Tripura University 5th Semester, Diploma Electrical Engg. Exam(Mid Term),Sub-TRANSMISSION & DISTRIBUTION OF POWER, Tripura Institute of Technology model question papers



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Posted Date: 19 Nov 2010      Posted By:: Rupesh Das    Member Level: Silver  Points: 5 (₹ 1)

2010 Tripura University Diploma Electrical Engineering (DEE) 5th Semester, Diploma Electrical Engg. Exam(Mid Term),Sub-TRANSMISSION & DISTRIBUTION OF POWER, Tripura Institute of Technology Question paper



Course: Diploma Electrical Engineering (DEE)   University/board: Tripura University





TRANSMISSION & DISTRIBUTION OF POWER
3rd Year (Electrical)
5th Semester, Diploma Electrical Engineering

Time- 1/2 Hrs Full Marks-15

Group-B
Answer Question No. 1 and two from the rest.

1. Choose the correct answer from the given alternatives[Marks 3x1=3]
a) Power factor is define as_________.
(i)KVA/KW, (ii)KW/KVA, (iii)VI/Watt, (iv)KVAR/KVA

b) Service connection is given from____
(i)feeder, (ii)Distribution, (iii)either from feeder or distribution, (iv)from none of them

c) The size of feeder is based on ____
(i)Voltage, (ii)Voltage drop, (iii)Current carrying capacity, (iv) frequency.


2. a) Discuss the term voltage regulation and transmission efficiency as applied to transmission line.[Marks 2]

b) An overhead 3-phase transmission line delivers 5000Kw and 22KV at 0.8 p.f. lagging. The resistance and reactance of each conductor is 4? and 6? respectively.Determine---
(i)Sending end voltage, (ii)Percentage regulation,(iii)Transmission efficiency.[Marks 4]

3. a) State the advantage of low power factor.
b) A synchronous motor improves the power factor of a load of 250KW from 0.8 lagging to 0.95 lagging.Simultaneously the motor carries a load of 80kw.Find-(i)the leading KVAR taken by the motor,(ii)KVA rating of the motor,(iii) power factor at which the motor operates.[Marks 2+4=6]

4. a) Define the terms----(i)feeder,(ii) Distribution.[Marks 2]
b) A single phase a.c distribution AB is fed from end A and has a total impedance of (0.1+j0.2)ohm. At the far end, the voltage Vb=240v and the currents 1000A at a p.f. of 0.8 lagging.At the mid point M, a current of 150A is tapped at a p.f of 0.6 lagging with reference to the voltage at the far end.Calculate the supply voltage VA and phase angle between VA and VB.[Marks 4]











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