6.5 ENERGY CONSERVATION TECHNOLOGIES FOR POWER
machines with a total demand of 400 kW should be used for one 8 h shift. The company has chosen to limit the use of the machines by changing one shift to two shifts. What could be the total cost change in the next bill? The electricity billing structure is
Customer cost: $21.00/month
•
Energy cost: $0.05/kWh
•
Demand cost: $6.50/kW month
•
Demand ratchet: 65%
•
Power factor: 75%
•
Taxes: 8%
•
Fuel adjustment: 1.15 ¢/kWh
•
The highest corrected demand in the previous year: 500 kW
•
Solution 6.3 1. Demand charge:
The ratchet clause is Ratch = 500 × 0.65 = 325 kW
From Equation 6.5, the billed demand prior to shift change is
Base power factor (0.8) Billed demand before change = Actual demand ×
Actual power factor
= 400 × 0.8 = 427 kW 0.75
The actually billed demand before change = max (427 kW, 325 kW) = 427 kW
Base power factor (0.8) Billed demand after change = Actual demand ×
Actual power factor
= 200 × 0.8 = 213 kW 0.75
The actually billed demand after change = max (213 kW, 325 kW) = 325 kW The saving due to demand charge is S1 = (427 – 325) × 6.50 = 663 $/month 2. Electricity consumption costs:
There is no change in energy consumption. The total cost of electricity is 400 kW × 8 h/day × 30 day/month × ($0.05/kWh + $0.0115/kWh) = 5904 $/month
S2 = 0
3. Total savings are
S = S1 + S2 = 663 $/month After tax
S′ = 663 × (1 + 0.08) = 716 $/month
This shift change may also reduce the ratchet clause next year.
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6.5.2 POWER FACTOR IMPROVEMENT
Induction motors are often the major cause of low power factor in food processing facilities as they are the main electrical loads. Power factor improvement can increase the facility capacity and reduce any power factor penalty from utility com-panies. Power factor improvement can improve the voltage characteristics of the supply and reduce power losses in distribution lines and transformers. A normal watt-hour meter shows only the real power delivered to the system, irrespective of power factor. If the electricity is billed by a watt-hour meter, improvement of power factor may not reduce the utility bill directly but it may reduce the bill indirectly because implicit energy losses due to a low power factor occur in utility lines and transformers.
Example 6.4
A 230 volt electrical distribution system has a lagging power factor of 0.7. If the power factor has to be increased to a lagging power factor of 0.9, what size capac-itor should be used to correct a 10 kW load?
Solution 6.4
Using Equation 6.7, the apparent power before installing the capacitor is kW 10 [kW]
kVA = = = 14.3 kVA
pf 0.7
Using Equation 6.6, the reactive power before installing the capacitor is
( ) ( )2− 2 2− 2
kVAR = kVA kW = 14.3 10 = 10.22 kVAR The apparent power after installing the capacitor is
′ kW 10 [kW]
kVA = = = 11.1 kVA
pf 0.9
The reactive power after installing the capacitor is thus
( )
( )′ 2 2 2 2
kVAR = kVA¢ - kW 11.1-10 = 4.82 kVAR
The decrease in reactive power caused by the addition of a capacitor is the difference between the reactive powers before and after installing the capaci-tor, which is
ΔkVAR = 10.22 4.82 = 5.4 kVAR−
Using Equation 6.3, the size of the capacitor should be
( ) Ω
Δ
2 2
c
230 V
= E = = 9.8
kVAR 5400 VAR X
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6.5.3 REPLACEMENTWITH HIGH-EFFICIENCY MOTORS
Replacement of a standard motor with an energy effi cient motor can decrease energy consumption. Potential economic savings, S, through improved effi ciency can be calculated by
l h
100 100 0.746 hp
S L C N ⎛ ⎞
= × × × × ×⎜⎝ h − h ⎟⎠ (6.9)
where
hp is the horsepower of the motor (hp) L is the percentage of rated load C is the electricity price ($/kWh) N is the life expectancy (h)
hl and hh are effi ciency of a low-effi ciency motor and high-effi ciency motor, respectively
Example 6.5
An old 5 hp electrical motor has an effi ciency of 77% and 40,000 h life expectancy.
If the electricity price is $0.05/kWh and a high-effi ciency motor at an effi ciency of 85% is used as an alternative, what could be the cost saving through the whole life of the motor? Suppose that the motor operates at 100% of the rated load.
Solution 6.5
Using Equation 6.9, the replacement will save
⎛ ⎞
= × × × × ×⎜⎝ − ⎟⎠
⎛ ⎞
= × × × × ×⎜⎝ − ⎟⎠
=
l h
100 100 0.746 hp
100 100 0.746 5 100% 0.05 40,000
77 85
$912
S L C N
h h
If the loads are sensitive to the motor speed, a correction factor of ⎛ ⎞
⎜ ⎟
⎝ ⎠
3 ee std
RPM should be used to correct the expected savings. RPM
6.5.4 REPLACEMENTWITH ELECTRONIC ADJUSTABLE SPEED MOTORS
In most motor installations, motors are sized to provide the maximum power output required. If the rotational speed is constant at its maximum value to provide the maxi-mum designed load, the power input to the motor remains constant at the maximaxi-mum value. However, if the load decreases, signifi cant energy savings can be achieved if the rotational speed of the motor is decreased to match the load requirement of the device driven by the motor. The Affi nity law is used to determine the performance of fans and blowers at different rotational speeds. There are three rules of the Affi nity law to determine the fan capacity, pressure produced by the fan, and the horsepower required to drive the fan with a change in the fan rotational speed:
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2 2
1 1
RPM RPM V
V =
(6.10)
2
2 2
1 1
RPM RPM P
P
⎛ ⎞
= ⎜⎝ ⎟⎠ (6.11)
3
2 2
1 1
RPM RPM W
W
⎛ ⎞
= ⎜⎝ ⎟⎠ (6.12)
It can be seen from Equation 6.12 that the power required to drive a load such as a fan and a pump is proportional to the cube of the rotational speed. A small change in motor speed can cause a signifi cant change in energy consumption.
Therefore, energy conservation can be achieved with energy effi cient motor retrofi ts.
Variable speed drive can change the speed of the motor by changing the voltage and frequency of the electricity supplied to the motor based on the load requirement.
This is accomplished by converting the AC to DC and then by inverting the DC to a synthetic AC output with controlled voltage and frequency based on various switch-ing mechanisms. Variable frequency drives change the speed of a motor by changswitch-ing the voltage and frequency of the power supplied to the motor (Hordeski, 2003).
A variable frequency drive motor can result in energy savings by eliminating throt-tling and frictional losses affi liated with mechanical or electromechanical adjustable speed technologies.
Example 6.6
A fan, which has a power input of 15 kW at the rated speed of 1800 RPM, is used to supply 1000 m3/min air for an air blast chiller at an operating speed of 1750 RPM. A new energy effi cient motor, which has a power output of 15 kW at the rated speed of 1800 RPM, is considered as an alternative. The new motor operates at 1790 RPM speed. Can this energy effi cient motor save energy?
Solution 6.6
The new motor will increase the volumetric fl ow rate of air supply:
= ×
= ×
=
2 1 2
1
3
RPM RPM 1000 1790
1750 1023 m /min
V V
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⎛ ⎞
= × ⎜⎝ ⎟⎠
⎛ ⎞
= × ⎜⎝ ⎟⎠
=
3 2
2 1
1 3
RPM RPM 15 1790
1750 16.05 kW
W W
Therefore, the replacement of a standard motor with an energy effi cient motor in the fan can result in a 7% increase in energy consumption if the energy effi cient motor operates at a higher RPM. However, the energy effi cient motor at a higher RPM can supply more air at a fl ow rate of 1023 m3/min than 1000 m3/ min for the old motor.
Example 6.7
A 400 kW compressor is used to supply air. If the pressure of the air can be reduced to one-half of the initial value, what could be the energy savings for 1 year? Sup-pose that the operating time of the compressor is 4000 h per year and the electric-ity price is $0.05/kWh.
Solution 6.7
According to Equation 6.11, the ratio of the rotational speed of the compressor before and after pressure reduction is
=⎛ ⎞⎜ ⎟⎝ ⎠ = =
1
2 0.5
2 2
1 1
RPM 0.5 0.707
RPM P P
The power ration is thus
⎛ ⎞
= ⎜⎝ ⎟⎠ = =
3
3
2 2
1 1
RPM 0.707 35.4%
RPM W
W
Therefore, the total electricity savings would be
=35.4%×400×4000×0.05=2832 $/year S
However, the actual savings would be much lower than this theoretical value.