Spring 2000 EE361: MIDTERM EXAM 1

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1 NAME: STUDENT NUMBER: Spring 2000 EE361: MIDTERM EXAM 1 This exam is open book and closed notes. Assume f=60 hz and use the constant µ o =4π 10-7 wherever necessary. Be sure to show all work clearly. 1. (20 points) Phasors - 3-φ fundamentals: A measurement of the voltage between phase a and b gives v ab (t)=163000cos(377t+π/3) with the phase shift π/3 relative to the line current i a (t). Assuming a balanced 3-φ system, answer the following: a) Find the phasor V ab b) Sketch a phasor diagram showing all line-line voltages and line-neutral voltages. Indicate clearly the magnitude and phase of each voltage. c) Find the power factor of the load.

2 2. 30 points - Magnetic Circuits/Solenoids: Two coils are wrapped around a core with a plunger as indicated in the sketch below. The plunger is touching the core on the sides but is free to move. The core and core are made of high permeability ferromagnetic material (you can assume µ is infinite). The dimensions are as given in the sketch below. (a) Sketch an equivalent magnetic circuit for this solenoid. Clearly show the mmf sources and the reluctances in your circuit. (b) The two windings are connected in parallel to a voltage source. Calculate the inductance seen by this voltage source. (c) Find the force on the plunger caused by the currents in the coils. N a =15 i a =10A N b =15 i b =10A 10mm 4mm 2mm air gap 5mm 10mm Plunger dimensions

3 3. 20 points - Impedance Diagram: You are given a single phase 200 kva 4800/480 V transformer where the primary winding impedance is Z = j4.0 Ω and the secondary winding impedance is Z2 = j0.04 Ω. Non-ideal core effects are represented by a shunt impedance of Rc = 50 kω and Xc = 10 kω on the primary side. In the following, you should use standard transformer modeling approximations wherever it simplifies calculations. a) Find the per unit values for all the impedances. b) Sketch an impedance diagram for the transformer using the per unit values calculated in (a). c) Find the winding losses at full load. Assume the secondary voltage is at rated value and the load has unity power factor.

4 4. 10 points - Autotransformer: A 1200/120 V transformer is to provide a small step down voltage for a load as sketched below. Find the power rating for a standard transformer so that in the autotransformer configuration, a 5 kva load can be supplied at the rated voltages shown in the sketch. Assume an ideal transformer model. Supply 5 kva 1320V Load 5 kva 1200 V

5 5. (20 points) Short Answer. (a) Name two benefits of providing electric power using a balanced three phase system? (b) As you may have observed, a transformer connected to a power supply will become warm even if not supplying any load. What non-ideal property(s) of a practical transformer causes this heating? (c) Why are transformer cores laminated? (d) Name one application for a high impedance transformer (a transformer with high leakage reactance)?

6 NAME: STUDENT NUMBER: FALL 1995 EE361: EXAM This exam is open book and open notes but all work must be your own. Be sure to show all work clearly points - 3-φ Power: Find the real and reactive power delivered to the unbalanced deltaconnected load sketched below. The 3 phase delta-connected source is a balanced set with lineline voltages of 12.5 kv. V ab V ca -j150ω 250Ω V bc j100ω 2. 7 points - Magnetic Circuit: Three coils are wrapped around a core with two air gaps as indicated in the sketch below. The core is made of high permeability ferromagnetic material (you can assume µ is infinite). The core has length 25mm and height 10mm when measured from center line to center line. The cross section dimensions are 1mm 10mm. Calculate the magnetic flux φ C seen in the leg C. 25mm 2mm 1mm φ C ia i B i C 10mm N A =10 i A =10sinωt N B =10 i B =15sinωt N C =20 i C =5sinωt 1mm 10mm Cross Section

7 3. 5 points - Autotransformer: A 2200/220 V transformer is used as an autotransformer to provide a small step down voltage. Find the transformer rating needed in order to have the capacity to deliver 50kVA. Assume an ideal transformer. Supply 50 kva 2420 V Load 50 kva 2200 V 4. 8 points - Transformer Performance: You are given a single phase 50 kva 2200/220 V transformer where the primary winding impedance is Z1 = 2+ j5ω and the secondary winding impedance is Z2 = j0.05 Ω. Non-ideal core effects are represented by a shunt impedance of R C = 50 kω on the primary side. You may use standard transformer modeling approximations but do not neglect core effects entirely. (a) Sketch an impedance diagram for the transformer showing the per unit values for all impedances. (b) Find the transformer efficiency when supplying rated load at unity power factor. (c) Find the voltage regulation when supplying rated load at 0.8 power factor lagging points - 3-φ Transformer Connections: A bank of three single phase 25kVA 2200/220 transformers are connected Υ in order to step down the voltage. Use an ideal transformer model for answering the following. (a) Sketch a phasor diagram of the primary and secondary line voltages showing phase and the magnitude if delivering rated voltage. (Magnitude need not be drawn to scale but must be identified). (b) Find the capacity of the transformer bank.

8 NAME: STUDENT NUMBER: Fall 2013 ECE 325: MIDTERM EXAM 1 This exam is closed book and closed notes. Assume f=60 hz and use the constant µ o =4p 10-7 H/m wherever necessary. Be sure to show all work clearly points Inductance calculation: In the ferromagnetic core sketched below there are two windings on the left leg, two gaps on the center leg filled with a stainless steel alloy (µ r =10 indicated by shading) and an air gap on the third leg, calculate the inductance seen by a source connected to winding A. You can assume the core is infinitely permeable. N A =10 N B =20 1mm 5mm Cross Section

9 2. 15 points Transformer performance: You are given an impedance diagram for a three phase transformer as sketched below with values shown in per unit. At full load with a power factor of 0.9 lagging, find the transformer efficiency. X m =50 p.u.

10 3. 15 points - Autotransformer: A 1200/120 V transformer is to provide a small step up voltage for a load as sketched below. Find the power rating for a standard transformer so that in the autotransformer configuration, a 5 kva load can be supplied at the rated voltages shown in the sketch. Assume an ideal transformer model. Supply 5 kva 1200V Load 5 kva 1320 V

11 4. 15 points - Phasors - 3-f fundamentals: A measurement of the line current in phase a is i a (t)=170cos(377t-p/4) A with the phase shift p/4 relative to the line to neutral voltage, v an (t)=12.2cos(377t) kv. Assuming a balanced 3-f delta connected load, find the following: a) the line current phasor I a b) the current through the load across ab i.e, the phasor I ab c) the power factor of the load, indicating leading or lagging.

12 5. 15 points Transformer impedance Diagram: You are given a single phase 500 kva 4800/480 V transformer where the primary winding impedance is Z = j4.0 W and the secondary winding impedance is Z2 = j0.04 W. Non-ideal core effects are represented by a shunt impedance of R c = 50kW and X = kw 1 m 10 on the primary side. 1 a) Find the per unit values for all the impedances. b) Sketch an impedance diagram for the transformer using the per unit values calculated in (a).

13 6. 20 points - Short Answer. (a) What is the primary method in the design of power transformer cores to reduce eddy currents? (b) In the transformer model, a resistor R c is used to model core losses. What non-ideal effects does this resistor represent? (c) Why is it advantageous to use a higher frequency power supply in aircraft? (d) Name two reasons why power is transmitted in three phases. (e) It is observed that the line current of a 4 wire three phase Y-connected system (the 4 th wire is a neutral return) has a 3rd harmonic with magnitude 4A. What is the magnitude of the 3 rd harmonic current in the neutral return?

86 chapter 2 Transformers

86 chapter 2 Transformers Wb 1.2x10 3 0 1/60 2/60 3/60 4/60 5/60 6/60 t (sec) 1.2x10 3 FIGURE P2.2 2.3 A single-phase transformer has 800 turns on the primary winding and 400 turns on the secondary winding.