Motor Starting Analysis

MOTOR STARTING ANALYSIS A. INTR INTROD ODUC UCTI TION ON Motor starting study should be done before purchasing power transformer and cables. In the course of the analysis, the motor starting current and method of  starting will be decided. Motor terminal voltage as well as torque requirement should be analyzed at the same time. A complete motor starting analysis consists of voltage drop and acceleration calculations. In the initial study, typical motor data shall be used. These data shall  be updated when vendor data are made available. B. WHEN TO CONDUCT CONDUCT A MOTOR STARTING STARTING ANALYSIS ANALYSIS 1) Transf Transform ormer er self-coo self-cooled led KVA rating rating is less than three three times times the larges largestt motor KVA rating. 2) Motor Motor feeder feeder reduces reduces the short short-ci -circu rcuit it KVA to less less than eight times times the motor-starting KVA. 3) Ratio of bus shortshort-circu circuit it KVA to motor-st motor-starti arting ng KVA is less than than 8 times. times. 4) Voltage Voltage drop restrict restriction ion imposed imposed by the Utility Utility (usually (usually limited limited at 2%). 2%). 5) Presen Presence ce of of high high iner inerti tiaa load. load. C. CORRECTING CORRECTING FOR EXCESSIVE EXCESSIVE VOLTAGE VOLTAGE DROP DROP 1) Use a motor motor with lower starting starting current. current. 2) Connect to a system system with a higher higher short-c short-circui ircuitt MVA. 3) Use a power transform transformer er with with lower lower impedance. impedance. 4) Use a reduced reduced voltage voltage starter starter or soft soft starte starter. r. 5) Instal Installl a capacit capacitor or bank. bank. D. FOR FORMULA MULAE E 1)

2)

0 1

 Z new

VA LR

=

 POWERBASE new  POWERBASE old 

=  I  LR ×V   ×  MOTOR

3) XL of Running Load,

0 1

V  L    V  L 

2

old 

0 1

Z old 

new

3

 X  L

=

VA BASE  VA L

4) Torque is is directly directly proporti proportional onal to the the square square of voltage. voltage.

E. SAMPLE PROBLEM NO. 1 Given:

In Fig. 1, determine the voltages at the utility and motor terminal at the instant when the motor is started.

FIG. 1: DIAGRAM FOR CAPTIVE LOAD

Solution: At 400MVA power base; 13.8KV, and 2.4KV Voltage bases; 1

 X U 

=1

1

 X T 

=

0

0

0 1

 X  M 

1

V TU 

=

0 1

1. 5

( 0.065) = 17.33 400 ×10

2

 2.3  =   = 49.846 1850 × 2300 × 3  2.4 

0

0

400

1  I ST 

V TOTAL

=

0 1

6

V TOTAL

∑

0 1

X 

=

1 1 + 17.33 + 49.846

=

0.01466

− 01  X U  ( 01 I ST  ) = 1 − 1( 0.01446 ) = 0.9853

Voltage level at the Utility = 0.9853(13.8) = 13.597KV

0 1

V TM 

=

0 1

V TOTAL

− 01  I ST  ( 01  X U  + 01 X T  ) = 1 − 0.01446(1 + 17.33) =

0.73128

Voltage level at the Motor Terminal = 0.73128(2.4) = 1.755KV

F. EXAMPLE NO. 2 For system with running load, the diagram is as shown below

The per unit impedance for the running load is shown in formula no. 3.