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vittal said:
Yoy may know
P=1.732(root 3) x VXIXPOWER FACTOR
AND 1HP=?wATTS
NOW YOU MAY FIND YOUR CONSUMPTION
By just knowing the connected load in HP you can not calculate what is the consumption in kWh.It depends on loading. Otherwise 1000HP = 746 kW and if it is fully loaded than it will consume 746kWh i.e 746 units in one hour
Hello, no it s not normal at all, try to consult with manufacturer there is some routine test that they could investigate your transformers during this tests.
Sincerely,
S.Solimani
It is physics and has do with 2nd year electrical vector diagrams on resistance inductance and capacitance
The statement is correct resistive is the bottom of the triangle inductive goes up and capacitive down which the total load determines the angle which is the power factor (PF).
The only way to reduce power factor (PF=cos(Pheta)) that is inductive (motors) is to add capacitors or ballast with capacitor in them as it is an additive relationship base on the angle of usage.
This is very complicated but I hope that put it in laymen’s terms.
hi erikench,
thanks for information. i want to know whether class of insulation will affect the overload setting .
anil i agree with you.
But i think to know the full load losses you have to go for open ckt test (i.e. iron losses) and short ckt test (i.e copper losses).then to know tfransformer loss at any load= iron loss + copper loss at that load.
e.g. copper loss at 50% load = ( 50% 0f full load current)2*R
=(I/2)2*R
=I2R/4
Hi,
Look for Transformer Test report. It will have Iron losses which will remain almost constant through out the loading cycle. Then look for efficiency on the name plate. This corresponds to full load. Now calculate the full load losses. Example if 1000KVA transformer efficiency is 98% then the losses are 2% i.e. 20KW. If the Iron Losses is indicated as 2KW, then the copper losses are 18KW. Now the copper losses are proportional to square of current. Therefore at 50% load copperlosses will be 0.5 sqare times the full load copper loss. i.e. 0.25X18=4.5KW. Therefore at 50% load you will have 6.5KW losses. If you dont know what is the Iron losses of the transformer, then see the noload power consumption of the transformer, it will be very close to iron losses.
just ask the technical question dear …..its depends on the motor location from power source.but I think u want what size of cable required for this rating ….may be …so the current consumption for 100 kw motor is
100= 1.732x415x0.8x I
I =100000/1.732x415x0.8
I= 100000/575.024=173A
main difference between three phase and single phase energy meter .The CT ratio and PT are connected in 3 phase system for measuring the current and voltage simultaniousely.whereas in single phase no need to connect such things.KVRAH is the reactive power taken by transformer and kiosk unit for itself.for the modifying of this power factor we are connected the capacitor bank to this equipments so it may inductive power or capacitive power so it should be the express in capacitive and inductive respectively.
If you have pf 0.99 or more that is mean you are improve the main electric network , and the company of electruc should give you the bonus for that .
and you have also avery emportant advantage , your load Ampere ,ll be decrease that is mean you can add more load and that is mean also your panel and circuit braker ,ll be work at the good status >
Eng. Ali Younes
MCB is miniature circuit breaker which is thermal oprated
and use for short circuit protection in small current
rating circuit. MCCB moulded case circuit breaker and is
thermal operated for over load current and magnetic
operation for instant trip in short circuit condition.under
voltage and under frequency may be inbuilt. Normally it is
used where normal current is more than 100A.
MCB – MINIATURE CIRCUIT BREAKER
MCCB – MODULE CASE CIRCUIT BREAKER
MCB RATING : 63 AMP (MAXIMUM) i.e. in the form of tpn(
THREE PHASE + NUETRAL)
MCCB _ 63AMP (min.)
MCB IS USED WHERE LOAD IS LOW BUT MCCB IS USED WHERE LOAD
IS HIGH AND USE AS INCOMER.WHILE THE MCB IS USED FOR
OUTGOING.
RCCB
Residual Current Devices (6GDRC2)
6GDRC2 RCCB/RCDThe flow of current through electrical facilities always involves risks. Poorly insulated equipment, faulty wires and incorrect use of an electrical devise cause currents to flow through the wrong path (i.e., through the insulation) to the earth. This current is called ‘Leakage Current’.
Earth Leakage is an electrical hazard and is responsible for electrical shocks and fire risk. Earth Leakage and its associated hazard can be prevented by residual current device (RCD), similar to that of a residual current circuit breaker (RCCB), also popularly known as Earth Leakage Circuit Breaker (ELCB).
Features
– Simple and Robust operating mechanism.
– Rotary handle with ON/Off indication
– Dual termination for Bus Bar as well as cable connection
– Advance Neautral
– Test Button for regular inspectionyou can get more at this link
http://www.mcbrccb.com/ProductShow/Residual-Current-Devices_ID21.html
power = 1.73 * V* I
V = 380 v
I = ?
power = 315 kva = 315 * 1000 = 315000 va
I = power / 1.37 * v = 315000/1.73 * 380 = 315000/657.4 = 479.16 A
and from my experience you can know the Ampere direct by this
if you have KVA and output volt is 380
you can do that
KVA * 1.5 = Ampere
if you have the power be KW you can do that
KW * 2 = Ampere
about your load and you hvae pool pump motor 3 phase
you must know her we have a power factor not unity
and you must make a palance on the three phase
Engniee
Ali younes
