An additional motor load of 20kW (including losses) is to be added to an existing load of 100 kVA power factor lagging. The new load my be:
a) An additional motor costing $ 200 which is fitted with loss factor capacitor costing $ 5 per kVA to improve its power its power factor form 0.8 to 0.95 lagging, or
b) A synchronous induction motor costing $ 10 per kVA which can be excited so that the overall total load kVA remains the same as be the extension of load.
Given the new load as in a) and b), calculate the total annual cost of each scheme if the tariff is $ 4 per kVA m.d 0.312 cent per kWh and the annual interest and depreciation charge for (a) is 8 per cent and for (b) is 12 per cent. Assume the loads to be constant for 3000 hours per annum and zero for the remainder of the year.
a)
Load = SL = 100 KVA with lagging power factor
an additional motor load of Pm = 20 KW is added costing $200
with loss factor capacitor to improve power factor from 0.8 to 0.95
as we know,
and
after adding motor load
which is due to additiion of capacitor
where QC is added capcitive KVAR
therefore
To calculate annual cost
Total annual cost = motor cost + KVA tariff + KWH tariff + Interest and depreciation cost
MOTOR COST
M = $200 + capacitor cost = 200 + 5*33.706 = 200 + 168.53 = $368.53
TARIFF
T1 = $4 per KVA = 4*105.263 = $421.052
and
T2 = 0.312 cent per KWh
Total hours = 3000
therefore
T2 = 0.312*100*3000 = 93600 cents = $936
INTEREST and DEPRECIATION
is 8% of motor
I&D = 0.08*368.53 = $29.4824
TOTAL ANNUAL COST
$
b)
Load = SL = 100 KVA with lagging power factor
an additional synchronus induction motor load of Pm = 20 KW is added costing $10 per KVA
as we know,
and
after adding motor load
therefore
To calculate annual cost
Total annual cost = motor cost + KVA tariff + KWH tariff + Interest and depreciation cost
MOTOR COST
M = $10 per KVA = 10*63.245 = $632.45
TARIFF
T1 = $4 per KVA = 4*100 = $400
and
T2 = 0.312 cent per KWh
Total hours = 3000
therefore
T2 = 0.312*100*3000 = 93600 cents = $936
INTEREST and DEPRECIATION
is 12% of motor
I&D = 0.12*632.45 = $75.894
TOTAL ANNUAL COST
$
An additional motor load of 20kW (including losses) is to be added to an existing load...
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