Question

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.

Answer 1

a)

Load = S_L = 100 KVA with lagging power factor

an additional motor load of P_m = 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 Q_C 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 = S_L = 100 KVA with lagging power factor

an additional synchronus induction motor load of P_m = 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

$