### Transmission - Scenario 1

This post is inspired by Problem 21, Chapter 18 (Transmission Lines) of "Textbook-Reviewer in Electrical Engineering, 1st Edition" by Professional Electrical Engr. (PEE) Marcialito M. Valenzona.

SITUATION

A manufacturing plant is supplied by a 3-phase transmission line 10 km long. The plant receives 2.5MVA at 6kV with power factor 0.8 lagging. With a phase impedance of (0.30 + j0.40) ohm per kilometer, what is the sending voltage between the lines?

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-> = (0.30 + j0.40) * 10km = (3 + j4) ohms

|Z,line| = sqrt(3^2 + 4^2) = 5 ohms

angle (b) = arctan(X/R) = arctan(4/2) = 53.13deg

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

V,load,LL-> = 6kV < 0deg

4.) Load voltage, line-neutral:

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

V,load,LN-> = (6kV < 0) / sqrt(3)

V,load,LN-> = 3,464.1V < 0

5.) 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 = (+)arccos(0.8) = 36.87deg

-- Power positive angle means "lagging".

6.) Load apparent power, 3-phase:

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

S,load,3ph-> = 2.5MVA < 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->)* = (2.5MVA<36.87) / (sqrt(3) 6kV<0)

(i,line->)* = 240.56A < 36.87deg

i,line-> = 240.56A < (-36.87deg)

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

8.) Sending voltage, line-neutral:

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

V,src-> = 240.56<(-36.87) (5<53.13) + 3,464.1<0

V,src-> = (240.56 * 5)<(-36.87 +53.13) + 3,464.1<0

V,src-> = 1,202.8<16.26 + 3,464.1<0

V,src-> = (1,154.69 + j336.78) + (3,464.1 + j0)

V,src,LN-> = (4,618.79 + j336.78) V

V,src,LN-> = 4,631.05V < 4.17deg

9.) Sending voltage, line-line:

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

V,src,LL-> = sqrt(3) * 4,631.05<4.17

V,src,LL-> = 8,021.21V < 4.17deg

CONCLUSION

The sending voltage between the transmission lines is 8.021 kiloVolts.

SITUATION

A manufacturing plant is supplied by a 3-phase transmission line 10 km long. The plant receives 2.5MVA at 6kV with power factor 0.8 lagging. With a phase impedance of (0.30 + j0.40) ohm per kilometer, what is the sending voltage between the lines?

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-> = (0.30 + j0.40) * 10km = (3 + j4) ohms

|Z,line| = sqrt(3^2 + 4^2) = 5 ohms

angle (b) = arctan(X/R) = arctan(4/2) = 53.13deg

Z,line-> = |Z,line|<b = 5 ohms < 53.13deg

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

V,load,LL-> = 6kV < 0deg

4.) Load voltage, line-neutral:

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

V,load,LN-> = (6kV < 0) / sqrt(3)

V,load,LN-> = 3,464.1V < 0

5.) 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 = (+)arccos(0.8) = 36.87deg

-- Power positive angle means "lagging".

6.) Load apparent power, 3-phase:

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

S,load,3ph-> = 2.5MVA < 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->)* = (2.5MVA<36.87) / (sqrt(3) 6kV<0)

(i,line->)* = 240.56A < 36.87deg

i,line-> = 240.56A < (-36.87deg)

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

8.) Sending voltage, line-neutral:

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

V,src-> = 240.56<(-36.87) (5<53.13) + 3,464.1<0

V,src-> = (240.56 * 5)<(-36.87 +53.13) + 3,464.1<0

V,src-> = 1,202.8<16.26 + 3,464.1<0

V,src-> = (1,154.69 + j336.78) + (3,464.1 + j0)

V,src,LN-> = (4,618.79 + j336.78) V

V,src,LN-> = 4,631.05V < 4.17deg

9.) Sending voltage, line-line:

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

V,src,LL-> = sqrt(3) * 4,631.05<4.17

V,src,LL-> = 8,021.21V < 4.17deg

CONCLUSION

The sending voltage between the transmission lines is 8.021 kiloVolts.

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