high-horsepower direct-current excited industrial sizes. In the
fractional horsepower range, most synchronous motors are used
where precise constant speed is required. In high-horsepower
industrial sizes, the synchronous motor provides two important
functions. First, it is a highly efficient means of converting AC energy to
work. Second, it can operate at leading or unity power factor and
thereby provide power-factor correction.
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7. More about Automatic Power Factor Controllers
• Voltage alone can be used as a source of
intelligence when the switched capacitors are applied
at point where the circuit voltage decreases as circuit
load increases. Generally, where they are applied
the voltage should decrease as circuit load increases
and the drop in voltage should be around 4 – 5 %
with increasing load.
Voltage is the most common type of intelligence
used in substation applications, when maintaining a
particular voltage is of prime importance. This type of
control is independent of load cycle. During light load
time and low source voltage, this may give leading
PF at the substation.
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• Kilovar sensitive controls are used at locations where the voltage
level is closely regulated and not available as a control variable. The
capacitors can be switched to respond to a decreasing power factor
as a result of change in system loading. This type of control can also
be used to avoid penalty on low power factor by adding capacitors in
steps as the system power factor begins to lag behind the desired
value. Kilovar control requires two inputs - current and voltage from
the incoming feeder, which are fed to the PF correction mechanism,
either the microprocessor or the relay.
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• Automatic Power Factor Control Relay controls the power factor of
the installation by giving signals to switch on or off power factor
correction capacitors. Relay is the brain of control circuit and needs
contactors of appropriate rating for switching on/off the capacitors.
There is a built-in power factor transducer, which measures the power
factor of the installation and converts it to a DC voltage of appropriate
polarity. This is compared with a reference voltage, which can be set
by means of a knob calibrated in terms of power factor. When the
power factor falls below setting, the capacitors are switched on in
sequence. The relays are provided with First in First out (FIFO) and
First in Last Out (FILO) sequence. The capacitors controlled by the
relay must be of the same rating and they are switched on/off in linear
sequence. To prevent over correction hunting, a dead band is
provided. This setting determines the range of phase angle over
which the relay does not respond; only when the PF goes beyond this
range, the relay acts.
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When the load is low, the effect of the capacitors is more pronounced
and may lead to hunting. Under current blocking (low current cut out)
shuts off the relay, switching off all capacitors one by one in
sequence, when load current is below setting. Special timing
sequences ensure that capacitors are fully discharged before they are
switched in. This avoids dangerous over voltage transient. The solid
state indicating lamps (LEDS) display various functions the operator
should know and also and indicate each capacitor switching stage.
• Intelligent Power Factor Controller (IPFC) determines the rating of
capacitance connected in each step during the first hour of its
operation and stores them in memory. Based on this measurement,
the IPFC switches on the most appropriate steps, thus eliminating the
hunting problems normally associated with capacitor switching.
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General Rules of Thumb:
• The average cost of electricity is $0.05/kWh ($15/MMBtu).
• There are 2000 hours per year per shift (based on the assumption
that one shift is 8 hours per day, 5 days per week, 50 weeks per
year).
• Switching from electric heat to natural gas or #2 fuel oil can reduce
heating costs by 78%.
• Average cost savings for demand reduction: By shifting an
operation to off-peak hours, the following savings are
achieved: $75/Hp/year.
• The average benefit of shifting other electrical equipment to offpeak hours is: $120/kW/year.