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# Transformer Riddle No. 27 – Power transformer bank capacity

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• #346
Alvina

Why cannot we use whole of parallel power transformer bank capacity for power transfer in practice?

Why cannot we use whole of parallel power transformer bank capacity for power transfer in practice?

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• #1653
Hamid

For determination of distributed load on many parallel transformers with different rating power and impedance voltage, we can use following equations: S = Total load S1 = Rated KVA of TR.1 S2 = Rated KVA of TR.2 S3 = Rated KVA of TR.3 Uk=Resultant impedance of power transformers bank Uk1 = Impedance voltage of TR.1 Uk2 = Impedance voltage of TR.2 Uk3 = Impedance voltage of TR.3 S/Uk = S1/Uk1 + S2/Uk2 + S3/Uk3 The power assumed by the individual transformers is : Sa1 = S1. Uk/Uk1 Sa2 = S1. Uk/Uk2 Sa3 = S1. Uk/Uk3 For example: S1 = 3500 KVA , Uk1= %6 S2 = 2500 KVA , Uk2= %5 S3 = 1500 KVA , Uk1= %4 S = 6750 KVA We can write: Uk= 6750 /( 3500/6 + 2500/5 +1500/4) = % 4.63 Sa1 = 3500 x 4.63/6 = 2700 KVA Sa2 = 2500 x 4.63/5 = 2315 KVA Sa3 = 1500 x 4.63/4 = 1736 KVA > 1500 KVA (The individual transformer should not be subjected to overload) Transformer No.3 is thus overloaded, therefore we cannot transfer whole of actual load with three mentioned transformers. Available transfer load is: 2700 + 2315 + 1500 = 6515 KVA

#1654
R K Mohapatra

Impedence implies regulations . Different impedence means different regulations i.e different output terminal voltages for same input voltage . When the 3 trfr.s with differnt impedence i.e different terminal volatges , there will be a circulating current bett the trfr.s that will reduce the outward power flow . If X MVA of power is fed to the individual trfr.s and circulating current owind to variation of terminal voltage calls for y MVA of power , then the total outward power flow will be X-Y MVA . Tht is why normally it is not advisable to put trfr.s with different % impedence in parallel . Morever , the circulating current is likely to increase the thermal loading and load loss of the trfr.s forming the loop through which the circulating current flows . In practice , there is always some difference in impedence of the transformers in parallel causing difference in power flow at input and output.

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