Output equation of DC machine

The output equation of a dc machine relates the power developed in the armature $P_a$ to the main dimensions ($D$, $L$), the specific loadings ($B_{av}$, $ac$) and the speed $n$.

Symbols

$P$ = poles, $\Phi$ = flux per pole, $Z$ = armature conductors, $a$ = parallel paths, $n$ = speed (rps), $I_a$ = armature current, $I_z = I_a/a$ = conductor current, $E$ = induced emf.

Derivation

Power developed in the armature:

$$P_a = E I_a \times 10^{-3}\ \text{kW}$$

Emf (with $n$ in rps):

$$E = \frac{\Phi Z n P}{a}$$

So

$$P_a = \frac{\Phi ZnP}{a} \times I_a \times 10^{-3} = (P\Phi)(I_z Z)\, n \times 10^{-3}$$

Total flux and specific magnetic loading:

$$B_{av} = \frac{P\Phi}{\pi DL} \Rightarrow P\Phi = B_{av}\pi DL$$

Total ampere-conductors and specific electric loading:

$$ac = \frac{I_zZ}{\pi D} \Rightarrow I_zZ = ac\,\pi D$$

Substituting:

$$\begin{aligned} P_a &= B_{av}\pi DL \times ac\,\pi D \times n \times 10^{-3} \\ &= \pi^2 B_{av}\, ac\, D^2 L n \times 10^{-3} \\ &= C_0 D^2 L n \end{aligned}$$

where the output coefficient is

$$C_0 = \pi^2 B_{av}\, ac \times 10^{-3}$$

Relation between $P_a$ and rated output $P$

Assuming armature copper loss is about one-third of total losses (Sawhney):

$$\text{Generator: } P_a = P\,\frac{1+2\eta}{3\eta}, \qquad \text{Motor: } P_a = P\,\frac{2+\eta}{3\eta}$$

For small machines $P_a$ may also be taken as $E I_a$ directly when $E$ and $I_a$ are known.

Use

$$D^2L = \frac{P_a}{C_0 n}$$

$D$ and $L$ are then separated using the pole proportions (square pole face: $L = \psi\tau = \psi\pi D/P$) and the peripheral speed limit (about 30 m/s).

Example: $B_{av} = 0.5$ Wb/m², $ac = 22000$ A/m gives $C_0 = \pi^2 \times 0.5 \times 22 = 108.6$.