Answer:
Check the explanation
Explanation:
Question 1.
The secondary current of 250/5 amps CT when 300 amps(rated current of transmission line ) flow in TL is
(5/250 ) X 300 = 6 amps
Question 2
The correct answer to this second question is yes, when Over current relay coil will operate and relay contacts gets close, if the pickup value( Ip) of relay is set as 6 amps in relay. ( because primary current of TL is 1.2 times of CT primary)
Question 3
Tap Block figure (Fig 1) is not available/uploaded in your question.
Answer:
The field strength needed is 0.625 T
Explanation:
Given;
angular frequency, ω = 400 rpm = (2π /60) x (400) = 41.893 rad/s
area of the rectangular coil, A = L x B = 0.0611 x 0.05 = 0.003055 m²
number of tuns of the coil, N = 300 turns
peak emf = 24 V
The peak emf is given by;
emf₀ = NABω
B = (emf₀ ) / (NA ω)
B = (24) / (300 x 0.003055 x 41.893)
B = 0.625 T
Therefore, the field strength needed is 0.625 T
In the United States, fire codes are developed primarily by two model code organizations, the International Code Council (ICC) and the National Fire Protection Association (NFPA).
s 0Miles (short), 150 Miles(medium), and 300 Miles (long).
Explanation:
Answer:
Explanation:
The turbine is modelled after the First Law of Thermodynamics:
The work done by the turbine is:
The properties of the water are obtained from property tables:
Inlet (Superheated Steam)
Outlet (Superheated Steam)
The work output is:
