Page 83 - Industrial Power Engineering and Applications Handbook

3/64 Industrial Power Engineering and Applications Handbook

5

4
Load
losses

Time - I 0

t, and tr, periods of rest.
@- Thermal curve of a motor with a rating of Peq.
@- Thermal curves for intermittent loads, the average heating not exceeding the permissible temperature rise 6,

Figure 3.16 Equivalent output of short-time duties

as shown in Figure 3.17. There is no rest period, but in one
. tl +-T: . t, -+ T: .t3 cycle, the motor runs idle twice at no load for 2 minutes each.
Teq (r.m.s.) = (3.12)
tl + t2 + t3 The cycle starts with a load requirement of 10.5 kW for 4
minutes followed by an idle running, a load of 7.5 kW for 2
and motor output minutes, again with an idle running and then the cycle repeats.
Solution
Assuming the no-load losses to be roughly 5% of the motor
rating of, say, 10 kW, then
If the cycle has a short-time rating, with a period of
energization and one of rest, the motor will obviously (10.5)' x 4 + (0.5)' x 2 + (7.5)' x 2 + (0.5)' x 2
cool during the de-energizing period, and depending upon peq =
the peak load and the rest periods, a comparatively lower 10
output motor can perform duties at higher loads. The _ _ 441 + 0.50 + 112.50 + 0.50
equivalent output for the load cycle of Figure 3.16 is 10
=.J5545
P? . t, + P; . t2 +pi . t3
Pq = t = 7.45 kW
Since total time tis more in this instance, the equivalent The nearest standard rating to this is 7.5 kW, and a motor of
power required will be less. this rating will suit the duty cycle. To ensure that it can also
meet the torque requirement of 10.5 kW, it should have a
Example 3.5 minimum pull-out torque of 10.5ff.5 or 140% with the slip at
Determine the motor rating, for a 10-minute cycle operating this point as low as possible so that when operating at 140%