WHAT IS WAVE TRAP AND USE OF WAVE TRAP

What is wave Trap and uses of wave trap ? Here is the answer to the question from our follower.

Wave trap is a device used in the substation at the incoming lines.Wave trap is also called as line trap.As the name indicates a wave trap, is used to trap/catch the wave.But what is that wave and why it should be trapped/caught by  a wave trap ?

Have you ever heard of Power Line Carrier Communication (PLCC) ? The substations uses the power line for communication purpose also. They use high frequency waves to communicate between the substations. Communication between the substation was done using the power lines in the days where communication through the telephone/communication lines were difficult. But still PLCC is adopted in the substations as it is faster, easier,  cost effective and also very safe to transmit the messages without any interference.

Are you confused, whether it is possible to send 2 waves in a single transmission line ? Yes, it is possible the power signal will be at the frequency of 50 Hz whereas the communication signal's  frequency would be in the range of kHz.

What happens when wave trap is not used at the entrance of substation ? If the communication frequencies are not filtered, then it would act as harmonics to the electrical equipments like transformers, circuit breakers and damage the equipments.

So in order to protect the electrical equipments and for communication purpose the wave trap is used in the substation. I have attended industrial training at a substation and I have seen these wave traps directly and how the maintenance works are done at substation. I would suggest you students also to attend training like this and gain knowledge. I am sure it would be a nice fun and good practical learning experience.

ONLINE ELECTRICAL ENGINEERING TEST - 4

Hi all.Please take up the Electrical Engineering Test-4.The test contains the questions which were asked in some competitive exams.Some questions are problematic and others are theoretical. Though most of the questions are theoretical it would be helpful in understanding the basic concepts of electrical engineering.We have been preparing questions ourselves, so that it is useful for all Electrical Engineers.Online Electrical Engineering team will be coming up with more tests :) Your valuable comments are welcome to improve the quality of the tests.

Cheers !!

TIME LEFT :

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1. A consumer with the industrial load has a daily load pattern of 2000 kW,.8 lag for 12 hours and 1000 kW at UPF for 12 hours.What is the load factor ?.

   A) .65

   B) .6

   C) .70

   D) .75

ANSWER :

A) .65
EXPLANATION :
Load Factor = (Average load * time) / (maximum load * time)
= ((2000*.8*12) + (1000*12)) / (2000*24) = 31200/ 48000 = .65

2. What is the maximum load which consume 500kWh per day at a load factor of .40, if the consumer increases the load factor of .50 without increasing the maximum demand ?

   A) 50.08 kW

   B) 52.08 kW

   C) 62.08 kW

   D) 60 kW

ANSWER :

B) 52.8.
EXPLANATION :
Load Factor = (Average load * time) / (maximum load * time)
.40 = ((500 * 1000) / (maximum load*24) maximum laod = 500000/ (.4*24)= 52.0833 kW

3. A consumer consumes 600 kWh per day at a load factor of .40. What is the energy comsumed by the consumer if the load factor is increased to .70 without increasing the maximum demand ?

   A) 1100 kWh

   B) 950 kWh

   C) 1050 kWh

   D) 1000 kWh

ANSWER :

C) 1050 kWh
EXPLANATION :
Load Factor = (Average load * time) / (maximum load * time)
.40 = ((600 * 1000) / (maximum load*24) maximum laod = 600000/ (.4*24)= 62.5 kW Energy consumed in kWh = Load Factor * Maximum Energy Consumed = .70 * 62.5 * 24 = 1050 kWh

4. A loss less line terminated with its surge impedance has

   A) transmision line angle is greater than actual length of line

   B) flat voltage profile

   C) transmission line angle is less than the actual length of line

   D) both A) and B)

ANSWER :

B) flat voltage profile

5. An RLC series circuit remains predominantly inductive

   A) above resonance frequency

   B) at resonance frequency

   C) below resonance frequency

   D) at the lower half power frequency

ANSWER :

A) above resonance frequency

6. If the excitation of the synchronous generator fails, it acts as a

   A) induction motor

   B) synchronous motor

   C) synchronous generator

   D) induction generator

ANSWER :

D) induction generator

7. The use of high speed circuit breakers

   A) increases system stability

   B) reduces short circuit current

   C) increases short circuit current

   D) decreases system stability

ANSWER :

A) increases system stability

8. Resistance switching is normally employed in ?

   A) bulk oil breakers

   B) minimum oil breakers

   C) all breakers

   D) air blast circuit breakers

ANSWER :

D) air blast circuit breakers

9. Plug setting of a relay can be altered by varying

   A) adjustable back up stop

   B) air gap of magnetic path

   C) number of ampere-turns

   D) none of these

ANSWER :

C) number of ampere-turns

10. If the time of operation of a relay for unity TMS is 12 secs, the time of operation for 0.6 TMS will be

   A) 20 secs

   B) 19.2 secs

   C) 8 secs

   D) 7.2 secs

ANSWER :

D) 7.2 secs

SUBMIT TEST

ELECTRICAL ENGINEERING QUESTIONS AND ANSWERS FOR INTERVIEW

Can I plug a 10.7kW cooker into a 13A socket ? is today's electrical engineering question asked by one our followers.Before answering this question let us see some of the basics concepts in Electrical Engineering which would obviously give answer to this question.

What is the voltage used in India for domestic purpose ? the answer to this is 240V with 50Hz frequency. So all the appliances/generators will be manufactured in such a way that it supports the 50hz frequency and the 240V.

What is the voltage used in USA for domestic purpose ? the answer to this question would be 120V with 60 Hz frequency. So the appliances manufactured should be supporting these specifications.

Having sorted out the voltage we use, now let us see how much power/ Current would our appliance draw based on our usage.The question states that the rated power of the cooker is 10.7kW.So let us assume it will work on 240V/50Hz supply.

Power = V* I 

10700 = 240 * I

I = 10700/240 =44. 583 A

So if we want to plug the cooker and use it in a 240V supply it would require a 45A socket or a 50A socket.What would happen if you plug it in a 13A socket ? the answer is simple but not the real outcome of it as the socket would blast because it cannot withstand only 45A drawn by the appliance. 

Have you ever seen the sockets used for the home appliances like TV, Refrigerators, Air conditioners, Geysers ? 

For TV mostly we use 5A socket, for Refrigerators we use 5A - 10A socket based on the capacity of the Refrigerators. For AC we use 20A socket. And for Geysers we use a 25A socket. Again remember that the Ampere rating of the socket may vary with respect to the capacity of the appliance. What are you waiting for ???!! Just jump up from your PC and check the Ampere rating of the socket you use in your home and post it in comment.. Any US visitors also can post you comment in the comments section to enlighten what is practically used in US. Cheers guys !!