# Basic principles of marine DC motors 1. Electromagnetic torque equation

Figure 1 is a schematic diagram of a shunt motor. It can be seen from the figure that when the power is turned on, the current If passes through the armature, while the current Ia passes through the shunt winding in parallel with the armature and generates a magnetic flux. Assuming that the current direction of the upper semiconductor of the armature is ⊕, and the current direction of the lower semiconductor is ⊙, it can be seen from the figure that due to the interaction between the armature current and the magnetic field, the conductor on the armature is subjected to the action of the electromagnetic force F, and its direction can be determined by the left-hand rule.

T=CTΦIa     (1-1)

The electromagnetic torque formula states that the electromagnetic torque T of the motor is proportional to the magnetic flux Φ and the armature current Ia.

When the motor runs stably with a load, the electromagnetic torque T is equal to the braking torque Tc, while

The braking torque includes the load braking torque T2 and the hysteresis and eddy due to mechanical friction and iron core.

The no-load braking torque T0 caused by the flow and other reasons, that is, Tc=T2+T0. Therefore, the torque relationship of the motor in steady state operation is:

T=CTΦIa=T2+T0=Tc     (1-2)

It can be seen from the above formula that when the mechanical load of the motor increases, that is, when T2 increases, the electromagnetic torque T must increase accordingly. For a shunt motor, Φ can be considered constant, so the armature current Ia must increase (ie Ia=Tc/CTΦ), when the driving torque T and the braking torque Tc=T2+T0 are not equal, the motor speed n At this time, the torque relationship of the unit is as follows:

Tc=Ti=J(dΩ/dt)=J(2π/60)(dn/dt)     (1-3)

This is the dynamic torque balance relationship of the motor, where Ti is called the dynamic torque, J is the moment of inertia of the rotating part of the unit (the motor rotor plus the dragged machine), and Ω=2πn/60 is the angular velocity of the rotor. Obviously, if T>Tc, then the angular acceleration dΩ/dt>0, the speed n will increase, on the contrary, if T<Tc, then dΩ/dt<0, the speed n will decrease.

2. Motor voltage equation

The back EMF of the motor is generated due to electromagnetic induction. Its calculation formula is

Ea=CeΦn     (1-4)

According to the armature circuit of the motor, as shown in Figure 2, the voltage balance equation can be listed as follows:

U=EaIaRa     (1-5)

where Ra is the internal resistance of the armature circuit.

It can be seen from the above formula that a small part of the supply voltage U falls on the internal resistance (IaRa) of the armature circuit, and most of it is used to overcome the back EMF to maintain the armature current Ia. After the voltage balance equation is transformed, we can get:

Ia=(U-Ea)/Ra=(U-CenΦ)/Ra     (1-6)

It can be seen that when U is constant, the armature current Ia of the motor will change with the change of the back electromotive force Ea. The larger the Ea, the smaller the Ia. If the magnetic flux Φ remains the same (the shunt motor can be regarded as this situation), the armature current Ia will change with the change of the rotational speed n. The higher the n, the larger the E, and thus the smaller the Ia.

3. Power equation

The electric power input by the motor from the power source is P1=UI, a small part of it is consumed by the excitation loss Pcuf =UIf and the copper loss PcuaIa2Ra of the armature circuit, and the rest of the electric power is converted into mechanical power. Because the power is converted by the action of current and magnetic field, it is also called electromagnetic power, which is expressed by PM, that is,

P1-(Pcuf+Pcua)=PM     (1-7)

The mechanical power cannot be fully utilized yet, and the output power on the shaft only needs to be deducted from the core loss PFe and the mechanical loss Pi, that is,

PM-(Pi+PFe)=PM-P0=P2     (1-8)

In the formula, P0=Pi+PFe. It is called no-load loss.

It can be obtained from the above two formulas:

P1-(Pcuf+Pcua+Pi+PFe)=P1-∑P     (1-9)

∑P=Pcuf+Pcua+Pi+PFe is the total power loss of the motor.

The efficiency of the motor is

η=P2/P1×100%=P2/P2+∑P     (1-10)

The above three relationships are interrelated, that is, the torque relationship can be obtained by dividing both sides of the power relationship PM=P0+P2 by the angular velocity Ω of the motor, that is,