Heat dessipation from motor.pdf

Heat Loss from Electrical and Control Equipment in Industrial Plants: Part ll-Results 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, 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 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 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.

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 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 well-defined, 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 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

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 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

Rated Voltage 0')

480D-208Y

480D-208Y

480D-208Y

480D-208Y

480D-208Y

480D-208Y

48OD-208Y

480D-208Y

480D-208Y

480D-208Y

480D-208Y

480D-208Y

480D-208Y

15kD-480Y

15kD-480Y

15kD-480Y

15kD-480Y

15kD-480Y

15kD-480Y

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 environ-

mental temperature; thus, the full-load and no-load 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-

Kilo-Volt- 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 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 = 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


(w)

507

85 1

810

959

1104

1577

1985

2127

2442

3255

3749

6008

9126

5600

8560

9644

12424

16420

19040


(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 Liquid-Filled Units

Rated Voltage Total Losses Total Losses

Substation

15kD-480Y 225 760 3400 4160 2936 1610

Table 3. NEMA TPI Efficiencies-q

Efficiency % Efficiency % Dry Type, Medium Voltage, Liquid Immersed, Medium

Kilo-Volt-Amps Dry Type, Low Voltage, 7S°C

15 97 96.8 98.0

The power factor in Equation 2 depends upon the electrical 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 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

require any temperature correction. The temperature correction 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 Dry-Type 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 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.

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.


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 three-phase 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 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.

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 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).

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 three-phase (as indicated 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 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

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

Table 5. Cable Tray Losses at Selected Voltages


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 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-

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 determined 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 cabinet, 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.


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 combination 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 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)

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 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)

N F M 1 4hp

W h pm.9 df-1.0

W 2 20 143 q=%?h p f 4 . 9 &lo

NEMA3 35 hp 1@5% pm.9 df-1.0

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 individual losses are totaled to provide the power loss of the motor control center.

NEMA 1 3hp

@So/, pm.9 df-1.0

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 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 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


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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 three-phase 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


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 24-hour 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 = 111-load 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 Three-Phase Battery Charger Comparison


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. Low-Voltage Breaker Losses and Resistance Values in Ohms

Rubin (1 979) used a relation similar to Equation 22 for-the commercial applications, 100% in full-load 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 Low-Voltage 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

low-voltage 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 24-hour-or document, all of the efficiencies for different motors of the perhaps longer-period. The margin for breakers is 80% in same horsepower rating were averaged to get one efficiency

2250 3375 0.00025


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 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,

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 calcula-


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 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 1 1 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.

$ 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


WATTS LOSS DATA 600 V Switchgear

Figure 6 Unit substation example spreadsheet.


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


Instr.

Instr.

Future

Future

800 amp breaker

350 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

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 Adjustable-speed drive comparison.

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%.

Adjustable-Speed Drives

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

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 = 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 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 adjustable-speed 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


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 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 adjustable- speed 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 (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 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 Indus- try Applications IA-2 l(4): 803-8 1 9.

Rubin, I.M. 1979. Heat losses from electrical equipment in generating stations. IEEE Transactions on Power Appa- ratus 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 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


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.


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