Formulas for Voltage Drop of Motors (Starting & Normal)

Have you asked the NEMA?

NEMA - National Electrical Manufacturers Association

Why would they not apply to other motors?

Check here http://cr4.globalspec.com/thread/87927/Voltage-Drop-Of-The-Motor-During-Normal-And-Starting-Conditions

I have sent query to Nema but still with no reply

The above link for other thread didnt conclude with an answer to this thread

The formulas shown in Nema ICS-14 ch. 2.3 TRANSFORMER SIZING are connected to the example presented further[ 2.3.3] This is a transformer dedicated for a pump and does not take into consideration the other load on this transformer. The rated current:

I=MFLA*1.25 as per NEC art. 695.5(A)

The general formula is S [apparent power] =sqrt(3)*VL_L*I[current in A]/1000 [kVA] by definition.

Accurate formula for voltage drop one can see in IEEE-141[Redbook]:

3.11 Calculation of voltage drops

3.11.1 General mathematical formulas

The arithmetic voltage drop V, as defined, is usually considerably less than the absolute value |I*ZT| of the vectorial voltage drop I*ZT.[For three phases system you have to multiply by sqrt(3).]

ZT=ZT%/100*VL_L^2/Srated [ohm] by definition.

Voltage drop[p.u.]=sqrt(3)*ZT*MFLA*k/VL_L= sqrt(3)*ZT%/100*VL_L^2/Srated*MFLA*k/VL_L

Voltage drop[p.u.]= sqrt(3)*ZT%/100*VL_L*MFLA*k /(Srated*1000) [1000*Srated [kVA]=Srated[VA].

During motor running k=1.15 according to NEMA MG-1 ch.12.51 SERVICE FACTOR OF ALTERNATING-CURRENT MOTORS Table 12.4:

For medium motors [ from 1.5 hp up to 500 hp] the service factor is 1.15.

During the start I=6*MFLA [k=6].

Thanks for your reply, however please advise on the below:

*Definition of MFLA , and its code source

* I didnt get your reply concerning the connection of IEEE-141 to the formulas in my query item (about voltage drops during normal and starting of motor); so please clarify (not in P.U system)

* I didnt find the reference for the factor 6 (K=6)

Thanks

1)Since the motor ratings are 300 Hp, 460 Volt, 60 hertz and the current mentioned is 361 A this matches the NEC Table 430.250" Full-Load Current, Three-Phase Alternating-Current Motors". NEC defines this current as "Full Load Current -FLC or Motor Full-Load Current [MFLC].

Usually FLA [MFLA] is referring to motor nameplate current.

The actual motor description is in ICS-14 ch.2.1.1 "Example" and in pos.a) the current is named "motor full-load current" but in pos.c) "motor full-load amperes" [it is the same 361 A].

2) In IEEE-141/1993 ch.3.11.1 General mathematical formulas:

"For exact calculations, the following formula may be used:

actual voltage drop= es+IRcosφ+IRsinφ-sqrt(es^2-(IXcosφ-IRsinφ)^2)

where the symbols correspond to those in figure 3-11."

3) In the same above pos. c) it is noted:

"Multiply motor full-load amperes 361 by 6, which equals 2,166 amperes (the locked rotor current of a NEMA Design B 3-phase squirrel cage induction motor is approximately six times the full load current of that motor, and this approximation is used throughout similar examples in this Guide."