DC Motor Power Flow

A separately excited DC motor has two input power sources, one to the armature circuit and one the field circuit. Shunt and series motors only have one power source, the armature circuit terminals. The input power is the sum of the power flowing at the electrical terminals of the machine. The power to the armature terminal is:

\[ P_T=V_T I_T \]

while the power input to the field circuit is

\[ \begin{aligned} P_F & =V_F I_F \\ P_F & = I_F^2 R_F \end{aligned} \]

Note that all of the power flowing in the field circuit is losses, there is no elecromechanical power transfer from the field circuit.

Consider now the losses in the armature circuit. For shunt and separate excitation: the armature losses are:

\[ P_{RA}=I_A^2 R_A \]

and for a series motor:

\[ P_{RA}=I_A^2 (R_A+R_S) \]

Power Converted and Output Power

In all machines, the electromagnetic torque developed gives the power that is converted between the electrical energy and mechanical energy. The electromagnetic torque developed and power converted differ from the external measurable torque and power due to mechanical rotational losses. The mechanical rotational losses in a machines are the contact friction and air friction, which are commonly called "Friction and Windage" \(P_{FW}\)

\[ \begin{aligned} P_{conv}&=\tau \omega_m\\ P_{out}&=P_{conv}-P_{FW} \end{aligned} \]

If the mechanical losses are negligible, then the output power will equal the power converted.

Considering the power converted to the mechanical system in DC machines, subsituting the armature voltge and torque equations gives:

\[ \begin{aligned} P_{conv}&=\tau \omega_m\\ P_{conv} &=(k \phi I_A) (\frac{E_A}{k \phi}) \\ P_{conv} &= E_A I_A \end{aligned} \]

Applying this information to the armature circuit: (\(R\) in these equations can be substitued for \(R_A\) in shunt and seperately excited machines, \(R_A+R_S\) in series excited machines).

\[ \begin{aligned} V_T&=E_A +I_A R\\ V_T I_A &= E_A I_A + I_A^2 R \\ V_T I_A &=P_{conv} + P_{RA} \end{aligned} \]

Using the above equations, it is possible to calculate the armature current requried to provide a desired output power, for a given supply voltage

Efficiency

The efficiency of a DC Motor is given by

\[ \begin{aligned} \eta & =\frac{P_{out} } {P_{in} } \\ \eta & =\frac{P_{conv}-P_{FW}} {P_{conv}+P_{RA}+P_{F}} \end{aligned} \]

It is important to understand that while sometimes \(P_{in}\) is given by \(V_T I_A\) that only applies to series connected machines and permanent magnet excited DC machines.

Summary

Expressions for power flow in the armature circuit are considered and an expression for mechanical power in terms of electrical circuit quanitites is derived