SOLUTIONS ELECTRICAL ENGINEERING (EE) PRACTICE PROBLEMS FOR TECHNICAL MAJORS

Transcription

1 SOUTIONS EECTRICA ENGINEERING (EE) PRACTICE PROBEMS FOR TECHNICA MAJORS Nte: The text entitled Applied Engineering Principles Manual is used as the reference fr the questins and prblems belw. Althugh nly Sectin 1.3- Three-Phase Systems and Transfrmers, Sectin 1.4- Generatrs, and Sectin 1.5- Mtrs are explicitly needed t answer these questins and prblems, students with weak r nn-existent backgrunds in EE may first need t review prtins f Sectin 1.1- Fundamentals f Electricity and Sectin 1.2- Alternating Current Thery. Sectin 1.3 Three-Phase Systems and Transfrmers 1. State the advantages f three-phase systems ver single-phase systems. The advantages include a reductin in the size f machinery fr the same capacity, simpler machine cnstructin, reduced cst, and increased reliability f pwer transmissin. 2. A balanced three-phase 60-Hz system is driven by a generatr with a line vltage f 450 V. a. If the generatr is wye-cnnected, what is the vltage acrss each f its phases? V 450 V = = = 60V. 3 3 VφY 2 b. If the generatr is delta-cnnected, what is the vltage acrss each f its phases? V V 450 V. = = φ c. If each phase f a delta-cnnected lad can be represented by a 100 Ω resistr in series with a 53 mh inductr, find the line current and ttal real pwer. (See figure at the tp f the next page.) Page 1 f 8

2 450V rad sec Z = Z + Z = 100 Ω+ j20 Ω = 102 Ω ZR = Z = jω = j (377 )(53x10 H) = j 20 R φ Iφ 4.41A V 450 V = = = Z 102 Ω I = 3(I ) = 3(4.41A) = 7.64 A φy 3V 3(450 V) Z 102 Ω 2 2 φ P= csθ = cs(11.3 ) = 5,840 W. Ω d. Repeat Part c. fr a wye-cnnected lad f the same phase impedance. 450V V 450 V VφY = = = 260V 3 3 V 260 V = = = Z 102 Ω 11.3 φ Iφ 2.55A I = I = 2.55A φy 3V 3(260 V) = = = W. Z φ P csθ cs(11.3 ) 1,950 Page 2 f 8

3 3. A three-phase 450V system is supplying a delta-cnnected 100hp lad (746W = 1hp) at a pwer factr f 0.92 lagging. Find: a. The real pwer 746W P = 100 hp = 74.6 kw. 1hp b. The apparent pwer P app P 74.6kW = = = pf kva r 81.1x10 VA. c. The line current P app = 3V I P = = = 04A. 3V 3(450V) 3 app 81.1x10 VA I 1 d. The phase current I 104 A I = φ 6 3 = 3 = 0A. Page 3 f 8

4 4. Given the fllwing balanced 3 system, find the apparent pwer delivered by the generatr. 450V 60Hz #1 100 kva.866 lead #2 100 kvar.866 lag OAD #1 θ = cs (pf) = cs (0.866) = 30 Z 1 app 1 1 P = P pf = (100 kw)(0.866) = 86.6 kw Q1= Pappsinθ = (100 kw)sin( 30 ) = 50 kvar. OAD #2 1 1 θz = cs (pf) = cs (0.866) = 30 Q 2 = 100 kvar Q2 100 kw P2 = = = kw. tanθ tan(30 ) TOTA P = P + P = 86.6 kw kW = kw T 1 2 Q = Q + Q = 50 kvar kvar = 50 kvar T 1 2 P = P + Q = (259.8) + (50) P app T T app = kva. Page 4 f 8

5 Sectin 1.4 Generatrs D-C Generatrs 1. Describe briefly the cnstructin and peratin f a d-c generatr. The field windings f a d-c generatr, lcated n the statr, generate a fixed and nearly unifrm magnetic field thrugh which the rtr rtates. As the armature windings (lcated n the rtr) pass thrugh the magnetic field, an a-c vltage is induced in each armature winding. The resulting current is effectively cnverted t direct current and carried t the lad by means f the cmmutatr and brushes. 2. Describe the prblems with cmmutatin in d-c generatrs. First, the rientatin f the "neutral plane," the plane in which an armature winding prduces zer utput vltage, is shifted frm the n-lad psitin. This results in sparking acrss the cmmutatr-brush cnnectin each time the armature cnnectin is reversed. Secnd, the distrted flux acrss the ple gap causes a gradual reductin in utput vltage with increasing generatr lad. 3. Describe hw prblems with cmmutatin are cmpensated fr in a d-c generatr. The cmmutatin prblem is usually slved by the additin f tw small ples called "interples" in the n-lad neutral plane. The armature current is passed thrugh the interple windings t prduce a flux exactly equal t the armature field but ppsite in directin. This technique assures that the neutral plane remains fixed and that cmmutatin always ccurs in the neutral plane regardless f lad. The distrted flux near the air gap between ple tips and rtr may be crrected, if desired, by the additin f cmpensating windings cnnected in series with the armature windings. The cmpensating windings develp a flux which ppses the armature flux in the air gap. This cmpensating flux varies with lad in the same manner as the armature reactin. Thus, the field flux in the air gap remains unifrm regardless f lad. Hwever, cmpensating windings add t the cst f a generatr and are nt usually emplyed. Page 5 f 8

6 4. What is meant by cmpunding in a d-c generatr? Cmpund generatrs emply a series field winding in additin t the shunt field winding. The series field cils are made f a relatively small number f turns f large cpper cnductr, either circular r rectangular in crss sectin, and are cnnected in series with the armature circuit. These cils cntribute a magnetic field that influences the main field flux f the generatr. 5. Explain the cncept f degree f cmpunding in a cumulatively cmpunded d-c generatr. If the ampere-turns f the series field act in the same directin as thse f the shunt field, the cmbined magnetic field is equal t the sum f the series and shunt field cmpnents. If lad is added t a cmpund generatr, armature current and thus series field circuit current increases. The effect f the additive series field is t increase field flux with increasing lad. The extent f the increased field flux depends n the degree f saturatin f the field irn as determined by the shunt field current. Thus, the terminal vltage f the generatr may increase r decrease with lad depending upn the influence f the series field cils. This influence is referred t as the degree f cmpunding. A flatcmpunded generatr is ne in which the n-lad and full-lad vltages have the same value. An under-cmpunded generatr is ne in which the full-lad vltage is less than the n-lad value. An ver-cmpunded generatr is ne in which the full-lad vltage is higher than the n-lad value. The way the terminal vltage changes with increasing lad depends upn the degree f cmpunding. A-C Generatrs 1. Describe the cnstructin and peratin f a simple a-c synchrnus generatr. The field winding is n the rtr and the armature winding n the statr. The rtating field winding is excited by direct current supplied thrugh slip rings and brushes. The effect f the rtating field is the same as if a bar magnet was cnnected t the generatr shaft. Rtating the field past the statinary armature winding causes a vltage t be induced in the armature. One cmplete cycle is generated fr each revlutin f the tw-ple rtr. Page 6 f 8

7 2. Hw fast shuld a 6-ple synchrnus generatr be rtated t prvide an utput frequency f 60 Hz? PN f = f 120(60) N= = =1200 rpm. P 6 3. ist the requirements fr paralleling a-c generatrs. Terminal vltages must be equal in magnitude. Terminal vltages must be in phase with each ther. Terminal vltages must have the same frequency. Bth generatrs must have the same phase sequence. Sectin 1.5 Mtrs Inductin Mtrs and Synchrnus Mtrs 1. Draw the trque-speed characteristic curve fr an inductin mtr. abel the hrizntal axis with tw physical quantities and give tw names fr the maximum trque the mtr can prduce. Describe the nrmal perating regin fr this mtr. The maximum trque is called the breakdwn trque r the pull-ut trque. The inductin mtr is nrmally perated t the right f the breakdwn trque with slip values in the range f 3 t 10 percent. Page 7 f 8

8 2. Describe the peratin f a synchrnus mtr. Include cnstructin and hw the mtr is started. A synchrnus mtr is the same in cnstructin as a three-phase generatr. An a-c current supplied t the statr winding f a synchrnus mtr prduces a rtating magnetic field the same as in an inductin mtr. A direct current is supplied t the rtr winding, thus prducing a fixed plarity at each ple. If it culd be assumed that the rtr had inertia and that n lad f any kind were applied, then the rtr wuld revlve in step with the revlving field immediately upn applicatin f pwer t bth windings. This, hwever, is nt the case; the rtr has inertia and, in additin, there is a lad. A synchrnus mtr has t be brught up t synchrnus speed by special means [by including a squirrel-cage r damper winding in the rtr ple heads]. If the statr and rtr windings are energized, then as the ples f the rtating magnetic field apprach rtr ples f ppsite plarity, the attracting frce tends t turn the rtr in the directin ppsite t that f the rtating field. As the rtr starts in this directin, the rtating-field ples are leaving the rtr ples, and this tends t pull the rtr ples in the same directin as the rtating field. Hwever, the rtr des nt reach synchrnus speed immediately because f its inertia: the rtating field passes it up and the rtr is again pulled in the reverse directin. Thus, the rtating field tends t pull the rtr ples first in ne directin and then in the ther, with the result that the average starting trque is zer. Page 8 f 8