DC Machine Design

The number of poles of a dc machine is chosen after the output equation gives $D^2L$. More poles reduce some material but increase other costs and losses, so a compromise is made.

1. Frequency of flux reversal

$$f = \frac{Pn}{2}$$

The armature iron sees alternating flux at this frequency. Iron loss rises with $f$, so $f$ is normally kept between 25 and 50 Hz. This limits the number of poles at high speed.

2. Weight of iron

Flux per pole $\Phi \propto 1/P$ for the same total flux. So the yoke and armature core depth (which carry $\Phi/2$) become thinner, and iron weight falls as $P$ increases.

3. Weight of copper

• Armature copper: the end connections are shorter for a smaller pole pitch, so armature copper decreases with more poles.
• Field copper: each pole needs its own coil; the total field copper is roughly constant or falls a little.

4. Armature reaction (ampere-turns per pole)

$$AT_a = \frac{ac\,\pi D}{2P}$$

More poles lower the armature mmf per pole, so distortion and the field mmf needed to overcome armature reaction are less. Typical limit: armature AT per pole below about 5000–10000 depending on output.

5. Current per brush arm / per path

For a lap winding, current per path $= I_a/P$. It should not exceed about 200 A (and about 400 A per brush arm), otherwise the commutator gets long and commutation becomes difficult. Large-current machines therefore need more poles.

6. Commutator length and brushes

More poles mean more brush arms with smaller current each, so a shorter commutator, but more brush gear.

7. Overall size, labour and cost

More poles give a smaller overall diameter and less material, but more coils, more parts and more labour. Frequency and losses also rise.

8. Flashover

More poles means less voltage between adjacent brush arms, but a smaller pole pitch and more crowded commutator; the voltage between segments must be checked.

Typical choice

Example: a 37 kW, 1400 rpm motor with 4 poles has $f = 4 \times 1400/120 = 46.7$ Hz and current per path about 45 A, so 4 poles is suitable; 6 poles would give 70 Hz (too high).

Specific electric loading ($ac$) of a dc machine is the total armature ampere-conductors per metre of armature periphery:

$$ac = \frac{I_z Z}{\pi D}$$

where $I_z$ is the current in each conductor ($I_a/a$) and $Z$ the total conductors. Usual values are 15000–50000 A/m. Its choice depends on:

1. Temperature rise

Higher $ac$ means more copper loss per unit armature surface. Since heat is removed through that surface, temperature rise increases. Machines with good cooling (forced ventilation) can use a higher $ac$.

2. Speed of machine

A high-speed armature is better ventilated (fan action), so a higher $ac$ can be used.

3. Voltage

High-voltage machines need thicker insulation, which leaves less slot space for copper. Hence a lower $ac$ is used for high voltage.

4. Size of machine

Larger machines have deeper slots and more room for copper, so they can use a higher $ac$.

5. Armature reaction

Armature mmf per pole $= ac\,\pi D/(2P)$. A high $ac$ means strong armature reaction, so more field mmf (more field copper, larger poles) is needed to avoid excessive distortion and loss of flux.

6. Commutation

Reactance voltage is proportional to the ampere-conductors per slot and $ac$. A high $ac$ makes commutation difficult and may need interpoles.

7. Copper loss and efficiency

A higher $ac$ increases copper loss and lowers efficiency.

8. Size and cost

From $P_a = C_0D^2Ln$ with $C_0 = \pi^2B_{av}\,ac\times10^{-3}$, a higher $ac$ gives a smaller, cheaper 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 $C_0 = \pi^2 \times 0.5 \times 22 = 108.6$.

Data: 7.5 kW, 220 V shunt motor, 4 poles, 1000 rpm ($n = 16.67$ rps), $\eta = 0.83$, $B_g = 0.9$ Wb/m², $ac = 30000$ A/m, field form factor $\psi = 0.7$, $I_f = 2.5\%$ of rated current, maximum efficiency at full load, square pole face.

Armature power $P_a$

Input and losses:

$$\begin{aligned} P_{in} &= \frac{7500}{0.83} = 9036.1\ \text{W}, \qquad \text{Losses} = 9036.1 - 7500 = 1536.1\ \text{W} \end{aligned}$$

Maximum efficiency at full load means variable (armature copper) loss = constant loss:

$$I_a^2R_a = \frac{1536.1}{2} = 768.1\ \text{W}$$

Currents:

$$\begin{aligned} I &= \frac{9036.1}{220} = 41.07\ \text{A}, \qquad I_f = 0.025 \times 41.07 = 1.03\ \text{A} \\ I_a &= 41.07 - 1.03 = 40.05\ \text{A} \end{aligned}$$

Back emf:

$$\begin{aligned} I_aR_a &= \frac{768.1}{40.05} = 19.18\ \text{V} \\ E_b &= 220 - 19.18 = 200.82\ \text{V} \\ P_a &= E_bI_a = 200.82 \times 40.05 = 8042\ \text{W} = 8.042\ \text{kW} \end{aligned}$$

Output coefficient

$$\begin{aligned} B_{av} &= \psi B_g = 0.7 \times 0.9 = 0.63\ \text{Wb/m}^2 \\ C_0 &= \pi^2 B_{av}\,ac \times 10^{-3} = \pi^2 \times 0.63 \times 30000 \times 10^{-3} = 186.54 \end{aligned}$$

$D^2L$

$$D^2L = \frac{P_a}{C_0 n} = \frac{8.042}{186.54 \times 16.67} = 2.587 \times 10^{-3}\ \text{m}^3$$

Square pole face

Pole arc $= L$, and pole arc $= \psi\tau = 0.7 \times \pi D/4$:

$$L = 0.5498D$$ $$D^3 = \frac{2.587\times10^{-3}}{0.5498} = 4.705\times10^{-3} \Rightarrow D = 0.1676\ \text{m}$$ $$L = 0.5498 \times 0.1676 = 0.0921\ \text{m}$$

Check: peripheral speed $= \pi \times 0.1676 \times 16.67 = 8.8$ m/s (acceptable).

Answer: $D \approx 16.8$ cm, $L \approx 9.2$ cm.

The output coefficient $C_0 = \pi^2B_{av}\,ac\times10^{-3}$ shows that high specific loadings give a small machine. Their upper limits are set by the factors below.

Specific magnetic loading ($B_{av}$), usually 0.4–0.8 Wb/m²

1. Flux density in teeth: the teeth carry the gap flux through a smaller area. Tooth density should not exceed about 2.1–2.2 Wb/m² in dc machines; otherwise the teeth saturate and need a large field mmf.
2. Frequency of flux reversal: iron loss grows with frequency and $B^2$. High-speed or many-pole machines (high $f$) need a lower $B_{av}$.
3. Iron loss and efficiency: higher $B_{av}$ raises iron loss, lowers efficiency and raises temperature.
4. Size of machine: higher $B_{av}$ means smaller $D^2L$ and lower cost.
5. Field mmf: higher gap density needs more field ampere-turns, so more field copper and larger poles.
6. Armature reaction: a higher $B_{av}$ (stronger main field) keeps the armature mmf relatively smaller, which helps against distortion.

Specific electric loading ($ac$), usually 15000–50000 A/m

1. Temperature rise: more copper loss per unit surface; good ventilation allows a higher value.
2. Speed: fast machines cool better and can take a higher $ac$.
3. Voltage: thicker insulation at high voltage leaves less room for copper, so a lower $ac$.
4. Size of machine: large machines can take higher $ac$.
5. Armature reaction: armature AT per pole $= ac\,\pi D/2P$; high $ac$ means strong armature reaction.
6. Commutation: high $ac$ gives high reactance voltage and poor commutation.
7. Copper loss and efficiency: increase with $ac$.

Summary

The ratio $B_{av}/ac$ also fixes the ratio of iron to copper in the machine, so the two are chosen together.

High specific loadings reduce the size and cost of a dc machine (since $D^2L = P_a/(\pi^2B_{av}\,ac\,n\times10^{-3})$), but they have the following disadvantages.

Higher specific magnetic loading ($B_{av}$)

• Higher iron loss: core loss rises roughly with $B^2$, so efficiency falls and temperature rises.
• Teeth saturation: flux density in the teeth may exceed about 2.2 Wb/m², needing a very large field mmf.
• More field copper: more ampere-turns are needed for the gap and teeth, so field copper loss, field coil size and pole size increase.
• Larger poles and yoke (for the same flux leakage), so the overall diameter may increase.

Higher specific electric loading ($ac$)

• Higher copper loss and lower efficiency.
• Higher temperature rise: more heat per unit surface of armature; may need forced cooling.
• Stronger armature reaction: armature AT per pole $= ac\,\pi D/2P$ increases, causing more field distortion, loss of flux and need of compensating winding.
• Poorer commutation: reactance voltage increases, causing sparking; interpoles may be needed.
• Less room for insulation: risky in high-voltage machines.

The number of commutator segments $C$ (= number of armature coils) is limited by the voltage between adjacent segments. If this voltage is too high, arcing between segments can lead to flashover.

Derivation

Let $E$ = terminal (induced) voltage, $P$ = poles. For a lap winding there are $a = P$ parallel paths, so the number of coils in series per path is:

$$\text{Coils per path} = \frac{C}{P}$$

Average voltage per coil, which is the average voltage between adjacent segments:

$$E_{c(av)} = \frac{E}{C/P} = \frac{EP}{C}$$

The flux density under the pole is not uniform: armature reaction distorts it, and the peak density is about twice the average. So the maximum voltage between adjacent segments is about:

$$E_{c(max)} \approx 2E_{c(av)} = \frac{2EP}{C}$$

For safe operation, the maximum voltage between segments is limited to about 30 V:

$$\frac{2EP}{C} \le 30$$

Hence

$$C_{min} = \frac{2EP}{30} = \frac{EP}{15}$$

Since each coil ends on two segments and each segment joins two coil ends, the number of commutator segments equals the number of coils.

Example

A 220 V, 4-pole machine:

$$C_{min} = \frac{220 \times 4}{15} = 58.7 \approx 59\ \text{segments}$$

A practical value is chosen above this, also matching the number of slots (coils per slot $\times$ slots).

Other limits on $C$

• Segment pitch (about 4 mm minimum) for mechanical strength.
• Commutator peripheral speed (not more than about 30 m/s).
• Reactance voltage: fewer turns per coil (more coils) gives better commutation.

Data: 37 kW, 230 V, 1400 rpm ($n = 23.33$ rps) shunt motor, $B_{av} = 0.5$ Wb/m², $ac = 22000$ A/m, $\eta = 0.9$, $\psi = 0.7$, square pole face.

Armature power

Taking armature copper loss as one-third of total losses (motor):

$$P_a = P\,\frac{2+\eta}{3\eta} = 37 \times \frac{2.9}{2.7} = 39.74\ \text{kW}$$

Output coefficient and $D^2L$

$$\begin{aligned} C_0 &= \pi^2B_{av}\,ac\times10^{-3} = \pi^2 \times 0.5 \times 22 = 108.57 \\ D^2L &= \frac{P_a}{C_0n} = \frac{39.74}{108.57 \times 23.33} = 0.01569\ \text{m}^3 \end{aligned}$$

Square pole face

$L = \text{pole arc} = \psi\tau = 0.7\pi D/P$, so $D^3 = D^2L \cdot P/(0.7\pi)$.

Choice of number of poles

Full-load current $\approx 37000/(0.9 \times 230) = 178.7$ A.

• 6 poles: frequency 70 Hz exceeds the usual 25–50 Hz limit (high iron loss).
• 2 poles: frequency 23.3 Hz is below the usual range; the core is long ($L > D$) and armature mmf per pole is high ($ac\,\pi D/2P \approx 4190$ A).
• 4 poles: frequency 46.7 Hz is within range, current per path 45 A, armature mmf per pole $= 22000 \times \pi \times 0.306/8 = 2640$ A, peripheral speed $= \pi \times 0.306 \times 23.33 = 22.4$ m/s (below 30 m/s).

Choose $P = 4$. Then

$$\begin{aligned} L &= \frac{0.7\pi D}{4} = 0.5498D \\ D^3 &= \frac{0.01569}{0.5498} = 0.02853 \Rightarrow D = 0.306\ \text{m} \\ L &= 0.5498 \times 0.306 = 0.168\ \text{m} \end{aligned}$$

Answer: 4 poles, $D \approx 30.6$ cm, $L \approx 16.8$ cm.

Data: 5 kW, 250 V, 4 poles, 1500 rpm ($n = 25$ rps) shunt generator, $\eta = 0.87$, $B_{av} = 0.42$ Wb/m², $ac = 15000$ A/m, $\psi = 0.66$, square pole face.

Armature power

Taking armature copper loss as one-third of total losses (generator):

$$P_a = P\,\frac{1+2\eta}{3\eta} = 5 \times \frac{1 + 2 \times 0.87}{3 \times 0.87} = 5 \times \frac{2.74}{2.61} = 5.249\ \text{kW}$$

Output coefficient

$$C_0 = \pi^2B_{av}\,ac\times10^{-3} = \pi^2 \times 0.42 \times 15000 \times 10^{-3} = 62.18$$

$D^2L$

$$D^2L = \frac{P_a}{C_0n} = \frac{5.249}{62.18 \times 25} = 3.377\times10^{-3}\ \text{m}^3$$

Square pole face

Pole arc $= L$ and pole arc $= \psi\tau = 0.66 \times \pi D/4 = 0.5184D$, so $L = 0.5184D$:

$$\begin{aligned} D^3 &= \frac{3.377\times10^{-3}}{0.5184} = 6.514\times10^{-3}\ \text{m}^3 \\ D &= 0.1868\ \text{m} \\ L &= 0.5184 \times 0.1868 = 0.0968\ \text{m} \end{aligned}$$

Checks

• Pole pitch $\tau = \pi D/4 = 0.1467$ m; pole arc $= 0.66\tau = 0.0968$ m $= L$ (square face).
• Peripheral speed $= \pi Dn = 14.7$ m/s (below 30 m/s).
• Frequency $= Pn/2 = 50$ Hz (upper limit of the normal range).

Answer: $D \approx 18.7$ cm, $L \approx 9.68$ cm.

Data: 5 kW, 250 V, 4 poles, 1500 rpm ($n = 25$ rps) shunt generator, $\eta = 0.87$, $B_{av} = 0.40$ Wb/m², $ac = 15000$ A/m, $\psi = 0.66$, square pole face.

Armature power

Taking armature copper loss as one-third of total losses (generator):

$$P_a = P\,\frac{1+2\eta}{3\eta} = 5 \times \frac{1 + 2 \times 0.87}{3 \times 0.87} = 5 \times \frac{2.74}{2.61} = 5.249\ \text{kW}$$

Output coefficient

$$C_0 = \pi^2B_{av}\,ac\times10^{-3} = \pi^2 \times 0.40 \times 15000 \times 10^{-3} = 59.22$$

$D^2L$

$$D^2L = \frac{P_a}{C_0n} = \frac{5.249}{59.22 \times 25} = 3.546\times10^{-3}\ \text{m}^3$$

Square pole face

Pole arc $= L$ and pole arc $= \psi\tau = 0.66 \times \pi D/4 = 0.5184D$, so $L = 0.5184D$:

$$\begin{aligned} D^3 &= \frac{3.546\times10^{-3}}{0.5184} = 6.840\times10^{-3}\ \text{m}^3 \\ D &= 0.1898\ \text{m} \\ L &= 0.5184 \times 0.1898 = 0.0984\ \text{m} \end{aligned}$$

Checks

• Pole pitch $\tau = \pi D/4 = 0.1491$ m; pole arc $= 0.66\tau = 0.0984$ m $= L$ (square face).
• Peripheral speed $= \pi Dn = 14.9$ m/s (below 30 m/s).
• Frequency $= Pn/2 = 50$ Hz (upper limit of the normal range).

Answer: $D \approx 19.0$ cm, $L \approx 9.84$ cm.

After the armature diameter is fixed, the number of armature slots $S$ is chosen to suit the winding, commutation, cooling and cost.

1. Slot pitch

Slot pitch $y_s = \pi D/S$ is normally 25–35 mm. Too small a pitch gives weak, narrow teeth; too large a pitch gives deep, wide slots with high leakage.

2. Cooling

More slots spread the copper over a larger surface and improve heat dissipation, so the temperature rise is lower.

3. Flux pulsation

As the armature turns, the number of teeth under a pole face changes, causing flux pulsation, iron loss and noise. To reduce this, the number of slots under a pole arc is made (integer + ½), and the slots per pole are kept at least about 9.

4. Commutation

Fewer conductors per slot means lower ampere-conductors per slot and lower reactance voltage. Slot loading is limited to about 1500 ampere-conductors. More slots therefore improve commutation.

5. Tooth flux density

More slots give narrower teeth, so tooth flux density rises. Teeth must not saturate beyond about 2.1–2.2 Wb/m².

6. Cost

More slots mean more coils, more insulation and more labour, so higher cost. Insulation also takes a larger fraction of slot space (poorer space factor).

7. Suitability for winding

• For a lap winding the slots per pole pair should suit equalizer connections.
• For a wave winding the number of coils must satisfy the winding rule ($C = (P/2)y_c \pm 1$).
• $S$ must be compatible with the number of commutator segments ($C = u \times S$, with $u$ coil sides per layer).

Example

For $D = 0.30$ m: slot pitch 25–35 mm gives $S = \pi \times 300/35 \approx 27$ to $\pi \times 300/25 \approx 38$; a 4-pole machine might use 33 slots (8.25 per pole) or 35 slots, checked against the winding rule and commutator segments.

Field form factor

The field form factor $\psi$ (or $K_f$) is the ratio of average gap flux density over a pole pitch to the maximum gap flux density:

$$\psi = \frac{B_{av}}{B_g} \approx \frac{\text{pole arc}}{\text{pole pitch}}$$

Its usual value is 0.64–0.72.

Factors in selecting the number of poles

1. Frequency: $f = Pn/2$ should usually be 25–50 Hz. More poles raise frequency and iron loss in the armature.
2. Weight of iron: flux per pole falls as $P$ increases, so the yoke and armature core become thinner; iron weight falls.
3. Weight of copper: armature end connections are shorter with smaller pole pitch, so armature copper falls; field copper changes little.
4. Armature reaction: armature AT per pole $= ac\,\pi D/2P$ falls with more poles, reducing distortion and the extra field mmf needed.
5. Current per brush arm: for a lap winding the current per path ($I_a/P$) should not exceed about 200 A; large-current machines need more poles.
6. Commutator length: more poles give more brush arms with less current each, so a shorter commutator.
7. Labour and cost: more poles mean more field coils and parts, so more labour.
8. Flashover: voltage between brush arms and the voltage between segments must stay within safe limits.

So the number of poles is chosen as a compromise. Usual choices: 2 poles up to about 5 kW, 4 poles for about 5–100 kW, and 6 or more for larger machines.

TRUE.

Specific electric loading $ac = I_zZ/(\pi D)$ is the ampere-conductors per metre of armature periphery. A higher $ac$ means more conductors (or more current) per unit surface of the armature, so the $I^2R$ copper loss per unit cooling surface increases. Since this heat must be removed through the same surface, the temperature rise increases. That is why $ac$ is limited by the cooling arrangement and the class of insulation, and machines with better ventilation (high-speed or forced cooled) can use a higher $ac$.

Example: if $ac$ is raised from 20000 to 30000 A/m with the same conductor current density, the copper volume per metre of periphery rises by 50%, so copper loss per unit surface (and roughly the temperature rise) also rises by about 50%. Typical values are 15000–50000 A/m.

TRUE.

The armature mmf per pole is

$$AT_a = \frac{I_zZ}{2P} = \frac{ac\,\pi D}{2P}$$

So, for a given diameter and number of poles, armature mmf is directly proportional to the specific electric loading $ac$. A large $ac$ produces strong armature reaction: more cross-magnetisation, more distortion of the gap flux, more demagnetising effect at saturation, and poorer commutation. A larger field mmf (or compensating winding) is then needed.

Example: with $ac = 22000$ A/m, $D = 0.306$ m and $P = 4$, $AT_a = 22000 \times \pi \times 0.306/8 \approx 2640$ A per pole; doubling $ac$ doubles this armature mmf. Hence designers limit $ac$, or add interpoles and compensating windings in heavily loaded machines.

TRUE.

The air gap needs most of the field ampere-turns ($AT_g = 800000\,B_g K_g l_g$). A longer gap therefore needs more field mmf, which means more field copper, larger field coils and poles, and a larger frame. Field copper loss also rises. All these increase the cost of the machine. (A longer gap does reduce armature reaction and noise, but it is kept just long enough to limit armature reaction because of this extra cost.)

Example: doubling $l_g$ from 3 mm to 6 mm roughly doubles the gap ampere-turns, which form the largest part of the field mmf per pole, so field copper and field loss rise almost in the same proportion. So the gap is not made longer than needed: in practice the gap ampere-turns are kept at about 0.5–0.7 times the armature ampere-turns per pole, just enough to control armature reaction.

FALSE.

The teeth carry the gap flux through an iron area smaller than the gap area (tooth width is less than slot pitch). Tooth flux density is roughly

$$B_t \approx B_g \times \frac{\text{slot pitch}}{\text{tooth width} \times k_i}$$

So a higher specific magnetic loading (higher $B_{av}$, hence higher $B_g$) raises the flux density in the teeth. This is why $B_{av}$ is limited: tooth density should not exceed about 2.1–2.2 Wb/m², otherwise the teeth saturate and a large field mmf is needed.

Example: with $B_g = 0.8$ Wb/m², slot pitch 30 mm, tooth width 15 mm and $k_i = 0.9$, $B_t \approx 0.8 \times 30/(15 \times 0.9) = 1.78$ Wb/m²; raising $B_g$ to 1.0 Wb/m² gives $B_t \approx 2.22$ Wb/m², which is already at the saturation limit.

The main dimensions come from the output equation of a dc machine (A.K. Sawhney):

$$P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}$$

where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps).

Step 1: Armature power (losses split at maximum efficiency)

$$\begin{aligned} \text{Input} &= \frac{7.5}{0.83} = 9.036\ \text{kW} \\ I_L &= \frac{9036}{220} = 41.07\ \text{A} \\ I_f &= 0.025 \times 41.07 = 1.027\ \text{A} \\ I_a &= I_L - I_f = 41.07 - 1.027 = 40.05\ \text{A} \\ \text{Total losses} &= 9036 - 7500 = 1536\ \text{W} \end{aligned}$$

At maximum efficiency, variable loss = constant loss, so armature copper loss $= 1536/2 = 768$ W. (Field copper, iron, friction and brush losses form the constant part, so all losses are counted.)

$$\begin{aligned} R_a &= \frac{768}{40.05^2} = 0.479\ \Omega \\ E_b &= V - I_aR_a = 220 - 40.05\times0.479 = 200.8\ \text{V} \\ P_a &= E_b I_a = 200.8 \times 40.05 = 8042\ \text{W} = 8.04\ \text{kW} \end{aligned}$$

Step 2: Specific magnetic loading and output coefficient

The field form factor $K_f = B_{av}/B_m = 0.7$; Sawhney takes $\psi$ (pole arc/pole pitch) $= K_f = 0.7$.

$$\begin{aligned} B_{av} &= K_f B_m = 0.7\times0.9 = 0.63\ \text{Wb/m}^2 \\ C_0 &= \pi^2 B_{av}\, ac\times10^{-3} = \pi^2\times0.63\times30000\times10^{-3} = 186.5 \\ n &= 1800/60 = 30\ \text{rps} \end{aligned}$$ $$D^2L = \frac{P_a}{C_0 n} = \frac{8.04}{186.5\times30} = 1.437\times10^{-3}\ \text{m}^3$$

Step 3: Square pole face

For a square pole face, core length = pole arc:

$$L = \psi\,\tau = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D$$ $$\begin{aligned} D^3 &= \frac{1.437\times10^{-3}}{0.5498} = 2.614\times10^{-3}\ \text{m}^3 \\ D &= 0.1378\ \text{m} \\ L &= 0.5498 \times 0.1378 = 0.0757\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi D/4 = 0.108$ m; pole arc $= 0.7\tau = 0.0757$ m $= L$ (square).
• Peripheral speed $= \pi D n = \pi\times0.1378\times30 = 13.0$ m/s (well below 30 m/s).
• Frequency $= pn/2 = 4\times30/2 = 60$ Hz (acceptable for a small machine).

Answer: Armature diameter $D \approx 0.138\ \text{m}\ (138\ \text{mm})$, armature core length $L \approx 0.076\ \text{m}\ (76\ \text{mm})$.

The main dimensions come from the output equation of a dc machine (A.K. Sawhney):

$$P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}$$

where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps).

Step 1: Armature power (losses split at maximum efficiency)

$$\begin{aligned} \text{Input} &= \frac{10}{0.80} = 12.5\ \text{kW} \\ I_L &= \frac{12500}{220} = 56.82\ \text{A} \\ I_f &= 0.03 \times 56.82 = 1.705\ \text{A} \\ I_a &= I_L - I_f = 56.82 - 1.705 = 55.11\ \text{A} \\ \text{Total losses} &= 12500 - 10000 = 2500\ \text{W} \end{aligned}$$

At maximum efficiency, variable loss (armature copper loss) = constant loss, so armature copper loss $= 2500/2 = 1250$ W.

$$\begin{aligned} R_a &= \frac{1250}{55.11^2} = 0.4115\ \Omega \\ E_b &= V - I_aR_a = 220 - 55.11\times0.4115 = 197.3\ \text{V} \\ P_a &= E_b I_a = 197.3 \times 55.11 = 10\,875\ \text{W} = 10.875\ \text{kW} \end{aligned}$$

Step 2: Specific magnetic loading and output coefficient

The field form factor $K_f = B_{av}/B_m = 0.7$; Sawhney takes $\psi$ (pole arc/pole pitch) $= K_f = 0.7$.

$$\begin{aligned} B_{av} &= K_f B_m = 0.7\times0.85 = 0.595\ \text{Wb/m}^2 \\ C_0 &= \pi^2 B_{av}\, ac\times10^{-3} = \pi^2\times0.595\times30000\times10^{-3} = 176.2 \\ n &= 900/60 = 15\ \text{rps} \end{aligned}$$ $$D^2L = \frac{P_a}{C_0 n} = \frac{10.875}{176.2\times15} = 4.115\times10^{-3}\ \text{m}^3$$

Step 3: Square pole face

For a square pole face, core length = pole arc:

$$L = \psi\,\tau = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D$$ $$\begin{aligned} D^3 &= \frac{4.115\times10^{-3}}{0.5498} = 7.485\times10^{-3}\ \text{m}^3 \\ D &= 0.1956\ \text{m} \\ L &= 0.5498 \times 0.1956 = 0.1075\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi D/4 = 0.154$ m; pole arc $= 0.7\tau = 0.1075$ m $= L$ (square).
• Peripheral speed $= \pi D n = \pi\times0.1956\times15 = 9.2$ m/s (safe).
• Frequency $= pn/2 = 4\times15/2 = 30$ Hz (below 50 Hz).

Answer: Armature diameter $D \approx 0.196\ \text{m}\ (196\ \text{mm})$, armature core length $L \approx 0.108\ \text{m}\ (108\ \text{mm})$.

The main dimensions come from the output equation of a dc machine (A.K. Sawhney):

$$P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}$$

where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps).

For a motor, Sawhney takes the armature copper loss as about one-third of the total losses, which gives

$$P_a = \frac{2+\eta}{3\eta}\,P$$

Step 1: Armature power and $D^2L$

$$\begin{aligned} P_a &= \frac{2+0.92}{3\times0.92}\times40 = 42.32\ \text{kW} \\ C_0 &= \pi^2 \times 0.5 \times 22000\times10^{-3} = 108.6 \\ n &= 1450/60 = 24.17\ \text{rps} \\ D^2L &= \frac{42.32}{108.6\times24.17} = 0.01613\ \text{m}^3 \end{aligned}$$

Step 2: Square pole face

Core length = pole arc: $L = \psi\pi D/p = 0.7\pi D/p$, so

$$D^3 = \frac{D^2L\; p}{0.7\pi}$$

Step 3: Choice of number of poles

Try common pole numbers and check frequency ($f = pn/2 \le 50$ Hz) and peripheral speed ($\le 30$ m/s):

• $p = 6$ gives $f = 72.5$ Hz, too high (large iron loss).
• $p = 2$ gives a long, narrow machine ($L > D$) with heavy armature reaction per pole and large yoke; not used for 40 kW.
• $p = 4$ keeps $f$ below 50 Hz with a reasonable $L/D$. Armature current $\approx 40000/(0.92\times240) = 181$ A, so with a lap winding the current per path ($\approx 45$ A) and current per brush arm ($\approx 90$ A) are small.

Choose 4 poles.

Step 4: Main dimensions for $p = 4$

$$\begin{aligned} D^3 &= \frac{0.01613\times4}{0.7\pi} = 0.02934\ \text{m}^3 \\ D &= 0.308\ \text{m} \\ L &= \frac{0.7\pi\times0.308}{4} = 0.170\ \text{m} \\ \tau &= \frac{\pi D}{p} = 0.242\ \text{m}, \quad \text{pole arc} = 0.7\tau = 0.170\ \text{m} \end{aligned}$$

Answer: Number of poles = 4, armature diameter $D \approx 0.308$ m, core length $L \approx 0.170$ m (pole face 0.170 m x 0.170 m).

The main dimensions come from the output equation of a dc machine (A.K. Sawhney):

$$P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}$$

where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps).

For a generator, taking armature copper loss as about one-third of total losses (Sawhney),

$$P_a = \frac{1+2\eta}{3\eta}\,P$$

Step 1: Armature power

$$P_a = \frac{1+2\times0.91}{3\times0.91}\times200 = \frac{2.82}{2.73}\times200 = 206.6\ \text{kW}$$

Step 2: Loadings and output coefficient

$$\begin{aligned} B_{av} &= \psi B_m = 0.67\times0.87 = 0.583\ \text{Wb/m}^2 \\ C_0 &= \pi^2\times0.583\times31000\times10^{-3} = 178.3 \\ n &= 1000/60 = 16.67\ \text{rps} \\ D^2L &= \frac{206.6}{178.3\times16.67} = 0.0695\ \text{m}^3 \end{aligned}$$

Step 3: Use $L/\tau = 0.75$

$$\tau = \frac{\pi D}{p} = \frac{\pi D}{6}, \qquad L = 0.75\tau = \frac{0.75\pi D}{6} = 0.3927D$$ $$\begin{aligned} D^3 &= \frac{0.0695}{0.3927} = 0.1770\ \text{m}^3 \\ D &= 0.561\ \text{m} \\ L &= 0.3927\times0.561 = 0.220\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi\times0.561/6 = 0.294$ m; pole arc $= 0.67\tau = 0.197$ m.
• Peripheral speed $= \pi D n = \pi\times0.561\times16.67 = 29.4$ m/s (within the usual 30 m/s limit).
• Frequency $= pn/2 = 6\times16.67/2 = 50$ Hz (acceptable).

Answer: Armature diameter $D \approx 0.561$ m, core length $L \approx 0.220$ m.

The main dimensions come from the output equation of a dc machine (A.K. Sawhney):

$$P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}$$

where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps).

The question does not state a length condition, so the usual assumption is made: square pole face (core length = pole arc). The full-load armature voltage drop is taken as 3% of terminal voltage.

Step 1: Armature power from currents and emf

$$\begin{aligned} I_L &= \frac{30000}{220} = 136.36\ \text{A} \\ I_f &= \frac{220}{120} = 1.833\ \text{A} \\ I_a &= I_L + I_f = 138.20\ \text{A} \\ E &= V + I_aR_a = 220 + 0.03\times220 = 226.6\ \text{V} \\ P_a &= E I_a = 226.6\times138.20 = 31.32\ \text{kW} \end{aligned}$$

Step 2: Loadings and $D^2L$

$$\begin{aligned} B_{av} &= \psi B_m = 0.7\times0.85 = 0.595\ \text{Wb/m}^2 \\ C_0 &= \pi^2\times0.595\times20000\times10^{-3} = 117.4 \\ n &= 900/60 = 15\ \text{rps} \\ D^2L &= \frac{31.32}{117.4\times15} = 0.01778\ \text{m}^3 \end{aligned}$$

Step 3: Square pole face

$$L = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D$$ $$\begin{aligned} D^3 &= \frac{0.01778}{0.5498} = 0.03233\ \text{m}^3 \\ D &= 0.3186\ \text{m} \\ L &= 0.5498\times0.3186 = 0.1751\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi D/4 = 0.250$ m; pole arc $= 0.7\tau = 0.1751$ m $= L$.
• Peripheral speed $= \pi Dn = 15.0$ m/s; frequency $= pn/2 = 30$ Hz. Both acceptable.

Answer: Armature diameter $D \approx 0.319$ m, armature core length $L \approx 0.175$ m.

The main dimensions come from the output equation of a dc machine (A.K. Sawhney):

$$P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}$$

where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps).

Here "armature voltage 3%" means the full-load armature resistance drop $I_aR_a = 3\%$ of terminal voltage. The ratio pole arc/pole pitch $\psi = 0.7$ is used as $B_{av}/B_m$.

Step 1: Armature power from currents and emf

$$\begin{aligned} I_L &= \frac{30000}{240} = 125.0\ \text{A} \\ I_f &= \frac{240}{120} = 2.0\ \text{A} \\ I_a &= I_L + I_f = 127.0\ \text{A} \\ E &= V + I_aR_a = 240 + 0.03\times240 = 247.2\ \text{V} \\ P_a &= E I_a = 247.2\times127.0 = 31.39\ \text{kW} \end{aligned}$$

Step 2: Loadings and $D^2L$

$$\begin{aligned} B_{av} &= \psi B_m = 0.7\times0.85 = 0.595\ \text{Wb/m}^2 \\ C_0 &= \pi^2\times0.595\times20000\times10^{-3} = 117.4 \\ n &= 900/60 = 15\ \text{rps} \\ D^2L &= \frac{31.39}{117.4\times15} = 0.01782\ \text{m}^3 \end{aligned}$$

Step 3: Square pole face

$$L = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D$$ $$\begin{aligned} D^3 &= \frac{0.01782}{0.5498} = 0.03241\ \text{m}^3 \\ D &= 0.3188\ \text{m} \\ L &= 0.5498\times0.3188 = 0.1753\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi D/4 = 0.250$ m; pole arc $= 0.7\tau = 0.1753$ m $= L$.
• Peripheral speed $= \pi Dn = 15.0$ m/s; frequency $= pn/2 = 30$ Hz. Both acceptable.

Answer: Armature diameter $D \approx 0.319$ m, armature core length $L \approx 0.175$ m.

The main dimensions come from the output equation of a dc machine (A.K. Sawhney):

$$P_a = C_0 D^2 L n, \qquad C_0 = \pi^2 B_{av}\, ac \times 10^{-3}\ \text{kW/(m}^3\text{ rps)}$$

where $P_a$ = armature power (kW), $D$ = armature diameter, $L$ = core length (m), $n$ = speed (rps).

The full-load armature voltage drop $I_aR_a$ is 3% of terminal voltage. Field form factor $K_f = B_{av}/B_m = 0.7$ (equal to $\psi$). Since no length ratio is given, the standard assumption of a square pole face (L = pole arc) is used.

Step 1: Armature power from currents and emf

$$\begin{aligned} I_L &= \frac{30000}{240} = 125.0\ \text{A} \\ I_f &= \frac{240}{120} = 2.0\ \text{A} \\ I_a &= I_L + I_f = 127.0\ \text{A} \\ E &= V + I_aR_a = 240 + 0.03\times240 = 247.2\ \text{V} \\ P_a &= E I_a = 247.2\times127.0 = 31.39\ \text{kW} \end{aligned}$$

Step 2: Loadings and $D^2L$

$$\begin{aligned} B_{av} &= \psi B_m = 0.7\times0.85 = 0.595\ \text{Wb/m}^2 \\ C_0 &= \pi^2\times0.595\times20000\times10^{-3} = 117.4 \\ n &= 900/60 = 15\ \text{rps} \\ D^2L &= \frac{31.39}{117.4\times15} = 0.01782\ \text{m}^3 \end{aligned}$$

Step 3: Square pole face

$$L = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D$$ $$\begin{aligned} D^3 &= \frac{0.01782}{0.5498} = 0.03241\ \text{m}^3 \\ D &= 0.3188\ \text{m} \\ L &= 0.5498\times0.3188 = 0.1753\ \text{m} \end{aligned}$$

Check

• Pole pitch $\tau = \pi D/4 = 0.250$ m; pole arc $= 0.7\tau = 0.1753$ m $= L$.
• Peripheral speed $= \pi Dn = 15.0$ m/s; frequency $= pn/2 = 30$ Hz. Both acceptable.

Answer: Armature diameter $D \approx 0.319$ m, armature core length $L \approx 0.175$ m.