08-17-2017, 12:40 AM
[attachment=14899]
FUSE RATING:
(a)For Motor- 125% of rated current
= 125% of 17A=21.25A=25A
(b)For Alternator- 125% of rated current
=125% of 4A= 5A
THEORY:
In a salient pole alternator, the reactance of magnetic circuit along is along its quad stator axis. The alternator is driven by auxiliary prime mover at a speed slightly less than the synchronous speed under these conditions. The armature current is when the armature current mmf is in line with the field poles. The reactance by the magnetic field current is minimum. The ratio of maximum voltage to minimum current gives the direct axis impedance and the ratio of minimum voltage to maximum current gives the armature axis impedance.
PRECAUTIONS:
1. The motor field rheostat should be kept in minimum.
2. The direction of the rotation due to prime mover and the alternator on the motor should be the same.
3. Initially all the switches are kept open.
PROCEDURE:
1. Note down the name plate details of motor and alternator.
2. Connections are made as per the circuit diagram.
3. Give the supply by closing the DPST switch.
4. Using the three point starter, start the motor to run at the synchronous speed by varying the motor field rheostat at the same time check whether the alternator field has been opened or not.
5. Apply 20% to 30% of the rated voltage to the armature of the alternator by adjusting the autotransformer.
6. To obtain the slip and the maximum oscillation of pointers the speed is reduced slightly lesser than the synchronous speed.
7. Maximum current, minimum current, maximum voltage and minimum voltage are noted.
8. Find out the direct and quadrature axis impedances.
PROCEDURE TO DRAW THE VECTOR DIAGRAM:
1. Draw the line OA that represents the rated voltage V.
2. Draw the line OB vector to represent the rated current I, which makes an angle (it may lag/lead/in phase) with the voltage.
3. Draw the line AC vector to represent IRa drop, which is parallel to OB vector.
4. Draw the perpendicular line CD to the line AC (IRa drop) that represents IXq drop.
5. Draw the line from the origin through the point D, which represents the no load voltage (Eo).
6. Draw the pole axis through origin, which should be perpendicular to vector OD.
7. Draw a perpendicular line to the pole axis from the same point E which should pass through the point B [where vector OE represents Direct Axis Current (Id) and Vector EB represents Quadrature Axis Current (Iq)].
8. Find out the reactive voltage drops IdXd and IqXq.
9. Draw a parallel line (ie perpendicular to Id) to OD vector from the point C, with the magnitude of the drop IdXd (Line CF).
10. Draw a parallel line (ie perpendicular to Iq) to OE vector from the point F, with the magnitude of the drop IqXq (Line FG).
11. Let the point at where the IqXq drop meets the OD line be G. here the vector OG represents the no load voltage (Eo).
12. Find out the voltage regulation by using the suitable formula.