Session 1 Review of Basics
Types of Loads ∗ Resistive ∗ Inductive ∗ Capacitive
Resistive Circuit
I
V
Current in Phase with Voltage
Inductive Loads V
I
Current Lagging Voltage by 900
Capacitive Loads I
V
Current Leading Voltage by 900
Power factor correction Some fundamental thoughts • • • • • •
What is Power factor? Why power factor is important? Why improve power factor? What is the power factor of various loads? What is the origin of power factor? How to improve power factor?
Definition of Power Factor • Power Factor = Active Power (kW)/ Apparent Power (kVA) • Power Factor can never be greater than 1.00 • Power Factor at best can be equal to 1.00 • Usually P.F is always “Lag” ( Inductive) • Some times P.F can be “Lead” ( Capacitive)
Origin of Low Power Factor • Electrical Equipment need Reactive Power • Inductive loads draw Reactive Power • Phase difference between current & Voltage reduces “Displacement PF”. • Reactive Power to maintain magnetic fields in Motors. • Non-Linear loads reduces “Distortion PF”. • True PF, being product of displacement and distortion PF is lower than both.
Power Factor Improvement Concept • Reactive Power flow analogy • Power Triangle analogy • Resonance analogy
Reactive Power Flow Analogy Voltage
+ V
Pi
-
Active power Reactive power
Inductive Load
Current
Inductive Load
I
Active power
Inductive Load
Inductive Load
Reactive Power Flow Analogy
Capacitor Reactive power
Power Triangle Analogy QC S(kVA) S(kVA)
φ1
Q(kVAr)
φ1 φ2 P(kW) -QC
P(kW)
kVA=√(kW)2 + (kVAr)2
QC = P (Tan φ 1 - Tan φ 2)
PF = kW/kVA = Cos φ 1 Q = P .Tan φ 1
Cos φ1 = Initial Power Factor Cos φ 2 = Final Power Factor
Uncompensated Load
Partially compensated Load
Power Factor Correction
Reactive Current (capacitive) IR(C)
I=Active Current
Ø1 Reactive Current (inductive) IR(L)
V= Line Voltage
Ø2 I2
I1
Resonance Analogy-1 Definition:Resonance is defined as a condition where Capacitive Reactance becomes equal to Inductive Reactance in magnitude. The frequency at which this occurs is called the Resonance Frequency. | XL|
=
|XC |
| XL|
Inductor |Z|
Inductor
=
=
|XC |
0
Zeq = Z1 + Z2
Series Resonance
|Z| = Zeq = Z1Z2 / (Z1+ Z2)
Parallel Resonance
Resonance Analogy-2
Resistor
If you make |XL| of Load = |XC| of Capacitor at Fundamental Frequency, then the PF will be Unity due to Parallel resonance b/w capacitor & load inductor.
Resistor
Resistor
Inductor
Inductive Load
Compensated Load
Inductor
Inductive Load
Uncompensated load
Practical Example 40 W Fluorescent Tube Light
Choke
P = 40W+10W = 50W Power Factor = 0.6
P
N
230 Volts 50 Hz.
Calculation of PF correction based on Power Triangle concept Active Power = 50 W. ;
Power Factor = 0.6
Apparent Power = Active Power/ PF = 83.33 VA. Reactive Pr.= √(VA)2-(W)2 = √(83.33)2-(50)2 = 66.67 VAr. Capacitive VAr. req. for UPF = 66.67=V2(2πf)C Hence Capacitor req. for UPF=106x66.67/2302/100π = 4.01 µF.
Calculation of PF correction based on Resonance concept Equivalent Circuit of Tube Light R = V2/W = 2302/50 = 1058 Ω Resistor Resistor
L
Inductor
230V C
Inductor
230V
L
R
XL = V2/VAr = 2302/66.67 = 793.5 Ω
L = XL/(2πf) = 793.5/100π = 2.526 H. The value of capacitive reactance required to Resonate with the inductive reactance at the R fundamental frequency is given by, |XC| = |XL| = 793.5 Ω = 1/100πC C = 106/(793.5x100π) = 4.01µF.
Types of Power Factors • “Displacement PF” is defined as the cosine of the angle between fundamental voltage and fundamental current of the load. • Presence of “harmonics” increases the RMS current and voltage relative to their respective fundamental values. This increases the kVA of the load. • The PF taking into account the effect of harmonics, called “True PF”, is lower than or at best equal to displacement PF. • The factor by which the displacement PF is related to true PF is called the “Distortion PF” True PF = Displacement PF x Distortion PF • Capacitors can only improve displacement PF.
Mathematical expression of PF Displacement Power factor = P/(V1I1) Where P = watts and V1 and I1 are fundamental voltage and current
√ (
)
√ (
√ (
) √ (
THDV 2 THDI 2 V = V rms 1 1+ 1+ 100 100 True Power factor = P/ (VrmsIrms) P = THDV 2 V1I1 1 + THDI 2 1+ 100 100 True Power factor = Displacement PF x Distortion PF Irms= I1
Where Distortion PF =
)
)
1
√ ( 1+
THDI 100
) √1 + ( 2
THDV 100
)
2
Effect of harmonics on PF %THD(V)
%THD(I)
Distortion PF
0
0
1.00
1
20
0.98
2
40
0.93
3
60
0.86
4
80
0.78
5
100
0.71
Three dimensional power triangle kW kdVA
kVA kVA= √ kW2+kVAr2+kdVA2 kVAr
Displacement PF = True PF =
kW
√ kW2+kVAr2
kW √ kW2+kVAr2+kdVA2
Electric Power
Po we r tiv e Re ac
ive t c A
r e w Po
Apparent Power kVA
Reactive Power
Power Triangle Ap
pa r
en tP
kVA = ow er
√kW2 + kVAr2 kW
P.F. = kVA
Active Power
PF of various Industries Industry Textiles Chemical Machine shop Arc Welding Arc Furnaces Coreless induction furnaces and heaters Cement plants Garment factories Breweries Steel Plants Collieries Brick Works Cold Storage Foundries Plastic moulding plants Printing Quarries Rolling Mills (i.e. ,Paper, Steel , etc.)
Power Factor 0.65/0.75 0.75/0.85 0.4 / 0.65 0.35/ 0.4 0.7 / 0.9 0.15/0.4 0.78/0.8 0.35/0.6 0.75/0.8 0.6 / 0.85 0.65/0.85 0.6 / 0.75 0.7 / 0.8 0.5 / 0.7 0.6 / 0.75 0.55/0.7 0.5 / 0.7 0.3 / 0.75
Inductive Loads Induction Motor
Arc Furnace
0.8 P.F
0.8 P.F
Floursent Lamp 0.5 P.F. Induction Furnace
Welding Transformer
0.8 P.F
0.5 P.F.
Session 2 Benefits of Power factor improvement
Advantages of P.F Correction
Reduction in KVAR Demand
Reduction in Switchgear Rating
Reduction in KVA Demand
Reduction in Transformer Rating
Reduction in Cable / Bus-bar size
Reduction in Line Current
Reduction in Line loss
REDUCTION IN KVA DEMAND LOAD - 900 KW EXISTING P.F. (COS Ø 1) - 0.6 DESIRED P.F. (COS Ø2 ) - 0.92 COSØ. = KVA
=
kW kVA kW cos Ø.
KVA 1 =
900 / 0.6 = 1500
KVA2 =
900 / 0.92 = 978
KV A
Ø. KW
Reduction in KVA 1500 - 978 = 522
REDUCTION IN KVAR DEMAND KVA1 - 1500 KVA2 - 978
kVA
=
KVAR1 = =
Reduction in KVAR 1200 - 382 = 818
KVAR 2 =
√KW2 + KVAR2 √KVA1 2
KVAR
KW - 900
KW2
√1500 2 - 900 2 = 1200 √978 2 - 900 2 =
382
KV A
Ø. KW
KVA1 - 1500 KVA2 - 978
REDUCTION IN LINE CURRENT KVA
=
I
=
I1
=
√3 V I 1000 KVA x 1000 √3 x 415 1500 x 1000
√3 x 415 =
2087 Amp
=
978 x 1000
Reduction in Current 2087 - 1361 = 726
I2
√3 x 415 =
1361 Amp
% Rise in Current w.r.t. decrease in Power Factor % Rise in I n 100 90 80 70 60 50 40 30 20 10 0
1
0.9
0.8
0.7
0.6
0.5
P.F .
Cable Losses % of saving in losses
=
1-
1-
2
CosØ1 CosØ2
0.6 0.92
= 57.46
X 100
2
X 100
Saving in Cable Losses 80 70 60 50 40
P.F.
30
1.0
20
0.95 0.9
10 0 0.5
0.8 0.6
0.7
0.85
0.8
Initial P.F.
0.9
1
Transformer Losses Saving in losses = Wr x K1
1 Cos Ø
1
1 Cos Ø
2
Wr = Full load copper loss of the transformer connected load in Kw k1
KVA rating of the transformer
Graph Transformer Losses
Transformer Losses 40000 35000
cu Loss
30000
Copper losses
25000 20000 15000 10000 5000 0 0
500
1000
1500
2000
Transformer KVA
2500
3000
3500
Transformer Losses Saving in losses = Wr x K1
1 Cos Ø
1
900 = 18000 x
x 1500
=
1 Cos Ø
2
1 0.6
1 0.92
6260 Watts
Annual Saving = 6260 x 300 x 12 = 22536 KWH 1000
Power Savings 3000 2500
2500
2086
2000 1500
1500
1500 1000
1360
1600
1000
978
500 0 KVA
X'MER
AT 0.6 PF
CURRENT AT 0.92 PF
ACB RATNG
Workshop - I 1.
Calculate the pf and kVA demand at the secondary of a 1000kVA, 11/0.44 kVA transformer supplying the following loads: – – – –
100kW 150kW 250kW 100kW
-
UPF 0.9 lag 0.8 lag 0.9 lead
2.
A 50HP 440V, 3ph, 50Hz, 1500rpm Induction motor has the following operating conditions: PF = 0.9lag, & efficiency = 90% under full load. PF = 0.6lag, & efficiency = 70% under half load. If a 3ph. 440V capacitor of rating 12.5kVAr is connected at the motor terminals, find the pf of the motor and capacitor combination, under: a) Full load b) Half load. (Use 1HP = 0.746 kW for HP to kW conversion).
Session 3 Evaluating PF from Electricity Bill
ELECTRICITY TARIFFS - I In India, there are broadly 4 types of Electricity Tariffs. – Single part tariff • Measurement of kWh only - Energy charges Generally applicable for LT installations only.
– Two part tariff • • • •
kWh - Energy charges kVA or kW - Maximum demand charges PF Penalty/Incentives - vary from region to region Applicable for HT installations
ELECTRICITY TARIFFS – II – Three part tariff • • • • •
kWh - Energy charges kVArh - Reactive Energy charges kVA or kW - Maximum demand charges PF Penalty- vary from region to region Applicable for HT installations
– Time of day tariff • Different charges for the various quantities mentioned above depending on the time of the day Analogous to STD (Telecom) tariff structure
PF PENALTIES – Different structures are followed - Some common features • Minimum monthly PF limit - Varies from 0.85 upwards • If PF falls below minimum limit then penalty is levied • The penalty is normally calculated as a %age of the Energy charges or the full value of the Electricity Bill. • The %age of penalty is normally linked to the difference between actual monthly PF as calculated by Electricity supply authorities and the minimum PF limit specified. – For Ex: Minimum PF limit - 0.90 – Actual monthly PF as per calculated by Electricity Supply authorities - 0.82 – Penalty is 1% for every 0.01 difference between above PF values – Hence penalty will be 8% in this case.
INCENTIVES FOR HIGH PF – Different structures are followed - Some common features • Minimum Upper monthly PF limit - from 0.92 upwards • If actual monthly PF exceeds this upper limit, then incentive is offered • The incentive is normally calculated as a %age of the Energy charges or the full value of the Electricity Bill. • The %age of incentive is linked to the difference between actual monthly PF as calculated by Electricity supply authorities and the minimum Upper PF limit specified. – For Ex: Minimum Upper monthly PF limit - 0.95 – Actual monthly PF as per Electricity bill - 0.98 – Incentive is 1% for every 0.01 difference between above PF values – Hence Incentive will be 3% in this case.
EVALUATING PF FROM ELECTRICITY BILL – The value of monthly PF is normally indicated on the Electricity bill – If this value is not shown on the Electricity bill, then the normal procedure is as follows • • • • •
Note down kWh consumed as given in the bill Note down kVAh consumed as given in the bill Divide kWh by kVAh This value should always be less than 1 This gives the monthly PF as considered by the Electricity Supply authorities.
ESTIMATE kVAr REQUIRED - I – From Electricity bill data – Note down the value of maximum demand in kVA as given in the Electricity bill – Convert this value to kW by multiplying the maximum demand kVA with the monthly PF • For ex: If maximum demand kVA is 375 and monthly PF is 0.80 then, kW = 375 x 0.80 = 300 kW
– Monthly PF should be assumed as “Initial PF” - 0.80 – Fix target PF as “Final PF” - Let us assume - 0.96 – Note down multiplying factor from table 4.2 on Page 6 of RPM catalogue – This multiplying factor is 0.458
ESTIMATE kVAr REQUIRED - II – Multiply the kW calculated earlier by this multiplying factor • kVAr = 300 x 0.458 = 137.4 • kVAr - rounded off to 150 kVAr, since this is easy to offer
– Always recommend fixed compensation for the transformer in the installation - the kVAr required can be estimated from table 4.3 on Page 7 of RPM Catalogue • For ex: If in the above installation the transformer is 500 kVA then fixed compensation required is 6% of 500 kVA which works out to 30 kVAr
– Out of 150 kVAr we can now subtract this 30 kVAr i.e., leaving a balance of 120 kVAr
ESTIMATE kVAr REQUIRED - III – Of this 120 kVAr we can recommend additional compensation only for base load.
Fixed
• If base load is given as 40% of the installation, 40% of the above kVAr can be provided as fixed compensation. • Therefore 40% of 120 kVAr = 48 kVAr - about 50 kVAr • Consequently, the balance kVAr can be as an APFC • This works out to 120 - 50 = 70 kVAr - about 75 kVAr
– The final compensation scheme customer can be • Total Compensation - 150 kVAr - of which
– Fixed compensation - 75 kVAr – APFC - 75 kVAr
– This procedure is common to Industrial and Commercial Installations.
Workshop-2 –
Calculate the kVAr required to improve the pf of an LT installation to 0.95 lag. You have the following details from the Bill. Billing date = 01.07.2001 to 31.07.2001 Units consumed = 13500 kWh. Avg. PF = 0.8 lag.
–
Calculate the kVAr required to improve the pf of an HT installation to 0.97 lag. You have the following details from the Bill. Contract Demand = 300 kVA Recorded Demand = 270 kVA Avg. PF = 0.8 lag. Units consumed = 75600 kWh
Session 4 Evaluating kVAr for new installations
ESTIMATE kVAr REQUIRED for New Electrical Installations - I Refer the SLD below 500kVA, 11kV/415V, %Impedance = 4.25%
M 30 HP, 415V, 3ph, motor
M 75 HP, Other loads, 415V, total of 25 3ph, kW compressor
M 50 kVA, 440V, 3ph, UPS
75 HP, 415V, 3ph, compressor
M
Resistive Load 30kW Lighting
20 HP, Load 10kW 415V, 3ph, Pump, PF =0.80 Lag
ESTIMATE kVAr REQUIRED for New Electrical Installations - I – We must draw up a load list of the Maximum operating load (including the supply transformer) & fix the target Power Factor as desired by the Customer. – Let us assume the load list as follows • Supply transformer - 3 Phase, 500 kVA, 11 kV/415V, • 3 Phase, 415 V, Induction motors - Totaling to 200 HP • 3 Phase, 415 V, UPS system - 50 kVA • Resistive heating load - 30 kW • Lighting load (Fluorescent) - 10 kW • Miscellaneous loads - 14 kW – Let us assume that the target Power Factor as desired by the Customer is 0.95.
ESTIMATE kVAr REQUIRED - for New Electrical Installations - II – The kVAr can be estimated as follows: • The kVAr required for the supply transformer can be estimated from Table 4.3 on pg 7 of RPM Cat. – For 500 kVA transformer kVAr = 30 kVAr
• Convert induction motor rating from HP to kW - 200 HP x 0.746 = 150 kW – Assume that initial PF of motors is likely to be around 0.7, because of the fact that motors are generally oversized due to other considerations. – Calculate kVAr by using multiplying factor as given in Table 4.2 on Page 6 of RPM Catalogue - Multiplying factor for initial PF of 0.7 and final PF of 0.95 = 0.692. – Hence, kVAr = 150 x 0.692 = 104 kVAr
ESTIMATE kVAr REQUIRED - for New Electrical Installations - III – The kVAr can be estimated as follows: • Convert the UPS system kVA to kW by assuming a PF of 0.7 to be on the safer side. Hence, kVAr required for the UPS is 25 kVAr. • The resistive heating loads are unity Power Factor loads and hence do not consume Reactive Power. Hence, kVAr compensation is not required. • For other loads i.e., fluorescent lighting - 10 kW and miscellaneous loads of 14 kW assume an average PF of 0.7. Hence kVAr works out to about 17 kVAr.
ESTIMATE kVAr REQUIRED - for New Electrical Installations - IV – The total kVAr can be estimated as follows: • Transformer - 30 kVAr , Induction motors - 104 kVAr, UPS - 25 kVAr, Other loads - 17 kVAr: Total kVAr = 176 • Round off on the higher side by about 15% since, significant assumptions have been made in the calculation. • Hence, total kVAr recommended can be 200 kVAr. • Capacitors can be connected at motor terminals.The total kVAr of such Capacitors may be subtracted from the figure of 200 kVAr. • For calculating the balance fixed compensation and APFC combination, the procedure given earlier applies.
HOW TO CALCULATE SAVINGS • If Power Factor is improved cash savings arise due to the following :– Reduction in kVA demand charges – Elimination of Penalties for low Power Factor – Incentives for maintaining high Power Factor
• Hence, we must calculate the above savings as given in the workshop 3 problems.
Workshop-3 1.
In problem 1 of workshop-2, if the PF penalty is 5% of kWh charges, calculate the savings when the PF is improved to 0.95 lag. Unit charges = Rs. 3.5/ kWh. Unit consumed = 13500 kWh.
2.
In problem 2 of workshop-2, the penalty clause is as follows: kVA charges = Rs. 180/kVA (if Demand < CD) = Rs. 180 x 3 ( if Demand > CD) Unit charges = Rs. 3.4 / kWh. Units consumed = 56700 kWh PF penalty = 5% on energy consumed
CLAUSES, TERMS & CONDITIONS • All calculations are done on assumptions of certain Electrical loading in the installation. • If any changes occur due to modifications/ revisions of load data and characteristics the desired PF may not be achieved • This is particularly important in Indian conditions, since data given by the Customer is not always accurate. • This issue must be kept in mind when dealing with customers.
Session 5 Methods of improving power factor
Methods of Improving Power Factor Fixed Compensation – For Steady Loads – No load compensation of Motors – No load compensation of Transformers
Variable Compensation For Varying Loads
Selection of Capacitor 1. Individual Compensation 2. Group Compensation 3. Central Compensation
Where to install Power Factor correction Equipment Individual Compensation • • • • •
Directly at the Load terminals Ex. Motors, Steady loads Gives maximum benefit to user Not recommended for Drives Costly solution
Where to install Power Factor correction Equipment-2 Group Compensation • • • • •
Single compensation for Group of Loads Ex. Group of Motors Gives moderate benefit to user Few Capacitor Banks Relatively easy to maintain
Where to install Power Factor correction Equipment-3 Central Compensation • • • • •
Directly connected at the incomer Improves PF at the metering point Line losses continue to prevail down stream Least beneficial to user Extremely easy to maintain
Central Compensation Providing compensation at the main incomer of the installation is called central compensation (pos. No. 1). This is suitable for installations where the loads are few and situated close to the main supply. (Refer Fig. 3.1) Supply Bus
Transformer Circuit Breaker No 1 M
M
M Fig. 3.1
M
Central,Group and Individual Compensation Providing compensation at •main incomer bus – central compensation. (pos. No 1) •At power distribution boards – group compensation (pos. No. 2). •At individual load terminals – individual compensation. (pos. No. 3) This is suitable for installations consisting of main receiving station, substations, several load feeders and a wide variety of loads. (refer fig 3.3)
Supply Bus Transformer Circuit Breaker No 1 No 2 No 3
No 2 M No 3 M
No 3 M No 3 M
Fig. 3.3
Session 6 Selection of capacitors
TYPES OF CAPACITOR TECHNOLOGIES • MPP - METALLISED POLYPROPYLENE • MD - MIXED DIELECTRIC • FF/ALL PP - FILM - FOIL OR ALL POLY PROPELENE • MD -XL - MIXED DIELECTRIC LOW LOSS
METALISED POLYPROPELENE CAPACITOR • MPP - METALLISED METALLISED LAYER
PP FILM
POLYPROPELENE • METALISATION HAS BEEN DONE ON ONE SIDE OF POLY PROPELENE FILM AND USED FOR CAPACITOR WINDING • ECNOMICAL AND COMPETITIVE DESIGN • MPP-S - NORMAL DUTY • MPP-H - MEDIUM DUTY
SELF HEALING DURING INTERNAL FAULTS
SELF HEALING IS PROTECTIVE FEATURE
AFTER SELF HEALING
PRESSURE SENSITIVE DEVICE SELF HEALING PRODUCES GASES, WHICH WILL INCREASE THE PRESSURE INSIDE THE CAN. THIS WILL CAUSE THE BELLOWS TO EXPAND. BEYOND A POINT POWER SUPPLY WILL BE CUT-OFF. THUS BURSTING OF CAPACITOR IS PREVENTED.
MIXED DIELECTRIC • MD - MIXED DIELECTRIC PAPER
• PP FILM, FOIL AND PAPER ARE USED TO PP FILM FOIL
FORM CAPACITOR WINDING
FILM FOIL OR APP • FILM FOIL OR APP ALL POLY PROPELENE
PP FILM
• METAL LAYER IS PLACED IN - BETWEEN PP FILM
FOIL
PP FILM TO FORM CAPACITOR WINDING
MIXED DIELECTRIC LOW LOSS • MD-XL - MIXED DIELECTRIC LOW LOSS DOUBLE SIDE METALLISED PAPER
• PP FILM AND DOUBLE SIDED METALISED FILM PP FILM
ARE USED TO FORM CAPACITOR WINDING
Film foil/APP verses Mixed dielectric comparison Film foil/APP
Mixed dielectric
• low dielectric watt loss
• High dielectric watt loss
• Film not impregnable
• Paper impregnable
• More prone to ‘Self healing’
• less prone to ‘Self healing’
• Inferior long term stability
• Superior long term stability
• Moderate harmonic overload
• Good harmonic overload
capability
capability
Mixed dielectric verses MDXL Comparison Mixed dielectric
MDXL
• High dielectric watt loss
• Lowest dielectric watt loss
• Paper impregnable
• Combines plus points of MD
• less prone to ‘Self healing’
and APP types
• Superior long term stability
• Excellent long term stability
• Good harmonic overload
• Superior harmonic overload
capability
capability
Comparison of Technologies MPP-S
Rating
MPP-H
Rating
MD
Rating
MD-XL
Rating
FF
Rating
Life
Optimum
3
Long life
5
Long life
10
Long life
10
Long life
10
Non-linear loads Capability
Up to 10 %
1
Up to 15 %
6
Up to 25 %
10
Up to 25 %
10
Up to 25 %
10
Initial cost
Lowest
10
Medium
4
Highest
1
High
1
Highest
1
Operating cost
Lowest
10
Lowest
10
Highest
1
Lowest
10
high
4
Total
24
25
22
31
25
Cylindrical verses stand-alone type Stand-alone Type
Cylindrical Type
• • • • • • •
Compact Size Better heat dissipation Discharge resistor in ∆ Minimal internal wires Suitable only for panels MPP-S,MPP-H,MDXL Not repairable
• • • • • • •
Bulky Inferior heat dissipation Two resistor configuration More Internal wires Robust construction Available in all types Elements can be replaced
Gas filled capacitors • • • • • • • • •
Only made by EPCOS & Electro-Nicon (Germany) Considered to be better heat dissipation than oil Debated by ABB, hence controversial Equivalent to MPP but EPCOS claiming as APP Inferior to MDXL (MKV of EPCOS) SF6 banned for LV application, hence nitrogen Leakage not noticeable and failure is sudden Lighter in weight Generally available in 10 to 25 kVAr. units
Gas filled capacitors from L&T Meher • To fill the Technological gap • Design improvement over EPCOS • Protective coating on element ensures longer life even after gas leakage. • Available in the financial year 2004-05
Launch of Resin filled Capacitors • Jelly Resin has much better di-electric properties compared to Gas. • Meher is switching to resin filled capacitors. • Oil filled capacitors will also be available on request. • However MDXL will continue with oil
Competitive edge of MEHER • • • • •
Comprehensive test facility in Meher works. Raw materials imported from premium source. Automatic element winding machine. Robot spraying machine. Only Indian capacitor company to transfer technology to Germany. • Joint Venture in Capacitor manufacturing in Germany through “MKS Technologies” • ISO revalidated by BVQI from 2004 to 2007. • On the verge of getting “UL” certification for marketing internationally.
Peak current measurement capability at Meher Works
Session 7 Some basic formulae – capacitance,capacitor currents
Capacitor Connection R
Y
B
Capacitor Rated Current KVAR =
IL
=
√3. VL IL 1000 KVAR .1000 √3 .VL
Change in Current Vs. Change in Voltage VOLTAGE
KVAR
CURRENT
440
28.10
36.88
415
25.00
34.78
400
23.23
33.52
380
20.96
31.85
360
18.81
30.17
Capacitance 6π f C∆ VL2 KVAR = C∆ CM
CM
9
= =
*
10 KVAR X 10 9 6πf 1.5. C∆
VL2
µF
C∆
Calculate Capacitance C∆ and CM for 25 KVAR, 415 V, 50 Hz. capacitor * C∆ in µF
and VL in Volts
Peak inrush current of capacitor Ip = Ir
√2
√.
MVA SC X 103
Ip
= Peak inrush Current in Amps
Ir
= Capacitor Rated Current in Amps.
MVAsc
= Short circuit MVA of the System
kVArc
= Capacitor Rating in kVAr.
kVArc
Fault Level Calculation Maximum Fault Current = Transformer % Z = 5 1000 KVA , 22.0kV/415V
ACB
1.1 VL √3 ZT 1.1 x 415
=
√ 3 x 0.00861
=
30607 Amp
Transformer = impedance
%Z x 10 x kV2 kVA
=
5 x 10 x 0.4152 1000
=
0.00861 Ohm
Short Circuit MVA of the System Short circuit MVA of the System =
√3 VL Isc 10 6
=
√3x 415 x 30607 10 6
=
22.0
Peak Current Calculation Capacitor Rating = 25 KVAR, 415V, 50 Hz. Ip = Ir . √2
Ip= 34.78.√2
√.
MVA SC X 103 kVARc
√.
= 1459.1 Amp
22.0 X 103 25
Parallel Switching of Capacitor
Ip =
√2 √3
1 VL
√ X C XL
Ip = Peak inrush current in Amps. VL = Line to Line Voltage in Volts XC = Capacitive Reactance in Ohms XL = Inductive Reactance Between the Capacitors in Ohms.
Voltage Rise Due To Capacitor Q V = S V V = Voltage Rise V = System Voltage Without Capacitors Q = Capacitors Rating in MVAR S = System Fault Level In MVA
Voltage Rise Due To Capacitor For a 25 kVAr, 415V capacitor & System fault level of 22 MVA.
Q V = S V V =
415 x 0.025 22
= 0.47 volts
Discharge Time
Discharge Time < = 60 sec for LT capacitors < = 10 min for HT capacitors Voltage at the end of Discharge time should be < = 50 volts taking into account the plus side tolerances of the Capacitance value and supply voltage. C
Discharge Time t R< K C log e
(
Vn.√2 VR
)
R =
Discharge Resistance in M Ohm
t
Discharge Time in Sec.
=
K =
1/3 or 1 or 3 depending upon discharge resistor Configuration.
C =
Capacitance in µF
Vn =
Capacitor Rated Voltage
VR =
Permissible Residual Voltage
Configuration of Discharge Resistors R R
K=1
K = 1/3
C
C
R K=1
C
K=3
R C
R
R
C
K=3
C R
C K=1
USEFUL FORMULAE AND TABLES 1. Capacitance in parallel C = C1 + C2 + C3 Where C = equivalent capacitance of parallel circuit. 2. Capacitance in Series 1 1 1 1 = + + C C1 C2 C3 Where C = equivalent capacitance of series circuit.
3. Calculation of Capacitor kVAr Required for PowerFactor Improvement
Capacitor kVAr
= kW (tanϕ1 - tanϕ2)
Where ϕ1
= Cos-1(PF1) and
ϕ2
= Cos-1(PF2)
PF1 and PF2 are initial and final power factor respectively.
Multiplying Factor Table to Calculate kVAr Present PF 0.9 0.91 0.92 0.93 0.94 0.4 1.807 1.836 1.865 1.896 1.928 0.45 1.500 1.529 1.559 1.589 1.622 0.5 1.248 1.276 1.306 1.337 1.369 0.55 1.034 1.063 1.092 1.123 1.156 0.6 0.849 0.878 0.907 0.938 0.970 0.65 0.685 0.714 0.743 0.774 0.806 0.7 0.536 0.565 0.594 0.625 0.657 0.75 0.398 0.426 0.456 0.487 0.519 0.8 0.266 0.294 0.324 0.355 0.387 0.85 0.135 0.164 0.194 0.225 0.257 0.9 0.029 0.058 0.089 0.121 0.91 0.030 0.060 0.093 0.92 0.031 0.063 0.93 0.032 0.94 0.95 0.96 0.97 0.98 0.99
Final PF 0.95 1.963 1.656 1.403 1.190 1.005 0.840 0.692 0.553 0.421 0.291 0.156 0.127 0.097 0.067 0.034
0.96 2.000 1.693 1.440 1.227 1.042 0.877 0.729 0.590 0.458 0.328 0.193 0.164 0.134 0.104 0.071 0.037
0.97 2.041 1.734 1.481 1.268 1.083 0.919 0.770 0.631 0.499 0.369 0.234 0.205 0.175 0.145 0.112 0.078 0.041
0.98 2.088 1.781 1.529 1.315 1.130 0.966 0.817 0.679 0.547 0.417 0.281 0.253 0.223 0.192 0.160 0.126 0.089 0.048
0.99 2.149 1.842 1.590 1.376 1.191 1.027 0.878 0.739 0.608 0.477 0.342 0.313 0.284 0.253 0.220 0.186 0.149 0.108 0.061
1.00 2.291 1.985 1.732 1.518 1.333 1.169 1.020 0.882 0.750 0.620 0.484 0.456 0.426 0.395 0.363 0.329 0.292 0.251 0.203 0.142
4. For Single Phase Capacitor 1
lC
2 V
kVAr =
2 πfCV2 109
kVAr . 103 IC =
XC =
V
C
= Measured capacitance across terminals 1 &2 in µF
XC
= Capacitive Reactance in ohms
V
= Voltage in Volts
kVAr
= Rated output of capacitor
IC
= Capacitor Current in Amps.
106 2 πfC
5. For a Balanced Three Phase Delta Connected Capacitor 1 V
C
C
Vph
C
C 2 3
CM
=
CM
µF
1.5
V = Vph
CM is the measured capacitance across any two terminals with the other terminal left open circuited.
106 XC / ph = 2πfc
2πfcV2 X 3 3 phase kVAr = 109
Where
IL =
3ph kVAr X 103 √3 V
XC
= Capacitive Reactance per phase in ohms
V
= Voltage (line to line) in volts
kVAr
= Rated output of capacitor
IL
= Line current in Amps
6. For a Balanced Three Phase Star Connected Capacitor IL
1
> C
CM
Vph
Vph = V/ √3
V C
C
C= 2CM
2 3
CM is the measured capacitance across any two terminals with the other terminal left open circuited.
XC/Ph
=
106 2πfc
Where XC = Capacitive Reactance per phase in ohms
2 3 Ph kVAr = 2πfcV x 3
V = Voltage (line to line) in volts
109
kVAr = Rated output of capacitor IL
=
3 ph kVAr X √3 V
103
IL = Line current in Amps
7 Inrush transient current 7.1 Switching in single capacitor
Ip = Ir . √2 . Where IP Ir MVASC kVArC
√
MVAsc . 103 kVArc
= The peak value of Inrush Capacitor current in Amps = The rated capacitor current (rms) in Amps = the short circuit power in MVA at the point where the capacitor is connected = kVAr of the capacitor
7.2 Switching of capacitors in parallel with energized capacitor
IP =
Where IP V
√2 . V √3 . √XcXL = the crest value of Capacitors inrush current in Amps = Rated voltage in volts (line to line)
XC = Capacitive reactance per phase in (ohms) XL = the inductive reactance per phase between the capacitors in (ohms)
7.3 Frequency of Inrush Current
fS = fN .
Where fS
√
Xc XL
= the frequency of inrush current in Hz
fN
= the Rated frequency in Hz
XC
= Capacitive Reactance per phase in (ohms)
XL
= the inductive reactance per phase between the capacitors in (ohms)
Measured verses cell capacitance C1,C2,C3 are the capacitance measured as indicated with the third line open-circuited.
Cc
Ca
C1 C2
Using following formulae we can calculate cell capacitance, without opening and de-soldering/cutting the capacitor units.
Cb C3
Ca =
1
(
1 1 1 + C1 C3 C2
)
Cb =
1
(
where =
C1+C2+C3 C1C2C3
-
1 2
(
Ca,Cb,Cc are the cell capacitance, internal to the three phase capacitor.
)
1 1 1 + + C12 C22 C32
1 1 1 + C1 C2 C3
)
Cc = 1
(
1 1 - 1 + C2 C3 C1
)
These are derived from the following basic relationships. C cC a CaCb C3 = Cb+ C 1 = C c+ C + C C c+ C a a b CbCc C2 = Ca+ Cb+ Cc
Measured verses cell capacitance 1 Cc
Ca
1 Cc
Ca
1 Cc
Ca C2
2 Cb
C1
2
2
Cb
3
Cb 3
C3 3
C1,C2,C3 are measured capacitances across any line and other two lines short circuited. Full line to line voltages are applied across the cells. C1 = Ca+Cb
C2 = Ca+Cc
C3 = Cb+Cc
The individual cell capacitances can be computed as follows. Ca =
1 (C +C -C ) ; 1 (C +C -C ) ; C = 1 3 2 b 2 1 2 3 2
Cc =
1 (C +C -C ) 2 3 1 2
Cost Based Selection The total cost of using a capacitor is a function of •Purchase Cost
• Operating Cost
•While purchase cost is easy to estimate it is necessary to also evaluate operating cost •The operating cost of a capacitor is a function of the total losses & the operating time of the capacitor. Ex: An installation requires 1000 kVAr which will be operated for about 6000 hrs per year. Calculate the operating cost of MD-XL Capacitors verses MD type Capacitors assuming a life expectancy of 15 years for the Capacitors. The total energy consumed by the Capacitors for its on operation is calculated as follows
Calculation of operating cost of capacitors An installation requires 1000 kVAr which will be operated for about 6000 hrs per year. Calculate the operating cost of MD-XL Capacitors verses MD type Capacitors assuming a life expectancy of 15 years for the Capacitors. The total energy consumed by the Capacitors due to its internal watt loss is calculated as follows.
Cost Based Selection • MD-XL Capacitors Energy Consumed = (Loss per kVAr x Total kVAr x Operating time)/1000 = (0.5 x 1000 x 6000 x 15) / 1000 = 45,000 kWh
• MD Capacitors Energy Consumed = (Loss per kVAr x Total kVAr x Operating time)/1000 = (1.5 x 1000 x 6000 x 15) / 1000 = 1,35,000 kWh Consequently, the excess energy consumption due to the MD Capacitor shall be = 1,35,000 – 45,000 = 90,000 kWh This energy consumed can be converted into cost using a weighted average cost of Rs.5 per kWh. Consequently, the extra cost shall be Rs.5 x 90,000 = Rs.4,50,000. On a per kVAr base this can work out to Rs.450/-per kVAr. It is obvious that operating cost must be evaluated carefully before taking the final decision on the type of capacitor to be used. It is also self explanatory that lower the losses, lower will be the operating cost.
Session 8 Automatic Power Factor Correction (APFC)
NEED FOR AUTOMATIC POWER FACTOR CORRECTION •Modern Power networks cater to a wide variety of electrical and power electronic loads, which create a varying power demand on the supply system. •In case of such varying loads, the power factor also varies as a function of the load requirements. •It therefore becomes practically difficult to maintain a consistent power factor by use of Fixed Compensation i.e. fixed capacitors which shall need to be manually switched to suit the variations of the load. • This will lead to situations where the installation can have a low power factor leading to higher demand charges and levy of power factor penalties.
NEED FOR AUTOMATIC POWER FACTOR CORRECTION •The use of fixed compensation can also result in leading power factor under certain load conditions. •This is also unhealthy for the installation, as it can result in over voltages, saturation of transformers, mal-operation of diesel generating sets, penalties by electric supply authorities etc. •It is therefore necessary to automatically vary, without manual intervention, the compensation to suit the load requirements. •This is achieved by using an Automatic Power Factor Correction(APFC) system which can ensure consistently high power factor. •In addition, the occurrence of leading power factor will be prevented.
Session 9 Intelligent APFC Relay
POWER FACTOR CONTROL SCHEME R Y B
PFCR MEASURING UNIT
TO LOAD
OUTPUT RELAYS
8 STAGE 14 STAGE
CAPACITOR BANKS
1 : 1 : 1 : 1 : 1 : ……. 1 : 2 : 2 : 2 : 2 : ……. 1 : 2 : 4 : 4 : 4 : ……. 1 : 2 : 4 : 8 : 8 : …….
Features • • • •
Controls power factor Protects capacitor banks Measures & displays various parameters Records
Controls Power Factor • Maintains system power factor at a set value • Under varying load conditions • Uses microprocessor techniques for measurement of reactive current & system power factor • Can control upto 8 capacitor banks
Optimization of Capacitor Banks • Constantly selects right combination of capacitor banks to ensure pf is kept very close to the desired value.
Top-up Facility • Constantly monitors the actual pf & compares with target value. • Any spare capacitor bank is utilized to push up the pf to unity even after the target value is met. • Ensures reactive power consumption is kept to the minimum. • Feature can be enabled or disabled.
Short Time Delay Facility • This is the time delay between the immediate OFF to ON of a capacitor bank. • Provides additional delay of 30msec between switching, if enabled • Ensures longer life of capacitors
Hunt Free Operation • Capacitor hunting is avoided by providing threshold values.
Automatic Disconnection of Faulty Capacitor Bank • If any capacitor bank is reduced to 60% of its original rating, it treats the bank as defective, after three successive switching
Protection Against Fault Conditions • Switches OFF all capacitor banks & provides a safety lockout period of 60 sec when power interruption occurs • Protects against – over voltage, – under voltage – Under current – Harmonic overload By switching of capacitor banks one after another
• Provides alarm for fault conditions & for 5th & 7th harmonics
Multi-parameter Display • • • •
Voltage Current Reactive power Accurate display of power factor even in presence of harmonics.
Records • Keeps updated (every two hours) records for each bank. • Number of times each bank is switched on for pf compensation. • Configuration with respect to lowest bank size.
Session 10 Issues in Power Quality
Momentary Voltage Sag
Momentary voltage sag , which is a momentary decrease in voltage outside normal tolerance.
Momentary Voltage Swell
Momentary Voltage swell , which is a corresponding voltage increase often caused by the sudden de – energizing of heavy equipment.
Voltage Loss
Voltage Transient / Impulse
Voltage transient or impulse , which is a very short duration voltage, whose amplitude will be in the range of several tens to thousand volts.
Voltage Spike
Voltage Spike ,which can destroy electronic equipment and damage transformer and motor insulation. They also cause failures in capacitors and indicators.
Notch in the Voltage Waveform
Liner Load Characteristics
Voltage Waveform
Current Waveform
Non-linear Load Characteristics
Voltage Waveform
Current Waveform
Scope Improvement of power quality in LV networks • To enhance network reliability • To reduce failure of electrical & electronic equipment • To increase profitability by saving on energy costs • To achieve energy conservation
Power Quality in LV Networks - I External power quality • Conditions arising from incoming power supply source – Voltage fluctuations • Steady state • Transients
– Frequency variations – Interruptions in power supply – Import of harmonics
Power Quality in LV Networks - II • Internal power quality – Function of conditions arising due to use of equipment to overcome external power quality problems • On-line UPS systems • Voltage regulating devices – AUTO TRANSFORMER, STABILISERS etc.
• Back-up power supply – DG Sets – Inverters etc.
Power Quality in LV Networks - III • Internal power quality – Is also a function of types of loads connected in the network • Non-linear loads – Rectifiers, converters, drives – Battery chargers, UPS systems – Modern lighting systems
• Rapidly fluctuating loads – Welding machines – Plastic extruders – High speed presses
• Single phase SMPS loads – PC’s, Servers, LAN Networks.
Power Quality Issues • • • • • • •
Reactive power flow Harmonic currents & voltages Voltage dips Voltage flicker Unbalanced load High neutral current Excessive neutral to earth voltage
Problems ! - I Reactive power flow – Lower PF & Increased kVA Demand. – Overloading of transformers, cables & switchgear. – Increased energy consumption due to higher losses. – Financial penalties for low PF. – Loss of financial incentives for high PF.
PROBLEMS ! - II Harmonic currents & voltages – Overheating & failure of • Electrical equipment – Motors, transformers, switchgear – Capacitors
• Power electronic equipment
– Malfunction/failure of • Protective relays • Control & Automation equipment
– Increased energy consumption. – Risks of resonance - current amplification extremely dangerous
PROBLEMS ! - III Voltage dips & flicker – Failure of power electronic equipment – Malfunction/failure of • Protective relays • Control & Automation equipment
– Increased strain on eyes. • Due to fluctuations in intensity of lighting systems.
PROBLEMS ! - IV Unbalanced loads – Over & under voltage in the network. – Increased energy consumption by motors • Due to reduced efficiency.
– Failure of power electronic equipment – Malfunction/failure of • Protective relays • Control & Automation equipment
PROBLEMS ! - V High neutral current – Overheating of neutral conductors – Increased energy consumption – If neutral becomes open high voltages will occur, resulting in • Malfunction/failure of – Single phase loads – Protective relays – Control & Automation equipment
– Risk of fire & destruction
Accumulation of 3rd harmonic current in neutral Three phase system
Wave forms of balanced three phase fundamental currents. Time. R - phase.
R-Phase current with its third harmonic component. Time. Y - phase.
Y-Phase current with its third harmonic component. Time. B - phase.
B-Phase current with its third harmonic component. Time. Addition of third harmonics in Neutral conductor
Time.
Third harmonic currents of R,Y&B phases are in phase with each other and hence adds up, without cancellation in the neutral conductor.
SOLUTIONS - I Networks with <20% non linear load • Improve PF, reduce voltage dips/flicker by the use of – Power capacitors – APFC systems • Contactor switched • Thyristor switched – Open loop systems – Closed loop systems
SOLUTIONS - II Networks with >20% non-linear load • Improve PF, reduce harmonics & voltage dips/flicker by the use of – Fixed detuned filters – Detuned filters + APFC systems • Contactor switched • Thyristor switched – Open loop systems – Closed loop systems
– AHF - Active Harmonic Filters
SOLUTIONS - III Networks with unbalanced loads. • Improve PF, reduce harmonics & voltage dips/flicker by the use of – Phase balancing circuits – Electronic VAr Compensation Systems. – AHF - Active Harmonic Filters
SOLUTIONS - IV • Networks with high neutral current / excessive neutral to earth voltages – Ensure proper EARTHING quality – Oversize all neutral conductors to reduce neutral heating – If OVERSIZING is not possible, reduce harmonics by the use of • AHF - active harmonic filters in 4 line configuration
Active filter schematic diagram Sin u so id al Su pp ly C u rren t
Lo ad C u rren t w ith H arm o n ics
Su pp ly Sy stem
C o m p ensatin g C u rren t A ctiv e F ilter
+
= ( T im e Do ma in )
+
= ( F requ en cy D oma in )
Benefits of POWER QUALITY MANAGEMENT • • • • • •
Improved reliability of installation Reduced energy consumption Reduced fuel consumption Better productivity Improved profitability Enhanced equipment life
Session 11 Dynamic Compensation
Scope • Need for Dynamic Compensation
• Applications
Need for DYNAMIC COMPENSATION • When load conditions demand rapidly fluctuating reactive power. – Due to nature of load. – Due to process requirements.
• When switching transients are to be eliminated. • For Optimizing performance & fuel consumption of DG sets.
THYRISTOR SWITCHED APFC There are certain loads which demand, under certain operating conditions, large amount of reactive power for very short duration of time. Typical examples are: •Welding equipment •Injection moulding equipment •Starting of large induction motors •Traction loads such as, lifting cranes, elevators, lifts, etc.
Thyristor Switched APFC The large demand for reactive power by such loads during operation can cause: •Rapid voltage fluctuation •System instability •Over sizing of electrical installation since kVA capacity will have to be provided for the maximum power demand. •Malfunctioning of sensitive electrical and electronic equipment such as relays, PLC’s etc. These ill-effects can be overcome by injecting into the network defined amount of reactive power at a very fast rate which can meet the demand of such loads.
RAPIDLY FLUCTUATING LOADS - I • Variations in current are sudden & high • Lower PF & voltage dip • Examples – – – – –
Motor starting EOT cranes, lifts Rolling mills Conveyors - Mining etc. Wind electric generators
RAPIDLY FLUCTUATING LOADS - II • Current drawn in repetitive pulses • Lower PF & voltage dips / flicker • Examples – High speed presses • Metal working • Plastic processing
– Balanced welding loads
SWITCHING TRANSIENTS • Capacitor switching by contactors results in transients • These transients may interfere with operation of modern relays, control & automation equipment – For ex: digital relays, PLC’s etc. can malfunction
• Thyristor switching is a must for eliminating switching transients • Hence, dynamic compensation
DG SET PERFORMANCE & FUEL CONSUMPTION • Use of dynamic compensation systems can – – – – –
Stabilize DG set output voltage. Reduce DG set rating for a given load. Enable better % loading of the DG set. Reduce fuel consumption. Enhance life of DG set.
APPLICATIONS - I • Industrial networks – Automobile & automobile component Mfrg. Plants – Metal working • Fabrication & press shops • Rolling mills • Forging
– Plastic extrusion & Moulding. – Mining • Extraction • Polishing, Crushing etc.
APPLICATIONS - II • Industrial networks – Paper, wood & particle board Mfrg. plants – Plants with CNC machines
• Other networks with sensitive loads – Hospitals – IT parks – Intelligent buildings
DYNAMIC COMPENSATION SYSTEMS Open loop systems – – – –
Suitable for dedicated loads. Fastest Response. (< 15 msec) Unique “EPS” logic. External signal from load can be used for switching on. – “On” time - externally settable. – Integrated protection for – Wrong phase sequence. – Phase fall out. – Over temperature.
DYNAMIC COMPENSATION SYSTEMS Closed loop systems – – – – –
For groups of diverse loads. Use advanced programmable controller. Fast Response. (< 60 msec) “EPS” logic Desired PF & specified harmonic distortion values are settable. – Integrated protection for – – – –
Wrong phase sequence. Phase fall out. Over temperature. Harmonic over load.
DYNAMIC COMPENSATION SYSTEMS Open & closed loop systems – The system consists of • Ergonomic metal enclosure. • Incoming switchgear & protection. • Modern Copper Busbar System. (upto 80 kA short circuit withstand capacity) • Power capacitors. • Harmonic Reactors. (if required) • Thyristor modules – Firing circuits – Electronic control modules – Protection fuses
• By-pass contactors.
THYRISTORISED SWITCHED APFC •Conventional power factor correction systems using contactors as switching devices cannot be used in sufficient speed of response to meet the reactive power demand imposed by such loads.
•It is necessary to use a dynamic power factor correction system in which the switching and controlling devices used have a response time in milliseconds.
DYNAMIC COMPENSATION SYSTEMS : Advantage - I LOSSES & TEMPRATURE RISE – Thyristor Loss = 2 x In watts Ex: 3 Phase, 440 V, 50 Hz. 50 kVAr capacitor has In = 65 A. – watt loss/thyristor module= 130 watts. – for three phase switching two thyristor modules are required. – total watt loss = 260 watts. – unit watt loss = 5.2 watts/kVAr. – Generally therefore cooling fans are required for each thyristor step. – Energy consumption is very high.
– INTELLVAr - D : No cooling fans for each Thyristor step.
DYNAMIC COMPENSATION SYSTEMS : Advantage - II THYRISTOR RELIABILITY – Use of bypass contactors reduces • • • •
Operating losses. Utilization time of Thyristors. PIV of Thyristors used = 1800 Volts. In a 415 V, 50 Hz. system – peak to peak voltage = 1174 V – Therefore Safety factor > 150% in Voltage
– Lower utilization time + high PIV results in enhanced Thyristor reliability.
DYNAMIC COMPENSATION SYSTEMS : Advantage - III SWITCHING LOGIC – ZERO CROSSOVER SWITCHING DESIGNS – Due to various factors exact zero crossover is not consistently achievable in practice. – consequently use of [di/dt] limiting inductors/coils is common in zero crossover switching designs.
– EQUI-POTENTIAL SWITCHING - EPS LOGIC – – – –
Enables continuous sensing of capacitor & line potential. Switching is done at equi-potential instant. This reduces the [di/dt] to very safe values. Hence no current limiting devices are needed.
– EPS LOGIC THUS INCREASES RELIABILITY
Session 12 Power factor improvement of DG sets
Capacitors with Generators
G
Diesel generator set G
Prime mover
Alternator
100 KVA 0.8 P.F. 80 kW
Connected load P.F. is 0.6 Case 1
100 kVA Shortfall of P.F. 0.6 20 kW 60 kW
100 kVA P.F. 0.8 80 kW
Case 2
P.F. 0.6 80 kW kVA 133.33
Alternator overloaded by 33.33 %
Load P.F. improved up to 0.98
100 kVA P.F. 0.8 80 kW
100 kVA P.F.0.98 98 kW
Prime mover overloaded
DG Set fundamentals •
As load kW (Output) increases, input power from prime mover has to be increased.
•
Diesel engines can be overloaded by 10 %, for half an hour, within a span of 12 hours.
•
Prime movers are matched with alternator to operate at specific P.F.
•
Lagging P.F. weakens the flux which links with alternator stator and leading P.F. strengthens it.
DG Set fundamentals • At low lagging P.F, it is not possible to reach the nominal operating voltage of the alternator, even at low load. • With leading P.F, even with low excitation, there could be rise in voltage, causing damages to the connected equipment. • At leading P.F. generator becomes unstable. • If generator is operated with purely capacitive load voltage increases by 33.33 %
Selection of DG set rating • Connected load and demand factor.
• Short duration peak loads like starting of induction motors. • Allowance for extra kVA for harmonic generating loads. • Allowance for accommodating future additional loads. Hence DG sets are always oversized for a given application and operate at relatively lower percentage loading.
Loading verses Yield curve
Operating at lower % of loading , result in poor yield from DG set.
How to improve % loading in DG •
Do not exceed the current rating of Alternator.
•
Do not exceed the BHP/kW rating of the prime mover (Engine).
Golden Rules for safe DG set operation :•
Load the Alternator by ‘Amperes’.
•
Load the diesel engine by BHP/kW.
Improve % loading by operating at higher power factor •
Higher PF reduces current output from DG at a given load.
•
Loads can now be added without violating the ‘Golden Rules’. Operating at highest feasible PF, enables higher loading, resulting in better yield from DG set.
PF CORRECTION IN INSTALLATION WITH CAPTIVE GENERATION BY DG SETS The DG set consists of a diesel engine, which is mechanically coupled to an alternator. The engine supplies the mechanical energy to the alternator and the alternator supplies the electrical energy to the load. The alternator is subjected to certain copper losses, which is proportional to the square of the current delivered by it. The diesel engine has to supply these losses in addition to supplying the load requirements. Thus by reducing the losses in the alternator the diesel consumption of the diesel engine can be brought down.
ALTERNATOR LOSS REDUCTION BY P.F IMPROVEMENT Reducing the current output from alternator without altering the loading conditions can reduce the alternator losses. Improving the power factor at the output of the alternator can conveniently do this. Hence, improvement in the Power Factor in alternator leads to reduction of fuel consumption in DG sets. DG sets operate at a relatively low power factor of 0.6 to 0.8. Conventionally capacitors are not used along with DG sets.
Effect of PF improvement OUTPUT CURRENT FROM DG WITH /WITHOUT COMPENSATION 600
CURRENT IN AMPS
500 400
WITHOUT COMPENSATION
300 WITH COMPENSATION
200 100 0 1
2
3
4
5
6
7
TIME IN SECONDS
8
9
10
Alternator loss reduction due to P.F improvement The following example gives an approximate calculation to show the impact of power factor improvement on reduction of alternator losses and accrued savings in Diesel consumption. Consider a 3 phase, 415V, 50Hz, 500 kVA DG set used in an industry for 6000 hours/year with an average load of approximately 250kW at 0.65 PF. What is the fuel saving if PF is improved to 0.93? The full load copper loss of the alternator is 12kW and average yield of the DG set is 3kWh/litre of fuel (HSD).
Alternator loss reduction by P.F. improvement Rated Current of Alternator
= 500000/(√3 x 415) = 695.60 A
Current at 0.65 PF
= 250000/ √3 x 415 x 0.65 = 535.08 A
Copper loss at this current
= (535.08)/695.6)2x12kW = 7.1 kW
Current at 0.93pf
= 250000/ √3 x 415 x 0.93 = 373.98A
Copper loss at this current
= (373.98/695.6)2x12kW = 3.47kW
Saving in Copper loss
= 7.1 – 3.47 kW = 3.63 kW
Alternator loss reduction by P.F. improvement
Energy saved for 6000 hour Generation DG set Yield Potential savings in HSD fuel
= 3.63 x 6000 kWh = 21780 kWh = 3 kWh / liter of HSD = 21780/3
= 7260 liters per year Potential savings in Rs @ Rs. 35/liter
= 2,54,100 per year
Method of P.F. improvement Conventional fixed capacitors, should not be used with DG sets. This is because, by using fixed capacitors, there is a danger that the PF can become leading under lightly loaded condition, which is highly undesirable in DG set operation. Hence only Automatic Power factor Correction system (APFC) should be used with DG sets. The target power factor can be set to 0.93 to 0.95 for optimum performance.
APFC Selection The appropriate APFC system can be selected based upon the harmonic content of the load. If the harmonic generating load is less than 20%, the APFC should be used as mentioned in section-8. If the Harmonic load is greater than 20% then reactor protected APFC should be used as mentioned in section 9.5 The rating of the APFC has to be selected depending upon the kW of the load connected and the minimum power factor in the installation.
F I G .3 - B L O C K D I A G R A M O F C O N T A C T O R S W I T C H E D A P F C S Y S T E M G
C T ..../ 5 A
~
V a r ia b l e L o a d s
F .S .U
f
Pow er F a c to r C o n tr o lle r
C 1......C n K 1 ......K n F 1......F n f2 F .S .U G CT
= = = = = = =
F 1 ......F
2
K 1 .....K
N o . o f C a p a c it o r S te p s N o . o f C a p a c it o r S te p C o n ta c to r s C a p a c i to r S t e p P r o te c tio n F u s e s C o n tr o l F u s e s F u s e S w itc h U n it ( O p tio n a l) D i e s e l G e n e r a to r M a in S u p p l y C u r r e n t T r a n s f o r m e r
n
C 1......C
n
n
FIG 5. BLOCK DIAGRAM OF DYNAMIC COMPENSATION SYSTEM IN CLOSED LOOP G
~
CT
Variable Load F.S.U
Reactive Power Controller
f2 F1
Fn
F2
Thy 2
Thy 1
S1
S
2
= No. of Steps S 1......S n Thy 1......Thy n = Thyristor Switches = Capacitor Step Protection Fuses F1......F n = Control Fuses f2 = Fuse Switch Unit (Optional) F.S.U = Diesel Generator G CT = Main Supply Current Transformer
Thy n
Sn
Conclusions • The efficiency of the DG set is maximum at UPF. • The yield of the DG set is maximum at its peak loading. • By proper use of reactive power management the efficiency of DG set can be improved. • The loads can be transferred so as to optimize the loading of DG for better yield output. • The output voltage of DG set can be stabilized under rapidly fluctuating loads by use of “Dynamic compensation systems” • Saving in fuel is possible resulting in economic benefit to user.
Session 13 Harmonics & Effect of Adding Capacitors in the System
POWER FACTOR CORRECTION IN HARMONIC RICH ENVIRONMENT •A harmonic rich environment is said to exist when the percentage of non linear loads in an installation becomes greater than 20% of transformer rating. •Power factor correction by the use of capacitors, in such an environment, must therefore be carried out with certain precaution. •This is due to the fact that parallel resonance conditions can occur, I.e. the magnitude of the Capacitive reactance of capacitors installed and the inductive reactance of the network can tend to be come equal. •If such resonance occurs near to a frequency which is present in the network, current amplification takes place.
POWER FACTOR CORRECTION IN HARMONIC RICH ENVIRONMENT
lh
XC
XL
POWER FACTOR CORRECTION IN HARMONIC RICH ENVIRONMENT •This current amplification can lead to overloading of capacitors and an increase of the voltage distortion in the network. •Capacitors drawing higher current i.e. more than the rated current at normal operating voltages is a typical indication of presence of harmonics. •While it is possible to design the capacitors to withstand the overload conditions, the increase in distortion will cause other ill effects such as : • Capacitors installed being subjected to severe harmonic overloading, leading to premature failure
POWER FACTOR CORRECTION IN HARMONIC RICH ENVIRONMENT
• Total harmonic distortion in the network increasing beyond the permissible levels, which is harmful to various equipments within the installation. •The use of capacitors in the conventional manner is therefore not recommended in such situations.
Technical problems experienced in industry
Case – 1 • Type of industry - paperboard Manufacturing industry. Brief description of installation. Primary power source = grid supply at 33kv. Distribution voltage = 440V. • Load details: The total induction motor load was 800HP.
Case – 1 • There were no non-linear loads installed in this plant. • 300kvar, 3ph, 440V MPP-H capacitor banks were installed for power factor correction. Some capacitors were connected across the motor terminals and remaining used as central compensation.
• Problem experienced: • Frequent failure of capacitors.
Case – 1- Analysis • The system voltage was around 435 – 440V. • The capacitors installed were subjected to severe and intermittent overload. • For ex:- A 25 kVAr, 440V capacitor was drawing a current of 90 - 120A for certain periods of time. Compared to its rated current of 33 amps, the capacitors were subjected to an overload greater than 250%. • This abnormal overloading resulted in frequent capacitor failure.
Case – 1- Analysis • Following observations were made on the 33kV grid :• Only two industries were found to be connected to 33kV line, one of them being the paperboard manufacturing industry and the other was a steel rolling mill. • The length of the 33 kV transmission line between the two plants was approximately10kms. • The steel rolling mill had installed a high frequency induction furnace.
Case – 1- Analysis • It was observed that, whenever the high frequency induction furnace installed in steel rolling mill was operated, the harmonic distortion on the grid abnormally increased and during this period the capacitors were getting severely over loaded.
Case – 1- Conclusion • Thus the use of conventional capacitors in the network, where the harmonic voltage distortion at the grid was abnormally high resulted in the following: • Over loading of capacitors due to series resonance. • Frequent failure of capacitor banks • Increased harmonic distortion at the LV bus • Inability of the customer to maintain the desired power factor. • Financial losses incurred by customer.
Case – 2 • Type of industry Cement industry • Brief description of installation. Primary power source = grid at 220kV. Distribution voltage = 6.6kv and 440V. • The LV power factor correction is done as follows: At 440V bus, around 2000 kVAr APFC panels with conventional capacitors were installed.The APFC panels were distributed on several 6.6/0.440 kV distribution transformers installed in the plant.
Case – 2 Problem experienced at LV bus: • Malfunction of 350kW, 440V DC drive used for kiln motor installed at cement plant substation, when capacitors are connected in the network.
Case – 2: Analysis • The cement plant sub station was fed by a 1600kVA, 6.6/0.440 kV transformer. • The 350kW, 440V DC drive, was connected to this transformer, consequently the % non-linear load exceeded 22%. • A 475kVAr, 440V APFC panel with conventional capacitors, was also connected to this transformer for power factor correction. Other linear loads such as compressors, pumps etc were also connected to this transformer. • Malfunctioning of the DC drive was co-related to a situation when specific combination of capacitor steps in the APFC were ON.
Case – 2: Conclusion • The use of conventional capacitor in harmonic rich environment led to high total harmonic distortion on the LV bus. • As the same distorted sine wave is applied to the 350kW DC drive, sensitive electronic devices used in this drive were mal-functioning.
Case – 3 • • • • •
Type of industry Steel rolling mill. Brief description on installation. Primary power source = grid at 33kV. Distribution voltage = 440V. The plant was installed with 2 x 2000 kVA + 1 x 1000 kVA, 33/0.440 kV distribution transformers. • At the secondary of one of the 2MVA transformer following loads were connected: • 1000HP AC induction motor. • 500HP DC drive.
Case – 3 • Around 900 kVAr of conventional capacitors were connected to this transformer for power factor improvement as shown in the fig.
Case – 3 33kV supply from Grid.
2000kVA, 33/0.44 kV Trafo.
Feeder for 1000kVA Trafo.
Feeder for other 2000kVA Trafo.
440V bus
M 1000HP Induction Motor
500HP DC Drive
900kVAr Capacitor
Case – 3 • Problem experienced: • Frequent failure of capacitors installed for power factor correction.
Case – 3- Analysis • A team of engineers from MEHER made an analysis at the site. The result of the analysis is as follows: • The capacitors installed were drawing more than it’s rated current. A 50 kVAr, 440V capacitor was drawing a current of 200A against the rated value of 65amps, thus constituting an overload greater than 300%. • This abnormal over load resulted in frequent capacitor failure. • The total harmonic voltage distortion at the secondary of this 2000 kVA transformer with all the capacitors switched on was more than 25%. This value is higher than acceptable levels.
Case – 3- Conclusion • Thus the use of conventional capacitor in harmonic rich environment resulted in: • Over loading of capacitors due to parallel resonance. • Increased harmonic distortion at the LV bus. • Frequent failure of capacitor banks. • Inability of the customer to maintain the desired power factor. • Financial losses incurred by customer.
What Are Harmonics ? • • • •
Distorted sine wave cause harmonics. Distorted current wave cause current harmonics. Distorted voltage wave cause voltage harmonics. Fourier expansion result in integral multiples of fundamental frequency components. • Nth order harmonics is of n.Fs frequency. • Generally odd harmonics are prevalent because of half wave symmetry.
How Harmonics Are Generated ? • • • • •
Non-linear loads generate current harmonics. Harmonic currents flow largely through capacitors. Harmonic currents also flows through network. The flow of harmonic currents cause voltage harmonics. Harmonics are thus injected to other linear loads connected in the same bus. • Harmonics injected into the network flow towards other users connected to the network.
What Loads Generate Harmonics • Equipment using switched mode power supply - Television - Computers, other IT loads • Equipment using power electronic devices - AC & DC drives - Frequency converters - Rectifiers - Arc & induction furnaces - UPS - Compact fluorescent & other discharge lamps
Sources of Harmonics Following are some of the non-linear loads which generates harmonics: •Static Power Converters and Rectifiers, which are used in UPS, Battery chargers, etc. •Arc furnaces. •Power Electronics for motor controls (AC/DC Drives) •Computers. •Television receivers •Saturated Transformers •Fluorescent Lighting with electronic ballast. •Telecommunication equipment.
Type of Harmonics Characteristic harmonics - Related to circuit configuration. - Fairly predictable frequency spectrum. - Frequency spectrum given by k*p+1 ; k = 1,2,3…. - For ex. 5&7 for 6 pulse, 11 & 13 for 12 pulse. - Magnitude inversely proportional to order.
Non-characteristic harmonics - Caused by frequency converters. - System imbalance (voltage & impedance)
Triplen harmonics - 3.(2n+1) order n = 0,1,2… i.E 3,9,15,21.. Etc. - Zero sequence in nature. - Accumulates as neutral current.
Harmonic Order & Phase Sequence • Each harmonic order has a particular phase sequence relationship with respect to fundamental. • By convention the fundamental is assumed to have positive phase sequence. • All higher order harmonics have either positive,negative or zero phase sequence with respect to fundamental.
How to Determine the Phase Sequence of Each Harmonics ? R
Positive Sequence
Phase Sequence of RYB (+ Seq.) For Fundamental Component
Y
R Fundamental +120o
Y 0o
B -120o
+240o -120o
0o 0o
-240o +120o
Second Harmonic
B B
Negative Sequence Y
Thus Second Harmonic Behaves As a Negative Sequence Component. R
How to determine the phase sequence of each harmonics ? Fundamental Third Harmonic
R Y +1200 00
+3600 00
Fundamental R
Positive Sequence
B -1200
00 -3600 00 00 Third Harmonics Zero Sequence
Y
R Y B
B
Thus third harmonic behaves as a zero sequence component.
Accumulation of 3rd harmonic current in the neutral Three phase system
Wave forms of balanced three phase fundamental currents. Time. R - phase.
R-Phase current with its third harmonic component. Time. Y - phase.
Y-Phase current with its third harmonic component. Time. B - phase.
B-Phase current with its third harmonic component. Time. Addition of third harmonics in Neutral conductor
Time.
Third harmonic currents of R,Y&B phases are in phase with each other and hence adds up, without cancellation in the neutral conductor.
How to Determine the Phase Sequence of Harmonics ? Harmonic order
1
2
3
4
5
6
7
8
9
Phase Sequence
+
-
0
+
-
0
+
-
0
Positive Sequence
Negative Sequence
Zero Sequence
Fundamental
2nd Harmonic
3rd Harmonic
4th Harmonic
5th Harmonic
6th Harmonic
7th Harmonic
8th Harmonic
9th Harmonic
10th Harmonic
11th Harmonic
12th Harmonic
3n+1
3n+2
3n+3
Div. by 3 Rem. 1
Div. by 3 Rem. 2
Divisible by 3
Characteristics of Harmonics Positive Sequence
Negative Sequence
Zero Sequence
Causes over heating due to ‘Skin effect’
Causes over heating due to ‘Skin effect’
Causes over heating due to ‘Skin effect’
Aids the fundamental
Opposes the fundamental
Accumulates in the neutral
Moderate heating
Excessive heating
Creates ‘hot neutral’
Most harmful
Responsible for neutral to earth voltage and open neutral condition.
Relatively less harmful
Skin effect Cross-section of current carrying conductor
DC current flow
Low frequency AC current flow
High frequency AC current flow
The effective area of the conductor, available for current flow, reduces as the frequency of the AC current increases. Hence, the resistance of the conductor increases, at higher frequencies, as it is inversely proportional to its area of cross-section.
R =
ρL A
Skin effect explanation Magnified view of current carrying conductor Enhanced impedance due to high mutual inductance. Hence least current flow. Moderate impedance due to medium mutual inductance. Hence moderate current flow. Least impedance due to low mutual inductance. Hence max. current flow.
Effect of Harmonics Type of equipment • Rotating machines
• Transformer, switch-gear, power cables • Protective relays • Power electronics • Control & automation • Power capacitors
Effect of harmonics • Increased losses, over heating due to skin effect. • Pulsating torque • Over heating, increased power consumption • Mal-operation, nuisance tripping • Mal-operation, failure • Erratic operation • High currents & failure due to overload
Effect of Harmonics on Protective Relays Mal-operation
Nuisance tripping
Trip level set lower than the fundamental value. The relay should trip as the fundamental value is higher than the trip level. But the presence of harmonics has reduced the peak value. Hence the protective relay will not trip.
Trip level set higher than the fundamental value. The relay should not trip as the fundamental value is lower than the trip level. But the presence of harmonics has increased the peak value. Hence the protective relay will trip.
Circuit configuration of six pulse drive
Current spectrum of six pulse drive for star-star & star-delta configuration
Twelve pulse drive configuration
Current spectrum of twelve pulse drive
How Capacitors & Harmonics Are Related -1 Network behaviour without capacitors • Network do not reveal harmonics. • Most of the harmonic currents internal to network go to the grid. • No resonance at harmonic frequencies. • Network power factor is unacceptably low.
Network Without Capacitors GRID
Harmonic currents flow towards Grid. Min. Import of Harmonics from Grid.
BUS
Hence least Harmonic Problem.
Non Linear Load
No Resonance at harmonic frequencies.
Power Factor Very Low.
M
How Capacitors & Harmonics Are Related -2 Network behaviour with capacitors • Network start revealing harmonics • Internally generated harmonic currents may amplify due to parallel resonance • Externally generated harmonics enter capacitors due to series resonance • May increase harmonic distortions. • Capacitors draw excessive currents & fail • Network power factor improves
Network With Capacitors GRID
ZN
Harmonic currents flow towards Capacitors , due to parallel resonance with load “ZL”
ZT BUS
Non Linear Load
M Equivalent Load Impedance “ZL”
Import of Harmonics from Grid towards Capacitors , due to series resonance with network & transformer impedances “ZN” &“ZT” Increase of THD(V) in the Bus Harmonic overloading of Capacitors, leading to its failure Improvement in Power Factor With Harmonic overload
Session 14 Harmonic Filters
Harmonic Mitigation Concept Fire Triangle
Harmonic Triangle
How to Improve Power Factor Without Causing Harmonic Problem ? • Conventional capacitors should not be used.
• Capacitors should be replaced by harmonic suppression filters (series combination of suitable series reactor & capacitors) so that, • It offers capacitive reactance at fundamental frequency for necessary power factor correction. • It offers inductive reactance at all higher order dominant harmonic frequencies to avoid resonance. • Its self series resonance frequency “fR” do not coincide with predominant harmonics.
Network With Harmonic Filters GRID
No resonance at harmonic frequencies as filter is inductive at such frequencies
ZN ZT
Harmonic currents flow towards Grid , as it offers least impedance compared to filter
BUS L
Non Linear Load
M Equivalent Load Impedance “ZL”
C
Predominantly fundamental current flows through Capacitors Moderate THD(V) in the Bus No harmonic overloading of Capacitors Improvement in Power Factor without Harmonic overload
Harmonic Filter • Harmonic filter comprises of a reactor (L) in series with a capacitor (C) • Such a filter has a unique self series resonance frequency fR at which inductive reactance of reactor equals capacitive reactance of capacitor. Fr = 1/(2π√LC) • Below fR the filter is capacitive • Above fR the filter is inductive
Characteristics of Harmonic Filter Capacitive
Inductive
fR= Resonant
Frequency
Impedance
f < fR - Capacitive f > fR - Inductive
fR
Frequency
Harmonic filters are classified based upon how close fR is to a Harmonic frequency
Classification of Harmonic Filters • Detuned or harmonic suppression filters • Resonance frequency fR< 90% of lowest dominant harmonic frequency.
• Tuned or harmonic absorption filters • Resonance frequency fR within 10% of the frequency of the harmonic to be absorbed.
Classification of Harmonic Filters Harmonic Filters
Passive Harmonic Filters
Detuned Filters
Tuned Filters
7%
14 %
Active Harmonic Filters
Hybrid Harmonic Filters
3Ф 3wire 3Ф 4wire Single Phase
Selection Criteria for Harmonic Filters • Detuned filters • Power factor correction is of paramount importance. • If ordinary capacitors draw > 130% of its rated current. • Reduction of THD(V) not relevant. • To prevent capacitors from harmonic overload • Harmonic study not required for installing standard detuned filters.
Selection Criteria for Harmonic Filters • Tuned filters • Power factor correction & reduction of THD(V) are of paramount importance. • Ordinary capacitors draw > 130% of its rated current. • Harmonic study required for installing tuned filters. • Specifically designed for each location. • More bulky, since it carries large amount of harmonic currents. Hence expensive.
Standard Detuned Filters-1 • Standard detuned filters have a fixed percentage tuning factor “p” • Percentage tuning factor is defined as Reactor reactance at system frequency p = Capacitor reactance at system frequency X 100 % • Standard detuned filters are available for 7% tuning factor • The resonant frequency of the filter fR is related to tuning factor “p” by Fr = Fs/ √(p/100) = 189 Hz for 7% filter
Standard Detuned Filters-2 • Standard 7% detuned filters are suitable for use in majority of installations where the dominant harmonics are higher than 189 Hz like 5th and higher. • 7% detuned filters should not be used in installations where predominant 3rd harmonics are present like “IT based” industries. • For “IT based” industries 14% detuned filters (fR=134 Hz) should be used.
Design Features of Detuned Filter • Detuned filter consists of matched pair of specially designed reactor and capacitor. • Detuned filter is designed to provide the rated kVAr at the rated voltage at the bus. • The reactor capacitor combination is designed for the rated tuning factor. • Standard detuned filters are available for 7% tuning factor rated for 12.5, 25, 50, 75 & 100 kVAr at 440 volts.
Design Features of Detuned Filter Reactor features. • Reactors are specially designed to carry wide spectrum of harmonic and fundamental currents without saturating. • They are rated for operation up to 160°C through use of class “F” insulation. • Over load thermal cut off provided to protect the reactor.
Design Features of Detuned Filter Capacitor features • Capacitor is specially designed to carry wide spectrum of harmonic and fundamental currents without overloading. • It is designed for higher voltage to allow for increased voltage due to introduction of series reactor. • The kVAr of the capacitor is suitably designed to deliver the rated kVAr of the filter at the bus.
Calculation to Estimate the Rated Voltage of the Filter Capacitor I = V/ Xeq = V/ (XC (1-p/100))
I L XL= (p/100) XC
V C
Voltage across Capacitor VC is given by
VC = I XC = V/ (1-p/100) Allowing 10% for over voltage, the rated voltage of the capacitor is given by
XC
1.1 VC = 1.1 V/ (1-p/100) Bus Voltage
Percentage Tuning Factor
Voltage Rating of Capacitor
Voltage Rating Rounded off
415
7%
490.86
500 V
415
14%
530.81
550 V
Note on Capacitor for Detuned Filter Application • It is seen that the voltage rating of the capacitor has to be higher than the system voltage. • Hence normal capacitor of 415/440 volts rating should never be used in series with reactor. • Any such attempt would be hazardous to the capacitor and the installation.
Analysis of Detuned Filters
L
L
L
C∆
Actual connection of Detuned Filter
L
CY
L
L
L
CY
SLD Representation Star equivalent of Detuned Filter of Detuned Filter
Analysis of Detuned Filters Analysis of Detuned Filter can be done by analysing its single line diagram representation as shown IL L XL Let the net available kVAr. at Bus = N C Let the System Line Voltage in Volts = V V/√ 3 Let the Tuning Factor in % =p XCY Line current of the Filter IL= V/√3/(XC - XL) = V/√3/XC/(1-p/100) C The 3 phase kVAr. At Bus = √3V IL/1000 i.e NC = V2/ XC/(1-p/100)/1000 The XCY of the star eq. Capacitor = V2/ (NC/1000/(1-p/100)) The XL of the Reactor = XCY p/100 = V2/(NC/1000/(100/p-1)) The kVAr of the Capacitor at its rated voltage VC and the inductance of the reactor can be computed from the above.
Analysis of Detuned Filters kVAr. of the Capacitor at its rated voltage VC = (VC/V)2 NC (1-p/100) IL L Inductance per phase of the 3 phase Reactor in mH. XL L = V2/N /(100/p-1)/100/π C V/√ 3 From these formulae we can calculate the Detuned C XC Filter elements for standard outputs as under. For 7%, 440 Volts Detuned Filters Available Inductance per Rated Voltage of kVAr.of the Bus kVAr. Phase in mH. the Capacitor VC Capacitor at VC 12.5 25 50 75 100
3.71 1.855 0.928 0.618 0.464
500 500 500 500 500
15.01 30.02 60.05 90.07 120.09
Section 15 Exercise
Parallel Resonance
Description of the Problem
• • • • • •
A 500kVA, 415V, 3 phase, 50Hz., 4% impedance drop transformer is feeding a 100kW, 6pulse DC drive. The PF of the DC drive under full load condition is 0.7 lag. Calculate the following :A) Calculate the harmonic current spectrum of 100kW DC drive. B) Calculate the kVAr required to improve the PF of the drive above 0.99 lag. C) Calculate the harmonic voltage distortion and harmonic overload with the following :1) Without any capacitors. 2) With conventional capacitor connected to the network. 3) With de-tuned filter connected to the network.
Harmonic Spectrum • Harmonic spectrum of 100kW, 6 pulse, 0.7 PF DC drive is given by the relation: Ih =i1/h, where h is the harmonic order. Since the drive is 6 pulse, h = 5,7,11,13,17,19. I1 = =
100x1000 / (√3 x 415 x 0.7) 200 A (Approx.).
Harmonic Spectrum h
Ih Amps
5 7 11 13 17 19
40 28.6 18.2 15.4 11.8 10.5
Estimation of kVAr • kVAr required to improve the PF above 0.99 lag. • kVAr =kW x (tan (cos-1 (PF1)) - tan (cos-1 (PF2)) =100 x (tan (cos-1 (0.7)) - tan (cos-1 (0.99)) =87.8 kVAr. • kVAr =100 kVAr. (standard available)
Network Without Capacitors: 500 kVA
100 kW, 3ph,415V, DC drive
Linear resistive Loads
Network Without Capacitors Equivalent Circuit
XT
Ih
Calculation of Transformer Impedance Xt • We use the following formula. % Z = Transformer Impedance / Base Imp. Base Imp.= Phase voltage / Full load current. ∴ Xt = % Z* Phase voltage / Full load current. = 0.04* 415/√3/(500*1000/(415*√3 )) ohm. = 0.013778 ohm.
Harmonic Simulation Without Capacitors h
Ih
5 7
40 28.6
Xth = Xt*h 0.0689 0.0964
Vh = √3*Ih*Xtotal 4.7728 4.7728
11
18.2
0.1516
4.7728
13
15.4
0.1791
4.7728
17
11.8
0.2342
4.7728
19
10.5
0.2618
4.7728
Total Harmonic Voltage Distortion % Without Capacitors Total harmonic voltage distortion THDV% = (√(V52+V72+V112+V132+V172+V192) / V1) x 100 = (√(4.772+4.772+4.772+4.772+4.772+4.772)/415)*100
THDV% = (11.69/415)*100 THDV% = 2.82%
Network With Capacitors: •
Calculation of harmonic voltage distortion and harmonic overload: • With conventional capacitor of rating 100 kVAr. Capacitive reactance Xc= V2/ (kVAr x 1000) ohms. = 4152 / (100 x 1000). Xc = 1.72225 ohms. Transformer reactance. Xt = V2 x (%Z / 100 ) / (kVA x 1000) ohms = 4152 x (4 / 100) / (500 x 1000). Xt = 0.013778 ohms.
Network With Capacitors 500 kVA
100 kW, 3ph,415V, DC drive
Linear resistive Loads
100 kVAr, 3ph,415V
Network With Capacitors: Equivalent Circuit
Ih
Xt
XC
Harmonic Simulation With Capacitors Xth=Xt*h Xch=Xc/h
h
Ih
5
40.0
0.0689
0.3445
0.08611 5.9659
10.00
7
28.6
0.0964
0.2460
0.15863 7.8501
18.42
11
18.2
0.1516
0.1566
4.73619 149.151
549.89
13
15.4
0.1791
0.1325
0.50885 13.5593
59.08
17
11.8
0.2342
0.1013
0.17853 3.6379
20.73
19
10.5
0.2618
0.0906
0.13866 2.5281
16.11
|Xtotal| = Xth*Xch |(Xth-Xch)|
Vh = √3*Ih*Xtotal
Ic = Vh/(√3* Xch)
Overloading of Capacitors Ich = 554 Amps. Ic1 =139 Amps. Iceff=571 Amps. Overload = 411 %
THDV% With Capacitors THDV% = (√(V52+V72+V112+V132+V172+V192) / V1) x 100 =(√(5.972+7.852+149.152+13.562+3.642+2.532)/415)*100 = (150.16)/415*100
= 36.18%
Calculation With Detuned Filters 500 kVA
XLF 100 kW, 3ph,415V, DC drive
Linear resistive Loads
112.5 kVAr, 3ph,440V,DF
Why 112.5 kVAr • The detuned filters, supplied by L&T/Meher are rated for a bus voltage of 440V. This is mainly done to provide a standardized solution. • The kVAr output of a 440V capacitor is reduced if connected to a 415 volts system, given by the relation. = (Vsystem/Vrated)2 * kVAr. = (415/440)2 * 100 kVAr. = 88.9 kVAr. Hence while suggesting a detuned filter for a 415 V system, additional kVAr has to be provided so as to provide the required 100 kVAr at system voltage. .
Why 112.5 kVAr • Hence if the 100 kVAr capacitors have to be supplemented by detuned filter, the detuned filter should be rated for 112.5 kVAr, calculated by the relation = (Vrated/Vsystem)2 * kVAr = (440/415)2 * kVAr = 1.124 * 100 = 112.5 kVAr. (approx.)
Network With Detuned Filters: Equivalent Circuit
Ih
Xt
XLF
XCF
Detuned Filter of 112.5 kVAr, 440V Capacitive kVAr
= = = = XC = XCF =
112.5 * (1 – p/100) 112.5* (1-7/100) 112.5 *0.93 104.625 kVAr 4402 / (104.625 x 1000) 1.8504 ohms.
Reactance of reactor XLF = p/100* Xc = 0.07*1.8504 = 0.129528 ohm.
Harmonic Simulation Transformer reactance. Xt
Xt
=
V2 x (%Z / 100 ) / (kVA x 1000) ohms.
=
4152 x (4 / 100) / (500 x 1000).
=
0.013778 ohms.
Harmonic Simulation With Detuned Filter h
Ih
Xth = Xt*h
XLFH = XLF*h
XCFH = XCF/h
XFilter = XLFH – XCFH
Xtotal = Xfilter*Xth ( XFilter+Xth)
Vh = √3*Ih* Xtotal
Ic= Vh/ (√3*XFilter)
5
40
0.0689 0.6476 0.3701 0.2776 0.0552 3.8238
7.95
7
28.6
0.0964 0.9067 0.2643 0.6424 0.0839 4.1498
3.73
11
18.2
0.1516 1.4248 0.1682 1.2566 0.1353 4.2591
1.96
13
15.4
0.1791 1.6839 0.1423 1.5415 0.1605 4.2760
1.60
17
11.8
0.2342 2.2020 0.1088 2.0931 0.2107 4.2925
1.18
19
10.5
0.2618 2.4610 0.0974 2.3636 0.2357 4.2969
1.05
Overloading of Detuned Filter Ifh = 9.27
Amps.
If1 =139.12 Amps. Ifeff=139.43 Amps. Overload = Negligible.
THDV% With Detuned Filters THDV% = (√(V52+V72+V112+V132+V172+V192) / V1) x 100 = (√(3.822 +4.152+4.262+4.282+4.292+4.302)/415)*100 THDV% = (10.25/415)*100
THDV% = 2.47%
Summary It is seen that the introduction of capacitors into a network with non-linear loads not only leads to very high overloading of the capacitors but also increases the harmonic voltage distortion in the network due to parallel resonance. If the same capacitors are supplemented with detuned filters, the overloading is prevented and also the voltage distortion in the network is reduced to acceptable limits.
Conclusion
Detuned filter is a safe and proven solution to improve power factor in harmonic rich environment.
Limitations of Passive filters •
Sensitive to system frequency change.
•
Sensitive to change in network parameters.
•
Has location limitation when ‘Drives’ exist in system.
•
Mixing of filters having different tuning factor generally not possible due to risk of resonance.
•
Not immune from harmonic overloading.
•
Generally cannot handle wide harmonic spectrum.
•
kVAr. output of filter variable only in steps.
Session 15 Active Compensation
SCOPE • ACTIVE COMPENSATION • NEED • PRINCIPLE • BENEFITS • SOLUTIONS • AHF - ACTIVE HARMONIC FILTER • INTELLVAr - E • Electronic VAr Compensation
Need for Active Compensation - I In networks where HARMONIC FILTERING is to be done • Independent of PF Improvement
• for complex harmonic frequencies • for fine control of THD(V):
<3%
Need for Active Compensation - II for Unsymmetrical Reactive Power Compensation (PF improvement) • in REAL TIME MODE • in STEPLESS MODE • where INFINITE CONTROL is needed •
for real time Voltage Support
Principle of Active Compensation Involves real time CURRENT INJECTION into a network • in Variable AMPLITUDE & PHASE ANGLE • in COMPLEX WAVE SHAPES • with INFINITE CONTROL • at any LOCATION (in shunt)
Active Compensation Benefits • Filtering upto the 50th HARMONIC including Inter-Harmonics • Unsymmetrical Compensation of Reactive Power • Real time response < 2 msec • Independent of network characteristics, voltage & frequency behavior • INFINITE CONTROL • Compatibility with conventional compensation installations
Active Compensation - BENEFITS Total PF Control
No risk of Resonance
Voltage Stabilty
Extremely Flexible
THD (V) Control
Plug & Play Solution
SOLUTIONS - I • ACTIVE HARMONIC FILTERS - for 3 Phase, 415/440 V, 50 Hz. NETWORKS - In Current Ratings from 32 to 630 Amps - Optional Reactive Power Compensation
SOLUTIONS - II • INTELLVAr - E - Hybrid ELECTRONIC VAr COMPENSATION - for 3 Phase, 415/440 V, 50 Hz. NETWORKS - Output from 50 to 1000 kVAr
Active filter schematic diagram Sinusoidal Supply Current
Load Current with Harm onics
Supply System
Compensating Current Active Filter
+
= (Time Domain)
+
= (Frequency Domain)
Waveform of current without Active filter
Waveform of current with Active filter
Current harmonics without active filter
Current harmonics with active filter
Cost-Technology Pyramid of Harmonic Filters
Suitable for installations having high harmonic distortion (cement, sugar & steel plants, etc.)
Active filters
Top end solution for wide spectrum of current harmonics & suitable for installations having sensitive equipment
Hybrid Filters
Tuned Filters
14% Detuned Filters Suitable for installations having 3rd harmonics and above (IT parks, corporate banks & establishments) 7% Detuned Filters Most common, Base end product suitable for majority of industries having 5th and above harmonics
