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What are the Output Powers of a Synchronous Motor?
Consider a synchronous motor is operating at lagging power factor. The voltage equation of a synchronous motor is given by,
$$\mathrm{V=E_{f}+I_{a}Z_{S}\:\:\:\:\:\:...(1)}$$
Where,
$$\mathrm{V=V\angle 0°\:and\:E_{f}=E_{f}\:\angle-δ}$$
$$\mathrm{\therefore\:I_{a}=\frac{V-E_{f}}{Z_{S}}\:\:\:\:\:\:...(2)}$$
$$\mathrm{\Longrightarrow\:I_{a}=\frac{V\angle 0°-E_{f}-δ}{Z_{S}\angleθ_{Z}}=\frac{V}{Z_{S}}\angle-θ_{Z}-\frac{E_{f}}{Z_{S}}\angle-(δ+θ_{Z})}$$
$$\mathrm{\therefore\:I^{*}_{a}=\frac{V}{Z_{S}}\angleθ_{Z}-\frac{E_{f}}{Z_{S}}\angle(δ+θ_{Z})\:\:\:\:\:\:...(3)}$$
Complex Power Output per Phase of a Synchronous Motor
The complex power output of a synchronous motor is given by,
$$\mathrm{S_{o}=E_{f}I^{*}_{a}=P_{o}+jQ_{o}\:\:\:\:\:\:...(4)}$$
$$\mathrm{\Longrightarrow\:S_{o}=E_{f}\:\angle-δ\left(\frac{V}{Z_{S}}\angleθ_{Z}-\frac{E_{f}}{Z_{S}}\angle(δ+θ_{Z})\right)}$$
$$\mathrm{\Longrightarrow\:S_{o}=\left(\frac{VE_{f}}{Z_{S}}cos(θ_{Z}-δ)+j\frac{VE_{f}}{Z_{S}}sin(θ_{Z}-δ)\right)-\left(\frac{E^{2}_{f}}{Z_{S}}cosθ_{Z}+j\frac{E^{2}_{f}}{Z_{S}}sinθ_{Z}\right)}$$
$$\mathrm{\therefore\:S_{o}=\left(\frac{VE_{f}}{Z_{S}}cos(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z_{S}}cosθ_{Z}\right)+j\left(\frac{VE_{f}}{Z_{S}}sin(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z_{S}}sinθ_{Z}\right)\:\:\:\:\:\:...(5)}$$
Real Power Output per Phase of the Synchronous Motor
By equating the real part of equation(5), we get the real power output of the synchronous motor, i.e.,
$$\mathrm{P_{o}=\frac{VE_{f}}{Z_{S}}cos(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z_{S}}cosθ_{Z}}$$
$$\mathrm{\because\:cosθ_{Z}=\frac{R_{a}}{Z_{S}}}$$
$$\mathrm{\therefore\:P_{o}=\frac{VE_{f}}{Z_{S}}cos(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z^{2}_{S}}R_{a}\:\:\:\:\:\:...(6)}$$
But,
$$\mathrm{θ_{Z}=(90°-α_{Z});cos(θ_{Z}-δ)=cos(90°-δ+α_{Z})=sin(δ+α_{Z})}$$
$$\mathrm{\therefore\:P_{o}=\frac{VE_{f}}{Z_{S}}sin(δ+α_{Z})-\frac{E^{2}_{f}}{Z^{2}_{S}}R_{a}\:\:\:\:\:\:...(7)}$$
Reactive Power Output per Phase of the Synchronous Motor
By equating the imaginary part of Equation(5), we obtain the reactive power output of the synchronous motor, i.e.,
$$\mathrm{Q_{o}=\frac{VE_{f}}{Z_{S}}sin(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z_{S}}sinθ_{Z}}$$
$$\mathrm{\because\:sinθ_{Z}=\frac{X_{S}}{Z_{S}}}$$
$$\mathrm{\therefore\:Q_{o}=\frac{VE_{f}}{Z_{S}}sin(θ_{Z}-δ)-\frac{E^{2}_{f}}{Z^{2}_{S}}R_{a}\:\:\:\:\:\:...(8)}$$
But,
$$\mathrm{θ_{Z}=(90°-α_{Z});sin(θ_{Z}-δ)=sin(90°-δ+α_{Z})=cos(δ+α_{Z})}$$
$$\mathrm{\therefore\:Q_{o}=\frac{VE_{f}}{Z_{S}}cos(δ+α_{Z})-\frac{E^{2}_{f}}{Z^{2}_{S}}X_{S}\:\:\:\:\:\:...(9)}$$
Also, for a synchronous motor, the output power available at the shaft is given by,
$$\mathrm{P_{sh}=P_{o}-Rotational\: losses\:\:\:\:\:\:...(10)}$$
Where,po is the mechanical power (or gross power) developed by the motor. The rotational losses include core losses, friction and windage losses.
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