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Question 1 - (a) A short 3-phase 11 kV line delivers a load of 4.1 MW, 2.6 MVAr lag. If the series impedance of the line is (72 + j12.1)Ω/ph, calculate the sending end voltage and load angle.

Answer - Take the reference voltage as

V_R = (33<0)/√3 = 19.07 < 0 KV/ph

Now power = √3 V_LI_L Cos∅

Power = 4.1MW

I_L = Power/(√3 V_L cos∅)

From the power triangle tan θ = MVAR/MW

Tanθ = 2.6/4.1

θ = tan^-1 2.6/4.1 = 32.8^o

So cos ∅ = 0.84

V_L = 11kv ×3 = 33×10³ V

Substituting these values

I_L = (4.1×10⁶)/(√3×33×10³×0.84) = 85.39 < cos^-1 0.84

= 71.78 - j46.26

Now IZ= (71.78 -j46.26) × (7.2+j12.1)

= 1076.562+j535.466

V_S = V_R+IZ

=19.07×10³ < 0 + 1076.562+j535.466

= 20146.562+j 535.466

= 20154 V/ph

And δ = 1.522

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(b) Develop the receiving end performance chart for the above line and load to a 3-phase power scale.

Answer - Graphical plot

As we know VR = 19.07KV/ph

Here it is given that 10mm = 1000V

So 19.07KV = (19.07KV×10mm)/1000V = 190.7mm

Similarly VS = 20154V/ph

Here it is given that 10mm = 1000V

So 19064 = (20154×10mm)/1000V = 201.540mm

Here 7.2+j12.1 = 1.08 +j59.25

So φ = 59.25

Question 2 - Using the chart you produced in Question 1, determine: (a) the real and reactive power when

(i) The transmission angle remains unaltered and the sending end voltage is increased by 28%.

Answer - When Vs is increased by 28%

201.540mm is extended by 28%

(28×201.50)/100= 56.43mm

MW becomes =

30mm = 4.1MW

26 mm = (26 ×4.1)/30 = 3.5MW

Real Power becomes = 4.1 +3.5 = 7.6MW

MVAR becomes = 12.5 M VAR

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(ii) The original sending-end voltage is reduced by 15% and the transmission angle increased by 9^o.

Answer - When Vs is reduced by 15%

Then VS will be 201.540 × 0.85 = 171.309 mm

MW becomes =

30 mm = 4.1 MW

15mm = (15 ×4.1)/30 =2.05MW

MVAR becomes = 3.5 M VAR

δ = 1.522 +9 = 9.522

(b) The real power limit for the sending voltage in (ii) above (i.e. when reduced by 15%).

Answer - When an arc is drawn the maximum limit will be 20MW.

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