Answer to Question #224140 in Mechanical Engineering for Martha

Question #224140

Problem 18.

A 400 kVA transformer has a

primary winding resistance of 0.5 and a

secondary winding resistance of 0.001. The iron

loss is 2.5 kW and the primary and secondary

voltages are 5 kV and 320 V, respectively. If the

power factor of the load is 0.85, determine the

efficiency of the transformer (a) on full load, and

(b) on half load.



1
Expert's answer
2021-08-11T07:28:30-0400

a) transformer rating =400kVA=V1I1=V2I2

Primary current,I1=

Secondary current,I2=

Total copper loss=I12R1+I22R2

Where R1= and R2=

=(80)2(0.5)+(1250)2(0.001)

=3200+1562.5

=4762.5watts

On full load, total loss=copper loss+iron loss

=4762.5+2500=7262.5W

=7.2625kW

Total output power on full load=

=400*103*0.85=340kW

Input power=output power+losses

=(340+7.2625)kW=347.2625kW

Efficiency, =

=

=97.91%


b)total copper loss on half load=4762.5*(0.5)2

=1190.625watts

Total loss on half load=(1190.625+2500)watts=3690.625watts

=3.691kW

Output power on half load=0.5*340=170kW

Input power on half load=output power on half load+losses

=(170+3.691)kW=173.691kW

Efficiency,

=

=97.87%


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