Answer to Question #224140 in Mechanical Engineering for Unknown346307

Question

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.

Answer 1

a) transformer rating =400kVA=V₁I₁=V₂I₂

Primary current,I₁=$\frac{400*10^{3}}{V_1}=\frac{400*10^{3}}{5000}=80A$

Secondary current,I₂=$\frac{400*10^{3}}{V_2}=\frac{400*10^{3}}{320}=1250A$

Total copper loss=I₁²R₁+I₂²R₂

Where R₁=$0.5 \Omega$ and R₂=$0.001\Omega$

=(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=$V_2I_2Cos({\phi_2})$

=400*10³*0.85=340kW

Input power=output power+losses

=(340+7.2625)kW=347.2625kW

Efficiency,$\eta$ =$\frac{Output power}{Input power}*100\%$

=$\frac{340kW}{347.2625kW}*100\%$

=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,$\eta=\frac{output power on half load}{input power on half load}*100\%$

=$\frac{170kW}{173.691kW}*100\%$

=97.87%