Output equation and separation of D, L

Output equation

Let $D$ = bore diameter, $L$ = core length, $P$ = poles, $n_s$ = synchronous speed (rps), $T_{ph}$ = turns per phase.

kVA input:

$$Q = 3E_{ph}I_{ph}\times10^{-3}, \qquad E_{ph} = 4.44 f\Phi T_{ph}K_w, \qquad f = \frac{Pn_s}{2}$$

Specific magnetic loading:

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

Specific electric loading (6$T_{ph}$ conductors):

$$ac = \frac{6I_{ph}T_{ph}}{\pi D} \Rightarrow I_{ph}T_{ph} = \frac{ac\,\pi D}{6}$$

Substituting:

$$\begin{aligned} Q &= 3 \times 4.44 \times \frac{Pn_s}{2} \times \frac{B_{av}\pi DL}{P} \times K_w \times \frac{ac\,\pi D}{6} \times 10^{-3} \\ &= 11 B_{av}\, ac\, K_w \times 10^{-3}\, D^2 L n_s = C_0 D^2 L n_s \end{aligned}$$

where $C_0 = 11B_{av}\,ac\,K_w\times10^{-3}$ is the output coefficient and $Q = \text{kW}/(\eta\cos\phi)$.

Separation of $D$ and $L$

The output equation gives only the product $D^2L$. To get $D$ and $L$ separately, a ratio $L/\tau$ (with $\tau = \pi D/P$) is chosen according to the design aim:

With $L = k\tau = k\pi D/P$:

$$D^3 = \frac{D^2L \cdot P}{k\pi}, \qquad L = \frac{D^2L}{D^2}$$

Checks after separation

• Peripheral speed: $v_a = \pi D n_s$ must not exceed about 30 m/s for normal rotors (higher only with special construction).
• Ventilating ducts: if $L$ exceeds about 100–125 mm, radial ducts (about 10 mm wide) are provided.
• Practical proportions: very long, thin machines are hard to cool and the shaft deflects; very short, large-diameter machines have a high end-winding share.

Example

For $D^2L = 0.0110$ m³, 6 poles, $L = \tau$: $D^3 = 0.0110 \times 6/\pi = 0.0210$, so $D = 0.276$ m and $L = 0.1445$ m, with $v_a = \pi \times 0.276 \times 16.67 = 14.4$ m/s (acceptable).