Heat Loss from Electrical and
Control Equipment in Industrial Plants:
Part ll-Results and Comparisons
Warren N. White, Ph.D.
Anil Pahwa, Ph.D.
ABSTRACT
Industrial plants use electrical power equipment to
distributepower for lighting, driving motorized devices, operating 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 information 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 resultsfor 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 surrounding environment from electrical power and control equipment
in industrial plants and large buildings need updated information. This paper provides updated heat loss information on
medium-voltage (5 to 15 kV) and low-voltage (below 5 kV)
power devices for HVAC load calculations. The equipment
covered includes both power and lighting transformers,
medium-voltage switchgear, electric cables and cable trays,
motor control centers and combination motor starters, inverters, battery chargers, low-voltage circuit breakers, electric
motors, unit substations, series reactors, and adjustable-speed
or variable-frequency 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.
-
-
-
-
-
-
-
-
Chris Cruz
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 determine 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 varying 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 unanticipated increase in demand and opportunity for future growth.
The level of design margin assigned by the engineer ranges
from 100% for well-defined, noncritical applications to 50%
for conceptual designs and highly critical applications. Typically 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 equipment. 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 Dry-Type Units Having an 80°C Temperature Rise
Average No
Load Losses
Average Full
Load Losses
100% Margin
Total Losses
80% Margin
Total Losses
50% Margin
Total Losses
(w)
(w)
(w)
(w)
(w)
330
277
607
507
399
25
530
502
1032
851
656
480D-208Y
30
415
616.5
1032
810
569
80
480D-208Y
37.5
530
671
1201
959
698
80
480D-208Y
45
487.5
963.5
1451
1104
728
80
480D-208Y
50
700
1371
2071
1577
1043
80
48OD-208Y
75
725
1969.5
2695
1985
1217
80
480D-208Y
1 12.5
700
2230
2930
2127
1258
80
480D-208Y
150
1075
2136
3211
2442
1609
80
480D-208Y
225
1450
2820.5
4271
3255
2155
80
480D-208Y
300
1650
3279
4929
3749
2470
80
480D-208Y
500
2900
4857
7757
6008
41 14
80
480D-208Y
750
3640
8572
12212
9126
5783
80
15kD-480Y
500
2400
5000
7400
5600
3650
80
15kD-480Y
750
2800
9000
11800
8560
5050
80
15kD-480Y
1000
3500
9600
13100
9644
5900
80
15kD-480Y
1500
5000
11600
16600
12424
7900
80
15kD-480Y
2000
6500
15500
22000
16420
10375
80
15kD-480Y
2500
7200
18500
25700
19040
11 825
Temperature
Rise ( O C )
Rated Voltage
0')
Kilo-VoltAmps
80
480D-208Y
15
80
480D-208Y
80
RESULTS
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 dry-type units, with
an 80°C temperature rise. Other units could have different
temperature rises. Table 2 contains data concerning general
purpose liquid-filled units. The full-load loss figures in Tables
1 and 2 correspond to rated current. The losses at any fractional load can be determined by
Total losses = no load losses
+ load losses x (~m',
(1)
where
LF = the load fraction, i.e., the fraction of full-load current
(between zero and one).
Transformer losses are not a strong function of environmental temperature; thus, the full-load and no-load losses can
be considered as constant regardless of the ambient tenperature.
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 percentASHRAE Transactions: Symposia
age of load. For low-voltage units (600, 208, 120 volts), the
given load percentage for peak efficiency is 35%, while for
medium-voltage units, the load value for peak efficiency is
50%. The efficiencies of these dry-type 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 temperature to which the losses are referred is listed at the top of Table
3.
Given the full capability of the unit in kVA, the full-load
losses for the NEMA TP1 units are approximately
pf x kVA x 1000(1-
Full load losses =
wn+,
&)
watts ,
(2)
where
pf
=
LF
=
power factor,
load fraction for peak efficiency (0.35 or 0.5), and
q
=
efficiency from Table 3 correspondingto kVA and unit
type.
Table 2.
General Purpose Liquid-Filled Units
Rated Voltage
Total Losses
Total Losses
2936
1610
Substation
15kD-480Y
225
760
Table 3.
3400
4160
NEMA TPI Efficiencies-q
Kilo-Volt-Amps
Efficiency %
Dry Type, Low Voltage, 7S°C
15
97
The power factor in Equation 2 depends upon the electrical load, e.g., 0.8 for motor loads or 1.0 for resistive loads.
Efficiency %
Dry Type, Medium Voltage,
Liquid Immersed, Medium
96.8
98.0
require any temperature correction. The temperature correction consists of multiplying the load losses by the factor of
The no load losses are approximately
Temperature correction factor =
No load losses =pf
2
- (LF)
x
kVA
x
100
IOOO(LF)(- - 1)
1
(3)
(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 dry-type units (general purpose
and TP 1) must be corrected for temperature according to
Table 4. The load losses for liquid immersed units do not
854
( 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.
ASHRAE Transactions: Symposia
Table 4.
Limits for Temperature Rises for Dry-Type Units
Class (OC) Average Winding Temperature Rise (OC) TREF- Standard Reference Temperature (OC)
130
75
Diversity is only appliedto the loadlosses, whereas the no
load losses remain constant as long as the unit is energized.
Modifying Equation 1 to account for diversity provides the
relation,
Average total losses
=
no load losses
2
+ load losses x (LF) x diversity factor.
(5)
95
Medium Voltage Switchgear
The heat loss calculation for medium voltage switchgear
is based on a spreadsheet that provides a menu of the various
equipment devices found in them. The losses from medium
voltage circuit breakers (including enclosure effects) can be
estimated for 15 kV breakers and 5 kV breakers as
15 kV breaker loss
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,
=
13800 volts x 1.73 x Irared
x (I/I,.~,,~)~x pf x 0.00006 watts ,
5 kV breaker loss
= 4160
(9)
volts x 1.73 x Iraled
x (~/l,,,~)~ x pf x 0.0001 watts ,
(10)
where
pf
where
TI
=
LFL
=
T2
=
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:
LFH = high load fraction, and
LFAvE = average load fraction.
The diversity factor is then given by
(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.
ASHRAE Transactions: Symposia
power factor (nominally 0.9),
Irated = the rated line current for the breaker in amps,
(I/Irated)= the load fraction of the breaker.
time spent operating with a low load fraction,
low load fraction,
time spent operating with a high load fraction,
Diversity factor = (LF~~~/LF)*,
=
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 determine 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
full-load 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.
855
WATTS LOSS DATA
MEDIUM VOLTAGE SWITCHGEAR
Enler /he icenr
quanlily in lhfs
~ulunrn
LOAD CURRENT
1
I
I
I
I
I
I
I
TOTAl.WATl'Srl
Welt lossfur
ilemr a p p a a in
(his column
I
ITEM
I
I
Figure 1 Medium voltage switchgear spreadsheet.
856
ASHRAE Transactions: Symposia
Cables and Cable Trays
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 (1985).
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 three-phase (as indicated by the factor of 3) cable bundle is
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 three-phase
cables and are of the same size.
3.
4.
Total height of the cable bundle does not exceed 3 in.
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 considered 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 cross-sectional 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 fdl-load 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 full-load current.
Sheath and armor loss factor of 0 for 600 V cables, 5% for
5 kV cables, and 10% for 15 kV cables are considered.
9.
watts losslft = ( I x 0.48)~x R11000 x 3 ,
where
I
=
rated current in amps, and
R
=
resistance value in ohms11000 ft.
The inclusion of diversity in the loss predictions is accomplished by multiplying the rated current by the diversity factor,
representing the fraction of the time the rated current is flowing 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 construction-related 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
10. An ambient temperature of 26°C is considered for calculations.
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).
Table 5.
(13)
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 components, 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
Cable Tray Losses at Selected Voltages
KSU
Rubin
KSU
Rubin
KSU
Rubin
Tray Size (in.)
600 V (Wlft)
600 V (Wlft)
5 kV (Wlft)
5 kV (Wlft)
15 kV (Wlft)
15 kV (Wlft)
6
10
-
4
-
3
-
12
23
23
8
26
7
26
18
36
35
12
39
11
42
24
47
47
16
53
15
55
30
61
58
22
65
19
68
ASHRAE Transactions: Symposia
857
Table 6.
Losses per Cable of Different Sizes at Selected Voltages
for Three Conductor, Three Phase Cables
KSU
I
M&H
KSU
M&H
KSU
M&H
I
I
Cable Size
600 V (Wlft)
600 V (Wlft)
5 kV (Wlft)
5 kV (Wlft)
15 kV (Wlft)
15 kV (Wlft)
# 8 AWG
1.49
1.63
-
-
-
-
# 6 AWG
1.73
1.90
-
-
-
-
4 AWG
1.69
1.93
-
-
-
-
# 2 AWG
2.02
2.28
2.38
2.64
2.68
2.83
#
350 kcmil
750 kcmil
cabinets. In each cabinet, asecondarybus carries current vertically 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 horsepower, 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 24-hour period when multiplied by the compartment 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-
sheet essentially performs the same calculation. Given the
motor horsepower and efficiency, q, the kW ofpower supplied
by the starter is
The compartment current flowing through the starter is
I =
kW * 1000
amp
line voltage * & * pf
The line voltage used in this example is 480 volt, three
phase, andpf is the power factor. The riser current is determined by the relation,
riser current for this compartment = compartment current * df
+ riser current from compartment below,
(16)
where
df
=
diversity factor.
As one steps upward through the compartments of a cabinet, it is seen that the riser current increases. The vertical bus
loss in the riser for each compartment is determined by
bus loss = (riser current /rated bus ~ u r r e n t ) ~
x rated bus loss per standard length x compartment height,
(17)
where
rated bus current
=
300 or 600 amps.
ASHRAE Transactions: Symposia
800 Anp
Blak
Circuit
Breaker
Blank
N W 1
5hp
y85%
pH9
dH.8
Blank
NEMb.3
45 hp
m h
pH.9
m.8
NEMA3
40hP
f l h
NEMA5
2Ki hp
m h
pm.9
W.8
N W 5
150 hp
y85%
pm.9
w.8
pm.9
e
l
NFM1
4hp
W h
pm.9
df-1.0
W
2
20 143
q=%?h
pf4.9
&lo
NEMA 1
3hp
@So/,
pm.9
df-1.0
NEM42
15 1p
m h
pW.9
W . 8
NEMA3
35 hp
1@5%
pm.9
df-1.0
Blank
NEMA4
75 hp
q=w?/o
pm.9
dH.8
NEM.44
mhp
m h
pm.9
m.8
Figure 2 Motor control center example setup.
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 convenience, the resistance value and height of each NEMA combination motor starter are also shown in Figure 3.
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 horizontal 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)
ASHRAE Transactions: Symposia
As one steps through the cabinets, it is seen that the horizontal bus current increases. Once the bus currents are determined, the losses are determined by the relation,
Bus loss = (bus current /rated bus currentl2
x rated bus loss per 20" cabinet width.
(19)
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 lowvoltage breaker loss is presented in another section. The individual losses are totaled to provide the power loss of the motor
control center.
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.
~ heating,
Since the starter losses are based on 1 2ohmic
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 full-load 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 exponentially. 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
859
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Table 7.
Inverter Efficiencies and Power Losses
DC Bus Volts
A
23270
50
6818
130
88
50
260
88
23270
6818
60
130
84
31214
9146
75
260
86
33346
9770
100
260
87
Three Phase
kVA
DC Bus Volts
Efficiency
40815
11959
-
Margin
100%
50%
Full Load Loss
(Btulh)
Full Load Loss
Load Loss
(w)
(w)
10
130
80
6850
2007
1004
15
130
81
9600
2813
1406
20
130
82
12000
3516
1758
30
130
82
17985
5270
2635
40
130
83
22375
6556
3278
50
130
84
26010
762 1
3810
60
130
85
28920
8473
4237
75
260
87
30600
8966
4483
100
260
87
40800
11954
5977
from manufacturer literature and Web sites. Table 7 provides
a list of single- and three-phase inverters with power loss specified in both watts and B t d h as a hnction of the kVA rating of
the inverter. The maximum incoming DC current determines
the kVA of the inverter.
ASHRAE Transactions: Symposia
Battery Chargers
It is important to realize how a battery charger is used.
Batteries are usually used in supplying emergency power
during infrequent situations. Once the emergency is over, the
batteries are recharged. During this recharging situation, the
861
battery charger will experience heavy usage and the dissipated
heat will be greatest during this time. During normal operation, the chargerwill be lightly loaded, maintaining the battery
charge and possibly supplying some DC load. Thus, the usage
norm for the charger could quite probably consist of a light
load. Other applications of battery chargers consist of forklifts, telephone systems, and golf carts, where it is seen that the
charger usage will vary over a 24-hour period. The full load
losses are roughly 2.5 times greater than the no load losses.
The heat loss for fractional loading of the battery charger is
determined by
Q = no load loss + X
supplying the DC load. Including the diversity factor, the
result of the previous equation is
Q = (QFL/2.5) * (1 + 1.5 * X * diversity factor) .
The margin for battery chargers in commercial applications is 100% and 50% in critical application. Thus, in
commercial application the loss figure obtained from Table 8
would not require any modification. If the load on the charger
can be estimated, then a smaller loss figure than the one
provided by the table could be used in the HVAC heat load
calculation. The effect of environmental temperature is
unknown due to data coming solely from manufacturers and
no testing being performed. Table 8 is representative ofbattery
charger losses. There are many combinations of input and
output voltage; Table 8 is a small sample of the data available.
The battery charger loss presented in Table 8 was computed by
* (QFL- no load loss) = (QFL/2.5) + X
* (QFL - (QFL'~.~))= ( Q F L ' ~ . ~*)
*3
+
(20)
where
Q
=
heat loss in watts at load fraction,
QFL
=
111-load heat loss, and
X
=
load fraction of charger.
Watts loss
VB x (1 - q1100) ,
(22)
60 Hz Three-Phase Battery Charger Comparison
Input (Three Phase)
862
= IBC x
where
IBC = Output culTent,
VB = output voltage, and
q
= battery charger efficiency, %.
If a diversity factor is included, this diversity factor should
be applied only to the fractional load loss and not to the no load
loss. The diversity factor is the fraction the charger is actively
Table 8.
(21)
Output (DC)
Efficiency
YO
Power Loss
(KSU)
Power Loss
(Rubin)
Volts
Amps
Volts
Amps
q
Watts
Watts
208
19.5
24
200
88
576
720
208
29.3
24
300
88
864
1080
208
39.1
24
400
88
1152
1440
240
16.9
24
200
88
576
720
240
25.4
24
300
88
8 64
1080
240
33.9
24
400
88
1152
480
8.5
24
200
88
576
480
12.7
24
300
88
864
1080
480
16.9
24
400
88
1152
1440
208
39.1
48
200
88
1152
1440
1728
2304
2880
208
78.1
48
400
88
240
33.9
48
200
88
1152
1440
240
50.8
48
300
88
1728
2160
240
67.7
48
400
88
2304
2880
480
16.9
48
200
88
1152
1440
480
25.4
48
300
88
1728
2160
480
33.9
48
400
88
2304
2880
ASHRAE Transactions: Symposia
Table 9.
Low-Voltage Breaker Losses and Resistance Values in Ohms
wlo Enclosure
Watts Loss-
Watts LossW I Enclosure
225
60.0
250
7 1.1
Frame Current (amps)
Resistance-
Resistance-
90.0
0.001 1852
0.00 17778
73.6
0.001 1376
0.001 1776
p
p
p
400
147
220.5
0.00091 88
0.0013781
600
215
322.5
0.0005972
0.0008958
800
330
440
0.0005 156
0.0006875
1200
865
1080
0.0006007
0.00075
1600
1000
1500
0.0003906
0.0005859
2000
1500
2250
0.000375
0.0005625
3000
2250
3375
0.00025
1
p
p
3200
2400
3600
0.0002344
0.00035 16
4000
3000
4500
0.0001875
0.0002813
5000
4700
7050
0.0001 88
0.000282
Rubin (1979) used a relation similar to Equation 22 for-the
calculation of battery charger heat loss but added a safety
factor of 1.25. Rubin's relation for determining battery charger
heat loss was used in the calculation of the comparison
numbers listed in Table 8 and is given as
Watts loss = 1.25 x IBC x VBx (1 - ~l1100).
Circuit breakers are segregated according to frame sizes.
The frame size is the maximum current the breaker is designed
to pass safely in normal operation. A breaker in a given frame
can be made to trip at lower current values by using a rating
plug that conesponds to those limits, Table 9 shows a range of
low-voltage circuit breaker frame sizes. For each frame size,
loss figures are provided for the cases of with and without an
enclosure. Measurements have shown that the breaker enclosure significantly influences the amount ofthe heat loss. These
loss figures represent the losses that occur when rated frame
current flows in the breaker, For smaller currents, the resistance values presented in Table 9 are used to predict the heat
loss through
" an I'R calculation.
~
Watts loss = I2 x R
(25)
The current value produced by Equation 25 is the value to
be used in the loss calculation. Tests of the influence of
breaker heat loss have
little correlation. In these tests, the environmental temperature
was varied from 25°C to 50°C. Values in Table 9
to 60, loo, 250,
and 1200
frames
are
within
were measured, and the uncertainty of the
+lo% of the reported loss values. Data for all of the other
frames were obtained from manufacturer literature with the
of the 3000 amp frame breaker where the loss
were determined by
between the 2000
and 3200 amp
For those breakers not
the
indicated losses are representative of the expected values. The
breaker losses may vary with manufacturer. Figure shows
the losses at different current levels in two frames for
comparison
(24)
where .
I = trip rating in amps and
R = resistance for the frame in ohms.
The inclusion of load diversity is accomplished by
performing the loss calculation with the RMS current
determined by averaging the current over a 24-hour-or
perhaps longer-period.
The margin for breakers is 80% in
ASHRAE Transactions: Symposia
Breaker current = (%margin1100)
x rating plug current value.
(23)
Low-Voltage Circuit Breakers
~
commercial applications, 100% in full-load applications, and
50% in critical applications. To determine the breaker current
for the loss calculation, perform the calculation of
Motors
All manufacturer data for motors were in the form of e f i ciencies. Efficiency values were collected for polyphase
elec~.
tric motors for horsepower ratings from 10 to 2000 hp. The
data were collected for a large number of motor frames. In this
document, all of the efficiencies for different motors of the
same horsepower rating were averaged to get one efficiency
-
863
-
- - - 225 Amp Frame- M&H
- - - 600 Amp Frame - M&H
-600
- - . - -225
. Amp Frame - Rubin
225 Amp Frame - KSU
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 Low-voltage 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) *(I-q/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 full-load applications, and 50% in critical 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,
864
thus, the conductor resistivity, significantly. Thus, the influence 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 averaging the nominal efficiencies from several different manufacturers. 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 low-voltage
switchgear that might include (in addition to a transformer)
circuit breakers, current transformers, control power transformers, auxiliary compartment, space heaters, circuit breakers, and high-current 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 calculaASHRAE Transactions: Symposia
Heat Loss (Watts) M 8 H
Loss (Watts) KSU
-Heat
- - - -Heat
. Loss (Watts) Rubin
160000
140000
.-.
-
..=--
.
120000 -
$ 1 ooooo
.'
1
3
0
i.2:
T
.-.-.
---.
-.-.-.
-.
200
400
.'
600
800
1000
1200
1400
1600
1800
2000
-
Motor Ratlng HP
Figure 5 Electric motor comparison.
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 low-voltage 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 24-hour--or
longer-period in the loss calculations. This calculation is
shown in Equations 11 and 12. The margin for unit substations
is 80% in commercial applications, 100% in full-load applications, 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.
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
= 234.S°C for copper windings and 225°C for
Tk
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
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
ASHRAE Transactions: Symposia
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
WATTS LOSS DATA
600 V Switchgear
Figure 6 Unit substation example spreadsheet.
ASHRAE Transactions: Symposia
instruments
800 amp
breaker
350 amp
Instr.
Instruments
800 amp
breaker
350 amp
Instruments
4000 amp
Main
800 amp
breaker
300 amp
Instr.
Fut ure
800 amp
breaker
300 amp
4000 amp
Main
800 amp
breaker
300 amp
Future
4000 amp Tie
800 amp
breaker
300 amp
800 amp
m e r
350 amp
Future
2300 amp
1600 amp
feeder
1oooamP
Figure 7
2300 amp
800 amp
breaker
350 amp
1600 amp
feeder
OoO aV
Unit substation example setup.
Table 10.
ASHRAE Transactions: Symposia
Reactor Power Loss in Watts at Rated Current and Room Temperature
867
--... 240 Volt - K
-600 Volt - K
Volt - K U
- - - 460
McDolald &%ickok
U
~ U
/
/
/
---
/
--#-
/
--
.
/
/
/
/
/
/
/
/
/
/
/
/
1-
-.
/
/
/
/
/
/
I._.--'
/
/
/
/
/
._.--
..-_.-_-.-.
..-a-
/
Horsepower
Figure 8 Adjustable-speed drive comparison.
spond to the rated current (horsepower), the power losses are
then predicted by
data appearing in Table 10 were obtained from manufacturers
and averaged. Some of the data going into the averaging calculation were verified experimentally. In general, the agreement
between the reported data points where testing was possible is
well within +lo%.
where
Adjustable-Speed Drives
P
=
power loss,
Losses in adjustable-speed drives 'vary linearly with
current. For a given line-to-line voltage, the rated current
varies linearly with the rated horsepower. Relations were
developed to describe the full-load power loss in watts in terms
of rated horsepower for different voltage levels. These relations are
hp
=
actual horsepower,
= hprafedx 25.6234 + 276.073 watts , (33)
240 V: Prated
= Prated
= hpratedx 14.55851 - 201.038 watts , (35)
600 V: Prated
where
Profed= full-load 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 corre868
('/'rated
3
(36)
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 workweek. The losses adjusted for diversity would be
= 'rated
460 V: Prated
= hpratedx 13.45435 + 363.7949 watts, (34)
and
(h~/h~rafed)
=
(h~/h~rated)
F~~~ = Prated
(Ilzrated) F
~
(37)
where
FAsD
=
fraction of time ASD is in use over standard work
period.
The margin for adjustable-speed drives in commercial
applications is 100% and 50% in critical application. Thus, in
commercial application, the loss figure obtained from Equations 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
ASHRAE Transactions: Symposia
~
~
temperature was not a factor in the measured power loss. The
data used to determine the slope and intercept values in Equations 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 information 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 adjustable-speed drives, so he
was not included in Figure 8. McDonald and Hickok (1985)
did provide information of percent efficiency for adjustablespeed drives based solely on horsepower rating. This efficiency 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 applications 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 (1985), yet Rubin's loss
tables show higher losses for the same equipment reported by
Hickok.
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 Electric, Rockwell International, Stanion Wholesale Electric of
Manhattan, KS, Tennessee Valley Authority, andU.S. Department of Energy.
REFERENCES
Hickok, H.N. 1978. Energy losses in electrical power systems. IEEE Transactions on Industry Applications
14(5): 373-387.
McDonald, W.J., and H.N. Hickok. 1985. Energy losses in
electrical power systems. IEEE Transactions on Industry Applications IA-2 l(4): 803-8 19.
Rubin, I.M. 1979. Heat losses from electrical equipment in
generating stations. IEEE Transactions on Power Apparatus and Systems PAS-98(4): 1149-1 152.
IEEE. 1996. IEEE Standard 112-1996, Test Procedure for
Polyphase Indzlction Motors and Generators. New
York: IEEE Press.
IEEE. 1996. IEEE Standard 115-1995, Guide: Test Procedures 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)
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
ASHRAE Transactions: Symposia
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 VBx (1 -?/loo)
where
IBC
=
output current
(22)
output voltage
= battery charger efficiency - %.
Since IBC is the output current and V, is the output voltage, the product ofthe two quantities is the outputpower. If the
percent efficiency, q, is replaced by the symbol eff, then Equation 22 can be written as
V,
=
Watts loss = Output power x ( I
-
effI100)
7
which is exactly the first of the two equations presented by Mr.
Elleson in the statement of his question and, also, deemed
correct by Mr. Elleson. Since our results agree, our definitions
of loss and efficiency concur.
ASHRAE Transactions: Symposia