Skip to main content

Nominal-T 100 km line, 50 MVA at 132 kV

Short answer

I_S = 201.07∠−29.78° A, V_S = 80.30∠1.67° kV per phase (139.09 kV line-to-line), efficiency ≈96.80%.

A 100 km long 3 phase 50 Hz overhead line delivers 50 MVA at 0.8 pf lagging and at 132 kV. The resistance and inductive reactance are 0.1 Ω per phase per km and j0.2 per phase per km respectively. The shunt admittance is j4 × 10⁻⁶ S/phase per km. Calculate sending end current, sending end voltage, and transmission efficiency using nominal T-model.

Line constants (total, per phase)

Z=(0.1+j0.2)×100=10+j20 Ω,Z/2=5+j10 ΩY=j4×10−6×100=j4×10−4 S\begin{aligned} Z &= (0.1 + j0.2)\times100 = 10 + j20\ \Omega,\quad Z/2 = 5 + j10\ \Omega \\ Y &= j4\times10^{-6}\times100 = j4\times10^{-4}\ \text{S} \end{aligned}

Receiving end

VR=132 000/3=76 210.2∠0∘ VIR=50×1063×132 000=218.69∠−36.87∘ A\begin{aligned} V_R &= 132\,000/\sqrt3 = 76\,210.2\angle0^\circ\ \text{V} \\ I_R &= \frac{50\times10^6}{\sqrt3\times132\,000} = 218.69\angle-36.87^\circ\ \text{A} \end{aligned}

Nominal-T calculation

 Vs   Z/2        V1        Z/2   Vr
 o---/\/\--+------+------/\/\---o
   Is      |      |  Ir
           |    Y (shunt at middle)
           |      |
 o---------+------+-------------o
V1=VR+IRZ2=78 404.8∠0.80∘ VIC=YV1=31.36∠90.80∘ AIS=IR+IC=201.07∠−29.78∘ AVS=V1+ISZ2=80 302.4∠1.67∘ V\begin{aligned} V_1 &= V_R + I_R\frac{Z}{2} = 78\,404.8\angle0.80^\circ\ \text{V} \\ I_C &= YV_1 = 31.36\angle90.80^\circ\ \text{A} \\ I_S &= I_R + I_C = 201.07\angle-29.78^\circ\ \text{A} \\ V_S &= V_1 + I_S\frac{Z}{2} = 80\,302.4\angle1.67^\circ\ \text{V} \end{aligned}

Sending-end line voltage =3×80.302=139.09= \sqrt3\times80.302 = 139.09 kV.

Efficiency

SS=3VSIS∗=41.32+j25.27 MVAPR=50×0.8=40 MWη=4041.32×100=96.80%\begin{aligned} S_S &= 3V_SI_S^* = 41.32 + j25.27\ \text{MVA} \\ P_R &= 50\times0.8 = 40\ \text{MW} \\ \eta &= \frac{40}{41.32}\times100 = 96.80\% \end{aligned}

(Check: loss =3(218.692+201.072)(5)=1.32= 3(218.69^2 + 201.07^2)(5) = 1.32 MW.)

Answer: ISI_S = 201.07∠−29.78° A, VSV_S = 80.30∠1.67° kV per phase (139.09 kV line), efficiency ≈ 96.80 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Power System Analysis I chapter 6 questions →