A single Phase 50kVA, 2400/240V, 60Hz distribution transformer is used as a step downtransformer at the load end of a 2400V feeder whose series impedance is (1.0 + j2.0) ohms. Theequivalent series impedance of the transformer is (1.0 + j2.5) ohms referred to the high voltage (i.e.primary) side. The transformer is delivering rated total power at 0.8 power factor lagging, and atrated secondary voltage. Neglecting the transformer excitation current, determine:(a) The voltage at the transformer primary terminals,(b) The voltage at the sending end of the feeder,(c) The real and reactive power delivered to the sending end of the feeder.Work in the Per Unit System, using the transformer ratings as base quantities

To solve this problem, we will use the per unit system and the transformer ratings as base quantities.

(a) The voltage at the transformer primary terminals:

First, we need to calculate the per unit impedance of the transformer and the feeder.

The base impedance (Zbase) on the primary side of the transformer is given by Vbase^2 / Sbase, where Vbase is the base voltage and Sbase is the base power.

So, Zbase = (2400)^2 / 50000 = 115.2 ohms.

The per unit impedance of the transformer (Zpu_t) is then the transformer impedance divided by the base impedance.

So, Zpu_t = (1.0 + j2.5) / 115.2 = 0.0087 + j0.0217.

Similarly, the per unit impedance of the feeder (Zpu_f) is (1.0 + j2.0) / 115.2 = 0.0087 + j0.0174.

The total per unit impedance (Zpu_total) is the sum of Zpu_t and Zpu_f, which is 0.0174 + j0.0391.

The per unit voltage at the primary terminals of the transformer (Vpu_t) is given by 1 - Ipu * Zpu_total, where Ipu is the per unit current.

Since the transformer is delivering rated total power at 0.8 power factor lagging, Ipu = 1 * 0.8 - j1 * 0.6.

So, Vpu_t = 1 - (0.8 - j0.6) * (0.0174 + j0.0391) = 0.970 - j0.027.

The actual voltage at the transformer primary terminals is then Vpu_t * Vbase = (0.970 - j0.027) * 2400 = 2328 - j64.8 V.

(b) The voltage at the sending end of the feeder:

The per unit voltage at the sending end of the feeder (Vpu_s) is given by Vpu_t + Ipu * Zpu_f.

So, Vpu_s = (0.970 - j0.027) + (0.8 - j0.6) * (0.0087 + j0.0174) = 0.977 - j0.044.

The actual voltage at the sending end of the feeder is then Vpu_s * Vbase = (0.977 - j0.044) * 2400 = 2344.8 - j105.6 V.

(c) The real and reactive power delivered to the sending end of the feeder:

The per unit power (Spu) is given by Vpu_s * Ipu*.

So, Spu = (0.977 - j0.044) * (0.8 - j0.6) = 0.7816 - j0.5864.

The actual power is then Spu * Sbase = (0.7816 - j0.5864) * 50000 = 39080 - j29320 VA.

So, the real power is 39080 W and the reactive power is -29320 VAR.