ENERGY_PERFORMANCE_ASSESSMENT_FOR_EQUIPMENT_AND_UTILITY_SYSTEMS_(Chapter-5: ENERGY PERFORMANCE ASSESSMENT OF MOTORS AND VARIABLE SPEED DRIVES)

ENERGY_PERFORMANCE_ASSESSMENT_FOR_EQUIPMENT_AND UTILITY SYSTEMS

Chapter-5: ENERGY PERFORMANCE ASSESSMENT OF MOTORS AND VARIABLE SPEED DRIVES

Introduction

The two parameters of importance in a motor are efficiency and power factor. The efficiencies of induction motors remain almost constant between 50% to 100% loading (Refer figure 5.1). With motors designed to perform this function efficiently; the opportunity for savings with motors rests primarily in their selection and use. When a motor has a higher rating than that required by the equipment, motor operates at part load. In this state, the efficiency of the motor is reduced. Replacement of under loaded motors with smaller motors will allow a fully loaded smaller motor to operate at a higher efficiency. This arrangement is generally most economical for larger motors, and only when they are operating at less than one-third to one-half capacity, depending on their size.

Performance Terms and Definitions

Efficiency Testing

While input power measurements are fairly simple, measurement of output or losses need a laborious exercise with extensive testing facilities. The following are the testing standards widely used.Europe: IEC 60034-2, and the new IEC 61972

US: IEEE 112 - Method B

Japan: JEC 37

Even between these standards the difference in efficiency value is up to 3%. For simplicity nameplate efficiency rating may be used for calculations if the motor load is in the range of 50 -100 %.

Field Tests for Determining Efficiency

(Note: The following section is a repeat of material provided in the chapter-2 on Electrical Motors in Book-3.)

No Load Test :

The motor is run at rated voltage and frequency without any shaft load. Input power, current, frequency and voltage are noted. The no load P.F. is quite low and hence low PF watt meters are  required. From the input power, stator I2R losses under no load are subtracted to give the sum of Friction and Windage (F&W) and core losses. To separate core and F & W losses, test is repeated at variable voltages. It is worthwhile plotting no-load input kW versus Voltage; the intercept is F & W kW loss component.

F&W and core losses = No load power (watts) – (No load current)2 x Stator resistance

Stator and Rotor I2R Losses :

The stator winding resistance is directly measured by a bridge or volt amp method. The resistance must be corrected to the operating temperature. For modern motors, the operating temperature is likely to be in the range of 100°C to 120°C and necessary correction should be made. Correction to 75°C may be inaccurate. The correction factor is given as follows :

The rotor resistance can be determined from locked rotor test at reduced frequency, but rotor I2R losses are measured from measurement of rotor slip.

Rotor I2R losses = Slip x (Stator Input - Stator I2R Losses - Core Loss)

Accurate measurement of slip is possible by stroboscope or non-contact type tachometer. Slip also must be corrected to operating temperature.

Stray Load Losses :

These losses are difficult to measure with any accuracy. IEEE Standard 112 gives a complicated method, which is rarely used on shop floor. IS and IEC standards take a fixed value as 0.5 % of output. It must be remarked that actual value of stray losses is likely to be more. IEEE - 112 specifies values from 0.9 % to 1.8 %.

Points for Users :

It must be clear that accurate determination of efficiency is very difficult. The same motor tested by different methods and by same methods by different manufacturers can give a difference of 2 %.

Estimation of efficiency in the field can be summarized as follows:

a) Measure stator resistance and correct to operating temperature. From rated current value, I2R losses are calculated.

b) From rated speed and output, rotor I2R losses are calculated

c) From no load test, core and F & W losses are determined for stray loss

The method is illustrated by the following example :

Example :

Motor Specifications

Rated power = 34 kW/45 HP

Voltage = 415 Volt

Current = 57 Amps

Speed = 1475 rpm

Insulation class = F

Frame = LD 200 L

Connection = Delta

No load test Data

Voltage, V = 415 Volts

Current, I = 16.1 Amps

Frequency, F = 50 Hz

Stator phase

resistance at 30°C = 0.264 Ohms

No load power, Pnl = 1063.74 Watts

a) Calculate iron plus friction and windage losses

b) Calculate stator resistance at 120°C

c) Calculate stator copper losses at operating temperature of resistance at 120°C

d) Calculate full load slip(s) and rotor input assuming rotor losses are slip times rotor input.

e) Determine the motor input assuming that stray losses are 0.5 % of the motor rated power

f) Calculate motor full load efficiency and full load power factor

Comments :

a)The measurement of stray load losses is very difficult and not practical even on test beds.

b)The actual value of stray loss of motors up to 200 HP is likely to be 1 % to 3 % compared to 0.5 % assumed by standards.

c)The value of full load slip taken from the nameplate data is not accurate. Actual measurement under full load conditions will give better results.

d)The friction and windage losses really are part of the shaft output; however, in the above calculation, it is not added to the rated shaft output, before calculating the rotor input power. The error however is minor.

e)When a motor is rewound, there is a fair chance that the resistance per phase would increase due to winding material quality and the losses would be higher. It would be interesting to assess the effect of a nominal 10 % increase in resistance per phase.

Determining Motor Loading

1. By Input Power Measurements

First measure input power Pi with a hand held or in-line power meter

Pi=Three-phase power in kW

Note the rated kW and efficiency from the motor name plate

The figures of kW mentioned in the name plate is for output conditions. So corresponding input power at full-rated load

The percentage loading can now be calculated as follows

Example

The nameplate details of a motor are given as power = 15 kW, efficiency n = 0.9. Using a power meter

the actual three phase power drawn is found to be 8 kW. Find out the loading of the motor.

Input power at full-rated power in kW,  Pir = 15/0.9

                                                                     = 16.7kW

                                     Percentage loading = 8/16.7

                                                                    = 48%

2. By Line Current Measurements

The line current load estimation method is used when input power cannot be measured and only amperage measurements are possible. The amperage draw of a motor varies approximately linearly with respect to load, down to about 75% of full load. Below the 75% load point, power factor degrades and the amperage curve becomes increasingly non-linear. In the low load region, current measurements are not a useful indicator of load. However, this method may be used only as a preliminary method just for the purpose of identification of oversized motors.

3. Slip Method

In the absence of a power meter, the slip method can be used which requires a tachometer. This method

also does not give the exact loading on the motors.

Example: Slip Load Calculation

Given: Synchronous speed in rpm = 1500 at 50 HZ operating frequency.

(Synchronous speed = 120f/P) ; f: frequency, P: Number of poles

Nameplate full load speed = 1450

Measured speed in rpm = 1480

Nameplate rated power =7.5 kW

Determine actual output power.

From the above equation, actual output power would be 40% x 7.5 kW =3 kW

The speed/slip method of determining motor part-load is often favored due to its simplicity and safety

advantages. Most motors are constructed such that the shaft is accessible to a tachometer or a strobe

light.

The accuracy of the slip method, however, is limited. The largest uncertainty relates to the accuracy with which manufacturers report the nameplate full-load speed. Manufacturers generally round their  reported full-load speed values to some multiple of 5 rpm. While 5 rpm is but a small percent of the full-load speed and may be considered as insignificant, the slip method relies on the difference between full-load nameplate and synchronous speeds. Given a 40 rpm “correct” slip, a seemingly minor 5 rpm disparity causes a 12% change in calculated load.

Slip also varies inversely with respect to the motor terminal voltage squared. A voltage correction factor can, also, be inserted into the slip load equation. The voltage compensated load can be calculated as shown

Performance Evaluation of Rewound Motors

Ideally, a comparison should be made of the efficiency before and after a rewinding. A relatively simple procedure for evaluating rewind quality is to keep a log of no-load input current for each motor in the population. This figure increases with poor quality rewinds. A review of the rewind shop’s procedure should also provide some indication of the quality of work. When rewinding a motor, if smaller diameter wire is used, the resistance and the I2R losses will increase.

Format for Data Collection

The motor loading survey can be performed using the format given below:

The monitoring format for rewound motor is given below:

Application of Variable Speed Drives (VSD)

Although there are many methods of varying the speeds of the driven equipment such as hydraulic coupling, gear box, variable pulley etc., the most possible method is one of varying the motor speed itself by varying the frequency and voltage by a variable frequency drive.

Concept of Variable Frequency Drive

The speed of an induction motor is proportional to the frequency of the AC voltage applied to it, as well as the number of poles in the motor stator. This is expressed by the equation:

RPM = (f x 120)/p

Where fis the frequency in Hz, and p is the number of poles in any multiple of 2.

Therefore, if the frequency applied to the motor is changed, the motor speed changes in direct proportion to the frequency change. The control of frequency applied to the motor is the job given to the VSD. 

The VSD’s basic principle of operation is to convert the electrical system frequency and voltage to the frequency and voltage required to drive a motor at a speed other than its rated speed. The two most basic functions of a VSD are to provide power conversion from one frequency to another, and to enable control of the output frequency.

VSD Power Conversion

As illustrated by Figure 5.2, there are two basic components, a rectifier and an inverter, to accomplish power conversion.

The rectifier receives the 50-Hz AC voltage and converts it to direct current (DC) voltage. A DC bus inside the VSD functions as a “parking lot” for the DC voltage. The DC bus energizes the inverter, which converts it back to AC voltage again. The inverter can be controlled to produce an output frequency of the proper value for the desired motor shaft speed.

Factors for Successful Implementation of Variable Speed Drives

a) Load Type for Variable Frequency Drives

The main consideration is whether the variable frequency drive application requires a variable torque or constant torque drive. If the equipment being driven is centrifugal, such as a fan or pump, then a variable torque drive will be more appropriate. Energy savings are usually the primary motivation for installing variable torque drives for centrifugal applications. For example, a fan needs less torque when running at 50% speed than it does when running at full speed. Variable torque operation allows the motor to apply only the torque needed, which results in reduced energy consumption.

Conveyors, positive displacement pumps, punch presses, extruders, and other similar type applications require constant level of torque at all speeds. In which case, constant torque variable frequency drives would be more appropriate for the job. A constant torque drive should have an overload current capacity of 150% or more for one minute. Variable torque variable frequency drives need only an overload current capacity of 120% for one minute since centrifugal applications rarely exceed the rated current.

If tight process control is needed, then you may need to utilize a sensor less vector, or flux vectorvariable frequency drive, which allow a high level of accuracy in controlling speed, torque, and positioning.

b) Motor Information

The following motor information will be needed to select the proper variable frequency drive:

Full Load Amperage Rating. Using a motor’s horsepower is an inaccurate way to size variable frequency drives.

Speed Range. Generally, a motor should not be run at any speed less than 20% of its specified maximum speed allowed. If it is run at a speed less than this without auxiliary motor cooling, the motor will overheat. Auxiliary motor cooling should be used if the motor must be operated at very slow speeds.

Multiple Motors. To size a variable frequency drive that will control more than one motor, add together the full-load amp ratings of each of the motors. All motors controlled by a single drive must have an equal voltage rating.

c) Efficiency and Power Factor

The variable frequency drive should have an efficiency rating of 95% or better at full load. Variable frequency drives should also offer a true system power factor of 0.95 or better across the operational speed range, to save on demand charges, and to protect the equipment (especially motors).

d) Protection and Power Quality

Motor overload Protection for instantaneous trip and motor over current.

Additional Protection: Over and under voltage, over temperature, ground fault, control or microprocessor fault. These protective circuits should provide an orderly shutdown of the VFD, provide indication of the fault condition, and require a manual reset (except under voltage) before restart. Under voltage from a power loss shall be set to automatically restart after return to normal. The history of the previous three faults shall remain in memory for future review.

If a built-up system is required, there should also be externally-operated short circuit protection, door-interlocked fused disconnect and circuit breaker or motor circuit protector (MCP)

To determine if the equipment under consideration is the right choice for a variable speed drive:

The load patterns should be thoroughly studied before exercising the option of VSD. In effect the load should be of a varying nature to demand a VSD ( refer figure 5.3 & 5.4).

The first step is to identify the number of operating hours of the equipment at various load conditions. This can be done by using a Power analyzer with continuous data storage or by a simple energy meter with periodic reading being taken.

Information needed to Evaluate Energy Savings for Variable Speed Application

1. Method of flow control to which adjustable speed is compared:

a) output throttling (pump) or dampers (fan)

b) recirculation (pump) or unrestrained flow (fan)

c) adjustable-speed coupling (eddy current coupling)

d) inlet guide vanes or inlet dampers (fan only)

e) two-speed motor.

2. Pump or fan data:

a) head v's flow curve for every different type of liquid (pump) or gas (fan) that is handled

b)  Pump efficiency curves.

3. Process information:

a) specific gravity (for pumps) or specific density of products (for fans)

b)system resistance head/flow curve

c) equipment duty cycle, 1.e. flow levels and time duration.

4. Efficiency information on all relevant electrical system apparatus:

a) motors, constant and variable speed

b) variable speed drives

c) gears

d) transformers.

If we do not have precise information for all of the above, we can make reasonable assumptions for points 2 and 4.

Solved Example:

a) Ina 75 kW four pole induction motor operating at 49.8 Hz and rated for 415 V and 1440 RPM, the actual measured speed is 1470 RPM. Find out the percentage loading of the motor if the voltage applied is 428 V.

b) A6 pole, 415 volt, 3 ©, 50 Hz induction motor delivers 22 kW power at rotor shaft at a speed of 950 rpm with PF of 0.88. The total loss in the stator including core, copper and other losses is 2 kW. Calculate the following.

i) Slip

i1) Rotor Copper Loss

11) Total Input to motor

iv) Line current at 415 V and motor pf of 0.88

v) Motor operating efficiency

--------------------

Chapter 4

Chapter 6