A 20 MVA- 11 kV- Y - connected- 3 - phase synchronous motor has a synchronous reactance of 5 -phase windage- The shaft load is 15 MW- The value of excitation emf line in kV is

V_t =11kV√(3) = 6.35 kV X_a =5 Ω P_m = 15 + 1.2 = 16.2 MW (3 phase) P _m (in) =P_m(out) = 16.2 MW (3 phase) I_a =16.2√(2)× 11 × 1 = 0.85 kA ( upf) E_f =V ...

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Standard X

Physics

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A 20 MVA, 11 ...

Question

A 20 MVA, 11 kV, Y - connected, 3 - phase synchronous motor has a synchronous reactance of $5 Ω$ /phase windage. The shaft load is 15 MW. The value of excitation emf (line) in kV is ______
13.23

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Solution

The correct option is A 13.23
$V_{t} = \frac{11 k V}{\sqrt{3}} = 6.35 k V$

$X_{a} = 5 Ω$

$P_{m} = 15 + 1.2 = 16.2 M W (3 - p h a s e)$

$P_{m} (i n) = P_{m (o u t)} = 16.2 M W (3 - p h a s e)$

$I_{a} = \frac{16.2}{\sqrt{2} \times 11 \times 1} = 0.85 k A (u p f)$

$_{f} =_{t} = j X_{s}_{a}$

$_{f} = 6.35 - j 5 \times 0.85 = 6.35 - j 4.25$

$|_{f} | = 7.64 k V (o r) 13.23 k V (l i n e)$

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A 20 MVA, 11 kV, Y - connected, 3 - phase synchronous motor has a synchronous reactance of 5Ω/phase windage. The shaft load is 15 MW. The value of excitation emf (line) in kV is ______13.23

The correct option is A 13.23Vt=11kV√3=6.35kV Xa=5Ω Pm=15+1.2=16.2MW (3−phase) Pm(in)=Pm(out)=16.2MW (3−phase) Ia=16.2√2×11×1=0.85 kA (upf) −→Ef=→Vt=jXs→Ia −→Ef=6.35−j5×0.85=6.35−j4.25 |−→Ef|=7.64kV (or) 13.23 kV (line)

A 20 MVA, 11 kV, Y - connected, 3 - phase synchronous motor has a synchronous reactance of $5 Ω$ /phase windage. The shaft load is 15 MW. The value of excitation emf (line) in kV is ______
13.23