Electrical Engineering Assignment Solution: Alternators and Circuits

This document presents a detailed solution to an electrical engineering assignment. The solution includes calculations for three-phase systems, analyzing currents and phase angles using polar and rectangular coordinates. It covers questions related to alternator design, including drawing stator coil interconnections, sketching output waveforms, and identifying system advantages. The assignment further addresses power calculations in circuits with resistive, inductive, and capacitive components, including determining currents, true power, reactive power, apparent power, and neutral current. It also involves applying relevant formulas for synchronous speed and frequency calculations, providing a comprehensive analysis of electrical power systems and circuit behavior.

ELECTRICAL L 31
ELECTRICAL L3
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Course
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Institution
Location
Date

ELECTRICAL L 32
QUESTION 19
From the following data
Red phase = 40A laggiing 450
White phase = 50 A lagging 300
Blue Phase = 60A lagging 200
( since the angles are lagging the currents are taken as negative)
X components
The value of X components can be calculated as below;
Red Phase
cos∅ × HYP

ELECTRICAL L 33
cos45 ×−40 = 28.284
White phase
cos∅ × HYP
cos30 ×−50 = 43.301
Blue Phase
cos∅ × HYP
cos20 ×−60 = 56.381
Y components
Red phase
sin∅ × HYP
sin45 ×−40 =-28.284
White phase
sin∅ × HYP
sin30 ×−50 =-25
Blue phase
sin∅ × HYP
sin20 ×−60 =−20.521
The above values can be presented in a table of polar and rectangular coordinate
Polar Coordinate Rectangular Coordinates
Phasor Phasor size Angle X value Y value
Red -40 A 45 28.284 -28.284
White -50A 30 43.301 -25
Blue -60A 20 56.381 -20.521
Neutral -147.7 A 29.9 127.966 -73.805
The X and Y components for the Neutral is obtained through addition of respective components
Y component = -28.284 -25-20.521
Y component = -73.805

ELECTRICAL L 34
X components = 28.284+43.301+56.381
X Component = 127.966
The value of current at nuetral is calculated using pythagorous theorem as below
I N = √ 127.9662+ 73.8052
I N =¿ -147.7 A ( lagging angles)
The angle at the neutral is calculated as below
Cos ∅ = 127.966
147.7 = 0.86639
∅ 29.860
QUESTION 20
Red phase = 60 Ω
White phase = 1000watts
Blue phase = 100Ω
Voltage = 400 and frequency = 50Hz
Current at the nuetral is obtained from the following equation;
I N = √ I R
2 + IB
2 +IW
2− ( IR I B )− ( IW IB ) −(I R IW )
Current on the Red Phase
P= V 2
R
P= 4002
60

ELECTRICAL L 35
P= 2666.6 W
And from
I= P
V
I= 2666.6
400
I= 6.6A
Current on White phase
From I = P
V
I = 1000
400
I= 2.5 A
Current on the Blue phase
From I = P
V
I = 400
100
I= 4 A
From I N = √ I R
2 +IB
2 + IW
2− ( IR IB )− ( IW IB ) −(I R IW )
I N = √ 6.62 +2.52 +42− ( 6.6 × 2.5 ) − ( 6.6 × 4 ) −(4 × 2.5)
I N = √ 65−16.5−10−26.4
I N = √ 12.1
I N =3.478 A

ELECTRICAL L 36
QUESTION 21
Red-Neutral = 60 Ω resistive
White-Neutral = 95mH plus 20Ω
Blue-Neutral = 100μF
Voltage = 415 and frequency = 50Hz
Total true power
On Red-Neutral
V 2
R = 4152
60 =2870.416Watts
On White-Neutral
V 2
R = 4152
20 =8611.25Watts
Total true power = 2870.416 Watts+ 8611.25 Watts
Total true power = 11481.66 W
Total reactive power
On White-Neutral
X L=2 πfL
X L=2× 3.142× 50× 95 ×10−3
29.849 Ω
Reactive power On White-Neutral

ELECTRICAL L 37
¿ 4152
29.844 =5769.875 Var
On Blue-Neutral
XC = 1
2 πfC
XC = 1
2× 3.142× 50× 100× 1 0−6
XC= 31.826 Ω
Reactive power On Blue-Neutral= 4152
31.826 = 5411.456 Var
Total reactive power = 5769.875 Var+¿5411.456 Var
Total reactive power = 11181.326 Var
Apparent power
From the pahsor diagram (power triangle) below;
S¿ √11481.662 +11181.3262

ELECTRICAL L 38
S=16026.558VA
Neutral Current
From the formula
I N = √ I R
2 + IB
2 +IW
2− ( IR I B )− ( IW IB ) −(I R IW )
Current on Red Phase
I= P
V
I= 2870.41
415
I=6.916 A
Current on the Blue phase
I= P
V
I= 5411.456
415
I=13.039 A
Current on White phase
In this case we have two different power (reactive and resistive power) thus we have to look for
the resultant power which will be employed in calculating the current
S= √5769.8752+ 8611.252
S= 10365.572VA
= S
V

ELECTRICAL L 39
I= 10365.572
415
I=24.977A
From I N = √ I R
2 + IB
2 +IW
2− ( IR I B )− ( IW IB ) −(IR IW )
I N = √ 6.9162 +13.0392+24.9772− ( 6.916 × 13.039 )− ( 24.977 ×6.916 )−(13.039× 24.977)
I N = √ 841.697−90.177−172.74−325.675
I N = √ 253.104897
I N =¿ 15.909 A