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Emf, regulation and loading with given load impedance

Short answer

With load impedance 80+j60 Ω, generated emf is 6546.2 V per phase (11.34 kV line), regulation 3.08%. The generator supplies 1210 kVA, so it is under-loaded by 19.33%.

A 3-phase, star-connected 3-phase synchronous generator is rated as 1500 kVA, 11 kV. The armature winding resistance is 0.8 ohm per phase and synchronous reactance is 4 ohms per phase. If the generator is supplying power to a three phase balanced load of (80+j60) ohm per phase at rated terminal voltage, calculate emf generated and voltage regulation. Is the generator overloaded OR under-loaded? Calculate the percentage by which it is overloaded OR under-loaded.

Given: star, 1500 kVA, 11 kV, Ra=0.8 ΩR_a = 0.8\ \Omega, Xs=4 ΩX_s = 4\ \Omega; load ZL=80+j60 ΩZ_L = 80 + j60\ \Omega per phase at rated voltage.

Load current

Vph=110003=6350.85 VZL=80+j60=100∠36.87∘ ΩI=6350.85100=63.51 A,pf=80100=0.8 lagging\begin{aligned} V_{ph} &= \frac{11000}{\sqrt3} = 6350.85\ \text{V} \\ Z_L &= 80 + j60 = 100\angle 36.87^\circ\ \Omega \\ I &= \frac{6350.85}{100} = 63.51\ \text{A},\quad \text{pf} = \frac{80}{100} = 0.8\ \text{lagging} \end{aligned}

EMF generated

Vcos⁡ϕ+IRa=5080.68+63.51×0.8=5131.49 VVsin⁡ϕ+IXs=3810.51+63.51×4=4064.55 VE=5131.492+4064.552=6546.20 V per phaseEL=3×6546.20=11338 V\begin{aligned} V\cos\phi + IR_a &= 5080.68 + 63.51\times0.8 = 5131.49\ \text{V} \\ V\sin\phi + IX_s &= 3810.51 + 63.51\times4 = 4064.55\ \text{V} \\ E &= \sqrt{5131.49^2 + 4064.55^2} = 6546.20\ \text{V per phase} \\ E_L &= \sqrt3\times6546.20 = 11338\ \text{V} \end{aligned}

Voltage regulation

%Reg=6546.20−6350.856350.85×100=3.08%\%\text{Reg} = \frac{6546.20 - 6350.85}{6350.85}\times100 = 3.08\%

Overloaded or under-loaded?

Irated=1500×1033×11000=78.73 ALoad kVA=3×6350.85×63.51×10−3=1210 kVAUnder-load=78.73−63.5178.73×100=1500−12101500×100=19.33%\begin{aligned} I_{rated} &= \frac{1500\times10^3}{\sqrt3\times11000} = 78.73\ \text{A} \\ \text{Load kVA} &= 3\times6350.85\times63.51\times10^{-3} = 1210\ \text{kVA} \\ \text{Under-load} &= \frac{78.73 - 63.51}{78.73}\times100 = \frac{1500 - 1210}{1500}\times100 = 19.33\% \end{aligned}

Answer: E = 6546.2 V/phase (11.34 kV line); regulation = 3.08%; the generator is under-loaded by 19.33% (supplies 1210 kVA, 968 kW).

✓ Answer checked by Claude Sonnet 5.5

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