Sizing of capacitor for power factor correction of system

Consider a system having following loads , we want to improve the system power factor to 0.95

Equipment	Output KW	Efficiency	Load KW	Power factor	KVA
Motor A	30	0.9	33.3	0.8	41.7
Motor B	90	0.92	97.8	0.85	115
Lighting	20	0.85	23.5	0.95	25
Heaters	10		10	1	10
Total	150		164.6		191.7

Total load in KW = 164.6

And total load in KVA = 191.7

Therefore , the average power factor of the system will be = 164.6/191.7 =0.86

As we want to improve this system power factor from 0.86 to 0.95.

Existing power factor = 0.86 = Cos?1 ; therefore tan ?1 =0.593

Required power factor = 0.95 = Cos?2; therefore tan ?2=0.328

So, amount of capacitors to be installed to improve power factor will be

= Total Load (tan ?1 – tan ?2)

= 164.6(0.593-0.328)

= 44 KVAr