12 | Electrical Power
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Electrical Power define
Electrical Power define
Electric power is define as;
the rate, per unit time, at which electrical energy is transferred by an electric circuit.
rate of doing work, measured in watts, and represented by the letter P.
The SI unit of power is the watt, one joule per second. The term wattage is used colloquially to mean "electric power in watts."
The electric power in watts produced by an electric current I consisting of a charge of Q coulombs every t seconds passing through an electric potential (voltage) difference of V is;
P = work done per unit time = W÷t = W÷Q , Q ÷ t = V x I
whereQ is electric charge in coulombst is time in secondsI is electric current in amperesV is electric potential or voltage in voltsWhat is True Power ?
What is True Power ?
Actual or real power consumed amount of power being used by a load equipment.
Unit is Watts (or ' W' ) and symbol commonly used is 'P'.
Formula: 1
True power formula
Formula: 2 AC Power formula
What is Apparent Power?
What is Apparent Power?
Design power or calculated power of a load equipment.
Unit is Volt-ampere (or VA ).
The symbol commonly used is ' S '.
Apparent Power Formula
S = V x I or S = P ÷ p.f
What is Reactive Power?
What is Reactive Power?
Reactive power is the resultant power in watts of an AC circuit when the current waveform is out of phase with the waveform of the voltage, usually by 90 degrees if the load is purely reactive, and is the result of either capacitive or inductive loads. Reactive power is used to provide the voltage levels necessary for active power to do useful work.
Unit is Volt-Ampere Reactive (or VAR)
Symbol commonly used for calculations is ' Q '.
Reactive Power Formula
Q = √ S² - P²
The Power Triangle
The Power Triangle
The relationship between real power, reactive power and apparent power can be expressed by representing the quantities as vectors. Real power is represented as a horizontal vector and reactive power is represented as a vertical vector. The apparent power vector is the hypotenuse of a right triangle formed by connecting the real and reactive power vectors.
This representation is often called the power triangle. Using the Pythagorean Theorem, the relationship among real, reactive and apparent power is:
Real and reactive powers can also be calculated directly from the apparent power, when the current and voltage are both sinusoidal with a known phase angle θ between them:
P = S x cos θ
Q = S x sin θ
Real power formula
The ratio of real power to apparent power is called power factor and is a number always between −1 and 1.
Where the currents and voltages have non-sinusoidal forms, power factor is generalized to include the effects of distortion.
What is Power Factor ?
What is Power Factor ?
In electrical engineering, the power factor of an AC power system is defined as the ratio of the real power absorbed by the load to the apparent power flowing in the circuit, and is a dimensionless number in the closed interval of −1 to 1.
Ideal power factor is 0.85 or greater.
When the angle cos θ becomes greater, this will draw more current to the connected load. When this happens, the power factor value will drop below 0.84 and this impact overall electrical system.
If the angle cos θ becomes smaller, then this helps improving the power factor value.
Power factor formula
p.f = P ÷ S
The effects of low power factor
The effects of low power factor
In an electric power system when overall power factor get below 0.85, this may cause series effects onto building or factory electrical system.
Example when electrical loads with a low power factor will draws more current than a load with a high power factor for the same amount of useful power transferred. The higher currents increase the energy lost in the distribution system, and require larger wires and other equipment. Because of the costs of larger equipment and wasted energy, electrical utilities will usually charge a higher cost to industrial or commercial customers where there is a low power factor.
Common causes is too many inductive loads.
Summary of low power factor:
current draws more from load
all conductors need to increase sized
all conductor temperature will rise than normal
consumer will get penalty 'penalty demand charges'
all circuits protection such as MCBs and MCCBs will cause fault
changes must get larger transformers
all changes on all switch gear
replacing all fuses to higher current rating
cable will suffer higher voltage drops
higher generating required by electricity provider
series copper losses (de-rating of conductors)
Solution to low power factor
Solution to low power factor
It is crucial to raise up the power factor value to achieve at least greater than 0.85 (better if >0.85). This helps to eliminate the problem encounter from effects of lower power factor. This commonly known as power factor correction (or PFC).
The only solution to raise up power factor is by using capacitor bank circuit.
When capacitor bank is install it will increase the power factor of a load, improving efficiency for the distribution system to which it is attached. Linear loads with low power factor (such as induction motors) can be corrected with a passive network of capacitors or inductors.
Finding cos θ angle in degree
Formula: cos θ angle = cos-1 pf = (°)
Example calculation
Example calculation
Example: True Power
1) Voltage is 24 volts, current I = 3A calculate total power, W.
Answer: P = V x I = 24v X 3A = 72 watts
2) The current is I = 10A and resistance is 65 ohm, calculate total power, W.
Answer: P = I^2 x R = (6A x 6A) X 65 = 2340 watts
3) Given V = 240 volts, current I = 20A, power factor is 0.88 on load connected, calculate power in single phase.
Answer: P = V x I x p.f = 240v X 20A x 0.88 = 4224.0 watts
4) Given V = 415 volts, current I = 30A, power factor is 0.86on load connected, calculate power in three phase.
Answer: P = √ 3 x V x I x p.f = √ 3 X 415v X 30A x 0.86 = 18,544.5 watts
Example: Apparent Power
1) Given voltage supply is 240 volts and current is 60 ampere, calculate the apparent power, S.
Answer: S = V x I = 240v x 60A = 14,400 VA (or 14.4 k VA)
2) Given power is 24650 watts and power factor is 0.87, calculate the apparent power, S.
Answer: S = P ÷ p.f = 24,650 W ÷ 0.87 = 28,333.33 VA (or 28.3 k VA)
Example: Reactive Power
1) Given value of S = 50 kVA and P = 38 kW, calculate the Q value.
Answer: Q =√S² - P² = √50² - 38² = 32.49 kVAR
To calculate the current shall apply this,
I = kVAR ÷ ( √ 3 X 415v )
Answer: I = 32.49 x 1000 VAR ÷ ( √ 3 X 415v ) = 45.20 ampere
Example: Angle of cos in degree
1) Given value of pf = 0.78, calculate the angle of cos θ in degree.
Answer: cos-1 pf = (°) = cos-1 0.78 = 38° (or 38.73)
Source
https://en.wikipedia.org/wiki/Power_factor
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