Electrical power 1.
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Electrical power 1
ELECTRICAL POWER
By Name
Course
Instructor
Institution
Location
Date
ELECTRICAL POWER
By Name
Course
Instructor
Institution
Location
Date
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Electrical power 2
Question 1
a)
XL= 2πfL
XL= 2*3.142*50*0.1592
XL=50.02
XC= 1
2 πfC
XC= 1
2∗3.142∗50∗0.00003
XC=106.089
Coil current
I= V
XL+R
I= 230
90.020
I=2.55A
Capacitor current
I= V
XC
I= 230
106.089
Question 1
a)
XL= 2πfL
XL= 2*3.142*50*0.1592
XL=50.02
XC= 1
2 πfC
XC= 1
2∗3.142∗50∗0.00003
XC=106.089
Coil current
I= V
XL+R
I= 230
90.020
I=2.55A
Capacitor current
I= V
XC
I= 230
106.089
Electrical power 3
Icap= 2.167
b) Total current
IS= I total+ I capacitor
I= 2.55+2.167
Is=4.1717A
Phase angle
X= XL-Xc = 106.08-50
X= 56.086
R=40
Therefore the phase angle θ= 56.086
40
Phase angle θ=54.500
Part d
Impedance
Z2=√ R2 +( Xl−Xc)2
Z2=√402 +(56.08)2
Z2=√4745.639
Z= 68.886Ω
Question 2
a
Fr= 1
2 π √LC
Fr= 1
2∗3.142 √0.1592∗0,00003
Icap= 2.167
b) Total current
IS= I total+ I capacitor
I= 2.55+2.167
Is=4.1717A
Phase angle
X= XL-Xc = 106.08-50
X= 56.086
R=40
Therefore the phase angle θ= 56.086
40
Phase angle θ=54.500
Part d
Impedance
Z2=√ R2 +( Xl−Xc)2
Z2=√402 +(56.08)2
Z2=√4745.639
Z= 68.886Ω
Question 2
a
Fr= 1
2 π √LC
Fr= 1
2∗3.142 √0.1592∗0,00003
Electrical power 4
Fr= 72.8Hz
Part b
Rd= V
Icap
Rd= 300
2.167
Rd= 138.44Ω
Part c
Is= V
Z
Is= 300
68.355
Is=4.355 A
d)
Q factor
Q= 1
R √ L
C
Q= 1
40 √ 0.1952
0.00003
Q= 1.8211
e) Circuit bandwidth
B= R
2 πL
B= 40
2∗3.142∗0.1592
Fr= 72.8Hz
Part b
Rd= V
Icap
Rd= 300
2.167
Rd= 138.44Ω
Part c
Is= V
Z
Is= 300
68.355
Is=4.355 A
d)
Q factor
Q= 1
R √ L
C
Q= 1
40 √ 0.1952
0.00003
Q= 1.8211
e) Circuit bandwidth
B= R
2 πL
B= 40
2∗3.142∗0.1592
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Electrical power 5
B= 39.98Hz
Part f
The usage of the higher power factor help in reducing the amount of reactive power which in most
cases result to wastage power for 3 phase power supply. And some prototype calculation includes the
following
B= 39.98Hz
Part f
The usage of the higher power factor help in reducing the amount of reactive power which in most
cases result to wastage power for 3 phase power supply. And some prototype calculation includes the
following
Electrical power 6
Question 3
Part a
Current drawn
From P=VI
3000= 300 I
300
I= 10 A
b).
Actual P.F = 0.8
Required P.F = 0.95
θ1= 36.8990
θ2= 18.1940
Intended capacitor
Tanθ 1 -Tanθ 2
0.795079 – 0.32866 = 0.4221
3000 (0.4221)
= 1.266 kVAR
C= kVAR
2 πf V 2
Question 3
Part a
Current drawn
From P=VI
3000= 300 I
300
I= 10 A
b).
Actual P.F = 0.8
Required P.F = 0.95
θ1= 36.8990
θ2= 18.1940
Intended capacitor
Tanθ 1 -Tanθ 2
0.795079 – 0.32866 = 0.4221
3000 (0.4221)
= 1.266 kVAR
C= kVAR
2 πf V 2
Electrical power 7
C= 1266
2∗3.142∗3002
C= 1266
28278000
C= 44.769 μF
Part c
(0.795079 – 0.32866) 3000 = 1266 kVAR
Original kVAR = (0.795079*3000) = 2252.38
Percentage reduction = 2252.38−1266
2252.38 × 100
Percentage reduction =43.792 %
Part d
C= 1266
2∗3.142∗3002
C= 1266
28278000
C= 44.769 μF
Part c
(0.795079 – 0.32866) 3000 = 1266 kVAR
Original kVAR = (0.795079*3000) = 2252.38
Percentage reduction = 2252.38−1266
2252.38 × 100
Percentage reduction =43.792 %
Part d
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Electrical power 8
Question 4
Advantages of 3 phase includes the following
Star connection
Diagram
L= 32mH
R=8Ω
V= 415 Hz
XL= 2*3.142*50*0.032
XL=10.0544 Ω
Vph= VL
√3
Question 4
Advantages of 3 phase includes the following
Star connection
Diagram
L= 32mH
R=8Ω
V= 415 Hz
XL= 2*3.142*50*0.032
XL=10.0544 Ω
Vph= VL
√3
Electrical power 9
Vph= 415
√ 3 = 239.6 V
Total power dissipated in star connection
P=3I2R
I= 239.6
18.05
I=13.274
P= 3*13.2742*8
P= 4228.77 Watts
Delta connection
VL=VPH =239.6
IL= √3 IPh
IL= √3∗13.274
IPh= 6.310A
Total dissipated power
P= √3 VL* IL*Cos θ
P= √3 239.6* 6.310*120
P= 1309.4 Watts
Vph= 415
√ 3 = 239.6 V
Total power dissipated in star connection
P=3I2R
I= 239.6
18.05
I=13.274
P= 3*13.2742*8
P= 4228.77 Watts
Delta connection
VL=VPH =239.6
IL= √3 IPh
IL= √3∗13.274
IPh= 6.310A
Total dissipated power
P= √3 VL* IL*Cos θ
P= √3 239.6* 6.310*120
P= 1309.4 Watts
Electrical power 10
References
Lam, C.-S., 2016. Parallel Power Electronics Filters in Three-Phase Four-Wire Systems: Principle, Control
and Design. 3rd ed. Liverpool: Springer.
Sensarma, P., 2017. Design and Control of Matrix Converters: Regulated 3-Phase Power Supply and
Voltage Sag Mitigation for Linear Loads. 2nd ed. Chicago: Springer.
References
Lam, C.-S., 2016. Parallel Power Electronics Filters in Three-Phase Four-Wire Systems: Principle, Control
and Design. 3rd ed. Liverpool: Springer.
Sensarma, P., 2017. Design and Control of Matrix Converters: Regulated 3-Phase Power Supply and
Voltage Sag Mitigation for Linear Loads. 2nd ed. Chicago: Springer.
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