# Calculation of motor capacity

Electric motor is the one of common device for rotating equipment and it’s useful for smooth operation and make our process more faster and efficiency.

But we must remember,we should sizing the motor capacity properly and must suitable with load,system and application for our process.

We can connect the electric motor with several types of mechanical load with gearbox,pulley,timing chain and etc.Proper calculation is must to avoid overload,bearing damage and motor burn due to wrong estimation about load.

## How to Measure motor capacity with gear head?

When a motor is coupled to the equipment, the speed is measured at output shaft section using a strobe light etc. In the case of a motor with a gear head, the speed is calculated from the following formula :-

### n = i x n1n : Motor speed (min–1) n1 : Speed of gear output shaft or pulley etc. attached to it i : Reduction ratio of gear head (e.g. i = 20 for 1/20)

When measuring the speed of a gear output shaft having a large reduction ratio, do not measure the number of revolutions per minute,

but measure the time taken for the gear output shaft to rotate 100 turns using a stopwatch after putting a mark on the shaft.

Then calculate the number of revolutions per minute from the measured time.

## Example of motor selection

Application : Driving of conveyor
Voltage : 100 V
Speed : 30 min–1
Working condition : Continuous
Frequency : 60 Hz

From this data,Select a motor that meet the requirement.

### 1) Set the suitable SPEED

The required speed is 30 min–1

The gear ratio that realizes a rated motor speed (60 Hz area) of 1500 min–1 is 1500/30 t = 50

Therefore use a gear ratio of 1/50.

### 2) Voltage calculation

Actual measurement of minimum starting voltage, minimum stable voltage and speed
Assume that the following are obtained as a result of actual measurement.
Voltage power supply = 100 V

Minimum starting voltage : 75 V
Minimum stable voltage : 55 V
Speed : 1700 min–1

### 3) Torque measurement

Measure the approximate load with a spring balance etc.

Assume that it is 2.65 N·m (375.27 oz-in).

It is for choose a right reduction gearbox.refer manual or catalog of gear head.

### 4) Set the speed torque

From speed-torque curve of 4-pole 25 W induction motor ( refer catalog from manufacture )

Example :
Ts : Starting torque Ts = 0.16 N·m (22.66 oz-in)
Tm : Stalling torque Tm = 0.25 N·m (35.4 oz-in)

The torque is proportional to the square of the voltage and the following values are obtained.

(Minimum starting torque)

0.16 x (75/100 )²

= 9 x 10 -² N·m (12.75 oz-in)
(Minimum required stalling torque)
0.25 x ( 55/100)²

= 7 x 10 -² N·m (9.91 oz-in)
(Torque at motor speed of 1700 min–1)
= 0.12 N·m (16.99 oz-in)

From the above, it can be seen that this application is a constant torque load and that the 4-pole 25 W induction motor still has a more than sufficientcapacity.

In addition, as is evident from the S-T curve of the attached S-T data, Ts and Tm of the 4-pole 15W induction motor are as follows:
Ts = 0.1 N·m (14.16 oz-in)
Tm = 0.15 N·m (21.24 oz-in)

Considering the voltage drop and variation when used for conveyors, Ts and Tm of the 4-pole 15 W induction motor at 90 V are assumed to be as follows:
Ts = 0.08 N·m (11.33 oz-in)
Tm = 0.12 N·m (16.99 oz-in)

When the voltage drop and variation or load variation is thought to be insignificant, the 4-pole 15 W induction motor and gear head ( refer manual ) can be used. When the voltage variation or load variation is significant, the 4-pole 25 W induction motor should be used.

### Lemau

Lemau is Owner of Electricneutron.com .Involved in Electrical & Automation Engineering since 2001. Passionate to shared all experience & knowledge about Electrical.