• awaiting
In summary, if you have another motor - you can use as a generator and connect a resistor as load. This allows you to change the load on the motor.
awaiting
Hi everyone,

I have implemented PI control of speed of a PMDC motor under no load condition. I now intend to implement the same with a load added to it.
I want to compare the results with simulation as well. Is there any way of adding a known load to a DC motor. In my set up the dc motor is mounted on a metallic box.

Thanks,

If you have another motor - you can use as a generator and connect a resistor as load. Then as you change resistances you can change the load on the motor.

If you have another motor - you can use as a generator and connect a resistor as load. Then as you change resistances you can change the load on the motor.
Good idea if he knows well enough characteristics of another motor

I was just thinking to use the same type motor - by monitoring the input and output electrical power - you can then "see" the characteristics of what is between the measurements and divide by 2 - roughly speaking a poor man's dyno.

I was just thinking to use the same type motor - by monitoring the input and output electrical power - you can then "see" the characteristics of what is between the measurements and divide by 2 - roughly speaking a poor man's dyno.
"Roughly" is a good word. If you have two identical machines operating like this, total loses (Pin-Pout) won't be shared in proportion 1:1 between them.To illustrate suppose measured input power is Pin=10 kW, and let the efficiency of the first machine η1=0.9. Loss of the machine is
P1=(1-η1)Pin=1000 W, and power throughput P'=η1⋅Pin=9 kW. Suppose now that efficency of the second machine is also η2=0.9. Then, the loss of the second machine is P2=(1-η2)⋅P'= 900 W, and Pout2⋅P'=8.1 kW.Thus, one makes about 5% error assuming that loss of the motor 1900/2 = 950 W. The error increases in the case of small power machines or/and if the operating point is far from nominal.

Hmmm.. somehow I think that 5% error for the OP would be pretty good... esp if he does a few boundary cases : no load on the motor and not linked to the load Generator, not linked to the Generator and the motor in stall (full torque) for a second, linked to the load Generator w/no electrical load - etc...lastly he can swap the two motors and compare the datapoints. I'll bet with some "good practices" he can determine the characteristics to less than 1%. In the real world - that is actually very good.

awaiting said:
Hi everyone,

I have implemented PI control of speed of a PMDC motor under no load condition. I now intend to implement the same with a load added to it.
I want to compare the results with simulation as well. Is there any way of adding a known load to a DC motor. In my set up the dc motor is mounted on a metallic box.

Thanks,

What is the watt rating of your motor?

Hmmm.. somehow I think that 5% error for the OP would be pretty good... esp if he does a few boundary cases : no load on the motor and not linked to the load Generator, not linked to the Generator and the motor in stall (full torque) for a second, linked to the load Generator w/no electrical load - etc...lastly he can swap the two motors and compare the datapoints. I'll bet with some "good practices" he can determine the characteristics to less than 1%. In the real world - that is actually very good.
In the real world in many cases 1% accuracy is acceptable even for methods with professional dyno equipment.
Simple Pout/Pin = ηtot = η1⋅η2 depends on assumption η12.
This may be preceise enough for a bulky machine operating near rated load, but for "toys" with Pnom<1 kW operating far from their nominal point, 5% preceison would be quite success.

Z85 -- I think we are on the same page -- but then in my "real world" there are kind of 2 real worlds... academic and industry... in an academic (lab) setting 1% is a good goal for real systems, in industry 5%... the challenge I face, that drives me crazy is academics (without much real lab experience ~1%) come into industry so used to "perfect" simulations and can not accept the 5% of industry ( paper engineers)...so to me this case is actually a very good exercise. However the question needs to asked at the outset - what are you trying to learn or achieve, including how accurate do you want to be.

## 1. How do I add a known load to a DC motor?

The first step in adding a known load to a DC motor is to determine the type of load you will be adding. This could be a mechanical load, electrical load, or a combination of both. Once you have determined the type of load, you will need to select the appropriate load device and connect it to the motor.

## 2. What are the benefits of adding a known load to a DC motor?

Adding a known load to a DC motor allows you to control and measure the motor's performance, such as its speed and torque. This can be useful for testing and optimizing the motor's efficiency, as well as for safety purposes.

## 3. How do I calculate the load to add to a DC motor?

The load to add to a DC motor can be calculated using the formula: Load = Force x Distance. You will need to know the required force and the distance at which the load will be applied. Additionally, you will need to consider the motor's specifications and capabilities to ensure the load is within its limits.

Not all DC motors are designed to handle added loads. It is important to refer to the manufacturer's specifications and guidelines to determine if your motor can handle the desired load. If you are unsure, it is best to consult a professional or use a motor with a higher output rating.

## 5. What precautions should I take when adding a known load to a DC motor?

When adding a known load to a DC motor, it is important to make sure the load is securely attached and will not come loose during operation. Additionally, you should monitor the motor's performance and make any necessary adjustments to ensure the load is not causing excessive strain on the motor. It is also important to follow all safety precautions and guidelines provided by the manufacturer.

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