5.4.1 ENERGY CONSERVATIONIN COMPRESSED AIR SYSTEMS
Compressed air is one of the major utilities in food processing facilities. The production of compressed air can be one of the most expensive processes in manufacturing facilities. There are several publications that describe the energy saving potential of compressed air systems (Talbott, 1993; Cerci et al., 1995; Risi, 1995; Terrell, 1999;
Kaya et al., 2002). The energy saving opportunities for compressed air can come from three main areas:
Compressors themselves
•
Compressed air distribution systems
•
Compressed air utilization units
•
Generally, energy saving for a compressed air system can be achieved by (Kaya et al., 2002)
Installing high-effi ciency motors (for compressors)
•
Reducing the average air inlet temperature by using outside air (for
•
compressors),
Repairing air leaks (for distribution systems)
•
Reducing compressor air pressure (for utilization units)
•
5.4.2 HIGH-EFFICIENCY MOTORS
Most industrial equipment in manufacturing facilities is powered by electric motors.
The electrical energy that a motor consumes to generate a specifi ed power output is inversely proportional to its effi ciency. Electric motors cannot completely convert the electrical energy consumed into mechanical energy. The ratio of the mechanical power supplied by a motor to the electrical power consumed during operation is called the effi ciency of the motor. Therefore, high-effi ciency motors cost less to operate than their standard counterparts. Motor effi ciencies range from about 70%
to over 96%.
Replacement of a standard motor with an energy effi cient motor can result in a decrease in energy consumption. Potential savings, S, through the improved effi ciency can be calculated by
l h
100 100 S= × × × ×p L C N ⎛⎜⎝ h − h ⎞⎟⎠
(5.1)
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where
p is the horsepower of the motor (kW) L is the percentage of full-specifi ed load C is the electricity price ($/kWh) N is the life expectancy (h)
hl and hh are the effi ciencies of a low-effi ciency motor and a high-effi ciency motor, respectively
Example 5.1
At a food processing facility, there is an 850 kW standard motor associated with a compressor to generate compressed air. The average price of electricity is 0.05 $/kWh.
The fraction of rated load is 0.9. At this load, the energy effi ciency of the motor is 77%.
The annual operating time is 7200 h/year. If a high-effi ciency motor at an effi ciency of 82% is used to replace the standard motor, what is the annual energy saving?
Solution 5.1
Using Equation 5.1, the annual energy saving with the high-effi ciency motor is
( )
l
⎛ ⎞ ⎛ ⎞
= × × × ×⎜⎝ − ⎟⎠= × × × × ×⎜⎝ − ⎟⎠
=
h
100 100 100 100
850 0.9 0.05 7200 1
77 82
$21,809/year
S p L C N
h h
The effi ciency of a motor depends on its load. To obtain optimal energy effi ciency, compressors should run at their full-specifi ed load. In addition, variable-speed motors can be used to meet varying air demands.
5.4.3 REPAIRINGOF AIR LEAKS
Air leaks are the greatest single cause of energy loss from a compressed air system in manufacturing facilities. The cost of compressed air leaks is the cost of the energy required to compress the lost air from the atmospheric pressure to the compressor operating pres-sure. Leaks often represent as much as 25% of the output of an industrial compressed air system (Terrell, 1999). Eliminating air leaks totally is impractical, and a leakage rate of 10% is considered acceptable in practice (Cerci et al., 1995). The cost of compressed air leaks increases exponentially with the increase in leak diameters, as shown in Figure 5.2.
When the ratio of the atmospheric pressure to the line pressure of compressed air is less than 0.5283, the air fl ow is considered to be choked (i.e., the air travels at the speed of sound). The volumetric fl ow rate of free air exiting the holes is dependent on the extent to which the fl ow is choked. The volumetric fl ow rate of free air under the choked condition, Vf, m3/h, exiting all the leaks of a given size can be calculated by (Cerci et al., 1995):
(
i)
l i 1 2 d 2f
3 l
NL 273.15 4
273.15
T P P C C C D
V C T
× + × × × × × π
= × + (5.2)
where
D is the leak diameter (mm) NL is the number of air leaks
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Tl and Ti are the average temperatures of the air in the pipeline and inlet point of the compressor (°C)
Pl and Pi are the pressures of the air in the pipeline and at the ambient (kPa) C1 is the isentropic sonic volumetric fl ow constant, C1 = 7.3587 m/sK0.5 C2 is the conversion constant, C2 = 3600 s/h
C3 is the conversion constant, C3 = 106 mm2/m2
Cd is the isentropic coeffi cient of discharge for square edges orifi ce, Cd = 0.8 The power loss from leaks can be estimated as the power required to compress the volume of air lost from the compressor inlet pressure, Pi, to the compressor discharge pressure, Pl. The power loss from a leak can be calculated by (Cerci et al., 1995)
( ) ( ) ( )
( ) (1 )i 2 f l i
a m
1 / 1 1
PL
k k N
P C V k k N P P
E E
− ×
⎡ ⎤
× × × − × ×⎣ − ⎦
= × (5.3)
where
PL is the power loss from a given air leak (kW) k is the specifi c heat ratio of air, k = 1.4 N is the number of compression stages
Pl and Pi are the compressor operating pressure and inlet air pressure, respectively (kPa)
Ea is the compressor isentropic effi ciency Em is the compressor motor effi ciency Example 5.2
In a food processing facility, fi ve air leaks are found and are estimated as 5 mm in diameter. The air compressor used in this facility operates at 800 kPa (absolute pressure) FIGURE 5.2 Power loss as a function of leak diameter at 600 kPa pressure. (Adapted from Kaya, D., Phelan, P., Chan, D., and Sarac, H.I., Int. J. Energy Res., 26, 837, 2002.
Copyright John Wiley & Son, Ltd. With permission.) 45
40 35 30 25
Power lost (kW)
20 15 10 5
0 1 3 5
Diameter of hole (mm)
10
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with 1 compression stage. The atmospheric pressure is 101.3 kPa (absolute pressure).
The average ambient temperature and compressed air temperature are 25°C and 32°C. Rotary screw compressor isentropic (adiabatic) effi ciency is 82% (Ea) and com-pressor motor effi ciency is 90%. Estimate the volumetric fl ow rate of air from the leaks and determine the power loss. If the compressor operates 24 h/day for 300 days per year and the electricity price is $0.05/kWh, what will be the energy saving if all leaks are repaired?
Solution 5.2
1. From Equation 5.2, the volumetric fl ow rate of air from all fi ve leaks is
( )
( )
( )
× + × × × × × π
= × +
× + × × × × × ×
= × +
=
2 2
1
l l i d
f
3 i
2 6
3
NL 273.15 4
273.15
5 273.15 32 800 101.3 7.3587 3600 0.8 3.14159 5 / 4 10 273.15 25
290.38 m /h
T P P C C C D
V C T
2. From Equation 5.3, the power loss is
( )
( )
( ) ( )( )
( )( )
( ) ( )− ×
− ×
⎛ ⎞ ⎡ ⎤
×⎝ ⎠× × − × ×⎢⎣ − ⎥⎦
= ×
⎡ ⎤
× × × − × ×⎣ − ⎦
= ×
=
1
i 2 f l i
a m
1.4 1 1.4 1
1 1 1
PL
101.3 1 3600 290.38 1.4 1.4 1 1 800 101.3 1
0.82 0.9 31.2 kW
k k N
P C V k k N P P
E E
3. The energy savings per year is
= × × ×
=
31.2 [kW] 24 [h/day] 300 [days/year] 0.05 $/kWh 11,232 $/year
S
Repairing of air leaks may involve replacement of couplings or hoses, replace-ment of seals around fi lters, shutting off air fl ow during break periods, or repair-ing breaks in lines. All these costs should be very low (e.g., $20/leak). Therefore, the payback period for the implementation cost is very short (Cerci et al., 1995).
5.4.4 REDUCED AIR PRESSURE
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 horsepower required to drive the fan with a change in the fan rotational speed, respectively:
2 2
1 1
RPM RPM V
V =
(5.4)
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2
2 2
1 1
RPM RPM P
P
⎛ ⎞
= ⎜⎝ ⎟⎠ (5.5)
3
2 2
1 1
RPM RPM W
W
⎛ ⎞
= ⎜⎝ ⎟⎠ (5.6)
A small change in motor speed can cause a signifi cant change in energy consumption according to Equation 5.6. Different tools and process unit operations may require compressed air at different pressures. Therefore, energy conservation can be achieved with energy effi cient motor retrofi ts.
Example 5.3
An 850 kW compressor is used to supply compressed air. Suppose the operating time is 7200 h/year and the electricity price is $0.05/kWh. If the pressure of the air can be reduced by 10% of the initial value, what would be the energy saving for 1 year?
Solution 5.3
According to Equation 5.5, if the air pressure is reduced by 10%, the rotational speed will be reduced to
⎛ ⎞ ( )
=⎜ ⎟⎝ ⎠ = =
1
2 0.5
2 2
1 1
RPM 0.9 0.95
RPM P P
From Equation 5.6, the power will be reduced to
⎛ ⎞
=⎜⎝ ⎟⎠ = =
3 2 3
1 1
RPM 0.95 85.4%
RPM W
W
Therefore, the total electricity savings would be
= −(1 85.4%) 850 7200 0.05× × × =44,676 $/year S
5.4.5 REDUCED AIR INLET TEMPERATURE
Compressors generate heat during operation. If they are located inside production facili-ties, the temperature of intake air drawn from the inside of a building is higher than that of outside air. The energy consumption increases with the increase in air intake tem-perature. Therefore, the air supplied directly from the outside of a building can reduce the energy consumption. The compressor work, W, is proportional to the absolute tem-perature of the intake air temtem-perature, which can be calculated by (Mull, 2001)
2 2
1 1
W T
W =T (5.7)
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According the above equation, if the air intake temperature increases by 10°C from the outside temperature of 25°C, the energy consumption of the compressor will increase by 3.35%.
For multi-stage compression, the heat generated by compression work may increase the air temperature to be as high as 205°C if no cooling unit is installed.
Cooling the air between stages can increase the density of the air and reduce the power required for compression (Mull, 2001).
5.4.6 WASTE-HEAT RECOVERY
Air compressors generate large amounts of heat as they compress air. Adiabatic compression of air to 690 kPa (gauge) results in air temperature between 175°C and 260°C. Approximately 80% of the energy used to compress air fi nally becomes heat stored in the compressed air. Heat can be recovered from the high-temperature compressed air. Waste-heat recovery is discussed in Chapter 8.