Question 4.AE.2: A p=2 pole, f=60 Hz synchronous machine (generator, alternat......

a) Calculation of the distribution (breadth) factor of the stator winding k_{stb}=k_{std} is as follows. The breadth factor is defined for one phase belt as

k_{s t b}=\frac{\text { geometric sum } \sum_{i=1}^8 F_i}{\text { algebraic sum } \sum_{i=1}^8 F_i}              (E4.2-1)

It is convenient to evaluate Eq. E4.2-1 graphically, as shown in Fig. E4.2.4. Note that one slot pitch corresponds to 360°/48=7.5°. Thus, one gets from Fig. E4.2.4 the distribution or breadth factor

k_{s t b}=0.96 \text {. }            (E4.2-2)

b) Calculation of the pitch factor of the stator winding k_{stp  1} is as follows. One stator turn spans 16 slots out of 24 slots, that is, it has a pitch of (2/3); therefore, one obtains for the fundamental n=1

k_{\text {stp } 1}=\left|\sin \left(\frac{n \beta}{2}\right)\right|=\left|\sin \left(\frac{(1)(120)}{2}\right)\right|=0.866                (E4.2-3)

or for the 5th harmonic n=5

The 5th harmonic is not attenuated; therefore, the choice of a 2/3 winding pitch is not recommended.
The total fundamental winding factor of the stator winding is now

c) Calculate the pitch factor of the rotor (r) winding k_{rp} as follows. The fundamental (n=1) pitch-factor of the rotor winding is correspondingly

d) Determination of the stator flux Φ_{stmax} is as follows. The induced voltage in the stator winding per phase is

E_{p h t}=4.44 \cdot f \cdot N_{s t p h t} \cdot k_{s t b} \cdot k_{s t p 1} \cdot \Phi_{s t \operatorname{tax}}          (E4.2-5)

or one obtains for the stator flux

\Phi_{s t \max }=\frac{E_{L-L}}{\sqrt{3} \cdot 4.44 \cdot f \cdot N_{s t p h l} \cdot k_{s t b} \cdot k_{s t p 1}}                (E4.2-6)

with E_{L-L}=V_{L-L}=13,800 V at no load, one gets the flux for no-load condition:

\Phi_{s t \text { max }}^{\text {no-load }}=\frac{13800}{\sqrt{3}(4.44)(60)(16)(0.96)(0.866)}=2.25 \frac{\mathrm{Wb}}{\text { pole }} .                    (E4.2-7)

e) Compute the area (area_p) of one pole:

\operatorname{area}_p=\frac{\Phi_{\text {stmax }}^{\mathrm{no}-\mathrm{load}}}{B_{\max }}=2.25 \mathrm{~m}^2 \text {. }            (E4.2-8)

f) Compute the field (f ) current I_{fo} required at no load:

I_{f o}=\frac{B_{s t \text { max }} \cdot p \cdot g \cdot \pi}{4 \cdot \mu_0 \cdot k_{r p} \cdot N_{\text {rot }}}=\frac{1(2)(0.0127)(\pi)}{4\left(4 \pi \times 10^{-7}\right)(0.751)(288)}=73.4 A \text {. }                (E4.2-9)

g) Determine the synchronous (s) reactance (per phase value) XS in ohms. The selfinductance of phase a is

\begin{gathered} L_{a a}=\frac{\lambda_{a a}}{i_a}=\frac{N_{s t} \cdot \varphi_{s t \max }}{i_a}=\frac{N_{s t} \cdot B_{s t \max } \cdot \text { area }_p}{i_a} \\ =\frac{\mu_0 \cdot 4 \cdot k_{s t b} \cdot k_{s t p} \cdot\left(\mathrm{N}_{\mathrm{stph}}\right)^2 \cdot \text { area }_p}{p \cdot g \cdot \pi} \\ L_{a a}=\frac{4 \pi \times 10^{-7}(4)(0.96)(0.866)(16)^2(2.25)}{2(0.0127)(\pi)}=30.16 \mathrm{mH} . \end{gathered}              (E4.2-10)

According to Concordia [7]

L_d=L_{a a}+L_{a b}+1.5 L_{a a 2}              (E4.2-11)

Neglecting L_{a a 2} \text { and } L_{a a} \approx L_{a b} \text { one gets for the synchronous inductance } L_d.

The synchronous reactance of the round-rotor machine is

X_S=\omega L_d=\omega L_S=22.74 \Omega              (E4.2-12)

The base voltage is

V_{\text {base }}=\frac{V_{L-L}}{\sqrt{3}}=V_{\text {phase }}=7,967 \mathrm{~V} \text {. }              (E4.2-13)

The base apparent power is per phase

S_{\text {base }}=S_{\text {phase }}=\frac{S_{\text {rat }}}{3}=5.33 \mathrm{MVA}              (E4.2-14)

and the base current is

I_{\text {base }}=I_{\text {phase }}=\frac{S_{\text {base }}}{V_{\text {base }}}=669 \mathrm{~A} .                (E4.2-15)

h) Find the base impedance Z_{base} expressed in ohms. The base impedance becomes

Z_{\text {base }}=\frac{V_{\text {base }}}{I_{\text {base }}}=11.9 \Omega .              (E4.2-16)

i) Express the synchronous reactance X_S in per unit (pu). The per-unit synchronous reactance is now

X_S^{p u}=\frac{X_S}{Z_{\text {base }}}=\frac{22.74}{11.9}=1.91 \mathrm{pu} .                (E4.2-17)