EBK ELECTRICAL TRANSFORMERS AND ROTATIN
4th Edition
ISBN: 9781337025867
Author: Herman
Publisher: VST
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Chapter 11, Problem 2RQ
What is one advantage of shell-type construction in small control transformers?
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In a 15KVA, 2000 / 200 V power transformer, the iron and full load copper losses are 370W and 415W
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Efficiency at Full load =
Efficiency at Half load =
Efficiency at Quarter load =
Discuss the importance of using (1
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electrical transformer in terms of power
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components, working principle, and
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In a 20KVA, 2000 / 200 V power transformer, the iron and full load copper losses are 355W and 405W respectively. Calculate the efficiency at (i) full load, 0.8 pf,
(i) half load, 0.65 pf, (ii) Quarter load, 0.55 pf
Efficiency at Full load =
Efficiency at Half load =
Efficiency at Quarter load =
Chapter 11 Solutions
EBK ELECTRICAL TRANSFORMERS AND ROTATIN
Ch. 11 - What are the three main sources of heat in a...Ch. 11 - What is one advantage of shell-type construction...Ch. 11 - How are larger control transformers supplying a...Ch. 11 - What is a way to increase cooling on a large...Ch. 11 - What is one special problem with dry-type...Ch. 11 - What are two types of cooling medium used in...Ch. 11 - Name two functions of transformer oil.Ch. 11 - What type of coolant/insulator is banned in the...Ch. 11 - Name two ways of increasing the cooling of...Ch. 11 - What hazard is involved in water-cooled...
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- What are two types of cooling medium used in transformers?arrow_forwardTwo transformers rated at 100 kVA each are connected in an open-delta connection. What is the total output power that can be supplied by this bank?arrow_forwardHow are larger control transformers supplying a motor control center usually cooled?arrow_forward
- 10. The output voltage of a turbine is 224 V at no load and 218 V at full load. How much is the copper loss in the transformer if the current is 1000 A at full load, assuming that the entire copper loss materializes inside the transformer?arrow_forwardExample 7-3. A 460-V, 25-hp. 60-Hz, four-pole, Y-connected induction motor has the following impedances in ohms per phase referred to the stator circuit: - 0.641n X,-1.106 N R2 - 0.332 N X2 = 0.464 XM - 26.3 N The total rotational losses are 1100 W and are assumed to be constant. The core loss is lumped in with the rotational losses. For a rotor slip of 2.2 percent at the rated voltage and rated frequency, find the motor's (a) Speed (b) Stator current (c) Povwer factor (d) Peony and Pout (e) Tiad and Toad Efficiencyarrow_forwardWhat factors govern the type of maintenance procedures and the maintenance intervals for transformers?arrow_forward
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