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adminKeymaster
I'd use a watt-hour meter.
adminKeymasterNFPA 780 would be another good start for informaton.
adminKeymasterTANMOY METE said:
jadeja parakramsinh said:
what is capacitorbank? and how to use capacitor bank in 100hp load?
If the number of capacitors are connected in parallel it forms capacitor bank. it is actually used for the power factor improvement.
here the capacitors should be connected in delta form so that the capacitance required for the capacitor could be 1/3 of that 100hp load.
i think it would not do. because if we place near to the load power factor will fall.
adminKeymasterjadeja parakramsinh said:
what is capacitorbank? and how to use capacitor bank in 100hp load?
If the number of capacitors are connected in parallel it forms capacitor bank. it is actually used for the power factor improvement.
here the capacitors should be connected in delta form so that the capacitance required for the capacitor could be 1/3 of that 100hp load.
adminKeymasterjadeja parakramsinh said:
what is capacitorbank? and how to use capacitor bank in 100hp load?
A capacitor bank is a group (or bank) of several capacitors, sometime of fixed value, sometimes of variable value depending on measured KVAR, that provides sufficient KVAR (capacitance) to lower reactive power and raise power factor for the utility when used on the line side of the incoming electrical service.
A capacitor bank usually would not be required to treat one 100HP load (which I assume is an inductive (standard) electric motor). This could be handled by installation of 1 or 2 capacitors very near the load, on the line side, in parallel with the load. The correct amount of KVAR to add would need to be determined by measuring the KVAR while the motor is run through its varying load conditions (if there are any — i.e. heavy loaded, lightly loaded), so you don't add too much KVAR, but a sufficient amount to get it to approximately 0.95.
Make sure you purchase high quality capacitors of the correct value (KVAR), voltage, and frequency (50hz or 60hz).
adminKeymasterMistake there,
Power factor = cos (x)^-1
adminKeymasterP = kW, Q = kVAr (inductive (+Q), capacitive (-Q), S = kVA.
S = (P^2 + Q^2)^-2
kW / kVA = cos (x), cos (x)^1 = power factor
adminKeymasterHello,
You should look for you answers in IEC 62305 2nd edition, parts 1 (for an introduction) & 3 (for detailed explanation).
Another source of info is the DEHN lightning protection guide which can be freely downloaded from their website.
adminKeymaster2011/08/25 at 6:04 am in reply to: Why Is Isolated Neutral Preferred In Some Turbines/Generators? #12415adminKeymasterYou can only have one neutral solidly earthed at each foltage level, w/o causing spurious tripping problems for sensitive earth fault relais systems.
adminKeymasterGood info, thanks for sharing.
I understand that the requirement in Brazil is 0.92, but I do not have access the the penalty.
Do you plan to occasionally update the spreadsheet?
“Knowledge not shared is knowledge wsted.”
adminKeymaster1 phase is electronagative body and neutral is electropossitive body
when we connect any electric load between them the electrons flowing through the load and u know
force x displacement =work than
flowing current x voltage across load = power
adminKeymasterHi,
1) Phase is which has electric power (electron or voltage) passing through
2) Neutral is drawn seperately, which doesnt have any current
3) 3 Phase are the phases to be drawn seperately which should not be merged
4) 11 kva means 11000 volts.
adminKeymasterif power factor increase Amp of motor decrease
adminKeymasterBhavik said:
how can we calculate that capacitor value of Motor for maintain of power factor
2- if power factor increase motor effecency is aiso increase
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