Question 4.AE.7: A P=20 kW, VL-L=337 V permanent-magnet generator has the cro......
A P=20 kW, V_{L-L}=337 V permanent-magnet generator has the cross sections of Figs. E4.7.1 and E4.7.2. The number of poles is p=12 and its rated speed is n_{rat}=n_S=60 rpm, where n_S is the synchronous speed. The B–H characteristic of the neodymium–iron–boron (NdFeB) permanent-magnet material is shown in Fig. E4.7.3.
a) Calculate the frequency f of the stator voltages and currents.
b) Apply Ampere’s law to path C (shown in Figs. E4.7.1 and E4.7.2) assuming the stator and rotor iron cores are idea \left(\mu_n \rightarrow \infty\right) .
c) Apply the continuity of flux condition for the areas A_m\ and\ A_g (representing half of the magnet area per pole and half of the air-gap area per pole, respectively) perpendicular to path C (two times the one-sided magnet (m) length l_m, and two times the one-sided air gap (g) length lg, respectively). Furthermore l_m=7 mm, l_g= 1 mm, A_m=11,668 mm² , and A_g=14,469 mm².
d) Provided the relative permeability of the stator and rotor cores is \mu_r \rightarrow \infty, compute the load line of this machine. Plot the load line in a figure similar to that of Fig. E4.7.3. Is the permanent-magnet material (NdFeB) operated at the point of the maximum energy product?
e) What is the recoil permeability μ_R of the NdFeB material?
f) Compute the fluxes Φ_{max}\ and\ Φ_{totalmax}=2Φ_{max}, provided there are N=360 turns per stator phase, the (pitch and distribution) winding factor is k_w=0.8, and the iron-core stacking factor is k_{fe}=0.94.
g) Determine the diameter D of stator wire for a copper-fill factor of k_{cu}=0.62 and a stator slot cross section of A_{st\_slot}=368 mm² .
a) Synchronous speed
n_s=\frac{120 f}{p} \text { or } f=\frac{n_s p}{120}=\frac{60(12)}{120}=6 \mathrm{~Hz} (E4.7-1)
b) Ampere’s law
H_m l_m+H_g l_g=0 . (E4.7-2)
c) Continuity of flux
\Phi=B_m A_m=B_g A_g (E4.7-3)
d) Constituent relation
B_g=\mu_0 H_g . (E4.7-4)
From Eq. E4.7-2
H_m l_m=-H_g l_g . (E4.7-5)
With Eq. E4.7-4
H_m l_m=-B_g l_g / \mu_0 (E4.7-6)
and with Eq. E4.7-3
B_g=B_m\left(\frac{A_m}{A_g}\right) (E4.7-7)
the load-line equation becomes
H_m=-\left(\frac{A_m}{A_g}\right)\left(\frac{l_g}{l_m}\right) \frac{B_m}{\mu_0} (E4.7-8)
Introducing the given numerical values into load-line equation gives
H_m=-91.7 B_m\left[\frac{k A}{m}\right] . (E4.7-9)
This equation is plotted in Fig. E4.7.4. Note that the magnet material is not operated in the point of the maximum energy product.
e) Recoil permeability of NdFeB (slope of the permanent-magnet characteristic)
\mu_R=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}=\frac{\left(B_r-0\right)}{\left(0-\left(-H_c\right)\right)}=\frac{1.25}{950\left(10^3\right)}=1.316 \times 10^{-6}(\mathrm{H} / \mathrm{m}) (E4.7-10)
or
\mu_R=1.047 \mu_0f) Induced voltage
\Phi_{\max }=B_m A_m (E4.7-12)
\Phi_{\text {totalmax }}=2 \Phi_{\max }=2 B_m A_m (E4.7-13)
The flux is defined as
\Phi(t)=\Phi_{\text {totalmax }} \cos \omega t (E4.7-14)
Therefore, the induced voltage is
E_{\max }=N(d \Phi / d t)=\omega N_{p h} k_w k_{f e} \Phi_{\text {totalmax }} (E4.7-15)
From load line and magnet characteristic one obtains the operating point:
B_{m 0}=1.1 \mathrm{~T} (E4.7-16a)
H_{m 0}=-100 \mathrm{kA} / \mathrm{m} (E4.7-16b)
Therefore,
\Phi=\Phi_{\text {totalmax }}=(2)(1.1)\left(11668 \times 10^{-6}\right) \mathrm{Wb}=0.0257 \mathrm{~Wb} (E4.7-17)
and the induced voltage is now
e(t)=(360)(0.0257)(2 \pi)(6)(0.8)(0.94) \sin \omega t=E_{\max } \sin \omega t (E4.7-18a)
or
E_{\max }=262.3 \mathrm{~V} (E4.7-18b)
E_{n u s}=E_{\max } / \sqrt{2}=185.48 \mathrm{~V} (E4.7-18c)
and the line-to-line terminal voltage is approximately
V_{L-L}=\sqrt{3} V_{L-N}=\sqrt{3}(1.05) E_{m s}=337.3 \mathrm{~V} (E4.7-19)
g) Determination of wire diameter of the stator winding. Figure E4.7.5 depicts the stator winding housed in 72 slots. It is a full-pitch, double-layer winding. Using a copper-fill factor of kcu=0.62 one obtains for a stator slot cross section of A_{st\_slot}=368 mm² and (360/12)=30 turns per stator pole and per phase with the copper cross section of one stator slot A_{st\_slot\_cu}=A_{st\_slot} . k_{cu}=368.0.62 mm² =30πD² /4 or a wire diameter of D=3.1 mm. For a rated current of I_{rat}=31 A one obtains a current density of J=31/7.6=4.07 A/mm² . This current density is acceptable for a ventilated machine.
