Occasionally we get asked to size power factor correction capacitors to improve the power factor of a single motor. Usually the requested improved power factor level is 90 or 95%. The necessary calculations to get the proper capacitor KVAR (Kilovolt Ampere Reactive) value are straightforward, but since we don’t do it often it is nice to have the method in writing.

PROCEDURE

The first thing needed is the full load power factor and efficiency information for the motor. On Baldor motors this can be found in either the 502 data section or the Baldor CD-ROM. Next, since most power factor tables are worked in terms of Kilowatts, it is necessary to convert the motor output rating into Kilowatts. The procedure for doing this is to take the motor HP multiplied by the constant for KW per HP (0.746). This will give Output KW. Then it is necessary to divide this by the efficiency of the motor (as a decimal) to get the Input KW at full load. Next, refer to power factor correction Table I going in from the left with the existing power factor and coming down from the top with the desired power factor. Where they intersect find the multiplier needed.

Next, multiply the motor Input Kilowatts by the appropriate multiplier from Table 1 to get the required KVAR of power factor correction. This value would be rounded out to match commercially available power factor correction capacitor ratings shown in Table 2.

TABLE 1
ORIGINAL FACTOR %	DESIRED POWER FACTOR %
	85	90	95
60	0.713	0.849	1.004
62	0.646	0.782	0.937
64	0.581	0.717	0.872
66	0.518	0.654	0.809
68	0.458	0.594	0.749
70	0.400	0.536	0.691
72	0.344	0.480	0.635
74	0.289	0.425	0.580
76	0.235	0.371	0.526
77	0.209	0.345	0.500
78	0.182	0.318	0.473
79	0.156	0.292	0.447
80	0.130	0.266	0.421
81	0.104	0.240	0.395
82	0.078	0.214	0.369
83	0.052	0.188	0.343
84	0.026	0.162	0.317
85	0.000	0.136	0.291
86		0.109	0.264
87		0.083	0.238
88		0.056	0.211
89		0.028	0.183
90		0.000	0.155
91			0.127
92			0.097
93			0.066
94			0.034
95			0.000

EXAMPLE:

To illustrate the procedure an example is worked as follows:

What is the KVAR of power factor correction capacitors needed to improve the power factor of a catalog number M2555T, 100 HP motor, to 95% at full load?

Step 1: Look up the existing power factor and efficiency
Efficiency = 94.1%
Power Factor = 85%

Step 2: Convert the HP to Kilowatts output. 100 HP x 0.746 = 74.6 KW

Step 3: Convert Kilowatts output to Kilowatts input by dividing by the full load efficiency.
(74.6)/(.941) = 79.3 KW Input

Step 4: Look in Table 1 to find the multiplier to achieve the desired 95% corrected power factor.
The multiplier is 0.291.

Step 5: Multiply Input KW by this multiplier.
79.3 x 0.291 = 23.1 KVAR
This gives the required Capacitor KVAR.

Step 6: Select closest value from Table 2.
22.5 KVAR

The voltage of the capacitor would also have to be specified. In this case it would be 480 volts.

CURRENT CORRECTION

In many cases when a single motor is being corrected, the capacitors are connected between the motor starter and the motor at the motor terminals as shown in Figure 1. With this being the case, the effect of the correction is to reduce the current flowing through the starter and overload relay. Since the overload heaters are selected (or adjusted) on the basis of the motor full load current, this means that the overloads will not correctly protect the motor unless the ampacity is reduced to reflect the reduced current now flowing as a result of the power factor improvement.

The motor itself will draw the same number of amps at full load as it would without the Power Factor Correction. However, the power factor correction capacitors will be supplying a portion of the current and the balance will be coming through the starter from the power line.

The new value of current passing through the overloads is given by the following formula:

Current_new = Motor Full Load (Nameplate) Amps X (Power Factor Original/Power Factor Corrected)

For example, in the case of the 100HP motor in the example, the heater size, which would normally be selected from the motor nameplate current at 118 amps would have to be adjusted as follows:

Current_new = 118 X (.85/.95) = 118 X .895 = 105.6 or approximately 106 amps

SUMMARY

A few words of caution might be appropriate. Usually it is desirable to “under correct” rather than “over correct”. If the capacitors chosen are too large there can be a number of problems, including high transient torques and overvoltage. Thus it is usually desirable not to attempt to improve power factor beyond 95%. It also usually becomes uneconomical to attempt improvements beyond 95%.

Please note: This type of power factor improvement should not be used in any situation where the motor is being controlled by a solid state device such as a soft start control or a variable frequency drive.

For more information on power factor improvement request Bulletin PF-2000F “Power Factor Correction — A Guide for the Plant Engineer” available from

Commonwealth Sprague Capacitor Inc.
Brown Street
North Adams, MA 01247