1. What is is the watts watts output output of of a 25-hp 25-hp motor motor at full full load? load? PO= hp x 746 = 25 x 746 =18 65 W = 18.65 !W 2. " 15-!w #e$era #e$erator tor has a$ e%&ie$& e%&ie$&' ' of (1 per&e$t per&e$t at full load. load. )al&ula )al&ulate te the i$put i$ !ilowatts a$d the power loss. PO = 15!w *%&ie$&' = (1+ Pi = PO , e = 15 , .(1 = 164.84 !W
Ploss = PO 1/ , + e/ - P O = 151/ , (1/ - 15 = 14.84 !W
0. " 15-hp motor operates at a$ e%&ie$&' of of 87.5 per&e$t at full load. f the stra' power loss is approximatel' o$e-fourth of the total loss &al&ulate the &opper loss. PO = 15hp *%&ie$&' = 87.5+ stra' power loss = 3 of total loss Pi = PO , e = 15746/,.875 = 12.7( !W otal otal P loss = Pi - Po = 12788.6 111( = 15(8.6 W tra' power loss = 1,4/15(8.6/ = 0((.6 W )opper loss = total loss stra' str a' power loss = 15(8.6 0((.6 = 11(8.(6 W = 12W 4. he rotatio$ rotatio$al al loss i$ a #e$erator #e$erator was was fou$d to e 78 78 watts whe$ whe$ the #e$erated emf was 102 olts. 9etermi$e the rotatio$al loss for #e$erated olta#es of 108 a$d 126 olts. :otatio$al loss = 78 W *;< = 102 olts f *;< = 108 olts rotatio$al loss =? f *;< = 126 olts rotatio$al loss =? 78,102 = x1,108 78,102 = x2,126 >1 = 815.45 olts x2=744.54 olts
5. :eferri$# :eferri$# to examples examples 0 a$d a$d 4 pa#es pa#es 18(-1( ma!e ma!e a tale similar to the o$e there a$d &al&ulate the e%&ie$&ies for li$e &urre$ts of 5 1 15 2 a$d 25 amp assumi$# the followi$# &ha$#es li$e olts = 5 shu$t-@eld
resista$&e = 5 ohms armature a$d series-@eld resista$&e = 1.6 ohms rush drop assumed/ = 0 olts. Ai$e olts )urre$t " Ai A i$e hu$t @eld "r "rmature Be$erated olts Aosses Watts tra' power "rmature eries @eld hu$t @eld Crush &o$ta&t TOTAL
5
5
5
5
5
5 1 6 512.6
1 1 11 52.6
15 1 16 528.6
2 1 21 506.6
25 1 26 544.6
510
522
501
541
548
58 58
1(0 1(0
41 41
75 75
182 182
5 18
5 00
5 48
5 60
5 78
1147
1442
18((
2510
02(
25 0647 68.5
5 6442 77.6
75 (0(( 7(.8
1 12510 7(.(
125 157( 7(.16
LOSSES:
Power output Power i$put + e%&ie$&'
6. he output output torDue torDue of a motor is 6(.2 6(.2 l-ft l-ft whe$ it operates operates at (5 (5 rpm. )al&ulate the losses i$ the ma&hi$e a$d its e%&ie$&' if u$der this &o$ditio$ the power i$put is 1( watts. = 6(.2 l ft ft :pm = (5 rpm Pi = 1 ( !W Ep = 2Frpm//,00 2F rpm//,00 =2F(5/6(.2/,00 =12.5 hp Po= 12.5 12.5 x 746 746 = (025 (025 W Ploss = Pi Po = 1( (025 = 1575 W * = Po,Pi x 1 = (025,1( x 1 = 85.55+ 7. What should should e the full-lo full-load ad horsepower horsepower rati$# rati$# of a motor motor that dries dries a 5-!w #e$erator whose e%&ie$&' is 8(.5 per&e$t? Be$erator Po = 5 !W * = 8(.5+ Pi = Po,e = 5,.8(5 = 55865.( W EPi = 55865.(,746 = 74.8( hp
8. f the oer-all oer-all e%&ie$&' e%&ie$&' of the motor-# motor-#e$erator e$erator set of Prolem 7 is 78 per&e$t &al&ulate a/ the e%&ie$&' of the motor / the total losses i$ the motor#e$erator set &/ the losses i$ ea&h ma&hi$e. a. * = 78+ Epi = 6412.6,746 = 85.(0 hp *=Po,Pi e m = 74.8(/746/,85.(0/746/ 74.8(/746/,85.(0/746/ x 1 .78 = 5,P i = 87.15+ Pi = 6412.6 W . *# = 8(.5+ * m= 87.15+ PoB = 5 W EPo; = 74.8( hp
*#e$ = Po,Po G Ploss#e$ Ploss#e$ = Po e #e$ #e$/Po//,e #e$ #e$ = 5/-.8(5/5//,.8(5 =82.8 W
otal otal Power Power loss = 82.8 G 5865.(2 5865.(2 = 1466.72 W &. Plosses ; = 82.8 W Plosses B = 5865.(2 W
(. " 1-!w 22-ol 22-oltt &ompou$d &ompou$d #e$erato #e$eratorr is operated operated at $o load at the proper proper armature olta#e a$d speed from whi&h the stra'-power loss &al&ulatio$s are determi$ed to e 75 watts. he shu$t-@eld resista$&e is 11 ohms the armature resista$&e is .265 ohm a$d the series-@eld resista$&e is .05 ohm. "ssume a 2-olt rush drop a$d &al&ulate the full-load e%&ie$&'. Bie$ Po = 1 W H = 22 olts tra' power loss = 75 W :E = 11 ohms :" = .265 ohm :* = .05 ohm *B = 2 olts A = 1,22 1,22 = 45.45 " " = * 47.45 " sh = 22,11 = 2" LOSSES
WATTS
tra' power loss hu$t @eld eries @eld
75
"rmature
22/2/ = 44 47.45/2.05/ = 78.8 47.45/2.265/ = 5(6.6
=
Crush
2/47.45/ = (4.( otal otal 1(15.0 W
* = 1- P loss/,Po G Ploss// x 1 = 1 1(15.0/,1G1(15.0// 1(15.0/,1G1(15.0// x 1 = 80.(2 + 1.f the maximum e%&ie$&' of the #e$erator of Prolem ( o&&urs whe$ the sum of the &opper loss i$ the armature a$d series @eld is eDual to the sum of the stra'-power loss a$d the shu$t-@eld loss &al&ulate a/ the armature &urre$t / the li$e &urre$t &/ the !ilowatt output d/ the maximum e%&ie$&'. )opper loss = total power loss stra' str a' power loss )opper loss = 1(15.0 75 = 121.0 W eries @eld = stra' power loss G shu$t @eld = 75 G 44 = 1145 W a. E = 22,11 = 2 amp P = 2: " = IP,: = I1145,.265 G .05/ = 61.78 amp . " = A G E A = " E = 61.7 2 = 5(.7 amp &. PO = i =22/5(.78/ = 10.1516 !W d. * = Pi,Po
11." 25-olt shu$t #e$erator has a rated armature &urre$t of 4 amp a$d the followi$# losses at full load fri&tio$ a$d wi$da#e = 2 watts &ore loss = 26 watts shu$t-@eld loss = 1 watts rush &o$ta&t loss = 8 watts armature &opper loss = 4 watts. "ssumi$# that the maximum e%&ie$&' o&&urs whe$ the approximated &o$sta$t losses are eDual to those losses that ar' as the sDuare of the load &al&ulate the armature &urre$t for the &o$ditio$ of maximum e%&ie$&'. Bie$ 25 olts 4 amp
Power i$put = P O G Plosses = 114 W ;ax e = Po,Pi Po = 114
Po = H" " = PO,H " = 114,25 = 442 amp
12." 25-!w series #e$erator has a$ e%&ie$&' of 85 per &e$t whe$ operati$# at rated load. f the stra'-power loss is 2 per &e$t of the full-load loss &al&ulate the e%&ie$&' of the #e$erator whe$ it is delieri$# a load of 15!w assumi$# that the stra'-power loss is susta$tiall' &o$sta$t a$d the other losses ar' as the sDuare of the load.
10.he followi$# i$formatio$ is #ie$ i$ &o$$e&tio$ with a lo$#-shu$t &ompou$d #e$erator * = 22 olts output = 2!w stra'-power loss = 75 watts : E = 11ohms :" = .265 ohm : * = .05 ohm rush drop = 0 olts. )al&ulate the e%&ie$&'. e%&ie$&'.
tra' power loss = 75 W :E = 11 ohms :" = .265 ohm :* = .05 ohm *C = 0 olt PO = 2 !W H = 22 olts A = 2,22 = (.(1 amp " = * = (2.(1 amp
E = 22,11 = 2 amp
LOSSES
WATTS
tra' power loss hu$t @eld eries @eld
75
"rmature Crus Crush h &o$ &o$ta ta&t &t TOTAL:
2/ 211/ = 44 (2.(1/2.05/ = 02.10 (2.(22/.265/ = 2287.6 0/ 0/(2 (2.( .(1/ 1/ = 278 278.7 .70 0 410.46
*%&ie$&' = 1-Ploss, Po G Ploss// x1 = 1-410.46,2 G 410.(6// x 1
= 80.2(+
14." 25-olt shu$t #e$erator has a full-load armature &urre$t of 4 amp u$der whi&h &o$ditio$ the losses are fri&tio$ G wi$da#e = 2 watts shu$t @eld = 1 watts &ore = 26 watts rush &o$ta&t = 12 watts armature &opper = 4 watts.
46
46
46
46
46
11 2 ( 46(.5
22 2 2 477.4
00 2 01 485.0
44 2 42 4(0.2
55 2 50 51.2
1168
1105
111
188
164
58 1
288 48
6(2 115
127 212
222 007
(2 27
(2 6
(2 (0
(2 126
(2 15(
2180
2451
2(0
0616
452
56 2877 56.86
112 766( 75.78
1518 1225 8.7
224 16624 82.10
250 27(8 82.2
LOSSES:
Power output Power i$put + e%&ie$&'
16." 2-hp motor has a$ e%&ie$&' of 88.5 per &e$t a$d operates &o$ti$uousl' to drie a e$tilator at full load. )al&ulate a/ the power loss / the e$er#' loss per mo$th assumi$# operatio$ for 2 hours duri$# that period &/ the &ost of the e$er#' loss at 1 K &e$ts per !ilowatt-hour ! ilowatt-hour.. PO = 2 hp x 746 = 14(2 W * = 88.5+ a. Ploss = Po/1/,e PO = 14(2/1/,88.5 14(2 = 1.(4 !W
. Eours Eours used used per per mo$th mo$th = 2 2 h hour ours s Lwh loss = Wlosses/hours/ = 1(08.8/2/ = 087.76 !wh &. )ost of of e$er#' e$er#' loss loss = 088 !wh/ !wh/.1 .15 5 &e$ts &e$ts , !wh / = M5.82 M5.82 17.What sai$# would e made per 'ear i$ e$er#' &ost if the motor i$ Prolem 16 had a$ e%&ie$&' of (1.5 per &e$t? Ploss = Po/1/,e P o = 14(2/1/,(1.5 14(2 = 1.086 !W Lwh = 10862/ = 277.2 !wh )ost of e$er#' loss = 277.2 !wh/.15 &e$ts,!wh/ = M4.158 M5.82 - M4.158 = M1.662 x 12mo$ths = M1(.(44 18."$ e$&losed motor has a rati$# of 5 hp. he &oer plates are remoed a$d the ma&hi$e is lo&ated where it is &apale of &ooli$# itself extremel' well. f tests show that it &a$ &arr' 28 per &e$t more load without ex&essie heati$# what rati$# should e #ie$ the motor? Epo = 5 hp 28+ more load :ati$# of motor = 5.28/ = 14 hp 14hp G 5 = 64 hp 1(." 20,115-olt 20,115-olt three wire #e$erator deliers the followi$# loads 65!w at 20 olts 05!w at 115 olts etwee$ the positie a$d $eutral li$es. )al&ulate a/ the total !ilowatt load deliered ' the #e$erator / the &urre$t i$ the positie li$e &/ the &urre$t i$ the $e#atie li$e d/ the &urre$t a$d its dire&tio$ i$ the $eutral li$e.
a. . &. d.
Bie$ 20 olts = 65 !W 115 olts = 05 !W 115 olts = 25 !W otal otal !W = 65 65 G 05 G 25 25 = 125 125 !W positie = 20 G 115 = 65 !W,20 olts G 05 !W, 20 olts = 586.(6 amp $e#atie = 20 G 115 = 65 !W,20 olts G 25 !W,115 olts = 5 amp $e#atie = positie $e#atie = 586.(6 5 = 86.(6 amp towards the &oil/
2.f the #e$erator i$ Prolem 1( deliers the same total load ex&ept that the two 115-olt loads are ala$&ed with respe&t to ea&h other &al&ulate the li$e a$d $eutral &urre$ts. 65 !W = 20 olts 05 !W = 115 olts etwee$ positie a$d $eutral li$es 25 !W = 115 olts etwee$ $e#atie a$d $eutral li$e 20 = 65,20 = 282.61 amp 115G = 05,115 = 04.05 amp
115- = 25,115 = 217.0( amp A = 115G G 115-/,2 G 20 = 04.05 G 217.0(/,2 G 22.61 = 540.48 amp N = amp 21.he #e$erator i$ Prolem 1( deliers a total of 16!w. 16!w. f the &urre$t i$ the positie wire is eDual to the sum of the &urre$ts i$ the $e#atie a$d $eutral wires &al&ulate the three !ilowatt loads. G = - G N eD$ 1 16 = 20 G G 115N eD$ 2 16 = 20 G G 20- eD$ 0
*D$ 2 eD$ 0 eD$ 5 16 = 20 G G 115N 115N 20 16 = 20 G G 2016 = -20 - - 045 N = 115 N 20 - eD$ 4 16 = 46N N = 047.80 amp 521.75 amp
eD$ 1 x 20 eD$ 2
eD$ 4
= 20 G - 20- G 20N
=
16 = 20 G G 115N
-
-16 = -20- - 045 N eD$ 5
- = 11 11504 5047 7.80/ .80/,2 ,20 0 = 170 170.( .(0 0 amp amp
G = 170.(0 G 047.80 =
si$# L)A G G 20 115G = 115G - 20 = 521.75 - eD$ a
115G - 115- - N = 115- - 20 = 115G - N = 047 047.8 .80 0 eD$ eD$ 115- - 20 = 170.(2 eD$ &
*D$ a eD$ 115G - 20 = 521.75 115170.(2 G 20/ 115G - N = 047.80 -20 G 115- = 170.(2
16 = 20 20 G 115115G G 11511516 = 20 20 G 115521.75 G 20/ G 16 = 46 20 G 8 20 = 16-8/,46 = 170.(1 amp
115G = 521.75 G 20 = 6(5.66 amp 115- = 120.(2 G 20 = 047.80 amp
P115G = 115115G = 1156(5.66/ = 8 W P115- = 115115- = 115047.80/ = 4 W
22." d'$amometer has a$ output rati$# of .16 amp at 5 olts u$der whi&h &o$ditio$ it ta!es 11 amp from a 12-olt stora#e atter'. )al&ulate the e%&ie$&'. = .16 ampH = 5 olts 11 amp from 12 olts PO = .16/5/ = 8 W Pi = 11/12/ = 102 W * = Po,Pi x 1 = 8,102 x 1 = 6.61+ 20." d'$amometer has low-olta#e i$put a$d hi#h-olta#e output wi$di$#s whose resista$&es are respe&tiel' .1 ohm a$d 4 ohms a$d the hi#holta#e wi$di$# has 7 K times as ma$' &o$du&tors as the low-olta#e wi$di$#. f the ma&hi$e deliers a &urre$t of .5 amp at 18 olts u$der whi&h &o$ditio$ the &ore G fri&tio$ G wi$da#e loss is 05 watts &al&ulate the i$put &urre$t a$d olta#e. Output : = 4 ohm $put : = .1 ohm = .5 amp =? H = 18 olts H=? P = 05 W H = * : = 18 .5/4/ = 178 olts P = H = 05,178 = .1(6627.5/ i$put = .1(662/7.5/ = 1.4746 amp H = : = 1.4746/.1/ 1.4746/.1/ = .14746 olts 24." 05-hp 20-olt shu$t motor has a full-load armature &urre$t rati$# of 105 amp a$d a$ armature resista$&e i$&ludi$# rushes of .125 ohm. he a&&elerati$# resistors i$ the automati& starter to whi&h the ma&hi$e is &o$$e&ted hae a total resista$&e of .70 ohm. )al&ulate the ohmi& alue of a plu##i$# resistor that should e pla&ed i$ series with the a&&eleratio$ resistors to limit the i$rush armature &urre$t &urre$t to 1.5 times its full-load alue at the i$sta$t the motor is plu##ed. "ssume that the &ou$ter emf is 8 per &e$t of the impressed olta#e.
:p = 20 G .8/20//,1.5/105/ = 2.44 .125 .70 = 1.1( ohms
25.
