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Armature size, 10 kW 900 rpm motor

Short answer

Armature power Pa found by loss split at max efficiency: Pa=10.875 kW. Output coefficient Co=π²Bav·ac·10⁻³=176.2; D²L=Pa/(Co·n)=4.115×10⁻³ m³. For square pole face, L=0.5498D, so D=0.196 m, L=0.108 m.

Calculate the diameter (D) and length (L) of armature for a 10 kW, 4 pole, 900 rpm, 220 V shunt motor. Given full load efficiency = 0.80, maximum gap flux density = 0.85 Wb/m², specific electrical loading = 30,000 ampere conductor/meter, field form factor (Kf) = 0.7. Assume that maximum efficiency occurs at full load and the field current is 3% of rated current. The pole face is square.

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

Pa=C0D2Ln,C0=π2Bav ac×10−3 kW/(m3 rps)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 PaP_a = armature power (kW), DD = armature diameter, LL = core length (m), nn = speed (rps).

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

Input=100.80=12.5 kWIL=12500220=56.82 AIf=0.03×56.82=1.705 AIa=IL−If=56.82−1.705=55.11 ATotal losses=12500−10000=2500 W\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= 2500/2 = 1250 W.

Ra=125055.112=0.4115 ΩEb=V−IaRa=220−55.11×0.4115=197.3 VPa=EbIa=197.3×55.11=10 875 W=10.875 kW\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 Kf=Bav/Bm=0.7K_f = B_{av}/B_m = 0.7; Sawhney takes ψ\psi (pole arc/pole pitch) =Kf=0.7= K_f = 0.7.

Bav=KfBm=0.7×0.85=0.595 Wb/m2C0=π2Bav ac×10−3=π2×0.595×30000×10−3=176.2n=900/60=15 rps\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} D2L=PaC0n=10.875176.2×15=4.115×10−3 m3D^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=ψ τ=ψπDp=0.7πD4=0.5498DL = \psi\,\tau = \psi\frac{\pi D}{p} = \frac{0.7\pi D}{4} = 0.5498D D3=4.115×10−30.5498=7.485×10−3 m3D=0.1956 mL=0.5498×0.1956=0.1075 m\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 τ=πD/4=0.154\tau = \pi D/4 = 0.154 m; pole arc =0.7τ=0.1075= 0.7\tau = 0.1075 m =L= L (square).
  • Peripheral speed =πDn=π×0.1956×15=9.2= \pi D n = \pi\times0.1956\times15 = 9.2 m/s (safe).
  • Frequency =pn/2=4×15/2=30= pn/2 = 4\times15/2 = 30 Hz (below 50 Hz).

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

✓ Answer checked by Claude Sonnet 5.5

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