Week 2 _ Unit 27 Practice Problems
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Unit 27 Practice Problems
1.
Refer to the circuit shown in Figure 27–19 to answer the following questions, but assume that the alternator has a line voltage of 240 V and the load has an impedance of 12 V per phase
. Find all the missing values
FIGURE Calculating three-phase values using a wye-connected power source and a delta-connected load (example circuit).
EL(A) = 240V
Z(Phase) = 12
Since the load is directly connected to the alternator, the line voltage supplied by the alternator is equal to the line voltage of the load.
EL(L) = 240V
The three resistors in load are connected in delta, the line voltage and phase voltage of the load is same
.
EP(L) = EL(L)
= 240V
Therefore, the phase voltage of the load is
, EP(L) = 240V
Step 4/9
We are provided with impedance of 12
for each phase on load side, so the phase current of the load can be calculated with the help of phase voltage and impedance.
IP(L) = EP(L)
, = 240V
, = IP(L) 20A is the phase current of the load
Z
12
Step 5/9
In a Delta connection, the line current is 1,732
times greater than the phase current. So the load in which the three resistors are connected in Delta,
IL(L) = IP(L) x 1,732, = (20A) (1,732), = IL(L) 34.64A is the line current of the load
Step 6/9
Usually, the alternator supply line current to the load or loads connected. Here the alternator is connected to only one load. Therefore, the Line current of the load is equal to the line current of the alternator
IL(L) = IL(A), = IL(A) 34.64A is the line current of the alternator
Step 7/9
In a wye connection, line current is equal to the phase current. So on the alternator side the phase windings are connected in Wye,
IL(A) = IP(A)
= IP(A) 34.64 A is the phase current of the alternator
Step 8/9
In a Wye connection, the phase voltage is lesser than the line voltage by a factor of 1.732
. The phase voltage of the alternator is therefore,
EP(A) = EL(A),
= 240V = EP(A) 138.586V is the phase voltage of the alternator
1,732
1,732
Step 9/9
In the given circuit, the load is purely resistive. The “POWER FACTOR” is one because the current and voltage are in phase with each other. Therefore, the true power can be calculated as below:
P = 1,732 X EL(A) X IL(A) X Power Factor
P = 1,732 (240) (34.64 A) (1)
P = 14,399.155 W is true power
2. Refer to the circuit shown in Figure 27–20 to answer the following questions, but assume that the alternator has a line voltage of 4160 V
and the load has a resistance of 60 V
per phase. Find all the missing values
FIGURE Calculating three-phase values using a delta-connected source and a wye-connected load (example circuit).
The alternator line voltage is
= EL(A) 4160V
The load resistance per phase is
= Z(Phase) 60
EL(L) = 4160 V
In a Wye connection, the phase voltage is lesser than the line voltage by a factor of 1,732
EP(L) = EL(L)
1,732
= 4160V
1,732
EP(L)= 2401.849 V is the phase voltage of the load
We are provided with impedance of 60
for each phase connected on load side in Wye connection, so the phase current of the load can be calculated with the help of phase voltage and impedance.
IP(L) = EP(L)
Z
= 2401.849 V
60
IP(L)= 40.031 A is the phase current of the load
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In a Wye connection, the line current is equal to the phase current
IL(L) = IP(L)
IL(L) = 40.031 A is the line current of the load
Usually, the alternator supply line current to the load or loads connected. Here the alternator
is connected to only one load. Therefore the Line current of the load is equal to the line current of the alternator
IL(A) = IL(L)
IL(A) = 40.031 A is the line current of the load
In a Delta connection, the Phase current is less than the Line current equal. So on the alternator side the phase windings are connected in Delta,
IP(A) = IL(A)
1,732
= 40.031 A
1,732
IP(A)= 23.112 A
In a Delta connection, the phase voltage is equal to the line voltage. The phase voltage of the Alternator is therefore,
EP(A) = EL(A)
EP(A) = 4160 V is the phase voltage of the alternator
Step 10/10
In the given circuit, the load is purely resistive. The “POWER FACTOR” is one because the current and voltage are in phase with each other. Therefore, the true power can be calculated
as below:
P = 1,732 X EL(A) X IL(A) X POWER FACTOR
= 1,732 (4160V) (40.031A) (1)
P = 288,428,159 W is true power
Refer to the circuit shown in Figure to answer the following questions, but assume that the alternator has a line voltage of 560 V
. Load 1 has a resistance of 5 Ω
per phase, and Load 2 has
a resistance of 8 Ω
per phase. Find all the missing values.
FIGURE Calculating three-phase values using a wye-connected source and two three phase loads (example circuit)
The line voltage of the alternator is EL(A) = 560V
The resistance of the load 1 per phase is = Z(PHASE) = 5
The resistance of the load 2 per phase is = Z(PHASE) = 8
Step 2/12
Since the load 1
and load 2
are directly connected to the output of the alternator, the line voltage supplied by the alternator is equal to the line voltage of the loads.
EL(L1) = 560V
EL(L2) = 560V
Step 3/12
Load 2
is connected in delta, so the phase voltage is same as the line voltage.
EP(L2) = EL(L2)
EP(L2) = 560V is the phase voltage of load 2
Step 4/12
In Load 2, each phase has an impedance of 8
, Therefore the phase current is
IP(L2) = EP(L2)
Z
= 560 V
8
IP(L2) = 70A is the phase current of the load 2
Step 5/12
Load 2
which is delta connected, in which the line current is 1,732
times greater than the phase current
.
IL(L2) = IP(L2) X 1,732
= 70A X 1,732
IL(L2) = 121.24A is the load current of load 2
Step 6/12
EP(L1) = EL(L1)
1,732
= 560V
1,732
EP(L1) = 323.326V is the phase voltage of load 1
Step 7/12
IP(L1) = EP(L1)
Z
= 323.326V
5
IP(L1) = 64.665A is the phase current of the load 1
Step 8/12
Load 1 is connected in wye, so the phase current is same as the line current
IL(L1) = IP(L1)
IL(L1) = 64.665A is the load current of load 1
Step 9/12
The alternator supply line currents to load 1 and load 2 and they are purely resistive. Alternator line current is the summation of line current of load 1 and load 2.
IL(A) = IL(L1) + IL(L2)
= 64.665A + 121.24A
IL(A) 185.905A is the alternator supply line
Step 10/12
But the phase windings of the alternator are connected in wye, the phase current is same as the line current.
IP(A) = IL(A)
= 185.905 A is the alternator supply phase current
Step 11/12
The alternator which is wye connected, so the phase voltage is 1,732
times lesser than the line voltage.
EP(A) = EL(A)
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= 560V
1,732
EP(A) = 323.326V is the alternator phase voltage
Step 12/12
In this problem, the load 1 and load 2 are purely resistive and the “power factor” is one. Therefore, the true power can be calculated as below:
P = 1,732 X EL(A) X IL(A) X PF
= 1,732 X 560V A 185.91A X 1
P = 180,317.83 W is true power
Refer to the circuit shown in Figure to answer the following questions, but assume that the alternator has a line voltage of 480 V. Load 1 has a resistance of 12 Ω per phase. Load 2 has an
inductive reactance of 16 Ω per phase, and Load 3 has a capacitive reactance of 10 Ω per phase. Find all the missing values.
FIGURE Calculating three-phase values with a wye-connected source supplying power to a resistive, inductive, and capacitive load (example circuit).
The line voltage of the alternator is EL(A)= 480V
The resistance of the load 1 per phase is R(Phase) = 12
The inductive reactance of the load 2 per phase is XL(PHASE)= 16
The capacitive reactance of the load 3 per phase is XC(PHASE) = 10
Step 1/14
The load 1, load 2 and load 3 are directly connected to the output of the alternator; the line voltage supplied by the alternator is equal to the line voltage of the loads.
EL(L1) = 480 V is load 1
EL(L2) = 480 V is load 2
EL(L3) = 480 V is load 3
Step 2/14
Calculating missing values for load 3:
Load 3 which is wye connected, and each phase has a capacitive reactance of 8
, so the phase voltage is 1,732
times lesser than the line voltage. So the phase voltage is
EP(L3) = EL(L3)
1,732
= 480 V
1.732
EP(L3) = 277.136 V is the phase voltage of load 3
Step 3/14
Now the current computed can be calculated by Ohm’s law, where the voltage applied to each capacitor is,
IP(L3) = EP(L3)
XC = 277.136 V
10
IP(L3) = 27.714A is the phase current of load 3
Step 4/14
In a wye connection, the phase current is equal to the line current. Therefore the line current of load 3 is,
IL(L3) = IP(L3)
IL(L3) = 27.714A is the line current of load 3
Step 5/14
Now, we have the below formula to calculate the capacitive power VARsc
VARsc =
3 EL(L3) IL(L3)
= (1,732) (480V) (27.714A)
VARsc = 23,040.311
Step 6/14
Calculating missing values for load 2:
Load 2 is Delta connected; therefore the phase voltage is equal to the line voltage.
EP(L2) = EL(L2)
= 480V
Step 7/14
To compute the Phase current, we have the phase voltage and an inductive reactance of 16
is present in each phase. Now by using Ohm’s law.
IP(L2) = EP(L2
)
XL
= 480V
16
IP(L2) = 30A is a phase current of load 2
Step 8/14
In a Delta connection, the line current is 1,732
times greater than the phase current. So the line current of Load 2 is:
IL(L2) = 1,732 P(L2)
= (30A) (1.732)
IL(L2) = 51.96A is the line current of load 2
Step 9/14
Load 2 is made up of Inductors; the reactive power VARsl
can be calculated using line values of voltage and current.
VARsl =
3 EL(L2) IL(L2)
= (1,732) (480V) (51.96V)
VARsl = 43,197.466
Step 9/14
Calculating missing values for load 1:
Load 1consist of three resistors with a resistance of 12
each, connected in wye. In wye connection, phase voltage is less than line voltage by a factor of 1,732
. The phase voltage of Load 1 is same as the phase voltage of Load 3.
EP(L1) = EL(L1)
1,732
= 480V
1,732
EP(L1) = 277.136V is the voltage phase of load 1
Step 10/14
With the provided resistance and phase voltage, the phase current can be calculated using Ohm’s law.
IP(L1) = EP(L1)
XL
= 277.136V
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12
IP(L1) = 23.095A IS THE PHASE CURRENT LOAD OF LOAD 1
Step 11/14
In a wye connection, the phase current is equal to the line current. Therefore the line current of load 1 is
IL(L1) = IP(L1)
IL(L1) = 23.095A is the line current of load 1
Step 12/14
Load 1 is purely resistive, so true power can be computed using the line value of current and voltage.
P = 1,732 EL(L1) IL(L1)
= (1,732) (277.136V) (23.095A)
P = 196,200.259W is true power
Alternator calculations:
The alternator must supply the line current for each of the loads. In this problem, the line currents are out of phase with each other.
The current flow in Load 1 is resistive and in phase with the line voltage.
The current flow in Load 2 is inductive and lags the line voltage by 90 degrees
.
The current flow in Load 3 is capacitive and leads the line voltage by 90 degrees
.
Therefore, on considering the above factors, to find the total current delivered by the alternator can be calculated a formula similar to find the current flow in an RLC
circuit can be employed.
IL(A) =
(IL (L1)2sq + (IL(L2) – IL(L3)2sq
=
(23.095 A)2 + (51.96A – 27.714A)2
=
1121.47541 A
IL(A) = 33.485A IS THE TOTAL LINE CURRENT DELIVERED BY ALTERNATOR
Step 13/14
Since the alternator windings are wye connected, the phase voltage of the alternator is times
1.732
lesser than the line voltage.
EP(A) = EL(A)
1.732
= 480V
1.732
EP(A) = 277.136V
IP(A) = IL(A)
IP(A) = 33.485A is the phase current of the alternator
Step 14/14
The apparent power can now be found using the line voltage and current values of the alternator:
VA = 1732 EL(A) IL(A)
=1.732 (480V) (33.485
)
VA = 27,838.09
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