# K13 T1 Sol

Description
modeling and analysis
Categories
Published

View again

All materials on our website are shared by users. If you have any questions about copyright issues, please report us to resolve them. We are always happy to assist you.
Related Documents
Share
Transcript
2014-15 Solution Set K13 – Power System Modeling and Analysis-Test –I - Max.marks=50; Time- 90 mins PART - A 5 marks A1. What are the components of power system? (2) Generators, Transformers, Transmission lines, Compensators, and Loads A2. What is the equation used to convert the p.u. impedance expressed in one base to another. (2) A3. Expand the abbreviation – ULTC (1) Under Load Tap Changing PART-B 15 marks B1. Obtain the z and y-parameters for the network which operating at ω  =100 rad/sec. (8) Impedances Admittances  2014-15 B2. Draw the equivalent circuit of a single phase transformer and explain. (7) PART-C 30 marks C1. (a) A single phase transformer rate 1.2 kV/120 V, 7.2 KVA has the following winding parameters r 1 =0.8 Ω , x 1 =1.2 Ω , r 2 =0.01 Ω  and x 2 =0.01 Ω . Determine the voltage regulation of the transformer when it is delivering 7.2 kVA to a load at 120 V and 0.8 power-factor lagging. (6) Turns ratio, a = 10 R1 = 0.8 + a 2 x 0.01 = 1.8 X1 = 1.2 + a 2 x 0.01 = 2.2 So, Z1= 1.8+j 2.2 I2(fl) = Magnitude of (S2/V2) ∠ - θ  = ( 7200/120) ∠ -36.9 = 60 ∠ -36.9 Amps I1(fl) = I2 (fl)/a = 6 ∠ -36.9 Amps a V2(fl) = 1200 Volts V1(fl) = a V2(fl) + I1(fl) * Z1 = 1216.57 ∠ -0.19 volts  2014-15 V2(fl)= 120 Volts V2(nl) = V1(fl)/a = 121.66 Volts Regulation = 1.38% (b) A 60 Hz three phase transmission is 175 km long. It has a total series impedance of 35+j140 Ω  and a shunt admittance of 930x10 -6  S. It delivers 40MW at 220kV, with 90% power factor lagging. Find the sending end voltage by the long-line equation. (9) I R = 40000/( √ 3 x 220 x 0.9) = 116.6 ∠ -25.84 A Zc = √  (144.3 ∠ 75.96 / 930x10 -6   ∠ 90) = 394. ∠ -7.02 Ohms γ  l = 0.3663 ∠ 83.0 cosh γ  l = 0.9354 + j 0.0160 sinh γ  l = 0.0419 + j 0.3565 Vs= 130153 ∠ 6.5 V per phase Vs= 225.4 ∠ 6.5 V OR C2. (a) A single-phase ac voltage of 240 V is applied to a series circuit whose impedance is 10 ∠ 60 o  Find R, X, P, Q and the power factor of the circuit. (6) I = V/Z = 24 ∠ -60 A 24 A at 0.5 lagging pf R = 5 Ω  and X = 8.66 Ω  P= 2880 Watts and Q = 4988.306 VAr (b) Three identical single-phase transformers, each rated 1.2kV/120V, 7.2kVA and having a leakage reactance of 0.05 p.u, are connected together to form a three-phase bank. A balanced Y-connected load of 5 Ω  per phase is connected across the secondary of the bank. Determine the Y-equivalent per phase impedance (in ohms and in per unit) seen from the primary side when the transformer bank is connected ∆ - ∆  (9) Base impedance = 1200 sqaure / (7200x3) = 66.67 ohms |VLL| = 1200 V R L ’ = 5 x square of (1200/120) = 500 ohms X l = 66.67 x 0.05 = 3.33 ohms Z L ’ = 500 + j 3.33 ohms C3. (a) Let a 5-kVA, 400/200-V transformer be approximately represented by a 2 Ω  reactance referred to the low-voltage side. (i) Considering the rated values as base quantities, express the transformer reactance as a per-unit quantity. (ii) Repeat the determination, expressing all quantities in terms of the high-voltage side. (6) (i) Base VA = 5000 VA and Base voltage = 200 V Base current = 25 A Base impedance = 8 ohms p.u referring to LV side = 2/8 = 0.25  2014-15 (ii) Base VA = 5000 VA and Base voltage = 400 V Base current = 12.5 A Base impedance = 32 ohms The transformer reactance referred to HV = 2(400/200) 2  = 8 ohms p.u referring to HV side = 8/32 = 0.25 (b) Obtain the nodal admittance matrices for each branch and then determine Y-bus for the network represented by the reactance diagram as below. Assume a mutual impedance of j0.2 between the branches (1)-(2) and (3)-(r). (9) 1, r -j0.833 1,2 and 3,r -j5.77 j5.77 j0.77  j5.77 -j5.77 -j0.77  j0.77 -j0.77 -j0.77 1,3 -j3.33 j3.33  j3.33 -j3.33 2,3 -j6.67 j6.67  j6.67 -j6.67 Y bus -j 9.933 j5.77 j4.1  j5.77 -j12.44 j5.9  j4.1 j5.9 -j10.77 OR C4. (a) Explain the modeling of synchronous machine and loads. (6) (b) For the given short transmission line of impedance 0.3+j0.4 ohms per phase, the receiving-end voltage is 6351 V per phase .If the regulation is not to exceed 5%, find the maximum power that can be transmitted over the line. (9) Vr = 6351 Volts 5% regulation implies Vs=6668.55 V The maximum power that can be transmitted = 36.3 MW/phase

Jul 23, 2017

#### For some

Jul 23, 2017
Search
Similar documents

View more...
Tags

## Quantity

Related Search
We Need Your Support
Thank you for visiting our website and your interest in our free products and services. We are nonprofit website to share and download documents. To the running of this website, we need your help to support us.

Thanks to everyone for your continued support.

No, Thanks