Short Single-phase Line Calculations Single-phase circuits are always short and work at relatively low voltages. In such cases, the line
41.18. Short Single-phase Line Calculations Single-phase circuits are always short and work at relatively low voltages. In such cases, the line
constants can be regarded as ‘lumped’ instead of being distributed and capacitance can be neglected. Let E S = voltage at sending end ;
E R = voltage at receiving end
I = line current cos φ R = power factor at receiving end R = resistance of both conductors;
X = reactance of both conductors = ω L Then, the equivalent circuit of a short single-phase line is as shown in Fig. 41.19. Resistive drop = IR
—in phase with I Reactive drop = IX
—in quadrature with I
The vector diagram is shown in Fig. 41.20. From the right-angled triangle OMA, we get
OM 2 2 2 2 = OA 2 +AM = (OK + K A) + (AC + CM)
E S = [( E R cos φ+ R IR ) 2 ( E )] + 2 R sin φ+ R IX An approximate expression for the voltage drop is as follows :
Fig. 41.19 Fig. 41.20
Draw the various dotted lines as in Fig. 41.20. Here, ML is ⊥ OF, CL is || OF and CD is ⊥ OF. As seen OM = OF (approximately).
E S –E R = IR cos φ R + IX sin φ R ∴ drop = I(R cos φ R + X sin φ R ) (approx.) Voltage regn. of line = [(E S –E R )/E R ] × 100
1620 Electrical Technology
Solution in Complex Notation
Let us take E R as reference vector as shown in Fig.
41.21 so that E R = (E R +j 0 )
As seen from ∆ OAB, E S is equal to the vector sum
of E R and IZ or E S =E R + IZ
Now,
I =I ∠−φ R = I (cos φ R – j sin φ R ) Similarly, Z =Z ∠θ = (R + j X) ∴
E S =E R +I Z ∠θ−φ R
Fig. 41.21
or
E S =E R +I (cos φ R − j sin φ R ) (R + j X)
R + (IR cos φ R + IX sin φ R ) + j (IX cos φ R − I R sin φ R ) ∴
2 E 2 R + IR cos φ+ R IX φ R ) + ( IX sin φ− R IR sin φ R ) If the p.f. is leading, then
sin
I =I ∠φ R = I (cos φ R + j sin φ R )
E S = E R + IZ ∠θ + φ R = E R + I (cos φ R + j sin φ R ) (R + j X) It may be noted that the sending end power factor angle is φ s =( φ R + α ).
A single-phase line has an impedance of 5 ∠ 60° and supplies a load of 120 A, 3,300 V at 0.8 p.f. lagging. Calculate the sending-end voltage and draw a vector diagram.
Example 41.8.
(City & Guides, London)
Solution. The vector diagram is similar to that shown in Fig. 41.21. Here
E R = 3,300 ∠ 0°, Z =5 ∠ 60° − Since 1 φ R = cos (0.8) = 36°52 ′ , ∴ I = 120 ∠− 36°52 ′
Voltage drop
= IZ = 120 × 5 ∠ 60º − 36°52 ′
= 600 ∠ 23°8 ′ = 600 (0.9196 + j0.3928) = 551.8 + j 235.7 ∴
E S = (3,300 + j 0 ) + (551.8 + j235.7) = 3,851.8 + j235.7
2 3851.8 2 + 235.7 = 3,860 V
Example 40.9. An overhead, single-phase transmission line delivers 10 kW at 33 kV at
0.8 p.f. lagging. The total resistance of the line is 10 Ω and total inductive reactance is 15 Ω . Deter- mine (i) sending-end voltage (ii) sending-end p.f. and (iii) transmission efficiency.
(Electrical Technology-I, Bombay Univ.)
Solution. Full-load line current is I = 1100/33 ×
0.8 = 41.7 A
Line loss
= 41.7 2 × 10 = 17,390 W = 17.39 kW
Transmission efficiency = output + losses 10 17.39 = = 98.5%
( i) Line voltage drop = IZ = 41.7(0.8 − j 0.6) (10 + j 15) = 709 + j 250 Sending-end voltage is
E S = E R + I Z = (33,000 + j 0) + (709 + j 250) = 33,709 + j 250 = 33,710 ∠ 25° Hence, sending-end voltage is 33.71 kV
( ii) As seen from Fig. 41.22, α = 0°25 ′ Sending-end p.f. angle is
p.f. = cos 37°17 ′ = 0.795 (lag).
Fig. 41.22
A.C. Transmission and Distribution
Note. As seen from Fig. 41.22, approximate line drop is
0.6) = 709 V ∴
I (R cos φ R + X sin φ R ) = 41.7 (10 ×
33,000 + 709 = 33,709 V —as above Example 41.10. What is the maximum length in km for a 1-phase transmission line having
copper conductors of 0.775 cm 2 cross-section over which 200 kW at unity power factor and at 30 V can be delivered ? The efficiency of transmission is 90 per cent. Take specific resistance as
(1.725 –8 × 10 ) Ω -m.
(Electrical Technology, Bombay Univ.)
Solution. Since transmission efficiency is 90 per cent, sending-end power is 200/0.9 kW. Line loss
= (200/0.9 – 200) = 22.22 kW = 22,220 W Line current
= 200,000/3300 × 1 = 60.6 A
If R is the resistance of one conductor, then
2I 2 R = line loss or
2 × 60.62 × R = 22,220 or R = 3.03 W Now, R = 1 4 ρ ∴ 3.03 = 1.725 × 10 − 8 × l/0.775 × 10 − ∴ l = 13,600, m =
13.6 km
A Example 41.11. An industrial load consisting of a group of induction motors which aggregate
500 kW at 0.6 power factor lagging is supplied by a distribution feeder having an equivalent imped- ance of (0.15 + j0.6) ohm. The voltage at the load end of the feeder is 2300 volts.
(a) Determine the load current. (b) Find the power, reactive power and voltampere supplied to the sending end of the feeder.
(c) Find the voltage at the sending end of the feeder.
(Electrical Technology, Vikram Univ., Ujjain) Solution. ( a)
I = 500 × 103/ √ 3 = 209 A
( c) Let
V R = (2300 + j 0); I = 209 (0.6 − j 0.8)
Voltage drop = IZ = 209 (0.6 – j 0.8) (0.15 + j 0.6) = 119 + j50
V S = V R + IZ = 2300 + 119 + j 50 = 2419 + j 50 = 2420 ∠ 1.2 ′ ( b) Sending power
× 3 2420 209 0.5835 × × = 511.17 kW Sending-end reactive power =
× 3 2420 209 0.8121 × × = 711.42 kVAR
Sending-end volt ampere kVA =
2 2 2 kW + kVAR 2 = 511.17 + 711.42 = 876