### Transmission - Scenario 2

SITUATION

A nominal 69-kiloVolt short transmission line has a length of 16 kilometers, with an impedance of (0.25 + j0.4375) ohm per kilometer. It supplies a 70MVA 3-phase load, with 80% lagging power factor, at 64 kiloVolts only. What is the line's transmission efficiency?

ANALYSIS

1.) One line diagram:

o|---V,src---Z,line---Z,load---|>

V,src-> = i,line-> (Z,line-> + Z,load->)

V,load-> = i,line-> Z,load->

V,src-> = i,line-> Z,line-> + V,load->

2.) Line impedance:

Z,line-> = (0.25 + j0.4375) * 16km

Z,line-> = (4 + j7) ohms

Z,line-> = 8.0623ohms < 60.2551deg

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

V,load,LL-> = 64kV < 0deg

4.) Load voltage, line-neutral:

V,load,LN-> = V,load,LL-> / sqrt(3)

V,load,LN-> = (64kV < 0) / sqrt(3)

V,load,LN-> = 36.95kV < 0

5.) Power factor angle:

S<a = P + j Q

S<a = S cos(a) + j S sin(a)

P = S cos(a)

cos(a) = P/S = pf = 0.8

a = (+)arccos(0.8) = 36.87deg

-- Power positive angle means "lagging".

6.) Load apparent power, 3-phase:

S,load,3ph-> = S<a

S,load,3ph-> = 70MVA < 36.87deg

7.) Line current:

S-> = V-> (i->)*

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

S,load,3ph-> = sqrt(3) V,load,LL-> (i,line->)*

(i,line->)* = S,load,3ph-> / (sqrt(3) V,load,LL->)

(i,line->)* = (70MVA<36.87) / (sqrt(3) 64kV<0)

(i,line->)* = 631.39A < 36.87deg

i,line-> = 631.39A < (-36.87deg)

-- Current negative angle means "current lags voltage".

8.) Source voltage, line-neutral:

V,src-> = i,line-> * Z,line-> + V,load->

V,src-> = 631.39<(-36.87) (8.0623<60.2551) + 36.95k<0

V,src-> = (631.39 * 8.0623)<(-36.87 +60.2551) + 36.95k<0

V,src-> = 5,090.46<23.3851 + 36.95k<0

V,src-> = (4.67k + j2.02k) + 36.95k

V,src,LN-> = (41.62k + j2.02k)V

V,src,LN-> = 41.67kV<2.78

9.) Source voltage, line-line:

V,src,LL-> = sqrt(3) * V,src,LN->

V,src,LL-> = sqrt(3) * 41.67kV<2.78

V,src,LL-> = 72.17kV < 2.78deg

10.) Source apparent power, 3-phase:

S-> = V-> (i->)*

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

S,src,3ph-> = sqrt(3) V,src,LL-> (i,line->)*

S,src,3ph-> = sqrt(3) (72.17kV<2.78) (631.39A < -(-36.87))

S,src,3ph-> = 78.93MVA < 39.65deg

11.) Transmission efficiency, scalar:

eff = S,load / S,src

eff = 70MVA / 78.93MVA = 0.8869

CONCLUSION

The line's transmission efficiency is at 88.69%.

A nominal 69-kiloVolt short transmission line has a length of 16 kilometers, with an impedance of (0.25 + j0.4375) ohm per kilometer. It supplies a 70MVA 3-phase load, with 80% lagging power factor, at 64 kiloVolts only. What is the line's transmission efficiency?

ANALYSIS

1.) One line diagram:

o|---V,src---Z,line---Z,load---|>

V,src-> = i,line-> (Z,line-> + Z,load->)

V,load-> = i,line-> Z,load->

V,src-> = i,line-> Z,line-> + V,load->

2.) Line impedance:

Z,line-> = (0.25 + j0.4375) * 16km

Z,line-> = (4 + j7) ohms

Z,line-> = 8.0623ohms < 60.2551deg

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

V,load,LL-> = 64kV < 0deg

4.) Load voltage, line-neutral:

V,load,LN-> = V,load,LL-> / sqrt(3)

V,load,LN-> = (64kV < 0) / sqrt(3)

V,load,LN-> = 36.95kV < 0

5.) Power factor angle:

S<a = P + j Q

S<a = S cos(a) + j S sin(a)

P = S cos(a)

cos(a) = P/S = pf = 0.8

a = (+)arccos(0.8) = 36.87deg

-- Power positive angle means "lagging".

6.) Load apparent power, 3-phase:

S,load,3ph-> = S<a

S,load,3ph-> = 70MVA < 36.87deg

7.) Line current:

S-> = V-> (i->)*

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

S,load,3ph-> = sqrt(3) V,load,LL-> (i,line->)*

(i,line->)* = S,load,3ph-> / (sqrt(3) V,load,LL->)

(i,line->)* = (70MVA<36.87) / (sqrt(3) 64kV<0)

(i,line->)* = 631.39A < 36.87deg

i,line-> = 631.39A < (-36.87deg)

-- Current negative angle means "current lags voltage".

8.) Source voltage, line-neutral:

V,src-> = i,line-> * Z,line-> + V,load->

V,src-> = 631.39<(-36.87) (8.0623<60.2551) + 36.95k<0

V,src-> = (631.39 * 8.0623)<(-36.87 +60.2551) + 36.95k<0

V,src-> = 5,090.46<23.3851 + 36.95k<0

V,src-> = (4.67k + j2.02k) + 36.95k

V,src,LN-> = (41.62k + j2.02k)V

V,src,LN-> = 41.67kV<2.78

9.) Source voltage, line-line:

V,src,LL-> = sqrt(3) * V,src,LN->

V,src,LL-> = sqrt(3) * 41.67kV<2.78

V,src,LL-> = 72.17kV < 2.78deg

10.) Source apparent power, 3-phase:

S-> = V-> (i->)*

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

S,src,3ph-> = sqrt(3) V,src,LL-> (i,line->)*

S,src,3ph-> = sqrt(3) (72.17kV<2.78) (631.39A < -(-36.87))

S,src,3ph-> = 78.93MVA < 39.65deg

11.) Transmission efficiency, scalar:

eff = S,load / S,src

eff = 70MVA / 78.93MVA = 0.8869

CONCLUSION

The line's transmission efficiency is at 88.69%.

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