The 42-kVA balanced load depicted in Fig. 13.50 is supplied by a three-phase transformer. (a) Determine the type of transformer connections. (b) Find the line voltage and current on the primary side. (c) Determine the kVA rating of each transformer used in the transformer bank. Assume that the

Question

The 42-kVA balanced load depicted in Fig. 13.50 is supplied by a three-phase transformer. (a) Determine the type of transformer connections. (b) Find the line voltage and current on the primary side. (c) Determine the kVA rating of each transformer used in the transformer bank. Assume that the transformers are ideal.

Accepted Answer

(a) A careful observation of Fig. 13.50 shows that the primary side is Y-connected, while the secondary side is Δ-connected. Thus, the three-phase transformer is Y-Δ, similar to the one shown in Fig. 13.48. (b) Given a load with total apparent power [latex]S_T= 42 kVA[/latex], the turns ratio n  =  5, and the secondary line voltage [latex]V_{Ls}  =  240 V[/latex], we can find the secondary line current using Eq. (13.69a), by

(13.69a): [latex] S_T=\sqrt{3} V_L I_L [/latex]

[latex] I_{L s}=\frac{S_T}{\sqrt{3} V_{L s}}=\frac{42,000}{\sqrt{3}(240)}=101 A [/latex]

From Eq. (13.71),

(13.71a): [latex] V_{L s}=\frac{n V_{L p}}{\sqrt{3}}[/latex]
(13.71b): [latex] I_{L S}=\frac{\sqrt{3} I_{L p}}{n} [/latex]

[latex] I_{L p}=\frac{n}{\sqrt{3}} I_{L s}=\frac{5 imes 101}{\sqrt{3}}=292 A \ \space \ V_{L p}=\frac{\sqrt{3}}{n} V_{L s}=\frac{\sqrt{3} imes 240}{5}=83.14 V [/latex]

(c) Because the load is balanced, each transformer equally shares the total load and since there are no losses (assuming ideal transformers), the kVA rating of each transformer is [latex]S = S_T/3 = 14 kVA[/latex]. Alternatively, the transformer rating can be determined by the product of the phase current and phase voltage of the primary or secondary side. For the primary side, for example, we have a delta connection, so that the phase voltage is the same as the line voltage of 240 V, while the phase current is [latex]I_{L_p}/\sqrt{3}= 58.34 A.[/latex] Hence, S = 240 × 58.34 = 14 kVA.

Question 13.12: The 42-kVA balanced load depicted in Fig. 13.50 is supplied ...

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