# Calculating Motor Start Time

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1 Calculating Motor Start Time Timothy L. O'Hearn, P.E. Course Outline In this course, a method is described to approximate the motor starting time using the characteristics provided by the manufacturer. The motor torque versus speed curve, the load torque versus speed curve, and the current versus speed curve are utilized along with the rotating inertia to calculate the acceleration time for the motor. The acceleration curve is plotted on a time versus current curve along with the motor s thermal limits to select the proper over current protection. A discussion of the results and the methodology for the calculation is included. An example calculation is included for discussion purposes. I. Introduction to calculating motor starting time II. Method used to calculate motor starting time III.Example motor characteristics and typical manufacturer supplied curves IV. An example motor starting calculation and a discussion of the results. V. Course Summary. This course includes a multiple-choice quiz at the end, which is designed to enhance the understanding of the course materials. Learning Objective At the conclusion of this course, the student will: Have an understanding the method used to calculate an approximate value for the motor s starting time for full voltage across the line starting. Be capable of using the motor and load characteristics obtain from the equipment manufacturer to calculate the motor s start time. Have an understanding of the effect of a voltage dip on the motor s starting time. Understand how to use this method to assist in the selection of the proper over current protection for the motor for starting conditions. Course Content Large motor over current protection is normally set to trip prior to the locked-rotor safe stall time provided by the motor manufacturer but after the calculated motor start time. The locked-rotor safe stall time is determined by the manufacturer based on the heating of the rotor

2 parts for this locked-rotor condition where the motor continuously requires a large value of inrush current. During the starting condition, an induction motor draws high values of current (the motor is a constant impedance device during the starting condition) that are very close to the motor s locked rotor value and remains at this value for the time required to start the motor. This is the reason why the locked-rotor safe stall time is used as an allowable time limit for starting the motor across the line, full voltage. For more information on modeling devices in power systems the student is referred to reference 2. The capability to calculate motor start time for the largest motor on the electrical distribution system is important in order to evaluate the relative strength of the power system. Typically the motor manufacturer may calculate the motor starting time at a couple of selected values such as 90% or 85% and then plot the results on the time versus current curves for the supplied motor. System designers can use these starting times, but it is sometimes necessary to calculate the motor starting times using the results of a power system studies for the maximum voltage dip at minimum power system conditions. This method can then be used, using the maximum voltage dip, to avoid application problems, set motor protective devices and perform coordination studies with other protective devices on the system. The method described is an approximation that assumes that the inertia of the power source is significantly greater (at least ten times or more) than the inertia of the driven load. The following discussion will describe a common method for calculating the starting time for an induction motor starting full voltage, across the line. The motor s starting time is also referred to the acceleration time. This time is important because it determines the length of time the motor must withstand stator current that exceeds its continuous rating. This time is based upon Newton s second law (force is equal to mass times acceleration) applied to rotation. This law is: Accelerating Torque = Inertia * Angular Acceleration The acceleration time can be calculated using the following formula: t a = Wk 2 * Delta N / 308 * T a (1) Where t a acceleration time (seconds); Wk 2 total connected inertia (lb * ft 2 ); Delta N speed change during time t a (rev/min); torque); T a average accelerating torque (lb * ft) (average motor torque average load

3 In order to obtain these values it is necessary to request this information from the equipment manufacturer. This information is typically provided as torque versus speed curves for the induction motor and for the driven load. These curves are typically plotted using the equipment full load rating, and or rated speed. These curves must be converted to a common base or common reference scales in order to calculate the torque available for acceleration. The following motor torque versus speed curve is an example of the type of curves available from motor suppliers: The necessary values can be determined from the motor torque versus speed curve. Note that the torque versus speed curve for the compressor is included. The driven equipment s manufacturer typically provides the torque versus speed curve for the load. The starting condition for the driven load should be checked for initial conditions and if possible the required torque should be minimized to reduce the starting condition on the motor. The example motor characteristics are: Total connected Wk lb-ft 2 (Motor) lb-ft 2 (Compressor) = lb-ft 2

4 Full load torque = lb-ft Synchronous speed = 1800 rpm Full load speed = 1785 rpm The moment of inertia of the load and the motor rotor are combined numerically to give the total lb-ft 2 of the mass to be accelerated. The moment of inertia is the weight in pounds of the mass times the square of its radius of gyration in feet. These values are available from the manufacturers and are usually requested of bidders as part of their proposal data. Any difference in speed between the motor and the load, due to a pulley or gear ratio, requires the following modifications on torque and inertia. T 2 = (rpm 1 /rpm 2 ) (T 1 ) Wk 2 2 = ((rpm 1 ) 2 /(rpm 2 ) 2 ) Wk 2 1 Where: T 2 = effective torque at motor shaft speed T 1 = torque at load shaft Wk 2 2 = Effective Wk 2 of load reflected to motor shaft speed Wk 2 1 = Wk 2 at load shaft, including inertia of any reciprocating member rpm 1 = load shaft speed rpm 2 = corresponding motor shaft speed For calculation purposes the time interval is arbitrarily selected, the smaller time interval the more accurate the approximation becomes, lets choose the following: t 1 defined as the time to change speed from 0 to 10 % speed (0 to 180 rpm) t 2 defined as the time to change speed from 10% to 20% t 3 defined as the time to change speed from 20% to 30% t 4 defined as the time to change speed from 30% to 40% t 5 defined as the time to change speed from 40% to 50% t 6 defined as the time to change speed from 50% to 60%

5 t 7 defined as the time to change speed from 60% to 70% t 8 defined as the time to change speed from 70% to 80% t 9 defined as the time to change speed from 80% to 90% t full load defined as the time to change speed from 90% to 99.2% The next step is to establish the average acceleration torque for each time interval. The torque delivered by the motor varies as the square of the applied voltage based upon nameplate voltage being 100%. For example, if the voltage dipped to 90% the torque delivered to the load would be or 0.81 of the rated voltage. The torque delivered by the motor under these conditions should not be reduced to values that are less than the required load torque. The inrush current for the motor varies proportionately with the applied voltage. Other calculations will determine the expected voltage dip when starting a large motor. The average acceleration torque is then the average motor torque minus the average load torque during each interval. The average is determine by evaluating the characteristic curve during each interval: Ta for interval 1 is T motor average minus T load average ( ) * lb-ft = lb-ft t 1 = (Wk 2 ) (change in speed per unit) (rated rpm) / 308 (T a ) = lb-ft 2 (0.1) (1800) / 308 (18132 lb-ft) = 0.76 seconds Ta for interval 2 is T motor average minus T load average ( ) * lb-ft = lb-ft t 2 = 0.72 seconds Ta for interval 3 is T motor average minus T load average ( ) * lb-ft = lb-ft t 3 = 0.73 seconds Ta for interval 4 is T motor average minus T load average ( ) * lb-ft = lb-ft t 4 = 0.73 seconds Ta for interval 5 is T motor average minus T load average ( ) * lb-ft = lb-ft

6 t 5 = 0.74 seconds Ta for interval 6 is T motor average minus T load average ( ) * lb-ft = lb-ft t 6 = 0.7 seconds Ta for interval 7 is T motor average minus T load average ( ) * lb-ft = lb-ft t 7 = 0.67 seconds Ta for interval 8 is T motor average minus T load average ( ) * lb-ft = lb-ft t 8 = 0.61 seconds Ta for interval 9 is T motor average minus T load average ( ) * lb-ft = lb-ft t 9 = 0.51 seconds Ta for interval 10 is T motor average minus T load average ( ) * lb-ft = lb-ft t 10 = 0.28 seconds The results can the be determined as follows: At time t 1 = 0.76 seconds the motor is at 10% speed and at a current of 540% (which is provided on the speed versus current curve for the motor). At time t 1 + t 2 = = 1.48 seconds the motor is at 20% speed and at a current of 530%. This can be repeated and plotted on a time current curve to establish the acceleration time versus current curve for the motor. All of the time intervals can be added together to obtain the starting time: = 6.45 An example time current curve is included that plots the results for 100% and 90% voltage. In this example the acceleration time is 6.52 seconds at 100% voltage, which agrees with the results of the approximation that was previously calculated. In this example the acceleration time is 8.47 seconds for 90% voltage and is accomplished by reducing the available motor torque by the square of the voltage. This technique can be used once the voltage dip is determined in order to calculate the motor starting time.

7 The time current curve also provides the thermal limits for the motor during acceleration initially at operating temperature. The motor over current protection can now be selected to operate between the acceleration curve and the thermal limits to properly protect the motor during starting conditions. The over current protection should also be compared to other over current protective devices in the electrical system to insure proper coordination. The starting of high inertia loads can sometimes lead to acceleration times that exceed the motor s thermal limits. In this case the motor manufacturer should be contacted to verify that the motor is designed for this condition and that other required protective devices are properly applied. In addition it should be verified that proper coordination with other over current devices are maintained. Course Summary A method is described to approximate the motor starting time using the characteristics provided by the manufacturer. The motor torque versus speed curve, the load torque versus speed curve, and the current versus speed curve are utilized along with the rotating inertia to calculate the acceleration time for the motor. The acceleration curve is plotted on a time versus current curve along with the motor s thermal limits to select the proper over current protection.

8 References 1. NEMA MG IEEE Std Power System Analysis. 3. IEEE Std. 620 IEEE Guide for the Presentation of Thermal Limit Curves for Squirrel Cage Induction Machines. Once you finish studying the above course content, you need to take a quiz to obtain the PDH credits. DISCLAIMER: The materials contained in the online course are not intended as a representation or warranty on the part of PDHonline.org or any other person/organization named herein. The materials are for general information only. They are not a substitute for competent professional advice. Application of this information to a specific project should be reviewed by a registered professional engineer. Anyone making use of the information set forth herein does so at their own risk and assumes any and all resulting liability arising therefrom.

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