Page 66 - Industrial Power Engineering and Applications Handbook

Motor torque, load torque and selection of motors 2/47

175

150
1 125

al
; 100
100
0.96T,
b?
75

50

35
25
Speed - ?

N,
torque
(1485 rpm) Figure 2.18 Determining the accelerating

Calculate the starting time and consecutive cold and hot
starts for which the motor will be suitable with a DOL starting. (e) Total GD: at motor speed = 30 + (4 x 9.81 x 2.5) ___
( ::::)z
Solution
To determine the exact accelerating torque, measure the where [GD: = 4. g. MK2] (at the compressor speed)
ordinates of torque as shown at different speeds and calculate
the average torque as follows: 1.e. GD: = 30 + 901
(a) Average load torque = 931 Kgm'
- 35+40+50+60+70+80+85+95+100 931 x 1485
i- (f) Stating time t, =
9 375 x 146.9
61 5
=- = 25.1 seconds
9 Take roughly 10% more to account for any tolerance and
= 68.33% variations,
(b) Average motor torque :. ts = 25.1 x 1.1
150+120+110+110+130+170+180+120+100 = 27.6 seconds
T=
9
This motor is therefore suitable for only one cold or one hot
=- 1190 start at a time until the temperature rise stabilizes again.
9 If this motor is started with an auto-transformer with a
2 132.2% tapping of 80%, the motor average torque will be
= 132.2 x 0.64 (curve 2, Figure 2.18)
(c) :. Average accelerating torque
or T= 84.6%
T, = 132.2 - 68.33
and acceleration torque T, = 84.6 - 68.33
= 63.87%
= 16.27%
(d) Motor rated torque
i.e. 230 x 0.1627 = 37.42 mkg.
- 350x974
I-
1485 931 x 1485
* 230 mkg and t, = 375 x 37.42
:. T, = 230 x 0.6387 = 98.52 seconds
= 146.9 mkg which is much more than the safe stall-withstand time.