It is estimated that in 2010 Kuwait will consume 4021 billio

It is estimated that in 2010, Kuwait will consume 40.21 billion kW-h of electricity, all of it coming from the 40%-efficient power plants the country currently has. Estimate the amount of thermal energy rejected by the power plants this year.

Solution

>> Energy Consumed, E1 = 40.21 billion kW-h = 40.21*10⁹ kW-h

>> As, Efficiency of power plant = 40 % = 0.4

So, Total Energy available, E = E1/0.4 = 100.525*10⁹ kW-h

So, 60 % Energy is rejected by thermal plant

>> So, Total Thermal Energy rejected by thermal plant = 0.6*E = 60.315*10⁹ kW-h

>> In Year 2010, Total No of Hours are = 365*24 = 8760 hr

So, Total Energy released by Thermal Power plants in Year 2010 = 60.315*10⁹ * 8760 = 5.284*10¹⁴ kW ..ANSWER...