A.C. Distribution

A.C. Distribution

In earlier days, d.c. system was used for the generation, transmission and distribution of electrical energy. But in case of d.c. system the voltage level cannot be changed easily unless we used rotating machinary which may not prove to be economical in many cases. This is the major disadvantage while working with d.c.

Later on with the development of transformer, a.c. system has become predominant. Now a days large power systems in the world are using a.c. system rather than d.c. because of many advantages of a.c. system.

The transmission of electrical energy generated in the power station is at very high voltage with the use of 3 phase, 3 wire system. These voltages are stepped down for distribution at the substations. There are mainly two parts of the distribution system. They are primary distribution and secondary distribution. The voltage level of primary distribution system is higher than general utilisation level. The secondary distribution systems receive power from primary distribution systems through distribution transformers. By distribution transformer voltage is stepped down to the normal working level and the consumers get the power with the voltage 400/230 V. The very commonly used a.c. distribution system is three phase four wire system as studied earlier.

 

1. A.C. Distribution Calculations

The A.C. distribution calculations and d.c. distribution calculations are different in the following respects :

1. In case of d.c. system, the voltage drop is due to resistance only which in a.c. system it is due to combined effect of resistance, inductance and capacitance.

2. The voltages or currents are added or subtracted arithmetically in case of d.c. system whereas they are added or subtracted vectorially in case of a.c. system. 

3. It is required to take into account the power factor while making calculations in a.c. system which is absent in d.c. system. The distributors are normally tapped at different points with the loads having different power factors.

There are two ways of referring the power factor.

a) The p.f. may be referred to receiving end voltage which is reference vector.

b) The p.f. may be referred to the voltage at load point itself.

By different methods the a.c. distribution problems can be solved.

The most convenient method is the symbolic notation method wherein voltages, currents and impedances are expressed in the complex notation and the calculations are similar to those in case of d.c. distribution. In a.c. calculations, addition and subtraction must be done by expressing various quantities in the rectangular form while the multiplication and the division must be done by expressing the various quantities in the polar form.

 

2. Methods of Solving A.C. Distribution Problems

As discussed in earlier section in case of a.c. distribution system we have to take into account the power factor. This power factor can be either considered with respect to receiving end voltage or with respect to load voltage itself. Let us consider each case separately.

a. Power Factors Referred to Receiving End Voltage

Consider an A.C. distribution PQ having concentrated loads of and I2 tapped off at point Q and R respectively. This is shown in the Fig. 7.13.1.

Let voltage V_Q which is the voltage at the receiving end be taken as reference vector.

The power factors at R and Q are cos ϕ₁ and cos ϕ₂ with respect to V_Q and they are lagging.

Let     R₁ = Resistance of section PR, X₁ = Reactance of section PR

R₂ = Resistance of section RQ, X₂= Reactance of section RQ

Impedance of section PR is given by, Z_PR = R₁ + j X₁

Impedance of section RQ is given by, Z_RQ = R₂ + j X₂

The corresponding diagram is shown in the Fig. 7.13.2.

As shown in the Fig. 7.13.2 the receiving end voltage V_Q is taken as reference vector. The currents I₁ and I₂ are lagging from V_Q by angles of ϕ₁ and ϕ₂ respectively. The vector sum of I₁ and I₂ gives current I_PR. The    drop I₂ R₂ is in phase with I₂ while I₂ X₂ is leading by 90°. The vector sum of V_Q, I₂ R₂ and I₂ X₂ gives V_R. The drop I_PR R₁ is in phase with current I_PR while IPR X₁ is leading by 90°. The vector sum of V_R, I_PR R₁ and I_PR X₁ gives the sending end voltage V_P.

 

3. Power Factors Referred to Respective Load Voltages

In previous section we have considered the load power factors with respect to receiving end voltage. Here we will consider these power factors with respect to their respective load voltages. Now ϕ₁ is the phase angle between VR and Ix while the angle ϕ ₂ is the phase angle between V_Q and I₂. 

The phasor diagram under this condition will be as shown in the Fig. 7.13.3.

Here again the receiving end voltage V_Q is the reference phasor. The vector sum of I₁ and I₂ gives the current I_PR.

The drop I₂ R₂ is in phase with I₂ while I₂ X2 is leading by 90°. The vector sum of V_Q,

I₂ R₂ and I₂ X₂ gives voltage V_R. The drop I_PR R₁ is in phase with current IPR while the drop IPR X-^ is leading by 90°. The vector sum of V_R, I_PR R₁ and I_PR X₁ gives the sending end voltage V_P.

Now voltage drop in section RQ is given by,

 

Example 7.13.1 A single phase distributor ’AB' 300 m long supplies a load of 200 A at 0.8 pf lagging at its far end 'B' and a load of 100 A at 0.707 pf lagging at 200 m from sending end point A. Both pf are referred to the voltage at the far end. The total resistance and reactance per km (go and return) of the distributor is 0.2 ohm and 0.1 ohm. Calculate the total voltage drop in the distributor. 

Solution : The arrangement is shown in the Fig. 7.13.4.

Magnitude of voltage drop in distributor = 17.841 V

 

Example 7.13.2 A single phase ring distributor ABC is fed at A. The loads at B and C are 40 A at 0.8 p.f. lagging and 60 A at 0.6 p.f. lagging respectively. Both power factors expressed are referred to the voltage at point A. The total impedance of sections AB, BC and CA are (2 + jl), (2 + j3) and (1 + j2) Q respectively. Determine current in each section.

Solution : The single phase ring distributor ABC is shown in the Fig. 7.13.5.

We use Thevenin's Theorem to solve this problem. First of all, let us remove section BC as shown in the Fig. 7.13.5.

Current in section AB = 40 (0.8 - j 0.6) = (32 - j 24) A 

Open circuit voltage between points B and C is given as,

Looking into network between points B and C we can find equivalent impedance.

 

Example 7.13.3 A 3 phase ring distribution ABCD fed at A at 11 kV supplies balanced loads of 40 A at 0.8 p.f. lagging at B, 50 A at 0.707 p.f. lagging at C and 30 A at 0.8 p.f. lagging at D. The load currents are referred to the supply voltage at A.

The impedance per phase of the various sections are,

Section AB = (1 + j2) Ω

Section CD = (1 + j1) Ω

Section BC = (2 + j3) Ω

Section DA = (3+ j4) Ω

Calculate the currents in various sections and station bus bar voltages at B, C and D.

Solution: The ring distributor ABCD

is shown in the Fig. 7.13.6. One phase of ring main is shown. The problem can be solved using Kirchhoff's laws. Let current in section AB be (x + jy).

Current in section BC,

Example 7.13.4 A 400 V, 3 phase 4 wire service mains supplies a star connected load. The resistance of each line is 0.1 ohm and that of the neutral 0.2 ohm. The load impedances are ZR = (6 + j 9), Zy = 8 ohms and ZB = (6 - j 8). Calculate the violtage across each load impedance and current in the neutral. Phase sequence RYB.

Solution : The arrangement is shown Fig. 7.13.7.

Total line impedances are,

Review Questions

1. Draw and explain the phasor diagram for an a.c. distributor with power factors referred to the receiving end voltage.

2. Draw and explain the phasor diagram for an a.c. distributor with power factors referred to the respective load voltages.

3. A single phase a.c. distributor AB has length cf 300 m and is fed from end A and is loaded as under :

i) 100 A at 0.707 pf. lagging 200 m from point A

ii) 200 A at 0.8 p.f. lagging 300 m from point A

The total resistance and reactance of the distributor is 0.2 Ω and 0.1 Ω per kilometer. Calculate the voltage drop in the distributor. The load power factors refer to the voltage at the far end.

[Ans.: 17.44 - j 3.81 volts]

4. A3 phase 400 V distributor AB is loaded as shown in the Fig. 7.13.8. The3 phase load at point C takes 5 A per phase at a p.f. of 0.8 lagging. At point B, a 3 phase 400 V induction motor is connected which has an output of 10 HP with an efficiency of 90 % and p.f. 0.85 lagging.

If voltage at point B is to be maintained at 400 V what should be the voltage at point A ? The resistance and the reactance of the line are 1 Ω and 0.5 Ω  per phase per km respectively.

[Ans.: 433 volts]

5. A 3 phase ring main ABCD fed at 11 kV is loaded at points B, C, D as shown in the Fig. 7.13.9. Calculate the currents in various sections and voltages at B, C and D.