Calculating Voltage Drop

British Standard 7671.

I under stand that yours is 300 kw motor in star delta formation.you have indicated that 2 runs of 185 sq mm copper cable is connected.

Apparently for star delta connected motor there will be 6 connections from the starter.the full load current phase current will be 328 amps and line current will be 1.732 x 328 amps.

since two runs of cables are run, I presume one run only since it is delta connected.in this case the 185 sq.mm copper cable ,buried under ground can carry only 290 amps.

Cable size is inadequate, therefore you get very high voltage drop.

Quote "I under stand that yours is 300 kw motor in star delta formation.you have indicated that 2 runs of 185 sq mm copper cable is connected.

Apparently for star delta connected motor there will be 6 connections from the starter.the full load current phase current will be 328 amps and line current will be 1.732 x 328 amps."

I am not used to referring to a motor in kw, but since the post stated 300 kw at 380 volts, 300 kw/(.380 x 1.73) = 456 amps. The amps would be higher if you use KVA.

Quote "since two runs of cables are run, I presume one run only since it is delta connected.in this case the 185 sq.mm copper cable ,buried under ground can carry only 290 amps."

Why would you presume only one run when the post states 2 runs?

Quote"Cable size is inadequate, therefore you get very high voltage drop."

Not if there are 2 runs. VD is about 16 volts L-L or 364 volts.

"I am not used to referring to a motor in kw, but since the post stated 300 kw at 380 volts, 300 kw/(.380 x 1.73) = 456 amps. The amps would be higher if you use KVA."

I am sorry to intervene, you have not considered the power factor,when the motor is rated for 300 kW,you have to take pf into conideration for calculating the ful load current.

Therefore the full load line current would be I=(power)/(pf*line voltage *1.732).

in this instance, I=(300*1000)/(1.732*0.8*380)=569.7 Amps.

The phase current would be=328 amps.

From the recent post by the member I understand there is a main panel from which the starter is fed by two runs of 185 sq.mm copper cable.From the starter panel to motor there should be cable which should have 6 core ( 2 x 3 core).(delta formation)

The current flow on the cable between main panel to starter panel would be 569.7 amps and the current flow on the cable between starter and the motor will be 328 amps.

The cable should be designed for the above current rating.

The member has not indicated the cable particulars between the starter and the motor , therefore I have to assume the cable as single run.

Voltage drop can be arrived at if proper and all information are made available.

The voltage drop indicated by the member is very high but it should not be attributed only to cable.

There are 6 terminations in the route from the main panel to motor.

Even if the terminations are bad or contacts are bad the voltage drop will be high.

One has to visit the site and examine the installation.

Ramvinod

Thanks for your comments. In the original post the writer said

Quote" i have connected from distribution breaker 2 cable of 3 1/2 c 185 sqmm copper cable to a 3 phase 380v ,300 kw motor star delta ". I took that to mean he had a star distribution panel but he was connecting the motor as a delta. I did not think he was referring to a star - delta starter. I agree that the PF should be included in the calculations and I made that point in my first post. Thanks for your added information.

DEAR FRIEND YOU USE THE FOLLOWING FORMULA FOR 3PH.POWER LOSSES.

= 3xCosΦxLxRxT ⁄ 1000.

:CosΦ is what you maintain your P.F

L = Length, R = Resistance of cable , T = Temperature

Quote "

DEAR FRIEND YOU USE THE FOLLOWING FORMULA FOR 3PH.POWER LOSSES.

= 3xCosΦxLxRxT ⁄ 1000."

That is not correct. If you are referring to line losses in watts it is

I² x resistance of the conductor. If you are speaking of voltage drop, the formula

is VD = IR(cosΦ)+ IX(sinΦ) where R is the resistance and X is the reactance of the conductor. I = KVA/ Z x 1.73 where Z is the impedance of the system. You are using KW and Resistance which will not give you the correct answer. IF you are referring to the motor in KW, the actual amps will be less than when you use KVA which is the actual load.

Kumar, I presume 380/3/50/579A load serviced by (2) 3½ c 185mm2 Cu XLPE in ferrous raceway

Voltage Drop... by circular mil (CM) method applied to metric cable »

k = Rdc for 1000 CM conductor 1000ft long say copper = 12.9 CM-ohm/ft [ft/.3048M] ~ 42.32 CM-ohm/M

Q = inductance adjustment factor [Rac/Rdc] applied to conductors greater than 50mm2

VD% = SQRT(3) * k * Q * I * D / E / CM

SQRT (3) * [42.32] * 1.196 *570A * 200M / 380v / 730,216 CM ~ 3.6% voltage drop