A 25 MVA, 13 kV, 50 Hz synchronous machine has a short-circuit ratio of 0.52. For rated induced voltage on no-load, it requires a field current of 250 A. (a) Calculate the adjusted (saturated) synchronous reactance of the machine. What is its pu value? (b) The machine is connected to 13-kV

Question

A 25 MVA, 13 kV, 50 Hz synchronous machine has a short-circuit ratio of 0.52. For rated induced voltage on no-load, it requires a field current of 250 A.

(a) Calculate the adjusted (saturated) synchronous reactance of the machine. What is its pu value?

(b) The machine is connected to 13-kV infinite mains and is running as a motor on no-load.

(i) Its field current is adjusted to 200 A. Calculate its power factor angle and torque angle. Ignore machine losses. Draw the phasor diagram indicating terminal voltage, excitation emf and armature current.

(ii) Does the machine act like a capacitor or an inductor to the 13 kV system? Calculate the equivalent capacitor/inductor value.

(iii) Repeat part (i) for a field current of 300 A.

Accepted Answer

(a) Refer Fig. 8.15

[latex] SCR = ext { of }^{\prime} / o f^{\prime \prime}[/latex]

[latex] o f^{\prime}=250 A \Rightarrow o f^{\prime \prime}=250 / 0.52=481 A[/latex]

This is the field current needed to produce short-circuit current equal to rated value.

[latex] I_{a} ext { (rated) }=25 /(\sqrt{3} imes 13) imes 10^{3}=1110 A[/latex]

[latex] I_{S C}\left( ext { at } I_{f}=250 A ight)=1110 imes 250 / 481=577 A[/latex]

[latex] X_{s}( ext { adjusted })=\frac{13 imes 1000 / \sqrt{3}}{577}=13 \Omega[/latex]

[latex] X( ext { Base })=\frac{(13 imes 1000) / \sqrt{3}}{1110}=6.76 \Omega[/latex]

[latex] X_{s}[/latex] (adjusted) = 13/6.76 = 1.92 pu = 1/SCR (checks)

(b) The circuit model of the machine is drawn in Fig. 8.115(a)

(i) [latex] I_{f}=200 A[/latex]

From the modified air-gap line

[latex] E_{f}=13 imes(200 / 250)[/latex] = 10.45 kV (line) or 6.0 kV (phase)

[latex] I_{a}=\frac{V_{t}-E_{f}}{X_{s}}=\frac{7.51 imes 6.0}{13} imes 1000=116 A[/latex]

[latex] \delta=0[/latex] ; no load; no losses

[latex] p f=\cos 90^{\circ}=0[/latex] ; lagging (as [latex] V_{t}>E_{f}[/latex] )

The phasor diagram is drawn in Fig. 8.115(b)

The machine appears as an inductor

[latex] \omega L=\frac{13 imes 1000}{\sqrt{3} imes 116}=2 \pi imes 50 L[/latex]

Or L = 0.206 H

(ii) [latex] I_{f}=300 A[/latex]

[latex] E_{f}=\frac{13 imes 300}{250}=15.6 kV[/latex] (line) or 9.0 kV (phase)

[latex] I_{a}=\frac{9-6}{13}=231 A[/latex]

[latex] \delta=0[/latex] ; no load, no losses

[latex] p f=90^{\circ}=0[/latex] ; leading (as [latex] V_{t}

The machine appears like a capacitor.

[latex] \frac{1}{\omega C}=\frac{13 imes 1000}{231}=\frac{1}{2 \pi imes 50 C}[/latex]

Or [latex] C=56.6 \mu F[/latex]

Question 8.43: A 25 MVA, 13 kV, 50 Hz synchronous machine has a short-circu...

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