3- The efficiency at unity power factor of a 6600/384V, 200KVA, 1-phase transformer is98% both at full load and half load. The power factor on no load is 0.2 and the fullload regulation at a lagging power factor of 0.8 is 4%. Draw the equivalent circuitreferred to the low voltage side and insert all values.[R,'=1082; X,'= 22.12; Reg2= 0.010; Xeq2 = 0.036Q]4- A 4KVA, 200/400V, 50HZ, single phase transformer gave the following test figures:No-load (low-voltage data): 200V, 0.7A and 60W.Short-circuit (high voltage data): 9V, 6A and 21.6W.Calculate:a- The magnetizing current and the component corresponding to iron loss at normalvoltage and frequency.b- The efficiency on full load at unity power factor.c- The secondary terminal voltage on full load at power of unity, 0.8 lagging and 0.8leading. [a/ 0.63A, 0.3A; b/97.100%; c/ 394V, 387V, 403.4V]

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The efficiency at unity power factor of a 6600/384V, 200KVA, 1-phase transformer is 98% both at full load and half load. The power factor on no load is 0.2 and the full load regulation at a lagging power factor of 0.8 is 4%. Draw the equivalent circuit referred to the low voltage side and insert all values. [R,'=1082; X,'= 22.10; Regz= 0.012; Xeq2 = 0.0362] %3D

The efficiency at unity power factor of a 6600/384V, 200KVA, 1-phase transformer is 98% both at full load and half load. The power factor on no load is 0.2 and the full load regulation at a lagging power factor of 0.8 is 4%. Draw the equivalent circuit referred to the low voltage side and insert all values. [R,'=1082; X,'= 22.10; Regz= 0.012; Xeq2 = 0.0362] %3D

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 3.57P: A two-winding single-phase transformer rated 60kVA,240/1200V,60Hz, has an efficiency of 0.96 when...

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A two-winding single-phase transformer rated 60kVA,240/1200V,60Hz, has an efficiency of 0.96 when operated at rated load, 0.8 power factor lagging. This transformer is to be utilized as a 1440/1200-V step-down autotransformer in a power distribution system. (a) Find the permissible kVA rating of the autotransformer if the winding currents and voltages are not to exceed the ratings as a two-winding transformer. Assume an ideal transformer. (b) Determine the efficiency of the autotransformer with the kVA loading of part (a) and 0.8 power factor leading.
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