4.2 Eskom uses a 1MVA transformer to supply electricity to Willows community. The electricity cost price is R2.50 per kWh. The load supplied by the transformer varies as shown below. The purchasing price of the transformer is R350 000. The transformer has a full load copper loss of 2.2 kW and the iron loss of 4.18kW. A certain load varies as follows for 269 days per annum:  550 kVA at a power factor of 0.819 lagging for eleven hours per day.  120 kVA at a power factor of 0.836 lagging for six hours per day.  450 kVA at unity power factor for remaining hours of a day.  For the remaining time of the year, the load varies as follows: ➢ 290 kVA at a power factor of 0.822 lagging for eight hours per day. ➢ 25 kVA at a power factor of 0.866 lagging for eight hours per day. ➢ No load for eight hours per day.   50.Determine the average value of the load per annum (kWh).

Power System Analysis and Design (MindTap Course List)
6th Edition
ISBN:9781305632134
Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma
Chapter3: Power Transformers
Section: Chapter Questions
Problem 3.10P: A single-phase step-down transformer is rated 13MVA,66kV/11.5kV. With the 11.5 kV winding...
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4.2 Eskom uses a 1MVA transformer to supply electricity to Willows community. The electricity cost price is R2.50 per kWh. The load supplied by the transformer varies as shown below. The purchasing price of the transformer is R350 000. The transformer has a full load copper loss of 2.2 kW and the iron loss of 4.18kW.

A certain load varies as follows for 269 days per annum:
 550 kVA at a power factor of 0.819 lagging for eleven hours per day.
 120 kVA at a power factor of 0.836 lagging for six hours per day.
 450 kVA at unity power factor for remaining hours of a day. 
For the remaining time of the year, the load varies as follows:
➢ 290 kVA at a power factor of 0.822 lagging for eight hours per day.
➢ 25 kVA at a power factor of 0.866 lagging for eight hours per day.
➢ No load for eight hours per day.

 

50.Determine the average value of the load per annum (kWh).

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