Download Module 5.3

1
2005 Professional Engineer Review Course
Unit V – Electrical;
Instructor: J. Hennings
Introduction and Assignment
Module 5.3: Electrical Energy Conversion
Single Phase A/C: Sinusoids and Phasors
Consider the sinusoidal voltage:
v t   Vm cos    
Figure 5.3.1
Where:
Vm
is
the wave amplitude

is
the angular frequency in radians/seconds

is
the phase angle
The frequency in hertz is:

f =
2
and the wave period is:
1 2
T =

f

Although the average value of the voltage over a full period is zero
Vavg = 0
the more important voltage is the root mean square (RMS) or effective value. It is this
RMS value that actually delivers power to a circuit.
For sinusoids:
2
VRMS = .707  Vm =
Vm
2
A/C currents can also be expressed as sinusoids:
I (t) = Im cos ( t + )
It is the RMS value of current that actually delivers power to a circuit.
IRMS = .707  Im
The RMS or the effective value of A/C current is the value of an equivalent value of D/C current
that would deliver the same average power to a resistor. See Reference 4, Chapter 10.
Example:
Given v(t) = 169.7 cos (377 t + 30) volts
Find VRMS ,  , f and T
Solution:
VRMS = .707  169.7V = 120V
radian

= 377
sec
377
 60 hz
f
=
2
1
 .0166 seconds  16.6 mSec
T
=
60
A/C circuit calculations are greatly simplified by the use of phasor math. A phasor is a
complex number that represents A/C voltages and currents by their amplitudes and phase angle.
They look like vectors and the math using them is quite similar. The amplitude that is used for
power calculations and other A/C circuit analysis is ALWAYS the RMS value of the voltage, and
the phase is the angle  where:
V(t) = Vm cos ( t + )
this sinusoidal expression has an
equivalent RMS phasor of
V = VRMS 
in polar form
Example:
Express v (t) = 169.7 cos (377 t + 30) volts as a phasor
Solution:
V  VRMS  120V calculated previously
  30
so V  120  30 volt (read 120 volts at  30)
Note that frequency content is “lost” in the phasor expression, but so long as frequency
remains constant, it will not effect calculations. Also note that the original phasor must be a cosine
wave. If the v(t) is a sine wave then a phase shift of 90 must be included.
Thus:
V(t) = Vm Sin ( t + ) has a phasor representation of
V  VRMS
  90 
3
Phasor relationships for circuit elements
Resistors oppose the flow of current through them as given by Ohm’s Law
VR = IR  R
Where: VR and IR are phasors
Because resistors instantaneously change electrical energy into heat, the voltage across
resistors is always in phase with the current through them. In A/C circuits resistors are given
complex or phasor notations just as are voltages and currents. As a complex number:
R = R 0
Example:
A 10 resistor has a current of I = 5  30  amps flowing through it. Calculate
the voltage (VR) across it.
Solution:
VR = I  R = (5  30  )  (10 0  )
VR = 50  30  volts.
Note that when polar phasors are multiplied together, their magnitudes are
multiplied as in simple arithmetic, but their angles are added.
Inductors (L) have units of Henries. Inductors oppose changes in current through them.
This leads to the A/C current through an inductor being 90 out of phase with the voltage across it.
The amount of opposition in ohms that an inductor has to A/C is directly proportional to its value
of L and the frequency of the changing A/C. The inductors opposition to A/C is called inductive
reactance (XL).
Where
XL = L = 2fL
Inductive reactance in polar form has an angle of + 90 because of the fact that its current
lags its voltage by 90.
Example:
An inductor L = .0265 Henries = 26.5mH
(m = 10-3) is operated in a 60 hz system
Calculate XL as a polar phasor.
XL = 2fL
90 
= 6.28  60  26.5  10-3  90 
XL = 10  90  
Capacitors have units of Farads. Capacitors oppose changes in voltage across them. This
leads to the A/C voltage across a capacitor being 90 out of phase with the capacitor current. The
4
amount of opposition in ohms that a capacitor has to A/C is inversely proportional to the product of
its value and the frequency of the changing A/C. The capacitors opposition to A/C is called
capacitive reactance XC.
1
1

Where:
XC =
 C 2 fC
Capacitance reactance in polar form has an angle of -90.
Example:
A capacitor C = 5.3  10-4 F = 530 F
(  = 10-6 ) is operated in a 60hz system.
Calculate: XC as a polar phasor.
1
1

XC
=
 90 
2 fC 6.28  60  530  10 6
XC
= 5  90  
Like inductors, capacitors also follow Ohm’s Law in A/C circuits.
Example:
If
XC = 10  90   and VC = 25 30  V
Find IC
Solution:

VC 25 3 0
IC =

 2.5 30  ( 90 )
XC 10  9 0
IC = 2.5
120  AMPS.
Note that when polar phasors are divided, their magnitudes are divided as in simple
arithmetic, but the angle of the denominator is subtracted from the angle of the numerator.
Example:
If
Find
Solution:
XC = 5  90   and
IC = 10
30 
Amps
VC
VC = 10
30   5
 90 
= 50  60  Volts
A/C Series Circuits
The solution of A/C series circuits follows all the rules for D/C series circuits but using
complex or phasor math (Reference 4, Chapter 11)
In a series A/C circuit the total opposition is called impedance (Z) and has units of ohms.
Impedance is also a complex value and can be expressed in either polar or rectangular form.
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Figure 5.3.2
As is always the case, the total opposition in any series circuit is the sum of the oppositions.
But R cannot simply be added to XL and XC because of the angular relationships. In complex
form R at zero degrees is all real, but XL at +90 is purely imaginary. In rectangular form the
+90 for the inductor is replaced by +j, and the -90 of the capacitor is replaced by –j. (Note: do
not confuse –j in phasor calculations with the unit vector j)
For the above R – L – C circuit where R = 4, XL = 6, and XC = 3 the
impedance is:
Z = R + j XL - jXC
or
Z = 4 + j6 - j3 = 4 + j3 
Using the Pythagorean Theorem, Z can also be expressed in polar form.
Z =
R2  X L  X C 
2
 X  XC 
  tan 1  L

R


And
Where:
 is the impedance angle
Z =
so:
4 2  32  5 
 = tan-1 (¾) = 36.9º
Z = 4 + j3 = 5 36.9  
Had XC = -j6 and XL = +j3 then the impedance would have been
Z = 4 - j3 = 5  36.9  
As is also the case in all series circuits the current is the same through all oppositions
connected in series (same magnitude and same angle).
Example:
For the single phase series A/C circuit shown, find Z, I, VR, VL, and VC
6
Figure 5.3.3
Solution: (using the discussed methods)
Z = 4 +j10 –j5 = 4 + j5 = 6.4  51.3  
from Ohm’s Law:
120 0
VS
I =

 18.75  51.3 A

Z 6.4 51.3
VR = 18.75  51.3  4 0  75  51.3 V
VL =
18.75  51.3  10  90  187.5  38.7 V
VC =
18.75  51.3  5  90  93.75  141.3 V
It should be noted that the phasor sum of VR + VL + VC must equal VS. This is called
Kirchhoff’s Voltage Law (KVL).
VS  VR  (VL  VC ) 2
2
A/C Parallel Circuits
The solution of A/C parallel circuits follows all the rules for D/C parallel circuits but using
complex or phasor math (Reference 4, Chapter 12). As is always the case in all parallel circuits,
the voltage across each parallel branch is the same as the source voltage (same amplitude and same
phase angle). The phasor sum of IR, IL, and IC will equal IT. (Kirchhoff’s Current Law, KCL).
Then the total impedance (ZT) will be the ratio of VS/IT .
Figure 5.3.4
7
For the above parallel circuit find: VR , VL , VC , IR , IL , IC and ZT .
VS = VR = VL = VC = 120
0 V

VR 120 0
IR =

 30 0 Amps

R
4 0

VL 120 0
IL =

 24  90 Amps

XL 5 9 0
IC =
IT =
120 0
10  90

 12  90 Amps
30 0  24  90  12 90  32.3  21.8 Amps
120 0
VS

 3.7  21.8 
ZT =

IT 32.3  21.8
It should be noted that the phasor sum of I R  I L  I C must equal IT . This is called
Kirchhoff’s Current Law (KCL).
I T  I R  (I L  I C ) 2
2
Three Phase Loads
Most of the loads aboard ship are 3 phase loads. See Reference 4, Chapter 21.
Three phase loads can be either delta or wye connected. We will consider them to be “balanced”
loads. That is to say that each side of the delta has the same magnitude and angle of impedance.
Each leg of the wye also has the same magnitude and angle of impedance. When a set of 3 phase
voltages is applied to the loads, currents flow in the lines to the loads. The phasor sum of all
balanced, 3 phase voltages and currents is always zero. Notice that the lines are lettered A, B and
C.
8
Figure 5.3.5 A
Figure 5.3.5 B
For both delta and wye, the three voltages between each pair of lines are called Line voltage
(VLine). VLine is always the value of rated or nameplate voltage for the load. The three line voltages
are:
VAB, VBC and VCA.
They are all equal in magnitude but mutually 120 out of phase with each other.
VAB
For a typical 440V, 3 phase load:
= 440 0  V , VBC = 440  120  V , and VCA = 440
Note
V
Line
0
 120  V ,
9
For both delta and wye, the balanced impedances will draw 3 Line currents when connected
to the 3 Line voltages. The 3 Line currents are:
IA, IB, and IC
They are all equal in magnitude but mutually 120 out of phase with each other for balanced loads.
For a typical 3 phase load drawing 50A:
IA = 50 0  A , IB = 50  120  A , and IC = 50
Note
I
Line
 120  A ,
0
For both the delta and wye, the voltage across and the current through one side of the delta
or one leg of the wye are called phase voltage and phase current. Phase voltage and phase current
are functions of Line voltage. There are three 3 phase voltages and three 3 phase currents for each
load. The relationships between Line and Phase values for each type of load are the following:
For delta:
VLine = Vphase
I Line =
3 Iphase
V phase
Iphase =
Z
V
phase
0
I
phase
0
I
phase
0
For wye:
3 Vphase
VLine =
ILine = Iphase
V phase
Iphase =
Zy
V
phase
0
Example:
A balanced 3 phase delta load is rated at Z   10 37  


Is connected to a 440V, 3 phase system.
Calculate VLine , Vphase , ILine , Iphase
VLine = 440V
always the system voltage
Vphase = VLine = 440V
440
 44 A
Iphase =
10
3  44 = 76.1 A
ILine =
The nameplate voltage and nameplate current for this load would be 440v and 76.1A
respectively.
10
Example:
A balanced 3 phase wye load is rated at Zy  10 37 


Is connected to a 440v, 3 phase system.
Calculate VLine , Vphase , ILine , Iphase
VLine = 440v
always the system voltage
440
Vphase =
 254.3 V
3
254.3
 25.4 A
Iphase =
10
ILine = Iphase = 25.4 A
The nameplate voltage and nameplate current for this load would be 440v and 25.4A
respectively.
A/C Power
As discussed in Module 5.2, there are three types of powers associated with A/C systems:
Real or Average power, P in KW
Apparent power,
S in KVA
Reactive power,
Q in KVAR
And power factor
pf in KW/KVA
For single phase circuits:
Real power (KW) is the power to resistors and can be found with:
P = I2 R
or
by P = VS  I  pf
where pf can also be calculated by pf = cos of the impedance angle.
Apparent power (KVA) is the power to the circuit impedance and can be found with:
S = I2 Z = VS  I
or
S = P/pf
Reactive power (KVAR) is the power to XL and XC and can be found with:
Q = I2 X
where: X = XL - XC
or
Q = S 2  P2
The circuit power factor is said to be “lagging” if the impedance angle is positive and
“leading” if the impedance angle is negative. See Reference 4, Chapter 13.
Example:
11
For the series single phase circuit solved previously, find P, S, Q and pf.
Figure 5.3.6
Pf = cos (= 51.3) = 0.625 lagging
Lagging because of the positive impedance angle
P = 120  18.75  .625 = 1406 W = 1.4 KW
or
2
P = (18.75)  4 = 1.4 KW
S = 120  18.75 = 2250 VA = 2.25KVA
or
1.4KW
 2.25 KVA
S = P/pf =
.625
Q = I2 X where X = +10 –5 = 5 
Q = (18.75)2  5 = 1758 VAR approx. 1.76KVAR
or Q =
2.252  1.412  1.76KVAR
Note that no angles are used with voltage or current for calculating powers.
For three phase circuits
P =
S =
And
3  VLine  ILine  pf (KW)
3  VLine  ILine (KVA)
S 2  P 2 (KVAR)
Q=
pf = P/S
The circuit power factor is said to be “lagging” if the impedance angle is positive and
“leading” if the impedance angle is negative. P, Q and S are the total values to the entire load.
Example:
12
For the balanced delta circuit previously solved, find P, Q, S and pf.
Figure 5.3.7
Solution:
pf = cos (+37) = .8 lagging
The pf is lagging because of the positive impedance angle.
P =
S =
Q =
3  440  76.1  .8 = 46.3 KW
3  440  76.1 = 57.9 KVA
57.92
 46.3  34.8KVAR
2
Example:
For the balanced wye circuit previously solved, find P, Q, S and pf.
Figure 5.3.8
Solution:
pf = cos (+37) = .8 lagging
the pf is lagging because of the positive impedance angle.
P =
S =
3  440  25.4  .8 = 15.5 KW
3  440  25.4 = 19.3 KVA
13
Q =
19.32  15.52
 11.5KVAR
Again, note that no angles for voltage and current are used in power calculations.
Three phase power calculations can also be done for loads when only nameplate values are
given. Consider the induction motor example of Module 5.2.
Example:
A 3 phase squirrel-cage induction motor has the following nameplate data:
440V, 50hp,  = 90%, pf = .83, 60hz
find P, S, Q, ILine and Z
50  .746
 41.4KW
.9
41.4KW
 49.9 KVA
.83
P =
S =
Q =
49.92  41.42  27.9KVAR
ILine =
49.9KVA
 .0656KA  65.6A
3  440V
If the motor is delta connected:
VLine = 440V ,
Vphase = 440V
ILine = 65.6A,
Iphase =
65.6
 37.9A
3
Vphase 440

 11.6  33.9 
Iphase 37.9
The impedance angle is cos-1 (pf)
Z
=
If the motor is wye connected
440V
 254.3V
3
= 65.6A
VLine = 440V ,
Vphase =
ILine = 65.6A,
Iphase
14
Zy
=
Vphase 254.3

 3.88   33.9 
Iphase
65.6
For both delta and wye calculations, the impedance angle is cos 1 (pf).
Generators
Ships generators are typically three phase 450V, 60hz, synchronous machines. Their power
ratings run from 500KVA to 1500KVA. They can be either driven by steam or diesel prime
movers. There are always an even number of magnetic poles on the rotor carrying a nominally
small amount of D/C current. This rotor circuit is called the field or excitor circuit. The 3 phase
winding in the stator is called the armature. Ship service generators are each capable of carrying
the total at sea load. For peak loads, two generators are usually put on line at once.
For a synchronous machine shaft speed nm is the same as synchronous speed nS (the speed
of the rotating magnetic field).
120 f
nm = nS =
where f is frequency, usually 60hz and P is the
P
number of poles, usually 2, 4 or 6.
Like the 3 phase loads they power, generators can be either delta or wye connected. In
either case the nameplate voltage and current are always VLine and ILine .
The power can be given in either KW or KVA. While the generator voltage and frequency are
kept relatively constant, the current and power outputs change according to load demand changes.
Figure 5.3.9 A
15
Figure 5.3.9 B
The time sequence in which the three Line Voltage (VAB, VBC and VCA) are induced as
phasors is called the phase sequence. The only two possible phase sequences are:
A
B
C
and
C
-
B
-
A
In order for generators to be paralleled they must have the same phase sequence as well as
frequency and voltage. The direction of rotation of 3 phase motors depends on the phase sequence
of the applied voltages. If a 3 phase motor is installed and is turning in the wrong direction,
reverse either of the two input leads to correct the problem.
Example:
A 3 phase, 4 pole, 60hz, 450V synchronous generator is rated at 1200A and .8 lagging
power factor with  = 92%.
Calculate the rated power out (P,S and Q) and the required primer mover speed and horse
power.
Solution:
The given voltage and current on the nameplate are always Line values
Shaft speed:
S =
3  450  1200  934.2 KVA
P =
3  450 1200 * .8  747.4KW
Q=
934.22  747.42
 560KVAR
16
nm = nS =
120 f 120  60

 1800RPM
P
4
The calculated P = 747.4KW is the power output of the generator. Therefore the power
required of the prime mover is
747.4
 812.4KW
.92
812.4KW
Pin =
 1089hp
.746KW / hp
If the generator is wye connected:
450V
VLine = 450V ,
Vphase =
 260.1V
3
ILine = 1200A,
Iphase = 1200A
Pin =
If the generator is delta connected:
VLine = 450V ,
Vphase =
ILine = 1200A,
Iphase =
450V
1200A
 693.6A
3
Transformers:
Transformers are input/output devices that allow A/C voltages to be transformed from one
voltage level to another. They are highly efficient (usually  > 90% and do not cause any change
in frequency and very minimal change from Power-in to Power-out. For the purpose of these
discussions, we will assume  = 100%.
Figure 5.3.10
Transformers are constructed by wrapping insulated turns of copper wire on a laminated
steel core. See Reference 2, Chapter 7. The number of turns is designated N. Primary quantities
are:
N1
# primary turns
V1
rated primary voltage
I1
primary current
S1
primary KVA
17
Secondary quantities are:
N2
V2
I2
S2
# secondary turns
rated secondary voltage
secondary current
secondary KVA
In both the primary and secondary, the voltage levels are fixed by the turns ratio (a). Turns
ratio determines the ratio of V2 to V1 as follows:
a =
N1 V1

N 2 V2
If N1 > N2 then V1 > V2 and the transformer is a step-down device.
If N1 < N2 then V1 < V2 and the transformer is a step-up device. For both step-up and
step-down transformers
S1 = S2 (KVAin = KVAout)
So,
V1 I1 = V2 I2
Therefore,
P1 = P2
So,
V1  I1  pf = V2  I2  pf
The power factor is the same on both primary and secondary and depends on the load
impedance connected to the secondary. Where:
Z2 = Zload =
V2
I2
Transformers, as is the case with most other electrical equipment do not necessarily run at
full-rated nameplate load. Although V1 and V2 are always fixed at rated values, I2 depends on the
power demand of the load and I1 depends on I2.
For step-down transformers:
I1 < I2
For step-up transformers:
I1 > I2
I2 can be zero (when the secondary is open circuited providing no power out). If I2 = 0
then I1 = 0 but both V1 and V2 are rated so long as the transformer is energized.
Note: S1 = S2
Example:
18
A single phase 2.5KVA, 60hz, step-down transformer is rated at 440V/120V. If it is
providing rated KVA at .8 lagging pf, find:
V1 ,
Solution:
V2 ,
the turns ration, I1
, I2 , P1 , P2 , and Zload
Because it is a step-down device
So:
V1
>
V2
V1
=
440V
V2
=
120V
=
440V
 3.7 to 1
120V
Turns ratio
If N1 = 370 turns then N2 = 100 turns.
I2 =
S2 2.5KVA

 20.8A
V2
120V
P2 = S2  pf
I1 
= 2.5KVA  .8 = 2.0KW
S1 2.5KVA

 5.7 A
V1
440V
P1 = S1  pf
= 2.5KVA  .8 = 2.0KW
Note from the current calcs
I 2 20.8A

 3.7 to 1
I1
5.7A
note: turns ratio is
Z2 = Zload =
V2
I2
V1 I 2

V2 I1
=
120V
 5.77 
20.8A
Z2 = 5.77  37  
Where: + 37 is cos-1 (pf)
19
Three phase transformers
Marine 3 phase transformers are made of 3 identical single phase transformers connected
together to form one 3 phase device. Typically the primaries are connected in delta and the
secondaries in either delta or wye.
Figure 5.3.11 A
Figure 5.3.11 B
Both the primary and secondary nameplate voltage are both always the line values, as are
the nameplate primary and secondary currents. The rated KVA on the nameplate is the addition of
the separate KVA values of each single phase device. The three phase turns ratio is always:
Turns ratio
=
VLine1
VLine2
Where
VLine 1
is the rated 3 phase primary Line voltage
And
VLine 2
is the rated 3 phase secondary Line voltage.
Calculations for 3 phase transformers are similar to single phase devices.
20
Example:
A 3 phase, 60hz, 440v/120v, step-down transformer is connected delta-wye.
The three phase transformer is providing rated KVA of 15KVA to a .8 lagging power factor load.
Find VLine1 , Vphase1 , VLine2, Vphase2 , ILine1 , ILine2.
Solution:
VLine1 = 440V and VLine 2 = 120V
given on nameplate.
Because the primary is delta: Vphase 1 = VLine 1 = 440V
Because the secondary is wye: Vphase 2
Turns ratio
So:
Similarly:
=
=
Vline1 440V

 3.7 to 1
Vline2 120V
S2
=
ILine 2 =
Note again that S1
3  VLine2  I Line2
15KVA
 72.3A
3  120V
15KVA
ILine 1 =
VLine2 120V

 69.4V
3
3
3  440V
 19.7A
= S2 and that
I2
is the turns ratio.
I1
Problems:
1)
Given the two sinusoids:
v1(t)
=
155.6 cos (377t + 30) v
v2(t)
=
325.3 sin (2512 t) v
find the RMS phasor equivalent of both
find the phase angle () for both
V1: a) 0 V
b) 110 V c) 120 V d) 155.6 V
V2: a) 0 V
b) 208 V c) 220 V d) 230 V
f1: a) 50 hz
b) 60 hz c) 377 hz d) 400 hz
f2: a) 50 hz
b) 60 hz c) 400 hz d) 2512 hz
21
2)
Φ1: a) 0o
b) 377o
Φ2: a) 0o
b) 2512o
c) 180o
d) 30o
c) 180o d) –90o
An inductor of L = 5.3mH and a capacitor of C = 100F are operated at 60hz.
find XL and XC. If the system frequency somehow increases what would happen to
XL and XC?
3)
XL: a) 5.3Ω │90o
b) 5.3 Ω │-90o
XC: a) 100Ω │-90o
b) 100Ω │90o
c) 2Ω │90o
d) 2Ω │-90o
c ) 26.5Ω │90o
d) 26.5Ω │-90o
For a single phase series A/C circuit
R = 4Ω , XL  2 , VS  100v 0v
find: Z, I, VR, VL, P, Q, S, and the pf.
Z:
a) 6Ω │37o
I:
a) 16.7A │37o
b) 16.7A │-37o
c) 22.4A │26.6o
d) 22.4A │-26.6o
VR: a) 66.8V │37o
b) 66.8V │-37o
c) 89.6V │26.6o
d) 89.6V│-26.6o
VL: a) 33.4V │53o
b) 33.4V│-53o
c) 44.8V 63.4o
P:
4)
a) 1 KW
b) 6Ω │-37o
b) 2 KW
c) 3 KW
Q: a) 1 KVAR
b) 2 KVAR
S:
b) 2.24 KVA
a) 1.1 KVA
c) 4.47Ω │26.6o
d) 4.47Ω │-26.6o
d) 44.8V │-63.4o
d) 4 KW
c) 3 KVAR
d) 4 KVAR
c) 3.36 KVA
For a single phase parallel A/C circuit
R = 10Ω , X C  20 , VS  100v 0v
Find: VR, VC, IR, IC, IT, Z, P, Q, S, and the pf.
d) 4.47 KVA
22
5)
VR: a) 50 V
b) 100 V
c) 150 V
d) 200 V
VC: a) 50 V
b) 100 V
c) 150 V
d) 200 V
IR:
a) 5A │0o
IC:
a) 5A │90o b) 5A │-90o
IT:
a) 5A │26.6o
Z:
a) 15Ω │26.6o
P:
a) .5 KW
Q:
a) .5 KVAR
S:
a) .5 KVA
c) 5A │26.6o
b) 10A│0o
c) 10A │26.6o
d) 10A│-26.6o
b) 11.2A -26.6o
c) 11.2A │26.6o
b) 15Ω │-26.6o
c) 8.9Ω │26.6o
b) 1.0 KW
c) 1.12 KW
b) 1.0 KVAR
b) 1.0 KVA
d) 5A │-26.6o
d) 8.9Ω │-26.6o
d) 1.5 KW
c) 1.12 KVAR
c) 1.12 KVA
d) 1.5 KVAR
d) 1.5 KVA
A balanced 3 phase delta load has phase current of Iphase = 30A at .8 lagging power
factor, and is connected to a 450V, 3 phase system.
Calculate or find: VLine , Vphase ,
6)
d) 10A │26.6o
ILine, Z , P, Q and S.
VLine: a) 150 V
b) 260.1 V
c) 450 V
d) 778.5 V
VPhase: a) 150 V
b) 260.1 V
c) 450 V
d) 778.5 V
ZΔ:
a) 5Ω │36.9o
b) 8.7Ω │36.9o
P:
a) 24.3 KW
Q:
a) 24.3 KVAR
S:
a) 24.3 KVA
b) 32.3 KW
c) 40.4 KW
b) 32.3 KVAR
b) 32.3 KVA
c) 15Ω │-36.9o
d) 15Ω │36.9o
d) 55.6 KW
c) 40.4 KVAR
c) 40.4 KVA
d) 55.6 KVAR
d) 55.6 KVA
A balanced 3 phase, wye load is connected to a 208V, 3 phase system. The load
draws 1.5KW and 0.8 KVAR. Calculate or find VLine , Vphase , ILine, Iphase , ZY ,
P, Q , S and the pf.
VLine: a) 69.3 V
b) 120 V
c) 208 V
d) 360 V
VPhase: a) 69.3 V b) 120 V c) 208 V d) 360 V
ILine:
a) 2.73 A
b) 4.72 A
c) 8.16 A
d) 14.2 A
IPhase: a) 2.73 A
b) 4.72 A
c) 8.16 A d) 14.2 A
23
7)
P:
a) 1.5 KW
b) .8 KW
Q:
a) 1.5 KVAR
S:
a) 1.5 KVA
c) 1.7 KW
b) .8 KVAR
b) .8 KVA
d) 2.3 KW
c) 1.7 KVAR
c) 1.7 KVA
d) 2.3 KVAR
d) 2.3 KVA
A 3 phase, 450V, wye connected synchronous generator is supplying rated
1500KVA to a .8 lagging power factor load. The prime mover is at rated speed of
900RPM and the generator is at rated frequency of 60hz.
Calculate or find:
VLine , Vphase ,
VLine: a) 150 V
ILine,
Iphase , P, Q , and the number of poles.
b) 260.1 V c) 450 V
d) 778 V
VPhase: a) 150 V b) 260.1 V c) 450 V d) 778 V
ILine:
a) 633 A
b) 1098 A
c) 1927 A
IPhase: a) 633 A
b) 1098 A
c) 1927 A d) 3295 A
P:
a) 900 KW
b) 1200 KW
Q:
a) 900 KVAR
Poles: a) 2
8)
b) 4
c) 1500 KW
b) 1200 KVAR
c) 6
d) 3295 A
d) 2700 KW
c) 1500 KVAR
d) 8
Repeat problem No. 7, but with a delta generator.
lagging pf, 60hz and 900RPM.
VLine: a) 150 V
d) 2700 KVAR
b) 260.1 V c) 450 V
Same 450V, 1500KVA, .8
d) 778 V
VPhase: a) 150 V b) 260.1 V c) 450 V d) 778 V
ILine:
a) 633 A
b) 1098 A
c) 1927 A
IPhase: a) 633 A
b) 1098 A
c) 1927 A d) 3295 A
P:
a) 900 KW
b) 1200 KW
Q:
a) 900 KVAR
Poles: a) 2
9)
b) 4
c) 1500 KW
b) 1200 KVAR
c) 6
d) 3295 A
d) 2700 KW
c) 1500 KVAR
d) 2700 KVAR
d) 8
A single phase, 60hz step-down transformer operates at rated load and voltage of
24
230V/115V, 10KVA
and .8 lagging power factor
Calculate or find
V1 ,
V2 , I1 , I2 , Zload , S2 , S1 , P2 , and P1 .
If this transformer has N1 = 200 turns, Calculate N2.
V1:
a) 115 V
b) 230 V
V2:
a) 115 V
b) 230 V
I1:
a) 43.5 A
b) 87 A
I2:
a) 43.5 A
b) 87 A
ZLoad:
10)
a) 2.64Ω │37o
b) 1.32Ω │37o
c) 2.64Ω │-37o
S2:
a) 5 KVA
b) 10 KVA
c) 20 KVA
S1:
a) 5 KVA
b) 10 KVA
c) 20 KVA
P2:
a) 4 KW
b) 8 KW
c) 10 KW
P1:
a) 4 KW
b) 8 KW
c) 10 KW
N2:
a) 100
b) 200
d) 1.32Ω │-37o
c) 300
A 3 phase delta to wye transformer is a step-down device. It operates at its rated
nameplate values of:
208V/120V and 12KVA at power factor .8 lagging.
Calculate or find for both the primary and secondary:
VLine , Vphase ,
ILine,
Iphase , P, S, and Q.
If the three phase transformer consists of 3 identical single phase devices, what is
the KVA rating of each single phase transformer?
25
For the primary:
VLine: a) 69.3 V b) 120 V c) 208 V
d) 360 V
VPhase: a) 69.3 V
b) 120 V
c) 208 V
ILine:
b) 19.2 A
c) 33.3 A
a) 11.1 A
S1: a) 7.2 KVA b) 9.6 KVA
P1: a) 7.2 KW
b) 9.6 KW
Q1: a) 7.2 KVAR
d) 360 V
d) 57.8 A
c) 12 KVA d) 16.8 KVA
c) 12 KW
b) 9.6 KVAR
d) 16.8 KW
c) 12 KVAR d) 16.8 KVAR
For the secondary
VLine: a) 69.3 V b) 120 V c) 208 V
d) 360 V
VPhase: a) 69.3 V
b) 120 V
c) 208 V
ILine:
b) 19.2 A
c) 33.3 A
a) 11.1 A
S2: a) 7.2 KVA b) 9.6 KVA
P2: a) 7.2 KW
b) 9.6 KW
Q2: a) 7.2 KVAR
d) 360 V
d) 57.8 A
c) 12 KVA d) 16.8 KVA
c) 12 KW
b) 9.6 KVAR
d) 16.8 KW
c) 12 KVAR d) 16.8 KVAR
KVA Per Each Single Phase: a) 4 KVA b) 6.94 KVA c) 12 KVA d) 20.8 KVA