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2005 Professional Engineer Review Course
Unit V – Electrical;
Instructor: J. Hennings
Introduction and Assignment
Module 5.2: Electrical Distribution
Notes
System Power Factor
System power factor is the ratio of the active power in KW to the apparent power
in KVA
Power factor pf
=
KW/KVA
Where:
KW
KVA
is
is
Kilowatts
Kilovolt-amps
Active power is also called the true or average power and represents the rate of
conversion of energy from one form to another. In the case of heaters, the conversion is
from electrical energy to heat. For motors, it is from electrical to primarily mechanical
and some heat, and for generators it is mechanical to primarily electrical and some heat.
The heat in both the motor and generator are associated with power loss.
Apparent power (KVA) is supplied by the system generator(s) operating at a
given system frequency and voltage level. Both frequency and voltage level are held at
hopefully near constant levels by governors and voltage regulators respectively.
Although both frequency and voltage levels are subject to short term changes due to
changes in load, they will both be considered constant for the purpose of this discussion.
A ships electrical distribution system is an “islanded system” that provides power
on demand. Islanded meaning a finite number of generators (one, two or possibly three)
operating separately or in parallel providing power as demanded (required) by the
connected load. The apparent power is a function of the constant voltage multiplied by
the current requirement of the load demand.
In the case of a purely resistive load (heating elements for example), the real
power (KW) and the apparent power (KVA) are the same, thus the power factor is unity.
Most electrical loads (motors) are heavily inductive and require a third type of
power called reactive power (KVAR). This reactive power represents the rate of energy
converted to magnetic fields required for motor operation. The energy associated with
these magnetic fields alternately builds up and decays in the motor as the A/C current
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changes in both magnitude and polarity. During field build up, KVARs flow to the
motor. During decay KVARs flow from the motor to the generator. This rate of energy
conversion increases the KVA demand on the generator and can be accounted for by the
following:
KVA = [KW2 + KVAR2]½
Most textbooks and references usually represent real power (KW) by the letter P,
apparent power (KVA) by the letter S, and reactive power (KVAR) by the letter Q.
*Caution - do not confuse the letter S when used in power calculations for slip.
Thus:
S = [P2 + Q2] ½
And
pf = P/S = [P2 + Q2] ½
Because the current through an inductive load lags the voltage across that load, an
inductive load is said to have a “lagging” power factor. Therefore, most ship’s electrical
distribution systems have lagging power factors because of the large induction motor
load. These inductive KVARs are given the notation +Q. Capacitors on the other hand,
while also reactive elements, have current to them lead the voltage across them.
Capacitors are therefore said to have a “leading” power factor, and capacitive KVARs are
given the notation –Q.
The real power (KW) output of the system generator(s) at any time is the sum of
all the (KW) demands of all connected loads plus line losses.
Ptotal  Pgen  Ploads  P1  P2  P3 ....  PN
The reactive power (KVA) output of the system generator(s) at any time is the
“net” sum of all the (KVAR) demands of all connected loads.
Qtotal  Qgen  Qloads  Q1   Q2   Q3  ...  Q N 
Note the inclusion of the parentheses when summing KVAR loads due to
the possibility of both +Q (inductive) loads and –Q (capacitive) loads. For the most part,
shipboard loads are inductive loads, thus +Q yielding lagging power factors. Care must
be taken when summing the apparent loads. If loads do not have exactly the same power
factors their apparent powers (KVA) can not be summed directly. Thus, for different
power factor loads:
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2
STotal  S gen  PTotal
 QTotal
Power factors can range from PF = 1 for purely resistive loads to PF = 0 for
purely reactive loads. Although purely resistive loads such as heaters do occur, motors in
fact are both resistive and inductive. Most of the apparent power (KVA) supplied to a
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motor is converted to mechanical power (1hp = .746KW), so that motor power factors are
usually 0.8 to 0.9.
Example:
A squirrel-cage induction motor provides 50hp to a load. The motor efficiency is
90% and the motor power factor is .83 lagging.
Calculate P, Q and S to the motor.
Note: the 50hp is power out of the motor, and we are asked to calculate
electrical power P (KW) into the motor.
Solution:
50hp .746
 41.4KW
.9
pf = P/S
so:
41.4KW
S
 49.9KVA
.83
P
S  S2  P2 
49.92  41.42  27.9 KVAR
Example:
Four loads are supplied from the same feeder. The load characteristics are
Load #1:
Load #2:
Load #3:
Load #4:
40KW @ pf = .8 lagging
80 KVA @ pf = .9 lagging
5KW @ unity power factor
3KVA @ pf = .83 leading
Find PT (total KW), QT (total KVAR), ST (total KVA), and pfT (total power
factor) that the four loads will draw from the feeder.
Solution:
P1
=
Q1
=
P2
=
Q2
=
P3
Q3
=
=
40KW
given
502  402   30 KVAR
80 KVA * .9 = 72KW
802  722   34.9 KVAR
5KW
given
0
because unity power factor
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P4
=
Q4
=
PT
QT
=
=
ST
=
3KVA * .83 = 2.5KW
32  2.52   1.7 KVAR
40 + 72 + 5 + 2.5 = 119.5KW
(30) + (34.9) + (0) + (-1.7) = +63.2 KVAR
119.52  63.22  135.2 KVA
119.5
 .884 lagging
135.2
Note that because QT was positive, the pfT is lagging. Had QT been a negative
value, then pfT would have been leading.
pfT
=
Distribution and Power Circuits:
Reference 1, Chapter XVII (Section 6.3) states that power distribution serves
motor-driven auxiliaries and heating equipment. These loads are supplied either
individually or in groups by feeders from a ship service distribution switchboard. Some
of the most important characteristics of all feeders are the current-carrying capacity,
resistance in ohms per 1000ft. and the type of insulation used. Because these conductors
are non-ideal, they will incurr power losses and voltage drops (%V.D.) across them when
they conduct current.
To determine the percent voltage drop in a 3 phase feeder carrying power, the
following formulae are used:
For cables 52,600 CM and smaller:
% V.D. =
3  12  I  L  100
CM  V
For cables 66,400 CM and larger:
% V.D. =
3  12  I  L  CF  100
CM  V
Where:
% V.D.
I
L
CM
V
CF
is the percent voltage drop
current in Amperes
is the load center length in feet
is the cross sectional conductor area in circular mils
Voltage in volts
Correction factor is a constant for a given size cable and
power factor and is used where applicable.
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Load center length is a combination of the physical length of the cable and the
amount of power in watts being carried.
The following example shows how to calculate the load center length (L) used in
the % V.D. equation.
Example:
Calculate the load center length (L) for the following feeder:
Figure 5.2.1
78ft * 200W = 15600 ft-W
104ft * 200W = 20800 ft-W
122ft * 100W = 12200 ft-W
134ft * 100W = 13400 ft-W
L 
62000 ft  W
 103.3 ft
600W
Example:
Calculate the percent voltage drop (%V.D.) for the following 212,000CM,
3 phase feeder:
Service:
Voltage
Current
Load center length
Correction factor
% V.D. =
vital auxiliaries
V = 440V
I = 194.1 A
L = 110 ft
CF = 1.16
3  12  194.1  110  1.16  100
 .55%
212,000  440
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Problems:
1)
A 440v, 60hz, 3 phase feeder supplies the following 4 loads.
Load #1:
A 20hp squirrel-cage induction motor with efficiency of
88% and a power factor of .85 lagging.
Load #2:
A heater load of 7.5KVA @ pf = 1
Load #3:
25 KVA @ pf = .8 lagging
Load #4:
12 KW @ pf = .9 leading
Calculate: PT (total KW), QT (total KVAR), ST (total KVA) and pfT (total power
factor) that the four loads represent to the electrical system. Is pf T leading or lagging?
Why?
PT: a) 56.5KW b) 65.5KW c) 75.5KW d) 85.5KW
QT: a) +14.2KVAR b) +17.8KVAR c) +19.8KVAR d) +21.3KVAR
ST: a) 52.1KVA b) 59.9KVA c) 63.2KVA d) 67.1KVA
pfT: a) .89 b) .92 c) .94 d) .97
2)
A 3 phase, 52,600 CM feeder cable has a load center length of 35 ft. The
feeder voltage is 208v and carries 40 amps of current.
Calculate the percent voltage drop (% V.D.).
% V.D.: a) .27% b) .30% c) .32% d) .37%