MGVCL Exam Paper (30-07-2021 Shift 1)
Case 1: The PV connected to the DC bus was injecting 20 kW of power constituting of 1 kW of loss Case 2: The same PV is connected to the AC bus constituting to 1.5 kW loss. Find out the change in efficiency from Case 1 to Case 2?
In DC microgrid system, Efficiency α 1/Pin For 1 kW loss injecting power is 20 kW For 1.5 kW loss injecting power will be proportional to 30 kW So, Efficiency will be reduced 0.025
Intern turn winding fault can be easily detect by differential protection relay. Also detect by buchholz relay (gas operated relay), cause buchholz relay is used to detect the internal fault of transformer.
Condition of good conductor: σ/ϵω >> 1 Equation is given by: The constant ratio = σ/ϵω = 44913.6 σ/ϵω >> 1. hence sea water is a acting as a good conductor Since attenuation is 90%, so transmission is 10% So, e^(-αx) = 0.1
Attenuation constant = √(σμϵ/2) = 0.702 -αx = ln(0.1) x = 3.27 m
Conductor with 75 °C temperature rating must have ampacity from the 60°C column or the 75°C column. It is permissible to use the 75°C rating if the installed conductor is rated at least 75 °C. Because all of the connection points in this example have at least a 75°C rating, the conductor’s ampacity can be based on the 75°C column
Bus Type - Known Parameter - Unknown Parameter Load Bus -P, Q - V, phase angle Generator Bus - P, V (magnitude) - Q, Voltage phase angle Slack Bus Voltage - magnitude and phase angle - P, Q
EXAMIANS