This document discusses power supply quality in buildings. It covers topics like generation and distribution of electricity, phases of electricity, power factor, power quality issues like harmonics, and electrical load management strategies. Case studies are provided on using capacitors and load scheduling to reduce peak demand and lower electricity costs. Transformers are described as devices that transform voltages efficiently with losses that vary based on loading.

1. Power supply quality
in buildings
RCREEE Energy Audit in Building
Tunis, 1st – 5th June 2010
Muhieddin Tawalbeh - National Energy Research Center

2. Content
1. Introduction to power systems
2. Generation & distribution
3. Phase of electricity
4. Power factor
5. Power Quality
6. Transformers
7. Load Management
8. Case Studies

3. introduction
• Development can be measured by a
nation’s electricity consumption
• Electricity usage is divided into:
a) Industrial
b) Commercial and residential
c) Agriculture and irrigation
• Electricity important input for
industry

4. International Energy Agency predicts
for 2030:
• 78% of population in developing
countries has access to electricity
• 1.4 billion people no access
• 665 billion US$ needed to overcome
this

5. • How can electricity supply shortage
be solved?
a) Renovation and modernization of plants,
transmission and distribution systems
b) Demand side management with the
utilization of energy efficiency
technologies
c) Awareness raising among energy users

6. Generation & Distribution
• Electricity generation: fossil fuels and
uranium
• Renewable energy is growing

7. Generation & Distribution
• AC generators (“alternators”) generate
electricity
• Electricity generated at 9-13 KV
• Power generated from 67.5 to 1000 MW
• Power stations: generating transformers
(GTs) to increase voltage to 132-400 KV
• Substations: step-down transformers to
reduce voltage before distribution

8. Phase of Electricity
Single phase AC circuit:
• Two wires connected
to electricity source
• Direction of current
changes many times
per second
Three phase systems:
• 3 lines with electricity from 3 circuits
• One neutral line
• 3 waveforms offset in time: 50-60 cycles/second

9. Power
Resistive
0
180 360 Time
Voltage
Current

10. Power
Inductive
0
180 360 Time
Voltage
Current

11. Power
Voltage Capacitive
Current
0
180 360 Time

12. Phase of Electricity
Star connection
Delta connection

13. Active and Reactive Power
• Active power (kW): real power used
• Reactive power (kVAR): virtual power that
determines load/demand
• Utility pays for total power (kVA)
kVA = √ (KW)22+ (KVAR)22
kVA = √ (KW) + (KVAR)

14. Power Factor Correlation

15. Power Factor Correction
A
kV
pow er kVAR
r
we
reactive pow er kVAR
t po
reactive
A r en
r kV ap
pa
owe
tp φ
en φ’
par
ap
pow er kVAR
capacitative
active pow er [kW]
φ
active pow er [kW]
kVAr Rating = kW [tan φ1 – tan φ2]

16. PF Correction: Capacitors
• kVAR demand should be as low as
possible for the same kW output

17. PF Correction:
Capacitors
• Act as reactive
power
generators
• Reduce reactive
power
• Reduce total
power generated
by the utilities

18. PF Correction: Capacitors
Advantages for company:
• One off investment for capacitor
• Reduced electricity costs:
• Total demand reduced
• No penalty charges
• Reduced distribution losses
• Increased voltage level at load end, improved
motor performance

19. PF Correction: Capacitors
Advantages for utility:
• Reduced reactive component of network
• Reduced total current in the system from
the source end
• Reduced I2R power losses
• Reduced need to install additional
distribution network capacity
A major concern arising from the use of capacitors in a
power system is the possibility of system resonance.

20. Original power factor
kW capacity increase(%)

21. •Location capacitor is energized only when motor is in circuit
•Maximum benefit of for C1C installation by starter or operation
C1B,C1A C1A, C1B and new are installation locating them as
C1A is recommended capacitors is derived
C1B is recommended for existing after motor
breaker where noto the load.
close as possible extra switches are needed.
C4
Breaker Fuse
Transformer
MAIN DISTRIBUTION
Switch
Local Distrib.
S S S
C1C
C2 Motor overload
Thermal protection
C1B
M M M
C1A C3
. Power Distribution Diagram illustrating Capacitor Location

22. Power Factor Cable Losses
* Cable losses are in the range of 2%- 4%
for most plants
*Invest in large size cables than required by
thermal capacities.
*Installation of capacitors at distribution panels
reduces cable losses.

23. ECONOMICS AND EFFICIENCY OF WIRE UPSIZING
•Wire upsizing is often overlooked as an electricity saving opportunity
•Electrical Codes specify only minimum wire sizing
•Allows for greater flexibility for future expansion
•Three factors effecting economics of wire upsizing
•Duty factor
•Electricity price
•Load factor(I2R)

24. Example
50 kW, 380V, 50 Hz, 3Phase
Power factor 0.65
feeder length 200 meter
cross section of cable 35mm
operating times 570 hrs/month
price of electricity 35fils
What is the saving potential if power factor is improved to .85 due to
heat loss
What is the kVar of the capacitors required. what is the payback
period

25. Solution
1) Cost of energy consumed = 50 × 570 ×12 × 0035 =11970JD
.
2) Savings in heat losses
50000
I (p.f=0.65) = 3 × 380 × 0.65 = 116.8 Amp
50000
I (p.f=0.85) =
3 × 380 × 0.85 =89.373 Amp
Cable Resistance per Phase:(for 35mm2 cable, R=.0005 Ω/m)
Total cable resistance = .0005 × 200= 0 .1 Ω

26. * When (p.f=.65) =3RI2 =3 × 0.1 × (116.8)2 = 4.0926 kW
* When (p.f=.85) =3RI2 =3 × 0 .1 × (89.373)2 =2.396 kW
savings from heat losses = 4.0926-2.396 = 1.6966 kW
Annual saving from heat losses = 1.6966 ×570 ×12 = 11604.744 kW
= 11604.744 × 0.035= 406.166 J.D
Which equals to 3.39% from the total energy consumption

27. Capacitor size required to give the needed P.F
correction (.65-.85):
C KVAR = KW (tan θ1 - tan θ 2 )
⇒
= Cos-1 (P.F) =Cos-1 (0.65)= 49.458 Tan Θ1 = 1.1691
= Cos-1(p.F) =Cos-1(0.85)=31.880 ⇒ TanΘ 2 = 0.6197
Cap. size = 50 (1.169 - 0.6197) = 27.5 kvar
If we assume the price of one kvar = 20 J.D then,
Total price of capacitors = 20 ×27 = 540 J.D
This means a simple payback period of only 2.36 month. And by all
economical measures, the use of capacitors to improve the power factor is
economically viable.

28. Power Quality
An ideal power system is one which has voltage
current and frequency at every point constant no
interference of one load with other, unity power
factor and sinusoidal voltage and current waves.
None of these conditions are fulfilled in practice

29. What is power quality?
Power quality broadly refers to the delivery of a
sufficiently high grade of electric service
In general, it involves maintaining a steady state
sinusoidal load bus voltage at stipulated magnitude
and frequency
While there is no implied limitation on the voltage
rating of the power system being monitored, signal
inputs to the instruments are limited to 1000 VAC
rms or less.
The frequency ratings of the ac power systems being
monitored are in the range of 45-450 Hz.

30. Why is power quality important?
It affects both utilities as suppliers and
customers as users.
Impact on customer side
Computers, servers, Transmitters and Receiver
are susceptible to power system disturbances
which can lead to loss of data and erratic
operation
Automated manufacturing process such as
paper-making machinery, chip-making assembly
lines, etc. can shutdown in case of even short
voltage sags.

31. Impact on utility side
Failure of power-factor correction capacitors
due to resonance conditions.
Interference with ripple control and power
line carrier systems used for remote
switching, load control, etc.,
Error in energy meters, which are calibrated
to operate under sinusoidal conditions.

32. The normal power quality problems are:
1 Transient
2 Flicker
3 Sag and Swell
4 Imbalance between the phases
5 Electromagnetic Interference
6 Waveform distortion

33. Typical Power Quality problems
Transients
•Capacitor Cycling
•Lighting
Harmonics
•Rectifiers
•Switching Power Supplies
•Fluorescent Lighting
Frequency Deviating
•System dynamics

34. Typical Power Quality problems
Sags
•Turn on of heavy Loads
•Brown- outs
Swells
•Turn-off of heavy loads
Outages
•Faults
34

35. 35

36. Harmonics
Harmonics are sinusoidal voltage or currents having
frequencies that are integer multiples of the
frequency at which the supply system in designed to
operate (50 Hz or 60 Hz).

37. Fundamental
Combined Waveform
5 th,7th,11th,and 13th harmonics

38. Harmonic Distortion
Caused by non-linear devices
current not proportional to applied voltage
Even for perfectly sinusoidal voltage, current is
distorted.

39. Harmonic Vs. transients
Harmonics are a steady state phenomenon where
as transient lasts less than a few cycles.
Transients and other short term disturbances like
over voltage surges and under voltage sags can be
mitigated by transient voltage surge suppressors,
line reactors or insulation transformers.

40. Harmonic Generating Sources
Non linear loads generate harmonics. A non linear load is a
circuit element that draws current in a non-sinusoidal
manner.
AC & DC Drives
Switch mode power supply
Rectifiers, inverters and large UPS system
Arc furnace, induction furnace and welding units
Saturated reactors and transformer
CNC machine and PLCs
Smoothing chokes, PCs, Copiers and all single phase
electronic equipment.

41. Energy losses due to Harmonics
Conductor losses : due
∗ High frequency harmonic current causes skin effect.
∗ Leads to increased resistance & losses.
∗ Reduces the capacity of utilization.
Heating effect in transformer, motor and other wound
equipments.
Iron losses.
Increases losses in ferromagnetic materials.
Increased aging.

42. Electrical Load Management
What is Load Management?
Action taken to improve power demand at time of
peak loads and to shift some of the load to off-
peak times.
Shifting use of electricity from periods of high
demand to periods of lower demand, when the cost
of electricity usually is lower.
load management attempts to shift load from peak
use periods to low use periods.

43. Electrical Load Management
• Goal: reduce maximum electricity demand
to lower the electricity costs
• Load curve predicts patterns in demand
KVA
Daily load curve of an
engineering industry
Hours

44. Electrical Load Management
Strategies to manage peak load demand:
• Shift non-critical / non-continuous process
loads to off-peak time
• Shed non-essential loads during peak time
• Operate in-house generation or diesel
generator (dg) sets during peak time
• Operate AC units during off-peak times and
utilize cool thermal storage
• Install power factor correction equipment

45. Peak Clipping
kw
‫واط‬ ‫آ‬
‫اﻝ‬
Time

46. Load shifting
kW
Time

47. Conservation
KW
Time

48. Load Management &
Maximum Demand Control
•Aims at Improving Load Factor
Average Demand
Load Factor (monthly and annual) = Peak Demand
Example:
Compute monthly Load Factor
Peak kW = 1250 kW
Energy Use = 500,000 kW
Time = 720 hr
LF = 500,000 kW = 55.6%
1250kW * 720hr
•To improve Load factor it is necessary to reduce
peak demand which may be occurring for short periods

49. Electricity Billing Mechanism
• Demand
measured in
time intervals
• Maximum
demand is
highest reading
• Customer
charged on
highest
maximum
demand value!
A Typical Demand Curve

50. P
i
60
pi
∑ 60
1

51. Power(kW)
1000
500
0
10 20 30 Time(m)
P = (20×1000) + (20×500) + 0 = 500
60

52. A) New Pulper T company newly installed line has a Pulper.
- he
By insuring that only one Pulper (the new or the old) is
operating during the peak period, about J.D 3250.00 annually
can be saved. Details of the savings are given below:
Measured Demand for Pulper: 90 kW
Annual PD Charges Savings: 3294.00 JD/yr.
(based on 3.05 JD/kW MD Charges)

53. Well Pump Motor-
The pump operate manually and without and control. The motor
was measured to consume 33.3 kWh and by insuring that the
motor will not run during the Peak Demand Period an estimated
1218.8 JD/yr.. can be saved:
Measured Demand for Pulper: 33.3 kW
Annual PD Charges Savings: 1218.00 JD/yr..
(based on 3.05 JD/kW MD Charges)

54. Transformer
• Static electrical device that
transforms electrical energy
from one voltage level to
another
• Two or more coils linked
magnetically but electrically
insulated
• Turns Ratio: turns on 2nd coil (connected to load)
turns on 1st coil (connected to power source)

55. Copper loss
Cu at any load= (x)2 × Cu at full load
where (x):ratio of actual load
Cu at half -load = (1/2) × Cu at full-load
2

56. Maximum efficiency occur only when
Iron loss = Copper loss
Iron loss
load at maximum efficiency = Full load ×
F . L. Cu loss
x × full load KVA× P .F
Efficiency at any load =
( x × full load KVA× P .F ) + Iron loss + Cu loss × ( x ) 2 × 100

57. Transformer Losses & Efficiency
PTOTAL ==PNO-LOAD++(% Load/100)2 2 xPLOAD
PTOTAL PNO-LOAD (% Load/100) x PLOAD
PTOTAL ==PNO-LOAD++(Load KVA/Rated KVA)2 2 xPLOAD
PTOTAL PNO-LOAD (Load KVA/Rated KVA) x PLOAD
Transformer loss versus percent loading
• Transformer losses: constant and variable
• Best efficiency: load where constant loss = variable loss

58. Thank You