1. Lecture 02Electro Mechanical System1 Assignment 2 Page 195, Problems: 9-4, 9-6, 9-10* Note: problem 9-10 is a design problem Due Date: Tuesday 8 th Feb, 2011 Quiz No.1 Next Week

2. Lecture 05Electro Mechanical System2 Example A single-phase power system consists of a 480-V 60-Hz generator supplying a load Z load = 4 + j3 Ω through a transmission line of impedance Z line = 0.18 + j0.24 Ω. Answer the following questions about this system: a) If the power system is exactly as described above. i) What will the voltage at the load be? ii) What will the transmission line losses be? b) Suppose a 1:10 step-up transformer is placed at the generator end of the transmission line and a 10:1 step-down transformer is placed at the load end of the line. i) What will the load voltage be now? ii) What will the transmission line losses be now?

3. Lecture 05Electro Mechanical System3 Example

4. Lecture 05Electro Mechanical System4 Example To analyze this system, it is necessary to convert it to a common voltage level. This is done in two steps: 1. Eliminate transformer T2 by referring the load over to the transmission line's voltage level. 2. Eliminate transformer T1 by referring the transmission line's elements and the equivalent load at the transmission line's voltage over to the source side.

5. Lecture 05Electro Mechanical System5 Example The total impedance at the transmission line level is: Z eq = Z line + Z’ load = 400.18Ω + j300.24Ω = 500.3  36.88 o Ω The total impedance at the transmission line level (Z line + Z’ load ) is now reflected across T 1, to the source's voltage level: Z’ eq = a 2 Z eq Now a = N P /N S = 1/10 Z’ eq = a 2 (Z line + Z’ load ) = (1/10) 2 (500.3  36.88 o Ω) = 5.003  36.88 o Ω

6. Lecture 05Electro Mechanical System6 Example The generator current is: I G = V / Z’ eq = 480  0 o / 5.003  36.88 o = 95.94  -36.88 o A Knowing the current I G, we can now work back and find I line and I Ioad, Working back through T 1 : N p1 I G = N s1 I line I line = (N p1 / N s1 ) I G = (1 /10) (95.94  -36.88 o A) = 9.594  -36.88 o A Working back through T 2 : N p2 I line = N s2 I load I load = (N p2 / N s2 )I line = (10/1) (9.594  -36.88 o A) = 95.94  -36.88 o A The Load Voltage is now given by: V load = I load Z load = (95.94  -36.88 o A)( 5.003  36.87 o Ω) = 479.7  -0.01 o Line losses are: P loss = (I line ) 2 R line = (9.594 A) 2 (0.18 Ω)= 16.7 W  Notice that raising the transmission voltage of the power system reduced transmission losses by a factor of nearly 90.  Also, the voltage at the load dropped much less in the system with transformers compared to the system without transformers.

7. Lecture 05Electro Mechanical System7 Chapter 10 Practical Transformers

8. Lecture 05Electro Mechanical System8 Introduction  In the real world, transformers are not ideal  Windings have resistance  The cores are not infinitely permeable  The flux produced by the primary is not completely captured by the secondary  Leakage flux must be accounted  Iron core produces eddy-current and hysteresis losses  What happens when an infinitely permeable core is replaced by an iron core having hysteresis and eddy- current losses?

9. Lecture 05Electro Mechanical System9 Imperfect Cores  Core imperfections are represented by R m and X m in parallel with the primary winding  R m models the iron losses  X m models the permeability  The current I m flowing in X m is the magnetizing current that creates the flux  m  The total current I O needed to produce the flux is called the exciting current  R m and X m can be measured experimentally by  The power values are measured under no-load conditions