motor starting analysis
TRANSCRIPT
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MOTOR STARTING ANALYSIS
A. INTRODUCTIONMotor starting study should be done before purchasing power transformer andcables. In the course of the analysis, the motor starting current and method of
starting will be decided. Motor terminal voltage as well as torque requirementshould be analyzed at the same time.
A complete motor starting analysis consists of voltage drop and accelerationcalculations. 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 A MOTOR STARTING ANALYSIS1) Transformer self-cooled KVA rating is less than three times the largest
motor KVA rating.2) Motor feeder reduces the short-circuit KVA to less than eight times the
motor-starting KVA.3) Ratio of bus short-circuit KVA to motor-starting KVA is less than 8 times.4) Voltage drop restriction imposed by the Utility (usually limited at 2%).5) Presence of high inertia load.
C. CORRECTING FOR EXCESSIVE VOLTAGE DROP1) Use a motor with lower starting current.2) Connect to a system with a higher short-circuit MVA.3) Use a power transformer with lower impedance.4) Use a reduced voltage starter or soft starter.5) Install a capacitor bank.
D. FORMULAE
1) old L
L
old
newnew
Z V
V
POWERBASE
POWERBASE Z
new
old
10
2
10 =
2) 3= MOTOR LR LR V I VA
3) X L of Running Load, L
BASE L
VA
VA X =10
4) Torque is directly proportional to the square of voltage.
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E. SAMPLE PROBLEM NO. 1
Given: In Fig. 1, determine the voltages at the utility and motor terminal at theinstant when the motor is started.
FIG. 1: DIAGRAM FOR CAPTIVE LOAD
Solution:
At 400MVA power base; 13.8KV, and 2.4KV Voltage bases;
( )
846.494.2
3.2
323001850
10400
33.17065.05.1
400
1
26
10
10
10
=
=
==
=
M
T
U
X
X
X
01466.0846.4933.171
1
10
10
10 =
++==
X V
I TOTALST
( ) ( )01446.01110101010 == ST U TOTALTU I X V V = 0.9853
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Voltage level at the Utility = 0.9853(13.8) = 13.597KV
( ) ( ) 73128.033.17101446.011010101010 =+=+= T U ST TOTALTM X X I V V
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.