Question 1 |
A three-phase cable is supplying 800 kW and 600 kVAr to an inductive load. It is intended to supply an additional resistive load of 100 kW through the same cable without increasing the heat
dissipation in the cable, by providing a three-phase bank of capacitors connected in star across the load. Given the line voltage is 3.3 kV, 50 Hz, the capacitance per phase of the bank, expressed in microfarads, is ________.
500.42 | |
19.25 | |
47.96 | |
98.32 |
Question 1 Explanation:
\text{KVA}_1=\sqrt{800^2+600^2}=100 \text{ KVA}
Without excessive heat dissipation means current should be constant (i.e.) KVA erating must be constatn.
In second case active power,
P=800+100=900 KW
Reactive power in second case,
Q_2=\sqrt{1000^2-900^2}=435.889 \text{ KVAR}
Reactive power supplied by the three phase bank = 600-435.889 =164.44 KVAR
\begin{aligned} Q_{bank}/ph&=\frac{164.11}{3}=54.7 \text{ KVAR} \\ V/ph&= \frac{3.3}{\sqrt{3}}=1.9051 \text{ KV}\\ Q_c/ph&=\frac{(V/ph)^2}{X_C} \\ X_C&=\frac{(1.9052 \times 10^3)^2}{54.7 \times 10^3} \\ &= 66.36\; W\\ C&= \frac{1}{2 \pi f X_C}\\ &=\frac{1}{2 \pi \times 50 \times 66.36}\\ &=47.96 \mu F \end{aligned}
Without excessive heat dissipation means current should be constant (i.e.) KVA erating must be constatn.
In second case active power,
P=800+100=900 KW
Reactive power in second case,
Q_2=\sqrt{1000^2-900^2}=435.889 \text{ KVAR}
Reactive power supplied by the three phase bank = 600-435.889 =164.44 KVAR
\begin{aligned} Q_{bank}/ph&=\frac{164.11}{3}=54.7 \text{ KVAR} \\ V/ph&= \frac{3.3}{\sqrt{3}}=1.9051 \text{ KV}\\ Q_c/ph&=\frac{(V/ph)^2}{X_C} \\ X_C&=\frac{(1.9052 \times 10^3)^2}{54.7 \times 10^3} \\ &= 66.36\; W\\ C&= \frac{1}{2 \pi f X_C}\\ &=\frac{1}{2 \pi \times 50 \times 66.36}\\ &=47.96 \mu F \end{aligned}
Question 2 |
A distribution feeder of 1 km length having resistance, but negligible reactance, is fed from both the ends by 400 V, 50 Hz balanced sources. Both voltage sources S_1 \; and \; S_2 are in phase. The feeder supplies concentrated loads of unity power factor as shown in the figure.
The contributions of S_1 \; and \; S_2 in 100 A current supplied at location P respectively, are
The contributions of S_1 \; and \; S_2 in 100 A current supplied at location P respectively, are75 A and 25 A | |
50 A and 50 A | |
25 A and 75 A | |
0 A and 100 A |
Question 2 Explanation:
Let the resistance of whole length of feeder be R\;\Omega, length of feeder =1000m.
\therefore \; Resistance per unit length =\frac{R}{1000}\Omega /meter
\therefore \; Resistance of 400 m length =\frac{R}{1000} \times 400 =\frac{2R}{5} \Omega
\therefore \; Resistance of 200 m length =\frac{R}{5} \Omega
Let the current supplied by the sources be I_1 \text{ and } I_2.
Applying KVL from source S_1 \text{ and } S_2, we have:
\begin{aligned} 400 &= \frac{2R}{5}I_1+\frac{R}{5}(I_1-200)\\ & -\frac{R}{5}(I_2-200)-\frac{R}{5}I_2+400\\ 0&= \frac{2}{5}I_1 +\frac{I_1-200}{5}+\frac{200-I_2}{5}-\frac{I_2}{5}\\ 0&=2I_1+I_1-200+200-I_2-I_2 \\ 0&=3I_1-2I_2\\ I_1&=\frac{2}{3}I_2\;\;\;...(i)\\ \text{also, }& I_1 +I_2-400=100\\ \text{or, }&I_1+I_2=500 \;\;\;...(ii)\\ &\text{Solvinf eq. (i) and (ii), we have}\\ I_1&=200A \text{ and }I_2=300A\\ &\text{Contribution of }S_1 \text{in 100 A at location}\\ P&=I_1-200=0A \\ &\text{Contribution of }S_2 \text{in 100 A at location}\\ P&=I_2-200=300-200=100A \end{aligned}
Therefore, S_2 alone supplies the total load at location P.
\therefore \; Resistance per unit length =\frac{R}{1000}\Omega /meter
\therefore \; Resistance of 400 m length =\frac{R}{1000} \times 400 =\frac{2R}{5} \Omega
\therefore \; Resistance of 200 m length =\frac{R}{5} \Omega
Let the current supplied by the sources be I_1 \text{ and } I_2.
Applying KVL from source S_1 \text{ and } S_2, we have:
\begin{aligned} 400 &= \frac{2R}{5}I_1+\frac{R}{5}(I_1-200)\\ & -\frac{R}{5}(I_2-200)-\frac{R}{5}I_2+400\\ 0&= \frac{2}{5}I_1 +\frac{I_1-200}{5}+\frac{200-I_2}{5}-\frac{I_2}{5}\\ 0&=2I_1+I_1-200+200-I_2-I_2 \\ 0&=3I_1-2I_2\\ I_1&=\frac{2}{3}I_2\;\;\;...(i)\\ \text{also, }& I_1 +I_2-400=100\\ \text{or, }&I_1+I_2=500 \;\;\;...(ii)\\ &\text{Solvinf eq. (i) and (ii), we have}\\ I_1&=200A \text{ and }I_2=300A\\ &\text{Contribution of }S_1 \text{in 100 A at location}\\ P&=I_1-200=0A \\ &\text{Contribution of }S_2 \text{in 100 A at location}\\ P&=I_2-200=300-200=100A \end{aligned}
Therefore, S_2 alone supplies the total load at location P.
Question 3 |
The undesirable property of an electrical insulating material is
high dielectric strength | |
high relative permittivity | |
high thermal conductivity | |
high insulation resistivity |
Question 4 |
Consider a three-core, three-phase, 50 Hz, 11 kV cable whose conductors are
denoted as R,Y and B in the figure. The inter-phase capacitance(C1) between each pair of conductors is 0.2 \muF and the capacitance between
each line conductor and the sheath is 0.4 \muF . The per-phase charging current is
2.0A | |
2.4A | |
2.7A | |
3.5A |
Question 4 Explanation:
\begin{aligned} Z'&=Z/3\\ \frac{1}{\omega C'}&=\frac{1}{3\omega C}\\ C'&=3C \end{aligned}
Equivalent capacitance (C_{eq}) between a phase and ground
\begin{aligned} C_{eq} &=C'+C_2=3C_1+C_2\\ C_{eq} &= 3 \times 0.2+0.4=1\mu F\\ &\text{Per phase voltage,}\\ V_P&=\frac{11}{\sqrt{3}}kV\\ &\text{Per phase charging current}\\ I&=j\omega C_{eq}V_P\\ |I_C|&=\omega C_{eq}V_P\\ &=2 \pi \times 50 \times 1 \times 1 \times 10^{-6}\\ & \times \frac{11}{\sqrt{3}} \times 10^3\\ &\approx 2A \end{aligned}
Question 5 |
Consider a three-phase, 50 Hz, 11 kV distribution system. Each of the conductors
is suspended by an insulator string having two identical porcelain insulators. The
self capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figures. The voltages across the two insulators are
e_{1}=3.74 kV,e_{2}=2.61 kV | |
e_{1}=3.46 kV,e_{2}=2.89 kV | |
e_{1}=6.0 kV,e_{2}=4.23 kV | |
e_{1}=5.5 kV,e_{2}=5.5 kV |
Question 5 Explanation:
Line to line voltage =V_{l-l}=11kV
\begin{aligned} V_P&=\text{Phase to ground voltage} \\ V_P&=\frac{V_{l-l}}{\sqrt{3}}=\frac{11}{\sqrt{3}}kV \\ e_1+e_2&= \frac{11}{\sqrt{3}}kV=6.35kV \;\;...(i)\\ I&=I_1+I_2 \\ (j\omega 5C) \times e_1 &= (j\omega C) \times e_2 +(j\omega 5C) \times C_2 \;\;..(ii)\\ 5e_1&=6e_2\\ \text{Solving } & \text{equation (i) and (ii), we get,} \\ e_1&=3.46kV\\ e_2&=2.89kV \end{aligned}
Question 6 |
Single line diagram of a 4-bus single source distribution system is shown below.
Branches e_1,e_2,e_3 \; and \; e_4 have equal impedances. The load current values indicated in the figure are in per unit.
Distribution company's policy requires radial system operation with minimum loss. This can be achieved by opening of the branch
Distribution company's policy requires radial system operation with minimum loss. This can be achieved by opening of the branch
e_{1} | |
e_{2} | |
e_{3} | |
e_{4} |
Question 6 Explanation:
Assuming impedance of each branch R
(i) if e_1 is opened
Total losses =8^2R+3^2R+1^2R=74R
(ii) if e_2 is opened
Total losses =8^2R+7^2R+5^2R=138R
(i) if e_3 is opened
Total losses =1^2R+7^2R+2^2R=54R
(i) if e_4 is opened
Total losses =5^2R+3^2R+1^2R=38R
Operation with minimum loss can be achieved by opening line e_4.
(i) if e_1 is opened
Total losses =8^2R+3^2R+1^2R=74R
(ii) if e_2 is opened
Total losses =8^2R+7^2R+5^2R=138R
(i) if e_3 is opened
Total losses =1^2R+7^2R+2^2R=54R
(i) if e_4 is opened
Total losses =5^2R+3^2R+1^2R=38R
Operation with minimum loss can be achieved by opening line e_4.
Question 7 |
A 110 kV, single core coaxial, XLPE insulated power cable delivering power at
50 Hz, has a capacitance of 125 nF/km. If the dielectric loss tangent of XLPE is
2 \times 10^{-4}, then dielectric power loss in this cable in W/km is
5 | |
31.7 | |
37.8 | |
189 |
Question 7 Explanation:
\begin{aligned} \text{V(Phase voltage)} &=\frac{110}{\sqrt{3}}kV \\ C&=125 nF/km \\ \tan \delta &= 2 \times 10^{-4} \end{aligned}
Dielectric power loss in cable
\begin{aligned} P&=V62 \omega C \tan \delta \\ P&=\left ( \frac{110 \times 10^3}{\sqrt{3}} \right )^2 \times 2 \\ & \times \pi \times 50 \times 125 \times 10^{-9} \times 2 \times 10^{-4}\\ &\approx 31.7 W/km \end{aligned}
Question 8 |
The phase sequences of the 3-phase system shown in figure is
RYB | |
RBY | |
BRY | |
YBR |
Question 8 Explanation:
The phase sequence of the given figure is RBY.
RYB, BRY, and YBR represent the same phase sequence of the figure (given below).
RYB, BRY, and YBR represent the same phase sequence of the figure (given below).
Question 9 |
The rated voltage of a 3-phase power system is given as
rms phase voltage | |
peak phase voltage | |
rms line to line voltage | |
peak line to line voltage |
Question 9 Explanation:
NOTE:
Generator are specified in 3-\phi MVA, line to line voltage and per-phase reactance (equivalent star).
Tranformers are specified in 3-\phi MVA, line to line transformation ratio and per phase (equivalent star) impedance on one side.
Load are specified in 3-\phi MVA, line to line voltage and power factor.
Generator are specified in 3-\phi MVA, line to line voltage and per-phase reactance (equivalent star).
Tranformers are specified in 3-\phi MVA, line to line transformation ratio and per phase (equivalent star) impedance on one side.
Load are specified in 3-\phi MVA, line to line voltage and power factor.
Question 10 |
A dc distribution system is shown in figure with load current as marked. The
two ends of the feeder are fed by voltage sources such that V_P - V_Q = 3 V. The
value of the voltage V_P for a minimum voltage of 220 V at any point along the
feeder is
225.89 V | |
222.89 V | |
220.0 V | |
228.58 V |
Question 10 Explanation:
Let current injected at point P=I_P
\begin{aligned} V_P-V_Q &=0.1(I_P-10)+0.15(I_P-30) \\ & +0.2(I_P-60) \\ 3&=0.45I_P-17.5 \\ I_P&=45.55A \end{aligned}
direction of current will be in opposite in section SQ, so, points will have minimum voltage,
\begin{aligned} V_s&=220V\\ V_p&=V_s+0.1(I_p-10)+0.15(I_P-30)\\ &=220+0.1 \times (45.55-10)\\ &+0.15(45.55-30)\\ &\approx 225.89V \end{aligned}
\begin{aligned} V_P-V_Q &=0.1(I_P-10)+0.15(I_P-30) \\ & +0.2(I_P-60) \\ 3&=0.45I_P-17.5 \\ I_P&=45.55A \end{aligned}
direction of current will be in opposite in section SQ, so, points will have minimum voltage,
\begin{aligned} V_s&=220V\\ V_p&=V_s+0.1(I_p-10)+0.15(I_P-30)\\ &=220+0.1 \times (45.55-10)\\ &+0.15(45.55-30)\\ &\approx 225.89V \end{aligned}
There are 10 questions to complete.