When the load on a motor, say DC motor, is fixed, what is the relationship between torque and the rotational speed of it?

Question

I have been working on a report about factors which can influence the speed of a DC motor. From research, it said that ,for a given voltage, torque and speed in a DC motor are said to be inversely proportional. Also, the motor reaches the maximum speed when there is no load. But I want to know what will happen to the speed when the load on the motor is fixed and the torque increases? Will the speed increase, because what I am thinking is the speed of a simple DC motor actually depends on the magnitude of the force acting on it.

Answer 1

what will happen to the speed when the load on the motor is fixed and the torque increases?

You need to define what you mean by "load" very precisely here to make sense of this question. If by load you mean a constant torque $\tau_L$ opposing the motor's rotation, then if the motor's torque $\tau_M$ output increases we have a nett unbalanced torque on the motor's shaft and angular speed will increase - theoretically without bound as then

$$\tau_M - \tau_L = J\,\mathrm{d}_t\omega\tag{1}$$

where $J$ is the mass moment of inertia of the driven system.

So you need to delve into the physics of both the motor and the load to answer these questions meaningfully.

When running at speed, a DC motor looks like a voltage source opposing the supply. This voltage $V$ is proportional to the motor's angular speed:

$$V = \kappa_V\,I_s\,\omega\tag{2}$$

and arises from the relative motion between the rotor and magnetic fields: it's a back EMF arising from Faraday's law. Here $I_s$ is the stator current: it sets the magnitude of the magnetic field the rotor rotates in.

From the supply's standpoint, the supply does work on the motor by inputting current into this voltage source. The motor's torque

$$\tau=\kappa_\tau\,I_s\,I_r\tag{3}$$

is proportional to the currents $I_s$ and $I_r$ that flow through the stator and rotor, respectively: this is simply a combination of Ampère's law and the Lorentz force law. The rate of work done on the back EMF is both $\tau\,\omega$ and $I_r\,V$, from the equation of these two and (2) and (3) we find that the proportionality constants in (2) and (3) are the same: $\kappa_V=\kappa_\tau=\kappa$.

We can instantly see why series connected DC motors are used in transport and are almost exclusively used in locomotives: even internal combustion locomotives almost always drive a DC generator supplying DC motors on the wheels. When the load is not moving, there's no back EMF and the current through both stator and rotor is limited only by their resistance. It's HUGE and the torques produced are the highest the motor can output. So you get high torque at low speeds - exactly when you need it to get heavy loads moving. As the motor gains speed, the back EMF increases and the current tapers off, as does the torque. You should be able to use (2) and (3) now to model this effect assuming a given source resistance.

Lastly, let's look at the load. You need a physical model: often motors work against dissipative friction that is, say, proportional to the angular speed. This is what would happen, for example, if you're using a motor to suck air or mix a cake with: the fluid's viscosity will lead to a load torque-speed relationship of something like:

$$\tau_L = \mu\,\omega$$

You can then use all of equations (1), (2), (3) and (4) together with a circuit model for how the motor is connected to find out what will happen to the speed as you adjust the parameters. (1) will actually define dynamics: a transient change in speed leading to a new steady state after parameters are changed.

Answer 2

If we ignore resistive losses, the electrical power input to the motor must equal the mechanical power generated by the motor. You state that the torque exerted by the motor increases, and the only way to increase the torque is to increase the voltage, which will increase the current and therefore increase the electrical power. This increase in electrical power in must be balanced by an increase in the mechanical power out.

The mechanical output power out is simply:

$$ P_{out} = \tau \omega $$

where $\tau$ is the load (the torque generated by the load) and $\omega$ is the angular velocity. If $P_{out}$ increases then $\tau \omega$ must increase. Since you have stated that $\tau$ is fixed that means $\omega$ must increase.

Answer 3

In a DC motor, the torque is proportional to current through the motor, and RPM is proportional to the voltage across the motor. from a dynamics modeling standpoint, a DC motor is referred to as a gyrator element because it relates a input flow variable (current) to an output effort variable (torque).