This document presents 4 cases analyzing power distribution systems and investment projects. Case 1 calculates losses from unbalanced and balanced loads on a feeder. Case 2 determines the negative phase sequence for unbalanced line voltages. Case 3 calculates demand charges, capacitor bank size, and costs for a consumer with low power factor. Case 4 uses net present value analysis to determine if a 3-year investment project yielding savings is acceptable.

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Power Distribution System loss reduction & Investment analysis: Some Cases
Case 1 : A 11 KV feeder of X Private Ltd. company has unbalance loads on R, Y, B phases as 120 A, 200 A, 180
A respectively. The line conductor resistance is 4 ohm. Calculate
a) Losses in unbalanced load condition of feeder.
b) Losses in balanced condition of feeder, if load is balanced.
c) Savings, if load is balanced.
Sol: a) Losses in unbalanced load condition= (120)2 x 4+ (200)2 x 4 + (180)2x4
= 4x (32400+40000+14400) =347.2 KW
b) On balancing each phase carries load current= (120+200+180) / 3=166.67 A
Losses in bal. condition= (166.67) 2 x 4 + (166.67)2 x 4 + (166.67)2 x 4
333.3 KW
Savings = (a)- (b) Watts= 347.2 KW - 333.3KW =13.9 KW
Case 2 : As a result of Unequal loads on the individual phases, negative and zero sequence components of
current causes overheating of Transformers, cables, Conductors, Motors. Losses in system also increase.
As per IS: 325-1996 (clause 4.2) NPS of 1.5 % voltage is the limit. Calculate NPS, if line voltages measured are 395
V, 415 V, and 435 V.
(Hint: NPS= (71/V av. ) x (Vhigh-1/4xVmiddle-3/4 V low)
Sol: % NPS= [71 / (395+415+435)/ 3 ] X [ (435)-1/4 x ( 415)- 3/4 x (395)]
=0.172 X 35
= 5.985 %
Case 3 : X distribution ltd. is supplying three phase power to an industrial consumer. Sanctioned load is
4577.00 KVA; Contract demand is 4577 KVA with the condition imposed by utility that consumer will maintain
the power factor of his Installation as 0.95 lag. The average power factor of consumer billed during month of
September 2017 was 0.82 due to improper maintenance of installation by consumer as a result, KVA demand
from utility grid increased to 4800 KVA.
Demand charges are INR 200 per KVA, Calculate:
a) Total KVA Charges
b) Capacity of Capacitor banks to be installed
c) Total Cost of Supply & Installation of capacitor banks, if cost Supply & Installation is INR 1000 per KVAR.
d) Simple payback period.
Sol: a) Total KVA charge=4800 x 200=Rs. 960000
b) Capacity of capacitor banks= KVA2 x Sin (phi2)-KVA1 x Sin (phi1)
=4800 x 0.572-4577 x 0.311=2745.6 KVAR - 1423.447 KVAR= 1322.153 KVAR

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(c)Total Cost of supply and installation of CB=1322.153 X 1000= Rs.1322153
(d)SPBP= (Total Cost of supply and installation of CB) / saving per month
= (1322153)/ (4800 x 200-4577 x 200) = 1322153 /44600=29.64 months
= 2.47 years
Case 4: What is the NPV of a Project, (life 3 years) which requires an investment of Rs.60, 000 & it investment
yields savings of Rs. 20,000 in first year, Rs. 30,000 in second year and Rs. 40,000 in the third year. Discount rate
is 10%.
Sol:
NPV= -CF0 / (1+d) 0+ NF1/ (1+d)1+NF2/ (1+d)2+NF3/ (1+d)3
=-60,000/ (1.1)0 +20,000/ (1.1)1+30,000/ (1.1)2+40,000/ (1.1)3
=-60,000+ 18181.818 + 24793.388 + 30052.592
= (+) Rs. 13027.798, Acceptable