Consider a single-phase electric system shown in Figure 3.33. Transformers are rated as follows: X − Y 15 MVA , 13 . 8 / 138 kV , leakage reactance 10 % Y − Z 15 MVA , 138 / 69 kV , leakage reactance 8 % With the base in circuit Y chosen as 15MVA , 138 kV determine the per-unit impedance of the 500 Ω resistive load in circuit Z, referred to circuits Z, Y, and X. Neglecting magnetizing currents, transformer resistances, and line impedances, draw the impedance diagram in per unit.
Consider a single-phase electric system shown in Figure 3.33. Transformers are rated as follows: X − Y 15 MVA , 13 . 8 / 138 kV , leakage reactance 10 % Y − Z 15 MVA , 138 / 69 kV , leakage reactance 8 % With the base in circuit Y chosen as 15MVA , 138 kV determine the per-unit impedance of the 500 Ω resistive load in circuit Z, referred to circuits Z, Y, and X. Neglecting magnetizing currents, transformer resistances, and line impedances, draw the impedance diagram in per unit.
Solution Summary: The author illustrates a per-unit impedance diagram for the transformer X-Y and Transformer Y-Z circuits.
Consider a single-phase electric system shown in Figure 3.33. Transformers are rated as follows:
X
−
Y 15 MVA
,
13
.
8
/
138
kV
,
leakage reactance
10
%
Y
−
Z 15 MVA
,
138
/
69
kV
,
leakage reactance
8
%
With the base in circuit Y chosen as
15MVA
,
138 kV
determine the per-unit impedance of the
500
Ω
resistive load in circuit Z, referred to circuits Z, Y, and X. Neglecting magnetizing currents, transformer resistances, and line impedances, draw the impedance diagram in per unit.
A linear electrical load draws 11 A at a 0.72 lagging power factor./1 153. When a capacitor is connected, the line current dropped to 122 A and the power factor
improved to 0.98 lagging. Supply frequency is 50 Hz.
a. Let the current drawn from the source before and after introduction of the capacitor be 11 and 12
respectively. Take the source voltage as the reference and express 11 and 12 as vector
quantities in polar form.
b. Obtain the capacitor current, IC = 12 - 11, graphically as well as using complex number
manipulation. Compare the results.
c. Express the waveforms of the source current before (11(t)) and after (12(t)) introduction of the
capacitor in the form Im sin(2лft + 0). Hand sketch them on the same graph. Clearly label your
plots.
d. Analytically solve i2(t) – i1(t) using the theories of trigonometry to obtain the capacitor current
in the form, ¡C(t) = ICm sin(2πft + OC). Compare the result with the result in Part b.
Transmission line data:Data:• Active power of the load (P): 50 kW• Power factor of the load (PF): 0.8 (lagging)• Line-to-line voltage at the load (V_LC): 13.8 kV• Line resistance (R): 2 Ω• Line inductance (L): 0.8 H• Line capacitance (C): 0.0003 F• Required series compensation: 60% of the line impedance.• Line length: 250 kmDetermine:1. Characteristic impedance and propagation constant.2. The generalized long line constants A, B, C, D.3. Total voltage, current and power at the generating end.4. Voltage regulation.5. Parameters A, B, C, D of the compensation circuit.6. New generalized constants of the power system afterseries compensation.7. Conclusion of the results obtained.
3.18
In a single-phase half-wave ac-dc converter, the average value of the load
current is 1.78 A. If the converter is operated from a 240 V, 50 Hz supply
and if the average value of the output voltage is 27% of the maximum
possible value, calculate the following, assume the load to be resistive.
(a) Load resistance
(b) Firing angle
(c) Average output voltage
(d) The rms load voltage
(e) The rms load current
(f) DC power
(g) AC power
(h) Rectifier efficiency
(i) Form factor
(j) Ripple factor
Chapter 3 Solutions
MindTap Engineering, 1 term (6 months) Printed Access Card for Glover/Overbye/Sarma's Power System Analysis and Design, 6th
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