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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

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)
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%
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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