Reference no: EM131203008
Question 1 - Three balanced loads are connected to three phase supply via a feeder line ZF = 0.021 +j 0.12 0 Ω. The three loads are connected across a 415V, 50 Hz three phase source (Take 415∠0o V as reference).
Load 1 Δ 5 + j12 Ω per phase;
Load 2 Y 3.39∠53.13o Ω per phase;
Load 3 25 kVA at PF 0.62 Lag.
(i) Draw a diagram of the load system.
(ii) Calculate the Total Complex Power Loads, SL, PL & QL.
(iii) The current drawn from the source IS.
(iv) The MAGNITUDE of the source Line Voltage.
(v) The input PF to the system as seen by the source.
(vi) If three capacitors each of 1200 μF are connected STAR across the source calculate the new source PF.
QUESTION 2 - A 30 kVA, 3.3kV/240V, 50 Hz single phase transformer as the following parameters.
R1 = 1.0Ω X1 = j2.0 Ω RC = 15 kΩ
R2 = 5.0mΩ X1 = j10 mΩ XM = 3 kΩ
When delivering 80% of FULL LOAD at 0.85 PF lag. The output voltage is the rated Voltage of 240V. For this load condition:
(i) Draw the approximate Equivalent circuit referred to the PRIMARY.
(ii) Determine the Secondary Load Impedance ZL in Polar Form.
(iii) Calculate the INPUT voltage and current, V1 & I1.
(iv) Calculate the Complex Power delivered by the source to the transformer S1, P1 & Q1.
(v) Calculate the efficiency of the transformer at this load.
QUESTION 3 - A three phase induction motor with "Design B" characteristics is used as a water pump in a desalination plant. The motor name plate details are as follows:
RATING 15 kW VOLTAGE 415 V
POLES 4 CURRENT 28.0
SPEED 1439 rev/min CONNECTION DELTA
FREQUENCY 50 Hz INSULATION E
The Friction and Windage Losses plus Core Losses PF/W + PCORE = 1.51 kW as used in the IEEE model and may be considered constant over the speed range of the motor.
The known motor parameters are:
R1 = 0.92Ω X1 = j3.22Ω
XM = j95Ω X'2 = j4.81Ω
When operating at a speed of 1452 rev/min (NOT FULL LOAD) the input Power Factor (PF) is 0.895 lagging and the load torque is 77.23 Nm.
(i) Draw the IEEE model of the induction motor with known parameters values shown on the model. Calculate for the new speed of 1452 rev/min:
(ii) The Output Power of the Motor driving the Load.
(iii) The Input UNE current drawn from the supply by the motor.
(iv) The Efficiency of the motor.
QUESTION 4 - A 500 V, d.c. shunt motor draws 122 A from the supply when operating at approximately half full load. The Motor name plate data are as followings:
RATING 115 kW VOLTAGE 500 Vdc
POLES 6 CURRENT 248 A
SPEED 450 rev/min CONNECTION SHUNT
The Motor design parameters are:
ARAMATURE-
Length - 25 cm
Diameter - 55 cm
Slots - 61
Conductors/slot - 8
Turns/coil - 1
Total Resistance - 0.0763Ω
Connection - Wave
Commutator segments - 244
SHUNT FIELD-
Turns/pole - 2240
Total Resistance - 230.5Ω
INTERPOLE
Turns/pole - 28
Total Resistance - 0.0132Ω
GENERAL
Connection - shunt
Flux/pole - 43.35 mWb
Friction and Windage Loss - 2.02 kW
Brush volt drop - 2.0 V
(i) Draw a diagram of the Shunt Connected Motor including the Field Rheostat.
(ii) Determine the speed of the motor if it draws 122 A at Half Full Load, assuming that the volt drop across the brushes is 2.0 Vat all load values.
If the resistance of the shunt field is increased by means of a Field Rheostat RH = 69.15 Ω the field flux (Φ) will decrease and the motor speed (N) will increase.
This speed increase results in a 20% increase in developed torque (TDEV) from the half load condition. For this new operating condition, determine:
(iii) the current drawn from the supply;
(iv) the new speed of the motor.
Assume Linear Magnetic Characteristics and neglect Armature Reaction.