Page 210 - Industrial Power Engineering and Applications Handbook

7/190 Industrial Power Engineering and Applications Handbook
considered in view of condition (c) which is more severe. Considering the permissible tolerances in the declared
Therefore, average loading on motor in one-hour cycle torque values, the safer starting time should be taken as
roughly 1&15% more than the calculated value. Therefore,
/(1.25Ir)’ x20+/: x40 consider the safe starting time as 16.25 x 1.15, i.e. 19.0
=I 60 seconds. For a single start under hot conditions the motor
should have a minimum thermal withstand time of 19.0
= I, \lm
seconds, and for three consecutive starts, a minimum
= 1.091, 57.0 seconds, which, for this size of motor, may not be
i.e. the rating of the motor should be higher by 9% to practicable in view of design economics. For this appli-
account for this stipulation. cation, therefore, a squirrel cage motor is not recom-
mended, unless a fluid coupling is employed to transmit
Note Normally, the heating-up time constant T for an HT the power.
motor varies between 0.9 to 1.5 hours, i.e. heating-up
time under normal running condition from 2.5 to 4.0 hours Corollary
approximately. In one hour, therefore, the motor will not The use of a squirrel cage motor is, however, inevitable
reach its thermal equilibrium and the equivalent rating as
determined above is in order. One should not consider a as discussed earlier particularly for a process plant or a
cycle longer than heating time, which may give incongruous power house application, where downtime for maintenance
results: of a slip-ring motor is unwelcome, or a chemical plant or
:. kW to be chosen, considering all deratings contaminated locations, where the application of a slip-
ring motor is prohibitive. To meet such load requirements
- 450 x 1.09 with a squirrel cage motor, the use of fluid couplings to
0.92 x 0.90 start the motor lightly and reduce starting time, is quite
= 600 kW common and economical, as discussed in Chapter 8, and
must be adopted in the above case.
A 1000 r.p.m. 600 kW motor should be an economical Let us use a fluid coupling and start the motor lightly,
choice. similar to Figure 8.3. The revised accelerating torque
(e) To meet the condition at 70% voltage, the motor should
have a T of more than 189% and should be minimum and approximate clutching sequence of the coupling is
go
1/(0.605) (Table 1.5) or 273% of 450 kW or 204.75% of illustrated in Figure 7.22.
600 kW, so that the motor does not stall.
(f) To consider the service factor (SF) the motor should have 1 Consider the clutching of the coupling with the load
a continuous reserve capacity as may be desired, for at roughly 0.8Nr, by which time the motor would be
example for a SF of 1.15, a motor of at least 1 .I5 x 450, operating near its Tpo region
i.e. a 517.5 kW rating must be required. We have already
selected a motor of 600 kW, therefore this factor need not :. N,, = 0.8N,
be considered again. Moreover, too large a motor than
the load requires would make it operate underloaded and = 800 r.p.m.
diminish its operatinq efficiency and P.f., which is a drain
on the usable energy and is not desirable. See also Sections Considering an average coupling opposing torque as
23.3 and 7.10. roughly O.4Tr and average motor torque as 1.35Tr,
between 0 to 0.8 Nl., the exact torque can be calculated
Checking the suitability of a squirrel cage motor by measuring the torque ordinates at various speeds
Assume that the motor is designed for an average speed- and then calculating the average. See Example 2.4
up torque of 135% and Tpo of 220% (Figure 7.21). If the for more details.
average load torque is assumed as 68%, the average ... Tal = 1.35Tr - 0.4Tr
accelerating torque, T,, available will be 67% on DOL
starting, i.e. = 0.95Tr
450 - 450 x 974 o,95
T, = 974 x 0.67 rnkg (considering N, = 980 r.p.rri.) -
980 980
- 300 mkg = 424.88 mkg
Say, the GD; = 200 kgm2
GD2 during a light start, considering the GD: of the
2 coupling (impeller) as equal to the motor (the exact
and GD; = 10000 (%) (at motor speed) value should be obtained from the manufacturer).
.. GD; = 200 + 200
-- 1666 kgm2
:. GD; = 1866 kgm2 = 400 kgm2
400 x 800
and starting time, and t,, up to 0.8Nr = 375 x 424.88
t, = 1866 x 980 = 2.01 seconds
375 x 300
2- 16.25 seconds 2 The coupling average accelerating torque after it has