Q1- the primary voltage of transformer is 6600 V and the turn ratio equal 10.6. If the primary and secondary impedances are Z₁=1.4+j5.2. Z-0.0117 +10.0465 respectively. Find the output power when the load draw 300 A at 0.8 lag p.f. ANS. 144 KW
Q1- the primary voltage of transformer is 6600 V and the turn ratio equal 10.6. If the primary and secondary impedances are Z₁=1.4+j5.2. Z-0.0117 +10.0465 respectively. Find the output power when the load draw 300 A at 0.8 lag p.f. ANS. 144 KW
Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter3: Power Transformers
Section: Chapter Questions
Problem 3.28MCQ
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![Q1- the primary voltage of transformer is 6600 V and the turn ratio equal 10.6. If the
primary and secondary impedances are Z₁=1.4+j5.2. Z-0.0117+10.0465 respectively.
Find the output power when the load draw 300 A at 0.8 lag p.f.
ANS. 144 KW
Q2-100 KVA. (6600/330) ,single-phase transformer, the primary connected to supply
voltage 100 V and the secondary side short-circuit. The transformer drawn 10 A and 436
watt. calculate the voltage applied to primary winding at full load with 0.8 lag p.f to make
secondary voltage 330 V.
ANS. 6734 V
Q3-A100 KVA transformer has its maximum efficiency of 0.98 at full load with unity
power factor During the day it is loaded as follows:
For 12 hours
20 Kw at 0.5 p.f lag
For 6 hours
45 Kw at 0.9 p.flag
80 Kw at 0.8 p.f lag
For 6 hours
Determine its all day efficiency.
Q4-A10 KVA transformer During the day it is loaded as follows:
For 6 hours
full load at 0.9 p.f lag
For 8 hours
halve load at 0.8 p.f lag
no load
For 10 hours
If the iron loss equal 60 watt and copper loss equal 100 watt, determine its all day
efficiency.
ANS. 97.4%
Q5-A 800 KVA transformer During the day it is loaded as follows:
For 4 hours
500 KW at 0.8 p.f lag
For 4 hours
700 KW at 0.9 p.flag
For 4 hours
300 KW at 0.95 p.f lag
If the iron loss equal 7.5Kw and copper loss equal 14.2 KW, determine its all day
efficiency.
ANS. 95.86%](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F86528ab3-021b-461b-b2fc-44377282e0d6%2Fcaab06a8-bcb9-461b-9efc-fa833f2ff592%2F0a8266s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Q1- the primary voltage of transformer is 6600 V and the turn ratio equal 10.6. If the
primary and secondary impedances are Z₁=1.4+j5.2. Z-0.0117+10.0465 respectively.
Find the output power when the load draw 300 A at 0.8 lag p.f.
ANS. 144 KW
Q2-100 KVA. (6600/330) ,single-phase transformer, the primary connected to supply
voltage 100 V and the secondary side short-circuit. The transformer drawn 10 A and 436
watt. calculate the voltage applied to primary winding at full load with 0.8 lag p.f to make
secondary voltage 330 V.
ANS. 6734 V
Q3-A100 KVA transformer has its maximum efficiency of 0.98 at full load with unity
power factor During the day it is loaded as follows:
For 12 hours
20 Kw at 0.5 p.f lag
For 6 hours
45 Kw at 0.9 p.flag
80 Kw at 0.8 p.f lag
For 6 hours
Determine its all day efficiency.
Q4-A10 KVA transformer During the day it is loaded as follows:
For 6 hours
full load at 0.9 p.f lag
For 8 hours
halve load at 0.8 p.f lag
no load
For 10 hours
If the iron loss equal 60 watt and copper loss equal 100 watt, determine its all day
efficiency.
ANS. 97.4%
Q5-A 800 KVA transformer During the day it is loaded as follows:
For 4 hours
500 KW at 0.8 p.f lag
For 4 hours
700 KW at 0.9 p.flag
For 4 hours
300 KW at 0.95 p.f lag
If the iron loss equal 7.5Kw and copper loss equal 14.2 KW, determine its all day
efficiency.
ANS. 95.86%
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