Test: Power Systems- 1 - Electrical Engineering (EE) MCQ

• Cylindrical rotor is also called as round rotor or smooth rotor, there is no projection, there a closed portion contain field winding which is distributed and unslotted portion also acts as poles .
• It has large axial length and small diameter, resulting high peripheral speed.
• As the air gap is uniform throughout the machine it is more suitable for high speed operation, employed in all thermal, nuclear power plants containing turbo alternators with two or maximum 4 poles only.
   

• If the wires are made up with some other ordinary steels, rust and corrosion attacked the wire after some time due to the environmental wet conditions.
• A layer of zinc oxide can be provided on the steel wire by some chemical process.
• This zinc oxide layer protects the steel wire from the rust and corrosion effects. That’s why galvanized steel is used in many places.

Advantages of higher transmission voltage are:

1. Power transfer capability of the transmission line is increased
   Pmax = Vs*Vr/Xs
   Where, Vs = Supply voltage
   Vr = Receiving end voltage
   If Vs = Vr,
   Pmax ∝ Vs²
2. Transmission line losses are reduced
   Power P = VI cosφ
   I = P/(V cosφ)
   Transmission line losses Pl = I²R
   Therefore, Transmission line losses Pl∝ 1/V²
3. Area of cross section and volume of the conductor is reduced
   Resistance R = ρl/A
   Where, ρ= resistivity
   l = length of the conductor
   A = Area of cross section
   Therefore area A ∝ 1/V²

P = WQHη KW

= 9.81 × 12 × 80 × 0.8

= 7534.08 KW = 7.53 MW

Given power system with these identical generators G₁ has Speed governor G₂ and G₃ has drop of 5% When load increased, in steady state generation of G₁ is only increased while generation of G₂ and G₃ are unchanged.

= 48000/60000

= 0.8

By above two equations,
Maximum demad = Installed capacity x 

Reserve capacity = Installed capacity − Maximum demand
= 420 – 360 = 60 MW

Penalty Factor = 20/15 = 4/3

Cost of received power = 

Overall efficiency, η_overall = 0.33 × 0.9 = 0.297
Coal used/annum = 
Heat of combustion = coal used/annum × calorific value

= 10⁷ × 5000 = 5 × 10¹⁰ kcal

Heat output = η_overall × heat of combustion

= (0.297) × (5 × 10¹⁰) = 1485 × 10⁷ kcal
Units generated per annum = 
Average load on station = 

At higher voltage, volume of conductor is reduced.

Area of conductor A ∝ 1/Vs²
Where Vs = Supply voltage

So that material requirement is reduced and hence cost of the material and transmission is reduced.