DISCUSSION

1) From the result obtained above, draw the equivalent circuit of the transformer referred to the primary side.

Req= 2.177 Ω, Xeq= j5.523 Ω

Rc= 1008 Ω, Xm = j66.67

2) Recalculate the efficiency of the transformer using the general formula that you had written down in section (6) of your pre-lab work for full-load, 75% load and half load.

At rated (Is = 5.25 A)

Copper Loss, Pcu = Pscϕ

= 60 W

Core Loss, Pc = Pocϕ

= 5.33 W

Full Load Efficiency

= PsPs+ Pcu+ Pc x 100 %

= 607607+60 +5.33 x 100 %

= 90.28 %

75 % Load Efficiency

= (0.75)Ps(0.75)Ps+ 0.75(0.75)Pcu+ Pc x 100 %

= 455.25455.25+ 33.75 +5.33 x 100 %

= 98.04 %

50% Load Efficiency

= (0.5)Ps(0.5)Ps+ 0.5(0.5)Pcu+ Pc x 100 %

= 303.5303.5+15+5.33 x 100 %

= 93.72 %

3) Compare the values of the efficiencies obtained in section (3) of your results with recalculated values in section (3) above. Comment on your findings.

There is a difference in the efficiency for all the three condition which are full-load, 75% load and half-load. From the graph, the efficiencies for the full-load, 75% load and half-load are 96%, 98.5% and 96.8% respectively. While the calculated efficiency for all these three conditions are 90.28%, 98.04% and 93.72% respectively. The calculated value for the efficiency is done in part 2) discussion and theoretically the calculated value is more precise than experimented value. This happened because there are some errors while conducting the experiment whether human error or environment error or there is an error at the equipment itself.

4) Determine the maximum efficiency of the transformer.

n= PcPcu

n= 5.33360=0.298

ŋmax= 1820(0.298)18200.298+0.298(0.298)60+0.298(5.333)×100%=98.74%

5) Calculate P∆ and PV. The values should be same as the wattmeter readings. Determine the ratio of PV/ P∆ .

Pv=2VIcos30°=2 ×127 ×2.5×cos30°=549.92 W W

PΔ = 3VI = 3× 127 × 2.5 = 925.5 W

PvPΔ= 549.92925.50 = 0.594

1) From the result obtained above, draw the equivalent circuit of the transformer referred to the primary side.

Req= 2.177 Ω, Xeq= j5.523 Ω

Rc= 1008 Ω, Xm = j66.67

2) Recalculate the efficiency of the transformer using the general formula that you had written down in section (6) of your pre-lab work for full-load, 75% load and half load.

At rated (Is = 5.25 A)

Copper Loss, Pcu = Pscϕ

= 60 W

Core Loss, Pc = Pocϕ

= 5.33 W

Full Load Efficiency

= PsPs+ Pcu+ Pc x 100 %

= 607607+60 +5.33 x 100 %

= 90.28 %

75 % Load Efficiency

= (0.75)Ps(0.75)Ps+ 0.75(0.75)Pcu+ Pc x 100 %

= 455.25455.25+ 33.75 +5.33 x 100 %

= 98.04 %

50% Load Efficiency

= (0.5)Ps(0.5)Ps+ 0.5(0.5)Pcu+ Pc x 100 %

= 303.5303.5+15+5.33 x 100 %

= 93.72 %

3) Compare the values of the efficiencies obtained in section (3) of your results with recalculated values in section (3) above. Comment on your findings.

There is a difference in the efficiency for all the three condition which are full-load, 75% load and half-load. From the graph, the efficiencies for the full-load, 75% load and half-load are 96%, 98.5% and 96.8% respectively. While the calculated efficiency for all these three conditions are 90.28%, 98.04% and 93.72% respectively. The calculated value for the efficiency is done in part 2) discussion and theoretically the calculated value is more precise than experimented value. This happened because there are some errors while conducting the experiment whether human error or environment error or there is an error at the equipment itself.

4) Determine the maximum efficiency of the transformer.

n= PcPcu

n= 5.33360=0.298

ŋmax= 1820(0.298)18200.298+0.298(0.298)60+0.298(5.333)×100%=98.74%

5) Calculate P∆ and PV. The values should be same as the wattmeter readings. Determine the ratio of PV/ P∆ .

Pv=2VIcos30°=2 ×127 ×2.5×cos30°=549.92 W W

PΔ = 3VI = 3× 127 × 2.5 = 925.5 W

PvPΔ= 549.92925.50 = 0.594