This document analyzes improving the power factor at a Bosch shop floor by installing a capacitor bank. It defines power factor and discusses causes of low power factor like induction motors and transformers. Installing static capacitors is identified as a remedy to compensate for reactive power consumption. Calculations show the capacitor bank improved the power factor from 0.763 to 0.96, reducing power losses by 2.621KW and saving 1179.45KWh per year, equating to cost savings of 84920.4 rupees annually. Installation work and its effects on various feeders are discussed. In conclusion, the power factor was improved from 0.65-0.85 to 0.95-0.99,
1. ANALYSIS AND IMPROVEMENT
OF POWER FACTOR AT
BOSCH SHOP FLOOR BY
INSTALLING CAPACITOR BANK
BY External Guide
Lingaraju.Ky(Sptd.Eng)
P.Sreelekha Internal Guide
Akash Roy A.M.Leela(Asct.Professor)
Sunith Kumar.A
5. Causes for Low Power Factor
Most A.C motors are of induction type
having low lagging power factor.
Magnetizing current drawn by transformers
lead to lagging current at low loads(specially
during morning & evening).
Typical characteristics of Arc lamp in Cinema
Induction furnace ,Short transmission lines.
Current limiting reactors used to minimize
fault currents work at low power factor.
6. Disadvantages of Low Power Factor
Large KVA rating of equipment .
Greater conductor size .
Large copper losses .
Poor voltage regulation .
Reduced handling capacity of system.
Unnecessary penalty is imposed (during
billing).
7. Different PF Correction Methods Available
Reduce the amount of reactive energy.
Compensate artificially for the consumption
of reactive energy.
ROTARY EQUIPMENT
STATIC CAPACITORS
12. Advantages Of Capacitors Over
Other Method(s)
Capacitor banks ensures very low initial cost.
The running costs are minimal and they
can be used with the same high efficiency
on all sizes of installation.
Compact, reliable, highly efficient &
convenient to install .
22. Savings in Power
• Resistance of the cable 0.15ohm / km load ; Length= 175mt.
• Resistance of the cable = 0.15 *0.175km = 0.026 ohm.
• P= √3 *VL*IL* Cos Φ1
• Current drawn by the load at 0.763 P.F , IL1= P/ (√3 *VL*Cos Φ1 )
=171.83/( √3 *430.52*0.763)
=302.009A.
• PF after correction Cos Φ2 = 0.96
• Current drawn by the load at 0.96 P.F , IL2 = P/ ( √3 *VL*Cos Φ2 )
= 171.83/( √3*430.52*0.96)
= 240.03A.
23. Savings in Power
• Power loss P1= 3*I1
2R = 7.114KW
• Power loss P2= 3*I2
2R = 4.493KW
• Saving in power = 7.114KW – 4.493KW = 2.621KW.
• Time= 18Hours * 25 Days = 450.
• KWh loss in P1 = 3201.3KWh.
• KWh loss in P2 = 2021.85KWh.
• Reduction in KWh (units) = 1179.45
• Cost saving = 1179.45 *12 months * Rs. 6/-unit =
RS.84920.4
29. ALGORITHM
• Identify the feeder to be worked based on the
difference of power factor
• p= Power in KW
• a= Power factor(current)
• b= Target power factor
• Find out the Reactive power Compensation
required for achieving target power factor
• Compare the current and targeted P.F
• Improvement done is displayed
Simulation using MATLAB
32. CONCLUSION
• Improved the power factor from the range of
0.65-0.85 to an existing range of 0.95-0.99
effectively.
• Significant decline in the rating of current flow &
reduced I2R losses also leading to reduced stress
on the cable there by increased durability of the
cable.
• Finally reduced on the KVA demand there by
reducing the penalty imposed ,also the total KWh
consumption as a whole has been reduced. Thus
creating savings of 27.5lakhs p.a