Motor starting operations
The problems connected to motor starting operations are fundamentally linked to the type of motor which a determined motor operational torque “CM” offers, to the starting modality and to the connected load which has a determined load torque “C ”.
Calculation of motor starting time as first approximation (photo credit: c-var.com)The necessary starting torque “Ca” can be expressed as:
Ca = CM – CL
and shall be well calibrated to prevent it from being either too low, so as starting is not too long and heavy – which causes risks of temperature rise for the motor – or from being too high on the joints or on the operating machines.
A generic curve of the above mentioned quantities is shown in the Figure 1 below.
The concept of motor starting time “ta” can be associated to this concept of properly calibrated starting and can be evaluated making reference to concepts linked to the motion dynamics, but also by introducing simplifying hypotheses which allows, however, an evaluation with a good approximation.
It is possible to relate the acceleration torque, expressed as a difference between the motor operational torque and the load torque, to the moment of inertia of the motor “JM”, of the load “JL” and to the motor angular speed, to obtain the following formula:
where the expression of “dω” assumes the following form:
and it is obtained by differentiating the well known expression for the motor angular speed:
Through simple mathematical operations and applying the method of integral calculus, it is possible to make the unknown quantity “ta” explicit by the following expression:
To express the value of the acceleration torque, it is necessary to introduce some simplifications:
The first one consists in considering an average value for the motor operational torque to be expressed as:
CM = 0.45 x (Cs + Cmax)
where CS represents the inrush torque and Cmax the maximum torque;
The second one concerns the torque due to the load and which can be correct by applying the multiplying factor KL linked to the load typology as in Table 1 below.
Table 1 – Values of factor KL
| Type of comparable loads | ||||
| Load Coefficient | Lift | Fans | Piston Pumps | Flywheel |
| KL | 1 | 0.33 | 0.5 | 0 |
In order to better understand the significance of the coefficient KL we associate to the type of load indicated in the table the torque characterizing the starting phase of the load by means of the following assumptions:
- Lift = load torque constant during acceleration
- Fans = load torque with square law increase during acceleration
- Piston pumps = load torque with linear increase during acceleration
- Flywheel = zero load torque.
With these assumptions, the acceleration torque can be expressed as:
These hypotheses allow to obtain the motor starting time with the aid of the following formula
As an example, Table 2 below shows the values that these parameters can take for three-phase asynchronous motors of common use and typically present on the market. Obviously the parameters relevant to the load characterize each single application and must be known by the designer.
Table 2 – Typical values of some electrical and mechanical parameters of a three-phase asynchronous motor
Calculation of the starting time of a motor
Making reference to the data of the above table, here is an example of calculation of the starting time of a motor, according to the theoretical treatment previously developed.
| Three-phase asynchronous motor – 4 poles Frequency | 160 kW |
| Frequency | 50 Hz |
| Rated speed | 1500 rpm |
| Speed at full load | 1487 rpm |
| Moment of inertia of the motor | JM = 2.9 Kgm2 |
| Moment of inertia of the load | JL = 60 Kgm2 |
| Load torque | CL = 1600 Nm |
| Rated torque of the motor | CN = 1028 Nm |
| Inrush torque | Cs = 2467 Nm (Cs = 2.4 x 1028) |
| Max. torque | Cmax = 2981 Nm (Cmax = 2.9 x 1028) |
| Load with constant torque | KL = 1 |
Cacc = 0.45 · ( CS + Cmax) – KL· CL = 0.45 · (2467 + 2981) – (1 · 1600) = 851.6 Nm
from which
ta = (2 · π · 1500 · (2.9 + 60)) / 60 · 851.6 = 11.6 s
Load with quadratic rising torque KL = 0.33
Cacc = 0.45 · ( CS + Cmax) – KL · CL = 0.45 · (2467 + 2981) – (0.33 · 1600) = 1923.6 Nm
from which
ta = (2 · π · 1500 · (2.9 + 60)) / 60 · 1923.6 = 5.14 s
Reference // Three-phase asynchronous motors: generalities and proposals for the coordination of protective devices – ABB
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Edvard Csanyi
Hi, I'm an electrical engineer, programmer and founder of EEP - Electrical Engineering Portal. I worked twelve years at Schneider Electric in the position of technical support for low- and medium-voltage projects and the design of busbar trunking systems.I'm highly specialized in the design of LV/MV switchgear and low-voltage, high-power busbar trunking (<6300A) in substations, commercial buildings and industry facilities. I'm also a professional in AutoCAD programming.
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Hi
I have some questions?
An electro motor with a liquid starter of 3 MW is available.
In the manual of equipping, Start method is Directly, Can You Get Started with Liquid Starter?
The type of load is the fan
There is also no information about the load torque
How is the startup time calculating?
Thank you
are we take the starting current of motors in total load calculations in projects?
Good information and
I just want to know that how to increase the starting torque of a 130kw dc motor (series and compound).thanx
thank you
I am working in a chilar plant there v use vfds
One problem occurs while motor starting
It is starting with a abnormal sounds and vibration ager sum time it is getting to normal…..
Wht is the train for it
thank you
If load torque is higher than motor torque will current taken by motor become higher and CB trip.
Starting current of a motor will certainly not depend on difference between motor torque and load torque. But definitely the starting time will be lower, larger the difference which is also known as accelerating torque. Now, if the circuit breaker (providing only short circuit protection) faces current beyond it’s time-current characteristics, definitely it will do it’s job and trip the circuit. This will defilitely happen if there is a locked rotor situation, when starting current will flow for longer duration (and not a normal start duration of say 5 to 7 secs). For extended starting time, you would need to choose a heavy duty starter, which gives a longer time to the motor to start successfully. I have a excel spreadsheet to calculate the starting time of a motor, with some data required from the driven load + motor as explained above. It’s a graphical cum spreadsheet method, triving to do exactly what has been explained above.
Hi Cherry. Can you please share the excel spreadsheet to calculate the starting time of a motor, with some data required from the driven load + motor as explained above?
Hi Cherry. Can you please share the excel spreadsheet to calculate the starting time of a motor,thank you, I write to you from Peru
please give starting time of induction motors of various kW & speed as we use mostly them not syn motors.
Hi Sri….the starting time is a calculation of rotational dynamics, involving the driving torque (motor torque), driven torque (load torque) and the rotating inertia (combined inertia of motor & load). As stated above, the calculation is for a asynchronous motor (which means this motor DOES NOT rotate at synchronous speed i.e it is an induction motor). Hope the above example is clear with you.