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Current Full-load

STARTING CURRENT FULL LOAD CURRENT HEIGHT AGITATOR LBS, ... [Pg.683]

This may be 25-50% of the full-load current and sometimes even up to 60%. The higher the rating, the... [Pg.17]

The declared efficiency and power factor of a motor are affected by its loading. Irrespective of the load, no-load losses as well as the reactive component of the motor remain constant. The useful stator current, i.e. the phase current minus the no-load current of a normal induction motor, has a power factor as high as 0.9-0.95. But because of the magnetizing current, the p.f. of the motor does not generally exceed 0.8-0.85 at full load. Thus, at loads lower than rated, the magnetizing current remaining the same, the power factor of the motor decreases sharply. The efficiency, however, remains practically constant for up to nearly 70% of load in view of the fact that maximum efficiency occurs at a load when copper losses (f R) are equal to the no-load losses. Table 1.9 shows an approximate variation in the power factor and efficiency with the load. From the various tests conducted on different types and sizes of motors, it has been established that the... [Pg.17]

The higher the full load slip, the higher will be the rotor losses and rotor heat. This is clear from the circle diagram and also from equation (1.9). An attempt to limit the start-up current by increasing the slip and the rotor resistance in a squirrel cage motor may thus jeopardize the motor s performance. The selection of starling current and rotor resistance is thus a compromise to achieve optimum performance. [Pg.20]

The current in the copper ring opposes the main flux in that area of the pole and behaves like an artificial second winding, and develops a rotating field. Although the torque so developed is extremely low, it is enough to rotate such small drives, requiring an extremely low starting torque, of the order of 40-50% of the full load torque. [Pg.28]

Star contactor C and A/Tcontactor Ci must be rated for the square of the percentage lapping. For a tapping of 80%., for instance, the rating of the contactors C and Ci will be (0.8) or 64% of the full-load current of the motor. The main contactor Ct, however, will be rated for the full-load current. [Pg.74]

For a 3.3 kV 450 kW motor, with a full-load current of 100 A and starting current on DOL as six times the rated current, the kVA rating of the transformer for 50% tapping will be... [Pg.74]

The rotor voltage, refers to the standstill secondary induced e.m.f.. between the slip-rings. Whereas the rotor current, refers to the full load rotor current, when the slip-rings are shon-circuiled. [Pg.85]

When the torque is determined by the above method, the voltage during the test should be so adjusted that the locked rotor current is approximately equal to the full load current. After the locked rotor test, the resistance of the stator windings should be measured and may also be considered for calculating the I R losses. [Pg.264]

If single phasing occurs at 50% of the full load, the rotor current will be 100%. See curve 4 of Figure 12.10. [Pg.282]

The thermal curve of the motor does not show any significant overload capacity and therefore the relay must be set as close to the full-load current as possible, say at a setting of 110%. Then... [Pg.301]

If/ , = minimum fault current through the primary (chosen on the basis of the rated full-load current of the machine or the system being protected) required to trip the relay. It may be termed the minimum primary operating current (POC) of the scheme, /pi. In term.s of the secondary... [Pg.483]

I No current chopping An induction motor or a transformer should be operating at near full load, to possess a high p.f. and carrying a near full load current... [Pg.569]

By selecting the rating of the interrupting device as close to the full-load current of the system or machine as possible. An excessive rating than necessary may have tendency towards current chopping. [Pg.579]

Full load current of the system Unit impedance of the system... [Pg.711]

Current or full-load current expressed as amps (amperes) and drawn by motor at a stated voltage. [Pg.615]

Figure 14-4A. Oscillogram shows variation of current to a synchronous motor driving a reciprocating compressor, The compressor is two-cylinder, horizontal, double-acting, and operates at 257 rpm. Line A is the envelope of the current wave. Difference B-C is current variation. Value B-C divided by the rated full load current is the percentage of current variation. (Used by permission Oscarson, G. L. E-M Synchronizer, 200 SYN 52, p. 11. Dresser-Rand Company.)... Figure 14-4A. Oscillogram shows variation of current to a synchronous motor driving a reciprocating compressor, The compressor is two-cylinder, horizontal, double-acting, and operates at 257 rpm. Line A is the envelope of the current wave. Difference B-C is current variation. Value B-C divided by the rated full load current is the percentage of current variation. (Used by permission Oscarson, G. L. E-M Synchronizer, 200 SYN 52, p. 11. Dresser-Rand Company.)...

See other pages where Current Full-load is mentioned: [Pg.108]    [Pg.2483]    [Pg.2484]    [Pg.2490]    [Pg.14]    [Pg.40]    [Pg.72]    [Pg.84]    [Pg.240]    [Pg.256]    [Pg.282]    [Pg.282]    [Pg.287]    [Pg.287]    [Pg.299]    [Pg.314]    [Pg.314]    [Pg.318]    [Pg.351]    [Pg.352]    [Pg.356]    [Pg.569]    [Pg.577]    [Pg.670]    [Pg.692]    [Pg.711]    [Pg.751]    [Pg.752]    [Pg.774]    [Pg.940]    [Pg.945]    [Pg.125]    [Pg.625]   
See also in sourсe #XX -- [ Pg.230 ]




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