Answer :
The power factor for a load when a 667 MVA transformer is operating at full power with an efficiency of 99.3% is 99.31%.
The question asks about the power factor of a load when a 667 MVA transformer is operating at full power with an efficiency of 99.3%. The power factor is a measure of how effectively the current is being converted into useful work output. It is defined as the ratio of real power (measured in watts) to apparent power (measured in volt-amperes), and can be expressed as a percentage or a decimal.
Since the transformer's rated capacity is 667 MVA (apparent power) and the efficiency is 99.3%, we can calculate the real power (P) delivered to the load by multiplying the apparent power by the efficiency:
P = apparent power imes efficiency = 667 MVA imes 0.993 = 662.01 MW
The real power in megawatts (MW) is the product of the apparent power in MVA and the power factor (pf):
Real Power (MW) = Apparent Power (MVA) imes Power Factor (pf)
Solving for the power factor, we get:
Power Factor (pf) = Real Power / Apparent Power = 662.01 MW / 667 MVA = 0.9931 or 99.31%
