Capacitor Bank For Chiller

You'll likely find this in your system documentation. But we weren't sure you were qualified to be able to read it so we sent an agent to smuggle it out of sight. Now that you found us out, we'll tell you that we hid it in the second file cabinet on the left, third drawer down filed under "M" for manual. There you'll find all that you'll need to know on the Ronco 350 tonne chiller.

Wait a minute, my records show me that you haven't paid your bill. As a matter of fact, it seems the check you sent us for the initial payment bounced. Well I'm just going to send that agent back out again and move that paperwork again. Until you've paid us in full, don't expect any useful information from us.

While I did answer this question in as silly an approach as I could. The OP does not provide enough information to even start an analysis to design the capacitor network to correct the power factor. So when no clear topic has been defined, how can my answer be off topic?

Based on limited information and WAGging, 75 (± 50) KVAR.

Please provide

actual pf make sure you have this

target pf

then use KVAR tables for suitable product to install

cheers :)

ok now in detail. Votage: 380 Runing amp : 400 present PF: 0.75 Target PF: 0.97

Hi,

Please next time make an effort (research) to understand your problem so that you can supply necessary data for evaluation, dole out is good but you will not retain or capture the essence of learning. (like redfred said)

Anyway, its a typical problem for big motors to improve its power factor to unity. Try browsing capacitor bank and power factor. Soft starters can also improve your power factor at any load condition.

Good Luck.

Hi the following site gives the best information required by you.

http://www.teknatool.com/products/Lathes/DVR/downloads/How%20Power%20Factor%20corection%20works.pdf

thanks seethasub really very useful information. and also thanks other friends who answer my Q. best regards

Dear friends that I inform you ago last month already I design one KVAR capacitor bank for this chiller which is working 100% properly and 35% saving bill. my calculation base was on: ruining load Amp x Volts x1.73 = KVA KVA x 0.750 KVAR x 18.37 = Microfarad capacitor. Connection in Delta. 360 Amp x 380V x 1.73 = 236 KVA 236 KVA x 0.750 = 177 KVAR 177 KVAR x 18.37 = 3,251.49 Microfarad capacitor / 3 = 1,083.83 UF -each line. 18.37 is factor which is converting KVAR to Microfarad, but when voltage 380v, for more detail please visit,Table (Calculation Table of KVAr and µF) http://www.ec21.com/product-details/Calculation-Table-of-KVAr--3936929.html

Dear member,

Useful formula to create and understanding on your own situations as below; Formula to estimate how much kVAR needed for installation (chiller)

kVAR = P (Tan Ø1 - Tan Ø2) where

kVar = Total size of Capacitor for the installation.

P = Total KiloWatt (Total Connection Load "TCL")

Tan Ø1 = 1st angle of power factor.

Tan Ø2 = Second angle of power factor.

e.g.

Say one installation of chiller have 350 tons capacity, running at 400Amp, where voltage at 415VAC. This new installation doesn't have any capacitor bank and need to install one unit of it, assume at rated given that power factor currently poor; say at CosØ 0.7, how much the total value of capacitor bank do need to install base on this installation figure;

P= √3 VI CosØ

P= 1.732 x 415 x 400 x 0.7

P= 201,000W = 201kW (TCL)

Therefore;

TCL (P) = 201kW

1st powr factor CosØ 0.7 = 45.57° = TanØ 1.02

Say power factor that need to repair to CosØ 0.85 = 31.79° = TanØ 0.619

So;

kVAR = P (Tan Ø1 - Tan Ø2)

kVAR = 201kW (TanØ1=1.02 - TanØ2=0.619)

kVAR = 201x10³ (1.02-0.619)

kVAR = 80601 VAR = 80.60kVAR

Therefore; total kVAR for this load is 80.60, corrective angle CosØ from 0.6 to 0.85.

Usefull formula;

Setting for capacitor bank with use of capacitor regulator for react to change and/or cooresponding to about 2/3 of the current at one capacitor step;

Setting = 2/3 ∫ P / (√3 V CTC)

Where;

P= Power rating of one capacitor (in kVAR)

V = Voltage rating

CTC = Currrent Transformer Rating e.g. for that installation is 500/5

Now I can reduce my WAG uncertainty band (post 2) to 75 (± 6) KVAR.

You need to know present power factor and then to what value u want to improve it.Incidentally when u say chiller i presume u are referring to its compressor motor

First - The present factor can be calculated and/or played through the same formula, its additional of √3 (1.732) present in the 3 phase calculation compared to single phase calculation, therefore;

CosØ= P/(√3 VI) 3 phase calculations

CosØ=P/(VI) single phase calculations

Second - The formula equally the same, what make it different? Chiller measure in tonnages where more to btuH and Hp of the motor itself load applications base on end user usage e.g. AHU, PAHU, FCU and etc. they wil play with water flow in GPM application and heat and cold exchanger between copper tubing inside evaporator or condenser, where this more applicable to superheat and T applications.

However, centrifugal motor for chillers, screw type motor for chiller, horizontal motor for chillers and etc. took electricity supply in the same source of incoming e.g feeders, transformer, generator set and etc. therefore;

1. Start-up chillers can use soft starter, auto transformer, direct module injections (DC motor applications), Variable speed drive (VSD) and etc.

2. To stabilized the angle between true power (kW) and apparent power (kVA) for chiller motor usage, the electrical formula still applied to the situation, additional efficiency is needed when you would like to calculate for it, e.g.;

P = √3 VI CosØ Eff - efficiency of 3 phase motor application.

is that your name or" Engineer" is your profession. You should have enough knowledge to calculate the power factor if you put the word engineer in front of you. Any way it is a typical middle east and arab style. The best answer is by redfred for you.

we are too stupid to judge someone else.... else! somebody wil judge us...