Question 4.AE.1: A three-phase (m=3), four-pole (p=4) nonsalientpole (round-r......
A three-phase (m=3), four-pole (p=4) nonsalientpole (round-rotor) synchronous machine has the parameters X_S=2\ pu\ and\ R_a=0.05 pu. It is operated at (the phase voltage) V_a=1 pu at an overexcited (lagging current based on the generator reference system, where the current flows out of the machine) displacement power factor of cos φ=0.8 lagging, and per-phase current \left|\tilde{I}_a\right|=1 \mathrm{pu} \text {. The base (rated phase) values are }\left|\tilde{V}_a\right|_{\text {rated }}=V_{\text {base }}=24 \mathrm{kV}, \quad\left|I_a\right|_{\text {rated }}=I_{\text {base }}=1.4 \mathrm{kA}, \text { and the base impedance is } Z_{\text {base }}=V_ {base} ^/ I_{\text {base }}=17.14 \Omega ..
a) Draw the per-phase equivalent circuit of this machine.
b) What is the total rated apparent input power S (expressed in MVA)?
c) What is the total rated real input power P (expressed in MW)?
d) Draw a per-phase phasor diagram with the voltage scale of 1.0 pu≡3 inches, and the current scale of 1.0 pu≡2.5 inches.
e) From this phasor diagram determine the per-phase induced voltage \left|\tilde{E}_a\right| and the torque angle δ.
f) Calculate the rated speed (in rpm) of this machine at f=60 Hz.
g) Calculate the angular velocity ω_S (in rad/s).
h) Calculate the total (approximate) shaft power P_{\text {shaft }} \approx 3\left(E_{\mathrm{a}} \cdot V_{\mathrm{a}} \cdot \sin \delta\right) / X_{\mathrm{S}}.
i) Find the shaft torque T_{shaft} (in Nm).
j) Determine the total ohmic loss of the motor P_{\text {loss }}=3 \cdot R_a\left|\tilde{I}_a\right|^2.
k) Repeat the above analysis for cos φ=0.8 underexcited (leading current based on generator notation) displacement power factor and \left|\tilde{I}_a\right|=0.5 \mathrm{pu}.
a) The per-phase equivalent circuit of a nonsalient-pole synchronous machine based on generator notation is shown in Fig. 4.15a. The parameters of the machine are in ohms:
\begin{aligned} & R_a=0.05 \cdot Z_{\text {base }}=(0.05)(17.14 \Omega)=0.857 \Omega, \text { and } X_S=2 \cdot Z_{\text {base }}=2(17.14 \Omega) \\ & =34.28 \Omega \end{aligned}b) S_{\text {rated }}=3 \cdot V_{a_{-} \text {rated }} \cdot I_{a_{-\text {rated }}}=3(24 \mathrm{kV})(1.4 \mathrm{kA})=100.8 \mathrm{MVA} (E4.1-1)
c) P_{\text {rated }}=S_{\text {rated }} \cdot \cos \varphi=80.64 \mathrm{MW} (E4.1-2)
d) The phasor diagram is given in Fig. E4.1.1.
\varphi=\cos ^{-1}(0.8)=0.6435 \mathrm{rad} \text { or } 36.87^{\circ} \text { lagging } (E4.1-3)
e) For the induced voltage one obtains
E_a=\sqrt{\begin{array}{l} \left(\left|\tilde{V}_a\right|+\left|\tilde{V}_R\right| \cos \varphi+\left|\tilde{V}_X\right| \sin \varphi\right)^2 \\ +\left(\left|\tilde{V}_X\right| \cos \varphi-\left|\tilde{V}_R\right| \sin \varphi\right)^2 \end{array}}=2.735 \mathrm{pu} (E4.1-4)
where
\left|\widetilde{V}_a\right|=1 \mathrm{pu},\left|\widetilde{V}_X\right|=2 \mathrm{pu},\left|\widetilde{V}_R\right|=0.05 \mathrm{pu}, \sin \varphi=0.6, \text { and } \cos \varphi=0.8The torque angle δ is
\delta=\cos ^{-1}\left\{\left(\left|\widetilde{V}_a\right|+\left|\widetilde{V}_R\right| \cos \varphi+\left|\widetilde{V}_X\right| \sin \varphi\right) / E_a\right\}=0.611 \mathrm{rad} \text { or } 35^{\circ} .f) Rated speed is
n_S=120 \mathrm{f} / \mathrm{p}=120(60) / 4=1800 \mathrm{rpm} (E4.1-5)
g) Angular velocity is
\omega_S=2 \pi n_S / 60=188.7 \mathrm{rad} / \mathrm{s} (E4.1-6)
h) Approximate shaft power is
\begin{aligned} P_{\text {shaft }} & \approx 3\left(E_a \cdot V_a \cdot \sin \delta\right) / X_S=3[(2.735)(24 \mathrm{k})(24 \mathrm{k}) \sin (0.611 \mathrm{rad})] /[(2)(17.14)] \\ & =79.1 \mathrm{MW} . \end{aligned}
(E4.1-7)
i) The shaft torque T_{shaft} is
T_{\text {shaft }}=P_{\text {shaft }} / \omega_S=419.14 \mathrm{kNm} (E4.1-8)
j) The total ohmic loss of the motor is
P_{\operatorname{loss}}=3 \cdot R_a\left|\widetilde{I}_{\mathrm{a}}\right|^2=3(0.05)(17.14)(1.4 \mathrm{k})^2=5.04 \mathrm{MW} (E4.1-9)
k) Repeatthe above analysis for underexcited (leading current based on generator reference system) displacement power factor \cos \varphi=0.8 \text { leading and }\left|\widetilde{I}_a\right|=0.5 \mathrm{pu}.
a) see Fig. 4.15a
b) S_{rated }=3 \cdot V_{a_rated } \cdot I_{a_rated}=3(24 \mathrm{kV})(0.7 \mathrm{kA})=50.4 \mathrm{MVA}
c) P_{\text {rated }}=S_{\text {rated }} \cdot \cos \varphi=40.32 \mathrm{MW}
d) The phasor diagram is given in Fig. E4.1.2.
\begin{aligned} & \varphi=\cos ^{-1}(0.8)=0.6435 \text { rador } 36.87^{\circ} \text { leading (underexcited), } \\ & \left|\widetilde{V}_{\text {a-rated }}\right|=1 \mathrm{pu}=24 \mathrm{kV} \equiv 3 \text { inches, } \\ & \mid \widetilde{I}_2-\text { rated } \mid=0.5 \mathrm{pu}=700 \mathrm{~A} \equiv 1.25 \text { inches, } \\ & \mid \tilde{I}_2-\text { rated } \mid \cdot R_{\mathrm{a}}=700 \mathrm{~A}(0.05)(17.141 \Omega)=599.9 \mathrm{~V} \equiv 0.075 \text { inches, } \\ & \left|I_{a-\text { rated }}\right| \cdot X_S=700 \mathrm{~A}(2)(17.14 \Omega)=23.996 \mathrm{kV} \equiv 3 \text { inches. } \end{aligned}e) For the induced voltage one obtains
\begin{aligned} E_a & =\sqrt{\begin{array}{l} \left(\left|\widetilde{V}_a\right|+\left|\widetilde{V}_R\right| \cos \varphi-\left|\widetilde{V}_X\right| \sin \varphi\right)^2 \\ +\left(\left|\widetilde{V}_X\right| \cos \varphi+\left|\widetilde{V}_R\right| \sin \varphi\right)^2 \end{array}} \\ & =0.917 \text { pu or } 22.004 \mathrm{kV}, \end{aligned}where
\left|\widetilde{V}_a\right|=1 \mathrm{pu},\left|\widetilde{V}_X\right|=1 \mathrm{pu},\left|\tilde{V}_r\right|=0.025 \mathrm{pu}, \sin \varphi=0.6, \cos \varphi=0.8
The torque angle δ is
f) Rated speed is
n_S=120 \mathrm{f} / \mathrm{p}=120 \cdot 60 / 4=1800 \mathrm{rpm}.
g) Angular velocity is \omega_S=2 \pi n_S / 60=188.7 \mathrm{rad} / \mathrm{s}
h) Approximate shaft power is
\begin{aligned} P_{\text {shaft }} & \approx 3\left(E_a \cdot V_a \cdot \sin \delta\right) / X_S=3[(0.917)(24 \mathrm{k})(24 \mathrm{k}) \sin (1.095 \mathrm{rad})] /[(2)(17.14)] \\ & =41.09 \mathrm{MW} \end{aligned}i) The shaft torque T_{\text {shaft }} \text { is } T_{\text {shaft }}=P_{\text {shaft }} / \omega_S=217.75 \mathrm{kNm}.
j) The total ohmic loss of the motor is
P_{\text {loss }}=3 \cdot R_a\left|\widetilde{I}_a\right|^2=3(0.05)(17.14)(0.7 \mathrm{k})^2=1.26 \mathrm{MW}