ELEN 3441 – Fundamentals of Power Engineering Lab # 11 Spring 2008 Lab 11: Capacitor start motor and Capacitor run motor.
Objective: to examine the construction of capacitor-start and capacitor-run motors; to determine
their running and starting characteristics and compare them with each other and with the
characteristics of split-phase motor. Equipment: Power Supply, DAI, Capacitor start motor (8251), Capacitor run motor (8253),
Electrodynamometer (8960), timing belt.
Theory:
When the split-phase rotating field was described, it was pointed out that the phase difference
between start and run winding currents falls far short of 90 degrees. The starting torque developed in
a motor that uses a split-phase stator also falls far short of the maximum that can be attained at an
ideal 90 degree phase difference.
A phase shift closer to the ideal 90 degrees is possible through the capacitor-start system for
creating a rotating stator field. This system, a modification of the split-phase system, uses a low
reactance capacitor placed in series with the start winding of the stator to provide a phase shift of
approximately 90 degrees for the start current. This results in greatly improved starting torque over
the standard split-phase system. Capacitor start motors have the same running characteristics as their
splitphase counterparts.
The capacitor and the start winding are disconnected by a centrifugal switch, just as in the case of
the standard split-phase motor. Reversing the direction of rotation of a capacitor start motor is the
same as in the case of the split-phase motor; that is, reverse the connections to the start or to the
running winding leads.
Single-phase motors are all rather noisy because they vibrate at 120Hz when operated on a 60Hz
power line. Various attempts to reduce this noise – such as resilient rubber mounting – are never
totally effective in eliminating this vibration, particularly when the motor is directly coupled to a
large resonant-prone fan.
The capacitor run motor is very useful in this type of application, because the motor can be
designed to have low vibration under full-load. The capacitor serves to shift the phase on one of the
windings so that the voltage across the winding is at 90° from the other winding, thus making the
capacitor run motor a truly two-phase machine at its rated load. Since the capacitor remains in the
circuit at all times, no centrifugal switch is required. When running at no-load, the motor is always
noisier than at full-load, because only under full load it runs as a true two-phase machine. If the
proper value of capacitance is chosen, the currents through each of the two equal stator windings
(under full-load) can be made such that the power factor is close to 100%. However, the starting
torque is rather low and the capacitor run motor is not recommended for severe starting conditions.
The power output (in horsepower) of the motor delivered to the load is defined as follows:
Pout ,hp =
1.4 ⋅ ωrpm ⋅ TNm
10 000
(11-1)
where ωrpm is the motor speed in revolutions per minute, TNm is its torque in Newton-meters.
Page | 1 ELEN 3441 – Fundamentals of Power Engineering Lab # 11 Spring 2008 Keep in mind that one horsepower equals approximately to 746 W. The reactive power [var] can be
computed as:
Q = S 2 − P2
(11-2)
where S is the apparent power [VA], P is the real power [W] consumed by the motor. The
efficiency of the motor is:
efficiency =
Pout ,W
P
⋅100 %
(11-3)
where Pout,W is the output power delivered to the load in Watts. The motor losses, therefore, are
estimated as:
Losses = P − Pout ,W
(11-4)
Experiment:
1) Notice that the module used for the capacitor start motor was previously used to study the
split-phase induction motor. Therefore, the design of the motor should be familiar. Construct
the circuit as shown in Figure 11-1.
Figure 11-1
Couple the dynamometer to the motor with a timing belt. Connect the dynamometer to the
fixed low voltage AC source by the grey cable. Using thin red wires, connect the “torque”
and “speed” outputs of the dynamometer to the “T” and “N” terminals of the DAI; connect
“ground” terminals of dynamometer and DAI. Set the “MODE” switch of the dynamometer
to the “DYN” position and the dynamometer load control switch to the “MAN”. Set the
dynamometer control knob at its utmost counter-clockwise position to provide the minimum
starting torque for the capacitor start motor.
2) Turn ON the PS and adjust for 120 V AC. Measure and record to a Data table the values for
the line voltage E1, current I1, apparent power S1 [VA], real power P1 [W], power factor
measured for E1 and I1, load torque, and the motor speed for the following values of the
Page | 2 ELEN 3441 – Fundamentals of Power Engineering Lab # 11 Spring 2008 output torque: 0, 0.2, 0.4, 0.6, 0.8, 1.0, and 1.2 Nm. Note: do not keep your motor running
under the load of 1.2 longer than is needed to take measurements. Disassemble your circuit.
3) Replace the Capacitor start motor by the Capacitor run motor module. Examine the
construction of the Capacitor run motor module (Figure 11-2).
Figure 11-2
Identify the two stator windings appearing to be identical. Note that there are four stator
poles. Observe the capacitor on the rear of the motor. Construct the circuit as shown in
Figure 11-3.
Page | 3 ELEN 3441 – Fundamentals of Power Engineering Lab # 11 Spring 2008 Figure 11-3
Couple the dynamometer to the motor with a timing belt. Make all the necessary
dynamometer’s connections and commutations as described in Part 1. Set the dynamometer
control knob at its utmost counter-clockwise position to provide the minimum starting torque
for the capacitor start motor.
4) Turn ON the PS and adjust for 120 V AC. Measure and record to a Data table the values for
the line voltage E1, current I1, apparent power S1 [VA], real power P1 [W], power factor
measured for E1 and I1, load torque, and the motor speed as indicated by the tachometer for
the following values of the output torque: 0, 0.2, 0.4, 0.6, 0.8, 1.0, and 1.2 Nm. Note: do not
keep your motor running under the load of 1.2 Nm. longer than is needed to take
measurements. Disassemble your circuit.
In your report:
1) Using Matlab and the data recorded in Part 2 for the Capacitor start motor, plot the
dependence of the real power on the load torque. On separate axes, plot the motor speed and
power factor as functions of the motor load. 2) For the data collected in Part 2 for the full-load operation (1 Nm) of the Capacitor start
motor, compute and report the power delivered by the motor, its efficiency, and the motor
losses. 3) Using Matlab and the data recorded in Part 4 for the Capacitor run motor, plot the
dependence of the real power on the load torque. On separate axes, plot the motor speed and
power factor as functions of the motor load. 4) For the data collected in Part 4 for the full-load operation (1 Nm) of the Capacitor run motor,
compute and report the power delivered by the motor, its efficiency, and the motor losses.
5) On new axes, plot the dependence of the power factor on the load (Nm) for both the
capacitor start and the capacitor run motors. Make these two curves distinguishable.
Compare the efficiency and the motor losses between the split-phase induction motor, the
capacitor start motor and the capacitor run motor. What are your conclusions? Which
motor’s (out of these three) speed is less affected by changes in its load?
Page | 4