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Alecsey [184]
3 years ago
6

Question 5 (20 pts) The rated current of a three-phase transmission line is 300 A. The currents flowing by the line are measured

with three current transformers each with ratio 250 : 5. What is the current you would measure at the secondary of each current transformer if rated current is flowing through each line. (of course the secondaries of the transformer must be part of closed circuits.) Consider the measured currents are exciting three over-current relays. Should the contacts of the relay close if rated current is circulating in the lines? If your answer is between no and yes, the current that the relay observes when rated current flows by the line must be the minimum pick-up current Ip. Hence, the tap position you should select, e.g., in the tap block of Fig. 1, must be equal o slightly greater than Ip for an over-current relay. Slightly greater when Ip is not an integer that matches the tap selection.
Engineering
1 answer:
prisoha [69]3 years ago
3 0

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.

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An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500
max2010maxim [7]

Answer: The exit temperature of the gas in deg C is 32^{o}C.

Explanation:

The given data is as follows.

C_{p} = 1000 J/kg K,   R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

P_{1} = 100 kPa,     V_{1} = 15 m^{3}/s

T_{1} = 27^{o}C = (27 + 273) K = 300 K

We know that for an ideal gas the mass flow rate will be calculated as follows.

     P_{1}V_{1} = mRT_{1}

or,         m = \frac{P_{1}V_{1}}{RT_{1}}

                = \frac{100 \times 15}{0.5 \times 300}

                = 10 kg/s

Now, according to the steady flow energy equation:

mh_{1} + Q = mh_{2} + W

h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}

C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}

(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}

(T_{2} - T_{1}) = 5 K

T_{2} = 5 K + 300 K

T_{2} = 305 K

           = (305 K - 273 K)

           = 32^{o}C

Therefore, we can conclude that the exit temperature of the gas in deg C is 32^{o}C.

7 0
3 years ago
QUESTION 3
lianna [129]
D D D D D D D D D D D D D D D DdDdddddf
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Scorpion4ik [409]
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