Can DC Geared Motors Be Used as Generators?

[auroraseerthi]

Homework Statement

in a real time scenario, when we buy a dc geared motor.could we run it as generator ?
how much current would a dc geared motor produce when run as generator? could anyone please provide me the equation to find the generated voltage in case the said motor is bought with the following known parameters: operating voltage,current drawn,rated rpm,rated power.

Homework Equations

Faraday's Law of induction, written for an alternator with a DC (permanent magnet) two-pole stator, is

V = - N A (2 pi RPM/60) B

where N = number of turns on the armature (example N= 50)
A = area of coil on the armature (example A = 0.1 meter times 0.05 meter = 0.005 m2.
RPM = armature rotation speed (example 3600 RPM)
B = DC magnetic field (e.g., 0.5 tesla)
V= volts out (example V = 50 x .005 x 2 pi x 3600/60 x 0.5 = 47.1 volts @ 60 Hz)

reference:physicsforums.com

The Attempt at a Solution

but the said motor doesn't come with DC magnetic field(B),area of coil on armature(A),number turns on the armature(N).

 

[berkeman]

Homework Statement

in a real time scenario, when we buy a dc geared motor.could we run it as generator ?
how much current would a dc geared motor produce when run as generator? could anyone please provide me the equation to find the generated voltage in case the said motor is bought with the following known parameters: operating voltage,current drawn,rated rpm,rated power.

Homework Equations

Faraday's Law of induction, written for an alternator with a DC (permanent magnet) two-pole stator, is

V = - N A (2 pi RPM/60) B

where N = number of turns on the armature (example N= 50)
A = area of coil on the armature (example A = 0.1 meter times 0.05 meter = 0.005 m2.
RPM = armature rotation speed (example 3600 RPM)
B = DC magnetic field (e.g., 0.5 tesla)
V= volts out (example V = 50 x .005 x 2 pi x 3600/60 x 0.5 = 47.1 volts @ 60 Hz)

reference:physicsforums.com

The Attempt at a Solution

but the said motor doesn't come with DC magnetic field(B),area of coil on armature(A),number turns on the armature(N).

Welcome to the PF.

Can you say more about what you mean by a "DC Geared Motor"? Do you mean a brushed DC motor with a speed reduction gearbox on the output shaft of the motor? What is the ratio of the reduction? The higher the gear ratio, the harder it will be to turn the gearbox and motor in reverse...

 

[CWatters]

in a real time scenario...

Do you mean... "Can a DC motor switch between motor and generator in the same system"?

Possibly, but the gearbox might cause problems. For example a worm drive cannot be back driven. Likewise if the motor has a big reduction ratio then it might be very inefficient when used in generator mode (eg What Berkeman said.. hard to turn in reverse).

 

[auroraseerthi]

for example let us assume that,
i am using 24V 10 Amp PMDC motor to drive
and
i need to charge a battery which has 24V rating, now
i would need a suggestion for a generator probably less weight and low torque to produce enough voltage to charge the same battery.

and the question i asked before was...
if i have a motor(A.k.A generator) PMDC type with voltage and current reading such as 180V and 0.2A, with RPM of 1700 max...
what will be the output voltage and ampere for the same motor if i use it as a generator ? ( am talking about two different motors manipulated at two different instances in different ways )

 

[CWatters]

I'm a bit rusty but...

The PMDC motor will have a motor constant that relates the applied voltage to the RPM. You mention 180V and 1700 rpm so the motor constant will be around 1700/180 = 9.4 rpm per volt. (Aside: That's quite a low figure. Is the 1700 rpm figure after a reduction gearbox?)

If you want to use the same motor as a 24V generator you would need to spin it at roughly...

24 * 9.4 = 225 rpm

You can't calculate the current for a number of reasons. In practice you would need to generate more than 24V and use some sort of charge/current limiter/controller between the generator and the battery. You can't calculate the max charging current available because you don't know (or haven't told us?) the available power to turn the generator. Nor do you know (or haven't told us?) the losses in the motor and any gearbox when back driven.

 

[CWatters]

You can't calculate the current for a number of reasons. In practice you would need to generate more than 24V and use some sort of charge/current limiter/controller between the generator and the battery.

Consider what would happen if the generator was spinning at say 450 rpm. The no load voltage would be..

450/9.4 = 48V

If you connected that to a 24V battery you would in effect be connecting a 48V source to a 24V source. The current and the load on the generator would be very high (especially if the 24V battery can sink a lot of current without it's voltage rising).

If this were in a car you would be in a situation where the car is going at a speed equivalent to say 450rpm with severe regenerative braking (in effect) trying to make the wheels spin at only 225rpm. It might cause the wheels to slip?

The solution is to limit and control the current and hence the load on the generator and the resulting braking effect.

Once the vehicle speed falls to below the equivalent of 225rpm you can no longer charge the 24V battery and you loose regenerative braking.

I'm no expert but I believe it's quite challenging (but not impossible) to make regenerative braking work over a wide speed range (eg down to zero rpm).