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Heat Loss from Electrical and Control Equipment in Industrial Plants: Part llResults and Comparisons
Warren N. White, Ph.D. Anil Pahwa, Ph.D. Chris Cruz
ABSTRACT
Industrial plants use electrical power equipment to distribute power for lighting, driving motorized devices, oper ating HVAC equipment, and control of equipment. The main focus of this paper is to provide updated information on heat losses by various types of electricpower equipment. The infor mation is organized by equipment type, andpracticalguidance on using this information to compute losses under dzfferent conditions is provided. The effect of loading margin used by designers in sizing the electric equipment, load diversity, and ambient temperature on heat loss is discussed. Uncertainties in the results for different pieces of equipment are presented. Also, a comparison of the results to the previouslypublished information is provided,
INTRODUCTION
Engineers wanting to estimate heat loss to the surround ing environment from electrical power and control equipment in industrial plants and large buildings need updated informa tion. This paper provides updated heat loss information on mediumvoltage (5 to 15 kV) and lowvoltage (below 5 kV) power devices for HVAC load calculations. The equipment covered includes both power and lighting transformers, mediumvoltage switchgear, electric cables and cable trays, motor control centers and combination motor starters, invert ers, battery chargers, lowvoltage circuit breakers, electric motors, unit substations, series reactors, and adjustablespeed or variablefrequency drives (ASD or VFD). The first part of this paper describes the types and varieties of information sources for equipment heat loss, how tests were conducted, and the uncertainties associated with the gathered data. The second part of this paper reports the study results.
Helpful information to guide designers on using this updated information to compute heat loss is provided for each piece of equipment in the form of load diversity, design margin, and the effect of the ambient environmental temperature. Load diversity assigns a fraction to a piece of equipment subjected to partial duty over a period of time. This fraction is used to deter mine the average power dissipated by the device over a period of time, while the load varies in a routine fashion. The diversity fraction definition varies according to the equipment. The reason for this variation stems from the dissipation ofheat vary ing either linearly with or as the square ofthe load current. The definition of the diversity factor or fraction for each equipment category will be presented. Design margin accounts for unan ticipated increase in demand and opportunity for future growth. The level of design margin assigned by the engineer ranges from 100% for welldefined, noncritical applications to 50% for conceptual designs and highly critical applications. Typi cally margins of 80% are used for most applications, providing a balance between reliability and initial costs. For example, if a piece of equipment was expected to require X amps with a 50% margin, the equipment would be selected for a maximum capacity of 2X amps (XJ0.5).
Uncertainties in the results are discussed for each piece of equipment. Many of these uncertainties stem from differences from manufacturer to manufacturer of the same type of equip ment. A comparison of the final results to previously published results is provided in the form of tables or graphs of data. The previously published data include Rubin (1979) and McDonald and Hickok (1985).
        
Warren N. White is an associate professor and Chris Cruz is a graduate student in the Mechanical and Nuclear Engineering Department, and Anil Pahwa is a professor in the Electrical and Computer Engineering Department, Kansas State University, Manhattan, Kans.
02004 ASHRAE.
Table I. General Purpose DryType Units Having an 80°C Temperature Rise
RESULTS
Temperature Rise ( O C )
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
80
Transformers
There are a variety of different transformer types. A small sample of the available data is presented here. Table 1 presents information concerning general purpose drytype units, with an 80°C temperature rise. Other units could have different temperature rises. Table 2 contains data concerning general purpose liquidfilled units. The fullload loss figures in Tables 1 and 2 correspond to rated current. The losses at any frac tional load can be determined by
Rated Voltage 0')
480D208Y
480D208Y
480D208Y
480D208Y
480D208Y
480D208Y
48OD208Y
480D208Y
480D208Y
480D208Y
480D208Y
480D208Y
480D208Y
15kD480Y
15kD480Y
15kD480Y
15kD480Y
15kD480Y
15kD480Y
Total losses = no load losses + load losses x (~m' , (1)
where LF = the load fraction, i.e., the fraction of fullload current
(between zero and one). Transformer losses are not a strong function of environ
mental temperature; thus, the fullload and noload losses can be considered as constant regardless of the ambient tenpera ture.
Those power and lighting transformers (and larger units) built and tested in accordance with the NEMA TP1 Standard (NEMA 1996) have maximum efficiencies that either exceed or meet those efficiencies shown in Table 3 at a given percent
KiloVolt Amps
15
25
3 0
37.5
45
50
7 5
1 12.5
150
225
300
500
750
5 00
750
1000
1500
2000
2500
age of load. For lowvoltage units (600, 208, 120 volts), the given load percentage for peak efficiency is 35%, while for mediumvoltage units, the load value for peak efficiency is 50%. The efficiencies of these drytype units are referred to an average winding rise temperature of 7S°C, while the liquid immersed efficiencies are referred to an average winding temperature rise of 85°C. Losses vary linearly with winding temperature. Referring the efficiencies to a particular winding temperature allows comparison between units. The tempera ture to which the losses are referred is listed at the top of Table 3.
Given the full capability of the unit in kVA, the fullload losses for the NEMA TP1 units are approximately
pf x kVA x 1000(1 &) Full load losses = watts , (2)
wn+,
Average No Load Losses
(w)
330
530
415
530
487.5
700
725
700
1075
1450
1650
2900
3640
2400
2800
3500
5000
6500
7200
where
pf = power factor,
LF = load fraction for peak efficiency (0.35 or 0.5), and
q = efficiency from Table 3 corresponding to kVA and unit
type.
ASHRAE Transactions: Symposia
Average Full Load Losses
(w)
277
502
616.5
67 1
963.5
1371
1969.5
2230
2136
2820.5
3279
4857
8572
5 000
9000
9600
11600
15500
18500
100% Margin Total Losses
(w)
607
1032
1032
1201
1451
207 1
2695
2930
3211
427 1
4929
7757
12212
7400
11800
13100
16600
22000
25700
80% Margin Total Losses
(w)
507
85 1
810
959
1104
1577
1985
2127
2442
3255
3749
6008
9126
5600
8560
9644
12424
16420
19040
50% Margin Total Losses
(w) 399
656
569
698
728
1043
1217
1258
1609
2155
2470
41 14
5783
3650
5050
5 900
7900
10375
11 825
Table 2. General Purpose LiquidFilled Units
Rated Voltage Total Losses Total Losses
Substation
15kD480Y 225 760 3400 4160 2936 1610
Table 3. NEMA TPI Efficienciesq
Efficiency % Efficiency % Dry Type, Medium Voltage, Liquid Immersed, Medium
KiloVoltAmps Dry Type, Low Voltage, 7S°C
15 97 96.8 98.0
The power factor in Equation 2 depends upon the electri cal load, e.g., 0.8 for motor loads or 1.0 for resistive loads.
The no load losses are approximately
100 No load losses =pf x kVA x IOOO(LF)(  1) 1 (3)
2  (LF) (load 1osses)watts .
The losses provided by Equations 2 and 3 correspond to the average winding temperature rises shown at the top of Table 3. The load losses for all drytype units (general purpose and TP 1) must be corrected for temperature according to Table 4. The load losses for liquid immersed units do not
require any temperature correction. The temperature correc tion consists of multiplying the load losses by the factor of
Temperature correction factor = ( TK + TREF) (TK + 75°C) '
where
TK = 234.5OC for copper windings and 225°C for aluminum windings, and
TREF = reference temperature shown in Table 4.
Once the temperature corrected load and no load losses are determined, Equation 1 can be used to calculate the losses at the given level of loading.
854 ASHRAE Transactions: Symposia
Table 4. Limits for Temperature Rises for DryType Units
Class (OC) Average Winding Temperature Rise (OC) TREF  Standard Reference Temperature (OC)
130 75 95
Diversity is only appliedto the loadlosses, whereas the no Medium Voltage Switchgear load losses remain constant as long as the unit is energized. The heat loss calculation for medium voltage switchgear Modifying Equation 1 to account for diversity provides the is based on a spreadsheet that provides a menu of the various relation, equipment devices found in them. The losses from medium
Average total losses = no load losses voltage circuit breakers (including enclosure effects) can be 2 (5) estimated for 15 kV breakers and 5 kV breakers as
+ load losses x (LF) x diversity factor.
In order to determine the diversity factor, the time average of the load fraction in the RMS (root mean square) sense has to be found by the relation,
where
TI = time spent operating with a low load fraction,
LFL = low load fraction,
T2 = time spent operating with a high load fraction,
LFH = high load fraction, and
LFAvE = average load fraction.
The diversity factor is then given by
Diversity factor = (LF~~~/LF)*, (7)
where
LF = load fraction corresponding to expected peak load of the unit (usually LFH).
The power loss, taking into account margin information, is
Power loss with margin = no load loss
+ load loss x ( ~ / 1 0 0 ) ~ , (8)
where
M = the margin.
The value of M is either 100% for full loading, 80% for expected (recommended) loading applications, or 50% for critical applications. Transformer literature states that power losses are only weakly influenced by the ambient temperature. The loss information presented came from manufacturers that followed relevant IEEE standards for measuring and reporting transformer losses. As a result, the uncertainty of the loss values reported by these manufacturers is +lo% or less.
15 kV breaker loss = 13800 volts x 1.73 x Irared
x (I/I,.~,,~)~ x pf x 0.00006 watts , (9)
5 kV breaker loss = 4160 volts x 1.73 x Iraled
x (~/l,,,~)~ x pf x 0.0001 watts , (10)
where
pf = power factor (nominally 0.9),
Irated = the rated line current for the breaker in amps,
(I/Irated) = the load fraction of the breaker.
The spreadsheet is shown in Figure 1. In order to account for diversity, the current flowing in a breaker or bus must be averaged using the following equation:
where
IL = low current flowing through breaker for time TI,
IH = high current flowing through breaker for time T2,
IAVE = average current in RMS sense.
A diversity factor can then be defined to be used to deter mine the current values to enter into the spreadsheet of Figure 1.
Diversity factor = IAVE/ZH (12)
The margins for medium voltage switchgear are 100% for fullload applications, 80% for commercial applications, and 50% for critical applications. These numbers were chosen as the result of discussions with industrial plant design engineers. The margin of 100% would indicate that equipment would be operated continuously at its rated capacity, whereas a margin of 80% would indicate that equipment is loaded to the point where 80% of the current capacity is utilized.
ASHRAE Transactions: Symposia 855
WATTS LOSS DATA MEDIUM VOLTAGE SWITCHGEAR
Enler /he icenr Welt loss fur quanlily in lhfs ilemr a p p a a in ~ulunrn (his column
LOAD CURRENT 1 I I I I
I I I I
T O T A l . W A T l ' S r l I T E M I I
Figure 1 Medium voltage switchgear spreadsheet.
856 ASHRAE Transactions: Symposia
Cables and Cable Trays
The bulk ofcable losses are the resistive losses in the main conductors. Thus, if the number of cables and their physical arrangement in the trays are known, total losses per foot at the room temperature of that specific cable tray can be computed. There are many combinations of loading, size, and packing of the cables in a tray. However, normal industry practice and the following simplifying assumptions make it possible to compute losses. Specific factors considered are:
1. A cable tray can have cables of only one voltage level at a time.
2. All of the cables in a tray are considered to be threephase cables and are of the same size.
3. Total height of the cable bundle does not exceed 3 in.
4. Only single layers are considered for 5 kV and 15 kV cables for all sizes because the height of the bundle even for the smallest size would exceed 3 in. Multiple layers are consid ered for 600 V cables.
5. A packing or fill factor of 40% is considered for the trays. In other words, the total area occupied by the cables in a layer does not exceed 40% of the total crosssectional area of that layer.
6. The cables are assumed to be stacked one on top of the other with airspace in between each stack.
7. The cables are sized such that normal fdlload current is 80% of the ampacity at 90°C.
8. A diversity factor of 60% is considered. This implies that the average current in a cable over a period of time is 60% of the fullload current.
9. Sheath and armor loss factor of 0 for 600 V cables, 5% for 5 kV cables, and 10% for 15 kV cables are considered.
10. An ambient temperature of 26°C is considered for calcula tions.
Table 5 shows average losses for trays of different size of the selected voltage levels. These losses are for the conditions specified above. If specific information, such as number of cables and their sizes, for a tray is not available, the loss values specified in Table 5 can be used. Results in Table 5 under the heading of KSU are results from the present study. Those results in Table 5 under the heading of Rubin are results from Rubin (1979).
If the number of cables and their sizes are known along with the size of the tray they are in and the voltage level, then a more accurate heat loss figure can be computed by summing up the product of the number of cables with the loss per cable from Table 6. Results in Table 6 under the heading of M & H come from McDonald and Hickok (1 985).
For comparison purposes, the values for cable tray loss in Table 5 were accompanied by losses reported by Rubin (1979). Table 6 compares individual cable loss to losses computed from resistance values for individual cables. These resistance values came from McDonald and Hickok (1985) where the loss per foot of cable length of a threephase (as indi cated by the factor of 3) cable bundle is
watts losslft = ( I x 0.48)~ x R11000 x 3 , (13)
where
I = rated current in amps, and
R = resistance value in ohms11000 ft.
The inclusion of diversity in the loss predictions is accom plished by multiplying the rated current by the diversity factor, representing the fraction of the time the rated current is flow ing through the cable. An ambient temperature of 32°C increased the losses by an average of 2.3% in comparison to the losses at 26°C. The differences in cables of different type but of the same size are mainly due to insulation thickness, presence of shield, and other constructionrelated details. Calculations performed to take these factors into effect showed that they do not have a significant impact on results. Overall changes in losses due to these variations are less than 10%.
Motor Control Centers
Motor control centers have losses associated with the components of their construction. These components include circuit breakers, motor starters, and horizontal and vertical buses. Since there are many combinations of these compo nents, a spreadsheet has been developed to determine the loss of the entire motor control center. Figure 2 shows an example of a motor control center consisting of five cabinet sections. The motor control center is fed through an 800 amp breaker. The main bus is arranged horizontally across the top of the
Table 5. Cable Tray Losses at Selected Voltages
ASHRAE Transactions: Symposia 857
Tray Size (in.)
6
12
18
24
3 0
KSU
600 V (Wlft)
10
23
3 6
47
6 1
Rubin
600 V (Wlft)

23
3 5
47
58
KSU
5 kV (Wlft)
4
8
12
16
22
Rubin
5 kV (Wlft)

26
3 9
53
65
KSU
15 kV (Wlft)
3
7
11
15
19
Rubin
15 kV (Wlft)

26
42
55
68
Table 6. Losses per Cable of Different Sizes at Selected Voltages for Three Conductor, Three Phase Cables
KSU
350 kcmil
750 kcmil
Cable Size
# 8 AWG
# 6 AWG
# 4 AWG
# 2 AWG
cabinets. In each cabinet, asecondarybus carries current verti cally to the various starters. Each cabinet is 20 in. wide and 72 in. high. These lengths are typical dimensions for motor control centers. In the example, the breaker is on the left. The horizontal bus in the cabinet immediately to the right of the breaker cabinet carries the entire current needed for all of the cabinets containing the motor starters. The horizontal bus in the center cabinet carries the current for the three cabinets on the right. It is seen that the current distributes itself among the cabinets as airflow would in a manifold. The current in the vertical bus also distributes itself among the compartments as air in a manifold. This distribution of the current is used in the example. Figure 2 shows each motor starter in the motor control center along with the size of the starter, the motor horsepower, the motor efficiency, the power factor of the load, and the diversity factor. The arrangement of the motor control center is a given fact for any application. The motor horse power, efficiency, and power factor together with the starter size are also given information. The diversity factor is a value that must be determined for the application. The diversity factor, cif; is the fraction that provides the RMS compartment current over a 24hour period when multiplied by the compart ment current (defined in Equation 11 and Equation 12).
Figure 3 shows the loss calculation. The calculation is organized according to cabinets. The "first cabinet" is the one on the far right in Figure 2. The cabinets are numbered in this example from right to left. The compartments in each cabinet are listed in order from top to bottom. Each row of the spread
I I I
M & H
sheet essentially performs the same calculation. Given the motor horsepower and efficiency, q, the kW ofpower supplied by the starter is
600 V (Wlft)
1.49
1.73
1.69
2.02
The compartment current flowing through the starter is
KSU
I = kW * 1000
amp line voltage * & * pf
600 V (Wlft)
1.63
1.90
1.93
2.28
The line voltage used in this example is 480 volt, three phase, andpf is the power factor. The riser current is deter mined by the relation,
M & H
riser current for this compartment = compartment current * df + riser current from compartment below, (16)
5 kV (Wlft)



2.38
where
df = diversity factor.
As one steps upward through the compartments of a cabi net, it is seen that the riser current increases. The vertical bus loss in the riser for each compartment is determined by
KSU
bus loss = (riser current /rated bus ~ u r r e n t ) ~ x rated bus loss per standard length x compartment height,
(17)
M & H
5 kV (Wlft)



2.64
where
rated bus current = 300 or 600 amps.
ASHRAE Transactions: Symposia
15 kV (Wlft)



2.68
15 kV (Wlft)



2.83
Figure 2 Motor control center example setup.
800 Anp Circuit Breaker
Note that the rated bus loss per standard length is listed in Figure 3. As an example, consider the losses in the vertical bus of the 35 hp starter in the second cabinet. The losses are (107.481300)~ x (50 watts16 ft) x 2 ft = 2.14 watts.
In calculating the starter loss, the diversity factor is applied to the current but not to the relay losses since the relay losses are present regardless of whether the motor is running or not. The starter loss would then be the relay loss plus R* (I* dn2, where R is the resistance of the starter circuit, I is the starter current in amps, anddf is the starter diversity factor. For the 80 hp starter in the first cabinet, the losses are 18.8 watts + 0.001488fl x (0.8 x 88.59 amps)' = 26.27 watts. For conve nience, the resistance value and height of each NEMA combi nation motor starter are also shown in Figure 3.
Blak
Blank
NEMb.3 45 hp m h pH.9 m. 8
NEMA5 2Ki hp m h pm.9 W . 8
To determine the losses in the cabinet, the riser and starter losses of each compartment are summed. The total current for the cabinet is also determined. In order to determine the hori zontal bus losses, the cabinets and cabinet currents are listed according to cabinet number. The horizontal bus current in the cabinet is calculated using the relation,
horizontal bus current of cabinet = cabinet riser current + horizontal bus current of previous cabinet. (18)
N W 1 5 h p
y85% pH9 dH.8 Blank
NEMA3 40hP f l h pm.9 w. 8
N W 5 150 hp y85% pm.9 el
As one steps through the cabinets, it is seen that the hori zontal bus current increases. Once the bus currents are deter mined, the losses are determined by the relation,
Bus loss = (bus current /rated bus currentl2 x rated bus loss per 20" cabinet width. (19)
N F M 1 4hp
W h pm.9 df1.0
W 2 20 143 q=%?h p f 4 . 9 &lo
NEMA3 35 hp 1@5% pm.9 df1.0
NEMA4 75 hp q=w?/o pm.9 dH.8
For instance, the horizontal bus loss of the first cabinet will be (87.67 amps 1800 amps12 x 40 watts = 0.48 watts.
All of the current flowing in the horizontal bus must also flow through the breaker. The breaker power loss is estimated and entered into the spreadsheet. The calculation of the low voltage breaker loss is presented in another section. The indi vidual losses are totaled to provide the power loss of the motor control center.
NEMA 1 3hp
@So/, pm.9 df1.0
NEM42 15 1p
m h pW.9 W . 8
Blank
NEM.44 mhp m h pm.9 m. 8
Sometimes the breaker might be placed in the middle of a set of cabinets. This avoids excessive horizontal bus currents and the corresponding losses. If this is the case, the losses of the cabinets on either side of the breaker can be determined separately and then added together. Adding the horizontal bus currents from each side together determines the breaker current.
Since the starter losses are based on 1 2 ~ ohmic heating, the means by which diversity is included is the same as the calculation of the RMS average current shown in Equation 10 for medium voltage switchgear.
The design margin for motor control centers is 80% in commercial applications, 100% in fullload applications, and 50% in critical application. However, since the loading of a given starter may not be indicative of the entire motor control center, the margin number should be used for motor control center bus work and breakers. No margin should be applied to the starter itself provided that a margin figure was used in determining the starter load. In the motor control example just presented, no margin information was applied to the starters since the starter loading was known. A margin of 80% was applied to the bus work and breaker in the example. Tests on NEMA size 1, 2, and 3 starters demonstrated that ambient temperature does not have a significant influence on heat losses. Loss expressions for the NEMA 1, 2, and 3 starters were derived from tests having an uncertainty of less than 10%.
Inverters
The peak of the inverter efficiency occurs at full load. From manufacturer data, the inverter efficiency remains close to the peak value when operated from about 50% to 100% full load. Below 50% full load, the efficiency falls almost expo nentially. The inclusion of diversity in the loss predictions is accomplished by multiplying the estimated losses occurring during normal operation by the diversity factor, representing the fraction of the time the inverter is driving the intended load. The margin for inverters in commercial applications is 100% and 50% in critical applications. No information has been located with regard to the influence of ambient temperature on inverter losses. The information presented here was obtained
ASHRAE Transactions: Symposia 859
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Table 7. Inverter Efficiencies and Power Losses
from manufacturer literature and Web sites. Table 7 provides Battery Chargers a list of single and threephase inverters with power loss spec It is important to realize how a battery charger is used. ified in both watts and Btdh as a hnction of the kVA rating of Batteries are usually used in supplying emergency power the inverter. The maximum incoming DC current determines during infrequent situations. Once the emergency is over, the the kVA of the inverter. batteries are recharged. During this recharging situation, the
DC Bus Volts
ASHRAE Transactions: Symposia 861
50
50
60
75
100
130
260
130
260
260
88  88
84
86
87
100%
Full Load Loss (w) 2007
2813
3516
5270
6556
762 1
8473
8966
11954
Margin
Full Load Loss (Btulh)
6850
9600
12000
17985
22375
26010
28920
30600
40800
Three Phase
23270
23270
31214
33346
40815
50%
Load Loss (w)
1004
1406
1758
2635
3278
3810
4237
4483
5977
kVA
10
15
20
3 0
40
5 0
60
75
100
6818 A 6818
9146
9770
11959 
DC Bus Volts
130
130
130
130
130
130
130
260
260
Efficiency
80
8 1
82
82
83
84
85
87
87
battery charger will experience heavy usage and the dissipated supplying the DC load. Including the diversity factor, the heat will be greatest during this time. During normal opera result of the previous equation is tion, the chargerwill be lightly loaded, maintaining the battery
Q = (QFL/2.5) * (1 + 1.5 * X * diversity factor) . (21) charge and possibly supplying some DC load. Thus, the usage norm for the charger could quite probably consist of a light The margin for battery chargers in commercial applica load. Other applications of battery chargers consist of fork tions is 100% and 50% in critical application. Thus, in lifts, telephone systems, and golf carts, where it is seen that the commercial application the loss figure obtained from Table 8 charger usage will vary over a 24hour period. The full load would not require any modification. If the load on the charger losses are roughly 2.5 times greater than the no load losses. can be estimated, then a smaller loss figure than the one The heat loss for fractional loading of the battery charger is provided by the table could be used in the HVAC heat load determined by calculation. The effect of environmental temperature is
unknown due to data coming solely from manufacturers and Q = no load loss + X * (QFL  no load loss) = (QFL/2.5) + X
no testing being performed. Table 8 is representative ofbattery * (QFL  (QFL'~.~)) = (QFL'~.~) * + * 3 (20) charger losses. There are many combinations of input and
where
Q = heat loss in watts at load fraction,
QFL = 111load heat loss, and
output voltage; Table 8 is a small sample of the data available. The battery charger loss presented in Table 8 was computed by
Watts loss = IBC x VB x (1  q1100) , (22)
X = load fraction of charger. where
If a diversity factor is included, this diversity factor should IBC = Output culTent,
be applied only to the fractional load loss and not to the no load VB = output voltage, and
loss. The diversity factor is the fraction the charger is actively q = battery charger efficiency, %.
Table 8. 60 Hz ThreePhase Battery Charger Comparison
862 ASHRAE Transactions: Symposia
Power Loss (Rubin)
Watts
720
1080
1440
720
1080
Input (Three Phase)
Volts
208
208
208
240
240
Amps
19.5
29.3
39.1
16.9
25.4
240
480
480
480
208
Output (DC) Efficiency
YO
q
88
8 8
8 8
8 8
8 8
33.9
8.5
12.7
16.9
39.1
Volts
24
24
24
24
24
Power Loss (KSU)
Watts
576
8 64
1152
576
8 64
Amps
200
300
400
200
3 00
24
24
24
24
48
1728
88
8 8
8 8
8 8
88
400
200
300
400
200
2880
1440
2160
2880
1440
2160
2880
208
240
240
240
480
480
480
1152
576
864
1152
1152
78.1
33.9
50.8
67.7
16.9
25.4
33.9
1080
1440
1440
48
48
48
48
48
48
48
400
200
300
400
200
300
400
8 8
88
8 8
88
88
88
8 8
 2304
1152
1728
2304
1152
1728
2304
Table 9. LowVoltage Breaker Losses and Resistance Values in Ohms
Rubin (1 979) used a relation similar to Equation 22 forthe commercial applications, 100% in fullload applications, and calculation of battery charger heat loss but added a safety 50% in critical applications. To determine the breaker current factor of 1.25. Rubin's relation for determining battery charger for the loss calculation, perform the calculation of heat loss was used in the calculation of the comparison numbers listed in Table 8 and is given as Breaker current = (%margin11 00)
x rating plug current value. Watts loss = 1.25 x IBC x VB x (1  ~l1100). (23)
(25)
The current value produced by Equation 25 is the value to LowVoltage Circuit Breakers be used in the loss calculation. Tests of the influence of
Watts Loss Watts Loss Resistance Resistance Frame Current (amps) wlo Enclosure WI Enclosure
225 60.0 90.0 0.001 1852 0.00 17778
250 7 1.1 73.6 0.001 1376 0.001 1776
Circuit breakers are segregated according to frame sizes. breaker heat loss have
The frame size is the maximum current the breaker is designed little correlation. In these tests, the environmental temperature
to pass safely in normal operation. A breaker in a given frame was varied from 25°C to 50°C. Values in Table 9
can be made to trip at lower current values by using a rating to 60, loo, 250, and 1200 frames
plug that conesponds to those limits, Table 9 shows a range of were measured, and the uncertainty of the are within
lowvoltage circuit breaker frame sizes. For each frame size, +lo% of the reported loss values. Data for all of the other
loss figures are provided for the cases of with and without an frames were obtained from manufacturer literature with the
enclosure. Measurements have shown that the breaker enclo of the 3000 amp frame breaker where the loss
sure significantly influences the amount ofthe heat loss. These were determined by between the 2000
loss figures represent the losses that occur when rated frame and 3200 amp For those breakers not the
current flows in the breaker, For smaller currents, the resis indicated losses are representative of the expected values. The
tance values presented in Table 9 are used to predict the heat breaker losses may vary with manufacturer. Figure shows
loss through an I'R calculation. the losses at different current levels in two frames for
400
600
800
1200
1600
2000
3000 pp
3200
4000
5000
~ ~ "
Watts loss = I2 x R
where . I = trip rating in amps and
R = resistance for the frame in ohms.
ppp
147
215
330
865
1000
1500
comparison
(24) Motors
All manufacturer data for motors were in the form of ef i  ciencies. Efficiency values were collected for polyphase elec
220.5
322.5
440
1080
1500
2250
 ~.
The inclusion of load diversity is accomplished by tric motors for horsepower ratings from 10 to 2000 hp. The performing the loss calculation with the RMS current data were collected for a large number of motor frames. In this determined by averaging the current over a 24houror document, all of the efficiencies for different motors of the perhaps longerperiod. The margin for breakers is 80% in same horsepower rating were averaged to get one efficiency
2250 3375 0.00025
ASHRAE Transactions: Symposia 863
0.00091 88
0.0005972
0.0005 156
0.0006007
0.0003906
0.000375
2400
3000
4700
0.0013781
0.0008958
0.0006875
0.00075 1 0.0005859
0.0005625
3600
4500
7050
0.0002344
0.0001875
0.0001 88
0.00035 16
0.0002813
0.000282
 225 Amp Frame  M&H  225 Amp Frame  KSU   .   . 225 Amp Frame  Rubin  600 Amp Frame  M&H 600 Amp Frame  KSU 600 Amp Frame  Rubin
Low Voltage Circuit Breaker Comparison
500 
400 
B 2
0 . 100 200 300 400 500 600
Heater Plug Amps
Figure 4 Lowvoltage circuit breaker comparison.
for each horsepower rating. Using the average efficiency, the rate of heat loss for the motor is
Watts loss = hp x 745.7 x (100/q) *(Iq/100) , (26)
where
Hp = delivered power, and
q = motor efficiency, %.
Equation 26 was used along with the average efficiency to compare data to McDonald and Hickok (1985).
Rubin (1979) used a slightly different equation and assumed all motors were 90% efficient. Rubin's loss equation is
Watts Loss = hp x 746 x (1  q/lOO). (27)
Diversity is accounted for by using the average (over time) input power delivered to the motor in the first part of Equation 1. The equation for input power is
Input power (watts) = hp x 745.7 x (100/q) . (28)
The margin for electric motors is 80% in commercial applications, 100% in fullload applications, and 50% in crit ical applications. To determine the heat loss, taking into consideration margin, the calculation is
Motor Heat Loss (watts) = (%margin/100) x Input power x (1  ~$100) x 745.7 wattskp. (29)
A variation of 20°K of the environmental temperature would not influence the conductor absolute temperature and,
thus, the conductor resistivity, significantly. Thus, the influ ence of environmental temperature on the motor losses is small. According to the pertinent IEEE standards (IEEE 1995, 1996), the instruments used in determining the efficiency of a motor must have an uncertainty of f0.2% of full scale or less. The standards describe many different ways of determining the motor efficiency, each of which having its own particular uncertainty. The efficiency averages are determined by aver aging the nominal efficiencies from several different manu facturers. The nominal efficiency represents the mean of a collection of identical motors (same frame and horsepower). This efficiency information is representative of the expected values. Efficiency values may change from manufacturer to manufacturer. Figure 5 compares heat loss data for motors ranging from 10 to 2000 horsepower.
Unit Substations
The unit substation can be thought of as lowvoltage switchgear that might include (in addition to a transformer) circuit breakers, current transformers, control power trans formers, auxiliary compartment, space heaters, circuit break ers, and highcurrent buses all arranged in a series of cabinets. To closely determine the power losses of such a device requires detailed knowledge of the construction, such as length of buses, losses of individual components, and loading information.
The most realistic way to estimate losses is to use a spreadsheet to model the unit substation. Figure 6 shows the spreadsheet. For the components included in the loss calcula
864 ASHRAE Transactions: Symposia
0 200 400 600 800 1000 1200 1400 1600 1800 2000
Motor Ratlng  HP
Heat Loss (Watts) M 8 H Heat Loss (Watts) KSU     . Heat Loss (Watts) Rubin
160000
140000 ..  . ..=
Figure 5 Electric motor comparison.
120000 
tion, the ability to include a partial load and enclosure effects is incorporated into the calculation. Figure 7 shows a sketch of the unit substation example. The lowvoltage switchgear consists of two 4000 amp main breakers, each having a 1600 amp feeder line and four 800 amp breakers. The current in each main incoming breaker is 2300 amps, while each of the 1600 amp feeders cany 1000 amps. The 800 amp breakers carry the currents indicated in the figure. Each set of four 800 amp breakers is supplied by the closest 4000 amp breaker. There are five instrument or auxiliary compartments. Figure 6 contains the loss calculation.
.' .'
Diversity is accounted for by using the RMS breaker and bus amperage determined by averaging over a 24houror longerperiod in the loss calculations. This calculation is shown in Equations 1 1 and 12. The margin for unit substations is 80% in commercial applications, 100% in fullload appli cations, and 50% in critical applications. However, in cases where the switchgear is double ended, such as the example in Figure 7, the margins are reduced by a factor of two.
$ 1 ooooo 1 .................T i .2 : 3
Reactors
Reactor power losses vary as the square of the current. Given the rated reactor winding temperature rise over room temperature (25"C), the reactor losses need to be corrected according to this value. For each voltage level, impedance, and current value, Table 10 lists the heat loss at 100% and 50% margins. The power loss calculation in watts for the reactor is
Power Loss = (reactor currentJrated current12 x Loss Value x temperature correction , (30)
where the temperature correction is given by
Temperature Correction = (rated T,,, + TJ(25 + T& , (31)
where
rated T ~ , = rated winding temperature rise (usually 115"C), and
Tk = 234.S°C for copper windings and 225°C for aluminium.
The loss value used in Equation 30 is obtained from Table 10 under the 100% margin heading for the corresponding rated current value and the appropriate voltage and impedance column. Diversity is accounted for by using the RMS reactor current obtained over a standard work period, e.g., day or week, in the loss calculations. If TI is the time the reactor is in use with current I, and T2 is the time that the reactor is not being used in the work period, then the RMS reactor current is
The margin for reactors in commercial applications is 100% and 50% in critical application. The values of heat loss for 50% margin were calculated using Equation 30 with a value of 0.5 substituted for the current ratio. Tests on reactors have shown that the reactor losses are neither a strong fimction of the enclosure nor the environmental temperature. All of the
ASHRAE Transactions: Symposia
WATTS LOSS DATA 600 V Switchgear
Figure 6 Unit substation example spreadsheet.
ASHRAE Transactions: Symposia
Figure 7 Unit substation example setup.
Table 10. Reactor Power Loss in Watts at Rated Current and Room Temperature
instruments
4000 amp Main
2300 amp
1600 amp feeder
1oooamP
ASHRAE Transactions: Symposia 867
Instr.
Instr.
Future
Future
800 amp breaker
350 amp
800 amp breaker 300 amp
800 amp breaker 300 amp
800 amp m e r
350 amp
Instruments
Fut ure
4000 amp Tie
800 amp breaker
350 amp
800 amp breaker
300 amp
800 amp breaker 300 amp
800 amp breaker
350 amp
Instruments
4000 amp Main
2300 amp
1600 amp feeder
OoO aV
... 240 Volt  K U  460 Volt  K U  600 Volt  K ~ U McDolald &%ickok /
/ /
/ /
/ /
/ /
 # / /
/ /
/ /
 .  1 .
/ /
/ /
/ /
/ /
/ / I . _ .   '
/ / _. _.. ..
/ / ..a ._.
Horsepower
Figure 8 Adjustablespeed drive comparison.
data appearing in Table 10 were obtained from manufacturers and averaged. Some of the data going into the averaging calcu lation were verified experimentally. In general, the agreement between the reported data points where testing was possible is well within +lo%.
AdjustableSpeed Drives
Losses in adjustablespeed drives 'vary linearly with current. For a given linetoline voltage, the rated current varies linearly with the rated horsepower. Relations were developed to describe the fullload power loss in watts in terms of rated horsepower for different voltage levels. These rela tions are
240 V: Prated = hprafed x 25.6234 + 276.073 watts , (33)
460 V: Prated = hprated x 13.45435 + 363.7949 watts, (34)
and
600 V: Prated = hprated x 14.5585 1  201.038 watts , (35)
where
Profed = fullload power loss at rated current and horsepower, and
hprakd = rated horsepower of ASD.
Equations 33 through 34 are curve fits of manufacturer data. These curve fits are valid for the horsepower range of 25 to 800 horsepower. If the current (horsepower) does not corre
spond to the rated current (horsepower), the power losses are then predicted by
= Prated (h~/h~rafed) = ('/'rated 3 (36)
where
P = power loss,
hp = actual horsepower,
I = actual current, and
Irofed = rated current.
In a given application, it is very likely that an ASD might only be used for a fraction of a standard workday or work week. The losses adjusted for diversity would be
= 'rated (h~/h~rated) F~~~ = Prated (Ilzrated) F ~ ~ ~ , (37)
where
FAsD = fraction of time ASD is in use over standard work period.
The margin for adjustablespeed drives in commercial applications is 100% and 50% in critical application. Thus, in commercial application, the loss figure obtained from Equa tions 33, 34, or 35 would not require any modification. No information has been found regarding the influence of ambient temperature on the power losses of ASD devices. Of the tests performed in the production of this document, the ambient
868 ASHRAE Transactions: Symposia
temperature was not a factor in the measured power loss. The data used to determine the slope and intercept values in Equa tions 33, 34, and 35 come from manufacturers' literature. It should be noted that the data from the different manufacturers are consistent and do exhibit a distinct linear trend. The infor mation presented is representative of expected losses, but it may vary from manufacturer to manufacturer. Of the ASD power loss tests performed in the production of this document, the measured values were consistent with Equation 34. Rubin (1979) had no data concerning adjustablespeed drives, so he was not included in Figure 8. McDonald and Hickok (1985) did provide information of percent efficiency for adjustable speed drives based solely on horsepower rating. This effi ciency was converted to watts loss by Equation 27, and all data are shown in Figure 8.
CONCLUSIONS
The ability to account for loading diversity and design margin are invaluable to the design engineer. It is unrealistic to assume every piece of equipment is going to operate at 100% load for 24 hours a day. This assumption could drastically overestimate the heat load in a mechanical room as a whole. Proper sizing of electrical equipment for particular applica tions to account for future growth and demand is also very beneficial. Knowing the heat loss for equipment that has higher capacity than needed presently ensures proper heat load calculations now and in the future.
Having data from direct measurements to compare to manufacturer data increases confidence in using manufacturer data to report losses for equipment sizes that were not tested due to time, money, or availability constraints. It is interesting to note that when comparing data to previously published results where the authors had access to equipment and testing facilities (McDonald and Hickok 1985), the results of the present study agree favorably with McDonald and Hickok. Since every piece of equipment compared could not be tested, it is not known that the favorable comparison would be the result in every case. Rubin reported that he used the results of Hickok (1978), which are almost identical to the published information of McDonald and Hickok (1 985), yet Rubin's loss tables show higher losses for the same equipment reported by Hickok.
ACKNOWLEDGMENTS
The authors would like to thank ASHRAE TC 9.2 for sponsoring this work and TC 9.1 and TC 9.8 for endorsing this effort. We are especially indebted to Mr. John Riley of Black and Veatch for serving as chair of the Project Monitoring Subcommittee of TC 9.2, for his guidance, and for his advice in the conduct of this investigation. Thanks are also in order for Mr. Deep Ghosh ofthe Southern Company, Mr. Dennis Wessel of Bacik, Karpinski Associates, Inc., and Mr. Dale Cagwin of Robson Lapina for serving on the Project Monitoring Subcommittee. We would also like to extend our thanks to the
chair of the TC 9.2 Research Committee, Mr. Wayne Lawton of Giffels Associates, for his interest and advice in this effort. It is difficult for the authors to express the full extent of their thanks to Dr. Gary Johnson, Professor Emeritus of Kansas State University, for his depth of knowledge and tremendous help and advice in this investigation. Finally, the authors would like to thank the following organizations that provided assistance, donated equipment for testing, andlor loaned us equipment for testing: ABB, Danfoss Graham, General Elec tric, Rockwell International, Stanion Wholesale Electric of Manhattan, KS, Tennessee Valley Authority, andU.S. Depart ment of Energy.
REFERENCES
Hickok, H.N. 1978. Energy losses in electrical power sys tems. IEEE Transactions on Industry Applications 14(5): 373387.
McDonald, W.J., and H.N. Hickok. 1985. Energy losses in electrical power systems. IEEE Transactions on Indus try Applications IA2 l(4): 8038 1 9.
Rubin, I.M. 1979. Heat losses from electrical equipment in generating stations. IEEE Transactions on Power Appa ratus and Systems PAS98(4): 11491 152.
IEEE. 1996. IEEE Standard 1121996, Test Procedure for Polyphase Indzlction Motors and Generators. New York: IEEE Press.
IEEE. 1996. IEEE Standard 1151995, Guide: Test Proce dures for Synchronous Machines. New York: IEEE Press.
NEMA. 1996. NEMA TPI, Guide for Determining Energy Eficiency for Distribution Transformers. Virginia: National Electrical Manufacturers Association.
DISCUSSION
Jim Elleson, University of Wisconsin, Madison, Wisc.: Equation 22 relates the battery charger losses to the efficiency and the product of output voltage and current, which is the output power. It appears to me that by the normal definition of efficiency, this equation is not correct. I would expect the equation to be
Loss = Output power x (1  eff / 100)
Loss = Input power x (100 / eff  1)
Could you comment on the derivation of Equation 22 and on your definitions of loss and efficiency? Warren N. White: I thank Mr. Elleson for his interest in this work and his question regarding Equation 22 which is, as stated in the paper,
Watts loss = IBC x VB x (1 ?/loo) (22)
where
IBC = output current
ASHRAE Transactions: Symposia
V, = output voltage Watts loss = Output power x ( I  effI100) 7 = battery charger efficiency  %.
Since IBC is the output current and V, is the output volt which is exactly the first of the two equations presented by Mr.
age, the product ofthe two quantities is the outputpower. If the Elleson in the statement of his question and, also, deemed
percent efficiency, q, is replaced by the symbol eff, then Equa correct by Mr. Elleson. Since our results agree, our definitions tion 22 can be written as of loss and efficiency concur.
ASHRAE Transactions: Symposia