A synchronous generator having a local load, represented by constant impedance, is connected to an infinite bus through a transformer and a double circuit transmission line. The direct axis generator synchronous reactance is 1.2 pu, the load impedance is Zload = 2 ∠ 36° pu and the rest of the

Question

A synchronous generator having a local load, represented by constant impedance, is connected to an infinite bus through a transformer and a double circuit transmission line. The direct axis generator synchronous reactance is 1.2 pu, the load impedance is [latex]Z_{load}= 2 \angle 36° pu[/latex] and the rest of the parameters are shown in Fig. 8.5. Check the steady-state stability of the given system, if the power transmitted to the systems is 0.5 pu and the generator terminal voltage is kept as 1.1 pu.

Accepted Answer

First we find the angle of the generator terminal voltage [latex]V_T = V_2.[/latex] The powerangle equation is

[latex]P=\frac{V_T V}{X_{23}} \sin\delta_{23} or 0.5=\frac{1.1\cdot 1}{0.6}\sin\delta_{23},[/latex]

where [latex]X_{23} = 0.1 + 1.0/2 = 0.6.[/latex] Then

[latex]\sin\delta_{23}=0.273 and \delta _{23}=15.8°. [/latex]

The current is found as

[latex]\widetilde{I}_{23}=\frac{\widetilde{V}_2-\widetilde{V_3} }{X_{23}}=\frac{1.1 \angle 15.8°-1}{0.6 \angle 90°} =0.5-j0.1 [/latex]pu.

The load current is found as

[latex]\widetilde{I}_{20}=\frac{1.1 \angle 15.8°}{2 \angle 36.8°}=0.55\angle -21°=0.513-j0.197 [/latex] pu,

and the generator current is

[latex]\widetilde{I}_G=\widetilde{I}_{20}+\widetilde{I}_{23}=0.513 − j0.197 + 0.5 − j0.1 = 1.05 \angle −16.13° [/latex] pu.

Then, the internal generator voltage (EMF) is

E = 1.1 ∠ 15.8°+(1.05 ∠ −16.3°)(1.2 ∠ 90°) = 1.41 + j1.51 = 2.07 ∠ 47.12° pu.

Hence, the angle between E and V is 47.12°. Since this angle is less than 90° with a safe area of about 40°, the system is stable. The active power produced by the generator is

[latex]P_G= EI \cos \varphi = 2.07·1.05 \cos[47.12°−(−16.3°)]\cong 1.0 [/latex] pu.

Question 8.1: A synchronous generator having a local load, represented by ...

Related Answered Questions

Examine the voltage stability of a non-static load supplied from a 275 kV infinite bus bar through a line of reactance 50 Ω per phase. The load consists of a constant active power of 200 MW and 200 MVAr rating reactive power, which is related to the voltage by the equation (in pu) ...

Verified Answer:

Apply the equal-area criterion to the system of Example 8.3. ...

Verified Answer:

Assume that, at the sending-end of one of the transmission lines in the system shown in Fig. 8.7, a three-phase fault occurs. Develop and solve the swing equation of the system, if the fault reactance is 0.07 pu. The inertia constant of the generator is H = 0.5 s and the frequency f = 60 Hz. ...

Verified Answer:

A synchronous generator of reactance 1.25 pu is connected to an infinite bus bar system of V =1 pu through a line and transformers of a total reactance of 0.5 pu. The generator’s inertia constant is H = 5 s and EMF is 2.5 pu, and it operates at a load angle of 47°. Find the expression of the ...

Verified Answer: