Answer / vishwas g. darne

You can use the the degree of P. F. Correction your load is
25 h.p. i.e. 25*746=18650-18.65kW present p.f.is .54 require p.f. is 0.9 degree of p.f. is 1.075 required kVAR per kW is 1.075 Therefore capacitor rating should be 1.075*18.65=20.04kVAR.  

Answer / muruganantham.n

decrease machine air gap density

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