Consider the transformer with data given in Example 3.7.
(a) With full-load on the LV side at rated voltage, calculate the excitation voltage on the HV side. The load power factor is (i) 0.8 lagging, (ii) 0.8 leading. What is the voltage regulation of the transformer in each case?
(b) The transformer supplies full-load current at 0.8 lagging power factor with 2000 V on the HV side. Find the voltage at the load terminals and the operating efficiency.

Question

Accepted Answer

(a) The HV side equivalent circuit of Fig. 3.26(a) will be used.

[latex] V_{L}=200V[/latex], [latex] l_{L}=\frac{200 imes 1000}{200} =100A[/latex]

[latex] V'_{L}=2000V[/latex], [latex] l'_{L}=10A[/latex]

Now [latex] V_{H}=V'_{L}l'_{L}(R_{H} \cos \phi \pm X_{H} \sin \phi ):[/latex] [latex] R_{H}= 30 \Omega[/latex]

[latex] X_{H}= 5.2 \Omega[/latex]

(i) [latex] \cos \phi =0.8[/latex] lagging, [latex] \sin \phi =0.6[/latex]

[latex] V_{H} =2000+ 10(3 imes 0.8 + 5.2 imes 0.6) = 2055.2 V[/latex]

Voltage regulation [latex] =\frac{2055.2-2000}{2000} imes 100= 2.76 \% [/latex]

(ii) [latex] \cos \phi =0.8[/latex] leading, [latex] \sin \phi =0.6[/latex]

[latex] V_{H} =2000+ 10(3 imes 0.8 - 5.2 imes 0.6) = 1992.8 V[/latex]

Voltage regulation [latex] =\frac{1992.8-2000}{2000} imes 100= -0.36 \% [/latex]

(b) [latex] l_{L(full-load)} =100A, 0.8[/latex] lagging pf

[latex] V'_{L}= V'_{H}-l'_{L}(R_{H} \cos \phi +X_{H} \sin \phi )[/latex]

[latex] =2000 -10(3 imes 0.8 +5.2 imes 0.6) =1944.8 V[/latex]

or [latex] V_{L}=194.48 V[/latex]

Efficiency

Output, [latex] P_{0}=V_{L}l_{L} \cos\phi[/latex]

[latex] = 194.48 imes 100 imes 0.8 = 15558.4 W[/latex]

[latex] P_{LOSS}= P_{i}+P_{c}[/latex]

[latex] P_{i}=120W(Ex. 3.7)[/latex]

[latex] P_{c}=(10)^{2} imes 3=300W[/latex]

[latex] P_{LOSS}=420W[/latex]

[latex] \eta =1-\frac{420}{15558.4 imes 420} =97.38 \% [/latex]

Example 3.14 is solved by writing the following MATLAB code

Question 3.14: Consider the transformer with data given in Example 3.7. (a)...

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