Topic 8 energy and power requirements of AC systems topic b8

Solve Problems in Single and Three
Phase Low Voltage Circuits

Topic 8: Energy and Power
Requirements of AC Systems


Three-Phase TRUE Power
True Power in ‘A’ Phase
PA = VAIACos Ø A
Total True Power in 3Ø Circuit
VA

ZA

ZB

P3Ø = 3x PA

ZC

Total True Power in 3Ø Circuit
P3Ø = 3x VAIACos Ø A
Total True Power in 3Ø Circuit

IA
L1

P3Ø = 3x VPIPCos Ø P
L2

L3

N

P3Ø = √3 VL IL Cos Ø

3Ø, 400 V
20/02/17

Total True Power in 3Ø
Circuit

Nathan Condie

2


Three-Phase Power Triangle

S
Apparent
Power

Reactive Power

(Ø)
Phase Angle

Q

P
True Power

20/02/17

Nathan Condie

3


Three-Phase Power Triangle

Apparent Power (S)
S = √3 VL IL
Ø

Reactive Power (Q)
Q = √3 VL I L sin Ø

True Power (P)
P = √3 V LI L cos Ø

20/02/17

Nathan Condie

4


Three-Phase Power

Exercises



A 3Ø, 415V motor with an impedance per phase of 15Ω at a pf
of 0.6 lag will consume how much true power when…

1.




A 3Ø, 400V motor rated at 12.5kW with a pf of 0.8 lag. If the
motor is delta-connected, determine:

2.




20/02/17

Connected in STAR configuration
Reconnected in DELTA configuration

Impedance per phase
True power if reconnected in STAR

Nathan Condie

5


Three-Phase Power
Answers Q.1



Star P = 6889 W

i.


Step 1: VL = 415V, so Vp = 239.6V



Step 2: Ip = 15.97A, so IL = 15.97A



Step 3: P= √3 V LI L cos Ø


Delta P = 20667 W

ii.


Step 1: VL = 415V, so Vp = 415V



Step 2: Ip = 27.67A, so IL = 47.92 A



Step 3: P= √3 V LI L cos Ø


20/02/17

P=√3x415x15.97x0.6 = 6889 W Star

P= √3 x 415 x 47.92 x 0.6 = 20667W

Nathan Condie

6


Three-Phase Power
Answers Q.2



Find ZP

i.


Step 1: Delta VL = 400 V, so Vp = 400 V



Step 2: IL = P/(√3xVLxCosØ), so IL = 22.55 A



Step 3: IP= I L /√3 = 13.02 A
ZP=400/13.02A = 30.72 Ω



Find Star P

ii.


Step 1: VL = 400 V, so Vp = 230.9 V



Step 2: IP = VP /ZP = 7.52 A, so IL = 7.52 A



Step 3: P= √3 V LI L cos Ø




20/02/17

P= √3 x 400 x 7.52 x 0.8 = 4166W
Note: 4166W x 3 = approx 12500W (delta power)
Nathan Condie

7


Three-Phase Power
Exercise 3

V

ZA

ZB

W

A

A balanced 3-phase load
has the measurements
recorded as per the
diagram. Determine the
following:

ZC

220V

2464W
L1

1) P consumed by load
16A

L2

2) Pf of the load
L3

N

3) Q of the load
4) Would a 10kVA
alternator be suitable to
supply this load?

20/02/17

Nathan Condie

8


Three-Phase Power

Answers Ex.3



P = 3 x P1Ø = 7392 W
Pf = 0.7

i.
ii.

Step 1:Vp = 220V, so VL = 381V
Step 2: Ip = 16A, so IL = 16A
Step 3: P= √3 V LI L cos Ø






Q = 7544 VARs

iii.

Step 1: Pf = cos Ø therefore Ø = cos-1(Pf)
Step 2: Ø = 45.60E
Step 3: Q = √3 V LI L sin Ø







20/02/17

Pf =7392 / (√3x381x16) = 0.7

Q= √3 x 381 x 16 x sin 45.6 = 7544 VARs

Nathan Condie

9


Methods of Correcting Power Factor:
Three Phase Installations





Ensuring that all motors and transformers
are correctly loaded
Over-exciting a synchronous motor
Adding capacitance to an installation (that is
inductive)

20/02/17

Nathan Condie

10


Energy and Power
Requirements of AC Systems
Power Measurement


Three-Phase Power Measurement


ONE-Wattmeter Method
PA

W

L1

3Ø,
fourwire

L2

W

L3

W

ThreePhase

Load

N

20/02/17

P3Ø = PA + P B + P C

Unbalanced Load

P3Ø = 3 x PA

Balanced Load Only
Nathan Condie

12


Three-Phase Power Measurement


TWO-Wattmeter Method
L1

3Ø,
threewire

W1
ThreePhase
Load

L2

L3

W2
P3Ø = W1 + W2

20/02/17

Balanced Load Only

Nathan Condie

13


Three-Phase Power Measurement


TWO-Wattmeter Method (cont)
 Power Factor Calculation


Step 1: Use Two-Wattmeter method





Step 2: Use Transposition to Find Ø




Ø = Tan-1(above number)

Step 3: Find Power factor


20/02/17

Tan Ø = √3 (W2 – W1)
(W2 + W1)

Pf = Cos Ø

Nathan Condie

14


Three-Phase Power Measurement


Exercise 1
+4.6kW
L1

3Ø,

threewire

Determine:

W1
ThreePhase
Load

L2

L3

W2

2. Pf of load
3. Apparent power
flowing from supply
4. Reactive power
oscillating in load

-1.1kW

20/02/17

1. True power
consumed by load

Nathan Condie

15

Three-Phase Power Measurement


Answers
1.
P = +4.6 + -1.1 = +3.5kW
2.
Pf = cos() = 0.33
1.
2.

3.
4.

20/02/17

TanФ = √3 (-5.7/3.5) = √ 3 (-1.629) = -2.82
Ф = Tan-1 (-2.82) = 70.470

S = 10.6 kVA
Q = 10 kVAR

Nathan Condie

16


Three-Phase Power Measurement



THREE-Wattmeter Method
L1
3Ø,
threeor
fourwire

W1

L2

W2

L3

W3

ThreePhase
Load

Artificial Star-point

N
P3Ø = W1 + W2 + W3
20/02/17

Balanced OR Unbalanced Load
Nathan Condie

17


Energy and Power
Requirements of AC Systems
Purpose for Energy Measurement


Three-Phase Energy and Power
Measurement


Purpose for Energy Measurement
 Revenue Raising




Planning and design requirements




Monitoring the total load required for a given area to
ensure sufficient supply is available

Power factor specifications


20/02/17

Monitoring and billing for the energy usage of the
customer

Ensuring the existing infrastructure is utilized to its
highest capacity

Nathan Condie

19


Three-Phase Energy and Power
Measurement


Types of Measurement
Instruments


Energy Meter




20/02/17

Measures true power
(electrical energy
consumed) over time (kWh

meter)
Induction-disc type or
Electronic

Nathan Condie

20


Three-Phase Energy and Power
Measurement


Types of Measurement
Instruments


Maximum Demand Indicator




Power Factor Meter


20/02/17

Measures the highest
apparent power occurring
within a set time frame – for

the purpose of determining
tariff level.
Measures phase relationship
between supply voltage and
current – for the purpose of
determine tariff level.

Nathan Condie

21


Three-Phase Energy and Power
Measurement

20/02/17

Nathan Condie

22


Three-Phase Energy and Power
Measurement

20/02/17

Nathan Condie

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