i'm in very much confusion :\ teach me masters!!!!

[thothset]

i want use a motor in a circuit. actually here motor is acts as a load. i just want to make a dummy load in order to replacement of motor.


Technical parameters:
Power supply voltage: 27 V DC V AC;
Rated power: 300 W;
Rated speed: 3400 rev / min;
Operation mode: continuous, 1.5 hours, by immersing the drive end of the body part by the water;
Connection scheme: two-wire;
my question is..
1. how to design a dummy load
2. what type of load should i use
3.i asked my friend he told me use a 1.1 ohms,23.5A,600W resistive load. how can i put 600W load but in specification motor rating is 300W. how??????

TEACH ME MASTERS

 

[Harald Kapp]

Power is P=V²/R.
You have V= 27 V and P = 300 W.
Now calculate R ...

 

[thothset]

Power is P=V²/R.
You have V= 27 V and P = 300 W.
Now calculate R ...

Thank you very much for your reply,

sorry for asking your again,
i found it on internet for motor specification (given below). my friend told me use R-load of 1.1 ohm,23.5Amps,600W. but how wattage becomes 600 Watts? is the specification found on internet is Nominal power or maximum power. what is full load current, starting current of this motor? how to find it?


DC motors, the mixed excitation are designed to drive the water pump.

Technical parameters:
Power supply voltage: 27 V DC V AC;
Rated power: 300 W;
Rated speed: 3400 rev / min;
Operation mode: continuous, 1.5 hours, by immersing the drive end of the body part by the water;
Connection scheme: two-wire;
Design: waterproof;
Operating position: vertical output shaft end down;
Rated torque: 0.83 N · m (0,058 kgf · m);
Direction of rotation: right;
Weight: not more than 9.4 kg;
Overall dimensions, mm (diam.) 112h164.

 

[(*steve*)]

Having a dummy load rated for a higher power than it will be called on to dissipate means you won't be running it right on the edge of failure.

If you need to reliably transport 10 tons of equipment, would you prefer to use a truck rated for a maximum load of 10 tons, or would you prefer one rated to carry 20 tons (consider that your load might only be "approximately 10 tons").

 

[Harald Kapp]

Having the resistor rated at >300 W is o.k., as Steve explained.
But 1.1 Ω on 27 V doesn't compute to 300 W. Use the equation in my post #2.

 

[thothset]

Having a dummy load rated for a higher power than it will be called on to dissipate means you won't be running it right on the edge of failure.

If you need to reliably transport 10 tons of equipment, would you prefer to use a truck rated for a maximum load of 10 tons, or would you prefer one rated to carry 20 tons (consider that your load might only be "approximately 10 tons").

but why double the power,current and resistance value for load. why exactly 600Watts? why not 350 or 400 Watts?

 

[(*steve*)]

Wide margin of safety? Relatively little difference in cost? Less risk of burning yourself? Less need to be perfect with your heatsinking? Allows higher ambient temperature? More rugged?

Probably one or more of them.

Doubling is a pretty common ballpark when you rate resistors. It helps make sure you don't have to worry to much about doing any exact calculations. If you get something rated at 500W, the margin is probably enough that you don't have to care. Likewise if some miscalculation results in higher dissipation the resistor is less likely to be immediately damaged.

You don't want an "oops" to immediately become a "#$@!"

 

[(*steve*)]

2.43Ω for 300W at 27V.

1.1Ω would be greater than 650W (and if recommend 1000W to 1200W rating of the resistor in question.

However, a motor draws more than the normal peak current on startup, so a simple resistor doesn't represent a load which is a lot like a motor (it also doesn't do the nasty inductive things a motor does)

 

[thothset]

2.43Ω for 300W at 27V.

1.1Ω would be greater than 650W (and if recommend 1000W to 1200W rating of the resistor in question.

However, a motor draws more than the normal peak current on startup, so a simple resistor doesn't represent a load which is a lot like a motor (it also doesn't do the nasty inductive things a motor does)

thanks for reply,
then please tell me an alternative solution

 

[Harald Kapp]

A resistor is the nearest simple solution.
A real motor is another possibility.
Simulating a motor by electronic means is possible but will require a much more elaborate setup.

Your alternatives depend on what specifically you want to simulate by the dummy load.

 

[thothset]

A resistor is the nearest simple solution.
A real motor is another possibility.
Simulating a motor by electronic means is possible but will require a much more elaborate setup.

Your alternatives depend on what specifically you want to simulate by the dummy load.

simply to say here motor load is a water pump. i am making a switching unit to control the on& off operation of pump using a relay. here i want to check the control unit not the pump. so i just need a dummy load instead of pump/motor. objective is to check whether the load is consuming from the source or not. depending upon the max load current taken by pump (max load current) i can design a dummy load from your idea. but how to calculate max load current from given specification

 

[duke37]

The motor is specified as 27 V DC V AC
I do not understand what this is. Is the motor DC or AC or both.
The speed is quoted as 3400 rpm. This does not match the frequency of the Indian supply (50Hz). Much more likely to be a motor made for the 60Hz USA supply.

It could be a universal motor, with carbon brushes, which will have a speed and power consumption dependant on the load. In any case, it will have an inductance which will be arduous to turn on and off. You cannot simulate this with a resistor.

 

[Harald Kapp]

but how to calculate max load current from given specification

I = P/V

objective is to check whether the load is consuming from the source or not

Why do you think you need a dummy at full load? Attach a lamp suitable for 27 V operation or something similar. This will allow you to check whether the relay operates on/off as intended or not.

 

[thothset]

this is the motor spec.
MU431 Specifications:
Power supply voltage: 24V
Rated power: 400W,
Rotational Speed: 5000 rev / min,..
Current: 32A,
load torque, Rated .: 0,78N th,
Brevity starting torque: 4.5
Dimensions - 143mm -80,5mm-80,5mm,
Weight, no more than 2,2kg.
Air motors are available in various types. MU series motors are designed for control plane mechanisms are reversible motors series excitation.


here voltage is 24V,current is 32 A, when i calculate power by using this formula-> P=V*I
i get 768 Watts. but they mention is 400W how they calculate?


http://vertex-rd.ru/catalog/9257/9458/

 

[duke37]

I do not know any russian.
The output power i calculate as 408W which matches the specification.
5000/60 *2*pi *0.78.
The motor needs more input power since it is not 100% efficient.
The starting torque and so current will be five times this.
I think that there is no way you can simulate a loaded motor. Use a real one loaded as it is normally used.

 

[thothset]

I do not know any russian.
The output power i calculate as 408W which matches the specification.
5000/60 *2*pi *0.78.
The motor needs more input power since it is not 100% efficient.
The starting torque and so current will be five times this.
I think that there is no way you can simulate a loaded motor. Use a real one loaded as it is normally used.

Thanks for the reply. cost of that motor is too high than we expected. because i'm testing a control unit which controls the ON & OFF operation of the motor. when i give 27V from power source to motor via EM relay'c coil point, and across the NO point Motor is connected. if the coil energize it closes the NO Point to NC hence motor gets started. is there is any alternative way to replace that motor with a R load or L load.
Thanks in advance.

 

[duke37]

It is easy to test the opersation of the relay but switching a motor needs the contacts to supply the current and to switch off when necessary.
The motor will be loaded and will have inertia so the start current will be high.
Turn off will generate a spark across the contacts due to the inductance.
To make a simulation circuit will be immensly complicated and much data on the motor and load will be necessary.

A simulation may be made and may work for a little while but may fail after a number of cycles.