**Medium transmission line**

Transmission line having length between 80 Km and 200 Km and line voltage between 20 KV and 100 KV fall under medium transmission lines. In medium transmission line with the increase in the length and the operating voltage shunt capacitance becomes more pronounced and the capacitive current is appreciable.Hence line capacitance is taken into account.The capacitive current is always flowing in the line while the supply end switches are closed even though the receiving end of the line may be open circuited.The capacitance of the line is uniformly distributed over its entire length.For the simple calculations,the capacitance of the system is assumed to be divided up and lumped in the form of capacitors shunted across the line at one or more point.

There are mainly three methods of representation called localized capacitance methods:

- end condenser method,
- nominal-T method(middle condenser method) and
- nominal π method(split-condenser method).

**1.End condenser method**

This method overestimates the effect of capacitance.The capacitance of the line assumed to be lumped at the load-end in end condenser method.

* fig:equivalent circuit for end condenser method*

Let the load current per phase IR lagging behind the load end phase voltage VR by an angle ΦR.

R=resistance per phase of the line in ohms

X=reactance per phase of the line ohms

c=capacitance per phase of the line in farads.

Taking receiving end phase voltage as refrence phasor,

receiving end or load end current

capacitive current

sending end current,

voltage drop in line(per phase)

phase voltage at the sending end,

**percentage voltage regulation**

**Transmission efficiency,**

* fig: phasor diagram for end condenser method*

*OA=voltage neutral at the receiving end*

*OD=load current IR that lags behind the receiving end voltage VR by ΦR.*

*DF=capacitive current Ic that leads the receiving end voltage VR by 90 degree.*

*OF=sending end current Is that flows through the line and is the phasor sum of receiving-end current IR and capacitive current IC.*

*AB=resistive voltage drop IsR that is in phase with Is.*

*BC=reactive voltage drop IsX that leads the Is by 90 degree.*

*OC=sending end voltage Vs that is the phasor sum of VR,IsR and IsX and*

*Φs=phase angle between the sending end voltage Vs and sending end current Is.*

**2.Nominal-T method**

* fig:Equivalent circuit for nominal T-method*

Line capacitance is assumed to be concentrated at the middle point of the line.Half the line resistance and reactance is lumped on the either side of the line.

V’=intermediates voltage between sending end voltage Vs and receiving end voltage VR(voltage at the terminals of condenser)

Ic=capacitive current which leads V’ by 90° flows across the first half of the line from the sending end.

IR=current in the receiving end half .

Is=current in sending end half of the linei.e phasor sum of the receiving end current (IR) and capacitance current Ic.

Taking receiving end phase voltage as reference phasor,

receiving end or load end current

voltage across the condenser,

capacitive current,

sending-end current,

sending end voltage,

**Transmission efficiency**,

* fig: phasor diagram for nominal T method*

*OA=voltage to neutral at the receiving end*

*OB=current in the receiving end half of the line*

*ΦR=load power factor angle*

*AC is drawn parallel to OB =ohmic voltage drop in the receiving end half of the line.*

*CD drawn 90 degree ahead of OB=reactive voltage drop in the receiving end half of the line.*

*OD =voltage across the capacitor.*

*BF drawn 90 degree ahead of OD=charging or capacitive current Ic.*

*OF=sending end current Is(phasor sum of OB and BF)*

*DG drawn parallel to OF =ohmic voltage drop in the sending-end half of the line.*

*GH drawn 90 degree ahead of OF=reactive voltage drop in the sending end half of the line.*

*OH=sending end voltage.*

*Φs=angle between OH and OF (phase angle at the sending voltage)*

### 3.Nominal-π method

The capacitance of each line conductor is assumed to be divided into two halves,one half being shunted between line conductor and neutral at the receiving end and the other half at the sending end.

IL=current flowing in the line at any point in between the two capacitor(phasor sum of load current IR and the current ICR drawn by the receiving end capacitor).

Taking receiving end phase voltage as reference phasor,

load end current

charging current at the receiving end,

Line current,

sending end voltage,

charging current at the sending end,

sending end current,

*fig: phasor diagram for nominal π method*

*OA=voltage to neutral at the receiving end VR*

*OB=load current IR drawn ΦR degrees behind OA*

*ΦR=load power factor angle*

*ICR leads the phasor VR by 90° and is equal to 1/2wcVR in magnitude.*

*OC=phasor sum of load current IR and the current ICR and represented .*

*AD=voltage drop in the line(phasor sum of resistive drop ILR in phase with phasor IL)*

*DF=reactive drop ILX in quadrature with IL.*

*OF=phase voltage Vs,phasor sum of receiving end phase voltage VR and the line drop.*

*OG=sending end current Is(phasor sum of IL and Ics where Ics equal to 1/2wcVs in magnitude and leads Vs by 90°)*

Good