Homework #2 (Due 2/18, Wednesday)
1. A single-phase 50-kVA, 2400/240-volt, 60-Hz distribution
transformer is used as a step-down transformer at the load end of a
2400-volt feeder whose series impedance is 1+j2 . The equivalent
series impedance of the transformer is 1+ j2.5 referred to the
high-voltage (primary) side. The transformer is delivering rated
load at 0.8 power factor lagging and at rated secondary voltage.
Neglect the transformer exciting current. (a) Find the voltage at
the transformer primary terminals. (b) Find the voltage at the
sending end of the feeder. (c) Find the real and reactive power
delivered to the sending end of the feeder.
2. Three single-phase transformers, each rated 25 MVA, 38.1/3.81
kV, are connected Y- with a balanced load of three 0.6-,
Y-connected resistors. Power base is 75 MVA and voltage base (base
VLL) on the high voltage side is 66 kV. For the following
questions, you need to understand the table with four connections
shown on page 29 of lecture note #3. (a) Explain why 66 kV is
chosen as the voltage base (base VLL) for the high voltage
side.
(b) Which is the voltage base for the low-voltage side, 3.81 kV
or . kV? (c) Find the per-unit resistance of the load on the base
for the low-voltage side. (d) Find the load resistance RL both in
and in per unit referred to the high-voltage side.
3. The below figures show a three-phase generator rated 300 MVA,
23 kV supplying a system load of 240 MVA, 0.9 power-factor lagging
at 230 kV through a 330-MVA 23/230Y-kV step-up transformer of
leakage reactance 11%. Neglect the transformer exciting current and
choose base values at the load of 100 MVA and 230 kV. (a) Find IA,
IB, and IC in per unit with VAN as reference. (b) Find Ia, Ib, and
Ic in per unit. (c) Find the transformer reactance in per unit. (d)
Find Van both in per unit and in actual value.
4. Two transformers are connected in parallel to supply an
impedance to neutral per phase of 0.8 + j0.6 per unit at
a voltage of V2 = 1.0 per unit. Transformer Ta has a voltage
ratio equal to the ratio of the base voltages on the two sides of
the transformer. Ta has an impedance of j0.1 per unit on the
appropriate base. The second transformer Tb also has an impedance
of j0.1 per unit on the same base but has a step-up toward the load
of 1.05 times that of Ta (secondary windings on 1.05 tap). The
below figure shows the equivalent circuit with transformer Tb
represented by its impedance and the insertion of a voltage V. (a)
Find the load current ILoad if the switch S is closed. (b) Find the
loop or circulation current Icirc if the switch S is open. (c) Find
the transformer Ta current ITa (= 0.5*ILoad - Icirc) and the
transformer Tb current ITb (= 0.5*ILoad + Icirc).
*The superposition principle is applied. (d) Find the complex
power transmitted to the load through each transformer of Ta and Tb
using V2, ITa, and ITb.
