2. Outline
In this presentation we are going to look at:
• Steady state description
• Phasors description
• Rectangular representation of phasors
• Polar representation
• Electromagnetism
• Coulomb’s law
• Ampere’s law
• Power factor
• Capacitor banks
• Sizing capacitor banks
• Conclusion
• references
3. STEADY STATE
Steady state refers to the position or state after a system becomes
settled and is at a ‘steady state’ or is working normally (Muppala, 2018)
4. Example of steady state
Before you throw stone in water it is in ‘equilibrium’. But when you throw
the stone, it collides with the surface of water. Therefore the momentum is
transferred to water which produces ripples. Eventually the Ripples die out
after some time and the stone sinks and the water is back to its stagnant
state.
• The Stone collides with the surface of water - forcing function which exists
only for a certain small time. Like an impulse.
• Water produces ripples - transient response.
• Ripples die out after some time and the water is back to its stagnant state -
Steady state response.
6. Electrical circuit continued
• The inductor has no energy stored initially. So it has no voltage across it.
• Therefore Inductor voltage = 0
• Rate of change of current = 0
• Initially current in inductor = 0 = Energy stored in inductor.
Eventually at t>0s; inductor stores current until coil voltage drop is equal to
applied voltage. Then when coil voltage drop is equal to applied voltage,
then current flowing through the circuit is constant as inductor voltage is
zero which means rate of change of current is zero. The state described is
called ‘steady state’. (Muppala, 2018)
10. Phasors continued
v (t)=sinθ
If the waveform has a period of 2π and frequency of f
Then
v (t)=sin2πft
If you consider a rotating rod and on the horizontal axis
v (t)=cos2πft
11.
12. For an amplitude of A and a lag of ф
The wave form can be represented as Acos(2πft+∅)
=Acos∅cos2πft-Asin∅sin2πft
Therefore at time t=0 the tip of the wave will have coordinates
(X,Y)=(Acos∅,Asin∅)
If we think of the plane as an Argand Diagram (or complex plane), then
the complex number X + jY corresponding to the tip of the rod at t = 0
is called a phasor (imperial college London, 2018)
Magnitude A= √(X²+Y²)
16. Electromagnetism
Magnetism is the phenomenon associated with magnetic fields which
arise from the movement of electric charges. And therefore movement
of electric current in a conductor or charged particles through space
(Eustace, Kashy, Robinson, Bleaney, & McGrayne, 2018)
And Electromagnetism is the science of charge and of the forces and
fields associated with charge. Electricity and magnetism are two
aspects of electromagnetism.
17. Refresher on magnetism
• From coulomb’s law, the electric field of a point charge is
𝐸 =
𝑞𝑅
4𝜋𝜀𝑅²
where ԑ is the permittivity of free space,
And the force between to charges can be got from F = qE
The flux from a surface is given as
Φ = ʃE.ds
The work done, W, in moving a particle of charge from point A to point
B along an arbitrary path in the electric field of charge located at r is
W = ʃQE(r) · dr
20. Ampere’s law
Ampere's Law states that for any closed loop path, the sum of the
length elements times the magnetic field in the direction of the length
element is equal to the permeability times the electric current enclosed
in the loop (hyperphysics, 2018).
𝐹𝑚𝑎𝑔 = 𝑑𝑄 𝑣 𝑥𝐵 = 𝐼𝑑𝑡 𝑥𝐵 = 𝐼𝑑𝑟 𝑥 𝐵
𝐹𝑚𝑎𝑔 = ʃ (𝐼𝑑𝑟𝑥𝐵)
23. Link it all to Electrical energy systems and renewable energy
24. POWER FACTOR
Most loads in modern electrical distribution systems are inductive.
Examples include motors, Transformers, gaseous tube lighting ballasts,
and induction furnaces. Inductive loads need a magnetic field to
operate. Inductive loads require two kinds of current:
• Working power (kW) to perform the actual work of creating heat,
light, motion, machine output, and so on.
• Reactive power (kVAR) to sustain the magnetic field (EATON, 2010)
25. Power factor continued
A high power factor signals efficient utilization of electrical power,
while a low power factor indicates poor utilization of electrical power.
26. How to measure your power factor
To determine power factor (PF), divide working power (kW) by
apparent power (kVA) (Rapid tables, 2018).
In AC systems, P = VI cos θ; where cos θ is the power factor.
PF = 𝑘𝑊
𝑘𝑉𝐴𝑅
27. Reading the power factor
It can be done
• Directly – through direct measuring or by reading off a power factor
meter
• Indirectly – through the reading of the active and reactive energy
meters (Csanyi, 2018)
28. Power factor continued
When apparent power (kVA) is greater than working power (kW), the
utility must supply the excess reactive current plus the working current.
Power capacitors act as reactive current generators.
32. CALCULATION
Assume an uncorrected 460 kVA demand, 480V, three-phase at 0.87 power factor.
Compare the bill for 0.87pf and 0.97pf.
Billing:
$4.75/kVA demand
Correct to 0.97 power factor
Solution:
kVA × power factor = kW
460 × 0.87 = 400 kW actual demand
kW/pf=kVA = 400/0.97 = 412 kVA
Bill for 0.97 pf
=412x4.75 = $1957
Bill for 0.87 pf
= 460x4.75 = $2,185
33. Selecting capacitor banks
When selecting capacitor banks, we are to choose matching the right
type size and number of capacitors.
There are two basic types of capacitor installations:
• individual capacitors on linear or sinusoidal loads,
• and banks of fixed or automatically switched capacitors at the feeder
or substation.
34. Advantages of individual capacitors at the load:
• Complete control; capacitors cannot cause problems on the line during light load conditions
• No need for separate switching; motor always operates with capacitor
• Improved motor performance due to more efficient power use and reduced voltage drops
• Motors and capacitors can be easily relocated together
• Easier to select the right capacitor for the load
• Reduced line losses
• Increased system capacity
• Advantages of bank installations at the feeder or substation:
• Lower cost per kVAR
• Total plant power factor improved—reduces or eliminates all forms of kVAR charges
• Automatic switching ensures exact amount of power factor correction, eliminates over-
capacitance and resulting over voltages (EATON, 2010)
35. Capacitor bank Selection criteria continuation
Consideration taking into account the Load type
• If your plant has many large motors, 50 hp and above, it is usually economical to install
one capacitor per motor and switch the capacitor and motor together. If your plant
consists of many small motors, 1/2 to 25 hp, you can group the motors and install one
capacitor at a central point in the distribution system. Often, the best solution for plants
with large and small motors is to use both types of capacitor installations. (EATON, 2010)
Load size
• Facilities with large loads benefit from a combination of individual load, group load, and
banks of fixed and automatically-switched capacitor units. A small facility, on the other
hand, may require only one capacitor at the control board. (EATON, 2010)
• Sometimes, only an isolated trouble spot requires power factor correction. This may be
the case if your plant has welding machines, induction heaters, or DC drives. If a
particular feeder serving a low power factor load is corrected, it may raise overall plant
power factor enough that additional capacitors are unnecessary. (EATON, 2010)
36. Sizing capacitors for individual motor loads
• To size capacitors for individual motor loads, use table below. Simply
look up the type of motor frame, RPM, and horsepower. The charts
indicate the kVAR rating you need to bring power factor to 95%. The
charts also indicate how much current is reduced when capacitors are
installed. (EATON, 2010)
37.
38. Sizing capacitors for entire plant loads
If you know the total kW consumption of your plant, its present power
factor, and the power factor you’re aiming for, you can use (EATON,
2010) Table 6, on Page 13 to select capacitors.
39.
40.
41. Example calculation & table source; (Csanyi,
2018)
In a plant with active power equal to 300 kW at 400 V and cosφ= 0.75,
we want to increase the power factor up to 0.90. In the table 1 above, at
the intersection between the row “initial cosφ” 0.75 with the column
“final cosφ” 0.9, a value of 0.398 for the coefficient K is obtained.
Therefore a capacitor bank is necessary with power Qc equal to:
Qc = K · P = 0.398 · 300 = 119.4 kvar
44. References
(2018, April 3). Retrieved from Ibiblio: https://www.ibiblio.org/kuphaldt/electricCircuits/AC/AC_6.html
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and-how-does-it-affect-us/
(2018, April 10). Retrieved from Wikipedia: https://en.wikipedia.org/wiki/Electromagnetic_interference
(2018, April 17). Retrieved from Electronics tutorials: https://www.electronics-tutorials.ws/accircuits/sinusoidal-waveform.html
(2018, April 17). Retrieved from Electronics tutorials: https://www.electronics-tutorials.ws/accircuits/phasors.html
Busbars and distribution. (2009). Power guide(12), 20 - 21.
Csanyi, E. (2018, March 16). Electrical Engineering Portal. Retrieved from http://electrical-engineering-portal.com/capacitor-bank-power-factor
EATON. (2010, November). Power factor correction: a guide for the plant engineer. EATON.
Eustace, S. E., Kashy, E., Robinson, F. N., Bleaney, B., & McGrayne, S. B. (2018, April 17). Britannica. Retrieved from
https://www.britannica.com/science/magnetism
Franceschetti, G. (1979). Fundamentals of Steady-State and Transient Electromagnetic Fields in Shielding Enclosures. IEEE, 335 - 348.
Franceschetti, G. (1979). Steady state and transient electromagnetic coupling through slabs. IEEE, 590 - 596.
imperial college London. (2018, April 17). Retrieved from http://www.ee.ic.ac.uk/hp/staff/dmb/courses/ccts1/01000_Phasors.pdf
Muppala, P. (2018, April 3). Retrieved from Quora: https://www.quora.com/What-is-meant-by-steady-state-Is-it-defined-for-ac-circuits-also-If-yes-how
Noto, J., Fenical , G., & Tong , C. (2010, April 1). Automotive EMI Shielding –Controlling Automotive Electronic Emissions and Susceptibility with Proper
EMI Suppression Methods. Laird Technologies.
Rapid tables. (2018, April 16). Retrieved from https://www.rapidtables.com/electric/Power_Factor.html
Schoofs, A. (2018, April 16). Analysing electrical demand and supply (kW, kVAR, kVA, PF, V). Retrieved from
https://docs.wattics.com/2016/03/03/analysing-electrical-demand-and-supply-kw-kvar-kva-pf-v/