Conductor Material Required in Underground 1-Phase 2-Wire AC System with Mid-Point Earthed

 

Conductor Material Required in Underground 1-Phase 2-Wire AC System with Mid-Point Earthed

Figure-2 shows the underground single-phase two-wire AC system with mid-point earthed.

Here,

• Maximumvoltagebetweenlinewires=Vm$\mathrm{<br>}$

• RMSvalueofvoltage=Vm2√$\mathrm{<br>}$

• Powerfactoroftheload=cosϕ$\mathrm{<br>}$

Thus, the load current is given by,

I2=PVm2√cosϕ=2–√PVmcosϕ$${\mathit{<br>}}_{}$$

If 𝑎2 is the cross-section area per conductor and R2 is the resistance per conductor. Then, the total power loss in the line conductors is

W=2I22R2=2×(2–√PVmcosϕ)2×(ρla2)$$\mathit{<br>}$$

⇒W=4P2ρlV2mcos2ϕa2$$<br>$$

∴Areaofcrosssection,a2=4P2ρlWV2mcos2ϕ$$<br>\frac{}{}$$

Hence, the volume of conductor material required in the underground single-phase two-wire AC system with mid-point earthed, say K1, is given by,

K1=2a2l=2×(4P2ρlWV2mcos2ϕ)×l$${\mathit{<br>}}_{}$$

∴K1=8P2ρl2WV2mcos2ϕ⋅⋅⋅(2)$$<br>$$

Hence, from equations (1) and (2), it is clear that the volume of conductor material required in the 1-phase 2-wire underground AC system with mid-point earthed is same as that required for the 1-phase 2-wire underground AC system.