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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:
In engineering design, failure is expected. It helps you find the best solutions before implementing them in the “real world”. Having a prototype fail is a GOOD thing, because that means you have learned something new about the problem and potential solutions.
Explanation:
Answer:
0.245 m^3/s
Explanation:
Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s