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Loads - Scenario 4

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This post is inspired by Problem 31, Chapter 14 (Synchronous Motor) of "Textbook-Reviewer in Electrical Engineering, 1st Edition" by Professional Electrical Engr. (PEE) Marcialito M. Valenzona.

SITUATION

A Y-configured, three-phase synchronous motor draws its rated current of 70 Amperes at a leading power factor of 0.8, with 6.6 kiloVolts at its terminals. If each phase has 2 ohms armature resistance and 20 ohms synchronous reactance, what is the induced electromotive force between the lines?


ANALYSIS

1.) One-line diagram:

o|---V,load,LN---R,ar,1ph---X,sy,1ph---EMF,load,LN---|>

-V,load,LN-> + i,load-> (R,ar,1ph +j X,sy,1ph) + EMF,load,LN-> = 0
EMF,load,LN-> = V,load,LN-> - [ i,load-> (R,ar,1ph +j X,sy,1ph) ]
EMF,load,LN-> = V,load,LN-> - [ i,load-> (Z,load,1ph) ]

2.) Armature resistance, 1-phase: R,ar,1ph = 2 ohms

3.) Synchronous reactance, 1-phase: X,sy,1ph = 20 ohms

4.) Load impedance, 1-phase:

Z,load,1ph-> = R,ar,1ph + j X,sy,1ph
Z,load,1ph-> = (2 + j20) ohms

|Z,load,1ph| = sqrt(2^2 + 20^2)
|Z,load,1ph| = 20.1 ohms

angle b = arctan( X,sy,1ph / R,ar,1ph )
angle b = arctan(20/2)
angle b = 84.29 deg

Z,load,1ph-> = |Z,load,1ph| < b
Z,load,1ph-> = 20.1 ohms < 84.29 deg

5.) Load voltage, line-line, as reference:

V,load,LL-> = 6.6kV < 0 deg

6.) Load voltage, line-neutral:

V,load,LN-> = V,load,LL-> / sqrt(3)
V,load,LN-> = (6.6kV < 0) / sqrt(3)
V,load,LN-> = 3.81kV < 0 deg

7.) Power factor angle:

S-> = P + j Q
S<a = S cos(a) + j S sin(a)

P = S cos(a)
cos(a) = P/S = pf = 0.8
a = (-1) * arccos(0.8) = -36.87deg
-- Power negative angle means "leading".

8.) Load current:

i,load-> = 70A < c

S-> = V-> (i->)*
-- Conjugate (i->)* means "to reverse angle sign".

S<a = V<b [i<(c)]*
S<a = V<b i<(-c)
Vi <a = Vi <(b-c)
Vi <(-36.87) = Vi <(0-c)
1<(-36.87) = 1 <(-c)
c = 36.87deg

i,load-> = 70A < 36.87deg
-- Current positive angle means "current leads voltage".

9.) Electromotive force, line-neutral:

EMF,load,LN-> = V,load,LN-> - [ i,load-> (Z,load,1ph) ]
EMF,load,LN-> = (3.81k < 0) - [ (70 < 36.87) (20.1 < 84.29) ]
EMF,load,LN-> = 3,810 - [ (70 * 20.1) < (36.87 + 84.29) ]
EMF,load,LN-> = 3,810 - (1,407 < 121.16)
EMF,load,LN-> = 3,810 - (-728.02 + j1,204)
EMF,load,LN-> = (4,538.02 - j1,204) V
EMF,load,LN-> = 4,695.02V < (-14.86deg)

10.) Electromotive force, line-line:

EMF,load,LL-> = sqrt(3) * EMF,load,LN->
EMF,load,LL-> = sqrt(3) * [ 4,695.02 < (-14.86) ]
EMF,load,LL-> = 8,132.01V < (-14.86deg)

CONCLUSION

The induced electromotive force between the lines is 8.132 kiloVolts.

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