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Q1: For the three-bus lossless power network shown in the figure, the voltage magnitudes at all the buses are equal to 1 per unit (pu), and the differences of the voltage phase angles are very small. The line reactances are marked in the figure, where α, β, γ and x are strictly positive. The bus injections P1 and P2 are in pu. If P1 = mP2, where m > 0, and the real power flow from bus 1 to bus 2 is 0 pu. then which one of the following options is correct?      (2024)
Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)(a) γ = mβ
(b) β = mγ
(c) α = mγ
(d) α = mβ
Ans:
(a)
Sol: Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)
Q2: A three phase star-connected load is drawing power at a voltage of 0.9 pu and 0.8 power factor lagging. The three phase base power and base current are 100 MVA and 437.38 A respectively. The line-to-line load voltage in kV is _____.      (SET-2(2014))
(a) 88.2
(b) 108.9
(c) 118.8
(d) 127.4
Ans:
(c)
Sol: Given,
Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)For a 3 − ϕ star connected systems,
Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)∴ Actual line to line load voltage in kV = 0.9 x 132 kV = 118.8 kV

Q3: For the power system shown in the figure below, the specifications of the components are the following :
G1: 25 kV, 100 MVA, X = 9%
G2: 25 kV, 100 MVA, X = 9%
T1: 25 kV/220 kV, 90 MVA, X = 12%
T2: 220 kV/25 kV, 90 MVA, X = 12%
Line 1: 200 kV, X = 150 ohms
Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Choose 25 kV as the base voltage at the generator G1, and 200 MVA as the MVA base. The impedance diagram is      (2010)
(a) Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)(b) Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)(c) Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)(d) Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Ans: 
(b)
Sol: Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)
Q4: A generator is connected through a 20 MVA, 13.8/138 kV step down transformer, to a transmission line. At the receiving end of the line a load is supplied through a step down transformer of 10 MVA, 138/69 kV rating. A 0.72 pu. load, evaluated on load side transformer ratings as base values , is supplied from the above system. For system base values of 10 MVA and 69 kV in load circuit, the value of the load (in per unit) in generator will be      (2006)
(a) 36
(b) 1.44
(c) 0.72
(d) 0.18
Ans:
(a)
Sol: Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Base value on load circuit,
Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Value of load in ohm = ZL 
= value of load in p.u. x Base impedance
Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)Base impedance,
Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)
Q5: The p.u. parameter for a 500 MVA machine on its own base are:
inertia M = 20 p.u.; reactance X = 2 p.u.
The p.u. values of inertia and reactance on 100 MVA common base, respectively, are     (2005)
(a) 4, 0.4
(b) 100, 10
(c) 4, 10
(d) 100, 0.4
Ans:
(d)
Sol: Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)
Q6: A 75 MVA, 10 kV synchronous generator has Xd = 0.4 p.u. The Xd value (in p.u.) is a base of 100 MVA, 11 kV is       (2001)
(a) 0.578
(b) 0.279
(c) 0.412
(d) 0.44
Ans:
(d)
Sol: Previous Year Questions- Per Unit System | Power Systems - Electrical Engineering (EE)

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FAQs on Previous Year Questions- Per Unit System - Power Systems - Electrical Engineering (EE)

1. What is the per unit system in electrical engineering and why is it used?
Ans. The per unit system is a method of normalizing values in electrical engineering to simplify calculations and comparisons. It expresses quantities as fractions of a defined base unit, making it easier to work with different voltage levels and system components. This system minimizes the complexity in calculations, particularly in power systems analysis, as it reduces the range of numbers and helps in identifying relationships between different variables.
2. How do you convert electrical quantities to the per unit system?
Ans. To convert electrical quantities to the per unit system, you need to divide the actual value of the quantity by its base value. For example, for voltage, you take the actual voltage and divide it by the base voltage. The base values are usually chosen based on the system's maximum capacity or nominal ratings. The formula is: \[ \text{Per Unit Value} = \frac{\text{Actual Value}}{\text{Base Value}} \]
3. What are the advantages of using the per unit system in power system analysis?
Ans. The advantages of using the per unit system include simplification of calculations, reduction of numerical values (which makes it easier to handle calculations), and the ability to easily compare different systems or components. Additionally, it helps in visualizing changes in system parameters and facilitates the analysis of fault conditions and system stability.
4. What are the base values typically used in the per unit system for power systems?
Ans. The base values in the per unit system typically include base power (usually in MVA), base voltage (in kV), and base current (in A). The base power is often chosen based on the largest transformer or generator capacity in the system. The base voltage can be selected based on the nominal voltage level of the system, and the base current can be calculated from the base power and voltage using the formula: \[ \text{Base Current} = \frac{\text{Base Power}}{\sqrt{3} \times \text{Base Voltage}} \] for three-phase systems.
5. How does the per unit system help in fault analysis in power systems?
Ans. The per unit system aids in fault analysis by allowing engineers to express all system impedances in a normalized form, making it easier to compute fault currents and voltages. Since all values are expressed in per unit, they can be easily compared and manipulated to determine the effects of faults on the system. This uniformity simplifies calculations and enhances understanding of how faults affect different components and the overall system performance.
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